THE FAMOUS WOOD’S EXPERIMENT FULLY EXPLAINED

Written by Alberto Miatello

miatellomoistair

(Why Wood and Nahle were correct and Pratt was in error.)

 

Introduction

The famous experiment by Robert W. Wood, at John Hopkins University, with two carton boxes/greenhouses, in 1909, is being mentioned everywhere, and on many websites,* as simple experimental evidence proving the fallacy of the greenhouse gas effect theory (GHE).

According to the GHE theory, the small greenhouse with a glass cover had to reach a temperature of nearly 15°C higher than the other small greenhouse with a salt rock (halite) ceiling. This is because salt rock is a material which is “neutral” to infra-red, while glass can theoretically “trap” almost 80-85% of infra-red outgoing from the heated bottom of the greenhouse, and significantly increase the temperature, by “backradiating” the infrared (IR) waves.

Nothing of that took place, and both greenhouses showed almost the same temperatures inside, with a discrepancy of “scarcely one degree”. For years this experiment was sufficient to dispel giving any scientific ground to the greenhouse gas effect theory. But several decades later, many GHE advocates “forgot” this experiment.

 

One hundred years on, in 2009, Professor Vaughan Pratt of Stanford University (Palo Alto, California) tried to replicate the Woods experiment using more modern materials (plastic plates and foils, along with the “old” glass plates).** Pratt came to the conclusion that Wood’s experiment was in error, because according to Pratt’s surveys the glass and acrylic greenhouses showed temperatures 15°C and 20° C higher than the one inside the other small greenhouse with a thin polyethylene film cover.

Thereafter, in 2011, Professor Nasif Nahle of Monterrey, Mexico performed his own very accurate repeat experiment using four small greenhouses under strict peer-reviewed control.*** Nahle came to the conclusion that Wood’s experiment was totally correct. Nahle’s findings were that in the three small greenhouses having covers of different materials (glass and plastic polymers) and upon one hour of solar exposure, the temperature differences were scarcely in the range of 1° to 1.5° C (as in the Wood’s experiment a century before). Nahle saw that the other “holed” greenhouse – more exposed to cooling convection and environment temperature – showed a lower temperature. This was compelling proof that a greenhouse is heated merely by the blocking of air convection with the outside environment and not by any specious mechanism(s) such as “backradiation” or the “trapping” of longwave outgoing infra-red radiations.

However, this article was written to address a “gap”; many sources and commentators mentioned the experiments above, but curiously nobody – as far as reasonably known – sought to explain (by rigorous physical and mathematical analysis) why Wood and Nahle were right yet Pratt was wrong.

Furthermore, it is usually said that Wood’s experiment is “easily replicable”, even by high school children. Although this assertion could be true, at least for the low cost and availability of the materials, it takes a lot of skill and attention to detail to carefully assemble them and conduct the experiment, as we shall see.

But it is always necessary for any researcher/physicist to try and turn into physical laws and mathematical formulas any physical experiments in order to get rigorous and scientific understanding – as well as acquire deeper insight of it.

Moreover, a technical physics analysis of those experiments, can better clarify the real mechanisms of heat transmission between soils and atmosphere, or solid and gaseous materials.

 

1) 1909: Wood’s experiment

 

Wood took two small carton greenhouses: the first with a halite ceiling, the other with a glass ceiling. We only know from his narration that the thickness of the glass and halite plates was the same, although Wood didn’t mention the exact size.

After some time (we don’t know how much) the interior temperature of the halite greenhouse rose to 65°C (338 K). This means that – according to the Stefan-Boltzmann equation:

 

W/m2 = σT4 (1)

5.67 * 10^-8* 3384 = 740 W/m2

Now, according to the Arrenhius’ GHE theory, the “backradiation” of the glass ceiling should have increased the temperature of the glass greenhouse 20% more than the temperature inside the halite greenhouse, i.e.: 740 + 740* 0.2 = 888 W/m2

Therefore:

888 W/m2 = σT4

T = (888/5.67* 10^-8) ^ 0.25

This means that – according to the Arrenhius theory – the glass greenhouse should have reached a temperature 15° C higher than the halite greenhouse.

However, Wood reported that the temperature of the halite greenhouse was “a little ahead” than the glass greenhouse with scarcely one degree of difference between them. Then, Wood filtered the Sun radiation entering both greenhouses with a plate of glass, and this had the effect of reducing the temperatures of both greenhouses to 55° C, again without significative difference between them.

For decades this experiment by Wood seemed to plant a definite gravestone on the greenhouse effect theory of Arrenhius.

 

2009: Pratt’s experiment

 

In November 2009 Professor Vaughan Pratt of Stanford University conducted another experiment to replicate Wood’s experiment using 3 small carton greenhouses. The sizes of these were 30 x 30 x 15 cm. (12” x 12” x 6” inches).

Box 1 was covered with a 0.015 mm. polyethylene film (thermal conductivity 0.5 W/mK)

Box 2 was covered with a window glass, thickness 2.38 mm. (3/32”), thermal conductivity 1W/mK

Box 3 was covered with an acrylic plate, thickness 9.5 mm. (3/8”), thermal conductivity 0.25 W/mK

 

Pratt reported:

“Very preliminary tests conducted on November 28, 2009 indicate that boxes 2 and 3 run respectively 15 and 20 degrees hotter than box 1, with a significantly larger drop across Box 3’s window than Box 2’s. While this is consistent with the expected differences obtained by theoretical considerations, it is so far removed from what Wood found as to raise the very interesting question of how he was unable to observe any significant difference between his two boxes. “

After one week Pratt conducted another experiment:

“We repeated the previous week’s experiment with boxes 1 and 3 increased from single glazing to double glazing. The main concern with box 1 had been that its thermal resistance was negligible. We dealt with this by adding a second Saran wrap window with a 6 mm (1/4″) air gap between the two windows, a standard for some greenhouses. Air has 40 times the thermal resistance or R-value as glass, and 10 times that of perspex, so this gap would correspond to a ten-inch glass slab. To be sure that there was no convection we divided this thin layer of air into 2″ x 2″ cells with 1/4 ” strips of acetate cut from blank overhead transparency slides.

……

At 2 pm on 12/03/09 the box with the double-glazed Saran wrap window reached 65 C while the box with the two 3/8″ perspex sheets reached 80 C. (We subseqently recorded 82 C with the latter.) During the previous hour the former climbed rapidly to its limiting temperature while the latter was much slower. Had we settled for an early measurement we would have concluded that box 1 was the hotter. However box 3 continued to rise and eventually overtook box 1 and went on to become 15 C higher. “

To summarize: in his first experiment Pratt’s carton greenhouses 2 and 3 reached temperatures 15° C and 20° C hotter than Box 1; even in the second experiment (by adding a second plate to the greenhouses) the temperature difference between Box 1 and Box 3 (no word regarding Box 2 with glass cover) was about 15° C.

Now, to better understand how those small greenhouses above got heated, we should calculate:

 

  1. The heat fluxes (q”) inside the greenhouses

  2. How much those heat fluxes could really heat the small greenhouses.

 

a) It is important to notice that – for technical physics and heat transmission purposes– the small carton greenhouses can be considered as multi-layers materials: a first 0.15 m. layer of air + a second plastic/glass layer having very small thickness.

To calculate the heat fluxes in a multi-layer material we have first to calculate the thermal resistance (R”) of the material, according to the eq.:

 

R” = L/ A*k (2)

Where L = thickness

A = area of the surface

K = thermal conductivity of the material

 

Now, for air we have:

L = 0.15 m

A = 0.09 m2

k = 0.027 W/mK (mean conductivity of air at 30°- 40° C), and so:

R” = 0.15/0.00243 = 61.7 mK/W

 

Whereas, with respect to the thermal resistance of thepolyethylene cover of Box 1, we have:

L = 0.000015 m.

A = 0.09 m2

k = 0.5 W/mK and so:

R” = 0.000015/0.045 = 0.0003 mK/W

 

The total thermal resistance (air layer + polyethylene layer) inside Box 1 is:

R” tot 1 = 61.7 + 0.0003 = 61.7003

 

Then, with respect to the Box 2, with glass cover, we have:

R” = 0.0023/ 0.09 = 0.025

 

Therefore, total thermal resistance inside box 2 is:

 R” tot. 2 = 61.7 + 0.025 = 61.725

 

And finally, with respect to the thermal resistance of the acrylic cover of Box 3, we have:

 R” = 0.0095/0.022 = 0.43

 

Hence, the total thermal resistance inside Box 3 is:

R” tot. 3 = 61.7 + 0.43 = 62.13

 

But our calculations above can immediately evidence that even the thick plastic cover of the box 3 has no significative influence on the thermal resistance of the heat fluxes inside the little greenhouses. Actually even in Box 3, 99.993% of thermal resistance was produced by the air layer, and just 0.007% by the acrylic cover.

 

Therefore, we can now calculate the heat fluxes inside the 3 boxes, according to the formula:

q” = U * Δ T (3)

where U is the global coefficient of thermal exchange, namely: 1/R” tot.

 

So for Box 1, considering a temperature interval (Δ T) to raise an outside temperature ambient around 32° C up to 60° C inside the boxes, we have Δ T = 28° C

q”1 = 0.0162 * 28 = 0.453 W

q” 2 = 0.0162 * 28 = 0.453 W

q” 3 = 0.0161 * 28 = 0.450 W

 

Now, as we can see the only small difference in the heat fluxes of the 3 boxes is between Box 1 and Box 3. It is:

Δ q” 1-3 at 60° C = 0.453 – 0.450 = 0.003 W

 

We can also calculate how large is the difference in heat fluxes of the boxes when the highest temperature (80° C) was reached, considering an ambient average temperature of 32° C, namely for a Δ T = 48° C (80 – 32)

q” 1 = 0.0162 * 48 = 0.777 W

q” 2 = 0.0162 * 48 = 0.777 W

q”3 = 0.0161 * 48 = 0.772 W

Δ q” 1-3 at 80° C = 0.777 – 0.772 = 0.005 W

 

b) At this point we can find how much thermal energy (Q) is requested to heat those small greenhouses:

The Volume of each greenhouse is: 30 * 30 * 15 = 13,500 cm3

namely 0.0135 m3(= 13,500/1000,000 cm3)

 

Now, according to the basic eq. of thermology and energy balance, we find the total thermal balance to heat a body:

Q = C * M * Δ T

 

Where

Q = total thermal energy

C = specific heat

M = mass

Δ T = temperature interval (T – T°)

 

We know that the specific heat of air at 30°-40° C is nearly 1,007 Joule/Kg.K, whereas mass of warm air is about 1.15 Kg/m3

Therefore, the energy we need to heat the air of those greenhouses for 1° C is:

Q = 1,007 * 0.0135 * 1.15 * 1

Q = 15.63 J/°C

 

Since we know that the heat flux difference between the polyethylene and the acrylic boxes, to raise the temperature up to 60° C is 0.003 W, i.e. 0.003 joule/sec., in 1 hour (3,600 sec.) time interval, we have a thermal energy flux difference for:

Δ q” 1-3 at 60° C = 0.003 * 3,600 = 10.8 Joule/hour

 

Whereas, to raise the temperature up to 80° C, it takes:

Δ q” 1-3 at 80° C = 0.005 * 3,600 = 18 Joule/hour

But, as we know that it takes 15.63 Joule to raise the temperature for just 1° C in those small greenhouses, then we can find that, the heat flux difference between the boxes, can raise the temperature for no more than:

  1. T =10.8/15.63 = 0.7° C (with fluxes at 60° C max)

    b) T = 18/15.63 = 1.15° C (with fluxes at 80° C max)

 

This means that a calculation according to the established laws of technical physics can evidence that Robert Wood was right, because the difference in covers of the greenhouses and in heat fluxes can scarcely produce a difference in the range of 0.7°/1.1° C, as Wood himself reported; but surely

not as great as 15° C or 20° C as Pratt found.

 

Now the question is: why did Pratt record so large a difference in the temperatures inside the boxes?

Let us address that by again examining Nahle’s experiment.

 

2011: Nahle’s experiment

 

The last experiment was conducted by Professor Nahle of Biology Cabinet (Monterrey, Mexico) on May 20, 2011, at San Nicolas de los Garza (Mexico). This was done in the morning at 10:00 a.m. until 11:00 a.m. Nahle very carefully arranged and described his work in order to give a full and very detailed account of it and leave no room for doubts and ambiguities.

On the Biology Cabinet weblink Professor Nahle makes it possible to find full details of the experiment; everything from the mode of construction of the cardboard boxes to the quality/brand names of the glue and painting, to the accuracy of thermometers, etc. As such it is possible to obtain greater certitude from his table of the changes of temperature inside the four small boxes (at intervals of 5-10 minutes) coupled with the additional visual verification of his photos.

We see Nahle used 3 different types of cover for the boxes:

 

  1. Polyethylene film, 0.3 mm. thickness

  2. Silica Glass panel, 3 mm. thickness

  3. Acrylic Plate, 3 mm. thickness

 

Box n°4 had a cover with acrylic holed plate, in order to allow some form of convection with the surrounding ambient. Note that the four boxes were of the same size: 30 x 30 x 20 cm making them slightly bigger than Pratt’s boxes.

It is also interesting to note from the table that in the hour of the experiment, at direct exposure of sunlight and ambient temperature from 32° C to 37° C, the boxes always showed very similar temperatures, in the range of 0.5° – 1.8° C. That is, apart from the first 10 minutes from starting, in which the polyethylene box was kept in the shadow. Thus starting from a temperature nearly 10°-12° lower than the ones of the other boxes.

This is in full accordance with the experiment conducted by Wood in 1909 and with the expected results of the experiment by Pratt in 2009, as calculated above.

Furthermore, considering the sizes of Nahle’s boxes and their thermal resistance, which was very similar (82.30 for polyethylene and 82.33 for glass box), we could have theoretically expected a very low difference in heat fluxes (Δ q” at 80° C = 0.0005 W). That is just 1.8 joules of energy in one hour inside the boxes. As such the temperatures should have been very similar with a range of less than 0.1° C of recorded temperatures.

The four “hand crafted” boxes were very similar but not identical. Thus even small differences ( e.g in the drying of paint, humidity of the cardboard, etc., plus accuracy of the thermometers (which was in the range of +/- 0.3° C for internal temperatures) can lead us to the conclusion that a difference of between 0.4° – 1.4° C in the temperatures of the air inside the boxes after one hour is is reasonable and correct. As such, we may come to the conclusion that both Wood’s and Nahle’s experiments have proved that infra red radiation is not “trapped” by glass, or by polyethylene molecules.

 

The importance of moist air inside the boxes.

 

But now, to better understand why the results from Pratt’s experiment showed such a wide difference between the boxes (at around 15° C) we have to consider another very important element: the percentage of humidity/moist air in the boxes.

Actually, even a little change in the content of water vapor/water molecules inside the boxes can dramatically change the % of humidity, from an almost dry to a very humid air, from a 10% humidity up to 90-100%.

We must not forget that a change from a dry to a moist air has also the consequence to change the thermal conductivity of air, sometimes very significantly.

In the diagram below we see what happens to the air conductivity when the temperature of air increases from dry (0% humidity) to very humid air (100% humidity). ****

 

miatellomoistair

 

Contrary to widespread but false belief, when the humidity of air increases then thermal conductivity decreases. This is particularly evident with temperatures above 40°-50° C, curiously these are the very same temperatures reached in the above experiments with the carton greenhouses.

Therefore, if the humidity inside Pratt’s acrylic greenhouse was much higher than that in the polyethylene greenhouse, this can easily explain why a difference of temperature up to 15° C , or even more, was recorded.

Indeed, if the cardboard boxes have been kept in different rooms, or somehow had varying solar exposure and shadow almost imperceptible differences in humidity could result and impact the final readings. Nahle himself was mindful to warn of this and urged caution to prevent condensation and moisture inside the boxes, otherwise the experimenter’s results would become “unpredictable”.

As such, we can easily show what happens when there exists moist air discrepancy inside the two small boxes in Pratt’s experiment. Now, we can suppose – according to the diagram above – that polyethylene Box 1, contained very dry air inside at around 10% humidity, whereas the acrylic box had moisture content around 90%.

This means that when the temperatures inside the two boxes attains 50° C then we can have a difference in thermal conductivity between humid and dry air for 0.00025 W/mK, on average.

Therefore, we can suppose that the air in the polyethylene Box 1 (dry) during the experiment had a mean thermal conductivity around 0.00295 W/mK, whereas the acrylic Box 3 (moist) had an average thermal conductivity around 0.0027 W/mK.

Now, if we introduce the new magnitudes in the parameters above, at point 2), we can find that Box 1 (dry) had a thermal resistance:

R” 1 = 0.15/0.09 * 0.0295

R” 1 = 56.6

R”tot 1 was: 56.6 + 0.0003 (polyethylene R”) = 56.6

 

The other, Box 3 (moist) had a thermal resistance:

R” 3 = 0.15/0.09 * 0.027

R” 3 = 61.72

R”tot 3 was : 61.72 + 0.43 (acrylic R”) = 62.15

 

Hence, the global coefficient of thermal exchange (U = 1/R”) was:

for Box 1: U =1/56.6 =0.0176

for Box 2: U = 1/62.15 =0.0160

 

And so, for a total heat exchange from 32° C to 60° C (Δ T = 28° C) in one hour, we have:

for Box 1, q” = 0.0176 * 28 =0.492 W

for Box 3, q” = 0.0160 * 28 =0.448 W

Δq” 60° C = 0.492 – 0.448 = 0.044 W

 

So,Δq” * 3,600” = 158.4 Joule thermal energy in one hour

 

But, as we know from point 2) above that it takes 15.63 Joules to raise the temperature for 1° C, then:

158.4J/15.63J = 10.1° C

 

which is the difference between Box 1 (dry, lower temperature) and Box 3 (moist, higher temperature)for a max. 60° C temperature. We can make the same calculation for a total heat exchange from 32° C to a max. 80° C temperature, i.e Δ T = 48° C, and we will have:

for Box 1, q” = 0.0176 * 48 =0.844 W

for Box 3, q” = 0.0160 * 48 =0.768 W

Δq” 80° C = 0.844 – 0.768 = 0.076 W

So,Δq” * 3,600” = 0.076 * 3,600 = 273.6 Joules

difference in thermal energy in one hour

 

And finally:

273.6J/15.63J = 17.5° C

 

This is a fairly precise calculation to explain the difference in temperatures between a dry and a humid box for a max 80° C temperature. Of course, in the real experiment we didn’t have a constant average ambient temperature (32° C). But we had a changing (probably growing) ambient temperature, and so it could have taken less than one hour to reach 80° C inside the hottest box.

As such, this fully explains why Pratt found such a large difference of 15° C and more between the temperatures of the boxes. Whereas Wood and Nahle found discrepancies in a very smaller range, from scarcely one degree to less than two degrees. This was not due to the fancy mechanism of “back radiation”. More likely it was simply a problem of difference in the thermal conductivity of dry and moist air, plus thermal resistance inside the carton boxes.

But that’s also what is constantly taking place in our atmosphere. There’s no need to resort to unscientific and false mechanisms such as the “backradiation” and the GHE. To explain the changes of temperatures between a dry and moist low layer of the atmosphere (apart from all the convective phenomena, and the water cycle) it is sufficient to calculate the change of thermal conductivity + lapse rate from dry to moist air or vice-versa.

 

Let us be clear: moist air is less thermally conductive. Thus outgoing heat fluxes from the Earth’s surface to the atmosphere find more thermal resistance than in dry air. By the same measure, this rule applies to experiments like those above if there exist risk of discrepancy in the humidity of such containers. This risk most obviously applied to Pratts’ experiment which was absent any such stated humidity controls (whereas Nahle these controls). As such Pratt’s experiment was liable to a conceivable contamination from a discrepancy induced by variance in the thermal energy accumulation in the boxes. Undoubtedly, this would cause error in Pratt’s results invalidating his conclusions whereby humid air would see a higher temperature rise compared with a dry air environment, despite receiving the same heat fluxes.

 

Alberto Miatello (May 2012)

 

References:

*One prominent example: O’Sullivan. J., ‘Greenhouse Gas Theory Trashed in Groundbreaking Lab Experiment,’ (July 18, 2011), climaterealists.com (accessed online: June 6, 2012).

** Pratt, V., ‘Failure to duplicate Wood’s 1909 greenhouse experiment, (January, 2012), http://boole.stanford.edu/ (accessed online: June 6, 2012).

***Nahle, Nasif S. Repeatability of Professor Robert W. Wood’s 1909 experiment on the Theory of the Greenhouse,June 12, 2011, Biology Cabinet Online, Academic Resources. Monterrey, N. L. **** Lasance, C.J.M., ‘The Thermal Conductivity of Moist Air,’ (November 1, 2003), http://www.electronics-cooling.com, (accessed online: June 6, 2012).

 

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Comments (1)

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    jackorr

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    In the calculation of thermal resistance of the 15cm of air, did you take into account convection within that air? I’d have expected it to be significant, dramatically reducing the figure of 61.7 mK/W, casting doubt on most of the conclusions stemming from that figure

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