PrintMany solar thermal energy conversion systems employ glass to reduce convective losses from the absorbing surface, increasing system efficiency. Glass is not perfectly transparent, with some absorption as well as reflective losses that are dependent on the incidence angle of the solar irradiation. The three useful metrics for understanding what the performance of a glass cover will be are transmittance (τ), reflectance (ρ), and absorptance (α).
It is important to realize that when a light beam hits a particular surface or material, radiation can be reflected from the surface of the material, transmitted through the bulk of the material, and absorbed in the material. There are no other options since the energy needs to be conserved. This concept is additionally illustrated below:
The key radiation properties can be derived for a particular material or surface using Fresnel equations. As we will see further, the transmittance, reflectance, and absorptance are dependent on the thickness, refractive index, and extinction coefficient of the material of interest. Before proceeding to the Fresnel derivation, it would be useful to quickly refresh our mind on the light reflection and refraction laws.
Refresher: Law of reflection and refraction
The following diagram illustrates the main directions of a light ray on a material surface. The key parameters to note:
- The angle of incidence (q1) - angle between the incoming ray and normal to the interface
- The angle of reflection (q3) - angle between the reflected ray and normal to the surface
- The angle of refraction (q2) - angle between the refracted ray and normal to the surface
- Index of Refraction (ni) - dimensionless number characterizing light propagation in a material. It is constant for every type of medium at a constant temperature. For more information, check the Wikipedia page for this property.
As you may recall from physics, by the Law of Reflection, the angle of incidence is equal to the angle of reflection:
θ1 = θ3
By Snell's Law, the light refraction process is described as follows:
n1sinθ1 = n2sinθ2
Now let us see how radiation characteristics can be derived.
Reading Assignment
Please read the following text to become familiar with the main characteristics of cover systems for solar thermal applications
Duffie, J.A., and Beckman, W.A., Solar Engineering of Thermal Processes, Wiley and Sons, 2013, Chapter 5, Sections 5.1-5.6 and 5.11.
Especially pay attention to examples, which are very helpful for understanding how the key equations are applied to practical problems. You will have a few problems on this topic in your homework assignment.
Example 5.3.1 is given below in a brief (14.31) video.
PRESENTER: Hello. This is example 5.3.1 on page 207 of the Duffie and Beckmann text. This problem is about a single glass cover system asking to calculate the transmissivity, reflectivity, and absorptivity of that class cover. So single glass cover. What that means is that we can look up in table 5.1.1 that the index of refraction is 1.526. That's the information that we gather from the fact that it's a single glass cover. The thickness-- L, length, is 2.3 millimeters. That's given, and that equals 0.0023 meters. The extinction coefficient, K, is equal to 32 units per meter. And the incidence angle, theta, is given as 60 degrees. And we are asked to find transmissivity-- tau; reflectivity-- rho; and absorptivity-- alpha. I've also added that I want to check that the sum of all of these-- tau plus rho plus alpha-- equals 1. Because it should be because the incident radiation is either transmitted through the material, reflected off the material, or absorbed in the material, and that's it. It's a one-sum game. Something's got to happen each photon, and those are the three things that can happen to them. So to kick this off, first we have to calculate the refraction angle. We do that using equation 5.1.4. We get theta 2 is equal to the sine inverse of the sine of the incidence angle divided by the index of refraction, and that results in theta 2 equal to 34.58 degrees. And so this allows us to calculate the extinction coefficient optical path product, which is KL over cosine theta 2. So K is 32 times 0.0023. We divide by the cosine of 34.58. So what we get is 0.0894. The next piece of the puzzle is to calculate tau sub a. So here, we're going to use equation 5.2.2. We get tau sub a equals exponential of negative 0.0894, the number we just calculated above. And what we get is 0.915. So note that this is not the transmissivity. This is tau sub a. We still have to average for the parallel and perpendicular components of polarization in the glass. So this is an intermediate step on our way to the solution. So in the next piece of the puzzle, we need to use equations 5.1.2 and 5.1.1 We're going to use 5.1.1 first. Calculate the perpendicular component of reflectance. That's equal to the sine squared of the angle we calculated above minus the incidence angle divided by sine squared of 34.58 plus the incidence angle. So we get 0.184 divided by 0.994, which is equal to 0.185. And then the parallel component is almost the same equation, but instead of sines, we use tangents. 34.58 minus 60 divided by the tan squared of 34.58 plus 60. And that gives us 0.226 divided by actually a very large number, 155.8. So what we end up with because we're dividing by such a large number is we end up with a very small number for the parallel component there. So what we end up with here is a way to calculate tau, transmissivity. So therefore tau is equal to 0.915, which is the tau sub a number, divided by 2 times 1 minus 0.185 divided by 1 plus 0.185, which is the perpendicular component, times 1 minus 0.185 squared divided by 1 minus 0.185 times 0.915 quantity squared. Close that parentheses. Plus now we have to do the parallel component. 1 minus 0.001 over 1 plus 0.001 multiplied by 1 minus 0.001 squared divided by 1 minus 0.001 times 0.915, quantity squared, and close all those parentheses and brackets. This gives us 0.5 times 0.625 plus 0.912, which is equal to 0.768-- final answer for tau. So that was definitely a long, drawn-out process, and now we have tau. The good part is that that's the hard part-- calculating reflectivity-- and absorptivity is relatively straightforward at this point. So in light of the lack of room in this little area here, we're going to move up into the empty space up here for the next part of the problem. So that was for calculating tau. So next, we're going to calculate reflectivity. So to do that, we use equation 5.3.2 for reflectance, and that gives us rho equals 0.5 times 0.185-- again, we reuse a lot of these same numbers that we calculated before, so we've already done the bulk of the work getting this far-- 1 plus 0.915 times 0.625 plus 0.001 times 1 plus 0.915 times 0.912. So when you crunch all those numbers, you end up with reflectivity equals to 0.147. Next, we're going to calculate absorptivity. I'm going to do that using equation 5.3.3. Alpha equals 1 minus 0.915 divided by 2 multiplied by 1 minus 0.185 which is, again, that perpendicular component of r over here, divided by 1 minus 0.185 times 0.915-- 0.915 is that tau sub a, again, over here-- plus 1 minus 0.001, which is the parallel component over here, divided by 1 minus 0.001 times 0.915-- that tau sub a again. Close that parentheses, crunch all those numbers, and you end up with alpha equals 0.085. So we've calculated each of these three pieces, and now it's good to just check it. So we're going to check that tau plus rho plus alpha equals 1. If you punch those numbers in your calculator, 0.768 plus 0.147 plus 0.085, you end up, indeed, that they do equal 1. Check. So we've accounted for all the photons essentially, and that's example 5.3.1. Thank you so much for your time and for listening.
Because of the complex intra-system interactions of incident solar energy in a combined cover-absorber system, the transmittance-absorptance product (τα) is defined. The τα parameter should be thought of a property of a cover-absorber combination, rather than the product of the two individual properties, which captures the essence of how clear (transmissive) is the cover as well as how absorbing is the absorber in the same system. Figure 5.5.1 in the D&B book schematically shows the reflection and absorption of light occurring at different material interfaces in a cover-absorber system. Example 5.5.1 showing the calculation of the τα parameter is given below in a brief (5:11) video.
PRESENTER: This is example 5.5.1 from page 214 of the Duffie and Beckman text. The previous problem we just did was for a single cover system. This problem is about two covers. And the glass panes that make up this two-cover system have a KL of 0.0370 per pane. And the absorber of the system has an absorptivity or alpha equal to 0.90. If you're not familiar with my notation, a w with a slash means "with" and alpha's the absorptivity there. We are asked to find the tau alpha product. Let me just back up a bit so that I can make that clear as mud-- the tau alpha product at 50 degrees angle of incidence. So this problem, the best way to start it off is using figure 5.3.1 from the text, where you can find from that figure that the transmissivity at that incidence angle of 50 is simply 0.75. So that's an approximation using that figure. Then, the next really important piece of the puzzle is to look at figure 5.5.1. It has some footnotes, so look at footnote 2 of figure 5.5.1. And this figure gives a really great representation of what's happening inside of a cover absorber system, where the light is reflecting back and forth within the system and you're trying to figure out what the overall transmissivity and absorptivity product is. So from that footnote, we can see that with a KL of 0.0370, rho sub d, the reflectivity. sub d is 0.22. So that's pretty important. Once we know those two pieces, this 0.75 and 0.22, we can then use equation 5.5.1 to calculate the tau alpha product using these values. So what you end up with is 0.75 times 0.90, which is the absorptivity, divided by 1 minus 1 minus 0.90 times 0.22, all of that in the denominator. And so you get a tau alpha product when you crunch those numbers of 0.69. And this is essentially a property of the overall system. It's not simply the product of tau times alpha. It's a little bit more involved and it has to do with the inter-reflections within the system, as well. But this is a very useful metric to help assess one system against another in light of the transmission properties and the absorption properties of the materials in the system. So thanks for listening. This has been example 5.5.1 from the text.
Self Check:
1. True or False? The reflectance of the glass surfaces increases with increasing angle of incidence
Correct! - increasing angle of incidence means "flatter" beam, which would be reflected off the surface more readily.
Incorrect :( Increasing angle of incidence means "flatter" beam, which indeed would be reflected off the surface more readily - that is the reflectance will increase.
2. What two types of losses should we consider in calculations of transmittance of cover materials?
ANSWER: (1) Reflection losses and (2) Absorption losses