# Estimating Thermal Performance of a Solar-thermal System

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## Introduction: Estimating Thermal Performance of a Solar-thermal System

Several people have written comments about the thermal characteristics of the receiver element and how it might be improved. Here, I will cover some basic thermodynamic principles that lead to designs that might show better thermal performance. This will be essential in connecting the system to a working steam engine to generate power. The goal of this exercise is not to build just any solar-thermal system, for many have already been built at great expense. The goal here is to understand what design elements are important in building such a system and what elements are not. In understanding these design elements, one might conceive a large scale solar-thermal system that can be built at a fraction of the cost of the ones being built today.

## Step 1: Understanding Basic Thermodynamics

Thermodynamics means "moving heat". The basic formula for thermal flow (the flow of heat) is:

dQ/dt = A/R (T - T_c_)

In this formula, dQ/dt is the amount of heat (energy) flowing per unit time. It can be measured in many ways, but the unit of measure we will use is the Watt since this is a common household term. Watts can also be converted to horsepower, another common measure. A is the area of the object (the receiver element, in this case) and R is a measure of thermal resistance. The "R" value of insulation is commonly given on household insulation products. The "R" value we will use here has different units than those quoted on household products, so you may have to convert from one to the other when you use this formula. The "T" value is the temperature of the receiver element and T_c_ is the "cold" temperature. In our case, T_c_ is ambient temperature.

What this formula tells us is that the amount of heat lost through radiation is proportional to the difference in temperature between the receiver element and its surroundings. The hotter it is, the more energy is lost. The more it is insulated (i.e. the higher the "R" value) the less heat is lost. Finally, the area of the receiver element is important. A larger surface area will radiate more heat than a smaller one.

## Step 2: Estimating the R Value of the Receiver.

Now that we have this formula, we can start putting in the things we know and calculating the things we don't. The big thing we don't really know for our receiver element is its "R" value. It turns out, however, that calculating it is no big deal.

We start by realizing that when we put the receiver in the heat source (in the focus of the concentrator) it will heat up. Eventually, the temperature will stop rising. Why is this? It stops rising because as the temperature increases, the amount of energy lost due to convection, radiation, and conduction starts to approach the amount of energy we pump in. This is what is called "thermodynamic equilibrium". This is a big term that simply means "energy in = energy out". The temperature at which this happens, we will call "Stasis temperature" which simply means the temperature at which the temperature stops changing.

So let's take this situation and plug it into our formula. We are collecting about 19 sq ft of sunlight and sunlight provides about 92 Watts per square foot. Our mirrors are only chrome, so we're really only collecting about 70% of that sunlight. All together, is gives us about 1223 Watts of energy going to the receiver element.

The receiver element measures 1ft by 1ft, front and back, so the total area of the collector is 2sq ft.

On a 100 degree day, we measured a stasis temperature of 325 degrees for a temperature difference of 225 degrees.

Plug this into the formula, and we get:

1223W = 2sqft / R * (225 deg)

Solve for R and we get:

R = 2sq ft / 1223 W * 225 deg

Here, we have found the "R" value to be 0.36

## Step 3: Total Energy Available.

Now that we have estimated the "R" value, we can use the formula again to find out how much energy is lost due to radiation and how much steam we can produce. Note that this is going to be an estimate. It should be good enough to design your system with, but there are other losses in the system, so bear in mind those losses when you design your own.

The next situation we'll use our formula in is the situation where we put water in the receiver and boil it. Because boiling is a constant temperature process, the receiver element will only get up to 212 degrees. Thermally, this is a good thing because it means the amount of energy lost due to radiation, conduction, convection, etc will be smaller. Boiling water is also good because it can be used to run engines, make tea, and distill things. Anyway, we want to estimate how much energy is usable in the form of steam and how much we are losing due to radiation.

We use the same formula again. This time, we know "R" since the "R" value doesn't change when we add water to the system. We also know "T", "T_c_" and "A". What we're going to calculate is dQ/dt.

dQ/dt = A/R (T - T_c_)

dQ/dt = 2sq ft / 0.36 (112Deg) = 608Watts.

This means that of the 1223 Watts we've put into the system, about 608 Watts are lost to the environment. This leaves us about 615 Watts not lost. What happens to that 615 watts? Well, since energy is conserved (i.e. cannot be created or destroyed), it must be going into boiling the water.

Converting Watts to horsepower, we get about 0.82 Horsepower that we can use.

## Step 4: Conclusions

Let's stress again that this method only yields an estimate. There are lots of factors I've ignored in this exercise. Reflection of sunlight off reflector, blackbody effects, etc. These are small in comparison with the thermodynamic effects, but they are all present. Taking these things into account is important in a large scale system.

What this does serve to teach, however, is the importance of understanding thermodynamic principles and how to apply them in designing a system. With this information in hand, for instance, we note that we could probably not even operate a 1hp engine with the steam yield of this device.

We also learned how much insulating the back of the device is likely to help us. By insulating the back, we would increase the "R" value of 1/2 of the surface area and thereby increase the stasis temperature and thus reduce the heat loss.

It also gives us insight into why some things burn up (temperatures exceed 450 degrees) and some thing only get warm. By looking at the broom, I can conclude that it has a very high "R" value.

Since it does not radiate much energy, it collects it all until the temperature exceeds its own flash point.

This in itself is an important lesson. It means that as insulation is added, great care must be taken that the insulation itself is never exposed directly to the sunlight. By its very nature, insulation is designed to have a very high "R" value. Directly exposed, the "R" value will cause the stasis temperature to rise extremely high. This will ultimately result in the breakdown (burning) of the insulation. For example, styrofoam will simply burn like the broom does if directly exposed. Fiberglass is a good candidate because it will only melt and will not actually burn The paper backing on the fiberglass is at risk, however.

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## 5 Discussions

I'm commenting as a person who is considering building a solar still solely to have some pure water at the end of summer, when the rain water is used up, to water tropical plants that don't like our municipal water's high dissolved solids content. It appears to me that although the boiling temperature rises as the salt concentration increases in a batch process, that I waste less water this way than with a reverse osmosis filter, and waste less water than I would recycling the polymer beads in a deionization filter process. This gives me some planning tools that I'd neglected to consider, so thanks!

while this is all well and good, and frankly a resource, you have to keep in mind that the numbers that count are BTU/dollar. Hell, when I bought an old Modine heater and put cold water from my well through it I was astounded at how well it would work as an air conditioning system.

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I prefer to use the Watt-Hour to measure energy, rather than the BTU, however, I do agree. I haven't worked on this for a while because I've been trying to make the energy/\$ calculation work. It's harder than it sounds. I've been thinking of having a well dug for exactly that reason. How deep's the well. Perhaps you could post an instructable describing your thermal transfer experiment. I'd love to do something like that, but am not sure how far I'd have to dig the well and how wide to make it in order for it to be practical here.

Mine is 80 ft., but the water level is less than 40 ft. down. Its the well I use for my house water supply. Standard casing. When it warms up a bit I'll do the numbers to develop the curve and post them.