Thermodynamics - Brain fusion in progress

[N_Molson]

Hello, :tiphat:

I'm desperatly trying to get an usable result from this equation, but until now I miserably failed. Can someone give it a try ? Thanks.To be useful, the result should be expressed in Watts/m², or Joules/s/m²

Net Radiation Loss Rate

If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as :

q = ε σ (Th4 - Tc4) Ac

where

Th = hot body absolute temperature (K)

Tc = cold surroundings absolute temperature (K)

Ac = area of the object (m2)

σ = Stefan-Boltzmann constant

ε = emissivity

My values :

Th = 439.45 Kelvins
Tc = 2.37 Kelvins
Ac = 42.4 m²
ε = 0.04

Thanks in advance, my brain begins to overheat wned::sos:

 

[Fizyk]

Well, you got the equation, the only thing you need to do is divide both sides by the area (to get the power flux, not the power):

[math]I = \epsilon \sigma \left( T_h^4 - T_c^4 \right) = 84.58 \frac{W}{m^2}[/math]

 

[tori]

The result is 3509.56 W.

You're probably looking for specific heat radiation rate, in which case drop the Ac from the equation (82.77 W/m²).

 

[N_Molson]

By that you mean that each square meter of the object is emitting 84.58 Joules each seconds ?

I'm trying to get how much energy the object radiates in space each second.

 

[tori]

Yes. (my result was slightly different probably due to [strike]lack of rounding[/strike]approximated SB constant)

 

[Fizyk]

The result without any rounding is [math]84.5885464220509168820224 \frac{W}{m^2}[/math] (assuming [math]\sigma = 5.6704 \times 10^{-8} \frac{W}{m^2 K^4}[/math] - the value from Wikipedia)

By that you mean that each square meter of the object is emitting 84.58 Joules each seconds ?

Yup.

EDIT:
tori: I know what you did wrong. You calculated [math]\left(T_h - T_c\right)^4[/math] instead of [math]T_h^4 - T_c^4[/math]

 

[N_Molson]

Thank you very much. So I guess I was completely wrong there :

(because if I'm not, there a serious inbalance between the energy input and output)

http://www.orbiter-forum.com/showthread.php?t=20465&page=4

Let's assume that the surface area of the spacecraft cabin is a 3*1.5 meters cylinder. (1.5 for radius)

I want to calculate the total amount of energy absorbed by the hull, when the spacecraft is pointed at 90° of pitch/yaw (reference : Sun).

The total surface area of the cylinder is 42.4 m². I remove the two "caps" on the top and the bottom ; and I get 28.3 m².

I assume that the hull is made of polished aluminium, which emissivity is 0.04 (almost a mirror), much like the Apollo spacecraft. Using the Stefan-Bolzmann equation, I get two results :

- The half of the hull exposed to the Sun has a temperature of 881.6 °K.
- The radiant flux of each exposed square meter of hull is 1370 W/m².

We will assume that the dark side of the hull is exposed to 2.7 K, the cosmic background radiation, and we'll neglect planets albedo for simplicity.

The average temperature of the cabin hull, assuming that we have an efficient heat transfer all over the hull (aluminium is highly conductive), is 439.45 K.

The part of the hull exposed to the Sun is 28.3/2 = 14.15 m².

Then the total radiant flux is 1370 W * 14.15 m² = 19385.5 W, or 19385.5 J/s.

In the case in which the spacecraft is facing the Sun (0° of pitch/yaw) we have : 1370 W * 7.05 m² = 9658.5 W or 9658.5 J/s.

These are the total energy inputs received from the Sun as heat

.

 

[Fizyk]

The average temperature of the cabin hull, assuming that we have an efficient heat transfer all over the hull (aluminium is highly conductive), is 439.45 K.

This is the part where it goes wrong. If you want to calculate total emitted energy, you need average T^4, not (average T)^4.

To present the difference, let's say we have 5 numbers: 1,2,3,4,5.
(Average n)^4: Average n = (1+2+3+4+5)/5 = 3, (Average n)^4 = 81
Average n^4 = (1^4 + 2^4 + 3^4 + 4^4 + 5^4)/5 = 195.8

The difference is quite large, and for the same reasons the energy output will be a lot bigger than what we calculated.

 

[N_Molson]

Now I get :

q = 0.04 * 5.6704×10^-8 W/(m^2 K^4) * (441.99 K^4 - 2.37 K^4) * 42.4 m²
q = 4.228×10e-5 Watts

That seems a very very small result... :blink:

or if I do that :

q2 = 0.04 * 5.6704×10^-8 W/(m^2 K^4) * (441.99^4 K^4 - 2.37^4 K^4) * 42.4 m²

q2 = 3670 W

or 86.56 W / m²

---------- Post added at 11:43 PM ---------- Previous post was at 09:42 PM ----------

It seems I found something, the way I calculated the average temperature was obviously completely wrong.

The whole object (cylinder) has a surface of 42.4 m².

We have an incoming energy flux of 1370 W/m² on the 14.15 m² exposed to the Sun.

The 28.25 m² that are in the shadow receive 0 W of energy.

I make the average of the energy each m² of the object receives :

e = (1370 W *14.15 + 0 W * 28.25) / 42.4

e = 457.205 W

From there, I get the average temperature on the surface of the object, using the Stefan-Bolzmann equation :

I get 670.1 K

I enter it in our "Net Radiation Loss Rate equation" :

q = 0.04 * 5.6704×10^-8 W/(m^2 K^4) * (670.1^4 K^4 - 2.37^4 K^4)

q = 457.3 W / m²

So we have an input of e = 457.205 W, and an output of q = 457.3 W. Seems a lot more credible to me ; the little difference is probably caused by decimal truncatures, what do you think of it ?

 

[Fizyk]

The little difference is because you didn't take the 2.37 K from CMB (actually it should be 2.73 K, IIRC) when calculating average temperature. But yeah, this looks ok now.

 

[N_Molson]

I found my error by reading this statement on the Web :

"Temperature is not energy, but a measure of it. Heat is energy."

(Caltech)

Edit :

actually it should be 2.73 K, IIRC

Right, that's the cosmic background radiation, another horrible typo ! :facepalm:

 

[RisingFury]

0 will be fine for the universe. Why?

100^4 - 10^4 = 10^8 - 10^3 roughly equals to 10^8. Spare yourself the pain of decimal places.