# Solar Panel temperature

1. Apr 28, 2010

### lgaston99

1. The problem statement, all variables and given/known data
Consider what would happen to the temp of a solar array when a solar spacecraft gets into an eclipse. Distance to the Sun is 1 AU and Fs = 1368 W/m2

Consider an infinitely thin flat panel thermally isolated from the spacecraft. Assume the Specific Heat Capacity is 8.0 Kj/K-m2. Also assume the pamel material has infinite thermal conductivity.

In daylight, the solar array is normally illuminated by the solar radiation, and it quickly reaches equilibrium temp.

Assume panel absorptivity $$\alpha$$ = 0.84 and it's IR emissivity $$\epsilon$$=0.74

(a) Calculate solar panel equilibrium temp under solar normal illumination

(b) Estimate solar panel temp by the end of the longest possible eclipse (71 min)

2. Relevant equations

for (a) T = ($$\alpha$$ * Fs) / (4*$$\epsilon$$*$$\sigma$$SB) --greek letters not to be superscripted (i.e. [4*e*sigSB])

I have no idea what equations should be used for (b) I have looked at
Q = m*c*$$\Delta$$T
Q = e*sigma*A*$$\Delta$$ T
Q = (A*$$\Delta$$ T) / H

3. The attempt at a solution
From my calculations I got that the temp = 287.65K
I am completely lost as to where to begin for the temp at the end of the eclipse. Any help or advice would be greatly appreciated.

2. Apr 28, 2010

### zachzach

Just a guess, but Newton's law of cooling?

3. Apr 29, 2010

### Gokul43201

Staff Emeritus
How did you arrive at this equation?

What thermal processes do you expect to be taking place when the panel is in direct sunlight? Now what happens when you take away the sunlight?

This is a section for getting help with homework/textbook problems. Please try to avoid making guesses, and instead leave the problem to people that do not need to do so.

4. Apr 29, 2010

### lgaston99

I arrived at the equation from a section in our notes identified as Passive thermal control. In looking at it again I did realize that the formula was incorrect. The formula should be

T = [(alpha * Fs) / (epsilon * sigmaSB)]1/4

derived from:

alpha * A * Fs = epsilon * A *sigmaSB * T4

I expect the heat to radiate.

5. Apr 29, 2010

### Gokul43201

Staff Emeritus
Correct. Now you can recalculate the equilibrium temperature for part (a). I assume here that you understand where the left side and right side of the above equation come from. Without that understanding, you can not solve part (b).

Also correct. So what expression would you use for the rate at which heat is radiated away? And what expression deals with the heat content of the panel? How are these related?

6. Apr 29, 2010

### lgaston99

Here is my current understanding and numbers.

(a) T = [(alpha * Fs) / (epsilon * sigmaSB)]1/4 = [(0.84*1368)/(0.74*5.67x10-8)]1/4 = 406.8K

(b) Here I think I have 2 equations

1) (panel_heat_capacity)*dT = (loss_per_unit_time)* dt

2) (loss_per_unit_time) = 2*epsilon*signma_SB*T^4

This yields (panel_heat_capacity)*dT = (2*epsilon*signma_SB*T^4)* dt

From there I believe i integrate...which requires pulling out some books so I thought I would check my rationale in the meantime.

Thank you all for the help.

7. Apr 30, 2010

### Gokul43201

Staff Emeritus
Haven't run the numbers myself, but going by your previous calculation (with the wrong equation), this value looks too large. Do check it.

Looking good! As for the integration, there's one tiny step before you have to do that (collect/separate variables). The integration itself is going to be relatively simple once you do this.