Direct Heat Removal



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Let's take a few minutes to estimate the heat sent to Outer Space using the Stefan-Boltzmann Law.

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Here is some Background for Stefan-Boltzmann Law

From JILA, a joint institute of University of Colorado Boulder and NIST

https://jila.colorado.edu/~ajsh/courses/astr1120_03/text/chapter1/SBLaw.htm

Portions (c) Copyright University of Colorado Boulder and NIST, All Rights Reserved

SBLaw

The Stefan-Boltzmann Law

The Stefan-Boltzmann Law was first discovered experimentally in 1879 by Stefan, then derived theoretically in 1884 by his student Boltzmann. It says:

L = As T4

Note the strong dependence on temperature: -- as the fourth power of the temperature. I.e., double the temperature, and the radiated power will increase by a factor 24 = 16. Triple the temperature, and the power will increase by a factor 34 = 81, etc. The Stefan-Boltzmann Law is valid only for perfect radiators (called "blackbodies"). Actual radiating surfaces are not perfect radiators, and will always radiate less than the luminosity given by the S-B Law -- typically some 10 - 80%. But the radiating surfaces (called "photospheres") of stars are fairly good approximations to black bodies, and typically radiate a luminosity of more than 90% of the value given by the S-B Law.

The value of the Stefan-Boltzmann constant, s = 5.67 x 10-5 ergs cm-2 s-1 K-4, is determined by experiments in Earth labs. We assume that the S-B Law, with the same value of s, is valid everywhere in the universe. That assumption is one example of the Principle of Universality of Physical Laws.

Examples:

1. The Earth: The average temperature of the Earth's surface must be about 10 C. We must first convert this to the Kelvin scale: T(K) = T(C) + 273 = 283 K. The Earth's surface area is about A = 4p R2 = 4p (6.4 x 108 cm)2 = 5.15 x 1018 cm2. Now we can calculate the radiated luminosity of the Earth: L = 5.15 x 1018 x 5.67 x 10-5 x (283)4 = 1.87 x 1024 ergs s-1. To change ergs s-1 to Watts, multiply by 10-7. So, L = 1.87 x 1017 Watts. (Actually, the Earth radiates only about half this much luminosity into outer space because it is not a black body.)

2. The Sun: The temperature of the Sun's photosphere is 5800 K and its radius is 7 x 1010 cm. So the Stefan-Boltzmann Law gives a luminosity L = 4 x 1033 erg s-1 = 4 x 1026 Watts.

3. Your body: (This calculation will be a bit more subtle because you must correct for the temperature of the environment.)

Your body temperature is about T(F) = 98.6 F. To use the S-B Law, we must convert this to the Kelvin scale. To do this, first convert your body temperature to Centigrade: T(C) = (5/9)[T(F) -32] = 37 C. Then, to convert this to Kelvin, add 273: T(K) = T(C) + 273 = 310 K. Now, let's estimate your body surface area. Approximate your body shape roughly by a rectangular block, of height 180 cm, width 30 cm, and depth 15 cm, giving a surface area A = 2[180 x 30 + 180 x 15 + 30 x 15] = 17,100 cm2. Now we can calculate the luminosity of your body: L = 17,100 x 5.67 x 10-5 x (310)4 = 8.95 x 109 ergs s-1. So, according to the S-B Law, your body could radiate a luminosity L = 895 Watts. That's a lot! But this would only be true if your body was in outer space where the temperature of the sky is very cold -- about 5 K -- and if you were naked. If you wore an insulating space suit, you would radiate much less than this because the surface temperature of the suit would be much colder than your body. On Earth, you won't lose nearly as much heat as this, because your body is absorbing heat from its surroundings at the same time it is radiating. To correct for this absorption, you should use a modified Stefan-Boltzmann Law:

L = As (T4 - Tenv4)

Where Tenv is the temperature of the environment. Let's say that Tenv = 290 K (about 63 F). Using this formula, the calculation gives L = 210 Watts. But, if you put your clothes on, you can cut this radiation down by a big factor -- say, a factor 4 with light clothes and a sweater. Then you are losing about 50 Watts, the equivalent of about 1 light bulb.

The Stefan-Boltzmann Law Worksheet Sample Values

To check the calculator (below) with the sample problems (above), use the following values:

1. The Earth:
Area (M2): 515000000000000 M2                  (5150000000000000000 cm2) (5.15 * 10**18) cm2
Temperature C: 10 C
Results (Watts): 187299246417160480

2. The Sun:
Area (M2): 6150000000000000000 M2          (61500000000000000000000 cm2) (4 * 3.14 * ((7*10 ** 10)**2) cm2)
Temperature C: 5527 C
Results (Watts): 3.94611873768e+26

3. Your body (First Example):
Area (M2): 1.71 M2                  (17,100 cm2)
Temperature C: 37
Results (Watts): 895


The Stefan-Boltzmann Law Worksheet Calculator
Instructions: Fill in Area (M2) and Temperature (C) variables, Click "calculate"

                   Area (M2): * >> cm2
Temperature (Deg. C): * >> K
                       Results: Watts MW

  * Required


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Direct Heat Removal with Stefan-Boltzmann Law

4. Direct Heat Removal Example
Use the Worksheet Calculator below to estimate Heat transmitted to outer space. Adjust scenarios as desired.

Description and Example: 6000 Acre Direct Heat Removal Solar and Emitter farm, size like Spotsylvania, VA.

Solar farm 617 MW (daytime), solar panels fold down at night and Emitters exposed (nighttime).
Emitters set at night and maintained at temperature of 38C (100F), close to desert high temperatures.
Desert temperature drops to -3.9C (about 25F) at night.

Area (M2): 24280000 M2          (6000 Acres)
Temperature C: 38 100 F
Results (Watts): 12878737681

The Direct Heat Removal Stefan-Boltzmann Law Worksheet Calculator
Instructions: Fill in Area (M2) and Temperature (C) variables, Click "calculate"

                   Area (M2): * >> acres
Temperature (Deg. C): * >> K
                       Results: Watts MW

  * Required

The Stefan-Boltzmann Law Worksheet Adjustments

You may feel that the Results do not represent the actual energy transferred to outer space. To make an adjustment, enter a percentage Adjustment and hit Adjust.

Adjusted Results: Watts MW

Adjusted Percentage:


Observations and Conclusions:

Scenario 1: During the daytime hours, the solar panels charge up the Storage Batteries for use during the night.

During the night, the Emitters are exposed to the nighttime sky and the emitter plates are brought to temperature and maintained at 38C (100F) emulating the highs of the desert, as long as battery power is available. Note that at 100% Adjustment, the Emitter's radiated energy to outer space could be as high as 12,879 MW. You will likely want to try some adjustments to judge the feasibility.

In nature, the temperature of the desert drops to as low as -3.9C (about 25F) at night.

The Emitter plates or the natural environment would radiate much less energy to the nighttime sky, if that lower temperature of the desert were maintained.

Scenario 2:

It has been suggested that maintaining the emitter plates at a constant 3C for the entire nighttime should be attempted. The user can work on this scenario.

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