Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm working through an example I have been given to study. Suppose I have a 2m X 4m photovoltaic panel on my roof that is irradiated with a solar flux of $G_s = 700W/m^2$.

Given:

$\alpha_s = 0.83$

$\eta = P/\alpha_sG_sA = 0.553-0.001T_p/K$

$\epsilon = 0.90$ $T_{sur} = T_\infty = 35^oC$

$h = 10W/m^2K$

I want to find how much electric power is generated. I start by using the energy balance equation.

$E_{in} - E_{out} + E_{generated} = E_{stored}$

$E_{in} = \alpha_sG_sA$

$E_{out} = \epsilon\sigma(T_s^4 - T_{sur}^4)A + h(T_s - T_\infty)$

$E_{generated} = -P_{elec}$

$E_{stored} = 0$

Okay, so here is my first question - why is $E_{generated} = -P_{elec}$ and not $E_{generated} = +P_{elec}$? Where did the negative come from?

After plugging it all into the energy balance equation, I get: $\alpha_sG_sA -[\epsilon\sigma(T_s^4 - T_{sur}^4)A + h(T_s - T_\infty)]-P_{elec} = 0$

$(0.83)(700)(2)(4) -[(0.90)(5.67X10^{-8})((T_p)^4 - (35+273)^4)(2)(4) + (10)(T_p - (35+273))]- (0.553-0.001T_p)(0.83)(700)(2)(4) = 0$

Plugging this into my TI-89 calculator, I get $T_p = -666.633$ or $T_p = 335.051$

Obviously, I take $T_p = 335$ since it probably shouldn't go below absolute zero :-) but since I am asking questions - what does the negative value represent? Does it mean anything at all?

Now that I have $T_p$, I plug that into $\eta = P/\alpha_sG_sA = 0.553-0.001T_p/K$ to solve for the power generated. If I do it this way, I get $P = 1010W$. Here's a funny problem, though - if I plug $T_p$ into $\alpha_sG_sA -[\epsilon\sigma(T_p^4 - T_{sur}^4)A + h(T_p - T_\infty)]-P_{elec} = 0$, I get $P = 1032W$. Why the difference? Does it have something to do with the way the calculator solves the problem? Or is there a mistake somewhere?

Thanks so much for looking at this and leading me in the right direction.

share|improve this question

1 Answer 1

You wrote effectively the following:

$$E_{in} - E_{out} - P_{elec} = 0$$

We just need to look at the balances. All of the variables in the above equation are positive. The $E_{in}$ term is energy coming in from the sun. You have energy entering the system and energy leaving the system. The electric power production is energy leaving the system that would have otherwise went to heat (and then radiating or convecting). It didn't, it left in the form of electrons crossing an electric potential. This same accounting goes on for all power producing technology. Electrical power is energy leaving the system.


I don't understand the use of $T_{sur}$ by the way. Looking at a balance of radiative heat transfer, you've accounted for radiative absorption from the sun in another term. So what's the use of this one? It seems like the other transfer would be from the sky to the panel, and I don't think it would be handled in this way.

There could be some transfer from the Earth's surface to the panel, but it wouldn't be the full angle. The panel points to the sky.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.