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.

Comments

[fieldman]

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.

[jarney1]

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.

[fieldman]

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.

[jarney1]

P.S. Here's some other information of mine, a later work, that describes why an accurate measure of stasis temperature is important to understanding the energy/$ calculation. The basic idea is that the higher the stasis temperature, the more energy you lose to your environment as waste heat. That energy cannot take part in any heat-engine you connect.

http://www.iwilltry.org/w/index.php?title=Design_elements_for_solar-thermal_systems