This is a simple solar collector being used to warm water. We get cool water entering the solar collector and warm water leaving it.

If we look at the interior of it we see that the pipe snakes back and forth throughout the collector collecting as much solar energy as we can. We're going to neglect any heat losses from the sides and the bottom of this collector and are really just gonna focus on the top of the collector and see how much sunlight is being absorbed.

So to do that, let's take a look at our solar collector from the side, here we've provided relevant temperatures; the temperature of the sun, the temperature of the sky above ambient temperature in the local vicinity of the plate, and the surface temperature of the plate itself. The area of the plate has the dimensions of one meter by one meter, so the area is one square meter.

So the primary means at which the plate gathers energy is from the sun itself. So there's some solar insulation, G, and only a fraction of that radiation actually is absorbed by the plate. So to account for that we're gonna multiply by the plate's absorptivity. In doing so, we can now neglect the amount of reflected radiation because it's all being accounted for in alpha. We might incorrectly assume that G is equal to sigma times the temperature of the sun to the fourth power. And in doing so, we would predict that the plate receives 63 megawatts of energy per square meter.

Of course, we're not at the surface of the sun, that's what this meter squared represents, the meter squared of the surface of the sun. If we were, we would burn our plate up, so that is not a reasonable value for the amount of irradiation from the sun, because we're so far away from the sun. The more reasonable value is 900 watts per square meter. Multiply that by the area, and we find that the sun shines about 900 watts of energy on the surface of the plate.

We also have irradiation from the sky and the sky is relatively cool and we can represent that as the absorptivity multiplied by sigma times T-sky to the fourth power. We also have energy loss in the form of diffuse radiation from the surface of the plate. It means the radiation comes off uniformly in all directions from the plate. So to calculate this, we'll say the emissivity of the plate multiplied by sigma times surface temperature to the fourth power. We've also got natural convection coming off the surface of the plate, using appropriate correlation for natural convection from a plate. We find that ℏ is about equal to 4.7 watts per square meter cal.

So we find out the amount of energy being transferred from the water is due to irradiation from both the sky and the sun minus the loss due to natural convection and radiation out into the surroundings. If we were to assume that the surface was perfectly black we would find that the absorptivity and the emissivity were all equal to one and now it becomes a matter of plugging in numbers to find the solution.

However, to complicate matters, it turns out the surface is not black, and in fact, the values of alpha, these two values are not the same. So we’re gonna call one alpha-sky and another alpha-sun. They aren't the same and we need to go about calculating them.

To be able to do that we first have to talk about what it means for a surface to be black or gray or semi gray. In the graph below, the spectral emissivity and the spectral absorptivity for the three types of surfaces are shown, and due to Kirchhoff's law, we find that the spectral emissivity and the spectral absorptivity will be equal in all cases because the radiation is diffuse. At the top is what's known as a black surface which means it absorbs 100% of the radiation for all values of λ for all wavelengths of radiation. The second surface is known as a gray surface and the value of ε and alpha are the same value for all values of λ and they're less than a hundred percent but the idea behind a gray surface is that ε and alpha are the same for all wavelengths. And in the third case is what's known as a semi-gray surface. Again, the value of ε and alpha and in this case the spectral emissivity and the spectral absorptivity are different values for different wavelengths. We're going to say that there's a step change either downward or upward at some critical value of λ. We'll call that some value λ-critical. And in this case, we’ve set λ-critical equal to five microns, and we've said the spectral emissivity and the spectral absorptivity are equal to 0.9 at short wavelengths and equal to 0.1 at longer wavelengths.

Recall that if the surface was black, we could have completed the problem by now because we would've set ε and alpha equal to one in all cases. If the surface was gray, we could have simply set ε and alpha equal to 0.6 in all cases, however, the surface we're dealing with here is semi gray, so the question becomes, well, what is ε itself or what is alpha itself? Because the wavelengths from the sun tend to be very short and the wavelengths of the emitting surface tend to be longer because the temperature of the surface is of course much cooler than the surface of the sun.

Here is a simulation of a gray surface in which the spectral emissivity and the spectral absorptivity are equal to 0.6 for all values of λ. And I'm keeping the temperature right now at the temperature of the sun. We see most of the energy occurs at short wavelengths.

However, if we take the surface and cool it down, we see a shift to higher and higher wavelengths. It's scaling by orders of magnitude as we decrease the temperature. So if we move from right to left, the temperature increases and the energy increases by orders of magnitude. The gray curve represents the emission if this was a black body and the red curve represents the case in which we have a gray surface in which we've set the emissivity equal to 60% across the board for all temperatures,which means that the area under the red curve is exactly 60% of the area under the gray curve.

This next simulation shows what would occur if the surface was black, we would find that the area under the red curve is exactly equal to the area under the gray curve because the surface is black. So in this case, ε and alpha are equal to a hundred percent no matter what the temperature is, even for the cool temperature of the plate it's equal to one and the temperature of the sun, it's also equal to one.

For the case of the semi gray surface however, we'll find that the area of the red curve is 90% of the area of the gray curve at short wavelength.

What we find is that as we decrease the temperature the fraction of radiation being emitted or being absorbed above five microns is substantially less that we find the right of that critical value of λ, the actual radiative power is only 10% of that of the black body and to the left it's only 90%. So we have to figure out what value of ε and what value of alpha to use, which is indicated on the title on the bottom. So what we find, if we increase the temperature back up to the temperature of the sun, we see ε and alpha are about equal to 90%, almost equal to 90%, very little of the radiation coming from the sun has wavelengths longer than five microns

As we cool it down though we're gonna continue cooling this down, pay attention to ε and alpha it's now at 68% at a temperature of 889° C. We're gonna continue cooling this down until we hit the temperature of the plate, which is at a temperature of 45° C. ε and alpha are at 0.12 at the temperature of the sky equal to minus 10° C, which is very cold.

If we were to continue to go even colder than that we would find that little or no radiation occurs at wavelength shorter than five microns. So at this temperature it would be safe to say that we were dealing with a gray surface because the step change in the spectral emissivity doesn't affect it.

If we'll heat this back up again. We'll find the emissivity and the absorptivity are approaching 90% at the temperature of the sun. So for the temperature of the sun, we'd say alpha-sun was equal to 0.89, for the surface temperature of 45° we'll use ε equal to 0.12, and for the sky we're going to use an absorptivity of 10%. So those are the numbers we're going to plug in to figure out the heat being delivered to the water.

We'll first note that the area under the red curve is the actual emissive power. The area under the gray curve is equal to the black body emissive power which is equal to sigma times T to the fourth. And the ratio of those two, E divided by sigma T to the fourth is equal to the value of ε that we're going to use. And in this case we've got the temperature 45° for the surface. Formally we'd write this, the value of ease being the integral of the red curve which isn't necessarily easy to integrate. So what we did was split it into two parts. We've got a part with ε one which is 90% and ε two, which is equal to 10%. And the fact that those are constants means that I can pull these out of both of those integrals.

So here we've rearranged those and solved for the value of ε that we wish to use. Now the tricky part is evaluating these integrals because we need to evaluate the integral of this function which isn't easy to do, there's no analytical expression for it. So what we did in our simulations was we used an ODE solver in MATLAB. Alternatively, we can recognize that this represents the fraction of the black body emissive power that occurs below our critical value of λ. And the term on the right represents the fraction of the black body emissive power that occurs above the critical value of λ.

Here we've represented the first integral as a fraction that is tabulated which we could look up. And the second integral is one minus the original fraction which we could look up, and because we know the temperature of the solar collectors is equal to 45°, we can find that its emissivity is equal to 0.12.

To go back to the original problem we find that alpha-sky is about equal to 0.104, we find that alpha-sun, because it's wavelengths were so short we find that that value is equal to 0.896 and we find the emissivity from the solar collector was equal to 0.115. Plugging in all of these numbers, we find that the amount of heat added to the water is about equal to 650 watts. Note that 900 watts of solar energy is being delivered to the surface, so we could find an efficiency of 650 divided by 900, which is equal to 72%. So 72% of the sunlight is being delivered to the water.