Solar Panel temperature

In summary: Thank you all for the help. Have you tried this equation for (b)? Without knowing the loss per unit time, it is difficult to estimate the temperature.
  • #1
lgaston99
3
0

Homework Statement


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 [tex]\alpha[/tex] = 0.84 and it's IR emissivity [tex]\epsilon[/tex]=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)

Homework Equations



for (a) T = ([tex]\alpha[/tex] * Fs) / (4*[tex]\epsilon[/tex]*[tex]\sigma[/tex]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*[tex]\Delta[/tex]T
Q = e*sigma*A*[tex]\Delta[/tex] T
Q = (A*[tex]\Delta[/tex] T) / H

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.
 
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  • #2
Just a guess, but Newton's law of cooling?
 
  • #3
lgaston99 said:

Homework Statement


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 [tex]\alpha[/tex] = 0.84 and it's IR emissivity [tex]\epsilon[/tex]=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)

Homework Equations



for (a) T = ([tex]\alpha[/tex] * Fs) / (4*[tex]\epsilon[/tex]*[tex]\sigma[/tex]SB) --greek letters not to be superscripted (i.e. [4*e*sigSB])
How did you arrive at this equation?

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

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.
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?

zachzach said:
Just a guess, but Newton's law of cooling?
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
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
lgaston99 said:
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
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).

I expect the heat to radiate.
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
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
lgaston99 said:
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
Haven't run the numbers myself, but going by your previous calculation (with the wrong equation), this value looks too large. Do check it.

(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.
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.
 

1. How does temperature affect the efficiency of solar panels?

As temperature increases, the efficiency of solar panels decreases. This is because the materials used in solar panels are most efficient at converting solar energy into electricity at lower temperatures. Higher temperatures also cause the electrical resistance of the materials to increase, resulting in a decrease in the overall efficiency of the solar panels.

2. Do solar panels work better in colder or warmer climates?

Solar panels actually work better in colder climates. This is because colder temperatures allow the materials in the solar panels to operate at their most efficient level. However, too much cold can also have a negative effect on the performance of solar panels, as extreme cold can cause the panels to freeze, reducing their efficiency.

3. How do solar panels regulate their temperature?

Solar panels have built-in temperature regulation systems, such as ventilation or cooling fans, to keep them from overheating. These systems work by dissipating excess heat and maintaining a consistent temperature to ensure optimal efficiency of the panels.

4. Can extreme temperatures damage solar panels?

Yes, extreme temperatures can damage solar panels. High temperatures can cause the materials to expand, potentially leading to cracks or other damage. Extreme cold can also cause the panels to contract and become brittle, leading to potential damage. It is important to choose solar panels that are designed to withstand a wide range of temperatures.

5. Are there any ways to mitigate the effects of temperature on solar panel efficiency?

Yes, there are several ways to mitigate the effects of temperature on solar panel efficiency. One way is to install the panels at an angle, allowing air to circulate underneath and help dissipate excess heat. Another method is to use shade or reflective materials to reduce the amount of direct sunlight hitting the panels. Additionally, regular maintenance and cleaning of the panels can also help maintain their efficiency in varying temperatures.

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