# Thermodynamics and Concentrated Solar Power

1. Jun 19, 2012

### Strat-O

I am interested in building a thermal solar collector. I'm mostly interested in using the heat directly but there's always a need for electricity. The trouble is converting heat into electricity is pretty inefficient no matter how you do it. Consider a peltier device. A typical one, depending on your delta T might convert the heat energy at an efficiency of 10%.

In the attached diagram There is a parabolic dish that is 70% efficient and a peltier device that is 10% efficient, so the efficiency of the system in generating electricity is 7%, right? (neglecting heat losses due to lousy insulation)

Someone suggested that because the heat is recirculated after the peltier device that it was somehow more efficient than 10%.

Any thoughts?

#### Attached Files:

• ###### ParaElec.png
File size:
10.9 KB
Views:
78
2. Jun 19, 2012

### willem2

What you haven't drawn is that the cold side of the peltier device has to be cooled.
The efficiency of the peltier device will depend on the temperature difference across it, so you'll need a radiator, or cooling water.
So there's a further loss, because the temperature difference across the peltier device itself will be less than the temperature difference between the tank and the outside air or cooling water.
Your efficiency also will go way down when the tank begins to cool if there's no more sun.
If the device gets 80C water from the tank and sends 60C water back to the tank, the heat from the return water isn't lost, but that wasn't included in the efficiency of the peltier device anyway. The reject heat on the cold side is probably lost, unless you can heat your house with it.

3. Jun 19, 2012

### haruspex

As perhaps you know, the theoretical maximum efficiency is determined by the temperatures involved. If the collector can get up to 530C (800K) and you have a cooling fluid available at 30C (300K) then the theoretical efficiency is 1 - 300/800 = 62.5%.

4. Jun 20, 2012

### Strat-O

Thanks for the responses.

So the 63.5 percent is from the formula for Carnot's theorem. I've seen it before but I think I've got it commited to memory now by seeing it stated so simply.

Marlin