Lesson 2: Materials for Optocaloric Performance

Lesson 2: Materials for Optocaloric Performance mjg8 Sat, 03/02/2013 - 15:49

The links below provide an outline of the material for this lesson.

2.0 Introduction

2.0 Introduction mxw142 Tue, 09/06/2016 - 15:36

In EME 810, you learned about the difference between optoelectronic and optocaloric solar energy conversion. As you may remember, optoelectronic refers to the photovoltaic effect of photons (radiation) being converted to electrons (electricity) while optocaloric refers to photons being converted to heat (thermal energy). In order to maximize solar energy gain and the subsequent conversion to useful thermal energy, it is necessary to understand how different materials interact with the sun’s electromagnetic radiation (photons) at different wavelengths. There are three key pieces to gaining a useful understanding of what is going on during the optocaloric conversion process:

1. A basic understanding of radiation heat transfer (D&B Chapter 3) is useful in light of the fact that solar energy, which passes through the atmosphere and reaches your collector surface, is trasferred via the process of radiation (not conduction or convection, though those heat transfer mechanisms can also play a part depending on the system configuration).

2. To convert photons to heat, the photons must be absorbed by the collector. This process requires opaque materials as opposed to reflective or transparent materials which would not absorb the energy but transfer it elsewhere. As such, an understanding of the radiation characteristics of opaque materials (D&B Chapter 4) is especially useful.

3. Many solar thermal energy conversion systems utilize a cover-absorber system to increase efficiency. Covers are typically glass, but can be made out of any material that transmit radiation while reducing losses from convection to the surroundings. As such, an understanding of the transmission of radiation through glazing (D&B Chapter 5) is useful.

These three main topics together provide important background for selection and use of materials in solar thermal energy conversion systems.

Lesson Objectives

Identify materials best suited as components of solar thermal collectors.
Apply selective surface characteristics in calculations.
Calculate transmittance, reflectance, absorptance, and tau-alpha for various scenarios.

What is due for Lesson 2?

This lesson will take us one week to complete. Please refer to the Course Calendar in Canvas for specific time frames and due dates. A brief list of the lesson assignments is provided in the table below. More specific directions can be found on the pages of this lesson and in Canvas.


Tasks	Assignment Details	Access/Directions
Readings	Required D&B Chapter 3 - sections 3.1 to 3.10 (12 pages) D&B Chapter 4 - sections 4.1 to 4.10 (23 pages) D&B Chapter 5 - sections 5.1 to 5.6, and 5.11 (15 pages) Supplementary Remaining sections in D&B Chapters 3-5	Registered students can use the following link to access the online textbook through the University Library.
Quiz	Lesson 2 Reading Quiz checks understanding of the material presented in required readings.	Access the quiz in the Lesson 2 Module in Canvas.
Assignment	Problem set - Radiation Characteristics	Specific directions for the assignment are provided on the Assignment page of this lesson.
Discussion	Understanding the radiation properties of materials	Read directions to this discussion and post you reflection in Lesson 2 Module in Canvas.

Questions?

If you have any questions, please post them to our Questions and Answers discussion forum, located under the Modules tab in Canvas. I will check that discussion forum regularly will do my best to respond. While you are on the forum, please feel free to post your own responses if you are able to help out a classmate.

2.1. Heat Transfer Considerations

2.1. Heat Transfer Considerations ksc17 Wed, 06/12/2013 - 10:27

This section focuses on the fundamentals of the radiation heat transfer, the nature of solar energy as electromagnetic radiation, and interactions of solar radiation with various materials. Radiation heat transfer is often addressed only briefly in any heat transfer course. However, In solar energy conversion systems, where the total energy flux is often orders of magnitude smaller than in conventional heat transfer systems, the contribution of the radiation heat transfer mechanism is significant. Conduction and convection also play a significant role in the performance of certain solar energy conversion systems.

Reading Assignment

To become introduced to the theory of the heat transfer as it applies to solar thermal systems, please read the following text:

Duffie, J.A., and Beckman, W.A., Solar Engineering of Thermal Processes, Wiley and Sons, 2013, Chapter 3, Sections 3.1-3.10.

Radiation heat transfer is dependent on the specific wavelength of the radiation. The distribution of wavelengths from a blackbody radiation source, such as the sun, is described by Planck’s Law (Equation 3.4.1 Duffie & Beckman, 2013). This equation can be integrated for a wavelength range of interest to find the total energy for different scenarios. The results of this integration are given in various simplified forms, which are convenient for practical use. Two important expressions derived from the Planck's law are Wien's displacement law (Equation 3.4.2 Duffie & Beckman, 2013) and Stefan-Boltzmann equation (Equation 3.5.1 Duffie & Beckman, 2013). Take a closer look at those expressions and understand what they are used for.

Another way Planck's law intergration data are often presented is Radiation Tables. Those become handy when the total emount of energy emitted by a blackbody source needs to be estimated for a specific wavelength interval. The example video (6:45) below specifically illustrates how such data are made useful to answer some practical questions.

Solution to example 3.6.1 from Solar Engineering of Thermal Processes (Duffie & Beckman, 2013)

Click for a transcript.

PRESENTER: Hello, this is example 3.6.1 from page 143 of the Duffie & Beckman text. So this problem is about the Sun, which is a 5,777 Kelvin black body radiator. And we are asked to find the wavelength or lambda, the wavelength at which the maximum monochromatic emissive power occurs. And this part of the question is solved by Wien's displacement law, which states that the maximum wavelength times the temperature of the black body radiator is equal to 2,897.8 micrometers Kelvin. So to divide by T on both sides, and therefore lambda max. This is equal to 2,897.8 divided by 5,777, which is 0.502 micrometers. So that is the wavelength at which maximum power occurs. So building into the second part of the problem, let's just draw what this looks like. This is wavelength, and this is power. The distribution of solar energy looks something like that. And so what it's saying is that at this point that happens, at 0.5 micrometers. So if this is problem A, problem B asks a second question about this same situation. It asks, what is the energy fraction from the source that's in the visible frequency range? So we find these fractional values in table 3.6.1A. And what this is asking, the visible frequency range is from 0.38 micrometers up to 0.78 micrometers. So in this graphic that's given by a range like that. And we're trying to figure out what is going on in between there. So from the table, we want to find what lambda T is. So 0.38 times 5,777 gives us 2,195 and 0.78 times the temperature of the black body radiator is 4,506. So with both of these numbers, you can now look it up in table 3.6.1A and obtain the fraction of energy from 0 up to that wavelength. So in the case of the first one, it's 10% percent. And in the case of the second one it's 56% from that table. So therefore, we can figure out our answer of fvis, the visible frequencies, which is simply 56% minus 10% 0.46, which is 46% of total energy in visible. And again, put the answer in a box so that it's clearer what we've done. And that's that. So thank you for listening. And that's problem 3.6.1.

Many solar thermal energy conversion systems use flat plate collectors, which are essentially two parallel plates (one transparent and the other absorptive) exchanging radiation. Calculating the radiation heat exchange between the two surfaces is a necessary aspect of understanding the energy balance of a system. Example 3.10.1 (Duffie & Beckman, 2013) is given below in a brief (6:19) video.

Solution to example 3.10.1 from Solar Engineering of Thermal Processes (Duffie & Beckman, 2013)

Click for a transcript.

PRESENTER: OK. Welcome back. This is example 3.10.1 from page 149 of the Duffy and Beckman text. And in this problem, we have two parallel plates-- an absorber plate and a cover. So this is a cover absorber system. And this could be like a solar hot water heater or an air solar heater. And these two plates are 25 millimeters apart. And some of the properties of each surface for the plate-- the emissivity, or epsilon, is equal to 0.15. And the temperature of the plate is 70 degrees Celsius, which we need to convert to Kelvin. So 70 plus 273. So 343 Kelvin. And the cover, similarly, has an emissivity. The cover is 0.88. And the temperature of the cover is a little bit cooler, at 50 degrees Celsius, which, converting that to Kelvin, we get 323 Kelvin. So in this problem, with this system, as defined above, we're asked to find two things. First, we're asked to find the radiation exchange between the two surfaces. And in this case, intuitively, it's helpful to think which direction the heat will flow. That way, you don't have some weird negative sign you're trying to deal with in the end. The heat flows from the hotter surface to the cooler surface. And the second, we're asked to find the radiation heat exchange coefficient under these conditions. So with all of that in mind, let's go ahead and solve the problem. , So Part A. To solve the amount of radiation heat exchange going on, we're going to use equation 3.8.4 from the text. And we solve this on a per unit area basis, because the size of the collector is not given. And so we solve it per meter squared, which then, if you had a, 3 meter squared collector, you would basically multiply this answer by three to obtain your final total heat exchange. So here's the Stefan-Boltzmann constant, sigma, in this heat exchange equation. And then, again, we use the hotter surface first so that we have a good sign convention. Temperature of that surface-- 343 minus 323, the temperature of the cooler surface, all divided by the emissivity. So 1 over 0.15 plus 1 over 0.88 minus 1, all in the denominator. And the Stefan-Boltzmann constant is 5.67 times 10 to the negative 8. If you've had a heat transfer class, you'll remember that number. It's pretty easy-- 5, 6, 7, 8. 5.67 times 10 to the negative 8. And when you run those numbers through your calculator, you get 24.6 watts per meter squared. So that's from absorber to the cover. Cool. And then Part B of this problem says, under these conditions, what's the radiation heat exchange coefficient? So here we're going to use equation 3.10.1, which is the definition of h sub r, the heat exchange coefficient. So h sub r equals this heat exchange we calculated in Part A-- 24.6 divided by the temperature difference, 70 minus 50. And you can do this in Kelvin or Celsius in the denominator. You would get the same number. And so what you end up with is 1.232 watts per meter squared Kelvin. And that's that. That's example 3.10.1. Thank you for listening.

Self Check:

1. What is the purpose of the following equations? (click on the name to see the answer)

Planck's Law

ANSWER: Planck's equation describes the distribution of the radiation emitted by a blackbody over the range of wavelengths.

Wien's Displacement Law

ANSWER: Wien's equation determines the wave length at which a blackbody of a certain temperature emits the maximum radiation.

Stefan-Boltzmann Equation

ANSWER: Stefan-Boltzmann equation is used to found the total energy flux emitted by a blackbody at a certain temperature T.

2. What is the purpose of the radiation tables?

Click here for answer

ANSWER: Radiation tables (Table 3.6.1a in Duffie & Beckman, 2013) are a simplified result of the integration of the Planck's equation - they allow determining the fraction of radiative energy emitted by a black body within a certain wave length range.

2.2. Radiation Characteristics of Opaque Materials

2.2. Radiation Characteristics of Opaque Materials mjg8 Sat, 03/02/2013 - 15:51

When solar radiation hits a surface, the photons can be absorbed, reflected, or transmitted. In the case of opaque (not transparent) materials, none of the photons are transmitted. If the material is dark and dull (not reflective or shiny), very few of the photons are reflected. As such, the majority of photons incident on dark opaque surfaces will be absorbed. As s result of absorption, the photons are converted to thermal energy (or heat). At the same time, because of the temperature of the material, the surface emits radiation back to its surroundings at a rate that is dependent on the emissivity of the material. Heat can also be lost to the surroundings by conduction and convection, but that is not the focus of this lesson.

Reading Assignment

To learn the interaction of the solar radiation with opaque materials and parameters that characterize heat transfer, please read the following text:

Duffie, J.A., and Beckman, W.A., Solar Engineering of Thermal Processes, Wiley and Sons, 2013, Chapter 4, Sections 4.1-4.10.

While reading especially work through the examples that illustrate the use of the main equations.

Some additional note on materials selection:

When designing a solar thermal conversion system, the selection of materials is critical. Choosing collector materials that are good absorbers (i.e. carbon black) will help your system to perform well. Sometimes the best material can be too costly to justify. As such, it becomes a balance of priorities towards an optimal system. If a material of high absorptance (α=0.95) costs $10/lb and a material of higher absorptance (α=0.98) costs $20/lb, the best option to achieve a desired solar gain may be to use the cheaper material and increase the system aperture or total collector area.

In addition to the cost and physical radiation properties of materials, we must be careful to select materials that will hold up under extreme climatic and environmental conditions. For example, in sandy desert environments, the abrasive sand can have a negative impact on the reflective properties of concentrating trough collector systems over time. Thus full understanding the in situ performance of a material over a period of decades is important to the design and optimization of solar thermal energy conversion systems.

Self Check:

1. How would you define absorptance?

Click here to see the definition

ANSWER: Absorptance is the ratio of the fraction of the incoming raditation that is absorbed by the material to the total incident radiation

2. How would you define emittance?

Click here to see the definition

ANSWER: Emittance is the ratio of the radiation that is emitted by a material surface to the radiation that would be emitted by a blackbody at the same temperature

3. How would you define reflectance?

Click here to see the definition

ANSWER: Reflectance of a surface is the ratio of the radiation that is reflected (i.e. not absorbed or transmitted) to the total incident radiation.

4. For best performance of flat-plate collectors, it is generally more important to maximize absorption of radiation rather than minimize emission of heat. If the highest temperature of the material surface is desired, which three options from the Table below would you pick?

Radiation Properties
Material	Type^a	$\frac{Emittance}{{Temperature}^{b}}$	Absorptance^c
Aluminum, pure	H	$\frac{0.102}{573}, \frac{0.130}{773}, \frac{0.113}{873}$	0.09-0.10
Aluminum, anodized	H	$\frac{0.842}{296}, \frac{0.720}{484}, \frac{0.669}{574}$	0.12-0.16
Aluminum, SiO₂ coated	H	$\frac{0.366}{263}, \frac{0.384}{293}, \frac{0.378}{324}$	0.11
Carbon black in acrylic binder	H	$\frac{0.83}{278}$	0.94
Chromium	N	$\frac{0.290}{722}, \frac{0.355}{905}, \frac{0.435}{1072}$	0.415
Copper, polished	H	$\frac{0.041}{338}, \frac{0.036}{463}, \frac{0.039}{803}$	0.35
Gold	H	$\frac{0.025}{275}, \frac{0.040}{468}, \frac{0.048}{668}$	0.20-0.23
Iron	H	$\frac{0.071}{199}, \frac{0.110}{468}, \frac{0.175}{668}$	0.44
Lampblack in epoxy	N	$\frac{0.89}{298}$	0.96
Magnesium oxide	H	$\frac{0.73}{380}, \frac{0.68}{491}, \frac{0.53}{755}$	0.14
Nickel	H	$\frac{0.10}{310}, \frac{0.10}{468}, \frac{0.12}{668}$	0.36-0.43
Paint - Parson's black	H	$\frac{0.981}{240}, \frac{0.981}{462}$	0.98
Paint - Acrylic white	H	$\frac{0.90}{298}$	0.26
Paint - White (ZnO)	H	$\frac{0.929}{295}, \frac{0.926}{478}, \frac{0.889}{646}$	0.12-0.18

^aH is total hemispheric emittance; N is total normal emittance.
^bThe numerator is the emittance at the temperature (K) of the denominator.
^cNormal solar absorptance.

Click for answer...

From quick look at the data, the three materials with the highest absorptance should provide the highest performance (see Section 4.10 D&B) :

Carbon Black
Lampblack
Parson’s Black

On the previous page of this lesson, we looked at how the wavelength of the incident radiation matters because the wavelength determines the amount of energy that is transmitted. Some specially selected or designed materials may absorb radiation in one range of wavelengths very efficiently while may be highly reflective in a longer wave length range. Such materials are referred to as selective surfaces. The concept of selectrive surface is discussed in Section 4.8. of D&B book, and Example 4.8.1. and 4.8.2. show how the radiation properties of such materials can be calculated. Please review those examples in detail. In this lesson assignment, you will be asked to perform a similar calculation.

2.3. Radiation in Cover-Absorber Systems

2.3. Radiation in Cover-Absorber Systems mjg8 Sat, 03/02/2013 - 15:51

Many 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:

Box w/ arrows = various types of radiation going in & out. Into the box is incident, inside: absorbed & out: transmitted & reflected

Firgure 2.1: Energy conservation principle for the radiation interacting with materials: τ + ρ + α = 1

Credit: Mark Fedkin

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 (n_i) - 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.

3 arrows 1 in & 2 out Incoming line is angle q1 between interface & perpendicular up leaving arrow is q3 down leaving arrow is q2 see text

Figure 2.2: Propagation of light rays at the interface between two media.

Credit: Mark Fedkin

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.

Solution to example 5.3.1 from Solar Engineering of Thermal Processes (Duffie & Beckman, 2013)

Click for transcript.

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.

Solution to example 5.5.1 from Solar Engineering of Thermal Processes (Duffie & Beckman, 2013)

Click for transcript.

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

TRUE

Correct! - increasing angle of incidence means "flatter" beam, which would be reflected off the surface more readily.

FALSE

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?

Click here to see the answer

ANSWER: (1) Reflection losses and (2) Absorption losses

2.4 Assignment

2.4 Assignment sxr133 Wed, 04/24/2013 - 13:54

This homework assignment consists of six quantitative problems that are closely tied with the readings. Of course these six problems selected from the D&B textbook do not make a complete assessment of how well versed you are with radiation property calculations, but they are good teasers challenging you to apply some of the key equations you see in the book. In that sense it is more of a learning activity rather than an assessment. The videos posted in the previous sections of this lesson give you an example how to approach this kind of calculations.

Deliverable - Lesson 2 Assignment

Problems: 3.1, 3.2, 4.1, 5.1, 5.2, and 5.3 from the D&B textbook (see Appendix A)

I ask you to complete these problems by hand, clearly showing your work step by step. While you should feel free to use any calculation software or spreadsheets behind scenes, I will only read and grade your hand-written solutions. The recommended format for hand-written problems should include underlined statements of:

knowns
assumptions
properties to find
analysis (including what equations used, numbers, and units)
solution (please draw a box around your final answer)

A sample hand-written problem is given in the Lesson 2 Module in Canvas.

Please create electronic images of your hand-written solutions (via scan or camera) and save them in a single PDF document. If you have series of images of your pages, you can first insert them into a MS Word document in proper order, and then save the file as PDF. Please note that It is your responsibility to make your submission legible. If I cannot read it, how can I give you credit?

The assignment is due by 11:55PM on Wednesday. Please see the Calendar in Canvas for specific due dates.

2.5 Summary and Final Tasks

2.5 Summary and Final Tasks sxr133 Mon, 04/15/2013 - 13:04

Summary

Materials for optocaloric performance vary significantly in physical properties. Additionally, new materials are being developed worldwide. Table 4.7.1 (in D&B book) gives an overview of some of the most common materials used for solar thermal reflection and absorption. However, in current project proposals in many settings, new materials will be used. It is important to be able to think critically about the physical behavior of materials in a broad sense so that new or unfamiliar materials can be put into perspective against available materials. Reading materials in this lesson provide you with the fundamental theory for material selection, which can be further used for practical purposes.

Reminder - Complete all of the Lesson 2 tasks!

Please double-check the to-do list on the Lesson 2 Introduction page to make sure you have completed all of the activities listed there before you begin Lesson 3.