PS:There are lots of skeptical blogs to vote for. Bloggies 2015 voting closes tonight 10pm Sunday EDT, which is 1pm Monday in Sydney. (And I had an old incorrect link in the post about it last week. Oops :- | ).
This is the third in a series of four comments I am posting on Joanne’s Unthreaded Weekend threads. Formatting issues prevent me from including figures and tables in this post. A PDF version of this post containing all figures and tables can be found at:
[Note: If when you “click” on either/both of the above links, instead of seeing an “opened PDF/Excel file” a new “window” pops up with the option to “download,” click on the “download” button. This should allow you to load the file onto your personal computer.]
In my first post [comment #2 at http://joannenova.com.au/2015/03/weekend-unthreaded-69/#comments%5D, I discussed the definitions and connotations of words and phrases. In my second post [comment #2 at http://joannenova.com.au/2015/03/weekend-unleaded/#comments%5D, I (a) challenged the AGW community to show that the claim if all greenhouse gases are removed from the atmosphere, the Earth’s surface temperature will be reduced by approximately 30 degrees Centigrade isn’t based on fatally flawed logic, and (b) reasoned that applying the backradiation argument commonly used by AGW proponents to “prove” greenhouse gases must increase the Earth surface temperature implies that when matter whose temperature is higher than the ambient background temperature is stored in a thermos bottle, a CO2 thermos bottle should “outperform” a vacuum thermos bottle. [For the storage of material hotter than the ambient background temperature, thermos bottle “A” “outperforms” thermos bottle “B” if for identical thermos bottle contents and identical thermos bottle background environments, the time it takes for the contents of thermos bottle “A” to decrease by a specified temperature is greater than the time it takes for the contents of thermos bottle “B” to decrease by the same temperature.] In this post I describe a thermos bottle experiment performed by Peter C. that shows vacuum thermos bottles that differ from CO2 thermos bottles only by the presence/absence of CO2 gas outperform CO2 thermos bottles. I assisted Peter in extracting derived information (primarily cooling rates) from Peter’s experimental data. The words “we”, “us”, and “our” are used throughout this post; but I want to acknowledge that Peter performed all of the experiments.
[Note: To many readers, a positive “cooling rate” implies a temperature that is “decreasing with time,” and a negative “cooling rate” implies a temperature that is “increasing with time.” In this post, however, the term “cooling rate” represents the time-rate-of-change of temperature. As such, in this post a positive “cooling rate” implies the temperature of an object is increasing with time, and a negative “cooling rate” implies the temperature of an object is decreasing with time. ]
As noted in my first two posts, it is often claimed that Carbon Dioxide (CO2) is a “heat-trapping gas” [Reference 1, http://www.ucsusa.org/global_warming/science_and_impacts/science/CO2-and-global-warming-faq.html.%5D. Once someone accepts this claim as fact, it is fairly easy to convince him/her that since CO2 gas is distributed throughout the Earth’s atmosphere, the presence of atmospheric CO2 gas should, via the trapping of heat, increase the Earth’s surface temperature above what the surface temperature would be in the absence of atmospheric CO2 gas.
As mentioned in my first post, a reasonable interpretation of “heat-trapping” is that a “heat-trapping anything” that surrounds an object hotter than its environment should prevent or at least retard heat loss from that object. Once someone accepts this “interpretation” and agrees that CO2 gas is a heat-trapping gas, it is fairly easy to convince him/her that since CO2 gas is distributed throughout the Earth’s atmosphere, the presence of atmospheric CO2 gas should, via the trapping of heat, increase the Earth’s surface temperature above what the surface temperature would be in the absence of atmospheric CO2 gas.
In my second post, I argued that this interpretation carries the implication that everything else being equal, when surrounded by a “heat-trapping anything” an object at a temperature higher than its surrounding environment should cool at a slower rate than in the absence of the “heat-trapping anything”. Thus, I argued that to claim that CO2 gas is a heat-trapping gas is to claim that when surrounded by CO2 gas, material at a temperature higher than its surrounding environment will retain its heat for an extended period of time relative to cooling times associated with the absence of the surrounding CO2 gas.
It occurred to us that if the above implication is valid, then at least for keeping liquids above ambient background temperatures (as opposed to keeping liquids below ambient background temperatures), CO2 thermos bottles should outperform vacuum thermos bottles. The goal of a thermos bottle is to keep material at a higher (or lower) temperature than the ambient background temperature for as long a time interval as possible. A vacuum thermos bottle accomplishes this goal in part by surrounding its inner chamber (the region in which the heated/cooled material is stored) with a vacuum. In this post, a CO2 thermos bottle differs from a vacuum thermos bottle only in that CO2 gas is injected into the vacuum region of a vacuum thermos bottle; and thus in place of a vacuum, CO2 gas surrounds both the inner chamber and the contents of the inner chamber. Thus in this post, with the exception of the presence of CO2 gas, the phrase “everything else being equal” applies for a vacuum thermos bottle versus a CO2 thermos bottle.
If the implication is valid, then a CO2 thermos bottle should keep “hot” material above the ambient background temperature for a longer period of time than a vacuum thermos bottle. We believe the opposite is true. That is, we believe vacuum thermos bottles will keep “hot” material above the ambient background temperature for a longer period of time than CO2 thermos bottles. One reason we hold this belief is that if CO2 thermos bottles outperform vacuum thermos bottles, thermos bottle manufacturers would flood the market with CO2 thermos bottles. Such is not the case. However, the lack of a plethora of commercial CO2 thermos bottles is not proof that vacuum thermos bottles outperform CO2 thermos bottles. We decided to run some admittedly simple experiments to see if we couldn’t quantify the relative cooling-rate performances of vacuum and CO2 thermos bottles. Our experiments are simple because we had limited access to sophisticated equipment. We encourage people with access to better equipment to perform similar experiments; and, thereby, either support or refute our results.
[For all figures referenced in this post, open URL https://www.dropbox.com/s/6b8j73qf34qq3ad/final_peter_reed_paper_to_Joanne_Nova_unthreaded_01_pdf.docx?dl=0 on your computer. The figure numbers in this post and the cited URL are the same. In fact, with the exceptions of tables and figures, this post is almost identical to the cited URL. As such, you might want to skip the remainder of this post and simply read the URL document.]
EXPERIMENTAL SET UP
Figure 1 depicts our laboratory.
Figure 1: Our Laboratory
The experimental apparatus consists of:
—-Vacuum Flask (dark green thermos bottle),
—-Alcohol Thermometer, -20 to 110 degrees Centigrade (protruding from the Styrofoam cap at the top of the Vacuum Flask),
—-Digital Thermometer with K type thermocouple (yellow object),
—-Wrist watch (not shown),
—-Pen and Paper (not shown).
VACUUM FLASK
Peter was able to purchase two flasks at a discount store for A$10 each. The primary temperature retaining feature of each flask is a double-walled, vacuum, glass insert mounted within a plastic outer case. The plastic outer case has a base that can be unscrewed to access the glass insert. The space between the inner and outer walls of the glass insert is evacuated. The glass insert is silvered on its inside surfaces.
Figure 2 shows the components of the flask: Outer Plastic Case, Removable Base, and Double-Walled Vacuum Glass Insert with the double-wall construction shown by removing a portion of the outer wall at the base of the glass insert.
Figure 2: Flask Construction Dismantled To Show The Double-Walled Vacuum Glass Insert (broken in this Figure)
Although the flasks looked the same, with flasks of this price the quality may be somewhat variable. We found this to be the case. We believe the performance difference was associated with the quality of the vacuum. The two flasks were tested by filling with near boiling water and measuring the cooling rate. Flask A, the flask with the better performance, was selected for all experiments reported in this post.
The original screw cap at the top of the flask was replaced by a homemade Styrofoam stopper with a hole to allow the thermometer access to the liquid inside the thermos. The Styrofoam stopper seemed to contain the heat better that the original plastic cap.
Five trials were run to quantify the flask’s rate of heat loss in its original form (vacuum). The vacuum was then destroyed by cutting a hole through the vacuum nipple. A thin plastic connecting tube was then attached through this hole and sealed with silicone. Four trials were run to quantify the flask’s rate of heat loss when the vacuum region was filled with air. Five trials were run to quantify the flask’s rate of heat loss when the vacuum region was filled with CO2 gas.
Plastic connecting tubing, three-way taps and syringes are easily obtained in hospitals. Using these items we were able to construct a simple vacuum pump which can create a vacuum of approximately 0.1 atmospheres. The vacuum was measured by attaching to the chamber of a Mercury Barometer. The vacuum was maintained over several days, demonstrating the sealing is very good.
THERMOMETERS
During each experimental run, we wanted to monitor the temperature of three objects: (1) the water inside the inner chamber, (2) the outermost surface of the glass insert, and (3) the ambient room temperature. We obtained four thermometers for this purpose: a Mercury thermometer (white), two alcohol thermometers (yellow and green), and a digital thermometer/thermocouple. On 17 April 2014 we measured the relative performance of the thermometers. [See the cited URL for Table 1. Basically, the thermometers agreed to within 2C.]
Table 1: Comparison Of Thermometers
Unfortunately, the mercury thermometer broke early on and was not used for the experiments described in this post. Because we thought that the digital thermometer might suffer from drift in the electronics, we decided to use the alcohol thermometers to measure the water temperature and the ambient temperature. Because it was the easiest to read, we used the green alcohol thermometer to measure the temperature of the water. We used the yellow alcohol thermometer to measure the ambient background temperature. We measured the ambient background temperature only at the start of the experiment. We taped the thermocouple to the outermost surface of the glass chamber. Monitoring the temperature of the outermost surface of the glass chamber provides additional insight into the heat transfer process. Although we did not estimate the cooling rate of the outermost surface of the glass chamber, we plotted those temperatures for all but our first (Run 3) vacuum experiment.
The green alcohol thermometer was calibrated by continuous insertion in a sauce pan of boiling water on a stove. It read 101C. The thermometer was then placed in a whisky glass containing ice blocks and cold water and read 1C. The temperature readings of our experiment were not corrected for these slight differences (101C vs 100C, and 1C vs 0C).
VACUUM, AIR AND CO2 GAS
All “vacuum measurements” reported in this post were made with the flask as provided by the manufacturer. After completing the vacuum measurements, we destroyed the as-purchased quality of the vacuum simply by cutting through the tip of the vacuum nipple. Air rushed into the original vacuum space. All “air temperature measurements” were made with the flask in this state.
When it came time to make “CO2 temperature measurements”, we wanted as high a concentration of CO2 gas as possible. CO2 gas was obtained from a cylinder of CO2 gas supplied by the British Oxygen Company, which is the local supplier for all our gases. The gas cylinder was labeled “BOC Medical EP Grade CARBON DIOXIDED Gas Code 530.” BOC specifies this gas as 99.95% CO2. A plastic party balloon was fitted with connecting tubing. The plastic balloon was initially completely emptied, then filled with CO2 gas from the cylinder. This procedure was repeated to purge any residual air from the balloon and tubing.
Using the syringe, the vacuum chamber was then evacuated to 0.1 atmospheres and refilled from the balloon via the three-way tap. This procedure was repeated four times. Although we did not test the purity of the CO2 gas in the vacuum chamber, we believe that for our “CO2 experiments” between 0.99 and 0.999 of the gas in the vacuum chamber was CO2.
Figure 3 depicts our set up for filling the original vacuum region with CO2 gas. Note the connecting tube sealed to the vacuum nipple with silicone; plastic balloon attached to three-way tap and syringe, K thermocouple taped to outside surface of glass vacuum chamber.
Figure 3: Flask With Equipment To Fill The Flask Vacuum Region With CO2 Gas
NOTE ON MEASUREMENTS
It was apparent that we could prioritize either the temperature or the timing measurements. Specifically, when the temperature reached an exact reading on the thermometer, we could record the time. Or when the watch reached an exact time, we could make our best reading of the thermometer. If we chose to record the temperature at what we judged to be the exact degree, the timing could be subject to an estimation error of 30 seconds or more. If the timing was prioritized, it could be accurate to within 5 seconds or less, but the temperature estimate accuracy was degraded. We elected to prioritize the timing—i.e., we would make the best temperature measurement we could at a time accurate to 5 seconds or less. We could read the temperature to 0.5 degrees or less. If the temperature was near the 0.5 degree mark, the temperature was recorded as the immediately lower integer plus 0.5. If the temperature was past the degree mark, but seemed a little less than 0.5 it was recorded as the immediately lower integer plus 0.2. If the temperature was nearly to the next degree mark, it was recorded as the immediately lower integer plus 0.8.
RESULTS
In total we made 14 experimental runs of the temperature of heated water in the flask as a function of time—five runs with a vacuum, four runs with air in the original vacuum region, and five runs with CO2 gas in the original vacuum region.
The experimental results are summarized below:
Results with VACUUM Thermos Bottle, Cooling Rate Average: -0.055612 C per minute
____Run #3, 10 June 2014, Ambient Temperature 19C
________Measurement Times (minutes) relative to time of first measurement 18:00
____________0, 30, 70, 90, 120, 150, 180, 220, 242, 245
________Measured temperatures (C)
____________98.0, 97.0, 94.5, 93.0, 91.2, 90.2, 88.5, 86.0, 85.2, 85.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.053834
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.000838
—-Run #4, 11 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 18:45
————0, 37, 53, 60, 90, 126, 150, 180, 210, 227, 240
——–Measured temperatures (C)
————98.0, 96.0, 95.0, 94.8, 92.5, 91.5, 89.0, 87.5, 85.8, 85.0, 84.5
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.057603
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001264
____Run #5, 12 June 2014, Ambient Temperature 18C
________Measurement Times (minutes) relative to time of first measurement 17:32
____________0, 13, 43, 62, 73, 103, 150, 171, 194, 223, 241, 258, 272, 303
________Measured temperatures (C)
____________98.0, 98.0, 96.0, 95.0, 94.2, 92.2, 90.0, 88.8, 87.5, 86.0, 85.0, 84.2, 83.5, 82.8
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.053578
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.000988
—-Run #6, 13 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 18:25
————0, 5, 12, 51, 82, 116, 140, 170, 200, 230
——–Measured temperatures (C)
————98.0, 98.2, 97.5, 95.5, 94.0, 91.2, 90.1, 89.0, 87.0, 85.2
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.056483
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001194
____Run #7, 15 June 2014, Ambient Temperature 19C
________Measurement Times (minutes) relative to time of first measurement of 17:40
____________0, 60, 90, 140, 210, 232
________Measured temperatures (C)
____________98.2, 95.0, 92.8, 90.0, 86.5, 85.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.056563
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001214
Results with AIR Thermos Bottle, Cooling Rate Average: -0.217540 C per minute
—-Run #10, 21 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 14:10
————0, 1, 9, 15, 20, 25, 30, 35, 40, 45, 51, 55, 60, 61, 65, 70, 77, 85
——–Measured temperatures (C)
————98.2, 98.0, 96.2, 95.0, 93.8, 93.0, 91.8, 90.5, 89.2, 88.5, 87.0, 86.2, 85.2, 85.0, 84.5, 83.5, 82.0, 80.5
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.210915
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.002071
____Run #11, 23 June 2014, Ambient Temperature 20C
________Measurement Times (minutes) relative to time of first measurement 7:10
____________0, 1, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 55.5, 60, 65, 70
________Measured temperatures (C)
____________96.0, 97.5, 96.0, 95.0, 94.2, 93.0, 92.0, 90.5, 89.5, 88.5, 87.2, 86.2, 85.2, 85.0, 84.2, 83.2, 82.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.213345
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.004074
—-Run #12, 28 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 9:00
————0, 5, 9, 10, 15, 20, 25, 31, 36, 40, 45, 50, 55, 60, 65, 72
——–Measured temperatures (C)
————97.2, 96.0, 95.0, 94.8, 93.8, 93.0. 91.8, 90.2, 89.2, 88.5, 87.2, 86.0, 85.0, 84.2, 83.0, 82.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.215037
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.002216
____Run #13, 26 June 2014, Ambient Temperature 20C
________Measurement Times (minutes) relative to time of first measurement 20:39
____________0, 1, 6, 11, 12, 16, 22, 26, 31, 38, 46, 51, 56, 57, 61, 66
________Measured temperatures (C)
____________98.2, 98.0, 97.5, 95.8, 95.0, 94.2, 93.2, 91.8, 90.8, 89.2, 87.2, 86.2, 85.2, 85, 84.2, 83.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.230865
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003741
Results with CO2 Thermos Bottle, Cooling Rate Average: -0.206672 C per minute
—-Run #30, 2 July 2014, Ambient Temperature Not Recorded
——–Measurement Times (minutes) relative to time of first measurement 18:13
————0, 1, 5, 12, 17, 18, 22, 28, 33, 41, 47, 58, 62, 67, 76, 85
——–Measured temperatures (C)
————98.2, 98.2, 97.5, 96.2, 95.2, 95.0, 94.2, 93.0, 92.0, 90.2, 89.0, 87.5, 86.2, 85.5, 83.5, 82.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.194215
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001631
____Run #31, 3 July 2014, Ambient Temperature Not Recorded
________Measurement Times (minutes) relative to time of first measurement 20:26
____________0, 5, 9, 13, 19, 24, 29, 34, 39, 46, 51, 55, 60, 61, 70, 74, 94
________Measured temperatures (C)
____________98.0, 97.5, 96.0, 95.0, 94.0, 93.0, 91.5, 90.5, 90.0, 88.0, 87.0, 86.2, 85.5, 85.0, 83.5, 82.5, 79.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.204740
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003633
—-Run #32, 5 July 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 15:02
————0, 1, 3, 8, 13, 19, 24, 29, 49, 54, 59, 64, 69, 74
——–Measured temperatures (C)
————97.5, 97.0, 97.0, 96.0, 95.0, 93.5, 92.5, 91.2, 87.0, 86.0, 85.0, 84.0, 83.0, 82.2
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.210518
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001785
____Run #33, 6 July 2014, Ambient Temperature 19.5C
________Measurement Times (minutes) relative to time of first measurement 9:50
____________0, 16, 24, 45, 64, 70, 76
________Measured temperatures (C)
____________97.2, 94.0, 92.0, 87.2, 83.2, 82.2, 81.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.214454
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003181
—-Run #34, 7 July 2014, Ambient Temperature 18C
——–Measurement Times (minutes) relative to time of first measurement of 20.26
————0, 4, 10, 15, 19, 25, 35, 40, 47, 55, 60, 65, 69, 75, 80
——–Measured temperatures (C)
————97.5, 96.5, 95.2, 94.5, 93.5, 92.2, 90.0, 89.2, 87.5, 86.0, 85.0, 83.5, 83.0, 81.5, 81.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.209431
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001747
For all but the first vacuum run, we simultaneously monitored the temperature of the outside of the glass vacuum flask. Figures 4 through 8 show the results for the “vacuum” thermos experiments. Figures 9 through 12 show the results for the “air” thermos experiments. Figures 13 through 17 show the results for the “CO2” thermos experiments.
Figure 4: 10 June 2014 Vacuum Thermos Figure 5: 11 June 2014 Vacuum Thermos
Figure 6: 12 June 2014 Vacuum Thermos Figure 7: 13 June 2014 Vacuum Thermos
Figure 8: 15 June 2014 Vacuum Thermos
Figure 9: 21 June 2014 Air Thermos Figure 10: 23 June 2014 Air Thermos
Figure 11: 28 June 2014 Air Thermos Figure 12: 26 June 2014 Air Thermos
Figure 13: 2 July 2014 CO2 Thermos Figure 14: 3 July 2014 CO2 Thermos
Figure 15: 5 July 2014 CO2 Thermos Figure 16: 6 July 2014 CO2 Thermos
Figure 17: 7 July 2014 CO2 Thermos
Table 2 summarizes the estimated cooling-rates and the estimated standard deviation associated with each estimated cooling rate. The cooling rates were generated by applying a Weighted-Least-Squares (WLS) estimator to a linear model of temperature as a function of time. The WLS process assumed each measurement error was a random sample from a zero-mean distribution. The standard deviations of the various measurement error distributions were assumed to be the same. The measurement errors were assumed to be uncorrelated. With these assumptions, WLS estimation produces a slope identical to the slope of a linear regression. In addition to providing an estimate of the slope, if the measurement error standard deviations are known, WLS theory also provides an estimate of the variance/covariance of the parameters being estimated (slope, offset).
Table 2: Summary Of Water Cooling Rates
Using the WLS slope estimate (cooling rate) and the WLS offset estimate (temperature at relative time 0), a temperature residual for each measured temperature can be generated. A temperature residual is the difference between the measured temperature and the corresponding model temperature. If the measurement error standard deviations are known, a weighted residual for each measurement can be computed. A weighted-residual is a residual divided by the corresponding measurement error standard deviation. If all of the above assumptions are valid and if the linear model is ideal, then the expected value of the sum-of-squares of weighted residuals will equal the number of measurements …minus… the number of degrees of freedom of the WLS estimation process. The number of degrees of freedom of a WLS estimation process is the number of variables to be estimated. For linear WLS estimation, there are two degrees of freedom (slope, offset). Since we assumed the measurement error standard deviations are equal, we can generate an estimate of that standard deviation by computing the square root of the ratio of (a) the sum-of-squares of weighted residuals, and (b) N minus 2, where N is the number of measurements. Using the estimated measurement error standard deviation, we can generate an estimate of the standard deviation of each measured slope (cooling rate). Those estimated standard deviations are given in Table 2.
The first observation is that the results are very good, given the cost and sophistication of the experimental equipment. The runs with the vacuum in the flask vary only in the third decimal place. Even at the higher rate of heat loss, with air and then CO2 gas as the insulating medium the variation is less than 0.03.
An examination of the results presented above shows that in the case of a vacuum thermos bottle versus a CO2 thermos bottle, CO2 gas acts to release heat, not to trap heat. From the perspective of a thermos bottle, CO2 gas is more correctly characterized as a “heat-releasing gas” than as a “heat-trapping gas.” Thus, if the term “heat-trapping gas” carries the connotation of “trapping heat”, CO2 gas cannot be called a “heat-trapping gas” because situations exist where CO2 gas doesn’t “trap heat”, CO2 gas “releases heat.” If one defines a “heat-trapping gas” to be any gas in the Earth’s atmosphere that causes an increase in the Earth surface temperature, then (a) it is a circular argument to use the “heat-trapping” nature of a gas to claim the presence of the gas in the Earth’s atmosphere will increase the surface temperature of the Earth, and (b) it must be shown that CO2 gas is a heat-trapping gas, not simply declared based solely on the observation that CO2 gas absorbs electromagnetic radiation in subbands of the IR band. After all, the IR absorption property exists for the CO2 gas in the thermos bottle and the CO2 thermos bottle loses heat faster than the vacuum thermos bottle. It may well be that atmospheric CO2 gas causes an increase in Earth surface temperature; but to make that claim based solely on the IR absorption property of CO2 gas is unwarranted.
Both Air and CO2 gas release heat from the inner surface of the flask to the environment, likely by increasing the rate of conduction. At first glance it would appear that CO2 gas is a better insulator than Air, and this difference is nearly but not quite significant at the 95% confidence level, which assuming a Gaussian distribution, we approximate to error bars of 2 standard deviations. That is an error of 2 standard deviations from the average below the air average to 2 standard deviations above the CO2 average overlap. Figure 18 depicts the Vacuum cooling rates with plus/minus 2 sigma error bars. Figure 19 depicts the “Air” and “CO2 gas” cooling rates with plus/minus 2 sigma error bars.
Figure 18: Vacuum Results With 2 Sigma Error Bars Figure 19: Air And CO2 Gas Results With 2 Sigma Error Bars
Also we can see from the results that Runs No 13 and 30 seem to be outliers in that they vary from the other runs by a noticeable amount. If we were to exclude these two results then the remaining runs have almost identical rates of heat loss for AIR and CO2 gas. Now one cannot arbitrarily exclude any result, without a proper reason, and hence these results have not been excluded.
If we were to repeat the experiments, we would try to control the ambient temperature during each experiment, or at least measure the ambient temperature during the course of the experiment, rather that just take it at the beginning. It may be that under those conditions the heat loss with Air and with CO2 gas would be even closer together. We would also use a digital clock as the timing device instead of an analog wrist watch . It did seem that some of the observations may have been out by a minute or even 2 minutes, suggesting that the watch might not have been read correctly. No observations were corrected or adjusted.
In addition to the numerical cooling rate results, it’s clear from the graphs of Figures 2 through 5 compared to the graphs of Figures 13 through 17 and Figures 30 through 34 that the temperature of the outermost wall of the glass insert behaves differently for a vacuum thermos bottle than it does for Air and CO2 thermos bottles. For a vacuum thermos bottle the temperature of the glass insert’s outermost wall never got above 30C, whereas for both the Air and CO2 thermos bottles this temperature got to 50C. Since the radiative rate of cooling is proportional to the difference in the fourth powers of the medium and the surrounding wall temperatures, the higher surrounding-wall temperature will result in a reduced radiative rate of cooling. That is, the radiative cooling rate will be larger for the vacuum thermos bottle than for the Air and CO2 thermos bottles. However, it’s clear from the data that the total cooling rate is larger for the Air and CO2 thermos bottles than for the vacuum thermos bottle. Obviously, for the thermos bottle example, the heat transfer processes of convection and conduction more than compensate for the decrease in radiative cooling rate.
This completes my third Unthreaded Weekend post. The experimental results described above show that situations exist where surrounding heated material with CO2 gas (a greehouse gas) hastens, not retards, heat loss. Although the thermos bottle experiment described in this post differs from CO2 gas in the atmosphere, the thermos bottle example brings into question the claim: Via backradiation to the Earth’s surface, atmospheric CO2 gas must increase the temperature of the Earth’s surface.
Next up (fourth Weekend Unthreaded post), both in the absence and presence of backradiation from an object devoid of an internal energy source (an inert object), a theoretical treatment of the temperature of a sphere possessing a constant-rate source of internal energy (an active object). The conclusion of the post will be that with no change to the active object’s internal rate of energy, backradiation from an inert object to an active object can simultaneously exist with lower, not higher, active object temperatures.
For those who have read Reed’s third comment post to the end, I would encourage you to look at the graphs which Reed has published in the PDF version. It was a great pleasure to see my pages of lab notes transformed into really elegant graphs. I was also most impressed by how close nearly all the measurements were to the line of best fit. Only the odd measurement fell far from the line, which likely meant that I had made a mistake recording the time.
I might consider performing more experiments if I can obtain some SFl6 gas. This is a heavy gas used occasionally in ophthalmology. It is said to be a greenhouse 25,000 times more powerful than CO2.
First of all, kudos for your compliance to the scientific method in all of it’s facets, especially in the presentation of the methods used and the raw data obtained.
Before I comment on my impressions of your experiment, I would first of all like to point out that anyone who has read my comments here will conclude that I am clearly a skeptic. The AGW meme, in my estimation, is just that, a meme. Cleverly contrived, as it is, for the promulgation of political and economic change and as such, it is as completely removed from all of accepted physics as oil seperates from water.
Having said that, I feel obliged to play the devil’s advocate and point out a flaw in your endeavor.
My understanding of the claim comimg from the AGW camp is that CO2 ‘traps’ radiative energy and not kinetic energy. The inner bottle in the thermos aparatus used in your experiment is ‘silvered on its inside surfaces’. I’m not a materials engineer, but my understanding is that the surface is silvered to prevent radiative energy from leaving the inner flask. How well the silvering accomplishes this task is, again, beyond my expertise but clearly the best cost effective process is used and so must perform fairly well in this respect.
Therefore, your experiment must, by construction, be measuring kinetic energy loss and not radiative heat loss.
Results with AIR Thermos Bottle, Cooling Rate Average: -0.217540 C per minute
and . . .
Results with CO2 Thermos Bottle, Cooling Rate Average: -0.206672 C per minute
From these quoted results, and correct me if I’m wrong, the conclusion must be that even when measuing kinetic energy loss (conduction), CO2 has a lower rate of heat loss than common air, (a combination of all the gases in the atmosphere).
Peter’s experiment measured the time-rate-of-change of temperature of the water in the thermos bottle, which is related to the time rate of change of the internal energy of the water in the thermos. In most macro thermodynamic treatments (especially gases), internal energy is associated with the average kinetic energy of the molecules comprising the material, which is related to temperature. In gases, heat transfer via conduction is modeled as elastic collisions of molecules that transfer the energy of high-velocity (i.e., high temperature) molecules to low-velocity (i.e., low temperature) molecules. So you’re correct, Peter’s experiment is indirectly measuring the rate of internal energy loss via all heat transfer mechanisms (radiation, conduction, convection, and change-of-state–evaporation). You’re also correct when you point out that Peter’s experiment shows that the rate of heat loss using CO2 is smaller than the rate of heat loss using air.
That brings up an interesting question. The heat conduction properties of gases at various temperatures and pressures have been measured. When reporting these conduction properties, are the conduction properties corrected for radiation effects; or do they represent heat transfer rates from all heat transfer mechanisms? It’s relatively simple to minimize heat transfer via convection and change-of-state; but for gases, it’s not so easy to separate conduction and radiation. At extremely low gas densities, I believe radiation dominates. At extremely high gas densities, I believe conduction dominates. Thus, although I agree Peter’s experiments indicate that CO2 is a better insulator than air (at least at atmospheric pressures) and this may be the result of absorbing IR radiation from the walls of the inner thermos bottle chamber, I don’t know why that is the case. Is it because of backradiation or different thermal conduction properties?
You wrote: “My understanding of the claim comimg from the AGW camp is that CO2 ‘traps’ radiative energy and not kinetic energy.” My exposure to the AGW position is slightly different from yours. Specifically, that in addition to claiming “CO2 traps radiative energy,”, they also claim “CO2 traps heat.” Three comments. First, the Earth surface temperature is affected by ALL heat loss mechanisms. It’s true that for the Earth/Earth-Atmosphere system as a whole, radiation (to space) is the overwhelming dominant means of energy loss. However, that is not the case for the Earth’s surface. Radiation, evaporation, convection, and conduction all play a role in transferring internal energy away from the Earth’s surface–likely in that order with radiation being the dominant mechanism. So attempting to predict the Earth’s surface temperature, or even changes to the Earth’s surface temperature, based solely on the electromagnetic radiation absorption/radiation properties of atmospheric gases is a non-starter in that it is bound to fail.
Second, it’s the total rate of energy transfer that affects temperature. Thus, it’s correct to say that CO2 absorbs (I hate the word ‘traps’) radiation emanating from the Earth’s surface. But such a statement, even coupled with backradiation, does not by itself “prove” the Earth’s surface temperature will increase. The statement provides plausibility, not proof. In my fourth and last post I will present a mathematical treatment showing that inactive material (i.e., material devoid of an internal source of energy): (a) completely surrounding an active object (i.e., an object with an internal source of energy), (b) absorbing ALL radiation frequencies, not just sub-bands of the IR, emitted from the surface of the active object, and (c) backradiating a portion of the absorbed energy to the active object can exist simultaneously with LOWER not HIGHER active object temperatures. If my mathematics is correct, then the existence of backradiation does not, by itself, guarantee increased active object temperatures.
Third, since the Earth/Earth-Atmosphere system overwhelmingly loses energy to space via radiation, it may be that atmospheric CO2 absorbs (“traps”) radiation emanating from the Earth’s surface; but it can’t be true that CO2 absorbs (“traps”) all radiation. If the latter were the case, the Earth/Earth-atmosphere system won’t lose energy to space and would eventually, how shall I say, fry.
Bottom line, I believe you’re correct when you say Peter’s experiment must “…be measuring kinetic energy loss…; but I believe you’re incorrect when you add “”…not radiative heat loss.”
Finally, don’t lose sight of the point I’m trying to make. I’m NOT arguing that thermos bottles behave similarly to the atmosphere. I’m addressing the question: “By any reasonable interpretation of “heat-trapping,” is CO2 a heat-trapping gas?” I believe this point is important because I believe this is the AGW community’s primary argument used to convince the average citizen that AGW is occurring is that CO2 is a “heat-trapping gas”–or as you put it “a radiative energy trapping gas.” If I can show a situation where CO2 gas acts more like a “heat-releasing” gas than a “heat-trapping” gas, then at a minimum I’ve given people something they can use to question the “heat-trapping” claim. And, if I can show a situation where just because CO2 ‘traps’ radiative energy, it doesn’t necessarily follow that the surface temperatures inside the “trap” must increase, then I’ve given people something to think about.
There’s a lot to be discussed in your reply so may I suggest we take it one point at a time, do you agree?
I’d like to start with this statement, If I may.
My exposure to the AGW position is slightly different from yours. Specifically, that in addition to claiming “CO2 traps radiative energy,”, they also claim “CO2 traps heat.”
Also, because of the close connection with this statement . . .
Peter’s experiment is indirectly measuring the rate of internal energy loss via all heat transfer mechanisms (radiation, conduction, convection, and change-of-state–evaporation).
. . . lets discuss them both together.
To the best of my knowledge, there are only three ways that energy can move from one location to another. They are, radiative flux, conduction, convection. In radiative flux, energy is emitted from a particle, (either an atom or a molecule), in discrete amounts called quanta. These quanta can travel through the vacuum and may eventually be absorbed by another particle. In conduction, one particle will bump into another particle. If one particle has a higher energy content than the other, then some energy will be transfered from the particle with the higher energy content to the particle with the lower energy content. In the case of convection, a collection of particles, say a body of water, will lose one or more of it’s particles through the process of evaporation, taking some of the energy contained in the collection with it.
If you’re with me so far, then we can proceed to the next point which concerns the bolded part of your statement above, they also claim “CO2 traps heat”. If, on the other hand, I’m missing something in my comprehension of energy transfer, let’s clear that up first.
Most “normal” glasses absorb LWIR energy. A glass thermos bottle traps heat two ways. First, it uses really thin glass so that the conductive area is very small from the inner portion of the glass envelope to the outer part. The silver coating is on the vacuum side of the glass wall. This is to prevent (slow) radiative transfer of the inner glass to the outer glass surface and visa versa when using it to keep things cold. Since the vacuum stops conductive heat transfer they silver the glass to slow the radiation to the outer surface.
Silvered or aluminized 2nd surface mirrors are crappy LWIR reflectors. They either need to be made as first surface mirrors or the glass substrate needs to be replaced by an LWIR transparent material like NaCl, ZnSe or Ge as 2nd surface mirrors depending on what wavelengths you need to reflect. So the silvering is not reflecting heat back to the contents, it’s attenuating radiative losses from the glass. It works better this way as the stainless vacuum bottle I used to have which does reflect radiatively was far poorer at keeping things hot since the conductive losses were a LOT higher.
I’d have to say the reason the heat transfer was slower in the all CO2 vs. all Air is that CO2’s heat properties are different. At 20°C, 1 atm normal Air has a density of 1.205 kg/m³ and CO2 is 1.842 kg/m³. Air’s Specific Heat (Cp) is 1.01 kJ/kg°K and CO2’s is .844 kJ/kg°K. So with equal volumes (and pressure) of CO2 and Air and equal input energy the air will get 28% hotter than the CO2. This according to Q = Cp * m * ΔT where Q is energy in kJ, Cp is specific heat, m is mass and ΔT is the temperature change.
Since moving energy in this system relies on conductive transfer we find that CO2’s Thermal Conductivity is 0.0146 W/(m°K) and Air’s is 0.024 W/(m°K). Since this system has a fixed surface area, again, the CO2 will conduct less heat according to Fourier’s Law since heat transferred is directly proportional to Thermal Conductivity here since everything else is equal (like a good experiment should be.) Data here was sourced from http://www.engineeringtoolbox.com/
The Climateers all say that CO2 “forces” other things in the climate system by “trapping” heat… but how can that be when it doesn’t get as hot with the same energy input as plain old air nor does it transfer what heat it has accumulated as well as plain old air. Is it creating energy all by itself? How can it force water vapor to be hotter when water vapor’s IR absorption swamps the tiny bit CO2 gets… and water vapor (being a vapor and not a gas) can actually radiate that energy at normal atmospheric pressures, CO2 doesn’t. It’s only radiative output is to space at 15µm (193°K) or the middle of Antarctica on June 21st.
Thank you so much for the comprehensive reply. Some years ago I had read that the reflective properties of mirrors weren’t able to contain radiative energy within an enclosed space very well, but that was only in passing and in a different context. I didn’t understand the reasoning back then and to tell you the truth it didn’t seem that important to me at the time. This explanation . . .
Most “normal” glasses absorb LWIR energy. A glass thermos bottle traps heat two ways. First, it uses really thin glass so that the conductive area is very small from the inner portion of the glass envelope to the outer part. The silver coating is on the vacuum side of the glass wall. This is to prevent (slow) radiative transfer of the inner glass to the outer glass surface and visa versa when using it to keep things cold. Since the vacuum stops conductive heat transfer they silver the glass to slow the radiation to the outer surface.
So the silvering is not reflecting heat back to the contents, it’s attenuating radiative losses from the glass.
is more clearly stated than what I heard back then so again, thanks for that.
If I’m comprehending what you’ve said, radiative energy is transfered (fluxed?) to the glass where it is absorbed and then, the silvering on the glass slows (attenuates) radiative energy transfer outward and away from the glass into the vacuum. So, in effect, the energy isn’t radiated back into the liquid inside the inner bottle it’s only slowed down in it’s rate of outward flux.
The Climateers all say that CO2 “forces” other things in the climate system by “trapping” heat… but how can that be when it doesn’t get as hot with the same energy input as plain old air nor does it transfer what heat it has accumulated as well as plain old air. Is it creating energy all by itself? How can it force water vapor to be hotter when water vapor’s IR absorption swamps the tiny bit CO2 gets…
I would just like to add that at 0.04% of normal atmosphere, one CO2 molecule would have to heat 2,500 other molecules composed of all the other gases out there. That’s one powerful little critter, yes? Atom Ant comes to mind!
If CO2 had all the magic properties they keep ascribing to it I don’t understand why we aren’t using it for all of our energy needs. If our contribution of 15ppmv can add that much energy to the atmosphere it must be powerful… considering that if we turned every single erg of energy we create and use in a year into heat at 100% efficiency, from nuclear to dung burning, and dumped it into the atmosphere all at once, we’d only raise the temperature about 0.07°C. (Based on IEA numbers for 2012) And one of the byproducts of some of that energy is a couple of orders of magnitude more powerful? We’re using the wrong stuff to power society.
The experiments did show that the rate of heat loss with CO2 was very slightly less than with common air, but the result was not (quite) significant. I think that the thermal conductivity of CO2 as measured by others is slightly less than Air. I think the value is given somewhere in the Engineering Toolbox.
You say that the silvering on the flask reduces the radiative heat loss to a very low level, hence we are measuring conductive loss mainly. I agree. I have subsequently removed the silvering from the flask and the rate of heat loss is then even greater.
I have not pursued that because Reed develops his argument further in Part 4, and it became unnecessary.
Surface Aluminium was removed by washing out the vacuum space with Green River Solution (Copper sulphate and Hydrochloric acid), then carefully drying out the chamber.
How much remained as increased conductivity of the fibre supports? No idea.
Peter, it would seem that having removed the reflective coating on the inner flask that now you have a warm body surrounded by the CO2 but it is still very different to the earth-atmosphere-space model to which you are trying to draw parallels because the heat transfer from the outer flask to the room is still very conductive and convective. Now if you could surround your flask with a vacuum and that by a very cooled radiation sink it might give some better idea of the ‘greenhouse’ properties of the gases which you are investigating.
I am told that an experiment similar to what you describe has been done. The person has not yet published his results. Therefore I cannot tell you what they were.
I had planned something similar but I was frustrated because I was unable to create a sufficient vacuum. The requirements are in fact exceedingly stringent. At 0.1 atmospheres the conductivity of gases is hardly affected. The vacuum has to be at least 2 orders of magnitude greater.
I was previously uncertain about the proposition that back radiation from a metal shell in a vacuum would slow the rate of cooling of a central sphere with a heat source. This is the steel greenhouse effect as described by Willis Eschenbach.
I am now persuaded that back radiation from a steel sphere can slow heat loss and hence the heated inner sphere will be hotter. By analogy the back radiation of greenhouse gases can slow heat loss from the Earth surface (slightly). But that is not their only property or effect.
As shown by the thermos experiment CO2 has a large conductive capacity, equal to (or almost equal to) Air.
When these other properties are accounted for it can be shown that under some circumstances a greenhouse gas can cause increased heat loss, not heat trapping.
Peter,
4 milli-torr even from a good roughing pump can make a lot of difference. 2 micron pressure with oil diffusion, takes weeks, or never, if one ant in your vacuum space.
The two T^4 terms in required parenthesis the S-B equation refer to opposing radiances (not flux) which limit radiative flux to only the difference in radiance, and only toward the lower radiance, as per Maxwell’s equations.
The Willis SGH proposes a perpetuum mobile of the second kind (spontaneity with no result).
The difference means not a wit, to flux through a vacuum, but makes all the difference in a dissipative media like this atmosphere. Again, This is the total demonstrated and perhaps intentional, incompetence of the entire climate science religion!
The reason that I am now persuaded about the Steel Greenhouse is that the result of experiment showed Willis was correct. So I am told. As I said the experimenter has not yet published. I hope he does soon because we NEED TO KNOW!
If I had access to a good vacuum pump I would like to do some more work. At present I do not.
Peter,
The temperature of the constant power inner must increase as the shell interferes with flux to 1 kelvin because of mass rethermalization. Radiative flux cannot get through opaque until opaque is at radiative equilibrium (radiative flux goes through opaque) with no change in sensible heat or temperature of opaque. If Willis is correct, the same fixed power applied only to the shell must via internal flux also increase temperature of sphere (with lower surface area) until net power transfer is zero. The lower area sphere, never can get above the temperature of the shell, radiating all available power outward, never inward, as inward has no lower radiance, and no internal flux at all. Projective geometry can demonstrate no opposing flux ever!
We do not NEED TO KNOW! GOD KNOWS! perhaps! Earthlings only need to think, then decide which way to run with pitchforks and torches against the not GOD. Long ago I was gifted with a lab full of VECO pumps and bell jars. For tight work Vac-Ion pumps. I are retired, sometimes, I can still remember for a while.
Peter,
Thank you, I will await part 4! Please suggest if I should be supportive, or combative?
I have done the physical proof of no opposing EM flux at any frequency!
The experiments did show that the rate of heat loss with CO2 was very slightly less than with common air, but the result was not (quite) significant.
And NielsZoo: (complete reply here)
I’d have to say the reason the heat transfer was slower in the all CO2 vs. all Air is that CO2′s heat properties are different.
.
.
heat transferred is directly proportional to Thermal Conductivity here since everything else is equal (like a good experiment should be.) Data here was sourced from http://www.engineeringtoolbox.com/
I’ll just repeat again, this time with corroborating evidence, kudos for your compliance to the scientific method in all of it’s facets. I think the fact that you got a slightly different result in the rate of heat loss between ‘standard air’ and CO2 alone is because of the fact that the experiment was carried out well enough to detect this difference.
Some commenters on Joanne’s blog believe the concept of “backradiation” leading to a “radiative greenhouse effect” is nonsense, and by discussing the physics (or lack thereof) of backradiation we skeptics are shooting ourselves in the foot. I disagree with that position. However, to get their way the forces aligned against us have certainly used questionable and offensive tactics. Thus, when addressing AGW from anything but a scientific perspective, I believe ridicule is a powerful tool and should be used. To that end, I submit the following.
If a conversation like the one below hasn’t taken place in a Climatology 101 classroom, it’s just a matter of time before it does:
Instructor: Carbon Dioxide or CO2, a heat-releasing gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of the CO2.
Student: Say what! Would you repeat that please?
Instructor: Carbon Dioxide or CO2, a heat-releasing gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of the CO2.
Student: Pardon me, but didn’t you mean to say “a heat-trapping gas” instead of “a heat-releasing gas?”
Instructor: No. I said it the way I wanted to. CO2 is a heat-releasing gas.
Student: But then what you said doesn’t make sense. It makes sense to say that trapping heat will cause an increase in temperature; but it makes no sense to say that releasing heat will cause an increase in temperature.
Instructor: You inferred from what I said that it is the heat-releasing nature of CO2 that causes the temperature increase. That inference is unwarranted. If I had said: “CO2, a colorless odorless gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of CO2,” you would not infer that it was the colorless/odorless nature of CO2 that caused the warming. You would simply have interpreted my statement to imply (a) CO2 is colorless and odorless, and (b) atmospheric CO2 causes an increase in Earth surface temperature.
Student: Now hold on. As an exercise in logic, you’re correct. But if your parenthetical clause “a heat-releasing gas” had been “a heat-trapping gas,” I’m not the only one who would infer that it is the heat-trapping nature of CO2 that causes Earth surface warming. In fact, I’d go so far as to say that way more than half the people would make that inference—especially lay people with little formal training in science or logic.
Instructor: I can’t help what people infer. If they read into my statement more than it actually contains, that’s their problem—not mine.
Student: Again, as a matter of logic, you’re correct. But haven’t I heard you demand mankind reduce its fossil fuel usage to pre-1980 levels because if we don’t the Earth is going reach a “tipping point” where unstoppable runaway global warming will commence to the detriment of all life on Earth?
Instructor: Yes, I have said that.
Student: And isn’t it true that to get the “world” to reduce its fossil fuel usage we must convince Joe-public that mankind’s use of fossil fuels at current or increased levels will be disastrous with the form of that disaster being runaway global warming?
Instructor: Unfortunately, yes. Without Joe public onboard we’ll never make headway in reducing fossil fuel usage. However, Joe public is incapable of understanding the science behind anthropogenic global warming (AGW); and since AGW is a fact, any method used to get Joe-public onboard is justified.
Student: In the case of mankind’s response to AGW, are you saying the ends justify the means?
Instructor: Yes.
Student: But if CO2 is a heat-releasing gas, it seems to me atmospheric CO2 should lower, not raise, the Earth’s surface temperature. I mean after all, if I said XYZ was a heat-trapping gas, then the two properties XYZ should have are: (a) XYZ is a gas, and (b) XYZ traps heat in the sense that if XYZ gas surrounds a hot object it should take the hot object a longer time to cool than in the absence of the surrounding XYZ gas. If calling XYZ a heat-trapping gas doesn’t assign those two properties to XYZ, then calling XYZ a heat-trapping gas is meaningless at best and deceptive at worst; and using the phrase “a heat-trapping gas” must be designed to mislead rather than educate. And by-the-way, CO2 is not always a gas. Isn’t dry ice CO2 in solid form?
Instructor: Yes, dry ice is CO2 in solid form. Calling CO2 “a heat-trapping gas” or a “heat-releasing gas” doesn’t assign the property of being a “gas” to CO2. Rather it defines the state of the CO2—in particular that CO2 is in gaseous form. So when I say “CO2 is a heat-releasing gas,” the only property I’m assigning to CO2 is the property of “heat-releasing,” and I’m caveating that the property only applies to CO2 in gaseous form.
Student: Back the truck up. If CO2 is a heat-releasing gas, then shouldn’t the presence of CO2 gas surrounding a hot object cause the hot object to cool faster than in the absence of CO2 gas?
Instructor: Yes.
Student: But you said CO2 was a heat-releasing gas and the presence of CO2 in the Earth’s atmosphere increases the temperature of the Earth’s surface. Now I’m really confused.
Instructor: Good. I’m using this conversation to assess its effect on the average Joe. A confused person is more likely than a knowledgeable person or even a thinking person to (a) believe there’s a crisis and (b) accept any solution proposed by those in authority.
Student: Let’s put that aside for a moment. Why do you say CO2 is a heat-releasing gas?
Instructor: Well, experiments run with thermos bottles show that CO2 injected into the thermos bottle’s vacuum region causes hot liquid in the thermos bottle to cool faster than it would without the CO2. As such, it can be shown that, everything else being equal, situations exist where CO2 acts to “release heat”.
Student: Okay. You’ve convinced me that there are situations where CO2 gas acts to increase the rate of heat release, and thus for some situations can properly be called a heat-releasing gas. Why, then, should I believe gaseous CO2 in the Earth’s atmosphere will increase the Earth’s surface temperature?
Instructor: Models tell us so.
Student: Models? What models?
Instructor: Models! They’re everywhere. You can’t get on the internet without tripping over a press release touting the latest man-made global warming doomsday model. Every university feasting at the public trough and every environmentalist group worth its salt has a global warming model. It’s your standard government-subsidized industry: produce results and only those results that support the government’s objectives. All the models predict—er, scratch that—all the models include among possible future scenarios a forthcoming manmade global warming crisis.
Student: Are those the models that (a) when plotted on a common graph look like a plate full of spaghetti, (b) predict a hot spot in the troposphere that no one can find via temperature measurements, and (c) didn’t predict the 17+ years of no temperature increase in the presence of ever increasing atmospheric CO2 levels?
Instructor: Yes.
Student: Okay. Let’s see if I have this right. First, fairly simple experiments that almost anyone can perform show that under some conditions CO2 gas acts to increase the rate of heat release and therefore to cause objects to cool faster than they would in the absence of the CO2 gas. Second, the models that foretell a coming CO2-driven manmade global warming crisis (a) predict wildly different future temperature scenarios, (b) predict a tropospheric hot spot that no one can find except via esoteric logic involving potentially increased wind speeds, and (c) didn’t predict the current 17+ year hiatus in global warming even in the presence of increasing atmospheric CO2 levels. Have I summarized the situation correctly?
Instructor: Yes.
Student: Given our conversation, why then should I believe what you say about reducing fossil fuel usage and climb on the AGW bandwagon?
Instructor: You want to pass the class, don’t you?
I really do not understand what you may by trying to say or demonstrate! Are you trying to ridicule AGW, how teachers brainwash students, that CO2 can absorb some properties of EMR at some frequencies under certain circumstances, or that opposing radiative flux can exist at any given frequency. One has nothing to do with the others. It seems to me that what you term ridicule is but encouragement to continue brainwashing innocent children, as it works so very well!
Reed/Peter C 1:45pm: “..in the case of a vacuum thermos bottle versus a CO2 thermos bottle, CO2 gas acts to release heat, not to trap heat….by increasing the rate of conduction”
This agrees with my (et. al) test results.
Unfortunately though – the poorly named “greenhouse” effect won’t be found in replicating Fourier conduction and Newton Law of Cooling tests since energy is not conducted from the atmosphere to deep space at about 3K like a thermos. There exists only radiative transfer from earth and atmosphere to deep space. Your “atmosphere” optical depth in the flask is only several mm thick whereas the earth’s atmosphere is several 10s of kilometers thick. The effect you seek isn’t measurable in your setup – the thermos CO2 atmosphere still has a nearly 0.0 optically thick atmosphere “greenhouse” as does the vacuum.
The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface. Step 2 of the scientific method is found in optical depth analysis coupled with 1st law energy balance & not Fourier conduction and Newton’s cooling law (the one as annunciated by Newton directly).
You have proven thermos manufacturers don’t fill with CO2 due to its resultant increased conduction cooling faster reducing the effectiveness of the replaced vacuum insulation effect cooling relatively slower.
“The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface.”
You claim a greenhouse effect. Then try to say that a poorly done radiometric measurement of the upper atmosphere coupled with a partially measured surface radiance at 8-13 microns compared to some averaged thermometric surface measurements is supposed to have any meaning like a greenhouse effect. Why such utter nonsense?
“Step 2 of the scientific method is found in optical depth analysis coupled with 1st law energy balance & not Fourier conduction and Newton’s cooling law (the one as annunciated by Newton directly).”
You try to do your fake Step 2 of the scientific method and tell us all of the nonsense of your analysis! You will not even be able to define any atmospheric optical depth analysis, over the range of wavelengths and pressures involved, as you cannot determine the exitance solid angle of gas cross sectional area, at any altitude! This is the total demonstrated incompetence of the entire climate science religion!
Will 1:46pm: “..a partially measured surface radiance at 8-13 microns..”
I make no claim just measurement. I wrote thermometers (some mercury type) which measure near surface temperature so they integrate over all wavelengths naturally. Your second paragraph isn’t even parsable.
(“you cannot determine the exitance solid angle of gas cross sectional area”)
“Doesn’t parse.”
Cannot you do geometry? A flat surface of specific area is limited to a solid angle of PI steradians because of cos(theta) an atmospheric cross sectional area radiates exitance to a maximum of 2 PI steradians.
“And atm. tau is well defined despite your claim.”
Yes indeed Miskolczi did a fine job of apparent radiance in a direction normal to some surface. He completely forgot about the solid angle.
Will 6:28am: Consider a cavity, spherical if you wish. The radiation inside is isotropic at temperature equilibrium. Imagine a flat plane surface within. Regardless of how your flat plane surface is oriented the same amount of radiant energy crosses that unit area in unit time. Planck considered the radiant energy (photons) propagating in a hemisphere of directions either above or below the surface. Thus, with additional considerations, atm. tau is well defined.
The “effect I seek” is that CO2 gas does not always “trap heat,” where the process of “trapping heat” implies a temperature increase to an object in the region where the heat is “trapped.” The physical reasons CO2 gas doesn’t “trap heat” are of interest and may very well be related to the optical depth of the CO2 gas. Therefore the effect of atmospheric CO2 gas on Earth surface temperature may very well be to increase the Earth surface temperature. But if so, the claim by itself that since CO2 gas is a “heat-trapping” gas, it must increase the Earth surface temperature doesn’t cut it with me. Additional arguments are necessary.
Look at it this way. Someone comes up to me and says:
(1) “The Earth’s surface temperature in the presence of greenhouse gases will be warmer than the Earth’s surface temperature in the absence of those gases.”
(2) I ask: “Why do you say that?”
(3) They answer: “Because greenhouse gases absorb some of the IR radiating from the Earth’s surface. Thus, greenhouse gases trap heat emitted from the Earth’s surface; and trapped heat increases the temperature of objects within the ‘trap’.”
(4) I ask: “Is CO2 a greenhouse gas?”
(5) They answer: “Yes.”
(6) I ask: “Since CO2 is a greenhouse gas and greenhouse gases trap heat, then it logically follows that CO2 gas traps heat. Is that correct?”
(7) What else can the answer but: “Yes, CO2 gas traps heat.”
(8) I then ask: “From what you said, it logically follows that everything else being equal the temperature of any object emitting heat via IR radiation and surrounded by CO2 gas must increase relative to its temperature without the greenhouse gas. Is that correct?
(9) Again, I don’t see how they can answer anything but “Yes.”
(10) I then ask: “If I build a thermos bottle whose inner chamber is transparent to radiation below the visible band and I fill the vacuum region of a vacuum thermos bottle with CO2 gas, does CO2 gas surround the contents of the thermos bottle?”
(11) They answer: “Not entirely. The thermos bottle chamber is kept separate from the thermos bottle by rigid material, so that for the most part, yes, the contents are surrounded by CO2 gas, but not entirely.”
(12) I then ask: “What if the rigid material keeping the chamber separated from the wall of the thermos bottle is (a) transparent to radiation below the visible band, and (b) bent such that the straight line path from any point within the thermos bottle chamber to any point outside the thermos bottle’s external wall must pass through CO2 gas. Is such a construction possible?”
(13) They answer: “I’m not sure, but for the sake of argument I’ll stipulate that such ‘rigid spacers’ can be constructed and if constructed, then yes CO2 gas surrounds the contents of the thermos bottle. What’s your point?”
(14) I answer: “My points are: (a) CO2 gas surrounds the contents of the thermos bottle. (b) The frequencies of radiation emanating from material at temperatures near 300K and stored in the inner chamber are for the most part below the visible band. (c) According to your claims [3 and 5], CO2 gas will ‘trap’ some of the heat emanating from the thermos bottle contents, and the trapping of that heat will result in an increase in the temperature of the thermos bottle contents. Thus, considering only those two claims and stipulating that with the exception of what resides in the vacuum region of a thermos bottle everything else is equal, the contents of a CO2 thermos bottle should be at a higher temperature than the contents of a vacuum thermos bottle. That conclusion contradicts experiment. How can that be?”
(15) At this point arguments such as optical depth, backradiation, convection, evaporation, surfaces painted with highly reflective materials, etc. may arise and justify the claim that atmospheric CO2 gas will increase the Earth’s surface temperature. But before I’m convinced that atmospheric CO2 must increase the Earth surface temperature, I want to hear those arguments. The “heat-trapping” nature of CO2 by itself doesn’t convince me!
As to the issue of optical depth. That may explain why CO2 gas in a thermos bottle has a different effect on the temperature of the thermos bottle’s contents than atmospheric CO2 gas has on the Earth’s surface temperature. But then, you are providing an additional argument over and above the simple “heat-trapping” gas argument.
Next, I’ll accept the fact that the IR “optical depths” of a thermos bottle and the atmosphere are many orders of magnitude different. But isn’t the optical depth in part a function of the density of the material through which the radiation is propagating? For a thermos bottle it may take enormous pressure in the volume occupied by the CO2 gas, but with sufficient pressure (assuming the pressure doesn’t liquefy or solidify the CO2 at room temperatures) the optical depth of the CO2 in a thermos bottle can be made arbitrarily large. It’s very likely that the optical depth of a real thermos bottle will never approach the optical depth of the atmosphere; but by making the distance between the inner chamber’s outer wall and the thermos bottles outer wall large and by increasing the pressure of the CO2, it seems to me we can construct a thermos bottle whose “optical depth” is approximately the same as the atmosphere’s “optical depth.” Neither I nor Peter has run such an experiment, but if I had to bet, I’d bet that if such a thermos bottle were constructed, the equivalent vacuum thermos bottle would outperform the CO2 thermos bottle.
Finally, your statement: “The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface” confuses me. Who said anything about measuring the 33K “greenhouse” effect by considering a “too thin thermos atm”? In a previous post, I argued that the 33K number comes from an illogical methodology for computing the Earth’s surface temperature in the absence of greenhouse gases.
I understand and at this time won’t disagree with the thermometer measurement of 288K. However, I want to discuss radiometer measurements of the Earth’s surface temperature. Radiometers measure the intensity of electromagnetic radiation. Thus, a radiometer above the Earth/Earth-atmosphere pointed towards the Earth measures the intensity of the electromagnetic radiation propagating in a direction away from the Earth. Some of this radiation is emitted from the Earth/Earth-atmosphere system and some of this radiation is reflected solar radiation. If a radiometer immediately above the Earth’s atmosphere and pointed towards the Earth doesn’t “see” reflected solar radiation, then where does reflected solar radiation go? If the Earth is in energy-rate-equilibrium, the total (emission and reflection) rate energy leaves the Earth/Earth-atmosphere system must equal the rate solar radiation is incident on the Earth/Earth atmosphere system. The latter rate is approximately 1,366 Watts per square meter. For a spherical object (absorbs as a disk, radiates as a sphere) whose surface emissivity is (a) not a function of frequency, (b) uniform over the surface, but (c) may have a value other than unity, the temperature associated with an incident power density of 1,366 Watts per square meter is approximately 278.6K, not 255K. This translates to a temperature difference of approximately 10K, not 33K.
If from the “total outgoing” radiation you subtract the portion from solar reflection, then yes, the temperature of a spherical object will be approximately 255K. But clouds reflect much if not most of the solar energy. If you remove all atmospheric greenhouse gases, you remove all water vapor, and if you remove all water vapor you remove all clouds. If you remove clouds, you remove the primary means of solar reflection. The Earth would then approximate a sphere with a uniform emissivity, and its surface temperature would be nearer 278K than 255K.
Bottom line, when computing (not measuring, but computing) the Earth’s surface temperature in the absence of greenhouse gases, it’s illogical to perform that computation using a value that depends on the presence of greenhouse gases. I don’t accept the statement that radiometers outside the Earth’s atmosphere and pointed towards the Earth measure radiation only emitted from the Earth’s surface. How can they not measure the sum of Earth-emitted radiation and reflected solar radiation? It’s only after you subtract out the reflected solar radiation that the intensity corresponds to a blackbody object at 255K. But without water vapor there would be no clouds and without clouds there would be no (or very little) reflected solar radiation, and without reflected solar radiation, there’s nothing to subtract from the radiometer measurements.
Reed 3:51pm: The “effect I seek” is that CO2 gas does not always “trap heat,” where the process of “trapping heat” implies a temperature increase to an object in the region where the heat is “trapped.”
The poorly named “green house” optical depth effect you seek really is operating in the few mm of the CO2 filled flask but is way overpowered by conduction so as to be not measurable.
“I argued that the 33K number comes from an illogical methodology..”
Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. 1st law and optical depth analysis though can be used to logically compute 33K in step 2 from measured input data, this work out is found in many modern text books & specialist papers. Look it up. The emissivity of any object can be routinely measured – this means it is routine to separate the amount of reflected radiation (for example earth albedo) and the amount of emitted radiation as they are LW and SW respectively. Reflectivity + absorptivity + transmissivity = 1.0.
——
Peter C 5:09pm: How much? Dunno offhand.
My going in approach would be find the precision calibration limits of today’s instruments and use grey IR differential optical depth analysis across a layer of atm. thickness to compute the optical depth (commonly tau) under hydrostatic condition that just becomes reasonably measurable above 0.0K by them at say 95% or 99% confidence interval given the partial pressures, mixing ratios and each of N constituents source of opacity (mass extinction coefficient) for earth (or any) atmosphere. Get a reasonable number somewhere between a few mm (0.0K) and 10’s of kilometers (33K for earth). A literature search may turn up this is already reasonably known.
You insist that the artificial 33 C difference in temperature has some possible meaning.
The 255 Kelvin is a computation of a black-body, never this Earth with its atmosphere There is.. Surface temperature 225-306 Kelvin as many measurements, all individual, all unique. There is no possible way that your 288 Kelvin has any meaning whatsoever! There is no possible (mass extinction coefficient) for this dynamic Earth atmosphere. Your whole approach is quiet Juvenile. Go somewhere and buy a clue!
Will 1:36am: The 255K is measured by precision calibrated instruments. So is the 288K. Both are meaningful in science traceable to reasonable CI – but do remember we are just practicing science, we haven’t perfected it yet. The patent office is still open.
” Will 1:36am: The 255K is measured by precision calibrated instruments. So is the 288K. Both are meaningful in science traceable to reasonable CI – but do remember we are just practicing science, we haven’t perfected it yet. The patent office is still open.”
The 215 Kelvin estimated at tropopause is reliable, never accurate, 255 Kelvin somewhere about 5 km is but your fantasy! Normal (vertical) surface radiometric, without clouds, in 8-13 microns goes between 245 Kelvin and 315 Kelvin. Why do you think your average of 288 Kelvin has any meaning for anything?
Will 4:38am: ..255 Kelvin somewhere about 5 km is but your fantasy!”
No fantasy, I looked up many precision calibrated sounding rocket data reports – their onboard thermometers measure the 255K around 5km depending on latitude.
“Why do you think your average of 288 Kelvin has any meaning for anything?”
288K is the value measured with precise enough calibrated instruments that are reasonably maintained.
Again Why do you think your average of 288 Kelvin has any meaning for anything?”
“288K is the value measured with precise enough calibrated instruments that are reasonably maintained.”
And again why do you think some average temperature of various mass with various specific heat, have any meaning whatsoever?
Will 5:56am: “..why do you think some average temperature..”
I didn’t. Please define what you mean by average. Do you mean arithmetic average (or arithmetic mean) or median (50% above & 50% below), most probable value (or mode), weighted average, root mean square (or nth root), so forth, the number of possible averages is boundless Will.
Or maybe you meant like page one of Maxwell’s Theory of Heat where he discussed hot and cold temperatures have an intermediate temperature like my hot and cold faucet running lukewarm out of the tap. What?
You wrote: “No fantasy, I looked up many precision calibrated sounding rocket data reports – their onboard thermometers measure the 255K around 5km depending on latitude.”
5km isn’t the surface of the earth or anywhere near it. The 33K claim has nothing to do with the temperature of the Earth’s atmosphere at an altitude of 5k. The claim is that in the absence of greenhouse gases the Earth’s surface temperature will be 33K cooler than the measured surface temperature of 288K. Since (a) temperature decreases with altitude by approximately 6.4C per kilometer and (b) the measured temperature at the surface of the Earth is approximately 288K, it’s not surprising that at an altitude of 5km the temperature is approximately 288 – 5×6.4 = 256K. But this has nothing to do with the 33K claim. Comparing the measured Earth surface temperature against the measured atmospheric temperature at an altitude of 5km, getting a temperature difference of 33K, and claiming that the 33K temperature difference applies to the Earth surface temperature difference with and without atmospheric greenhouse gases is just plain silly, or deliberately deceptive.
Later in a comment time stamped 12:46 am, 24 March, 2015, you wrote: “Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. 1st law and optical depth analysis though can be used to logically compute 33K in step 2 from measured input data, this work out is found in many modern text books & specialist papers. Look it up.”
The last time you requested I look something up (the meaning of Newton’s law of cooling), I did and you chided me for not going to Newton’s original wording. Suppose we avoid a repeat of my finding the wrong reference and you give me the reference where the 33K number is derived from 1st law and optical depth analysis.
Next, if you have a highly thermally conducting sphere of radius r=6,371,000 meters (the approximate mean radius of the Earth) and surface temperature “Tes” with a surface emissivity that is independent of frequency and is everywhere epsilon (a real number greater than 0 and less than or equal to 1) and you place that sphere a distance, 150,000,000,000 meters from the sun (the approximate distance from the sun to the Earth), then to a good approximation
(a) for a solar radius of 695,000,000 meters and a solar surface temperature of 5,778K, the solar power density at the sphere is approximately 1,359 watts per square meter.
(b) the rate solar energy is incident on the sphere is pi x r^2 x 1,359 = 1.733 x 10^17 watts–that is, the sphere absorbs energy as if its shape were a flat circular disk oriented perpendicular to the solar radiation
(c) the rate the sphere absorbs solar energy is “epsilon x 1.733 x 10^17 watts”
(d) the rate the sphere radiates energy is “epsilon x 4 x pi x sigma x r^2 x (Tes)^4”, where sigma is the Stefan-Boltzmann constant–i.e., the sphere radiates energy from its entire surface
If the rate the sphere absorbs solar energy is equal to the rate the sphere radiates energy, then the temperature of the sphere is (approximately)
Tes = 278.2K
“Tes” corresponds to the energy-rate-equilibrium temperature of an Earth-like object (devoid of an atmosphere) at an Earth-like distance from the sun. Provided the sphere’s surface emissivity equals the sphere’s surface absorptivity (which is true for black and gray bodies), the temperature of 278.2K is independent of the emissivity/absorptivity. Assuming negligible energy is absorbed/radiated by non-greenhouse gases in the Earth’s atmosphere, the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere.
Next, you wrote: “The emissivity of any object can be routinely measured – this means it is routine to separate the amount of reflected radiation (for example earth albedo) and the amount of emitted radiation as they are LW and SW respectively.” It is NOT routine (or rather if it is routine, then the routine is misleading) to use the albedo of an object in “State 1” to compute the rate the object absorbs energy, and then use that absorption rate in conjunction with the albedo of the object in “State 2” to compute the “State 2” temperature of the object. “State 1” is the Earth with a greenhouse gas atmosphere. The Earth’s State 1 albedo is approximately 0.3. “State 2” is the Earth devoid of all atmospheric greenhouse gases. The Earth’s State 2 albedo is approximately 0. If you believe it’s okay to use the Earth’s albedo (0.3) in the presence of atmospheric greenhouse gases to compute the rate the Earth absorbs solar energy and to use that absorption rate in conjunction with the Earth’s albedo (0) in the absence of atmospheric greenhouse gases to compute the Earth’s surface temperature in the absence of a greenhouse gas atmosphere, then we will likely never reach agreement on this topic. Furthermore, if the AGW community defends such a procedure, then you can think whatever you want; I think the AGW community is telling the public the emperor is wearing a beautiful suit.
Finally, in your 5:34 am, 24 March 2015 comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.”
Say What? Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df from a planar differential unit area, dA, at temperature T, into a differential solid angle domega defined with respect to a coordinate system at dA. There are no terms in Planck’s blackbody radiation law for the presence/absence of any other surfaces. In fact, it is a geometrical impossibility for a planar differential surface area to radiate into a differential angle that contains any portion of the radiating differential area. Provided the radius of curvature of an object is finite and non-zero, ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.
For the interior surface of a closed object (like the interior surface of a spherical shell), you can use Planck’s law to compute the power radiated from any finite size area of the interior surface to any other finite size area of the interior surface. It turns out that if no objects reside in the interior surface of a closed object, the total rate energy is radiated from the entire interior surface is equal to the total rate the interior surface absorbs radiated energy. As such, the net rate of energy exchange of the interior surface is zero. However, if an object (Object A) exists within the interior region of a closed surface, you can use Planck’s law to determine how much energy is radiated from the interior wall of the enclosed surface to Object A and how much of the energy is radiated from the interior wall to other portions of the interior wall.
For the rate energy leaves an object, the Stefan-Boltzmann Law, which contains a term for the total surface area of an object, doesn’t apply if portions of the object’s surface radiate to other portions of the object’s surface. So what you warn against applies to the Stefan-Boltzmann law of object radiative cooling, but not to Planck’s blackbody radiation law.
Reed 10:35am: “5km isn’t the surface of the earth or anywhere near it…”
I was responding to Will. Agree with your 1st paragraph.
“Suppose we avoid a repeat of my finding the wrong reference..”
Just make sure any ref. you use traces accurately to the original author, or chiding is fair game. You will have to do your own work here, otherwise if I suggest a modern text author, you won’t have anything invested.
“Tes = 278.2K”
Yes, for a no-atmosphere sphere situated as indicated with emissivity of 1.0, rounded up say from natural 0.95 and albedo of 0.0.
“..the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere.”
Not according to the 1st law. When you add the current atmosphere pressure with almost 0 optical thickness i.e. very optically thin atm. and make an earth out of the sphere, albedo goes to 0.3 as solar reflected is measured by radiometers (surface liquid/ice water & dirt & gas now reflect solar) and the Tes goes to 255K by 1st law.
If you want to argue that albedo, know that the albedo for NH and SH is measured the same despite much difference in land & water surface area. If you want to argue, pick any number for albedo 0 to 0.3 with optically thin atmosphere, then your Tes will be between 255K and 278.2K depending on your choice, one is as good as any other, just specify your choice of albedo.
“Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df”
Where did you get that from? No. Chide you again for not going to the ultimate source. Planck’s 1912 paper on-line gives the ideal specific intensity from an object at each frequency and temperature. For a range of frequencies, a curve can be constructed at each temperature. This was from his testing and many others illuminating an object with cavity BB radiation. His paper limits the object to positive radii and negligible diffraction, look up the paper. Invest in it.
“..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.”
Not according to Planck’s own writing in the 1912 paper introduction. If you want I can explain how to prove that to yourself by simple experiment with IR camera.
Your 2nd to last paragraph is agreed generally, as the interior body is illuminated with perfect BB radiation. This illumination won’t exist anywhere outside the cavity.
Your last paragraph is wrong, the limitation applies to Planck specific intensity law, which you can easily find reading the 1912 paper general introduction & eqn. 274. The same limitation then applies to S-B.
LtCusper, 2:02 am, 24 March 2015; and 12:46 am, 24 March 2015
In your 2:02 am comment, you wrote: “The 255K is measured by precision calibrated instruments. So is the 288K.” In your 12:48 am comment you wrote: “Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. ” Unless you’ve either (a) devised a way to remove all greenhouse gases from the Earth’s atmosphere, or (b) been able to construct a replica of the Earth whose atmosphere is devoid of greenhouse gases, it’s impossible to make measurements of the Earth surface temperature in the absence of greenhouse gases. All measurements will be made for the atmospheric conditions that exist at the time of the measurements, which means all measurements are for an atmosphere that contains greenhouse gases. Thus, the 255K measured temperature, the 288K measured temperature, and the temperature obtained by differencing these measurements are for an Earth with a greenhouse gas atmosphere. To represent any of these temperatures as being a direct measurement of the Earth surface temperature in the absence of greenhouse gases is simply wrong.
Reed 3:10pm: “it’s impossible to make measurements of the Earth surface temperature in the absence of greenhouse gases….To represent any of these temperatures as being a direct measurement of the Earth surface temperature in the absence of greenhouse gases is simply wrong.”
I agree. Wouldn’t want that to happen. Still, what I wrote is true. Both 255K and 288K are measured with precision calibrated instruments.
This is the response I said I would post. Before getting into the body of the response, a few introductory remarks are in order. First, when I quote Max Planck’s 1912 writing, I’m not quoting the German original, I’m quoting Morton Masius’s 1914 translation of Planck’s writing—see
Second, in the Preface to the Second Edition (pages viii and ix) of Masius’s 1914 translation, Planck wrote: “In contrast thereto I have now attempted to treat the subject from the very outset in such a way that none of the laws stated need, later on, be restricted or modified. This presents the advantage that the theory, so far as it is treated here, shows no contradiction in itself, though certainly I do not mean that it does not seem to call for improvements in many respects, as regards both its internal structure and its external form.”
And in the last paragraph of that Preface (page ix) Planck wrote: “Hence it follows from the nature of the case that it will require painstaking experimental and theoretical work for many years to come to make gradual advances in the new field.”
I interpret these words to mean that Planck expected work, both experimental and theoretical, in the field of Heat Radiation to continue into the future. Let’s call the current state of that work “Our Modern Understanding of the Theory of Heat Radiation.” Our Modern Understanding of the Theory of Heat Radiation may in no way conflict with Planck’s 1912 paper. If it does conflict, adjudicating which one, if either, is correct is beyond my paygrade. However, if a conflict arises and I was forced to choose between the two, I’d choose the modern version for no other reason than authors following Planck had access to his writing and thus could analyze an apparent discrepancy from two perspectives, theirs and Planck’s; whereas Planck didn’t have access to all modern texts and thus couldn’t analyze any apparent discrepancy. In any event, since Planck published his 1912 paper, many authors have written texts on the subject. Include those texts in Our Modern Understanding of the Theory of Heat Radiation. You ask, what is the point of this remark? The point is that if you decide to answer some of the questions I pose in this response, please do so from two perspectives—the perspective of the knowledge of physics as put down in Planck’s 1912 paper (chiding allowed and even encouraged:-), and the perspective of Our Modern Understanding of the Theory of Heat Radiation.
Third, in the lifetime of our interaction on this thread, I have no hope of reading and understanding all of Planck’s 1912 paper. Thus, relative to this response to your comment, my understanding/interpretation of Planck’s 1912 paper must be hit and miss. Be patient with me.
Fourth, in this response (and if I can remember, in all future responses on Joanne Nova threads), unless explicitly stated otherwise, I assume (a) at all locations on the Earth’s surface and for all frequencies, the transmissivity is zero, (b) at all locations on the Earth’s surface, emissivity is (i) independent of frequency, (ii) independent of temperature, (iii) greater than 0 and less than or equal to unity, and (iv) equal to the absorptivity, and (c) the Earth’s emissivity/absorptivity is uniform over the surface of the Earth.
Fifth, I hope to respond to all your 2:50 pm, 24 March 2015 points; but the order of my responses will not match the order of your points.
Now to my responses.
(1) I argued that the temperature of an Earth-like sphere positioned at an Earth-like distance from the sun and in energy-rate-equilibrium (ERE) with solar energy would be approximately Tes = 278.2K. To that argument you responded: “Yes, for a no-atmosphere sphere situated as indicated with emissivity of 1.0, rounded up say from natural 0.95 and albedo of 0.0.” Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes. I say this because the rate energy is absorbed by a surface is directly proportional to the absorptivity of the surface; and the rate energy is radiated from a surface is directly proportional to the emissivity of the surface. Thus, when writing an ERE equation for a surface, absorptivity appears as a multiplying factor on one side of the equation and emissivity appears as a multiplying factor on the other side. If absorptivity equals emissivity, then provided these terms are not zero, the resulting ERE temperature of the object will be independent of the value of emissivity/absorptivity. So for a no-atmosphere sphere and equal non-zero absorptivity/emissivity, the ERE surface temperature of an Earth-like sphere is approximately 278.2K independent of the absorptivity/emissivity of that Earth-like sphere. Do you agree or disagree?
(2) To compute the ERE temperature, I use the relationship that the rate energy is radiated from a surface is directly proportional to the fourth power of the surface temperature, T, in Kelvin. The fourth power relationship can be derived by integrating the “spectral radiance, Bf(f,T), function–see
over all frequencies, f, from zero to infinity. The T^4 relationship is not in general valid for an emissivity that changes with frequency. By using the T^4 relationship to generate ERE temperature estimates, I am implicitly assuming the emissivity is a non-zero constant at all finite frequencies. Now it may be that at sufficiently low temperatures, the contribution to the integrated spectral radiance from frequencies above a cutoff frequency is negligible (e.g., the emissivity may be a step function: a non-zero constant below a cutoff frequency and zero above the cutoff frequency) so that the T^4 relationship is a good approximation. Bottom line, if you employ the T^4 relationship to compute ERE temperatures for situations where the emissivity is frequency dependent, please so note. For your comments to date, I assume you arrive at ERE temperatures using the T^4 relationship. Is this assumption correct?
(3) I wrote: “..the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere” to which you responded:
“Not according to the 1st law. When you add the current atmosphere pressure with almost 0 optical thickness i.e. very optically thin atm. and make an earth out of the sphere, albedo goes to 0.3 as solar reflected is measured by radiometers (surface liquid/ice water & dirt & gas now reflect solar) and the Tes goes to 255K by 1st law.
“If you want to argue that albedo, know that the albedo for NH and SH is measured the same despite much difference in land & water surface area. If you want to argue, pick any number for albedo 0 to 0.3 with optically thin atmosphere, then your Tes will be between 255K and 278.2K depending on your choice, one is as good as any other, just specify your choice of albedo.”
I’ll take your word for it that radiometers have to a high degree of precision measured the rate of Earth/Earth-Atmosphere reflected solar energy, and have determined the reflected rate to be 30% of the rate solar energy is incident on the Earth/Earth-Atmosphere. Since the “average Earth albedo” is the ratio of “the rate of reflected solar energy” …to… “the rate of incident solar energy,” I agree that the average Earth albedo is 0.3. An average Earth albedo of 0.3 means the Earth absorbs solar energy at a rate of approximately 1.21 x 10^17 watts. Assuming energy-rate-equilibrium and applying the “1st law,” the Earth/Earth-Atmosphere must then radiate energy to space at rate of 1.21 x 10^17 watts. The rate energy is radiated from the Earth/Earth-Atmosphere to space is NOT a direct measure of the Earth’s surface temperature for either a greenhouse gasless Earth or a greenhouse gas Earth. For both kinds of atmospheres, the radiated energy rate must be converted to a temperature. If (a) the Earth’s surface is treated as a graybody (emissivity, e, independent of frequency and location) and (b) atmospheric gases neither absorb nor radiate energy, then via the equation
Radiated Power = e*sigma*A*(Tes)^4, where sigma is the Stefan-Boltzmann constant and “A” is the surface area of a smooth Earth,
the temperature Tes (assumed uniform) of a blackbody Earth surface (e=1) that produces an output energy rate of 1.21 x 10^17 watts is approximately 254.3K. I believe this is the methodology used to arrive at the 255K Earth surface temperature in the absence of atmospheric greenhouse gases.
[Note: As mentioned in (2) above, for an Earth with unity emissivity at “low” frequencies and less than unity emissivity at “high frequencies,” the factor of (Tes)^4 in the above equation is not theoretically valid. This factor is theoretically valid only for an emissivity that is independent of frequency. If the emissivity changes with frequency, the above equation may be a good approximation, but it is not exact.]
If this is the methodology used to arrive at the 255K temperature, then there are two problems: a minor problem and a major problem. The minor problem is the computation of Tes using a blackbody model. From the above equation, the temperature Tes is inversely proportional to the fourth root of e. If an emissivity (e=0.7) consistent with the albedo of 0.3 is used, Tes would be approximately 278K. To me, the use of an emissivity of unity with an albedo of 0.3 is at most a minor problem because energy is radiated from the Earth’s surface predominately over a frequency range (the emission power frequency, EPR, range) where the emissivity is close to unity. If the emissivity drops outside the EPR range, then the (Tes)^4 term in the above equation is not rigorously valid, but it is a good approximation. And because a significant portion of solar radiation exists at frequencies outside the EPR range, an emissivity near unity over the EPR range but less than unity outside the EPR range can easily produce an average albedo of 0.3. If that is the situation for the real Earth, then this “minor problem” is not an important issue with me.
The major problem, which is an issue with me, is that of using a value of 1.21 x 10^17 watts for the rate energy is radiated away from a greenhouse gasless Earth. The measured average Earth albedo of 0.3 can be segmented into two additive parts: (a) one part associated with solar reflection from atmospheric greenhouse gases and their artifacts (greenhouse gas artifacts are objects that are present in a greenhouse gas atmosphere but absent in a non-greenhouse gas atmosphere), and (b) one part associated with solar reflection from everything else.
[Note: The division of the measured average Earth albedo into the two additive parts described immediately above is not rigorously valid. Such a division assumes solar radiation reflected by one “part” will, in the absence of that “part,” not be reflected by the other “part.” This is obviously not rigorously true for the Earth. Some of the solar radiation reflected by clouds (a greenhouse gas artifact) will, in the absence of clouds, be reflected by something in the “everything else” category. Call solar energy that (a) is reflected to space by atmospheric greenhouse gases and/or their artifacts, and (b) in the absence of atmospheric greenhouse gases and their artifacts, would be reflected to space by something in the “everything else” category duplicate reflection. What follows is highly dependent on the amount of duplicate reflection. A small level of duplicate reflection, increases the relevance of the following argument. A large level of duplicate reflection, decreases the relevance of the following argument. As several references I have come across point out: Earth albedo is a complex phenomenon.]
Consider, for example, the situation where the measured albedo of 0.3 comes only from atmospheric greenhouse gases and their artifacts—i.e., the Earth’s albedo would be zero in the absence of greenhouse gases and their artifacts. For this situation, the Earth absorbs solar energy at an approximate rate of 1.73 x 10^17 watts; and for an Earth surface having an emissivity of unity, the greenhouse gasless Earth surface temperature would be approximately 278.2K. At the other extreme, if none of the measured albedo of 0.3 comes from atmospheric greenhouse gases or their artifacts—i.e., the Earth’s albedo would be 0.3 in the absence of greenhouse gases and their artifacts, a greenhouse gasless Earth would absorb solar energy at an approximate rate of 1.21 x 10^17 watts, and for an Earth surface having an emissivity of unity, the greenhouse gasless Earth surface temperature would be approximately 254K. Thus ignoring duplicate reflection, depending on what portion of the 0.3 measured albedo comes from atmospheric greenhouse gases and their artifacts and what portion comes from “everything else,” the greenhouse gasless Earth surface temperature can vary between approximately 255K and 278K. Or if duplicate reflection is included, then depending on the amount of duplicate reflection, the greenhouse gasless Earth surface temperature can vary between approximately 255K and 278K.
So it’s not so much that I want to argue about the measured value of the albedo (I’ll accept a value of 0.3), as it is that I want to argue about using the measured value to compute the surface temperature of a greenhouse gasless Earth. Specifically, what are (a) the contribution to the measured albedo of atmospheric greenhouse gases and their artifacts, and (b) the contribution to the measured albedo of everything else? From http://en.wikipedia.org/wiki/Albedo: “The average overall albedo of Earth, its planetary albedo, is 30 to 35% because of cloud cover, but widely varies locally across the surface because of different geological and environmental features [bold emphasis mine].
Clouds are a greenhouse gas artifact in that in the absence of water vapor (the dominant atmospheric greenhouse gas) clouds would not exist. Based on the above Wikipedia quote, a large part of the measured 0.3 albedo comes from atmospheric greenhouse gases and their artifacts. If Wikipedia is correct, a greenhouse gasless Earth surface temperature would be closer to 278K than to 255K.
To use your phraseology, if you want to argue about the contribution clouds make to the measured albedo and more importantly to what the albedo would be in the absence of clouds, that’s fine. But know that Wikipedia says the contribution from clouds is significant. When it comes to anything related to global warming, I’m not a big fan of Wikipedia; but there it is.
In summary, computation of a greenhouse gasless Earth surface temperature from measurements of the Earth’s albedo, no matter how accurate those measurements are, requires an algorithm to estimate the Earth’s albedo in the absence of atmospheric greenhouse gases and their artifacts. Wikipedia may be wrong when it implies clouds (greenhouse gas artifacts) are a major contributor the Earth’s albedo; but I’ve seen/heard/read nothing that says clouds contribute negligibly to the Earth’s albedo. As I see it, the oft quoted greenhouse gasless Earth surface temperature of approximately 255K relies heavily on the albedo of a greenhouse gasless Earth being nearly the same as the measured (with a greenhouse gas atmosphere) Earth albedo. If true, I have no problem with the 255K temperature. If not true, I do have a problem.
(4) I wrote: “Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df.” You responded: “Where did you get that from?” Here’s the answer to your question.
Assume you have a variable “A” that is a function of a second variable “r”. Someone gives you a formula, not for A(r) itself, but rather for the value of A(r) per unit length of “r”. For example, you’re interested in the area of a closed, planar, geometrical shape. Someone gives you the formula Y(r)=2*pi*r, and tells you that the formula represents the area of your geometrical shape per unit length of “r”. Provided the formula meets all the mathematical requirements (differentiable, integrable), it is routine to say: The area of the geometrical argument between r and r+dr is given by: Y(r)*dr. The area of the geometrical shape between r1 and r2 can be computed by integrating Y(r)*dr between the limits r1 to r2.
Item 6, page 6, of Planck’s 1912 document, uses just such terminology. Specifically, Planck wrote:
“Summing up everything said so far, we may equate the total energy in a range of frequency between f and f+df emitted in the time dt in the direction of the conical element dW by a volume element dV, [by] dt*dV*dU*df*2*e. The finite quantity e is called the coefficient of emission of the medium for the frequency f.”
Planck goes on to say: “The total emission of the volume element dV may be obtained from this by integrating over all directions and all frequencies. Since e is independent of the direction, and since the integral over all conical elements dW is 4*pi, we get: dt*dV*8*pi*(Integral from zero to infinity of e*df).”
From Wikipedia [http://en.wikipedia.org/wiki/Planck’s_law] I got an expression for the spectral radiance of a body as a function of frequency. According to Wikipedia, “The spectral radiance of a body, Bf, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body at absolute temperature T is given by:” and then Wikipedia gives a formula. It is not easy to write equations in a format compatible with comments on this blog. Suffice it to say the formula given by Wikipedia contains a factor of f^3 and a factor of 1/(e^K – 1), where K=h*f/(kB*T).
I admit that as of today (a cursory reading of Planck’s 1912 paper) I haven’t found an equivalent formula in Planck’s original paper. Maybe one doesn’t exist, in which case I won’t be able to give a reference directly attributable to Planck. However, the above discussion references modern texts and should answer your question: “Where did you get that from?”
(5) I wrote: “..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.” To which you responded: “Not according to Planck’s own writing in the 1912 paper introduction. If you want I can explain how to prove that to yourself by simple experiment with IR camera. I’ll describe a situation, ask you a question, and wait for your answer.
Assume you have a solid rectangular bar of uniform material and dimensions h x h x (2h+d). Use a saw with a blade width of exactly d to cut the bar into two identically oriented, identically sized cubes each having dimensions h x h x h and whose “cut faces” are separated by a distance d. Identify these cubes with the letters “L” and “R.” Assume all surfaces of both cubes are blackbody surfaces. Finally, assume the temperature of cube L is everywhere TL and the temperature of cube R is everywhere TR. I want to compute the rate energy is radiated from cube L to cube R. The geometry of the cubes is such that all points on the surface of cube L that have line-of-sight visibility to any point on the surface of cube R reside on the “cut face” of cube L; and vice-versa for points on the surface of cube R having line-of-sight visibility to any point on the surface of cube L. I’m going to describe a procedure for computing the rate energy radiates from cube L to cube R. I would like to hear your opinion of that procedure. Note: According to Planck (at the bottom of page 6), the emission from cube L to cube R is independent of the TL and TR, so that with the obvious substitutions, if the procedure I describe is valid for radiation from cube L to cube R, it is also valid for radiation from cube R to cube L.
(A) Define a rectangular, Cartesian coordinate system (a) whose origin is at the center of the “cut face” of cube L,” (b) whose z axis is perpendicular to the “cut face” of cube L in the direction to cube R, (c) whose x axis is parallel to any “side” of the cube L “cut face,” and (d) whose y axis completes a right-handed rectangular Cartesian coordinate system.
(B) Select a differential area, dSL=dxL*dyL at position xL, yL on the “cut face” of cube L.
(C) Select a differential area, dSR=dxR*dyR, at position xR, yR on the “cut face” of cube R.
(D) For those two differential surface areas, compute the angle, B, between (a) the normal to the dSL …and…(b) the vector from dSL to dSR. The tangent of B is given by:
TANGENT(B) = Sqrt[(xR – xL)^2 + (yR – yL)^2] / d
Note: (a) B is a function of xL, yL, xR, yR and the constant d; and (b) because the normal to the “cut face” of cube L points in the opposite direction to the normal to the “cut face” of cube R, the angle between the normal to dSR and the vector from dSR to dSL is also B.
(E) As seen from dSL, the solid angle, dW, subtended by dSR is
dW = dxR*dyR*COS(B)/[ (xR – xL)^2 + (yR – yL)^2 + d^2 ]
(F) From Equation 6, page 13 of Planck’s paper, the rate energy is radiated from dSL in the direction of dSR is directly proportional to dSL*COS(B)*dW. Thus, the rate energy is radiated from dSL towards dSR is proportional to
dxL*dyL*COS(B)*dxR*dyR*COS(B) / [ (xR – xL)^2 + (yR – yL)^2 + d^2 ]
(G) Integrate the Equation in (6) above over the face of cube R—i.e., integrate dxR from –h/2 to +h/2 and dyR from –h/2 to +h/2.
(H) Integrate the equation in (7) above over the face of cube L—i.e., integrate dxL from –h/2 to +h/2 and dyL from –h/2 to +h/2. The rate energy is radiated from cube L to cube R will be directly proportional to this value of this integral.
I believe the above procedure is consistent not only with the modern theory of physics, but with Planck’s 1912 description of radiative heat. Do you agree or disagree? Continue with this response only if you agree. If you disagree, please explain why.
Okay. Now reshape the “cut face” of cube L so that instead of being planar, the “cut face” of cube L is slightly concave. By doing this, the “cube” L cut face that radiates towards cube R is no longer planar, but is shaped such that portions of the “cut face” of cube L can now see other portions of the “cut face” of cube L. I believe you characterize such a shape as having a “negative radius of curvature.” I believe that if I make the necessary corrections to (a) the differential surface area on the “cut face” of cube L (the differential area is no longer dxL*dyL), (b) the angle between the normal to the new cube L differential surface area and the vector from that surface area to dSR, and (c) as seen from the cube L differential area, the solid angle subtended by dSR, the above procedure is valid. Do you agree or disagree? If you disagree, please explain why.
I point out that even with a concave cube L “cut face,” because the “cut face” of cube R is planar (i.e., has no “negative radius of curvature”), I see no reason that with the proper “integral corrections” the above procedure doesn’t apply to radiation from cube R to cube L. If the two cubes are at the same temperature, then the net rate of radiation between the cubes will be zero. This implies that for equal temperature cubes, the rate energy radiates from cube L to cube R must equal the rate energy radiates from cube R to cube L. I haven’t actually performed the integration for cube R to cube L radiation using a concave “cut face” surface for cube L, but there will be symmetry for each differential area on one face to a differential area on the other face; so it seems to me the procedure is valid for radiation from the concave surface of cube L to the planar surface of cube R.
Reed 3:20pm: “Our Modern Understanding of the Theory of Heat Radiation may in no way conflict with Planck’s 1912 paper.”
There is no conflict in the principles Reed. There IS nowadays better equipment to more closely determine, say, Planck’s constant which is only very slightly different in modern times than 1912. Those geezers could do great experiments.
“..I have no hope of reading and understanding all of Planck’s 1912 paper.”
You do need to understand when eqn. 274 is applicable and when it is not.
“Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes.”
This is incorrect Reed since tests show as the emissivity reduces below ideal 1.0 for the airless ideal sphere, reflectivity increases above 0.0. With your (a) transmissivity = 0.0 then reflectivity + emissivity = 1.0. The natural earth surface emissivity is measured around 0.95 so the natural surface reflectivity is about 0.05. Because some incident solar is reflected to deep space, never absorbed, from way above the surface, earth albedo measures out at about 0.3. Only about 0.70 incident solar is absorbed into the system.
“So for a no-atmosphere sphere and equal non-zero absorptivity/emissivity, the ERE surface temperature of an Earth-like sphere is approximately 278.2K independent of the absorptivity/emissivity of that Earth-like sphere. Do you agree or disagree?”
I agree if a sphere in earth’s orbit from 2000 to 2010 with albedo 0.0 and ideal emissivity 1.0, then annualized Tmedian over the surface would be 278.3K.
“The T^4 relationship is not in general valid for an emissivity that changes with (incident) frequency.”
True, the emissivity of the same satellite face rotating between incident solar and incident starlight will vary. Thus your resultant Tes will vary. This is accounted in the full S-B theory. For the earth, this is why the annual period is needed, for all incident angles and all incident frequency content. This should be so noted as you write, few do.
“I believe this is the methodology used to arrive at the 255K Earth surface temperature in the absence of atmospheric greenhouse gases.”
It is not Reed, the work out goes much deeper but is still introductory level. If you want to carry on a discussion as to how to find the 288K and the 255K at surface from 1st principles, I can help, but once again a modern text book will be more helpful. I can add in what I have read from them.
“If this is the methodology used to arrive at the 255K temperature..”
It is not, one needs to go deeper (pre-req.s: understanding of atm. optical depth principles & atm. emissivity) but still the methodology used is basic (1st law) to understanding your Tes = 255K and the 288K. Your minor and major problems are then not entirely applicable though there are still limits.
“I want to argue about using the measured value to compute the surface temperature of a greenhouse gasless Earth.”
Better to understand this by keeping the pressures of the atmosphere in place as well as its affect on incident asteroids. Then just vary the emissivity of the atmosphere down (reducing opacity) using the text book 1st law work out. Once you get that, you can discover even more based on moon observations. For example, your (iii) & Planck law eqn. 274 does not hold for around 10-20% of the moon surface.
“..the oft quoted greenhouse gasless Earth surface temperature of approximately 255K relies heavily on the albedo of a greenhouse gasless Earth being nearly the same as the measured (with a greenhouse gas atmosphere) Earth albedo. If true, I have no problem with the 255K temperature. If not true, I do have a problem.”
Ok, yes, the 255K 1st law text book workout relies on a steady 0.3 albedo. As I wrote before, as the albedo assumption is varied from 0.0 to 0.30, surface temperature will vary from 255K to 278.3K assuming the atmosphere is ideal transparent (no opacity). So one needs to state the albedo used.
“I admit that as of today (a cursory reading of Planck’s 1912 paper) I haven’t found an equivalent formula in Planck’s original paper.”
See eqn. 274 in 1912 paper. Any wiki equation needs to be traceable to it. You have to be careful with units for all the constants (Planck uses cgi IIRC) as well as whether the formula is for wavelength or frequency which means the pi term could look different.
Reed 3:20pm: “I believe the above procedure is consistent not only with the modern theory of physics, but with Planck’s 1912 description of radiative heat. Do you agree or disagree?”
You have changed nomenclature from Planck’s and my head hurts. For example, I think your angle B is Planck’s theta but not sure. You do not place a limit on B as Planck does for his theta (between 0 and pi). So if you go back and use Planck’s theta and azimuth phi (limited between 0 and 2pi) I might have some hope of a disagree or agree.
“Okay. Now reshape the “cut face” of cube L so that instead of being planar, the “cut face” of cube L is slightly concave…it seems to me the procedure is valid..”
This is where you stumble without limits on the integrals and I can’t be sure since my head hurts; here I think your angle B must need fully go outside Planck’s theta limits of 0 to pi.
You write “slightly” concave. Make it “hugely” concave” and I think you will begin to see the problem. As an example, there are internet IR camera views of a say 6″ by 6″ flat steel plate maybe 1/4″ thick at a certain temperature correctly registered by the camera. Then the experimenter drills two round (say 1/4″ drill by 1/8″ deep) holes. Those holes interior surfaces register a warmer color on the camera not because they are necessarily any appreciably amount warmer but because the surfaces are now radiating to themselves & Planck’s eqn. 274 which the camera works off is no longer applicable. This is a well known problem in being careful when using an IR camera to inspect electrical panels and circuits for unwanted high temperature resistances.
Your two blocks are like Planck’s bodies A at 100C, B at 0C, and B’ at 1000C bottom p. 6 to 7. I think Planck writes there what you are in part seeking to learn.
LtCusper, 5:44 pm 31 March 2015 and 6:52 pm 31 March 2015.
I’ll need time to digest and respond to what you wrote. However, off the top of my head, two comments.
First, Planck uses a polar coordinate system to expresses the differential solid angle into which energy is radiated. The origin of Planck’s polar coordinate system is the differential radiating plane. The polar axis of Planck’s polar coordinate system is the vector perpendicular to the differential radiating plane. In that coordinate system, the differential solid angle dW is SIN(theta)*dtheta*dphi, where theta is the polar angle and phi is the azimuthal angle. [Note: by integrating the solid angle over the limits theta from 0 to pi and phi from 0 to 2*pi you get the total solid angle of 4*pi steradians.] If you want to determine the energy radiated into the solid angle subtended by the “top” of a cone whose apex is at the differential planar surface, whose axis is coincident with the coordinate system’s polar axis, and whose cone half-angle is alpha, then (a) the expression you must integrate is COS(theta)*dW = COS(theta)*SIN(theta)*dtheta*dphi, and (b) the limits on the integration are theta from 0 to alpha, and phi from 0 to 2*pi.
However, if you’re interested in determining the rate energy is radiated into the solid angle subtended by a rectangular surface whose plane is parallel to the plane of the differential radiating surface, then in polar coordinates the limits of integration for theta and phi are (a) not simple, and (b) functions of one another–i.e., the limits of integration of theta are functions of the value of phi. So in a polar coordinate system, you’ve simplified the integrand [COS(theta)*SIN(theta)*dtheta*dphi] at the expense of making the limits of integration complex.
If you use a rectangular coordinate system whose origin is at the differential planar radiating surface and whose z axis is perpendicular to that surface, then the solid angle subtended by a planar differential receiving area dx*dy is simply dx*dy*COS(B)/r^2, where B is the angle between (a) the normal to the receiving differential area …and… the line from the receiving differential area to the transmitting differential area, and r is the distance between the two differential planar areas. If the plane of the receiving differential area is parallel to the plane of the radiating differential surface area, then B is also the angle between (a) the radiating differential area …and…(b) the vector from the radiating differential area to the receiving differential area. Then as Planck did using polar coordinates, a term COS(B) must be included to account for radiation that propagates away from the differential area at an angle B with respect to the normal to the differential radiating plane. Thus, the integrand, which is the product of dW and COS(B), becomes dx*dy*COS(B)*COS(B)/r^2, with limits of integration for x and y that define the rectangular area that energy is being radiated into.
So for computing the radiation emitted into the solid angle of a finite-size plane, in the one case (polar coordinates) you keep the integrand simple, but make the limits of integration complex. In the other case (rectangular coordinates), you complicate the integrand but make the limits of integration simple. Which integral is easier to perform is a function of the user’s knowledge and access to tables of indefinite integrals. So, for my procedure, yes B is Planck’s polar coordinate system variable theta, but that is only true for a planar receiving surface that is parallel to the planar differential radiating surface. Furthermore, you commented that I didn’t define the integration limits for B. I don’t have to. The integration limits are expressed in terms of x and y.
Second, I’m going to assume in your plate/plate-with-holes that all surfaces are blackbodies–i.e., no reflection occurs. A camera that is imaging a planar surface is measuring the rate energy is radiated from points (differential areas) on the surface of the plane in the direction of the camera lens. The lens focuses the energy that passes through the lens from a point on the radiating surface to a point on a plate. The greater the energy, the darker the point (or maybe the lighter the point depending on the material absorbing the energy). Thus, an image of a radiating plane doesn’t represent the total energy radiated from differential points on the plane; but rather represents the radiation from points on the radiating plane that strikes the focusing lens. When you drill a small hole in the surface of the plane you both remove some areas that were radiating before the hole was drilled, and you expose some new radiating areas that didn’t exist before the hole was drilled. For a finite radius, finite depth hole, it wouldn’t surprise me if the exposed radiating surface of the hole is different from the exposed radiating surface without the hole. However, the total exposed radiating surface area isn’t what counts. What matters is the relative rate (a) energy originating from within the hole is directed towards the camera lens compared to (b) the energy rate from the flat surface of the soon-to-be hole is directed towards the camera lens. It wouldn’t surprise me one bit that these energy rates are different. If so, the image of a hole would appear darker (or lighter depending) than the surrounding image. To me this only implies that you must be careful when applying Planck’s radiation law for differential planar surfaces to real-world surfaces, not that Planck’s law doesn’t apply.
Reed 10:44am: “Second, I’m going to assume in your plate/plate-with-holes that all surfaces are blackbodies–i.e., no reflection occurs.”
The steel plates are real but not polished enough to fake out the IR camera since it correctly registers the temperature measured by thermocouple on the surface. There is reflection.
“The integration limits are expressed in terms of x and y.”
There are no limits that I can see in your comment. The limits of integration, in any coordinate system chosen, need to conform the surfaces to Planck writing p. 2 “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.” Since wavelengths are always positive, the radii of curvature are ruled always positive, ruling out surfaces not applicable Planck eqn. 274 that are even “slightly” concave and ensuring diffraction is negligible before applying eqn. 274. The radii of the plate hole surfaces are concave, Planck eqn. 274 is not applicable.
LtCusper, 2:50 pm, March 24, 2015; 5:45 am, 31 March 2015; 6:52 am, 31 March 2015; 1:55 pm, 31 March 2015
First, I want to respond to the issue of “integration limits” in my procedure. You wrote: “There are no limits that I can see in your comment. The limits of integration, in any coordinate system chosen, need to conform the surfaces to Planck writing p. 2 “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.”
(A) In the case of a cube, all surfaces are planes so all surfaces have infinite radii of curvature, and as such all radiating surfaces have radii of curvature are much larger than any finite wavelength. The cube edges have zero radius of curvature in one direction and infinite radius of curvature in the orthogonal direction. I guess the corners of the cubes have zero radius of curvature in all directions. I’m not sure how to handle the edges and corners. In addition, there is still the issue of diffraction at the cube edges; but all cube faces conform to Planck’s p. 2 restriction “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.”
(B) In Step (G) of my procedure I wrote: “ Integrate the Equation in (6) above over the face of cube R—i.e., integrate dxR from –h/2 to +h/2 and dyR from –h/2 to +h/2.” The words …the Equation in (6) above… should have been …the Equation in (F) above… When I originally wrote the procedure I used numbers, not capital letters, to delineate the procedure steps. Because I had used numbers in my comment to delineate points I was trying to make, to avoid confusion I later changed the procedure steps from numbers to letters. I forgot to correct the wording of the procedure steps so that the wording still used numbers.
In any event, the origin of the x,y,z coordinate system is at the geometric center of the “cut face” of cube L. The “cut face” of cube L is defined by the coordinates: -h/2 less than x less than +h/2, -h/2 less than y less than +h/2, z = 0. The coordinate system’s z axis points to the geometric center of the “cut face” of cube R. The “cut face” of cube R is defined by the coordinates: -h/2 less than x less than +h/2, -h/2 less than y less than +h/2, z = d. The first integration (Procedure Step G) selects a fixed differential area, dxL x dyL, on the “cut face” surface of cube L and integrates over the entire “cut face” of cube R, which as I said in Procedure Step H is the integration over dxR from –h/2 to +h/2 and the integration of dyR from –h/2 to +h/2. The largest possible polar angle occurs for a differential area at one corner x=-h/2, y=-h/2, z=0 of cube L …to… the diagonally opposite corner x=h/2, y=h/2, z=d of cube R. Thus, the magnitude of the largest polar angle is Inverse_Tangent{h x [2^(1/2)]/d], which is less than pi/2. For all dxL x dyL not on an edge of the cut face of cube L, the azimuthal angle can range from 0 to 2*pi. The second integration (Procedure Step H) integrates over all differential areas on the “cut face” of cube L. The limits of this integration are also –h/2 less than x less than +h/2, and –h/2 less than y less than +h/2.
Second, you wrote: “You do need to understand when eqn. 274 is applicable and when it is not.” I agree. However, as best I can discern, Equation 274 is never mathematically applicable. Equation 274 gives the “specific intensity of a monochromatic plane polarized ray of frequency f…” According to what Planck wrote on page 2, the equation only applies for radii of curvature large compared to the wavelengths of rays considered. Since Equation 274 doesn’t restrict the range of frequencies, for any object having a finite radius of curvature, wavelengths exist that are long compared to the radius of curvature. Thus, Equation 274 mathematically applies only for an infinite radius of curvature—i.e., a planar surface. Even then, unless the plane is infinite in extent, diffraction will occur at the edge of the plane—another issue raised by Planck. Thus, Equation 274 mathematically applies only to an infinite plane, which means that mathematically it can never be applied to a real-world surface. If you’re going to apply Equation 274 to any real-world situation, you’re going to have to caveat the application with words to the effect that Planck’s distribution function is only an approximation. Then the issue becomes: “How good an approximation?” I have insufficient knowledge to answer that question for most real-world situations. What I have observed is that people use Planck’s specific intensity distribution for many situations without the caveat mentioned above. Although I may caveat things I write in the future, I still plan to use Planck’s Equation 274 as the behavior of the “specific intensity of a monochromatic plane polarized ray of frequency f…”
Bottom line, I would slightly change your admonition to: “You do need to understand when eqn. 274 is a good approximation to real-world situations and when it is not.” With that change, I agree with you; and any improper use of Equation 274 that I employ is my error.
Three other points. The T^4 law comes from integrating Equation 274 over all frequencies. Since for any real-world situation, Equation 274 is at best an approximation, then the T^4 dependence on temperature is also at best an approximation.
Next, in a previous comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.” Given that (a) an enclosed shell must at some point have surfaces with negative radii of curvatures—at least as you characterize radii of curvature as being negative, and (b) by poking a hole in an enclosed shell radiation escaping the hole undergoes diffraction, how can you claim “out comes BB radiation?” Don’t Planck’s radii of curvature and diffraction caveats contradict the BB claim? Or at a minimum raise the issue of how Planck’s caveats don’t seem to apply to a hole in an enclosed shell whose interior is at a single temperature?
Finally, at one point in the thread we had the interchange–me: “..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface”; you: Not according to Planck’s own writing in the 1912 paper introduction. In the sense that Planck’s Equation 274 might not apply, I agree with you. What I was referring to was that a planar differential surface cannot radiate energy to itself; whereas a finite sized object can. Consider a thin hemispherical bowl. The area of the bowl that has line-of-sight visibility to the outside world is the same as the area of an equal radius sphere. If you use the equation that the rate of cooling of an object is proportional to the product of the surface area of the object and the difference of the fourth powers of the object’s temperature and the background temperature, then I believe you can use the area of the sphere exposed to the background, but you can’t use the area of the bowl exposed to the background. Portions of the concave side of the bowl will radiate energy to other portions of the concave side of the bowl. Maybe inconsistent with Planck’s Equation 274, but transfer of energy non-the-less.
Third. We had the exchange: me: “Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes.”
You: “This is incorrect Reed since tests show as the emissivity reduces below ideal 1.0 for the airless ideal sphere, reflectivity increases above 0.0. With your (a) transmissivity = 0.0 then reflectivity + emissivity = 1.0. The natural earth surface emissivity is measured around 0.95 so the natural surface reflectivity is about 0.05. Because some incident solar is reflected to deep space, never absorbed, from way above the surface, earth albedo measures out at about 0.3. Only about 0.70 incident solar is absorbed into the system.
Now I’m confused. In my statement I caveated that the emissivity equaled the absorptivity. That may not be the case for an airless ideal sphere, but that was my caveat. Assuming transmissivity is zero, then absorptivity + reflectivity = 1, which says absorptivity = 1 minus reflectivity, which says emissivity = 1 minus reflectivity, which says emissivity = 1 minus albedo. So once you set the albedo, you are setting both the emissivity and the absorptivity and they will be equal. For a given albedo, the sphere will absorb in direct proportion to 1 minus the albedo, and the sphere will radiate in direct proportion to 1 minus the albedo. Maybe I’m missing something, but how can the temperature of inert matter in energy-rate-equilibrium that absorbs with the same proportionality it emits be a function of that quantity?
Fourth. On page 2, Planck wrote: “Throughout the following discussion it will be assumed that the linear dimensions of all parts of space considered, as well as the radii of curvature of all surfaces under consideration, are large compared with the wavelengths of the rays considered.”
You wrote: “Since wavelengths are always positive, the radii of curvature are ruled always positive, ruling out surfaces not applicable Planck eqn. 274 that are even “slightly” concave and ensuring diffraction is negligible before applying eqn. 274. ” I’m a little confused by what you wrote. If you’re saying that the radius of curvature of a concave surface is negative, and because it is negative (a) the radius of curvature can’t be greater, much less ‘much greater’, than the wavelengths of rays, and (b) thus Planck’s Equation 274 can’t be applied to concave surfaces, then I don’t agree. First, radii of curvature are distances and distances are always positive. The radius of curvature of a concave spherical mirror is often characterized using a positive number.
Furthermore, consider the “top of a cone” whose cone half angle is 0.5 degrees and whose radius is 1,000,000,000 meters. From a visual perspective, a person walking on the cone top would have a hard time deciding if he was on the concave side or the convex side of the cone top. For a very thin “cone top,” the total area of each side (concave and convex) of the “cone top” is approximately 2.39 x 10^14 square meters. If the temperature of the surface was everywhere say 300K, it’s hard, no it’s close to impossible, for me to believe the total rate energy is radiated by the concave surface of the cone top is appreciably different from the total rate energy is radiated by the convex surface of the cone top. So, it makes sense to me that the radius of curvature should be large compared to the wavelengths of the rays under consideration; but it makes little sense to treat the radii of a concave surface as being negative in Planck’s comparison of wavelength and radius of curvature.
Fifth. Your comments have both informed me and given me a lot to think about. For that I am grateful. Thank you for taking the time to respond to my comments.
Reed 2:18pm: “Thus, Equation 274 mathematically applies only for an infinite radius of curvature—i.e., a planar surface.”
Look up a paper ref.d by Planck for the test methods. Available on line. Eqn. 274 is for BB radiation, was developed experimenting with heated/cooled cavity BB radiation emitted from the hole & directed into a box to reduce convection. The box had some conditioning mirrors to eliminate unwanted signal with the intended signal hitting a thermopile attached to galvanometer to read the intensity. There is no such restriction you impose. This can be verified with an IR thermometer temperature reading the same as mercury thermometer temperature of a flat table or curved water glass, ice cube tray, etc.
Diffraction can occur but it needs to be negligible for the IR thermometer correct reading.
“So once you set the albedo, you are setting both the emissivity and the absorptivity and they will be equal..”
Only for the same content of frequencies. For earth system, albedo is SW, emissivity is LW.
“The radius of curvature of a concave spherical mirror is often characterized using a positive number…it makes little sense to treat the radii of a concave surface as being negative..”
The sense is the vector is opposite direction, positive radii point away from the object; concave radii point toward the object thus radiating to itself. Planck et. al. recognized through experiment a concave object surface situation was trouble so ruled it out. While the trouble may not be very large for “slightly concave” surfaces, it certainly will be to some extent – thus needing experiment to be definitive.
I read your 2:50 pm, 24 March 2015 comment. It will take me some time to respond–I need to give what you wrote some thought; but I will respond. Since you have elected to forego an email exchange, you’ll just have to intermittently visit this thread.
BTW, It won’t do any good for me to access Planck’s original paper–my German is inadequate. I did, however, find online
a translation by Morton Masius of a Max Planck paper entitled: The Theory of Heat Radiation. The translation was published in 1914. Examining the “Prefaces” in the published translation, it looks like the translation is of the second edition of a paper written by Max Planck in 1912. Is this the document you recommend I “invest in?”
Reed,
“Bottom line, when computing (not measuring, but computing) the Earth’s surface temperature in the absence of greenhouse gases, it’s illogical to perform that computation using a value that depends on the presence of greenhouse gases. I don’t accept the statement that radiometers outside the Earth’s atmosphere and pointed towards the Earth measure radiation only emitted from the Earth’s surface. How can they not measure the sum of Earth-emitted radiation and reflected solar radiation? It’s only after you subtract out the reflected solar radiation that the intensity corresponds to a blackbody object at 255K. But without water vapor there would be no clouds and without clouds there would be no (or very little) reflected solar radiation, and without reflected solar radiation, there’s nothing to subtract from the radiometer measurements.”
This is true but very misleading. The total radiative intensity Watts/steradian of this Earth cannot be measured at a distance less than our moon. Much better would be satellites at L1 and L2. both are looking the other way. Why?
The Earth with its atmosphere absorbs and reflects some power from the 68 micro-steradians Sun. then re-radiates all into 4 PI steradians space. If not the expression for accumulated power, as sensible heat, TEMPERATURE must go up or down! This planet, finely controls atmospheric WV so that temperature at any location remains comfortable. This is called CLIMATE, whatever you wish! The Wind determines atmospheric WV at every location. What determines wind direction and velocity at every location.
Again This is the total demonstrated and perhaps intentional, incompetence of the entire climate science religion!
MY knowledge is limited to “beats the shit out of me”!! 🙂
Reed 3:51pm: “But isn’t the optical depth in part a function of the density of the material through which the radiation is propagating?”
In original formulation, yes. Hydrostatic equilibrium is known good by experiment for large parts of the atm., and that realization takes density term out of the differential optical depth replacing with pressure. So the atm. layer differential grey opacity becomes a function of mixing ratio, mass extinction coefficient for the ith constituent and differential pressure summed over N constituents. Then for total optical depth integrate down from top of atmosphere to the reference pressure for tau.
So you would have to do this and obtain proper tau over a few mm at the required pressure for CO2 mass extinction coefficient and 100% mixing ratio to get the 33K result of the normal atmosphere. It should be evident you would now have an unweildly thermos bottle just from general experience.
As an alternative, look at the Mars atmosphere which is physically high but very thin optically causing more like a few Kelvin of the effect you seek to learn about which causes about 33K at earth pressures and constituent mass extinction coefficients.
NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.
IMO Part of the bigger problem is that we have such a poor understanding of what happens in our skys with it’s mix of IR and movement of variable dimensioned vapor droplets, the exchanges that happen between them, and layered structure.
If you can find it also refer to
JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 104, NO. D2, PAGES 2059-2066, JANUARY 27, 1999
Absorption of solar radiation by the cloudy atmosphere:
Further interpretations of collocated aircraft measurements
Robert D. Cess, 1 Minghua Zhang, 1 Francisco P. J. Valero, 2 Shelly K. Pope, 2 Anthony Bucholtz, 2 Brett Bush, 2 Charles S. Zender, 3 and John Vitko Jr. 4
Where they say in the conclusion –
_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_.-_,-_
5. Conclusions
This investigation augments the conclusions made in the two prior ARESE studies [Valero et al., 1997b; Zender et al., 1997]. Specifically, it suggests that the rather large 500 nm column absorptance for clear skies (October 11, Figure 1) is caused by some temporally variable absorber, most likely aerosols. Likewise, aerosols are most plausibly the cause of the comparable 500 nm absorptances when clouds are present (Figure 1).
Next, we have extended the prior investigation of sampling errors [Valero et al., 1997b] by employing the procedure of Marshak et al. [1997] to minimize sampling errors through use of the 500 nm measurements, and we conclude the enhanced cloud absorption is not likely an artifact of such errors, nor can it logically be explained by instrument calibration errors. We have empirically determined that at least some of the enhanced cloud absorption (_<20%) occurs at wavelengths <680nm, but the cause of this absorption, both at wavelengths <680nm and in the near infrared, remains unknown; we find no evidence for its existence at 500 nm.
As far as I am aware this matter has never been cleared up.
Nicely done,
Yes reflected SW cannot be distinguished from emmitted. It is all just “radiance”. The radiance of the Earth at the location of the Sun is very tiny so does not limit the radiant intensity or flux from the Sun.
(Finally, in your 5:34 am, 24 March 2015 comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.”)
“Say What? Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df from a planar differential unit area, dA, at temperature T, into a differential solid angle domega defined with respect to a coordinate system at dA. There are no terms in Planck’s blackbody radiation law for the presence/absence of any other surfaces.”
This is somewhat correct except for terms. Plank’s specific intensity is now termed spectral radiance an indication of electromagnetic field strength at a particular wavelength interval and direction. Any opposing spectral radiance must limit any actual spectral flux.
“In fact, it is a geometrical impossibility for a planar differential surface area to radiate into a differential angle that contains any portion of the radiating differential area. Provided the radius of curvature of an object is finite and non-zero, ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.”
Only for radiance never flux!
“For the interior surface of a closed object (like the interior surface of a spherical shell), you can use Planck’s law to compute the power radiated from any finite size area of the interior surface to any other finite size area of the interior surface. It turns out that if no objects reside in the interior surface of a closed object, the total rate energy is radiated from the entire interior surface is equal to the total rate the interior surface absorbs radiated energy. As such, the net rate of energy exchange of the interior surface is zero.”
The energy exchange is always zero as internal opposing radiance always cancels any emission in any direction. Any other interpretation defines a perpetuum mobile of the second kind. I look forward to #4! 🙂
The effect you seek isn’t measurable in your setup – the thermos CO2 atmosphere still has a nearly 0.0 optically thick atmosphere “greenhouse” as does the vacuum.
5mm of CO2 at 100% concentration is equivalent to 15m of atmosphere with a CO2 concentration of 0.03%. How much do you need to create a measurable effect?
At 100 kPa pressure, 350 ppmv CO2, the .5 micron band about 15 microns has an optical depth of less than 2 meters. 37% transmission of amplitude modulation at 300 Hz. The 300Hz is well above the low pass cutoff of even 1 meter of such gas. This in no way indicates that such gas at or above equilibrium temperature, would attenuate any DC 15 micron flux. It is the aluminum surfaces that prevent 15 micron EMR. The good vacuum prevents conductive and convective loss! Try 30 layers of super insulation for spendy things like liquid Neon! 🙂
You all also forget the emissivity of CO2 as a gas. The “back radiation” meme that supposedly heats the planet is dependent on CO2 radiating energy it absorbs. Problem, CO2’s emissivity is less than 0.0017 so for all practical purposes it doesn’t radiate. (Gases as a whole are really lousy radiators.) If it did it’s peak line is at 15µm which Wein’s Law tells us is ~193°K. That means that CO2 radiating at its fundamental frequency CANNOT make anything hotter than -80°C. In order to get gases to radiate look at what we do in the real world. They are contained at near vacuum pressures. You have to keep the gas molecule in a high energy state long enough for it to drop to ground and emit. Pressures above 1 Torr make this almost impossible as energetic gas molecules run into each other and transfer that energy kinetically before the can radiate.
I’ve said this a few times so my apologies for repeating it. Gas lasers and neon tubes are all extremely low pressure and extremely high energy density. CO2 lasers are even more complex since CO2 does not readily get to an emission energy level. There’s a mix of exotic gases necessary to get CO2 to radiate on top of either high frequency high energy RF or high voltage DC power supplies. Conditions that don’t readily occur in our atmosphere. If CO2 acted like the AGW crowd said it does all we’d need to build a CO2 laser would be one partial LWIR mirror, one full LWIR mirror and a heating element… that of course doesn’t work. The only places in our atmosphere where gases radiate, since the right conditions exist, are at the poles above a minimum of 60km and most above 100km driven by high energy particles from the sun funneled by magnetic flux lines… we call them Auroras.
Sorry, there’s also radiation from the TOA at pressures below 1 Torr as well from the rest of the atmosphere. The auroras are just a good example of the low pressure necessary since you’ll see a drastic cut off on their lower edge where the pressure gradient gets too high for them to continue to radiate.
“If it did it’s peak line is at 15µm which Wein’s Law tells us is ~193°K. That means that CO2 radiating at its fundamental frequency CANNOT make anything hotter than -80°C. In order to get gases to radiate look at what we do in the real world. They are contained at near vacuum pressures.”
Kind of correct. The thermal EM absorption or emission of matter does not depend on its peak wavelength or temperature. Thermal Radiance is proportional to T^4 not T or some linear delta T. Anything interfering with your ability to dispatch to space, or even conducting to cold “must” result in your higher temperature to compensate for such interference. Most times in this atmosphere, this interference does not happen! All molecules just keep on dooin whatever! 🙂
The thermal EM absorption or emission of matter does not depend on its peak wavelength or temperature.
It absolutely does. That’s basic. All materials emit, absorb and reflect radiation depending on their atomic and molecular structure and the amount of energy present keeping the electrons, atoms and molecules moving. All those frequencies and harmonics are the absorption/emission/reflection spectra and the sum of all the energies contained in the material is its temperature. Temperature is a way for us to define how much internal energy is present whether it’s atoms vibrating in a solid, bouncing around as a liquid, gas or plasma. As for your thermal radiance proportionality you are improperly using S-B math for a black body on a single wavelength emitter that is not even remotely like a black body… emissivity of 0.0017… remember? A line emission is not a Planck black body curve. If you don’t like Wein’s Displacement Law, take it up with Wilhelm’s ghost. That’s another AGW failing in treating atmospheric gases like black body radiators.
I don’t understand what the rest of your comment means with the “interference” references and “conducting to cold” so I’m not even going to hazard a guess as to what you’re trying to say there.
CO2 gas can, and will, via spontaneous EMR flux, transfer energy at 15 microns to any mass absorbing such waveband because of its lower temperature. Wien’s displacement law has nothing to do with any ability to radiate or absorb at any frequency.
Emissivity of CO2 gas at 15 microns depends on the density and depth of generation. STP air of 5 meters has emissivity of g.t. 98% just because of the 320 ppm of CO2. That same air without “any” CO2 would have such emissivity at 25 meters just because of WV. These are from “my” measurements (1969).
Australian Conservation Foundation says US-style regulation should be introduced to force the worst polluting plants to close
…
“In Australia there are taxpayer subsidies to keep these plants open, whereas in the US, China and parts of Europe, the government is taking actual direct action to close them down,” Cousins said.
…
“We have got to close down the worst polluting plants in Australia,” said Geoff Cousins, ACF’s president. “At the moment the government is offering no incentives for companies to get off fossil fuels.”
Total detachment from reality.
Gross misrepresentations of what is going on in China (new coal-fired power station every other week or thereabouts) and the German government of prohibiting the shutdown of unprofitable, coal-fired power stations whose operators are not being compensated for being a spinning reserve for wobbly power from the unreliables.
What the lunatics don’t recognize is that the investment climate was changed dramatically with the introduction of RET’s and carbon taxes, dissuading investment in upgrading coal-fired power plants to burn the fuel more efficiently, cleanly and to generate electricity more efficiently.
… What the lunatics don’t recognize is that the investment climate was changed dramatically with the introduction of RET’s and carbon taxes
Oh yes, they do recognise this. That’s their point …
Previously commented, but needs to be emphasised – I had not believed this could happen in a democracy, but I was hopelessly wrong. All it needs is for all major political groups to agree, then whatever you vote just doesn’t matter
The NHS could be landed with a £22bn bill as cold homes kill 100,000 vulnerable people over the next 15 years, a charity warns today.
National Energy Action reckons at least 30,000 vulnerable people have perished prematurely over this Parliament due to an inability to adequately heat their homes, and Treasury energy taxes will mean many more will die in the next decade and a half unless the next government takes positive action.
Look closely at the list on the image at this link.
It indicates what are astonishingly called Australia’s Worst Polluters. Keep in mind that this is for what is referred to as Carbon Dioxide ….. Pollution. The list was brought out by The Australian Conservation Foundation as a vehicle to launch their new president Geoff Cousins, and was featured on Thursday Night’s Lateline with a compliant Tony Jones interviewing him. That interview was just amazing, so full of inaccuracy I just sat there with my mouth open. Cousins said a number of things, one of them this:
And what the Government ought to be doing right at this minute is putting together a plan to start closing down some of these plants.
That list has 10 names on it and 7 of them are the electrical power generation companies operating the largest of the coal fired power plants in Australia.
Those 7 power generating entities generate around 75% of Australia’s electrical power needs, probably more even.
This is what is actually keeping Australia going. Take them away, and Australia just grinds to a halt ….. in an instant. That’s not shutting them all down, as it would only take two of them to close and the Country would just stop dead, say Bayswater in New South Wales and Loy Yang in Victoria
There’s nothing to replace those coal fired plants, not only on a power delivery basis, but on a time delivery basis, as each of those plants on that list operates on a 24/7/365 basis, something no renewable can do.
The total wind power output to all the Australian grids comes in at around the equivalent of 1200MW, and that’s not even enough to supply the absolute Base Load requirement for just Sydney alone.
Here’s a scenario then.
You now have a Labor State Government in Victoria and also in South Australia. Labor is typically the party of reducing CO2 emissions, believing that The Science is in, that coal fired plants should be closed, and replaced by renewables. When you add together all the wind plants in South Australia and Victoria, the Nameplate Capacity is similar to the Nameplate for Loy Yang, owned by AGL.
Okay then, let’s say that AGL Energy goes to the Victorian Labor Government and says that in the best interests of Victoria and Australia, they are going to close down their Loy Yang power plant , reducing that (CO2) pollution by millions of tons.
So then, what would be the reply from that grateful Victorian Government, who would then be able to say with pride that they lead Australia in reducing Carbon pollution.
The Victorian Government’s reply would go something along these lines.
You have a contract to supply electricity from Loy Yang. If you close Loy Yang, we will sue your ar$e so far into the future, your Company will be sent to the wall.
There would be a similarly worded reply from the Labor Government in South Australia, because without that power from the Vic-SA Interconnector, (all of it brown coal fired power) South Australia would also grind to a halt.
That story would NEVER get to the media.
Have you ever wondered why, with all the environmental noise about Hazelwood, Labor governments in that State have not just closed that plant down. They know, hand on heart, that there is nothing which can replace the electricity it provides, electricity which keeps Victoria from going dark, literally.
Cousins then went on to say later:
….. the concept that you would take shareholders’ money and invest it in old technology – and some of these plants are over 40 years old – technology that has been rapidly overtaken by renewable energy – that you would take shareholders’ money and invest it in those kind of plants is just crazy.
Now, AGL has just purchased Bayswater and Liddell from Macquarie Generation.
Why would they do that?
I mean Liddell is 44 years old and Bayswater is 30 years old. How stupid is it to sink shareholder money into this old technology, something which has no future, according to Cousins.
The future of this investment lies in the return from the sale of electricity to the NSW grid, and that is around $900 Million a year, so if Liddell gets run for its full 50 years, and the same 50 years for Bayswater, then the return from the sale of that electricity is $15.6 Billion, and that’s just at the contracted price of around $30 per MWH, the lowest cost electricity in the Country. Trust me, there is every chance both plants will be operating beyond 50 years.
If coal fired power was on the way out, do you think AGL would have sunk so much money to buy into these two plants?
Worst Polluters in Australia. Really!
Then, just shut them down. See what happens then. It would only take three days, and the whole CO2 emissions argument would just stop. Pity that the whole Country had to stop as well, just to prove the point that we cannot live without all that coal fired power.
I seriously wonder what’s going to happen to all these people hyping Wind and Solar Power to replace coal fired power when the truth gets out that it can’t, There’s nowhere to hide from all the things they’ve so publicly said.
Tony I read this over at wattsup a couple of months ago and I imagine you did to.
I haven’t read any more about it.
But what your writing about here is happening in the “jewel” of the global warmists renewable energy mis information campaign.
I might add, if I hadn’t read it at wattsup I probably would never have.
Mainstream wasn’t exactly “all over it”.
Anybody heard anymore?
“The led to an event that can be likened to the proverbial iceberg unexpectedly popping up right in front of the German state ship while it was plowing through the waves on its climate-saving mission at full-steam. With just a 48-hour notice delivered by a personal phone call to Ms. Merkel on a Saturday, the CEO of E.ON, the largest German and European power producer, let it be known that the company had decided to split itself in two, one part grouping fossil and nuclear power generation and a second part encompassing the “politically correct” activities in the field of “renewable” energies. Sort of a “Bad E.ON” / “Good E.ON” move. The intention is to get rid of the “bad” part as soon as possible by putting it up for sale. At the same time, this also means the “good” part will cease to be duty bound to ensure a stable power supply under all circumstances. Obviously, such a liability is not enforceable from an entity whose only power sources are unstable wind and solar power plants. In a nutshell, the message behind this move is that the silverback of the “big four” German energy producers who group the bulk of the country’s conventional and nuclear power production is about to close shop at short notice. The others will probably follow suit.”
Just adding a little-known piece of factual information here
Loy Yang had been owned by an American group whose business was buying and selling power utilities and other infrastructure worldwide. They put a huge amount of capital into the purchase of Loy Yang on the expectation that the Kennett Govt would keep its’ word and deregulate wholesale/retail prices. Kennett unexpectedly (for him,at least) lost power shortly thereafter and Bracks, succeeding him, promptly reneged on this. Loy Yang went into economic freefall and twice broke its’ bank loan covenants
Panic ensued, since Loy Yang could not be allowed to just stop producing electricity (for the reasons that TonyOz describes above). What to do, what to do ?
Sell at a distressed price of course, but to whom ? AGL was the only player both big enough and interested enough to make a reasonable offer. But AGL ownership of Loy Yang would contravene the Federal monopoly legislation (Alan Fels’ ACC). Bracks pushed through legislation in both Vic Houses in most unseemly, panicky haste to except the sale from the monopoly requirements
All of that to prevent the horror described above
Now the greenies want to reverse all that. I agree with TonyOz that this won’t happen just yet but Vic’s new ALP Govt is even crazier than Bracks. Already it has reneged on a perfectly legitimate, binding road-building contract and is threatening sovereign legislation in order to avoid the penalty for this. Forcing the closure of a major power station on “environmental” grounds and then suing the hapless owner for the immense public damage that would ensue seems not as unlikely as I had imagined only a few years ago
No one (with any authority/influence) wants to talk abut the fact that most power plants in Australia have had their maintenance schedules slashed and burned in order to save money. When these plants start needing large scale emergency repairs rather than preventative maintenance, then the owners will cry poor, and, in order to prevent widespread blackouts, various governments will have to fork out taxpayers money to fix them. The same can be said for other power infrastructure as well.
I won’t even get into the discussion about shutting down pretty much all the oil refineries in Australia… that’s another problem no one wants to face, but then, that’s ‘big oil’, so no reasonable public discussions can be had any more, as the imbecilic Greens will just throw another tantrum, as they always do.
There’s no doubt about the distain Labor and the greenies have for the health and welfare of the Victorian community.
In Victoria we see this week that the newly elected union controlled Labor government is to put at risk the health and welfare of the community by what it thinks will “make the state more attractive for building wind farms” and that somehow that will “help to unlock billions of dollars in investment”.
Not with my dollars they won’t. And given the sovereign risk associated with doing business in Victoria I suspect there will need to be huge taxpayer subsidies to get any sensible business person involved in these uneconomic schemes.
But, nevertheless, we have the Clean Energy Council advising us that: “’It is fantastic to see the Andrews Government recognising the need for change…”.
Under the changes, the 2 km setback distance between houses and wind turbines will be reduced to 1 km.
So much for the Pacific Hydro supported research that demonstrated that:
“…a pattern of high severity of disturbance to be associated with four different operating scenarios of the windfarm being:
• when the turbines were seeking to start….
• an increase in the power output of the windfarm in the order of 20%…
• A decrease in the power output of the windfarm in the order of 20%…
• The situation when the turbines were operating at maximum power and the wind increased above 12m/s.
There were at times other instances of high severity disturbance not fitting the above four scenarios.”
If anyone thinks the Andrews’ Labor government’s reneging on a multi-billion dollar freeway and the consequential likelihood of it having to pay $100s of millions in compensation lifts businesses’ sovereign risk calculations, just wait until the compo is calculated in relation to all aspects of wind-farms construction and operation under this proposed change in the setback distance starts to hit home.
Look again at the list of Worst Polluters, and keep in mind where I said that:
There is nothing to replace any of those coal fired power plants.
Now look at that same statement in another way.
The time will come when those plants have reached their end of life.
There are no renewables which can supply those huge amounts of electricity, and do it on the 24/7/365 basis which those plants do provide.
However, now they have reached the end of their life, just where will that replacement power come from. The State Governments sold off all the plants. The now private company owners have only purchased the plant for the electricity they provide until the end of the plant’s life. That’s ll their contract stipulates, that they provide electricity from that plant for as long as the plant can survive. There’s no contractual agreement to build a replacement plant once the existing plant expires. They’ll just flog the dead horse until there is no life left in it, and that will be the end of it.
There is no replacement for that power.
Macquarie Generation tried to actually do that with its Bayswater plant. They submitted plans for an upgrade, and one of 2 options was for a new technology USC coal fired plant which would have gone a long way towards securing the future for that huge supply for NSW. They submitted that proposal in 2009, six years ago. Bayswater knew that a new plant would last 50 years, so there must have been enough coal to cover that at the existing site, coal to supply both Bayswater and the nearby Liddell plant.
The Mt. Piper plant also submitted plans for a similar Upgrade to USC coal fired power technology.
So then, where did both of those proposals go?
You guess!
It went so well, that Macquarie Generation have now sold off Bayswater and Liddell to AGL.
When Bayswater reaches the end of its life, there’s nothing in the pipeline to replace it, and the same applies for all the plants on that list. AGL will then just sell the remaining huge amounts of coal to China.
They should have been starting this replacement program, not just for Bayswater, because it is one of the most recent plants on that list, even at 30 years of age itself, but for all of them.
There will come a time when those old plants will struggle to keep providing, and there will be nothing to replace them. Renewables won’t be able to replace that power, and no Government has the cojones to approve a new plant which actually can replace those large scale coal fired power plants.
That time has almost come, or may have even passed now.
But I’m very optimistic and look forward to the time when that which you forecast comes about – as it most certainly will, all other things staying the same.
You see, at that point the voters will well understand that they’ve been defrauded by the greenies and their lefty Labor comrades. The backlash will be massive.
Only if you’re near a port since you need mains power to pump that diesel around the country. I have a vision of the 2nd Mad Max movie in my head… fighting for fuel.
AGL will then just sell the remaining huge amounts of coal to China
Sorry, TonyOz, but 1) AGL does not own the mines; 2) China does not import much thermal coal from Aus in any case
In the event that Bayswater and Liddell just shut down, the mines will severely reduce production or simply put them on care and maintenance for a duration
Just saw a post on judith curry blog about homogenised temp in Australia by euan mearns . He finds that the homogenised nasa data do absolutely nothingoverall. But the biggest surprise was that there was absolutely no warming since 1880. Interesting to see how this analysis marries up with marohasys work and f others.
We all know that the BBC is unable to hide its political bias, and its love of all things ‘green’ (just like the ABC), but I have to say my faith in public broadcasters, humanity, and in science education has been, for the last 5 years been restored ever so slightly, albeit temporarily by the ‘Stargazing Live‘ series, even if it’s only over 3 days each year. I have to admit that The Sky at Night is also a bit of a favourite, but sadly, once a month is not nearly enough, in my opinion. I think the success of the Live version is that a lot of it is done live, and it involves ‘normal’ people.
I fear that science education in Australia is now at the point where an equivalent programme made in Australia would probably be more astrology than astronomy. I would very very much love to be proven wrong about this, though. The Science Show and Catalyst are very very pale shadows of their former selves, and are, very sadly, often almost unwatchable or unlistenable unless you like a torrent of sanctimonious climate change propaganda.
For a government which was supposedly ‘better’ for science than the current one, the previous Labor government refused to keep funding the Siding Springs Observatory programme, (initially started via the ANU, using NASA money, which was cut by ‘progressive’ Obama) an AU$140K/year one-man/one telescope operation dedicated to detecting Near Earth Orbit objects (the only one in the southern hemisphere), and determining whether or not they may at some point collide with earth ( a la Chelyabinsk). Much better to funnel money into something CO2 related, rather than something actually useful from both a scientific, and a humanitarian perspective. One would think that if the incredible complexities of climate science had been ‘settled’, then there should be an awful lot of funding available to other branches of science…? Needless to say the funding decision was made a month or 2 before the 2013 election so the person who was laughably referred to as the science minister was nowhere to be seen…
In the final analysis we all recognize that global warming, if it is happening, isn’t nearly as much of a threat as the measures proposed for mitigating it.
That would be bad enough if there really was good evidence for a significant warming trend. But there isn’t any such evidence. It’s all trumped up numbers, wild speculation or worse. And so far there’s no credible link to a human cause of any warming that has been shown to have happened by good evidence.
Does that summarize the situation accurately?
I’ve understood and practiced conservation since my Boy Scout days. I’ve run fluorescent lights for years, not only because they were less expensive to run but because they last 10 to 100 times longer than an incandescent bulb. Now we have changes to fluorescent technology that not only makes them less bright but makes them quit much too soon. It’s all downhill when regulation is driven by politics an I draw the line at being told I need to save the planet when it clearly doesn’t need saving.
It’s time for resistance to all this nonsense. If we can do nothing more than speak out and tell the truth we must get busy doing that. And I hate to say it but if we must engage in civil disobedience then we need to be doing that.
Pick battles we can hope to win and start to fight them before it’s too late.
‘And I hate to say it but if we must engage in civil disobedience then we need to be doing that.’
I buy 100 watt incandescent bulbs and everyone is horrified.
‘Pick battles we can hope to win and start to fight them before it’s too late.’
Global cooling will come as a shock, but under the present circumstances there is little that can be done to turn the world upside down. Contrarian blogs are the best form of civil disobedience, far better than standing outside the ASX with a sign around my neck stating that CO2 does not cause global warming.
Didn’t know you could still buy them—except the stock that remained after the ban. I stockpiled them. I use them to keep my well from freezing in cooler weather. My other choice is a 1300 watt milkhouse heater, which my math says takes more energy but apparently the government’s math is different.
I should add that I am very familiar with the general properties of hydrogen peroxide, but am just not currently able to do an adequate search for anything resembling papers published on its supposed carcinogenic effects…
From a quick look at Google scholar, hydrogen peroxide accumulates in cancer cells (Activated granulocytes and granulocyte-derived hydrogen peroxide are the underlying mechanism of suppression of t-cell function in advanced cancer patients;and Dual role of hydrogen peroxide in cancer: possible relevance to cancer chemoprevention and therapy; and Increased Nox1 and hydrogen peroxide in prostate cancer, etc.).
It would seem to indicate that hydrogen peroxide anomalous generation was a symptom not a cause of cancer.
.
And there is a theory paper (pdf) on ‘Aging: a theory based on free radical and radiation chemistry’
Thanks very much! Just to be difficult, I just emailed the Friends of the earth Adelaide office to ask them for the evidence to support their claim. Sadly the ABC article doesn’t have an author attributed to it.
No doubt it will be straight to the recycle bin for daring to question the Church of Green about anything they say, but still, it’s worth a try…
Good luck in getting a reply from Anon the writer of such foolishness.
Saying peroxide, or most anything that is not immediately toxic is ‘bad’ for people, is IMO foolish.
These absolutist rules never work. Our bodies, like all of nature’s processes, are too complex. The simplistic mechanical model of our bodily functions is not such a good model. These models rely on degrees of stasis and conformity between individuals that often they are just too unreasonable.
As a point of note my mouthwash contains hydrogen peroxide, also hydrogen peroxide is a very safe antibacterial/anti-fungal for topical use. Women have bleached their hair with it for a century or more. It will also remove – bleach out – rust and blood stains from linens (chlorine bleach will not).
back in the olden days, well, 20 years ago, I often used a 30% concentration to remove organic carbon from clay samples before preparing them for XRD and XRF analysis – works a treat, as you’d expect… You too, can turn your black, nasty looking soil into pale brown or white clay in a matter of minutes! Buy now before stocks are sold out!
Well, since it’s really unstable water with and extra oxygen molecule I don’t see how it stays bound on its way into the organism especially once dumped into open water. It’s action as an antiseptic is as an oxidizer. Since it’s also photoreactive dumping it in a lake breaks it down fast and simply over oxygenates the water. A standard way of controlling algae that’s usually done with aerating pumps and fountains.
the lake in question is a man-made lake of moderate size, and has almost always suffered from various algal problems, controlled slightly with aeration and fountains, but obviously they’ve decided to scare the generally scientifically illiterate populace half to death by using ‘chemicals’.
probably a good thing they don’t see what goes into the tap water they drink then…
Organize for Action, Obama’s propaganda wing, held a “contest” they called Climate Change Fantasy Tournament. Senator Inhofe was the winner. I would like to encourage all of you in the USA (and elsewhere, if you think it would help) to let the Senator know how proud you are of him for his acccomplishment. This is what I sent:
“Organize for Action has been running a Climate Change Fantasy Tournament and you have been declared the winner and deemed the worst climate change denier in the country. Congratulations! We need more individuals such as yourself who stand up to the political dogma associated with the global warming scam. Please wear the title proudly!”
The more I see of Obama, the more I dislike his reign. Of course, I say ‘reign’, because really, he, and his non-entity wife are treated just like royalty – approaching divine representatives on earth, as opposed to an elected president, and his non-elected and politically irrelevant wife
As an aside, I saw Michelle Obama, the Whitehouse equivalent of a Kardashian (serves no useful purpose to anyone, and has no role to play, but still gets press coverage) recently parading around Cambodia telling the locals how to live their lives – not actually giving them money or resources, just telling people that they should aspire to have things they cannot yet afford (at least this is the idea i get from the coverage I have seen, I could be wrong).
I only hope that for the sake of the American taxpayers, she is paying for this herself, and not sponging off the US government, but I guess that’d be too much to hope for…
I’d be extremely annoyed if the spouse of any Australian politician decided to fly around the world interfering in matters in which they have absolutely no business, and no authority. We have (and i am sure the USA has) diplomats and elected representatives perfectly capable and duly authorised to wander around being sanctimonious and shallow…
No, we’re stuck with the bill. She would (supposedly) pay back the equivalent of a first class airline ticket towards the millions of dollars her little jaunts cost. We pick up the rest plus security. Don’t get me wrong, she needs the best protection we can give her but she’s been a pretty sorry excuse for a First Lady. I thought Bubba and Hillary were bad… these two take the cake when it comes to being king and queen wannabes. I don’t know if we survive another 22 months of their Progressive reign of terror.
I think what’s equally annoying is that the Obamas seem to genuinely think they are making some sort of positive impact on the world… Arguably, people who suffer from such delusions should not have lots of power, lots of money, and control over nuclear weapons.
‘The hockey stick graph dealt with all those by eliminating the MWP and the LIA. It inappropriately tacked on, as the blade of the stick, an upturn in temperature in the 20th century. Phil Jones produced the upturn that claimed a 0.6°C ±0.2°C increase in 120 years. They claimed this rate of increase was beyond any natural increase, conveniently ignoring the ±33% error factor.’
James Hansen took Venus with 96% CO2 atmosphere, and closer to the sun than earth, very hot, and then “modeled” the effects of CO2 on earth rising from 0.03% (that’s not 3%, it’s one-hundredth of 3%) to 0.06%, which is to say, he thought 96% CO2 on a closer to the sun planet had lessons for a planet with 99.04% NOT C02. Massively dominant gas fraction compared to lower than minuscule gas fraction. Conclusion: the earth is going to become uninhabitable.
And skeptics, remember, remember, remember: the AGWer “scientists” refused to release their reports’ raw data, computer codes for analyzing their data, but insisted that their finished product results were unimpeachable, and all that anybody needed to see.
As in Michael Mann’s exponential 20th century temp rise, imputed tone caused by linear CO2 rise, but the exponential hockey stick should have been a logarithmic curve from 260 ppm CO2 to 350. And nobody even challenged the incongruity, Nobel Prize Winner Arrhenius, and genius Josef Fourier, versus the guy who didn’t get his PhD until age 32–straight study from BA. which took 5 years, not 4, Mann not somebody who took some years off working outside academia. It took 9 years enrolled in grad school to earn his PhD. And the IPCC’s TAR showcased his work!
Okay, Dr. Mann, your research was paid for by USA taxpayers, who gave you grants and jobs. Show us, your employers, your raw data and codes, an explain why you decided to ignore raw data, and why, using bristlecone pine tree rings you didn’t examine samples from the complete bristlecone pine biome ranging from Sheep Mountain, CA to Nevada and Utah, to determine old climate across-the-western-US high-altitude “climate”. A real scientist would have taken the Sheep Mountain tree rings and said, “Let’s look at the rest of the bristlecone areas.let’s see what they tell us.”
And this doesn’t even tell the tale of why Mr. Mann, a third-class-honors student didn’t get placed at first-rate physics programs: MIT, Harvard, Caltech, Princeton, Stanford,Cambrid Berkeley, but at second-rate Yale (zero Nobel laureates, only 4 NAS Physics Section members, now down to 2). Berkeley profs decided, “He’s not one of our brilliant minds, he’s not even one of our very bright minds who do not score in the top 3 of his classes, but nevertheless consistently hit top-10. percentile. Mr. Mann was in the bright category, by taking light loads, top 20th-30th percentile.
Top Berkeley grads have A/A+ grades in Honors Math and Physics lower division courses. Let Mr. Mann release his courses and grades. If he refuses, then you know where he’s resting his “laurels”. I mean, A/A+ grades in Berkeley lower division math and physics courses require hard study, no pot smoking. If Mann can demonstrate these, let’s hear him out.
Also, why he was recommended, but turned down opportunities to go to MIT, Harvard, Stanford, Princeton, Cambridge, Caltech, Berkeley, Chicago, Cornell, UIUC, for his PhD, except he wasn’t recommended and didn’t get in.
Just a belated footnote to say that, although I’ve been happy to vote for our famous hostess in previous years, unfortunately her credibility took a turn for the worse in 2014. She continues to claim it is possible “the ocean is causing the rise in CO2”, therefore I have not voted in the Bloggies for her blog and neither will I be tipping the choccy jar for the foreseeable future. Counterfactual conclusions supported only by easily disproven arguments, such as Salbyism, have no place in the skeptic’s toolbox.
Andrew,
Interesting She continues to claim it is possible “the ocean is causing the rise in CO2″!
If for unknown reasons ocean surface temperature increases by 0.03 kelvin would there not be a measurable increase in atmospheric CO2? This is not only possible it is proven measurable, with no foolish anthropoids being involved.
There is absolutely no warming sensitivity for any IR-active gases like water vapor, carbon dioxide and methane. The force of gravity produces a temperature gradient in a planet’s troposphere which, for an Earth with dry air, would raise the mean surface temperature to about 25°C to 28°C. But then inter-molecular radiation between water vapor molecules has a temperature leveling effect and thus reduces the surface temperature because it makes the gradient less steep, as is well known.
What is the sensitivity for each 1% of water vapor in the atmosphere?
We know that water vapor reduces the magnitude of the so-called lapse rate, which does not need a special name because it is just the temperature gradient in the troposphere. Now, we also know that radiative balance at the top of the atmosphere is virtually always close, in fact to within ±0.5%. The inward radiation is based on the so-called “Solar Constant” (although that does vary, especially due to variations in Earth’s eccentricity in a ~100,000 year cycle that regulates glacial periods) and the outward radiation (broadly speaking) increases if the whole temperature plot rises, making the area under that plot greater.
Now, what the AGW crowd want you to be gullible enough to believe is that (as the percentage of water vapor increases) the thermal plot can rise at the surface end whilst at the same time acquiring a less steep gradient. Any secondary student with a knowledge of coordinate geometry would know that the area under the new (higher and less steep) thermal plot would be far greater than that under the original plot for a world with less water vapor. So how could that happen? It can’t, because the whole plot would then fall to regain radiative balance, and in fact it would have just rotated downwards at the surface end in the first place. So how could the sensitivity for each 1% of water vapor be positive causing warming of the surface?
This is the third in a series of four comments I am posting on Joanne’s Unthreaded Weekend threads. Formatting issues prevent me from including figures and tables in this post. A PDF version of this post containing all figures and tables can be found at:
https://www.dropbox.com/s/6b8j73qf34qq3ad/final_peter_reed_paper_to_Joanne_Nova_unthreaded_01_pdf.docx?dl=0
In addition, a spreadsheet containing the raw measurements and most calculations can be found at:
https://www.dropbox.com/s/cyykdqo4c8uflrr/June_10_through_July_7_2014_examples_for_paper.xlsm?dl=0
[Note: If when you “click” on either/both of the above links, instead of seeing an “opened PDF/Excel file” a new “window” pops up with the option to “download,” click on the “download” button. This should allow you to load the file onto your personal computer.]
In my first post [comment #2 at http://joannenova.com.au/2015/03/weekend-unthreaded-69/#comments%5D, I discussed the definitions and connotations of words and phrases. In my second post [comment #2 at http://joannenova.com.au/2015/03/weekend-unleaded/#comments%5D, I (a) challenged the AGW community to show that the claim if all greenhouse gases are removed from the atmosphere, the Earth’s surface temperature will be reduced by approximately 30 degrees Centigrade isn’t based on fatally flawed logic, and (b) reasoned that applying the backradiation argument commonly used by AGW proponents to “prove” greenhouse gases must increase the Earth surface temperature implies that when matter whose temperature is higher than the ambient background temperature is stored in a thermos bottle, a CO2 thermos bottle should “outperform” a vacuum thermos bottle. [For the storage of material hotter than the ambient background temperature, thermos bottle “A” “outperforms” thermos bottle “B” if for identical thermos bottle contents and identical thermos bottle background environments, the time it takes for the contents of thermos bottle “A” to decrease by a specified temperature is greater than the time it takes for the contents of thermos bottle “B” to decrease by the same temperature.] In this post I describe a thermos bottle experiment performed by Peter C. that shows vacuum thermos bottles that differ from CO2 thermos bottles only by the presence/absence of CO2 gas outperform CO2 thermos bottles. I assisted Peter in extracting derived information (primarily cooling rates) from Peter’s experimental data. The words “we”, “us”, and “our” are used throughout this post; but I want to acknowledge that Peter performed all of the experiments.
[Note: To many readers, a positive “cooling rate” implies a temperature that is “decreasing with time,” and a negative “cooling rate” implies a temperature that is “increasing with time.” In this post, however, the term “cooling rate” represents the time-rate-of-change of temperature. As such, in this post a positive “cooling rate” implies the temperature of an object is increasing with time, and a negative “cooling rate” implies the temperature of an object is decreasing with time. ]
As noted in my first two posts, it is often claimed that Carbon Dioxide (CO2) is a “heat-trapping gas” [Reference 1, http://www.ucsusa.org/global_warming/science_and_impacts/science/CO2-and-global-warming-faq.html.%5D. Once someone accepts this claim as fact, it is fairly easy to convince him/her that since CO2 gas is distributed throughout the Earth’s atmosphere, the presence of atmospheric CO2 gas should, via the trapping of heat, increase the Earth’s surface temperature above what the surface temperature would be in the absence of atmospheric CO2 gas.
As mentioned in my first post, a reasonable interpretation of “heat-trapping” is that a “heat-trapping anything” that surrounds an object hotter than its environment should prevent or at least retard heat loss from that object. Once someone accepts this “interpretation” and agrees that CO2 gas is a heat-trapping gas, it is fairly easy to convince him/her that since CO2 gas is distributed throughout the Earth’s atmosphere, the presence of atmospheric CO2 gas should, via the trapping of heat, increase the Earth’s surface temperature above what the surface temperature would be in the absence of atmospheric CO2 gas.
In my second post, I argued that this interpretation carries the implication that everything else being equal, when surrounded by a “heat-trapping anything” an object at a temperature higher than its surrounding environment should cool at a slower rate than in the absence of the “heat-trapping anything”. Thus, I argued that to claim that CO2 gas is a heat-trapping gas is to claim that when surrounded by CO2 gas, material at a temperature higher than its surrounding environment will retain its heat for an extended period of time relative to cooling times associated with the absence of the surrounding CO2 gas.
It occurred to us that if the above implication is valid, then at least for keeping liquids above ambient background temperatures (as opposed to keeping liquids below ambient background temperatures), CO2 thermos bottles should outperform vacuum thermos bottles. The goal of a thermos bottle is to keep material at a higher (or lower) temperature than the ambient background temperature for as long a time interval as possible. A vacuum thermos bottle accomplishes this goal in part by surrounding its inner chamber (the region in which the heated/cooled material is stored) with a vacuum. In this post, a CO2 thermos bottle differs from a vacuum thermos bottle only in that CO2 gas is injected into the vacuum region of a vacuum thermos bottle; and thus in place of a vacuum, CO2 gas surrounds both the inner chamber and the contents of the inner chamber. Thus in this post, with the exception of the presence of CO2 gas, the phrase “everything else being equal” applies for a vacuum thermos bottle versus a CO2 thermos bottle.
If the implication is valid, then a CO2 thermos bottle should keep “hot” material above the ambient background temperature for a longer period of time than a vacuum thermos bottle. We believe the opposite is true. That is, we believe vacuum thermos bottles will keep “hot” material above the ambient background temperature for a longer period of time than CO2 thermos bottles. One reason we hold this belief is that if CO2 thermos bottles outperform vacuum thermos bottles, thermos bottle manufacturers would flood the market with CO2 thermos bottles. Such is not the case. However, the lack of a plethora of commercial CO2 thermos bottles is not proof that vacuum thermos bottles outperform CO2 thermos bottles. We decided to run some admittedly simple experiments to see if we couldn’t quantify the relative cooling-rate performances of vacuum and CO2 thermos bottles. Our experiments are simple because we had limited access to sophisticated equipment. We encourage people with access to better equipment to perform similar experiments; and, thereby, either support or refute our results.
[For all figures referenced in this post, open URL https://www.dropbox.com/s/6b8j73qf34qq3ad/final_peter_reed_paper_to_Joanne_Nova_unthreaded_01_pdf.docx?dl=0 on your computer. The figure numbers in this post and the cited URL are the same. In fact, with the exceptions of tables and figures, this post is almost identical to the cited URL. As such, you might want to skip the remainder of this post and simply read the URL document.]
EXPERIMENTAL SET UP
Figure 1 depicts our laboratory.
Figure 1: Our Laboratory
The experimental apparatus consists of:
—-Vacuum Flask (dark green thermos bottle),
—-Alcohol Thermometer, -20 to 110 degrees Centigrade (protruding from the Styrofoam cap at the top of the Vacuum Flask),
—-Digital Thermometer with K type thermocouple (yellow object),
—-Wrist watch (not shown),
—-Pen and Paper (not shown).
VACUUM FLASK
Peter was able to purchase two flasks at a discount store for A$10 each. The primary temperature retaining feature of each flask is a double-walled, vacuum, glass insert mounted within a plastic outer case. The plastic outer case has a base that can be unscrewed to access the glass insert. The space between the inner and outer walls of the glass insert is evacuated. The glass insert is silvered on its inside surfaces.
Figure 2 shows the components of the flask: Outer Plastic Case, Removable Base, and Double-Walled Vacuum Glass Insert with the double-wall construction shown by removing a portion of the outer wall at the base of the glass insert.
Figure 2: Flask Construction Dismantled To Show The Double-Walled Vacuum Glass Insert (broken in this Figure)
Although the flasks looked the same, with flasks of this price the quality may be somewhat variable. We found this to be the case. We believe the performance difference was associated with the quality of the vacuum. The two flasks were tested by filling with near boiling water and measuring the cooling rate. Flask A, the flask with the better performance, was selected for all experiments reported in this post.
The original screw cap at the top of the flask was replaced by a homemade Styrofoam stopper with a hole to allow the thermometer access to the liquid inside the thermos. The Styrofoam stopper seemed to contain the heat better that the original plastic cap.
Five trials were run to quantify the flask’s rate of heat loss in its original form (vacuum). The vacuum was then destroyed by cutting a hole through the vacuum nipple. A thin plastic connecting tube was then attached through this hole and sealed with silicone. Four trials were run to quantify the flask’s rate of heat loss when the vacuum region was filled with air. Five trials were run to quantify the flask’s rate of heat loss when the vacuum region was filled with CO2 gas.
Plastic connecting tubing, three-way taps and syringes are easily obtained in hospitals. Using these items we were able to construct a simple vacuum pump which can create a vacuum of approximately 0.1 atmospheres. The vacuum was measured by attaching to the chamber of a Mercury Barometer. The vacuum was maintained over several days, demonstrating the sealing is very good.
THERMOMETERS
During each experimental run, we wanted to monitor the temperature of three objects: (1) the water inside the inner chamber, (2) the outermost surface of the glass insert, and (3) the ambient room temperature. We obtained four thermometers for this purpose: a Mercury thermometer (white), two alcohol thermometers (yellow and green), and a digital thermometer/thermocouple. On 17 April 2014 we measured the relative performance of the thermometers. [See the cited URL for Table 1. Basically, the thermometers agreed to within 2C.]
Table 1: Comparison Of Thermometers
Unfortunately, the mercury thermometer broke early on and was not used for the experiments described in this post. Because we thought that the digital thermometer might suffer from drift in the electronics, we decided to use the alcohol thermometers to measure the water temperature and the ambient temperature. Because it was the easiest to read, we used the green alcohol thermometer to measure the temperature of the water. We used the yellow alcohol thermometer to measure the ambient background temperature. We measured the ambient background temperature only at the start of the experiment. We taped the thermocouple to the outermost surface of the glass chamber. Monitoring the temperature of the outermost surface of the glass chamber provides additional insight into the heat transfer process. Although we did not estimate the cooling rate of the outermost surface of the glass chamber, we plotted those temperatures for all but our first (Run 3) vacuum experiment.
The green alcohol thermometer was calibrated by continuous insertion in a sauce pan of boiling water on a stove. It read 101C. The thermometer was then placed in a whisky glass containing ice blocks and cold water and read 1C. The temperature readings of our experiment were not corrected for these slight differences (101C vs 100C, and 1C vs 0C).
VACUUM, AIR AND CO2 GAS
All “vacuum measurements” reported in this post were made with the flask as provided by the manufacturer. After completing the vacuum measurements, we destroyed the as-purchased quality of the vacuum simply by cutting through the tip of the vacuum nipple. Air rushed into the original vacuum space. All “air temperature measurements” were made with the flask in this state.
When it came time to make “CO2 temperature measurements”, we wanted as high a concentration of CO2 gas as possible. CO2 gas was obtained from a cylinder of CO2 gas supplied by the British Oxygen Company, which is the local supplier for all our gases. The gas cylinder was labeled “BOC Medical EP Grade CARBON DIOXIDED Gas Code 530.” BOC specifies this gas as 99.95% CO2. A plastic party balloon was fitted with connecting tubing. The plastic balloon was initially completely emptied, then filled with CO2 gas from the cylinder. This procedure was repeated to purge any residual air from the balloon and tubing.
Using the syringe, the vacuum chamber was then evacuated to 0.1 atmospheres and refilled from the balloon via the three-way tap. This procedure was repeated four times. Although we did not test the purity of the CO2 gas in the vacuum chamber, we believe that for our “CO2 experiments” between 0.99 and 0.999 of the gas in the vacuum chamber was CO2.
Figure 3 depicts our set up for filling the original vacuum region with CO2 gas. Note the connecting tube sealed to the vacuum nipple with silicone; plastic balloon attached to three-way tap and syringe, K thermocouple taped to outside surface of glass vacuum chamber.
Figure 3: Flask With Equipment To Fill The Flask Vacuum Region With CO2 Gas
NOTE ON MEASUREMENTS
It was apparent that we could prioritize either the temperature or the timing measurements. Specifically, when the temperature reached an exact reading on the thermometer, we could record the time. Or when the watch reached an exact time, we could make our best reading of the thermometer. If we chose to record the temperature at what we judged to be the exact degree, the timing could be subject to an estimation error of 30 seconds or more. If the timing was prioritized, it could be accurate to within 5 seconds or less, but the temperature estimate accuracy was degraded. We elected to prioritize the timing—i.e., we would make the best temperature measurement we could at a time accurate to 5 seconds or less. We could read the temperature to 0.5 degrees or less. If the temperature was near the 0.5 degree mark, the temperature was recorded as the immediately lower integer plus 0.5. If the temperature was past the degree mark, but seemed a little less than 0.5 it was recorded as the immediately lower integer plus 0.2. If the temperature was nearly to the next degree mark, it was recorded as the immediately lower integer plus 0.8.
RESULTS
In total we made 14 experimental runs of the temperature of heated water in the flask as a function of time—five runs with a vacuum, four runs with air in the original vacuum region, and five runs with CO2 gas in the original vacuum region.
The experimental results are summarized below:
Results with VACUUM Thermos Bottle, Cooling Rate Average: -0.055612 C per minute
____Run #3, 10 June 2014, Ambient Temperature 19C
________Measurement Times (minutes) relative to time of first measurement 18:00
____________0, 30, 70, 90, 120, 150, 180, 220, 242, 245
________Measured temperatures (C)
____________98.0, 97.0, 94.5, 93.0, 91.2, 90.2, 88.5, 86.0, 85.2, 85.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.053834
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.000838
—-Run #4, 11 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 18:45
————0, 37, 53, 60, 90, 126, 150, 180, 210, 227, 240
——–Measured temperatures (C)
————98.0, 96.0, 95.0, 94.8, 92.5, 91.5, 89.0, 87.5, 85.8, 85.0, 84.5
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.057603
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001264
____Run #5, 12 June 2014, Ambient Temperature 18C
________Measurement Times (minutes) relative to time of first measurement 17:32
____________0, 13, 43, 62, 73, 103, 150, 171, 194, 223, 241, 258, 272, 303
________Measured temperatures (C)
____________98.0, 98.0, 96.0, 95.0, 94.2, 92.2, 90.0, 88.8, 87.5, 86.0, 85.0, 84.2, 83.5, 82.8
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.053578
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.000988
—-Run #6, 13 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 18:25
————0, 5, 12, 51, 82, 116, 140, 170, 200, 230
——–Measured temperatures (C)
————98.0, 98.2, 97.5, 95.5, 94.0, 91.2, 90.1, 89.0, 87.0, 85.2
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.056483
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001194
____Run #7, 15 June 2014, Ambient Temperature 19C
________Measurement Times (minutes) relative to time of first measurement of 17:40
____________0, 60, 90, 140, 210, 232
________Measured temperatures (C)
____________98.2, 95.0, 92.8, 90.0, 86.5, 85.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.056563
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001214
Results with AIR Thermos Bottle, Cooling Rate Average: -0.217540 C per minute
—-Run #10, 21 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 14:10
————0, 1, 9, 15, 20, 25, 30, 35, 40, 45, 51, 55, 60, 61, 65, 70, 77, 85
——–Measured temperatures (C)
————98.2, 98.0, 96.2, 95.0, 93.8, 93.0, 91.8, 90.5, 89.2, 88.5, 87.0, 86.2, 85.2, 85.0, 84.5, 83.5, 82.0, 80.5
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.210915
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.002071
____Run #11, 23 June 2014, Ambient Temperature 20C
________Measurement Times (minutes) relative to time of first measurement 7:10
____________0, 1, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 55.5, 60, 65, 70
________Measured temperatures (C)
____________96.0, 97.5, 96.0, 95.0, 94.2, 93.0, 92.0, 90.5, 89.5, 88.5, 87.2, 86.2, 85.2, 85.0, 84.2, 83.2, 82.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.213345
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.004074
—-Run #12, 28 June 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 9:00
————0, 5, 9, 10, 15, 20, 25, 31, 36, 40, 45, 50, 55, 60, 65, 72
——–Measured temperatures (C)
————97.2, 96.0, 95.0, 94.8, 93.8, 93.0. 91.8, 90.2, 89.2, 88.5, 87.2, 86.0, 85.0, 84.2, 83.0, 82.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.215037
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.002216
____Run #13, 26 June 2014, Ambient Temperature 20C
________Measurement Times (minutes) relative to time of first measurement 20:39
____________0, 1, 6, 11, 12, 16, 22, 26, 31, 38, 46, 51, 56, 57, 61, 66
________Measured temperatures (C)
____________98.2, 98.0, 97.5, 95.8, 95.0, 94.2, 93.2, 91.8, 90.8, 89.2, 87.2, 86.2, 85.2, 85, 84.2, 83.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.230865
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003741
Results with CO2 Thermos Bottle, Cooling Rate Average: -0.206672 C per minute
—-Run #30, 2 July 2014, Ambient Temperature Not Recorded
——–Measurement Times (minutes) relative to time of first measurement 18:13
————0, 1, 5, 12, 17, 18, 22, 28, 33, 41, 47, 58, 62, 67, 76, 85
——–Measured temperatures (C)
————98.2, 98.2, 97.5, 96.2, 95.2, 95.0, 94.2, 93.0, 92.0, 90.2, 89.0, 87.5, 86.2, 85.5, 83.5, 82.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.194215
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001631
____Run #31, 3 July 2014, Ambient Temperature Not Recorded
________Measurement Times (minutes) relative to time of first measurement 20:26
____________0, 5, 9, 13, 19, 24, 29, 34, 39, 46, 51, 55, 60, 61, 70, 74, 94
________Measured temperatures (C)
____________98.0, 97.5, 96.0, 95.0, 94.0, 93.0, 91.5, 90.5, 90.0, 88.0, 87.0, 86.2, 85.5, 85.0, 83.5, 82.5, 79.5
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.204740
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003633
—-Run #32, 5 July 2014, Ambient Temperature 20C
——–Measurement Times (minutes) relative to time of first measurement 15:02
————0, 1, 3, 8, 13, 19, 24, 29, 49, 54, 59, 64, 69, 74
——–Measured temperatures (C)
————97.5, 97.0, 97.0, 96.0, 95.0, 93.5, 92.5, 91.2, 87.0, 86.0, 85.0, 84.0, 83.0, 82.2
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.210518
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001785
____Run #33, 6 July 2014, Ambient Temperature 19.5C
________Measurement Times (minutes) relative to time of first measurement 9:50
____________0, 16, 24, 45, 64, 70, 76
________Measured temperatures (C)
____________97.2, 94.0, 92.0, 87.2, 83.2, 82.2, 81.0
________WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.214454
________Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.003181
—-Run #34, 7 July 2014, Ambient Temperature 18C
——–Measurement Times (minutes) relative to time of first measurement of 20.26
————0, 4, 10, 15, 19, 25, 35, 40, 47, 55, 60, 65, 69, 75, 80
——–Measured temperatures (C)
————97.5, 96.5, 95.2, 94.5, 93.5, 92.2, 90.0, 89.2, 87.5, 86.0, 85.0, 83.5, 83.0, 81.5, 81.0
——–WLS-Linear-Fit Estimated Cooling Rate (C per minute): -0.209431
——–Estimated Standard Deviation Of Cooling Rate Estimate (C per minute): 0.001747
For all but the first vacuum run, we simultaneously monitored the temperature of the outside of the glass vacuum flask. Figures 4 through 8 show the results for the “vacuum” thermos experiments. Figures 9 through 12 show the results for the “air” thermos experiments. Figures 13 through 17 show the results for the “CO2” thermos experiments.
Figure 4: 10 June 2014 Vacuum Thermos Figure 5: 11 June 2014 Vacuum Thermos
Figure 6: 12 June 2014 Vacuum Thermos Figure 7: 13 June 2014 Vacuum Thermos
Figure 8: 15 June 2014 Vacuum Thermos
Figure 9: 21 June 2014 Air Thermos Figure 10: 23 June 2014 Air Thermos
Figure 11: 28 June 2014 Air Thermos Figure 12: 26 June 2014 Air Thermos
Figure 13: 2 July 2014 CO2 Thermos Figure 14: 3 July 2014 CO2 Thermos
Figure 15: 5 July 2014 CO2 Thermos Figure 16: 6 July 2014 CO2 Thermos
Figure 17: 7 July 2014 CO2 Thermos
Table 2 summarizes the estimated cooling-rates and the estimated standard deviation associated with each estimated cooling rate. The cooling rates were generated by applying a Weighted-Least-Squares (WLS) estimator to a linear model of temperature as a function of time. The WLS process assumed each measurement error was a random sample from a zero-mean distribution. The standard deviations of the various measurement error distributions were assumed to be the same. The measurement errors were assumed to be uncorrelated. With these assumptions, WLS estimation produces a slope identical to the slope of a linear regression. In addition to providing an estimate of the slope, if the measurement error standard deviations are known, WLS theory also provides an estimate of the variance/covariance of the parameters being estimated (slope, offset).
Table 2: Summary Of Water Cooling Rates
Using the WLS slope estimate (cooling rate) and the WLS offset estimate (temperature at relative time 0), a temperature residual for each measured temperature can be generated. A temperature residual is the difference between the measured temperature and the corresponding model temperature. If the measurement error standard deviations are known, a weighted residual for each measurement can be computed. A weighted-residual is a residual divided by the corresponding measurement error standard deviation. If all of the above assumptions are valid and if the linear model is ideal, then the expected value of the sum-of-squares of weighted residuals will equal the number of measurements …minus… the number of degrees of freedom of the WLS estimation process. The number of degrees of freedom of a WLS estimation process is the number of variables to be estimated. For linear WLS estimation, there are two degrees of freedom (slope, offset). Since we assumed the measurement error standard deviations are equal, we can generate an estimate of that standard deviation by computing the square root of the ratio of (a) the sum-of-squares of weighted residuals, and (b) N minus 2, where N is the number of measurements. Using the estimated measurement error standard deviation, we can generate an estimate of the standard deviation of each measured slope (cooling rate). Those estimated standard deviations are given in Table 2.
The first observation is that the results are very good, given the cost and sophistication of the experimental equipment. The runs with the vacuum in the flask vary only in the third decimal place. Even at the higher rate of heat loss, with air and then CO2 gas as the insulating medium the variation is less than 0.03.
An examination of the results presented above shows that in the case of a vacuum thermos bottle versus a CO2 thermos bottle, CO2 gas acts to release heat, not to trap heat. From the perspective of a thermos bottle, CO2 gas is more correctly characterized as a “heat-releasing gas” than as a “heat-trapping gas.” Thus, if the term “heat-trapping gas” carries the connotation of “trapping heat”, CO2 gas cannot be called a “heat-trapping gas” because situations exist where CO2 gas doesn’t “trap heat”, CO2 gas “releases heat.” If one defines a “heat-trapping gas” to be any gas in the Earth’s atmosphere that causes an increase in the Earth surface temperature, then (a) it is a circular argument to use the “heat-trapping” nature of a gas to claim the presence of the gas in the Earth’s atmosphere will increase the surface temperature of the Earth, and (b) it must be shown that CO2 gas is a heat-trapping gas, not simply declared based solely on the observation that CO2 gas absorbs electromagnetic radiation in subbands of the IR band. After all, the IR absorption property exists for the CO2 gas in the thermos bottle and the CO2 thermos bottle loses heat faster than the vacuum thermos bottle. It may well be that atmospheric CO2 gas causes an increase in Earth surface temperature; but to make that claim based solely on the IR absorption property of CO2 gas is unwarranted.
Both Air and CO2 gas release heat from the inner surface of the flask to the environment, likely by increasing the rate of conduction. At first glance it would appear that CO2 gas is a better insulator than Air, and this difference is nearly but not quite significant at the 95% confidence level, which assuming a Gaussian distribution, we approximate to error bars of 2 standard deviations. That is an error of 2 standard deviations from the average below the air average to 2 standard deviations above the CO2 average overlap. Figure 18 depicts the Vacuum cooling rates with plus/minus 2 sigma error bars. Figure 19 depicts the “Air” and “CO2 gas” cooling rates with plus/minus 2 sigma error bars.
Figure 18: Vacuum Results With 2 Sigma Error Bars Figure 19: Air And CO2 Gas Results With 2 Sigma Error Bars
Also we can see from the results that Runs No 13 and 30 seem to be outliers in that they vary from the other runs by a noticeable amount. If we were to exclude these two results then the remaining runs have almost identical rates of heat loss for AIR and CO2 gas. Now one cannot arbitrarily exclude any result, without a proper reason, and hence these results have not been excluded.
If we were to repeat the experiments, we would try to control the ambient temperature during each experiment, or at least measure the ambient temperature during the course of the experiment, rather that just take it at the beginning. It may be that under those conditions the heat loss with Air and with CO2 gas would be even closer together. We would also use a digital clock as the timing device instead of an analog wrist watch . It did seem that some of the observations may have been out by a minute or even 2 minutes, suggesting that the watch might not have been read correctly. No observations were corrected or adjusted.
In addition to the numerical cooling rate results, it’s clear from the graphs of Figures 2 through 5 compared to the graphs of Figures 13 through 17 and Figures 30 through 34 that the temperature of the outermost wall of the glass insert behaves differently for a vacuum thermos bottle than it does for Air and CO2 thermos bottles. For a vacuum thermos bottle the temperature of the glass insert’s outermost wall never got above 30C, whereas for both the Air and CO2 thermos bottles this temperature got to 50C. Since the radiative rate of cooling is proportional to the difference in the fourth powers of the medium and the surrounding wall temperatures, the higher surrounding-wall temperature will result in a reduced radiative rate of cooling. That is, the radiative cooling rate will be larger for the vacuum thermos bottle than for the Air and CO2 thermos bottles. However, it’s clear from the data that the total cooling rate is larger for the Air and CO2 thermos bottles than for the vacuum thermos bottle. Obviously, for the thermos bottle example, the heat transfer processes of convection and conduction more than compensate for the decrease in radiative cooling rate.
This completes my third Unthreaded Weekend post. The experimental results described above show that situations exist where surrounding heated material with CO2 gas (a greehouse gas) hastens, not retards, heat loss. Although the thermos bottle experiment described in this post differs from CO2 gas in the atmosphere, the thermos bottle example brings into question the claim: Via backradiation to the Earth’s surface, atmospheric CO2 gas must increase the temperature of the Earth’s surface.
Next up (fourth Weekend Unthreaded post), both in the absence and presence of backradiation from an object devoid of an internal energy source (an inert object), a theoretical treatment of the temperature of a sphere possessing a constant-rate source of internal energy (an active object). The conclusion of the post will be that with no change to the active object’s internal rate of energy, backradiation from an inert object to an active object can simultaneously exist with lower, not higher, active object temperatures.
90
For those who have read Reed’s third comment post to the end, I would encourage you to look at the graphs which Reed has published in the PDF version. It was a great pleasure to see my pages of lab notes transformed into really elegant graphs. I was also most impressed by how close nearly all the measurements were to the line of best fit. Only the odd measurement fell far from the line, which likely meant that I had made a mistake recording the time.
I might consider performing more experiments if I can obtain some SFl6 gas. This is a heavy gas used occasionally in ophthalmology. It is said to be a greenhouse 25,000 times more powerful than CO2.
60
Reed Coray & Peter C,
First of all, kudos for your compliance to the scientific method in all of it’s facets, especially in the presentation of the methods used and the raw data obtained.
Before I comment on my impressions of your experiment, I would first of all like to point out that anyone who has read my comments here will conclude that I am clearly a skeptic. The AGW meme, in my estimation, is just that, a meme. Cleverly contrived, as it is, for the promulgation of political and economic change and as such, it is as completely removed from all of accepted physics as oil seperates from water.
Having said that, I feel obliged to play the devil’s advocate and point out a flaw in your endeavor.
My understanding of the claim comimg from the AGW camp is that CO2 ‘traps’ radiative energy and not kinetic energy. The inner bottle in the thermos aparatus used in your experiment is ‘silvered on its inside surfaces’. I’m not a materials engineer, but my understanding is that the surface is silvered to prevent radiative energy from leaving the inner flask. How well the silvering accomplishes this task is, again, beyond my expertise but clearly the best cost effective process is used and so must perform fairly well in this respect.
Therefore, your experiment must, by construction, be measuring kinetic energy loss and not radiative heat loss.
and . . .
From these quoted results, and correct me if I’m wrong, the conclusion must be that even when measuing kinetic energy loss (conduction), CO2 has a lower rate of heat loss than common air, (a combination of all the gases in the atmosphere).
Abe
60
Just-A-Guy.
Peter’s experiment measured the time-rate-of-change of temperature of the water in the thermos bottle, which is related to the time rate of change of the internal energy of the water in the thermos. In most macro thermodynamic treatments (especially gases), internal energy is associated with the average kinetic energy of the molecules comprising the material, which is related to temperature. In gases, heat transfer via conduction is modeled as elastic collisions of molecules that transfer the energy of high-velocity (i.e., high temperature) molecules to low-velocity (i.e., low temperature) molecules. So you’re correct, Peter’s experiment is indirectly measuring the rate of internal energy loss via all heat transfer mechanisms (radiation, conduction, convection, and change-of-state–evaporation). You’re also correct when you point out that Peter’s experiment shows that the rate of heat loss using CO2 is smaller than the rate of heat loss using air.
That brings up an interesting question. The heat conduction properties of gases at various temperatures and pressures have been measured. When reporting these conduction properties, are the conduction properties corrected for radiation effects; or do they represent heat transfer rates from all heat transfer mechanisms? It’s relatively simple to minimize heat transfer via convection and change-of-state; but for gases, it’s not so easy to separate conduction and radiation. At extremely low gas densities, I believe radiation dominates. At extremely high gas densities, I believe conduction dominates. Thus, although I agree Peter’s experiments indicate that CO2 is a better insulator than air (at least at atmospheric pressures) and this may be the result of absorbing IR radiation from the walls of the inner thermos bottle chamber, I don’t know why that is the case. Is it because of backradiation or different thermal conduction properties?
You wrote: “My understanding of the claim comimg from the AGW camp is that CO2 ‘traps’ radiative energy and not kinetic energy.” My exposure to the AGW position is slightly different from yours. Specifically, that in addition to claiming “CO2 traps radiative energy,”, they also claim “CO2 traps heat.” Three comments. First, the Earth surface temperature is affected by ALL heat loss mechanisms. It’s true that for the Earth/Earth-Atmosphere system as a whole, radiation (to space) is the overwhelming dominant means of energy loss. However, that is not the case for the Earth’s surface. Radiation, evaporation, convection, and conduction all play a role in transferring internal energy away from the Earth’s surface–likely in that order with radiation being the dominant mechanism. So attempting to predict the Earth’s surface temperature, or even changes to the Earth’s surface temperature, based solely on the electromagnetic radiation absorption/radiation properties of atmospheric gases is a non-starter in that it is bound to fail.
Second, it’s the total rate of energy transfer that affects temperature. Thus, it’s correct to say that CO2 absorbs (I hate the word ‘traps’) radiation emanating from the Earth’s surface. But such a statement, even coupled with backradiation, does not by itself “prove” the Earth’s surface temperature will increase. The statement provides plausibility, not proof. In my fourth and last post I will present a mathematical treatment showing that inactive material (i.e., material devoid of an internal source of energy): (a) completely surrounding an active object (i.e., an object with an internal source of energy), (b) absorbing ALL radiation frequencies, not just sub-bands of the IR, emitted from the surface of the active object, and (c) backradiating a portion of the absorbed energy to the active object can exist simultaneously with LOWER not HIGHER active object temperatures. If my mathematics is correct, then the existence of backradiation does not, by itself, guarantee increased active object temperatures.
Third, since the Earth/Earth-Atmosphere system overwhelmingly loses energy to space via radiation, it may be that atmospheric CO2 absorbs (“traps”) radiation emanating from the Earth’s surface; but it can’t be true that CO2 absorbs (“traps”) all radiation. If the latter were the case, the Earth/Earth-atmosphere system won’t lose energy to space and would eventually, how shall I say, fry.
Bottom line, I believe you’re correct when you say Peter’s experiment must “…be measuring kinetic energy loss…; but I believe you’re incorrect when you add “”…not radiative heat loss.”
Finally, don’t lose sight of the point I’m trying to make. I’m NOT arguing that thermos bottles behave similarly to the atmosphere. I’m addressing the question: “By any reasonable interpretation of “heat-trapping,” is CO2 a heat-trapping gas?” I believe this point is important because I believe this is the AGW community’s primary argument used to convince the average citizen that AGW is occurring is that CO2 is a “heat-trapping gas”–or as you put it “a radiative energy trapping gas.” If I can show a situation where CO2 gas acts more like a “heat-releasing” gas than a “heat-trapping” gas, then at a minimum I’ve given people something they can use to question the “heat-trapping” claim. And, if I can show a situation where just because CO2 ‘traps’ radiative energy, it doesn’t necessarily follow that the surface temperatures inside the “trap” must increase, then I’ve given people something to think about.
20
Reed Coray,
There’s a lot to be discussed in your reply so may I suggest we take it one point at a time, do you agree?
I’d like to start with this statement, If I may.
Also, because of the close connection with this statement . . .
. . . lets discuss them both together.
To the best of my knowledge, there are only three ways that energy can move from one location to another. They are, radiative flux, conduction, convection. In radiative flux, energy is emitted from a particle, (either an atom or a molecule), in discrete amounts called quanta. These quanta can travel through the vacuum and may eventually be absorbed by another particle. In conduction, one particle will bump into another particle. If one particle has a higher energy content than the other, then some energy will be transfered from the particle with the higher energy content to the particle with the lower energy content. In the case of convection, a collection of particles, say a body of water, will lose one or more of it’s particles through the process of evaporation, taking some of the energy contained in the collection with it.
If you’re with me so far, then we can proceed to the next point which concerns the bolded part of your statement above, they also claim “CO2 traps heat”. If, on the other hand, I’m missing something in my comprehension of energy transfer, let’s clear that up first.
Abe
10
Most “normal” glasses absorb LWIR energy. A glass thermos bottle traps heat two ways. First, it uses really thin glass so that the conductive area is very small from the inner portion of the glass envelope to the outer part. The silver coating is on the vacuum side of the glass wall. This is to prevent (slow) radiative transfer of the inner glass to the outer glass surface and visa versa when using it to keep things cold. Since the vacuum stops conductive heat transfer they silver the glass to slow the radiation to the outer surface.
Silvered or aluminized 2nd surface mirrors are crappy LWIR reflectors. They either need to be made as first surface mirrors or the glass substrate needs to be replaced by an LWIR transparent material like NaCl, ZnSe or Ge as 2nd surface mirrors depending on what wavelengths you need to reflect. So the silvering is not reflecting heat back to the contents, it’s attenuating radiative losses from the glass. It works better this way as the stainless vacuum bottle I used to have which does reflect radiatively was far poorer at keeping things hot since the conductive losses were a LOT higher.
I’d have to say the reason the heat transfer was slower in the all CO2 vs. all Air is that CO2’s heat properties are different. At 20°C, 1 atm normal Air has a density of 1.205 kg/m³ and CO2 is 1.842 kg/m³. Air’s Specific Heat (Cp) is 1.01 kJ/kg°K and CO2’s is .844 kJ/kg°K. So with equal volumes (and pressure) of CO2 and Air and equal input energy the air will get 28% hotter than the CO2. This according to Q = Cp * m * ΔT where Q is energy in kJ, Cp is specific heat, m is mass and ΔT is the temperature change.
Since moving energy in this system relies on conductive transfer we find that CO2’s Thermal Conductivity is 0.0146 W/(m°K) and Air’s is 0.024 W/(m°K). Since this system has a fixed surface area, again, the CO2 will conduct less heat according to Fourier’s Law since heat transferred is directly proportional to Thermal Conductivity here since everything else is equal (like a good experiment should be.) Data here was sourced from http://www.engineeringtoolbox.com/
The Climateers all say that CO2 “forces” other things in the climate system by “trapping” heat… but how can that be when it doesn’t get as hot with the same energy input as plain old air nor does it transfer what heat it has accumulated as well as plain old air. Is it creating energy all by itself? How can it force water vapor to be hotter when water vapor’s IR absorption swamps the tiny bit CO2 gets… and water vapor (being a vapor and not a gas) can actually radiate that energy at normal atmospheric pressures, CO2 doesn’t. It’s only radiative output is to space at 15µm (193°K) or the middle of Antarctica on June 21st.
60
NielsZoo,
Thank you so much for the comprehensive reply. Some years ago I had read that the reflective properties of mirrors weren’t able to contain radiative energy within an enclosed space very well, but that was only in passing and in a different context. I didn’t understand the reasoning back then and to tell you the truth it didn’t seem that important to me at the time. This explanation . . .
is more clearly stated than what I heard back then so again, thanks for that.
If I’m comprehending what you’ve said, radiative energy is transfered (fluxed?) to the glass where it is absorbed and then, the silvering on the glass slows (attenuates) radiative energy transfer outward and away from the glass into the vacuum. So, in effect, the energy isn’t radiated back into the liquid inside the inner bottle it’s only slowed down in it’s rate of outward flux.
I would just like to add that at 0.04% of normal atmosphere, one CO2 molecule would have to heat 2,500 other molecules composed of all the other gases out there. That’s one powerful little critter, yes? Atom Ant comes to mind!
Abe
10
If CO2 had all the magic properties they keep ascribing to it I don’t understand why we aren’t using it for all of our energy needs. If our contribution of 15ppmv can add that much energy to the atmosphere it must be powerful… considering that if we turned every single erg of energy we create and use in a year into heat at 100% efficiency, from nuclear to dung burning, and dumped it into the atmosphere all at once, we’d only raise the temperature about 0.07°C. (Based on IEA numbers for 2012) And one of the byproducts of some of that energy is a couple of orders of magnitude more powerful? We’re using the wrong stuff to power society.
30
Thanks Just-a-Guy,
The experiments did show that the rate of heat loss with CO2 was very slightly less than with common air, but the result was not (quite) significant. I think that the thermal conductivity of CO2 as measured by others is slightly less than Air. I think the value is given somewhere in the Engineering Toolbox.
You say that the silvering on the flask reduces the radiative heat loss to a very low level, hence we are measuring conductive loss mainly. I agree. I have subsequently removed the silvering from the flask and the rate of heat loss is then even greater.
I have not pursued that because Reed develops his argument further in Part 4, and it became unnecessary.
40
How did you remove the surface aluminum between the two glass surfaces? How much of that remained as increased conductivity about the fiber supports?
20
Surface Aluminium was removed by washing out the vacuum space with Green River Solution (Copper sulphate and Hydrochloric acid), then carefully drying out the chamber.
How much remained as increased conductivity of the fibre supports? No idea.
30
Thank you!
30
If there is a tiny effect but not significant then is it because:
a. experimental design
b. instrument sensitivity
c. insufficient replicates
d. wrong stats used
As above for no affect.
20
ps. Do a power analysis. Even presenting a bootstrap is better than saying, “the result was not (quite) significant”
20
You left out;
e. there is no effect.
20
no I didn’t.
10
Peter, it would seem that having removed the reflective coating on the inner flask that now you have a warm body surrounded by the CO2 but it is still very different to the earth-atmosphere-space model to which you are trying to draw parallels because the heat transfer from the outer flask to the room is still very conductive and convective. Now if you could surround your flask with a vacuum and that by a very cooled radiation sink it might give some better idea of the ‘greenhouse’ properties of the gases which you are investigating.
20
Joe,
I am told that an experiment similar to what you describe has been done. The person has not yet published his results. Therefore I cannot tell you what they were.
I had planned something similar but I was frustrated because I was unable to create a sufficient vacuum. The requirements are in fact exceedingly stringent. At 0.1 atmospheres the conductivity of gases is hardly affected. The vacuum has to be at least 2 orders of magnitude greater.
I was previously uncertain about the proposition that back radiation from a metal shell in a vacuum would slow the rate of cooling of a central sphere with a heat source. This is the steel greenhouse effect as described by Willis Eschenbach.
I am now persuaded that back radiation from a steel sphere can slow heat loss and hence the heated inner sphere will be hotter. By analogy the back radiation of greenhouse gases can slow heat loss from the Earth surface (slightly). But that is not their only property or effect.
As shown by the thermos experiment CO2 has a large conductive capacity, equal to (or almost equal to) Air.
When these other properties are accounted for it can be shown that under some circumstances a greenhouse gas can cause increased heat loss, not heat trapping.
10
Peter,
4 milli-torr even from a good roughing pump can make a lot of difference. 2 micron pressure with oil diffusion, takes weeks, or never, if one ant in your vacuum space.
The two T^4 terms in required parenthesis the S-B equation refer to opposing radiances (not flux) which limit radiative flux to only the difference in radiance, and only toward the lower radiance, as per Maxwell’s equations.
The Willis SGH proposes a perpetuum mobile of the second kind (spontaneity with no result).
The difference means not a wit, to flux through a vacuum, but makes all the difference in a dissipative media like this atmosphere. Again, This is the total demonstrated and perhaps intentional, incompetence of the entire climate science religion!
12
Actually Will,
The reason that I am now persuaded about the Steel Greenhouse is that the result of experiment showed Willis was correct. So I am told. As I said the experimenter has not yet published. I hope he does soon because we NEED TO KNOW!
If I had access to a good vacuum pump I would like to do some more work. At present I do not.
30
Peter,
The temperature of the constant power inner must increase as the shell interferes with flux to 1 kelvin because of mass rethermalization. Radiative flux cannot get through opaque until opaque is at radiative equilibrium (radiative flux goes through opaque) with no change in sensible heat or temperature of opaque. If Willis is correct, the same fixed power applied only to the shell must via internal flux also increase temperature of sphere (with lower surface area) until net power transfer is zero. The lower area sphere, never can get above the temperature of the shell, radiating all available power outward, never inward, as inward has no lower radiance, and no internal flux at all. Projective geometry can demonstrate no opposing flux ever!
We do not NEED TO KNOW! GOD KNOWS! perhaps! Earthlings only need to think, then decide which way to run with pitchforks and torches against the not GOD. Long ago I was gifted with a lab full of VECO pumps and bell jars. For tight work Vac-Ion pumps. I are retired, sometimes, I can still remember for a while.
22
Thanks Will,
You do not need physical proof of the Steel Greenhouse, because it is obvious to you.
I prefer experimental evidence,
However it does not matter. Please read Reed ‘s part 4 next week.
10
Peter,
Thank you, I will await part 4! Please suggest if I should be supportive, or combative?
I have done the physical proof of no opposing EM flux at any frequency!
12
Peter C,
And NielsZoo: (complete reply here)
(NielsZoo’s complete reply here)
I’ll just repeat again, this time with corroborating evidence, kudos for your compliance to the scientific method in all of it’s facets. I think the fact that you got a slightly different result in the rate of heat loss between ‘standard air’ and CO2 alone is because of the fact that the experiment was carried out well enough to detect this difference.
Abe
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Some commenters on Joanne’s blog believe the concept of “backradiation” leading to a “radiative greenhouse effect” is nonsense, and by discussing the physics (or lack thereof) of backradiation we skeptics are shooting ourselves in the foot. I disagree with that position. However, to get their way the forces aligned against us have certainly used questionable and offensive tactics. Thus, when addressing AGW from anything but a scientific perspective, I believe ridicule is a powerful tool and should be used. To that end, I submit the following.
If a conversation like the one below hasn’t taken place in a Climatology 101 classroom, it’s just a matter of time before it does:
Instructor: Carbon Dioxide or CO2, a heat-releasing gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of the CO2.
Student: Say what! Would you repeat that please?
Instructor: Carbon Dioxide or CO2, a heat-releasing gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of the CO2.
Student: Pardon me, but didn’t you mean to say “a heat-trapping gas” instead of “a heat-releasing gas?”
Instructor: No. I said it the way I wanted to. CO2 is a heat-releasing gas.
Student: But then what you said doesn’t make sense. It makes sense to say that trapping heat will cause an increase in temperature; but it makes no sense to say that releasing heat will cause an increase in temperature.
Instructor: You inferred from what I said that it is the heat-releasing nature of CO2 that causes the temperature increase. That inference is unwarranted. If I had said: “CO2, a colorless odorless gas, in the Earth’s atmosphere causes the surface temperature of the Earth to be higher than it would be in the absence of CO2,” you would not infer that it was the colorless/odorless nature of CO2 that caused the warming. You would simply have interpreted my statement to imply (a) CO2 is colorless and odorless, and (b) atmospheric CO2 causes an increase in Earth surface temperature.
Student: Now hold on. As an exercise in logic, you’re correct. But if your parenthetical clause “a heat-releasing gas” had been “a heat-trapping gas,” I’m not the only one who would infer that it is the heat-trapping nature of CO2 that causes Earth surface warming. In fact, I’d go so far as to say that way more than half the people would make that inference—especially lay people with little formal training in science or logic.
Instructor: I can’t help what people infer. If they read into my statement more than it actually contains, that’s their problem—not mine.
Student: Again, as a matter of logic, you’re correct. But haven’t I heard you demand mankind reduce its fossil fuel usage to pre-1980 levels because if we don’t the Earth is going reach a “tipping point” where unstoppable runaway global warming will commence to the detriment of all life on Earth?
Instructor: Yes, I have said that.
Student: And isn’t it true that to get the “world” to reduce its fossil fuel usage we must convince Joe-public that mankind’s use of fossil fuels at current or increased levels will be disastrous with the form of that disaster being runaway global warming?
Instructor: Unfortunately, yes. Without Joe public onboard we’ll never make headway in reducing fossil fuel usage. However, Joe public is incapable of understanding the science behind anthropogenic global warming (AGW); and since AGW is a fact, any method used to get Joe-public onboard is justified.
Student: In the case of mankind’s response to AGW, are you saying the ends justify the means?
Instructor: Yes.
Student: But if CO2 is a heat-releasing gas, it seems to me atmospheric CO2 should lower, not raise, the Earth’s surface temperature. I mean after all, if I said XYZ was a heat-trapping gas, then the two properties XYZ should have are: (a) XYZ is a gas, and (b) XYZ traps heat in the sense that if XYZ gas surrounds a hot object it should take the hot object a longer time to cool than in the absence of the surrounding XYZ gas. If calling XYZ a heat-trapping gas doesn’t assign those two properties to XYZ, then calling XYZ a heat-trapping gas is meaningless at best and deceptive at worst; and using the phrase “a heat-trapping gas” must be designed to mislead rather than educate. And by-the-way, CO2 is not always a gas. Isn’t dry ice CO2 in solid form?
Instructor: Yes, dry ice is CO2 in solid form. Calling CO2 “a heat-trapping gas” or a “heat-releasing gas” doesn’t assign the property of being a “gas” to CO2. Rather it defines the state of the CO2—in particular that CO2 is in gaseous form. So when I say “CO2 is a heat-releasing gas,” the only property I’m assigning to CO2 is the property of “heat-releasing,” and I’m caveating that the property only applies to CO2 in gaseous form.
Student: Back the truck up. If CO2 is a heat-releasing gas, then shouldn’t the presence of CO2 gas surrounding a hot object cause the hot object to cool faster than in the absence of CO2 gas?
Instructor: Yes.
Student: But you said CO2 was a heat-releasing gas and the presence of CO2 in the Earth’s atmosphere increases the temperature of the Earth’s surface. Now I’m really confused.
Instructor: Good. I’m using this conversation to assess its effect on the average Joe. A confused person is more likely than a knowledgeable person or even a thinking person to (a) believe there’s a crisis and (b) accept any solution proposed by those in authority.
Student: Let’s put that aside for a moment. Why do you say CO2 is a heat-releasing gas?
Instructor: Well, experiments run with thermos bottles show that CO2 injected into the thermos bottle’s vacuum region causes hot liquid in the thermos bottle to cool faster than it would without the CO2. As such, it can be shown that, everything else being equal, situations exist where CO2 acts to “release heat”.
Student: Okay. You’ve convinced me that there are situations where CO2 gas acts to increase the rate of heat release, and thus for some situations can properly be called a heat-releasing gas. Why, then, should I believe gaseous CO2 in the Earth’s atmosphere will increase the Earth’s surface temperature?
Instructor: Models tell us so.
Student: Models? What models?
Instructor: Models! They’re everywhere. You can’t get on the internet without tripping over a press release touting the latest man-made global warming doomsday model. Every university feasting at the public trough and every environmentalist group worth its salt has a global warming model. It’s your standard government-subsidized industry: produce results and only those results that support the government’s objectives. All the models predict—er, scratch that—all the models include among possible future scenarios a forthcoming manmade global warming crisis.
Student: Are those the models that (a) when plotted on a common graph look like a plate full of spaghetti, (b) predict a hot spot in the troposphere that no one can find via temperature measurements, and (c) didn’t predict the 17+ years of no temperature increase in the presence of ever increasing atmospheric CO2 levels?
Instructor: Yes.
Student: Okay. Let’s see if I have this right. First, fairly simple experiments that almost anyone can perform show that under some conditions CO2 gas acts to increase the rate of heat release and therefore to cause objects to cool faster than they would in the absence of the CO2 gas. Second, the models that foretell a coming CO2-driven manmade global warming crisis (a) predict wildly different future temperature scenarios, (b) predict a tropospheric hot spot that no one can find except via esoteric logic involving potentially increased wind speeds, and (c) didn’t predict the current 17+ year hiatus in global warming even in the presence of increasing atmospheric CO2 levels. Have I summarized the situation correctly?
Instructor: Yes.
Student: Given our conversation, why then should I believe what you say about reducing fossil fuel usage and climb on the AGW bandwagon?
Instructor: You want to pass the class, don’t you?
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I really do not understand what you may by trying to say or demonstrate! Are you trying to ridicule AGW, how teachers brainwash students, that CO2 can absorb some properties of EMR at some frequencies under certain circumstances, or that opposing radiative flux can exist at any given frequency. One has nothing to do with the others. It seems to me that what you term ridicule is but encouragement to continue brainwashing innocent children, as it works so very well!
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Reed/Peter C 1:45pm: “..in the case of a vacuum thermos bottle versus a CO2 thermos bottle, CO2 gas acts to release heat, not to trap heat….by increasing the rate of conduction”
This agrees with my (et. al) test results.
Unfortunately though – the poorly named “greenhouse” effect won’t be found in replicating Fourier conduction and Newton Law of Cooling tests since energy is not conducted from the atmosphere to deep space at about 3K like a thermos. There exists only radiative transfer from earth and atmosphere to deep space. Your “atmosphere” optical depth in the flask is only several mm thick whereas the earth’s atmosphere is several 10s of kilometers thick. The effect you seek isn’t measurable in your setup – the thermos CO2 atmosphere still has a nearly 0.0 optically thick atmosphere “greenhouse” as does the vacuum.
The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface. Step 2 of the scientific method is found in optical depth analysis coupled with 1st law energy balance & not Fourier conduction and Newton’s cooling law (the one as annunciated by Newton directly).
You have proven thermos manufacturers don’t fill with CO2 due to its resultant increased conduction cooling faster reducing the effectiveness of the replaced vacuum insulation effect cooling relatively slower.
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“The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface.”
You claim a greenhouse effect. Then try to say that a poorly done radiometric measurement of the upper atmosphere coupled with a partially measured surface radiance at 8-13 microns compared to some averaged thermometric surface measurements is supposed to have any meaning like a greenhouse effect. Why such utter nonsense?
“Step 2 of the scientific method is found in optical depth analysis coupled with 1st law energy balance & not Fourier conduction and Newton’s cooling law (the one as annunciated by Newton directly).”
You try to do your fake Step 2 of the scientific method and tell us all of the nonsense of your analysis! You will not even be able to define any atmospheric optical depth analysis, over the range of wavelengths and pressures involved, as you cannot determine the exitance solid angle of gas cross sectional area, at any altitude! This is the total demonstrated incompetence of the entire climate science religion!
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Will 1:46pm: “..a partially measured surface radiance at 8-13 microns..”
I make no claim just measurement. I wrote thermometers (some mercury type) which measure near surface temperature so they integrate over all wavelengths naturally. Your second paragraph isn’t even parsable.
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What words do you complain of?
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Will 3:47pm: “you cannot determine the exitance solid angle of gas cross sectional area”
Doesn’t parse. And atm. tau is well defined despite your claim.
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(“you cannot determine the exitance solid angle of gas cross sectional area”)
“Doesn’t parse.”
Cannot you do geometry? A flat surface of specific area is limited to a solid angle of PI steradians because of cos(theta) an atmospheric cross sectional area radiates exitance to a maximum of 2 PI steradians.
“And atm. tau is well defined despite your claim.”
Yes indeed Miskolczi did a fine job of apparent radiance in a direction normal to some surface. He completely forgot about the solid angle.
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Will 6:28am: Consider a cavity, spherical if you wish. The radiation inside is isotropic at temperature equilibrium. Imagine a flat plane surface within. Regardless of how your flat plane surface is oriented the same amount of radiant energy crosses that unit area in unit time. Planck considered the radiant energy (photons) propagating in a hemisphere of directions either above or below the surface. Thus, with additional considerations, atm. tau is well defined.
No idea who Miskolczi is or what he forgot.
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LtCusper, 12:24 pm, 23 March 2015
The “effect I seek” is that CO2 gas does not always “trap heat,” where the process of “trapping heat” implies a temperature increase to an object in the region where the heat is “trapped.” The physical reasons CO2 gas doesn’t “trap heat” are of interest and may very well be related to the optical depth of the CO2 gas. Therefore the effect of atmospheric CO2 gas on Earth surface temperature may very well be to increase the Earth surface temperature. But if so, the claim by itself that since CO2 gas is a “heat-trapping” gas, it must increase the Earth surface temperature doesn’t cut it with me. Additional arguments are necessary.
Look at it this way. Someone comes up to me and says:
(1) “The Earth’s surface temperature in the presence of greenhouse gases will be warmer than the Earth’s surface temperature in the absence of those gases.”
(2) I ask: “Why do you say that?”
(3) They answer: “Because greenhouse gases absorb some of the IR radiating from the Earth’s surface. Thus, greenhouse gases trap heat emitted from the Earth’s surface; and trapped heat increases the temperature of objects within the ‘trap’.”
(4) I ask: “Is CO2 a greenhouse gas?”
(5) They answer: “Yes.”
(6) I ask: “Since CO2 is a greenhouse gas and greenhouse gases trap heat, then it logically follows that CO2 gas traps heat. Is that correct?”
(7) What else can the answer but: “Yes, CO2 gas traps heat.”
(8) I then ask: “From what you said, it logically follows that everything else being equal the temperature of any object emitting heat via IR radiation and surrounded by CO2 gas must increase relative to its temperature without the greenhouse gas. Is that correct?
(9) Again, I don’t see how they can answer anything but “Yes.”
(10) I then ask: “If I build a thermos bottle whose inner chamber is transparent to radiation below the visible band and I fill the vacuum region of a vacuum thermos bottle with CO2 gas, does CO2 gas surround the contents of the thermos bottle?”
(11) They answer: “Not entirely. The thermos bottle chamber is kept separate from the thermos bottle by rigid material, so that for the most part, yes, the contents are surrounded by CO2 gas, but not entirely.”
(12) I then ask: “What if the rigid material keeping the chamber separated from the wall of the thermos bottle is (a) transparent to radiation below the visible band, and (b) bent such that the straight line path from any point within the thermos bottle chamber to any point outside the thermos bottle’s external wall must pass through CO2 gas. Is such a construction possible?”
(13) They answer: “I’m not sure, but for the sake of argument I’ll stipulate that such ‘rigid spacers’ can be constructed and if constructed, then yes CO2 gas surrounds the contents of the thermos bottle. What’s your point?”
(14) I answer: “My points are: (a) CO2 gas surrounds the contents of the thermos bottle. (b) The frequencies of radiation emanating from material at temperatures near 300K and stored in the inner chamber are for the most part below the visible band. (c) According to your claims [3 and 5], CO2 gas will ‘trap’ some of the heat emanating from the thermos bottle contents, and the trapping of that heat will result in an increase in the temperature of the thermos bottle contents. Thus, considering only those two claims and stipulating that with the exception of what resides in the vacuum region of a thermos bottle everything else is equal, the contents of a CO2 thermos bottle should be at a higher temperature than the contents of a vacuum thermos bottle. That conclusion contradicts experiment. How can that be?”
(15) At this point arguments such as optical depth, backradiation, convection, evaporation, surfaces painted with highly reflective materials, etc. may arise and justify the claim that atmospheric CO2 gas will increase the Earth’s surface temperature. But before I’m convinced that atmospheric CO2 must increase the Earth surface temperature, I want to hear those arguments. The “heat-trapping” nature of CO2 by itself doesn’t convince me!
As to the issue of optical depth. That may explain why CO2 gas in a thermos bottle has a different effect on the temperature of the thermos bottle’s contents than atmospheric CO2 gas has on the Earth’s surface temperature. But then, you are providing an additional argument over and above the simple “heat-trapping” gas argument.
Next, I’ll accept the fact that the IR “optical depths” of a thermos bottle and the atmosphere are many orders of magnitude different. But isn’t the optical depth in part a function of the density of the material through which the radiation is propagating? For a thermos bottle it may take enormous pressure in the volume occupied by the CO2 gas, but with sufficient pressure (assuming the pressure doesn’t liquefy or solidify the CO2 at room temperatures) the optical depth of the CO2 in a thermos bottle can be made arbitrarily large. It’s very likely that the optical depth of a real thermos bottle will never approach the optical depth of the atmosphere; but by making the distance between the inner chamber’s outer wall and the thermos bottles outer wall large and by increasing the pressure of the CO2, it seems to me we can construct a thermos bottle whose “optical depth” is approximately the same as the atmosphere’s “optical depth.” Neither I nor Peter has run such an experiment, but if I had to bet, I’d bet that if such a thermos bottle were constructed, the equivalent vacuum thermos bottle would outperform the CO2 thermos bottle.
Finally, your statement: “The way to measure earth’s approx. 33K “greenhouse” effect is not by too thin thermos atm. but to put up satellites with radiometers above most of the earth’s atmosphere measuring global about 255K annually and thermometers measuring about global 288K near the surface” confuses me. Who said anything about measuring the 33K “greenhouse” effect by considering a “too thin thermos atm”? In a previous post, I argued that the 33K number comes from an illogical methodology for computing the Earth’s surface temperature in the absence of greenhouse gases.
I understand and at this time won’t disagree with the thermometer measurement of 288K. However, I want to discuss radiometer measurements of the Earth’s surface temperature. Radiometers measure the intensity of electromagnetic radiation. Thus, a radiometer above the Earth/Earth-atmosphere pointed towards the Earth measures the intensity of the electromagnetic radiation propagating in a direction away from the Earth. Some of this radiation is emitted from the Earth/Earth-atmosphere system and some of this radiation is reflected solar radiation. If a radiometer immediately above the Earth’s atmosphere and pointed towards the Earth doesn’t “see” reflected solar radiation, then where does reflected solar radiation go? If the Earth is in energy-rate-equilibrium, the total (emission and reflection) rate energy leaves the Earth/Earth-atmosphere system must equal the rate solar radiation is incident on the Earth/Earth atmosphere system. The latter rate is approximately 1,366 Watts per square meter. For a spherical object (absorbs as a disk, radiates as a sphere) whose surface emissivity is (a) not a function of frequency, (b) uniform over the surface, but (c) may have a value other than unity, the temperature associated with an incident power density of 1,366 Watts per square meter is approximately 278.6K, not 255K. This translates to a temperature difference of approximately 10K, not 33K.
If from the “total outgoing” radiation you subtract the portion from solar reflection, then yes, the temperature of a spherical object will be approximately 255K. But clouds reflect much if not most of the solar energy. If you remove all atmospheric greenhouse gases, you remove all water vapor, and if you remove all water vapor you remove all clouds. If you remove clouds, you remove the primary means of solar reflection. The Earth would then approximate a sphere with a uniform emissivity, and its surface temperature would be nearer 278K than 255K.
Bottom line, when computing (not measuring, but computing) the Earth’s surface temperature in the absence of greenhouse gases, it’s illogical to perform that computation using a value that depends on the presence of greenhouse gases. I don’t accept the statement that radiometers outside the Earth’s atmosphere and pointed towards the Earth measure radiation only emitted from the Earth’s surface. How can they not measure the sum of Earth-emitted radiation and reflected solar radiation? It’s only after you subtract out the reflected solar radiation that the intensity corresponds to a blackbody object at 255K. But without water vapor there would be no clouds and without clouds there would be no (or very little) reflected solar radiation, and without reflected solar radiation, there’s nothing to subtract from the radiometer measurements.
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Reed 3:51pm: The “effect I seek” is that CO2 gas does not always “trap heat,” where the process of “trapping heat” implies a temperature increase to an object in the region where the heat is “trapped.”
The poorly named “green house” optical depth effect you seek really is operating in the few mm of the CO2 filled flask but is way overpowered by conduction so as to be not measurable.
“I argued that the 33K number comes from an illogical methodology..”
Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. 1st law and optical depth analysis though can be used to logically compute 33K in step 2 from measured input data, this work out is found in many modern text books & specialist papers. Look it up. The emissivity of any object can be routinely measured – this means it is routine to separate the amount of reflected radiation (for example earth albedo) and the amount of emitted radiation as they are LW and SW respectively. Reflectivity + absorptivity + transmissivity = 1.0.
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Peter C 5:09pm: How much? Dunno offhand.
My going in approach would be find the precision calibration limits of today’s instruments and use grey IR differential optical depth analysis across a layer of atm. thickness to compute the optical depth (commonly tau) under hydrostatic condition that just becomes reasonably measurable above 0.0K by them at say 95% or 99% confidence interval given the partial pressures, mixing ratios and each of N constituents source of opacity (mass extinction coefficient) for earth (or any) atmosphere. Get a reasonable number somewhere between a few mm (0.0K) and 10’s of kilometers (33K for earth). A literature search may turn up this is already reasonably known.
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You insist that the artificial 33 C difference in temperature has some possible meaning.
The 255 Kelvin is a computation of a black-body, never this Earth with its atmosphere There is.. Surface temperature 225-306 Kelvin as many measurements, all individual, all unique. There is no possible way that your 288 Kelvin has any meaning whatsoever! There is no possible (mass extinction coefficient) for this dynamic Earth atmosphere. Your whole approach is quiet Juvenile. Go somewhere and buy a clue!
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Will 1:36am: The 255K is measured by precision calibrated instruments. So is the 288K. Both are meaningful in science traceable to reasonable CI – but do remember we are just practicing science, we haven’t perfected it yet. The patent office is still open.
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LtCusper March 24, 2015 at 2:02 am · Reply
” Will 1:36am: The 255K is measured by precision calibrated instruments. So is the 288K. Both are meaningful in science traceable to reasonable CI – but do remember we are just practicing science, we haven’t perfected it yet. The patent office is still open.”
The 215 Kelvin estimated at tropopause is reliable, never accurate, 255 Kelvin somewhere about 5 km is but your fantasy! Normal (vertical) surface radiometric, without clouds, in 8-13 microns goes between 245 Kelvin and 315 Kelvin. Why do you think your average of 288 Kelvin has any meaning for anything?
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Will 4:38am: ..255 Kelvin somewhere about 5 km is but your fantasy!”
No fantasy, I looked up many precision calibrated sounding rocket data reports – their onboard thermometers measure the 255K around 5km depending on latitude.
“Why do you think your average of 288 Kelvin has any meaning for anything?”
288K is the value measured with precise enough calibrated instruments that are reasonably maintained.
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Again Why do you think your average of 288 Kelvin has any meaning for anything?”
“288K is the value measured with precise enough calibrated instruments that are reasonably maintained.”
And again why do you think some average temperature of various mass with various specific heat, have any meaning whatsoever?
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Will 5:56am: “..why do you think some average temperature..”
I didn’t. Please define what you mean by average. Do you mean arithmetic average (or arithmetic mean) or median (50% above & 50% below), most probable value (or mode), weighted average, root mean square (or nth root), so forth, the number of possible averages is boundless Will.
Or maybe you meant like page one of Maxwell’s Theory of Heat where he discussed hot and cold temperatures have an intermediate temperature like my hot and cold faucet running lukewarm out of the tap. What?
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LtCusper, 5:05 am, 24 March 2015.
You wrote: “No fantasy, I looked up many precision calibrated sounding rocket data reports – their onboard thermometers measure the 255K around 5km depending on latitude.”
5km isn’t the surface of the earth or anywhere near it. The 33K claim has nothing to do with the temperature of the Earth’s atmosphere at an altitude of 5k. The claim is that in the absence of greenhouse gases the Earth’s surface temperature will be 33K cooler than the measured surface temperature of 288K. Since (a) temperature decreases with altitude by approximately 6.4C per kilometer and (b) the measured temperature at the surface of the Earth is approximately 288K, it’s not surprising that at an altitude of 5km the temperature is approximately 288 – 5×6.4 = 256K. But this has nothing to do with the 33K claim. Comparing the measured Earth surface temperature against the measured atmospheric temperature at an altitude of 5km, getting a temperature difference of 33K, and claiming that the 33K temperature difference applies to the Earth surface temperature difference with and without atmospheric greenhouse gases is just plain silly, or deliberately deceptive.
Later in a comment time stamped 12:46 am, 24 March, 2015, you wrote: “Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. 1st law and optical depth analysis though can be used to logically compute 33K in step 2 from measured input data, this work out is found in many modern text books & specialist papers. Look it up.”
The last time you requested I look something up (the meaning of Newton’s law of cooling), I did and you chided me for not going to Newton’s original wording. Suppose we avoid a repeat of my finding the wrong reference and you give me the reference where the 33K number is derived from 1st law and optical depth analysis.
Next, if you have a highly thermally conducting sphere of radius r=6,371,000 meters (the approximate mean radius of the Earth) and surface temperature “Tes” with a surface emissivity that is independent of frequency and is everywhere epsilon (a real number greater than 0 and less than or equal to 1) and you place that sphere a distance, 150,000,000,000 meters from the sun (the approximate distance from the sun to the Earth), then to a good approximation
(a) for a solar radius of 695,000,000 meters and a solar surface temperature of 5,778K, the solar power density at the sphere is approximately 1,359 watts per square meter.
(b) the rate solar energy is incident on the sphere is pi x r^2 x 1,359 = 1.733 x 10^17 watts–that is, the sphere absorbs energy as if its shape were a flat circular disk oriented perpendicular to the solar radiation
(c) the rate the sphere absorbs solar energy is “epsilon x 1.733 x 10^17 watts”
(d) the rate the sphere radiates energy is “epsilon x 4 x pi x sigma x r^2 x (Tes)^4”, where sigma is the Stefan-Boltzmann constant–i.e., the sphere radiates energy from its entire surface
If the rate the sphere absorbs solar energy is equal to the rate the sphere radiates energy, then the temperature of the sphere is (approximately)
Tes = 278.2K
“Tes” corresponds to the energy-rate-equilibrium temperature of an Earth-like object (devoid of an atmosphere) at an Earth-like distance from the sun. Provided the sphere’s surface emissivity equals the sphere’s surface absorptivity (which is true for black and gray bodies), the temperature of 278.2K is independent of the emissivity/absorptivity. Assuming negligible energy is absorbed/radiated by non-greenhouse gases in the Earth’s atmosphere, the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere.
Next, you wrote: “The emissivity of any object can be routinely measured – this means it is routine to separate the amount of reflected radiation (for example earth albedo) and the amount of emitted radiation as they are LW and SW respectively.” It is NOT routine (or rather if it is routine, then the routine is misleading) to use the albedo of an object in “State 1” to compute the rate the object absorbs energy, and then use that absorption rate in conjunction with the albedo of the object in “State 2” to compute the “State 2” temperature of the object. “State 1” is the Earth with a greenhouse gas atmosphere. The Earth’s State 1 albedo is approximately 0.3. “State 2” is the Earth devoid of all atmospheric greenhouse gases. The Earth’s State 2 albedo is approximately 0. If you believe it’s okay to use the Earth’s albedo (0.3) in the presence of atmospheric greenhouse gases to compute the rate the Earth absorbs solar energy and to use that absorption rate in conjunction with the Earth’s albedo (0) in the absence of atmospheric greenhouse gases to compute the Earth’s surface temperature in the absence of a greenhouse gas atmosphere, then we will likely never reach agreement on this topic. Furthermore, if the AGW community defends such a procedure, then you can think whatever you want; I think the AGW community is telling the public the emperor is wearing a beautiful suit.
Finally, in your 5:34 am, 24 March 2015 comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.”
Say What? Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df from a planar differential unit area, dA, at temperature T, into a differential solid angle domega defined with respect to a coordinate system at dA. There are no terms in Planck’s blackbody radiation law for the presence/absence of any other surfaces. In fact, it is a geometrical impossibility for a planar differential surface area to radiate into a differential angle that contains any portion of the radiating differential area. Provided the radius of curvature of an object is finite and non-zero, ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.
For the interior surface of a closed object (like the interior surface of a spherical shell), you can use Planck’s law to compute the power radiated from any finite size area of the interior surface to any other finite size area of the interior surface. It turns out that if no objects reside in the interior surface of a closed object, the total rate energy is radiated from the entire interior surface is equal to the total rate the interior surface absorbs radiated energy. As such, the net rate of energy exchange of the interior surface is zero. However, if an object (Object A) exists within the interior region of a closed surface, you can use Planck’s law to determine how much energy is radiated from the interior wall of the enclosed surface to Object A and how much of the energy is radiated from the interior wall to other portions of the interior wall.
For the rate energy leaves an object, the Stefan-Boltzmann Law, which contains a term for the total surface area of an object, doesn’t apply if portions of the object’s surface radiate to other portions of the object’s surface. So what you warn against applies to the Stefan-Boltzmann law of object radiative cooling, but not to Planck’s blackbody radiation law.
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Reed 10:35am: “5km isn’t the surface of the earth or anywhere near it…”
I was responding to Will. Agree with your 1st paragraph.
“Suppose we avoid a repeat of my finding the wrong reference..”
Just make sure any ref. you use traces accurately to the original author, or chiding is fair game. You will have to do your own work here, otherwise if I suggest a modern text author, you won’t have anything invested.
“Tes = 278.2K”
Yes, for a no-atmosphere sphere situated as indicated with emissivity of 1.0, rounded up say from natural 0.95 and albedo of 0.0.
“..the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere.”
Not according to the 1st law. When you add the current atmosphere pressure with almost 0 optical thickness i.e. very optically thin atm. and make an earth out of the sphere, albedo goes to 0.3 as solar reflected is measured by radiometers (surface liquid/ice water & dirt & gas now reflect solar) and the Tes goes to 255K by 1st law.
If you want to argue that albedo, know that the albedo for NH and SH is measured the same despite much difference in land & water surface area. If you want to argue, pick any number for albedo 0 to 0.3 with optically thin atmosphere, then your Tes will be between 255K and 278.2K depending on your choice, one is as good as any other, just specify your choice of albedo.
“Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df”
Where did you get that from? No. Chide you again for not going to the ultimate source. Planck’s 1912 paper on-line gives the ideal specific intensity from an object at each frequency and temperature. For a range of frequencies, a curve can be constructed at each temperature. This was from his testing and many others illuminating an object with cavity BB radiation. His paper limits the object to positive radii and negligible diffraction, look up the paper. Invest in it.
“..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.”
Not according to Planck’s own writing in the 1912 paper introduction. If you want I can explain how to prove that to yourself by simple experiment with IR camera.
Your 2nd to last paragraph is agreed generally, as the interior body is illuminated with perfect BB radiation. This illumination won’t exist anywhere outside the cavity.
Your last paragraph is wrong, the limitation applies to Planck specific intensity law, which you can easily find reading the 1912 paper general introduction & eqn. 274. The same limitation then applies to S-B.
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LtCusper, 2:02 am, 24 March 2015; and 12:46 am, 24 March 2015
In your 2:02 am comment, you wrote: “The 255K is measured by precision calibrated instruments. So is the 288K.” In your 12:48 am comment you wrote: “Reed, the 33K is measured from precision calibrated instruments & not computed using illogical methodology – as I wrote. ” Unless you’ve either (a) devised a way to remove all greenhouse gases from the Earth’s atmosphere, or (b) been able to construct a replica of the Earth whose atmosphere is devoid of greenhouse gases, it’s impossible to make measurements of the Earth surface temperature in the absence of greenhouse gases. All measurements will be made for the atmospheric conditions that exist at the time of the measurements, which means all measurements are for an atmosphere that contains greenhouse gases. Thus, the 255K measured temperature, the 288K measured temperature, and the temperature obtained by differencing these measurements are for an Earth with a greenhouse gas atmosphere. To represent any of these temperatures as being a direct measurement of the Earth surface temperature in the absence of greenhouse gases is simply wrong.
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Reed 3:10pm: “it’s impossible to make measurements of the Earth surface temperature in the absence of greenhouse gases….To represent any of these temperatures as being a direct measurement of the Earth surface temperature in the absence of greenhouse gases is simply wrong.”
I agree. Wouldn’t want that to happen. Still, what I wrote is true. Both 255K and 288K are measured with precision calibrated instruments.
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LtCusper, 2:50 pm, 24 March 2015
This is the response I said I would post. Before getting into the body of the response, a few introductory remarks are in order. First, when I quote Max Planck’s 1912 writing, I’m not quoting the German original, I’m quoting Morton Masius’s 1914 translation of Planck’s writing—see
https://archive.org/stream/theoryofheatradi00planrich#page/n5/mode/2up
Second, in the Preface to the Second Edition (pages viii and ix) of Masius’s 1914 translation, Planck wrote: “In contrast thereto I have now attempted to treat the subject from the very outset in such a way that none of the laws stated need, later on, be restricted or modified. This presents the advantage that the theory, so far as it is treated here, shows no contradiction in itself, though certainly I do not mean that it does not seem to call for improvements in many respects, as regards both its internal structure and its external form.”
And in the last paragraph of that Preface (page ix) Planck wrote: “Hence it follows from the nature of the case that it will require painstaking experimental and theoretical work for many years to come to make gradual advances in the new field.”
I interpret these words to mean that Planck expected work, both experimental and theoretical, in the field of Heat Radiation to continue into the future. Let’s call the current state of that work “Our Modern Understanding of the Theory of Heat Radiation.” Our Modern Understanding of the Theory of Heat Radiation may in no way conflict with Planck’s 1912 paper. If it does conflict, adjudicating which one, if either, is correct is beyond my paygrade. However, if a conflict arises and I was forced to choose between the two, I’d choose the modern version for no other reason than authors following Planck had access to his writing and thus could analyze an apparent discrepancy from two perspectives, theirs and Planck’s; whereas Planck didn’t have access to all modern texts and thus couldn’t analyze any apparent discrepancy. In any event, since Planck published his 1912 paper, many authors have written texts on the subject. Include those texts in Our Modern Understanding of the Theory of Heat Radiation. You ask, what is the point of this remark? The point is that if you decide to answer some of the questions I pose in this response, please do so from two perspectives—the perspective of the knowledge of physics as put down in Planck’s 1912 paper (chiding allowed and even encouraged:-), and the perspective of Our Modern Understanding of the Theory of Heat Radiation.
Third, in the lifetime of our interaction on this thread, I have no hope of reading and understanding all of Planck’s 1912 paper. Thus, relative to this response to your comment, my understanding/interpretation of Planck’s 1912 paper must be hit and miss. Be patient with me.
Fourth, in this response (and if I can remember, in all future responses on Joanne Nova threads), unless explicitly stated otherwise, I assume (a) at all locations on the Earth’s surface and for all frequencies, the transmissivity is zero, (b) at all locations on the Earth’s surface, emissivity is (i) independent of frequency, (ii) independent of temperature, (iii) greater than 0 and less than or equal to unity, and (iv) equal to the absorptivity, and (c) the Earth’s emissivity/absorptivity is uniform over the surface of the Earth.
Fifth, I hope to respond to all your 2:50 pm, 24 March 2015 points; but the order of my responses will not match the order of your points.
Now to my responses.
(1) I argued that the temperature of an Earth-like sphere positioned at an Earth-like distance from the sun and in energy-rate-equilibrium (ERE) with solar energy would be approximately Tes = 278.2K. To that argument you responded: “Yes, for a no-atmosphere sphere situated as indicated with emissivity of 1.0, rounded up say from natural 0.95 and albedo of 0.0.” Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes. I say this because the rate energy is absorbed by a surface is directly proportional to the absorptivity of the surface; and the rate energy is radiated from a surface is directly proportional to the emissivity of the surface. Thus, when writing an ERE equation for a surface, absorptivity appears as a multiplying factor on one side of the equation and emissivity appears as a multiplying factor on the other side. If absorptivity equals emissivity, then provided these terms are not zero, the resulting ERE temperature of the object will be independent of the value of emissivity/absorptivity. So for a no-atmosphere sphere and equal non-zero absorptivity/emissivity, the ERE surface temperature of an Earth-like sphere is approximately 278.2K independent of the absorptivity/emissivity of that Earth-like sphere. Do you agree or disagree?
(2) To compute the ERE temperature, I use the relationship that the rate energy is radiated from a surface is directly proportional to the fourth power of the surface temperature, T, in Kelvin. The fourth power relationship can be derived by integrating the “spectral radiance, Bf(f,T), function–see
http://en.wikipedia.org/wiki/Planck's_law
over all frequencies, f, from zero to infinity. The T^4 relationship is not in general valid for an emissivity that changes with frequency. By using the T^4 relationship to generate ERE temperature estimates, I am implicitly assuming the emissivity is a non-zero constant at all finite frequencies. Now it may be that at sufficiently low temperatures, the contribution to the integrated spectral radiance from frequencies above a cutoff frequency is negligible (e.g., the emissivity may be a step function: a non-zero constant below a cutoff frequency and zero above the cutoff frequency) so that the T^4 relationship is a good approximation. Bottom line, if you employ the T^4 relationship to compute ERE temperatures for situations where the emissivity is frequency dependent, please so note. For your comments to date, I assume you arrive at ERE temperatures using the T^4 relationship. Is this assumption correct?
(3) I wrote: “..the temperature of 278.2K, not 255K, represents the Earth surface temperature in the presence of a non-greenhouse gas atmosphere” to which you responded:
“Not according to the 1st law. When you add the current atmosphere pressure with almost 0 optical thickness i.e. very optically thin atm. and make an earth out of the sphere, albedo goes to 0.3 as solar reflected is measured by radiometers (surface liquid/ice water & dirt & gas now reflect solar) and the Tes goes to 255K by 1st law.
“If you want to argue that albedo, know that the albedo for NH and SH is measured the same despite much difference in land & water surface area. If you want to argue, pick any number for albedo 0 to 0.3 with optically thin atmosphere, then your Tes will be between 255K and 278.2K depending on your choice, one is as good as any other, just specify your choice of albedo.”
I’ll take your word for it that radiometers have to a high degree of precision measured the rate of Earth/Earth-Atmosphere reflected solar energy, and have determined the reflected rate to be 30% of the rate solar energy is incident on the Earth/Earth-Atmosphere. Since the “average Earth albedo” is the ratio of “the rate of reflected solar energy” …to… “the rate of incident solar energy,” I agree that the average Earth albedo is 0.3. An average Earth albedo of 0.3 means the Earth absorbs solar energy at a rate of approximately 1.21 x 10^17 watts. Assuming energy-rate-equilibrium and applying the “1st law,” the Earth/Earth-Atmosphere must then radiate energy to space at rate of 1.21 x 10^17 watts. The rate energy is radiated from the Earth/Earth-Atmosphere to space is NOT a direct measure of the Earth’s surface temperature for either a greenhouse gasless Earth or a greenhouse gas Earth. For both kinds of atmospheres, the radiated energy rate must be converted to a temperature. If (a) the Earth’s surface is treated as a graybody (emissivity, e, independent of frequency and location) and (b) atmospheric gases neither absorb nor radiate energy, then via the equation
Radiated Power = e*sigma*A*(Tes)^4, where sigma is the Stefan-Boltzmann constant and “A” is the surface area of a smooth Earth,
the temperature Tes (assumed uniform) of a blackbody Earth surface (e=1) that produces an output energy rate of 1.21 x 10^17 watts is approximately 254.3K. I believe this is the methodology used to arrive at the 255K Earth surface temperature in the absence of atmospheric greenhouse gases.
[Note: As mentioned in (2) above, for an Earth with unity emissivity at “low” frequencies and less than unity emissivity at “high frequencies,” the factor of (Tes)^4 in the above equation is not theoretically valid. This factor is theoretically valid only for an emissivity that is independent of frequency. If the emissivity changes with frequency, the above equation may be a good approximation, but it is not exact.]
If this is the methodology used to arrive at the 255K temperature, then there are two problems: a minor problem and a major problem. The minor problem is the computation of Tes using a blackbody model. From the above equation, the temperature Tes is inversely proportional to the fourth root of e. If an emissivity (e=0.7) consistent with the albedo of 0.3 is used, Tes would be approximately 278K. To me, the use of an emissivity of unity with an albedo of 0.3 is at most a minor problem because energy is radiated from the Earth’s surface predominately over a frequency range (the emission power frequency, EPR, range) where the emissivity is close to unity. If the emissivity drops outside the EPR range, then the (Tes)^4 term in the above equation is not rigorously valid, but it is a good approximation. And because a significant portion of solar radiation exists at frequencies outside the EPR range, an emissivity near unity over the EPR range but less than unity outside the EPR range can easily produce an average albedo of 0.3. If that is the situation for the real Earth, then this “minor problem” is not an important issue with me.
The major problem, which is an issue with me, is that of using a value of 1.21 x 10^17 watts for the rate energy is radiated away from a greenhouse gasless Earth. The measured average Earth albedo of 0.3 can be segmented into two additive parts: (a) one part associated with solar reflection from atmospheric greenhouse gases and their artifacts (greenhouse gas artifacts are objects that are present in a greenhouse gas atmosphere but absent in a non-greenhouse gas atmosphere), and (b) one part associated with solar reflection from everything else.
[Note: The division of the measured average Earth albedo into the two additive parts described immediately above is not rigorously valid. Such a division assumes solar radiation reflected by one “part” will, in the absence of that “part,” not be reflected by the other “part.” This is obviously not rigorously true for the Earth. Some of the solar radiation reflected by clouds (a greenhouse gas artifact) will, in the absence of clouds, be reflected by something in the “everything else” category. Call solar energy that (a) is reflected to space by atmospheric greenhouse gases and/or their artifacts, and (b) in the absence of atmospheric greenhouse gases and their artifacts, would be reflected to space by something in the “everything else” category duplicate reflection. What follows is highly dependent on the amount of duplicate reflection. A small level of duplicate reflection, increases the relevance of the following argument. A large level of duplicate reflection, decreases the relevance of the following argument. As several references I have come across point out: Earth albedo is a complex phenomenon.]
Consider, for example, the situation where the measured albedo of 0.3 comes only from atmospheric greenhouse gases and their artifacts—i.e., the Earth’s albedo would be zero in the absence of greenhouse gases and their artifacts. For this situation, the Earth absorbs solar energy at an approximate rate of 1.73 x 10^17 watts; and for an Earth surface having an emissivity of unity, the greenhouse gasless Earth surface temperature would be approximately 278.2K. At the other extreme, if none of the measured albedo of 0.3 comes from atmospheric greenhouse gases or their artifacts—i.e., the Earth’s albedo would be 0.3 in the absence of greenhouse gases and their artifacts, a greenhouse gasless Earth would absorb solar energy at an approximate rate of 1.21 x 10^17 watts, and for an Earth surface having an emissivity of unity, the greenhouse gasless Earth surface temperature would be approximately 254K. Thus ignoring duplicate reflection, depending on what portion of the 0.3 measured albedo comes from atmospheric greenhouse gases and their artifacts and what portion comes from “everything else,” the greenhouse gasless Earth surface temperature can vary between approximately 255K and 278K. Or if duplicate reflection is included, then depending on the amount of duplicate reflection, the greenhouse gasless Earth surface temperature can vary between approximately 255K and 278K.
So it’s not so much that I want to argue about the measured value of the albedo (I’ll accept a value of 0.3), as it is that I want to argue about using the measured value to compute the surface temperature of a greenhouse gasless Earth. Specifically, what are (a) the contribution to the measured albedo of atmospheric greenhouse gases and their artifacts, and (b) the contribution to the measured albedo of everything else? From http://en.wikipedia.org/wiki/Albedo: “The average overall albedo of Earth, its planetary albedo, is 30 to 35% because of cloud cover, but widely varies locally across the surface because of different geological and environmental features [bold emphasis mine].
Clouds are a greenhouse gas artifact in that in the absence of water vapor (the dominant atmospheric greenhouse gas) clouds would not exist. Based on the above Wikipedia quote, a large part of the measured 0.3 albedo comes from atmospheric greenhouse gases and their artifacts. If Wikipedia is correct, a greenhouse gasless Earth surface temperature would be closer to 278K than to 255K.
To use your phraseology, if you want to argue about the contribution clouds make to the measured albedo and more importantly to what the albedo would be in the absence of clouds, that’s fine. But know that Wikipedia says the contribution from clouds is significant. When it comes to anything related to global warming, I’m not a big fan of Wikipedia; but there it is.
In summary, computation of a greenhouse gasless Earth surface temperature from measurements of the Earth’s albedo, no matter how accurate those measurements are, requires an algorithm to estimate the Earth’s albedo in the absence of atmospheric greenhouse gases and their artifacts. Wikipedia may be wrong when it implies clouds (greenhouse gas artifacts) are a major contributor the Earth’s albedo; but I’ve seen/heard/read nothing that says clouds contribute negligibly to the Earth’s albedo. As I see it, the oft quoted greenhouse gasless Earth surface temperature of approximately 255K relies heavily on the albedo of a greenhouse gasless Earth being nearly the same as the measured (with a greenhouse gas atmosphere) Earth albedo. If true, I have no problem with the 255K temperature. If not true, I do have a problem.
(4) I wrote: “Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df.” You responded: “Where did you get that from?” Here’s the answer to your question.
Assume you have a variable “A” that is a function of a second variable “r”. Someone gives you a formula, not for A(r) itself, but rather for the value of A(r) per unit length of “r”. For example, you’re interested in the area of a closed, planar, geometrical shape. Someone gives you the formula Y(r)=2*pi*r, and tells you that the formula represents the area of your geometrical shape per unit length of “r”. Provided the formula meets all the mathematical requirements (differentiable, integrable), it is routine to say: The area of the geometrical argument between r and r+dr is given by: Y(r)*dr. The area of the geometrical shape between r1 and r2 can be computed by integrating Y(r)*dr between the limits r1 to r2.
Item 6, page 6, of Planck’s 1912 document, uses just such terminology. Specifically, Planck wrote:
“Summing up everything said so far, we may equate the total energy in a range of frequency between f and f+df emitted in the time dt in the direction of the conical element dW by a volume element dV, [by] dt*dV*dU*df*2*e. The finite quantity e is called the coefficient of emission of the medium for the frequency f.”
Planck goes on to say: “The total emission of the volume element dV may be obtained from this by integrating over all directions and all frequencies. Since e is independent of the direction, and since the integral over all conical elements dW is 4*pi, we get: dt*dV*8*pi*(Integral from zero to infinity of e*df).”
From Wikipedia [http://en.wikipedia.org/wiki/Planck’s_law] I got an expression for the spectral radiance of a body as a function of frequency. According to Wikipedia, “The spectral radiance of a body, Bf, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. Planck showed that the spectral radiance of a body at absolute temperature T is given by:” and then Wikipedia gives a formula. It is not easy to write equations in a format compatible with comments on this blog. Suffice it to say the formula given by Wikipedia contains a factor of f^3 and a factor of 1/(e^K – 1), where K=h*f/(kB*T).
I admit that as of today (a cursory reading of Planck’s 1912 paper) I haven’t found an equivalent formula in Planck’s original paper. Maybe one doesn’t exist, in which case I won’t be able to give a reference directly attributable to Planck. However, the above discussion references modern texts and should answer your question: “Where did you get that from?”
(5) I wrote: “..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.” To which you responded: “Not according to Planck’s own writing in the 1912 paper introduction. If you want I can explain how to prove that to yourself by simple experiment with IR camera. I’ll describe a situation, ask you a question, and wait for your answer.
Assume you have a solid rectangular bar of uniform material and dimensions h x h x (2h+d). Use a saw with a blade width of exactly d to cut the bar into two identically oriented, identically sized cubes each having dimensions h x h x h and whose “cut faces” are separated by a distance d. Identify these cubes with the letters “L” and “R.” Assume all surfaces of both cubes are blackbody surfaces. Finally, assume the temperature of cube L is everywhere TL and the temperature of cube R is everywhere TR. I want to compute the rate energy is radiated from cube L to cube R. The geometry of the cubes is such that all points on the surface of cube L that have line-of-sight visibility to any point on the surface of cube R reside on the “cut face” of cube L; and vice-versa for points on the surface of cube R having line-of-sight visibility to any point on the surface of cube L. I’m going to describe a procedure for computing the rate energy radiates from cube L to cube R. I would like to hear your opinion of that procedure. Note: According to Planck (at the bottom of page 6), the emission from cube L to cube R is independent of the TL and TR, so that with the obvious substitutions, if the procedure I describe is valid for radiation from cube L to cube R, it is also valid for radiation from cube R to cube L.
(A) Define a rectangular, Cartesian coordinate system (a) whose origin is at the center of the “cut face” of cube L,” (b) whose z axis is perpendicular to the “cut face” of cube L in the direction to cube R, (c) whose x axis is parallel to any “side” of the cube L “cut face,” and (d) whose y axis completes a right-handed rectangular Cartesian coordinate system.
(B) Select a differential area, dSL=dxL*dyL at position xL, yL on the “cut face” of cube L.
(C) Select a differential area, dSR=dxR*dyR, at position xR, yR on the “cut face” of cube R.
(D) For those two differential surface areas, compute the angle, B, between (a) the normal to the dSL …and…(b) the vector from dSL to dSR. The tangent of B is given by:
TANGENT(B) = Sqrt[(xR – xL)^2 + (yR – yL)^2] / d
Note: (a) B is a function of xL, yL, xR, yR and the constant d; and (b) because the normal to the “cut face” of cube L points in the opposite direction to the normal to the “cut face” of cube R, the angle between the normal to dSR and the vector from dSR to dSL is also B.
(E) As seen from dSL, the solid angle, dW, subtended by dSR is
dW = dxR*dyR*COS(B)/[ (xR – xL)^2 + (yR – yL)^2 + d^2 ]
(F) From Equation 6, page 13 of Planck’s paper, the rate energy is radiated from dSL in the direction of dSR is directly proportional to dSL*COS(B)*dW. Thus, the rate energy is radiated from dSL towards dSR is proportional to
dxL*dyL*COS(B)*dxR*dyR*COS(B) / [ (xR – xL)^2 + (yR – yL)^2 + d^2 ]
(G) Integrate the Equation in (6) above over the face of cube R—i.e., integrate dxR from –h/2 to +h/2 and dyR from –h/2 to +h/2.
(H) Integrate the equation in (7) above over the face of cube L—i.e., integrate dxL from –h/2 to +h/2 and dyL from –h/2 to +h/2. The rate energy is radiated from cube L to cube R will be directly proportional to this value of this integral.
I believe the above procedure is consistent not only with the modern theory of physics, but with Planck’s 1912 description of radiative heat. Do you agree or disagree? Continue with this response only if you agree. If you disagree, please explain why.
Okay. Now reshape the “cut face” of cube L so that instead of being planar, the “cut face” of cube L is slightly concave. By doing this, the “cube” L cut face that radiates towards cube R is no longer planar, but is shaped such that portions of the “cut face” of cube L can now see other portions of the “cut face” of cube L. I believe you characterize such a shape as having a “negative radius of curvature.” I believe that if I make the necessary corrections to (a) the differential surface area on the “cut face” of cube L (the differential area is no longer dxL*dyL), (b) the angle between the normal to the new cube L differential surface area and the vector from that surface area to dSR, and (c) as seen from the cube L differential area, the solid angle subtended by dSR, the above procedure is valid. Do you agree or disagree? If you disagree, please explain why.
I point out that even with a concave cube L “cut face,” because the “cut face” of cube R is planar (i.e., has no “negative radius of curvature”), I see no reason that with the proper “integral corrections” the above procedure doesn’t apply to radiation from cube R to cube L. If the two cubes are at the same temperature, then the net rate of radiation between the cubes will be zero. This implies that for equal temperature cubes, the rate energy radiates from cube L to cube R must equal the rate energy radiates from cube R to cube L. I haven’t actually performed the integration for cube R to cube L radiation using a concave “cut face” surface for cube L, but there will be symmetry for each differential area on one face to a differential area on the other face; so it seems to me the procedure is valid for radiation from the concave surface of cube L to the planar surface of cube R.
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Reed 3:20pm: “Our Modern Understanding of the Theory of Heat Radiation may in no way conflict with Planck’s 1912 paper.”
There is no conflict in the principles Reed. There IS nowadays better equipment to more closely determine, say, Planck’s constant which is only very slightly different in modern times than 1912. Those geezers could do great experiments.
“..I have no hope of reading and understanding all of Planck’s 1912 paper.”
You do need to understand when eqn. 274 is applicable and when it is not.
“Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes.”
This is incorrect Reed since tests show as the emissivity reduces below ideal 1.0 for the airless ideal sphere, reflectivity increases above 0.0. With your (a) transmissivity = 0.0 then reflectivity + emissivity = 1.0. The natural earth surface emissivity is measured around 0.95 so the natural surface reflectivity is about 0.05. Because some incident solar is reflected to deep space, never absorbed, from way above the surface, earth albedo measures out at about 0.3. Only about 0.70 incident solar is absorbed into the system.
“So for a no-atmosphere sphere and equal non-zero absorptivity/emissivity, the ERE surface temperature of an Earth-like sphere is approximately 278.2K independent of the absorptivity/emissivity of that Earth-like sphere. Do you agree or disagree?”
I agree if a sphere in earth’s orbit from 2000 to 2010 with albedo 0.0 and ideal emissivity 1.0, then annualized Tmedian over the surface would be 278.3K.
“The T^4 relationship is not in general valid for an emissivity that changes with (incident) frequency.”
True, the emissivity of the same satellite face rotating between incident solar and incident starlight will vary. Thus your resultant Tes will vary. This is accounted in the full S-B theory. For the earth, this is why the annual period is needed, for all incident angles and all incident frequency content. This should be so noted as you write, few do.
“I believe this is the methodology used to arrive at the 255K Earth surface temperature in the absence of atmospheric greenhouse gases.”
It is not Reed, the work out goes much deeper but is still introductory level. If you want to carry on a discussion as to how to find the 288K and the 255K at surface from 1st principles, I can help, but once again a modern text book will be more helpful. I can add in what I have read from them.
“If this is the methodology used to arrive at the 255K temperature..”
It is not, one needs to go deeper (pre-req.s: understanding of atm. optical depth principles & atm. emissivity) but still the methodology used is basic (1st law) to understanding your Tes = 255K and the 288K. Your minor and major problems are then not entirely applicable though there are still limits.
“I want to argue about using the measured value to compute the surface temperature of a greenhouse gasless Earth.”
Better to understand this by keeping the pressures of the atmosphere in place as well as its affect on incident asteroids. Then just vary the emissivity of the atmosphere down (reducing opacity) using the text book 1st law work out. Once you get that, you can discover even more based on moon observations. For example, your (iii) & Planck law eqn. 274 does not hold for around 10-20% of the moon surface.
“..the oft quoted greenhouse gasless Earth surface temperature of approximately 255K relies heavily on the albedo of a greenhouse gasless Earth being nearly the same as the measured (with a greenhouse gas atmosphere) Earth albedo. If true, I have no problem with the 255K temperature. If not true, I do have a problem.”
Ok, yes, the 255K 1st law text book workout relies on a steady 0.3 albedo. As I wrote before, as the albedo assumption is varied from 0.0 to 0.30, surface temperature will vary from 255K to 278.3K assuming the atmosphere is ideal transparent (no opacity). So one needs to state the albedo used.
“I admit that as of today (a cursory reading of Planck’s 1912 paper) I haven’t found an equivalent formula in Planck’s original paper.”
See eqn. 274 in 1912 paper. Any wiki equation needs to be traceable to it. You have to be careful with units for all the constants (Planck uses cgi IIRC) as well as whether the formula is for wavelength or frequency which means the pi term could look different.
“Assume you have a solid rectangular bar..”
Will take some time to read & absorb.
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Reed 3:20pm: “I believe the above procedure is consistent not only with the modern theory of physics, but with Planck’s 1912 description of radiative heat. Do you agree or disagree?”
You have changed nomenclature from Planck’s and my head hurts. For example, I think your angle B is Planck’s theta but not sure. You do not place a limit on B as Planck does for his theta (between 0 and pi). So if you go back and use Planck’s theta and azimuth phi (limited between 0 and 2pi) I might have some hope of a disagree or agree.
“Okay. Now reshape the “cut face” of cube L so that instead of being planar, the “cut face” of cube L is slightly concave…it seems to me the procedure is valid..”
This is where you stumble without limits on the integrals and I can’t be sure since my head hurts; here I think your angle B must need fully go outside Planck’s theta limits of 0 to pi.
You write “slightly” concave. Make it “hugely” concave” and I think you will begin to see the problem. As an example, there are internet IR camera views of a say 6″ by 6″ flat steel plate maybe 1/4″ thick at a certain temperature correctly registered by the camera. Then the experimenter drills two round (say 1/4″ drill by 1/8″ deep) holes. Those holes interior surfaces register a warmer color on the camera not because they are necessarily any appreciably amount warmer but because the surfaces are now radiating to themselves & Planck’s eqn. 274 which the camera works off is no longer applicable. This is a well known problem in being careful when using an IR camera to inspect electrical panels and circuits for unwanted high temperature resistances.
Your two blocks are like Planck’s bodies A at 100C, B at 0C, and B’ at 1000C bottom p. 6 to 7. I think Planck writes there what you are in part seeking to learn.
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LtCusper, 5:44 pm 31 March 2015 and 6:52 pm 31 March 2015.
I’ll need time to digest and respond to what you wrote. However, off the top of my head, two comments.
First, Planck uses a polar coordinate system to expresses the differential solid angle into which energy is radiated. The origin of Planck’s polar coordinate system is the differential radiating plane. The polar axis of Planck’s polar coordinate system is the vector perpendicular to the differential radiating plane. In that coordinate system, the differential solid angle dW is SIN(theta)*dtheta*dphi, where theta is the polar angle and phi is the azimuthal angle. [Note: by integrating the solid angle over the limits theta from 0 to pi and phi from 0 to 2*pi you get the total solid angle of 4*pi steradians.] If you want to determine the energy radiated into the solid angle subtended by the “top” of a cone whose apex is at the differential planar surface, whose axis is coincident with the coordinate system’s polar axis, and whose cone half-angle is alpha, then (a) the expression you must integrate is COS(theta)*dW = COS(theta)*SIN(theta)*dtheta*dphi, and (b) the limits on the integration are theta from 0 to alpha, and phi from 0 to 2*pi.
However, if you’re interested in determining the rate energy is radiated into the solid angle subtended by a rectangular surface whose plane is parallel to the plane of the differential radiating surface, then in polar coordinates the limits of integration for theta and phi are (a) not simple, and (b) functions of one another–i.e., the limits of integration of theta are functions of the value of phi. So in a polar coordinate system, you’ve simplified the integrand [COS(theta)*SIN(theta)*dtheta*dphi] at the expense of making the limits of integration complex.
If you use a rectangular coordinate system whose origin is at the differential planar radiating surface and whose z axis is perpendicular to that surface, then the solid angle subtended by a planar differential receiving area dx*dy is simply dx*dy*COS(B)/r^2, where B is the angle between (a) the normal to the receiving differential area …and… the line from the receiving differential area to the transmitting differential area, and r is the distance between the two differential planar areas. If the plane of the receiving differential area is parallel to the plane of the radiating differential surface area, then B is also the angle between (a) the radiating differential area …and…(b) the vector from the radiating differential area to the receiving differential area. Then as Planck did using polar coordinates, a term COS(B) must be included to account for radiation that propagates away from the differential area at an angle B with respect to the normal to the differential radiating plane. Thus, the integrand, which is the product of dW and COS(B), becomes dx*dy*COS(B)*COS(B)/r^2, with limits of integration for x and y that define the rectangular area that energy is being radiated into.
So for computing the radiation emitted into the solid angle of a finite-size plane, in the one case (polar coordinates) you keep the integrand simple, but make the limits of integration complex. In the other case (rectangular coordinates), you complicate the integrand but make the limits of integration simple. Which integral is easier to perform is a function of the user’s knowledge and access to tables of indefinite integrals. So, for my procedure, yes B is Planck’s polar coordinate system variable theta, but that is only true for a planar receiving surface that is parallel to the planar differential radiating surface. Furthermore, you commented that I didn’t define the integration limits for B. I don’t have to. The integration limits are expressed in terms of x and y.
Second, I’m going to assume in your plate/plate-with-holes that all surfaces are blackbodies–i.e., no reflection occurs. A camera that is imaging a planar surface is measuring the rate energy is radiated from points (differential areas) on the surface of the plane in the direction of the camera lens. The lens focuses the energy that passes through the lens from a point on the radiating surface to a point on a plate. The greater the energy, the darker the point (or maybe the lighter the point depending on the material absorbing the energy). Thus, an image of a radiating plane doesn’t represent the total energy radiated from differential points on the plane; but rather represents the radiation from points on the radiating plane that strikes the focusing lens. When you drill a small hole in the surface of the plane you both remove some areas that were radiating before the hole was drilled, and you expose some new radiating areas that didn’t exist before the hole was drilled. For a finite radius, finite depth hole, it wouldn’t surprise me if the exposed radiating surface of the hole is different from the exposed radiating surface without the hole. However, the total exposed radiating surface area isn’t what counts. What matters is the relative rate (a) energy originating from within the hole is directed towards the camera lens compared to (b) the energy rate from the flat surface of the soon-to-be hole is directed towards the camera lens. It wouldn’t surprise me one bit that these energy rates are different. If so, the image of a hole would appear darker (or lighter depending) than the surrounding image. To me this only implies that you must be careful when applying Planck’s radiation law for differential planar surfaces to real-world surfaces, not that Planck’s law doesn’t apply.
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Reed 10:44am: “Second, I’m going to assume in your plate/plate-with-holes that all surfaces are blackbodies–i.e., no reflection occurs.”
The steel plates are real but not polished enough to fake out the IR camera since it correctly registers the temperature measured by thermocouple on the surface. There is reflection.
“The integration limits are expressed in terms of x and y.”
There are no limits that I can see in your comment. The limits of integration, in any coordinate system chosen, need to conform the surfaces to Planck writing p. 2 “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.” Since wavelengths are always positive, the radii of curvature are ruled always positive, ruling out surfaces not applicable Planck eqn. 274 that are even “slightly” concave and ensuring diffraction is negligible before applying eqn. 274. The radii of the plate hole surfaces are concave, Planck eqn. 274 is not applicable.
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LtCusper, 2:50 pm, March 24, 2015; 5:45 am, 31 March 2015; 6:52 am, 31 March 2015; 1:55 pm, 31 March 2015
First, I want to respond to the issue of “integration limits” in my procedure. You wrote: “There are no limits that I can see in your comment. The limits of integration, in any coordinate system chosen, need to conform the surfaces to Planck writing p. 2 “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.”
(A) In the case of a cube, all surfaces are planes so all surfaces have infinite radii of curvature, and as such all radiating surfaces have radii of curvature are much larger than any finite wavelength. The cube edges have zero radius of curvature in one direction and infinite radius of curvature in the orthogonal direction. I guess the corners of the cubes have zero radius of curvature in all directions. I’m not sure how to handle the edges and corners. In addition, there is still the issue of diffraction at the cube edges; but all cube faces conform to Planck’s p. 2 restriction “the radii of curvature of all surfaces under consideration are large compared with the wave lengths of the rays considered.”
(B) In Step (G) of my procedure I wrote: “ Integrate the Equation in (6) above over the face of cube R—i.e., integrate dxR from –h/2 to +h/2 and dyR from –h/2 to +h/2.” The words …the Equation in (6) above… should have been …the Equation in (F) above… When I originally wrote the procedure I used numbers, not capital letters, to delineate the procedure steps. Because I had used numbers in my comment to delineate points I was trying to make, to avoid confusion I later changed the procedure steps from numbers to letters. I forgot to correct the wording of the procedure steps so that the wording still used numbers.
In any event, the origin of the x,y,z coordinate system is at the geometric center of the “cut face” of cube L. The “cut face” of cube L is defined by the coordinates: -h/2 less than x less than +h/2, -h/2 less than y less than +h/2, z = 0. The coordinate system’s z axis points to the geometric center of the “cut face” of cube R. The “cut face” of cube R is defined by the coordinates: -h/2 less than x less than +h/2, -h/2 less than y less than +h/2, z = d. The first integration (Procedure Step G) selects a fixed differential area, dxL x dyL, on the “cut face” surface of cube L and integrates over the entire “cut face” of cube R, which as I said in Procedure Step H is the integration over dxR from –h/2 to +h/2 and the integration of dyR from –h/2 to +h/2. The largest possible polar angle occurs for a differential area at one corner x=-h/2, y=-h/2, z=0 of cube L …to… the diagonally opposite corner x=h/2, y=h/2, z=d of cube R. Thus, the magnitude of the largest polar angle is Inverse_Tangent{h x [2^(1/2)]/d], which is less than pi/2. For all dxL x dyL not on an edge of the cut face of cube L, the azimuthal angle can range from 0 to 2*pi. The second integration (Procedure Step H) integrates over all differential areas on the “cut face” of cube L. The limits of this integration are also –h/2 less than x less than +h/2, and –h/2 less than y less than +h/2.
Second, you wrote: “You do need to understand when eqn. 274 is applicable and when it is not.” I agree. However, as best I can discern, Equation 274 is never mathematically applicable. Equation 274 gives the “specific intensity of a monochromatic plane polarized ray of frequency f…” According to what Planck wrote on page 2, the equation only applies for radii of curvature large compared to the wavelengths of rays considered. Since Equation 274 doesn’t restrict the range of frequencies, for any object having a finite radius of curvature, wavelengths exist that are long compared to the radius of curvature. Thus, Equation 274 mathematically applies only for an infinite radius of curvature—i.e., a planar surface. Even then, unless the plane is infinite in extent, diffraction will occur at the edge of the plane—another issue raised by Planck. Thus, Equation 274 mathematically applies only to an infinite plane, which means that mathematically it can never be applied to a real-world surface. If you’re going to apply Equation 274 to any real-world situation, you’re going to have to caveat the application with words to the effect that Planck’s distribution function is only an approximation. Then the issue becomes: “How good an approximation?” I have insufficient knowledge to answer that question for most real-world situations. What I have observed is that people use Planck’s specific intensity distribution for many situations without the caveat mentioned above. Although I may caveat things I write in the future, I still plan to use Planck’s Equation 274 as the behavior of the “specific intensity of a monochromatic plane polarized ray of frequency f…”
Bottom line, I would slightly change your admonition to: “You do need to understand when eqn. 274 is a good approximation to real-world situations and when it is not.” With that change, I agree with you; and any improper use of Equation 274 that I employ is my error.
Three other points. The T^4 law comes from integrating Equation 274 over all frequencies. Since for any real-world situation, Equation 274 is at best an approximation, then the T^4 dependence on temperature is also at best an approximation.
Next, in a previous comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.” Given that (a) an enclosed shell must at some point have surfaces with negative radii of curvatures—at least as you characterize radii of curvature as being negative, and (b) by poking a hole in an enclosed shell radiation escaping the hole undergoes diffraction, how can you claim “out comes BB radiation?” Don’t Planck’s radii of curvature and diffraction caveats contradict the BB claim? Or at a minimum raise the issue of how Planck’s caveats don’t seem to apply to a hole in an enclosed shell whose interior is at a single temperature?
Finally, at one point in the thread we had the interchange–me: “..ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface”; you: Not according to Planck’s own writing in the 1912 paper introduction. In the sense that Planck’s Equation 274 might not apply, I agree with you. What I was referring to was that a planar differential surface cannot radiate energy to itself; whereas a finite sized object can. Consider a thin hemispherical bowl. The area of the bowl that has line-of-sight visibility to the outside world is the same as the area of an equal radius sphere. If you use the equation that the rate of cooling of an object is proportional to the product of the surface area of the object and the difference of the fourth powers of the object’s temperature and the background temperature, then I believe you can use the area of the sphere exposed to the background, but you can’t use the area of the bowl exposed to the background. Portions of the concave side of the bowl will radiate energy to other portions of the concave side of the bowl. Maybe inconsistent with Planck’s Equation 274, but transfer of energy non-the-less.
Third. We had the exchange: me: “Provided the emissivity of the Earth-like object is non-zero and equal to its absorptivity, the emissivity/absorptivity of the Earth-like object has no effect on the temperature Tes.”
You: “This is incorrect Reed since tests show as the emissivity reduces below ideal 1.0 for the airless ideal sphere, reflectivity increases above 0.0. With your (a) transmissivity = 0.0 then reflectivity + emissivity = 1.0. The natural earth surface emissivity is measured around 0.95 so the natural surface reflectivity is about 0.05. Because some incident solar is reflected to deep space, never absorbed, from way above the surface, earth albedo measures out at about 0.3. Only about 0.70 incident solar is absorbed into the system.
Now I’m confused. In my statement I caveated that the emissivity equaled the absorptivity. That may not be the case for an airless ideal sphere, but that was my caveat. Assuming transmissivity is zero, then absorptivity + reflectivity = 1, which says absorptivity = 1 minus reflectivity, which says emissivity = 1 minus reflectivity, which says emissivity = 1 minus albedo. So once you set the albedo, you are setting both the emissivity and the absorptivity and they will be equal. For a given albedo, the sphere will absorb in direct proportion to 1 minus the albedo, and the sphere will radiate in direct proportion to 1 minus the albedo. Maybe I’m missing something, but how can the temperature of inert matter in energy-rate-equilibrium that absorbs with the same proportionality it emits be a function of that quantity?
Fourth. On page 2, Planck wrote: “Throughout the following discussion it will be assumed that the linear dimensions of all parts of space considered, as well as the radii of curvature of all surfaces under consideration, are large compared with the wavelengths of the rays considered.”
You wrote: “Since wavelengths are always positive, the radii of curvature are ruled always positive, ruling out surfaces not applicable Planck eqn. 274 that are even “slightly” concave and ensuring diffraction is negligible before applying eqn. 274. ” I’m a little confused by what you wrote. If you’re saying that the radius of curvature of a concave surface is negative, and because it is negative (a) the radius of curvature can’t be greater, much less ‘much greater’, than the wavelengths of rays, and (b) thus Planck’s Equation 274 can’t be applied to concave surfaces, then I don’t agree. First, radii of curvature are distances and distances are always positive. The radius of curvature of a concave spherical mirror is often characterized using a positive number.
Furthermore, consider the “top of a cone” whose cone half angle is 0.5 degrees and whose radius is 1,000,000,000 meters. From a visual perspective, a person walking on the cone top would have a hard time deciding if he was on the concave side or the convex side of the cone top. For a very thin “cone top,” the total area of each side (concave and convex) of the “cone top” is approximately 2.39 x 10^14 square meters. If the temperature of the surface was everywhere say 300K, it’s hard, no it’s close to impossible, for me to believe the total rate energy is radiated by the concave surface of the cone top is appreciably different from the total rate energy is radiated by the convex surface of the cone top. So, it makes sense to me that the radius of curvature should be large compared to the wavelengths of the rays under consideration; but it makes little sense to treat the radii of a concave surface as being negative in Planck’s comparison of wavelength and radius of curvature.
Fifth. Your comments have both informed me and given me a lot to think about. For that I am grateful. Thank you for taking the time to respond to my comments.
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Reed 2:18pm: “Thus, Equation 274 mathematically applies only for an infinite radius of curvature—i.e., a planar surface.”
Look up a paper ref.d by Planck for the test methods. Available on line. Eqn. 274 is for BB radiation, was developed experimenting with heated/cooled cavity BB radiation emitted from the hole & directed into a box to reduce convection. The box had some conditioning mirrors to eliminate unwanted signal with the intended signal hitting a thermopile attached to galvanometer to read the intensity. There is no such restriction you impose. This can be verified with an IR thermometer temperature reading the same as mercury thermometer temperature of a flat table or curved water glass, ice cube tray, etc.
Diffraction can occur but it needs to be negligible for the IR thermometer correct reading.
“So once you set the albedo, you are setting both the emissivity and the absorptivity and they will be equal..”
Only for the same content of frequencies. For earth system, albedo is SW, emissivity is LW.
“The radius of curvature of a concave spherical mirror is often characterized using a positive number…it makes little sense to treat the radii of a concave surface as being negative..”
The sense is the vector is opposite direction, positive radii point away from the object; concave radii point toward the object thus radiating to itself. Planck et. al. recognized through experiment a concave object surface situation was trouble so ruled it out. While the trouble may not be very large for “slightly concave” surfaces, it certainly will be to some extent – thus needing experiment to be definitive.
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LtCusper,
I read your 2:50 pm, 24 March 2015 comment. It will take me some time to respond–I need to give what you wrote some thought; but I will respond. Since you have elected to forego an email exchange, you’ll just have to intermittently visit this thread.
BTW, It won’t do any good for me to access Planck’s original paper–my German is inadequate. I did, however, find online
https://archive.org/stream/theoryofheatradi00planrich#page/viii/mode/2up
a translation by Morton Masius of a Max Planck paper entitled: The Theory of Heat Radiation. The translation was published in 1914. Examining the “Prefaces” in the published translation, it looks like the translation is of the second edition of a paper written by Max Planck in 1912. Is this the document you recommend I “invest in?”
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Yes Reed 5:49pm, well done, that’s the one; Planck writing dated Nov. 1912.
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Reed,
“Bottom line, when computing (not measuring, but computing) the Earth’s surface temperature in the absence of greenhouse gases, it’s illogical to perform that computation using a value that depends on the presence of greenhouse gases. I don’t accept the statement that radiometers outside the Earth’s atmosphere and pointed towards the Earth measure radiation only emitted from the Earth’s surface. How can they not measure the sum of Earth-emitted radiation and reflected solar radiation? It’s only after you subtract out the reflected solar radiation that the intensity corresponds to a blackbody object at 255K. But without water vapor there would be no clouds and without clouds there would be no (or very little) reflected solar radiation, and without reflected solar radiation, there’s nothing to subtract from the radiometer measurements.”
This is true but very misleading. The total radiative intensity Watts/steradian of this Earth cannot be measured at a distance less than our moon. Much better would be satellites at L1 and L2. both are looking the other way. Why?
The Earth with its atmosphere absorbs and reflects some power from the 68 micro-steradians Sun. then re-radiates all into 4 PI steradians space. If not the expression for accumulated power, as sensible heat, TEMPERATURE must go up or down! This planet, finely controls atmospheric WV so that temperature at any location remains comfortable. This is called CLIMATE, whatever you wish! The Wind determines atmospheric WV at every location. What determines wind direction and velocity at every location.
Again This is the total demonstrated and perhaps intentional, incompetence of the entire climate science religion!
MY knowledge is limited to “beats the shit out of me”!! 🙂
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Reed 3:51pm: “But isn’t the optical depth in part a function of the density of the material through which the radiation is propagating?”
In original formulation, yes. Hydrostatic equilibrium is known good by experiment for large parts of the atm., and that realization takes density term out of the differential optical depth replacing with pressure. So the atm. layer differential grey opacity becomes a function of mixing ratio, mass extinction coefficient for the ith constituent and differential pressure summed over N constituents. Then for total optical depth integrate down from top of atmosphere to the reference pressure for tau.
So you would have to do this and obtain proper tau over a few mm at the required pressure for CO2 mass extinction coefficient and 100% mixing ratio to get the 33K result of the normal atmosphere. It should be evident you would now have an unweildly thermos bottle just from general experience.
As an alternative, look at the Mars atmosphere which is physically high but very thin optically causing more like a few Kelvin of the effect you seek to learn about which causes about 33K at earth pressures and constituent mass extinction coefficients.
NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.
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Reed Coray
IMO Part of the bigger problem is that we have such a poor understanding of what happens in our skys with it’s mix of IR and movement of variable dimensioned vapor droplets, the exchanges that happen between them, and layered structure.
Please look here if you think I’m wrong. It is and old paper about the anomalous reading that were present when measuring cloud dynamics.
http://web.archive.org/web/20060908155431/http://www.cgd.ucar.edu/cms/wcollins/papers/cess.pdf
If you can find it also refer to
JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 104, NO. D2, PAGES 2059-2066, JANUARY 27, 1999
Absorption of solar radiation by the cloudy atmosphere:
Further interpretations of collocated aircraft measurements
Robert D. Cess, 1 Minghua Zhang, 1 Francisco P. J. Valero, 2 Shelly K. Pope, 2 Anthony Bucholtz, 2 Brett Bush, 2 Charles S. Zender, 3 and John Vitko Jr. 4
Where they say in the conclusion –
As far as I am aware this matter has never been cleared up.
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Nicely done,
Yes reflected SW cannot be distinguished from emmitted. It is all just “radiance”. The radiance of the Earth at the location of the Sun is very tiny so does not limit the radiant intensity or flux from the Sun.
(Finally, in your 5:34 am, 24 March 2015 comment you wrote: “NB: As you prepare your steel green house comment, realize ideal Planck formula is not applicable to objects with any negative radii, i.e. objects that radiate to themselves. Like the inside surface of a steel shell. The amazing thing is though, at uniform temperature, poke a hole in it and out comes BB radiation.”)
“Say What? Planck’s law for cavity radiation (sometimes called Planck’s blackbody radiation law) quantifies the power emitted in the frequency interval between f and f+df from a planar differential unit area, dA, at temperature T, into a differential solid angle domega defined with respect to a coordinate system at dA. There are no terms in Planck’s blackbody radiation law for the presence/absence of any other surfaces.”
This is somewhat correct except for terms. Plank’s specific intensity is now termed spectral radiance an indication of electromagnetic field strength at a particular wavelength interval and direction. Any opposing spectral radiance must limit any actual spectral flux.
“In fact, it is a geometrical impossibility for a planar differential surface area to radiate into a differential angle that contains any portion of the radiating differential area. Provided the radius of curvature of an object is finite and non-zero, ideal Planck’s law can be applied to differential planar areas of a surface that radiate to other portions of that surface.”
Only for radiance never flux!
“For the interior surface of a closed object (like the interior surface of a spherical shell), you can use Planck’s law to compute the power radiated from any finite size area of the interior surface to any other finite size area of the interior surface. It turns out that if no objects reside in the interior surface of a closed object, the total rate energy is radiated from the entire interior surface is equal to the total rate the interior surface absorbs radiated energy. As such, the net rate of energy exchange of the interior surface is zero.”
The energy exchange is always zero as internal opposing radiance always cancels any emission in any direction. Any other interpretation defines a perpetuum mobile of the second kind. I look forward to #4! 🙂
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5mm of CO2 at 100% concentration is equivalent to 15m of atmosphere with a CO2 concentration of 0.03%. How much do you need to create a measurable effect?
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At 100 kPa pressure, 350 ppmv CO2, the .5 micron band about 15 microns has an optical depth of less than 2 meters. 37% transmission of amplitude modulation at 300 Hz. The 300Hz is well above the low pass cutoff of even 1 meter of such gas. This in no way indicates that such gas at or above equilibrium temperature, would attenuate any DC 15 micron flux. It is the aluminum surfaces that prevent 15 micron EMR. The good vacuum prevents conductive and convective loss! Try 30 layers of super insulation for spendy things like liquid Neon! 🙂
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You all also forget the emissivity of CO2 as a gas. The “back radiation” meme that supposedly heats the planet is dependent on CO2 radiating energy it absorbs. Problem, CO2’s emissivity is less than 0.0017 so for all practical purposes it doesn’t radiate. (Gases as a whole are really lousy radiators.) If it did it’s peak line is at 15µm which Wein’s Law tells us is ~193°K. That means that CO2 radiating at its fundamental frequency CANNOT make anything hotter than -80°C. In order to get gases to radiate look at what we do in the real world. They are contained at near vacuum pressures. You have to keep the gas molecule in a high energy state long enough for it to drop to ground and emit. Pressures above 1 Torr make this almost impossible as energetic gas molecules run into each other and transfer that energy kinetically before the can radiate.
I’ve said this a few times so my apologies for repeating it. Gas lasers and neon tubes are all extremely low pressure and extremely high energy density. CO2 lasers are even more complex since CO2 does not readily get to an emission energy level. There’s a mix of exotic gases necessary to get CO2 to radiate on top of either high frequency high energy RF or high voltage DC power supplies. Conditions that don’t readily occur in our atmosphere. If CO2 acted like the AGW crowd said it does all we’d need to build a CO2 laser would be one partial LWIR mirror, one full LWIR mirror and a heating element… that of course doesn’t work. The only places in our atmosphere where gases radiate, since the right conditions exist, are at the poles above a minimum of 60km and most above 100km driven by high energy particles from the sun funneled by magnetic flux lines… we call them Auroras.
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Sorry, there’s also radiation from the TOA at pressures below 1 Torr as well from the rest of the atmosphere. The auroras are just a good example of the low pressure necessary since you’ll see a drastic cut off on their lower edge where the pressure gradient gets too high for them to continue to radiate.
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Niels,
Thank you for your comments.
“If it did it’s peak line is at 15µm which Wein’s Law tells us is ~193°K. That means that CO2 radiating at its fundamental frequency CANNOT make anything hotter than -80°C. In order to get gases to radiate look at what we do in the real world. They are contained at near vacuum pressures.”
Kind of correct. The thermal EM absorption or emission of matter does not depend on its peak wavelength or temperature. Thermal Radiance is proportional to T^4 not T or some linear delta T. Anything interfering with your ability to dispatch to space, or even conducting to cold “must” result in your higher temperature to compensate for such interference. Most times in this atmosphere, this interference does not happen! All molecules just keep on dooin whatever! 🙂
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It absolutely does. That’s basic. All materials emit, absorb and reflect radiation depending on their atomic and molecular structure and the amount of energy present keeping the electrons, atoms and molecules moving. All those frequencies and harmonics are the absorption/emission/reflection spectra and the sum of all the energies contained in the material is its temperature. Temperature is a way for us to define how much internal energy is present whether it’s atoms vibrating in a solid, bouncing around as a liquid, gas or plasma. As for your thermal radiance proportionality you are improperly using S-B math for a black body on a single wavelength emitter that is not even remotely like a black body… emissivity of 0.0017… remember? A line emission is not a Planck black body curve. If you don’t like Wein’s Displacement Law, take it up with Wilhelm’s ghost. That’s another AGW failing in treating atmospheric gases like black body radiators.
I don’t understand what the rest of your comment means with the “interference” references and “conducting to cold” so I’m not even going to hazard a guess as to what you’re trying to say there.
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CO2 gas can, and will, via spontaneous EMR flux, transfer energy at 15 microns to any mass absorbing such waveband because of its lower temperature. Wien’s displacement law has nothing to do with any ability to radiate or absorb at any frequency.
Emissivity of CO2 gas at 15 microns depends on the density and depth of generation. STP air of 5 meters has emissivity of g.t. 98% just because of the 320 ppm of CO2. That same air without “any” CO2 would have such emissivity at 25 meters just because of WV. These are from “my” measurements (1969).
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Grauniad/ACF went full tilt opposing useful energy production this past week:
Total detachment from reality.
Gross misrepresentations of what is going on in China (new coal-fired power station every other week or thereabouts) and the German government of prohibiting the shutdown of unprofitable, coal-fired power stations whose operators are not being compensated for being a spinning reserve for wobbly power from the unreliables.
What the lunatics don’t recognize is that the investment climate was changed dramatically with the introduction of RET’s and carbon taxes, dissuading investment in upgrading coal-fired power plants to burn the fuel more efficiently, cleanly and to generate electricity more efficiently.
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Oh yes, they do recognise this. That’s their point …
Previously commented, but needs to be emphasised – I had not believed this could happen in a democracy, but I was hopelessly wrong. All it needs is for all major political groups to agree, then whatever you vote just doesn’t matter
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A few years back I thought Geoff Cousins had a few brains. I amazed at how much my judgement has improved over the years.
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Ditto 🙂
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I’m not sure if this was posted anywhere, but I haven’t seen it in the MSM whatsoever: http://www.independent.co.uk/money/fuel-poverty-could-claim-100000-lives-over-next-15-years-warns-energy-charity-10072222.html. If it was about heat kills, then it’d be plastered everywhere.
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Good catch, bemused.
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Look closely at the list on the image at this link.
It indicates what are astonishingly called Australia’s Worst Polluters. Keep in mind that this is for what is referred to as Carbon Dioxide ….. Pollution. The list was brought out by The Australian Conservation Foundation as a vehicle to launch their new president Geoff Cousins, and was featured on Thursday Night’s Lateline with a compliant Tony Jones interviewing him. That interview was just amazing, so full of inaccuracy I just sat there with my mouth open. Cousins said a number of things, one of them this:
That list has 10 names on it and 7 of them are the electrical power generation companies operating the largest of the coal fired power plants in Australia.
Those 7 power generating entities generate around 75% of Australia’s electrical power needs, probably more even.
This is what is actually keeping Australia going. Take them away, and Australia just grinds to a halt ….. in an instant. That’s not shutting them all down, as it would only take two of them to close and the Country would just stop dead, say Bayswater in New South Wales and Loy Yang in Victoria
There’s nothing to replace those coal fired plants, not only on a power delivery basis, but on a time delivery basis, as each of those plants on that list operates on a 24/7/365 basis, something no renewable can do.
The total wind power output to all the Australian grids comes in at around the equivalent of 1200MW, and that’s not even enough to supply the absolute Base Load requirement for just Sydney alone.
Here’s a scenario then.
You now have a Labor State Government in Victoria and also in South Australia. Labor is typically the party of reducing CO2 emissions, believing that The Science is in, that coal fired plants should be closed, and replaced by renewables. When you add together all the wind plants in South Australia and Victoria, the Nameplate Capacity is similar to the Nameplate for Loy Yang, owned by AGL.
Okay then, let’s say that AGL Energy goes to the Victorian Labor Government and says that in the best interests of Victoria and Australia, they are going to close down their Loy Yang power plant , reducing that (CO2) pollution by millions of tons.
So then, what would be the reply from that grateful Victorian Government, who would then be able to say with pride that they lead Australia in reducing Carbon pollution.
The Victorian Government’s reply would go something along these lines.
You have a contract to supply electricity from Loy Yang. If you close Loy Yang, we will sue your ar$e so far into the future, your Company will be sent to the wall.
There would be a similarly worded reply from the Labor Government in South Australia, because without that power from the Vic-SA Interconnector, (all of it brown coal fired power) South Australia would also grind to a halt.
That story would NEVER get to the media.
Have you ever wondered why, with all the environmental noise about Hazelwood, Labor governments in that State have not just closed that plant down. They know, hand on heart, that there is nothing which can replace the electricity it provides, electricity which keeps Victoria from going dark, literally.
Cousins then went on to say later:
Now, AGL has just purchased Bayswater and Liddell from Macquarie Generation.
Why would they do that?
I mean Liddell is 44 years old and Bayswater is 30 years old. How stupid is it to sink shareholder money into this old technology, something which has no future, according to Cousins.
The future of this investment lies in the return from the sale of electricity to the NSW grid, and that is around $900 Million a year, so if Liddell gets run for its full 50 years, and the same 50 years for Bayswater, then the return from the sale of that electricity is $15.6 Billion, and that’s just at the contracted price of around $30 per MWH, the lowest cost electricity in the Country. Trust me, there is every chance both plants will be operating beyond 50 years.
If coal fired power was on the way out, do you think AGL would have sunk so much money to buy into these two plants?
Worst Polluters in Australia. Really!
Then, just shut them down. See what happens then. It would only take three days, and the whole CO2 emissions argument would just stop. Pity that the whole Country had to stop as well, just to prove the point that we cannot live without all that coal fired power.
I seriously wonder what’s going to happen to all these people hyping Wind and Solar Power to replace coal fired power when the truth gets out that it can’t, There’s nowhere to hide from all the things they’ve so publicly said.
Tony.
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Tony I read this over at wattsup a couple of months ago and I imagine you did to.
I haven’t read any more about it.
But what your writing about here is happening in the “jewel” of the global warmists renewable energy mis information campaign.
I might add, if I hadn’t read it at wattsup I probably would never have.
Mainstream wasn’t exactly “all over it”.
Anybody heard anymore?
“The led to an event that can be likened to the proverbial iceberg unexpectedly popping up right in front of the German state ship while it was plowing through the waves on its climate-saving mission at full-steam. With just a 48-hour notice delivered by a personal phone call to Ms. Merkel on a Saturday, the CEO of E.ON, the largest German and European power producer, let it be known that the company had decided to split itself in two, one part grouping fossil and nuclear power generation and a second part encompassing the “politically correct” activities in the field of “renewable” energies. Sort of a “Bad E.ON” / “Good E.ON” move. The intention is to get rid of the “bad” part as soon as possible by putting it up for sale. At the same time, this also means the “good” part will cease to be duty bound to ensure a stable power supply under all circumstances. Obviously, such a liability is not enforceable from an entity whose only power sources are unstable wind and solar power plants. In a nutshell, the message behind this move is that the silverback of the “big four” German energy producers who group the bulk of the country’s conventional and nuclear power production is about to close shop at short notice. The others will probably follow suit.”
http://wattsupwiththat.com/2014/12/10/the-unsinkable-german-anti-co2-titanic-just-found-its-iceberg/
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Just adding a little-known piece of factual information here
Loy Yang had been owned by an American group whose business was buying and selling power utilities and other infrastructure worldwide. They put a huge amount of capital into the purchase of Loy Yang on the expectation that the Kennett Govt would keep its’ word and deregulate wholesale/retail prices. Kennett unexpectedly (for him,at least) lost power shortly thereafter and Bracks, succeeding him, promptly reneged on this. Loy Yang went into economic freefall and twice broke its’ bank loan covenants
Panic ensued, since Loy Yang could not be allowed to just stop producing electricity (for the reasons that TonyOz describes above). What to do, what to do ?
Sell at a distressed price of course, but to whom ? AGL was the only player both big enough and interested enough to make a reasonable offer. But AGL ownership of Loy Yang would contravene the Federal monopoly legislation (Alan Fels’ ACC). Bracks pushed through legislation in both Vic Houses in most unseemly, panicky haste to except the sale from the monopoly requirements
All of that to prevent the horror described above
Now the greenies want to reverse all that. I agree with TonyOz that this won’t happen just yet but Vic’s new ALP Govt is even crazier than Bracks. Already it has reneged on a perfectly legitimate, binding road-building contract and is threatening sovereign legislation in order to avoid the penalty for this. Forcing the closure of a major power station on “environmental” grounds and then suing the hapless owner for the immense public damage that would ensue seems not as unlikely as I had imagined only a few years ago
140
No one (with any authority/influence) wants to talk abut the fact that most power plants in Australia have had their maintenance schedules slashed and burned in order to save money. When these plants start needing large scale emergency repairs rather than preventative maintenance, then the owners will cry poor, and, in order to prevent widespread blackouts, various governments will have to fork out taxpayers money to fix them. The same can be said for other power infrastructure as well.
I won’t even get into the discussion about shutting down pretty much all the oil refineries in Australia… that’s another problem no one wants to face, but then, that’s ‘big oil’, so no reasonable public discussions can be had any more, as the imbecilic Greens will just throw another tantrum, as they always do.
50
So let us start a campaign to close the power station and have the Labour govt’s have to explain why they cannot allow it
50
There’s no doubt about the distain Labor and the greenies have for the health and welfare of the Victorian community.
In Victoria we see this week that the newly elected union controlled Labor government is to put at risk the health and welfare of the community by what it thinks will “make the state more attractive for building wind farms” and that somehow that will “help to unlock billions of dollars in investment”.
Not with my dollars they won’t. And given the sovereign risk associated with doing business in Victoria I suspect there will need to be huge taxpayer subsidies to get any sensible business person involved in these uneconomic schemes.
But, nevertheless, we have the Clean Energy Council advising us that: “’It is fantastic to see the Andrews Government recognising the need for change…”.
Under the changes, the 2 km setback distance between houses and wind turbines will be reduced to 1 km.
http://www.cleanenergycouncil.org.au/media-centre/media-releases/march-2015/Vic-wind.html#sthash.wK0dwCmD.dpuf
So much for the Pacific Hydro supported research that demonstrated that:
“…a pattern of high severity of disturbance to be associated with four different operating scenarios of the windfarm being:
• when the turbines were seeking to start….
• an increase in the power output of the windfarm in the order of 20%…
• A decrease in the power output of the windfarm in the order of 20%…
• The situation when the turbines were operating at maximum power and the wind increased above 12m/s.
There were at times other instances of high severity disturbance not fitting the above four scenarios.”
http://www.pacifichydro.com.au/files/2015/01/Cape-Bridgewater-Acoustic-Report.pdf
If anyone thinks the Andrews’ Labor government’s reneging on a multi-billion dollar freeway and the consequential likelihood of it having to pay $100s of millions in compensation lifts businesses’ sovereign risk calculations, just wait until the compo is calculated in relation to all aspects of wind-farms construction and operation under this proposed change in the setback distance starts to hit home.
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Look again at the list of Worst Polluters, and keep in mind where I said that:
Now look at that same statement in another way.
The time will come when those plants have reached their end of life.
There are no renewables which can supply those huge amounts of electricity, and do it on the 24/7/365 basis which those plants do provide.
However, now they have reached the end of their life, just where will that replacement power come from. The State Governments sold off all the plants. The now private company owners have only purchased the plant for the electricity they provide until the end of the plant’s life. That’s ll their contract stipulates, that they provide electricity from that plant for as long as the plant can survive. There’s no contractual agreement to build a replacement plant once the existing plant expires. They’ll just flog the dead horse until there is no life left in it, and that will be the end of it.
There is no replacement for that power.
Macquarie Generation tried to actually do that with its Bayswater plant. They submitted plans for an upgrade, and one of 2 options was for a new technology USC coal fired plant which would have gone a long way towards securing the future for that huge supply for NSW. They submitted that proposal in 2009, six years ago. Bayswater knew that a new plant would last 50 years, so there must have been enough coal to cover that at the existing site, coal to supply both Bayswater and the nearby Liddell plant.
The Mt. Piper plant also submitted plans for a similar Upgrade to USC coal fired power technology.
So then, where did both of those proposals go?
You guess!
It went so well, that Macquarie Generation have now sold off Bayswater and Liddell to AGL.
When Bayswater reaches the end of its life, there’s nothing in the pipeline to replace it, and the same applies for all the plants on that list. AGL will then just sell the remaining huge amounts of coal to China.
They should have been starting this replacement program, not just for Bayswater, because it is one of the most recent plants on that list, even at 30 years of age itself, but for all of them.
There will come a time when those old plants will struggle to keep providing, and there will be nothing to replace them. Renewables won’t be able to replace that power, and no Government has the cojones to approve a new plant which actually can replace those large scale coal fired power plants.
That time has almost come, or may have even passed now.
Tony.
171
Hi Tony,
All good points.
But I’m very optimistic and look forward to the time when that which you forecast comes about – as it most certainly will, all other things staying the same.
You see, at that point the voters will well understand that they’ve been defrauded by the greenies and their lefty Labor comrades. The backlash will be massive.
Goodnight greenies.
Goodnight socialist Labor.
Ah! Roll on the day.
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Home diesel generators.
70
Only if you’re near a port since you need mains power to pump that diesel around the country. I have a vision of the 2nd Mad Max movie in my head… fighting for fuel.
60
The biggest ship ever built, Shell’s “Prelude”, actually a floating (FLNG) platform bound for the Browse gas field off Broome has just been launched. Pictured is Shell’s new Search and Rescue helicopter, which just arrived here in Broome.
100
Sorry, TonyOz, but 1) AGL does not own the mines; 2) China does not import much thermal coal from Aus in any case
In the event that Bayswater and Liddell just shut down, the mines will severely reduce production or simply put them on care and maintenance for a duration
61
Just saw a post on judith curry blog about homogenised temp in Australia by euan mearns . He finds that the homogenised nasa data do absolutely nothingoverall. But the biggest surprise was that there was absolutely no warming since 1880. Interesting to see how this analysis marries up with marohasys work and f others.
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Presumably you mean this one:
http://judithcurry.com/2015/03/17/temperature-adjustments-in-australia/
70
euanmearns has something further to say in the comments, which is pretty instructive.
He worries about us losing the raw data and mystified that temperatures have been flat for over a century around the Alice.
60
Weekend Unloaded?, sorry I don’t go anywhere without my “skepti-cal” ammunition Jo. 🙂
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What is EDT then?
I thought Sydney did keep Australian Eastern Daylight Saving Time!
50
For those interested in sea levels, King tides are happening here in Broome, with tomorrow’s a few cm larger, with both days encountering plus 10 metres of movement in the 6 hour period between high and low. This link provides several pics of both. This year was about 10 cm less than 2011 biggest tides.
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Sea levels are unchanged in decades.
70
The Aurora Australis, half buried in ice, at -15C, is ploughing it’s way back to the Davis base to pick up a sick expeditioner.
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Australian carbon sink to remain intact under AGW conditions.
http://www.co2science.org/articles/V18/mar/a19.php
40
An article on prof. Murry Salby’s EIKE lecture, of 13-3-2015.
It is in German though.
http://www.eike-klima-energie.eu/climategate-anzeige/neue-studie-zur-co2-konzentration-anthropogener-anteil-irgendwo-zwischen-0-und-max-30-vortrag-von-prof-murry-salby-am-13315-in-essen/
30
We all know that the BBC is unable to hide its political bias, and its love of all things ‘green’ (just like the ABC), but I have to say my faith in public broadcasters, humanity, and in science education has been, for the last 5 years been restored ever so slightly, albeit temporarily by the ‘Stargazing Live‘ series, even if it’s only over 3 days each year. I have to admit that The Sky at Night is also a bit of a favourite, but sadly, once a month is not nearly enough, in my opinion. I think the success of the Live version is that a lot of it is done live, and it involves ‘normal’ people.
I fear that science education in Australia is now at the point where an equivalent programme made in Australia would probably be more astrology than astronomy. I would very very much love to be proven wrong about this, though. The Science Show and Catalyst are very very pale shadows of their former selves, and are, very sadly, often almost unwatchable or unlistenable unless you like a torrent of sanctimonious climate change propaganda.
For a government which was supposedly ‘better’ for science than the current one, the previous Labor government refused to keep funding the Siding Springs Observatory programme, (initially started via the ANU, using NASA money, which was cut by ‘progressive’ Obama) an AU$140K/year one-man/one telescope operation dedicated to detecting Near Earth Orbit objects (the only one in the southern hemisphere), and determining whether or not they may at some point collide with earth ( a la Chelyabinsk). Much better to funnel money into something CO2 related, rather than something actually useful from both a scientific, and a humanitarian perspective. One would think that if the incredible complexities of climate science had been ‘settled’, then there should be an awful lot of funding available to other branches of science…? Needless to say the funding decision was made a month or 2 before the 2013 election so the person who was laughably referred to as the science minister was nowhere to be seen…
60
In the final analysis we all recognize that global warming, if it is happening, isn’t nearly as much of a threat as the measures proposed for mitigating it.
That would be bad enough if there really was good evidence for a significant warming trend. But there isn’t any such evidence. It’s all trumped up numbers, wild speculation or worse. And so far there’s no credible link to a human cause of any warming that has been shown to have happened by good evidence.
Does that summarize the situation accurately?
I’ve understood and practiced conservation since my Boy Scout days. I’ve run fluorescent lights for years, not only because they were less expensive to run but because they last 10 to 100 times longer than an incandescent bulb. Now we have changes to fluorescent technology that not only makes them less bright but makes them quit much too soon. It’s all downhill when regulation is driven by politics an I draw the line at being told I need to save the planet when it clearly doesn’t need saving.
It’s time for resistance to all this nonsense. If we can do nothing more than speak out and tell the truth we must get busy doing that. And I hate to say it but if we must engage in civil disobedience then we need to be doing that.
Pick battles we can hope to win and start to fight them before it’s too late.
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‘And I hate to say it but if we must engage in civil disobedience then we need to be doing that.’
I buy 100 watt incandescent bulbs and everyone is horrified.
‘Pick battles we can hope to win and start to fight them before it’s too late.’
Global cooling will come as a shock, but under the present circumstances there is little that can be done to turn the world upside down. Contrarian blogs are the best form of civil disobedience, far better than standing outside the ASX with a sign around my neck stating that CO2 does not cause global warming.
20
Didn’t know you could still buy them—except the stock that remained after the ban. I stockpiled them. I use them to keep my well from freezing in cooler weather. My other choice is a 1300 watt milkhouse heater, which my math says takes more energy but apparently the government’s math is different.
20
Is Hydrogen peroxide a “known animal carcinogen”, or is this just the ABC believing whatever the Friends of The Earth say (yet again)…?
50
I should add that I am very familiar with the general properties of hydrogen peroxide, but am just not currently able to do an adequate search for anything resembling papers published on its supposed carcinogenic effects…
40
From a quick look at Google scholar, hydrogen peroxide accumulates in cancer cells (Activated granulocytes and granulocyte-derived hydrogen peroxide are the underlying mechanism of suppression of t-cell function in advanced cancer patients;and Dual role of hydrogen peroxide in cancer: possible relevance to cancer chemoprevention and therapy; and Increased Nox1 and hydrogen peroxide in prostate cancer, etc.).
It would seem to indicate that hydrogen peroxide anomalous generation was a symptom not a cause of cancer.
.
And there is a theory paper (pdf) on ‘Aging: a theory based on free radical and radiation chemistry’
50
And on the other side, we have vitamin C using a hydrogen peroxide to selectively kill cancer cells.
http://www.pnas.org/content/102/38/13604.full
50
Thanks very much! Just to be difficult, I just emailed the Friends of the earth Adelaide office to ask them for the evidence to support their claim. Sadly the ABC article doesn’t have an author attributed to it.
No doubt it will be straight to the recycle bin for daring to question the Church of Green about anything they say, but still, it’s worth a try…
60
Good luck in getting a reply from Anon the writer of such foolishness.
Saying peroxide, or most anything that is not immediately toxic is ‘bad’ for people, is IMO foolish.
These absolutist rules never work. Our bodies, like all of nature’s processes, are too complex. The simplistic mechanical model of our bodily functions is not such a good model. These models rely on degrees of stasis and conformity between individuals that often they are just too unreasonable.
As a point of note my mouthwash contains hydrogen peroxide, also hydrogen peroxide is a very safe antibacterial/anti-fungal for topical use. Women have bleached their hair with it for a century or more. It will also remove – bleach out – rust and blood stains from linens (chlorine bleach will not).
50
back in the olden days, well, 20 years ago, I often used a 30% concentration to remove organic carbon from clay samples before preparing them for XRD and XRF analysis – works a treat, as you’d expect… You too, can turn your black, nasty looking soil into pale brown or white clay in a matter of minutes! Buy now before stocks are sold out!
40
Well, since it’s really unstable water with and extra oxygen molecule I don’t see how it stays bound on its way into the organism especially once dumped into open water. It’s action as an antiseptic is as an oxidizer. Since it’s also photoreactive dumping it in a lake breaks it down fast and simply over oxygenates the water. A standard way of controlling algae that’s usually done with aerating pumps and fountains.
50
the lake in question is a man-made lake of moderate size, and has almost always suffered from various algal problems, controlled slightly with aeration and fountains, but obviously they’ve decided to scare the generally scientifically illiterate populace half to death by using ‘chemicals’.
probably a good thing they don’t see what goes into the tap water they drink then…
30
Organize for Action, Obama’s propaganda wing, held a “contest” they called Climate Change Fantasy Tournament. Senator Inhofe was the winner. I would like to encourage all of you in the USA (and elsewhere, if you think it would help) to let the Senator know how proud you are of him for his acccomplishment. This is what I sent:
“Organize for Action has been running a Climate Change Fantasy Tournament and you have been declared the winner and deemed the worst climate change denier in the country. Congratulations! We need more individuals such as yourself who stand up to the political dogma associated with the global warming scam. Please wear the title proudly!”
The link is http://www.inhofe.senate.gov/contact.
Let’s get behind those who stand up to Obama and his global warming schemes.
90
Done.
80
The more I see of Obama, the more I dislike his reign. Of course, I say ‘reign’, because really, he, and his non-entity wife are treated just like royalty – approaching divine representatives on earth, as opposed to an elected president, and his non-elected and politically irrelevant wife
As an aside, I saw Michelle Obama, the Whitehouse equivalent of a Kardashian (serves no useful purpose to anyone, and has no role to play, but still gets press coverage) recently parading around Cambodia telling the locals how to live their lives – not actually giving them money or resources, just telling people that they should aspire to have things they cannot yet afford (at least this is the idea i get from the coverage I have seen, I could be wrong).
I only hope that for the sake of the American taxpayers, she is paying for this herself, and not sponging off the US government, but I guess that’d be too much to hope for…
I’d be extremely annoyed if the spouse of any Australian politician decided to fly around the world interfering in matters in which they have absolutely no business, and no authority. We have (and i am sure the USA has) diplomats and elected representatives perfectly capable and duly authorised to wander around being sanctimonious and shallow…
80
No, we’re stuck with the bill. She would (supposedly) pay back the equivalent of a first class airline ticket towards the millions of dollars her little jaunts cost. We pick up the rest plus security. Don’t get me wrong, she needs the best protection we can give her but she’s been a pretty sorry excuse for a First Lady. I thought Bubba and Hillary were bad… these two take the cake when it comes to being king and queen wannabes. I don’t know if we survive another 22 months of their Progressive reign of terror.
60
I think what’s equally annoying is that the Obamas seem to genuinely think they are making some sort of positive impact on the world… Arguably, people who suffer from such delusions should not have lots of power, lots of money, and control over nuclear weapons.
40
She was probably comfortable in Cambodia.
They are not “whites”.
She hates “whites”.
“Hates” is a very strong/emotional word, isn’t it?
She hates “whites”.
10
‘The hockey stick graph dealt with all those by eliminating the MWP and the LIA. It inappropriately tacked on, as the blade of the stick, an upturn in temperature in the 20th century. Phil Jones produced the upturn that claimed a 0.6°C ±0.2°C increase in 120 years. They claimed this rate of increase was beyond any natural increase, conveniently ignoring the ±33% error factor.’
Tim Ball
40
James Hansen took Venus with 96% CO2 atmosphere, and closer to the sun than earth, very hot, and then “modeled” the effects of CO2 on earth rising from 0.03% (that’s not 3%, it’s one-hundredth of 3%) to 0.06%, which is to say, he thought 96% CO2 on a closer to the sun planet had lessons for a planet with 99.04% NOT C02. Massively dominant gas fraction compared to lower than minuscule gas fraction. Conclusion: the earth is going to become uninhabitable.
And skeptics, remember, remember, remember: the AGWer “scientists” refused to release their reports’ raw data, computer codes for analyzing their data, but insisted that their finished product results were unimpeachable, and all that anybody needed to see.
As in Michael Mann’s exponential 20th century temp rise, imputed tone caused by linear CO2 rise, but the exponential hockey stick should have been a logarithmic curve from 260 ppm CO2 to 350. And nobody even challenged the incongruity, Nobel Prize Winner Arrhenius, and genius Josef Fourier, versus the guy who didn’t get his PhD until age 32–straight study from BA. which took 5 years, not 4, Mann not somebody who took some years off working outside academia. It took 9 years enrolled in grad school to earn his PhD. And the IPCC’s TAR showcased his work!
Okay, Dr. Mann, your research was paid for by USA taxpayers, who gave you grants and jobs. Show us, your employers, your raw data and codes, an explain why you decided to ignore raw data, and why, using bristlecone pine tree rings you didn’t examine samples from the complete bristlecone pine biome ranging from Sheep Mountain, CA to Nevada and Utah, to determine old climate across-the-western-US high-altitude “climate”. A real scientist would have taken the Sheep Mountain tree rings and said, “Let’s look at the rest of the bristlecone areas.let’s see what they tell us.”
And this doesn’t even tell the tale of why Mr. Mann, a third-class-honors student didn’t get placed at first-rate physics programs: MIT, Harvard, Caltech, Princeton, Stanford,Cambrid Berkeley, but at second-rate Yale (zero Nobel laureates, only 4 NAS Physics Section members, now down to 2). Berkeley profs decided, “He’s not one of our brilliant minds, he’s not even one of our very bright minds who do not score in the top 3 of his classes, but nevertheless consistently hit top-10. percentile. Mr. Mann was in the bright category, by taking light loads, top 20th-30th percentile.
Top Berkeley grads have A/A+ grades in Honors Math and Physics lower division courses. Let Mr. Mann release his courses and grades. If he refuses, then you know where he’s resting his “laurels”. I mean, A/A+ grades in Berkeley lower division math and physics courses require hard study, no pot smoking. If Mann can demonstrate these, let’s hear him out.
Also, why he was recommended, but turned down opportunities to go to MIT, Harvard, Stanford, Princeton, Cambridge, Caltech, Berkeley, Chicago, Cornell, UIUC, for his PhD, except he wasn’t recommended and didn’t get in.
40
March in the US normally has tornado activity, but its been quiet this year. There have been suggestions that the cold winter may have had an impact.
Anyway, the climate trend indicates that they are on the wane.
https://notalotofpeopleknowthat.files.wordpress.com/2014/10/image119.png
10
Did we unload, unlead or unthread?
20
`Yes!
20
Just a belated footnote to say that, although I’ve been happy to vote for our famous hostess in previous years, unfortunately her credibility took a turn for the worse in 2014. She continues to claim it is possible “the ocean is causing the rise in CO2”, therefore I have not voted in the Bloggies for her blog and neither will I be tipping the choccy jar for the foreseeable future. Counterfactual conclusions supported only by easily disproven arguments, such as Salbyism, have no place in the skeptic’s toolbox.
10
Andrew,
Interesting She continues to claim it is possible “the ocean is causing the rise in CO2″!
If for unknown reasons ocean surface temperature increases by 0.03 kelvin would there not be a measurable increase in atmospheric CO2? This is not only possible it is proven measurable, with no foolish anthropoids being involved.
22
There is absolutely no warming sensitivity for any IR-active gases like water vapor, carbon dioxide and methane. The force of gravity produces a temperature gradient in a planet’s troposphere which, for an Earth with dry air, would raise the mean surface temperature to about 25°C to 28°C. But then inter-molecular radiation between water vapor molecules has a temperature leveling effect and thus reduces the surface temperature because it makes the gradient less steep, as is well known.
What is the sensitivity for each 1% of water vapor in the atmosphere?
We know that water vapor reduces the magnitude of the so-called lapse rate, which does not need a special name because it is just the temperature gradient in the troposphere. Now, we also know that radiative balance at the top of the atmosphere is virtually always close, in fact to within ±0.5%. The inward radiation is based on the so-called “Solar Constant” (although that does vary, especially due to variations in Earth’s eccentricity in a ~100,000 year cycle that regulates glacial periods) and the outward radiation (broadly speaking) increases if the whole temperature plot rises, making the area under that plot greater.
Now, what the AGW crowd want you to be gullible enough to believe is that (as the percentage of water vapor increases) the thermal plot can rise at the surface end whilst at the same time acquiring a less steep gradient. Any secondary student with a knowledge of coordinate geometry would know that the area under the new (higher and less steep) thermal plot would be far greater than that under the original plot for a world with less water vapor. So how could that happen? It can’t, because the whole plot would then fall to regain radiative balance, and in fact it would have just rotated downwards at the surface end in the first place. So how could the sensitivity for each 1% of water vapor be positive causing warming of the surface?
11
China is supposedly running a temperature.
http://www.chinadaily.com.cn/china/2015-03/24/content_19894676.htm
Its plausible, methinks its got something to do with UHI.
20
el gordo
Have they over-homogenized?
10
A few years ago the Klimatariat was complaining about China’s UHI, they weren’t playing ball. Forget the details, but what we see maybe unadjusted.
10
Infrastructureinfrastructureinfrastruucture
http://www.chinadaily.com.cn/china/2015-03/28/content_19934711.htm
Austria, South Korea and Britain intend signing up and Abbott’s diffidence will soon be swept aside.
Its Keynesian.
10