|Jul10-12, 01:31 AM||#1|
Trying the crunch some numbers into the thermopile ballpark
After searching the net for a few days and not finding anything that would tell me what to expect from a very crude thermopile, I crunched a few numbers and wound up with an absurd number of amperes coming from my calculations--I doubt my copper/nickel thermopile will produce 320 amperes.
I realize this post is super long, but it should be pretty simple and I am guessing I made some rather silly rookie move somewhere in here. Thanks for your feedback.
This isn't an attempt to get super close, just some where near the ballpark. I chose 9 volts at 350 mili-amperes as the specs for my radio out of the air, but I don't think they are too far away from that.
I used this page to calculate the resistance of my elements:
measured in ohm meters
R=rL/A, R is resistance in ohms, r is resistance coefficient in ohm meters, L is length in meters, A is area of cross-section in meters squared.
And this page for the seebeck coefficients:
Net: 21.5 micro-volts per degree of Celsius or Kelvin difference between the two junctions.
I sandwiched my junctions between temperatures of 350* and 50* to make sure they were in the realm of possibility, difference of 300* and got:
21.5 uv/k times 300k=6.45 miv, where uv is microvolts.
To power my radio I'll need 9-volts:
9=g(6.45), where g is the number of thermocouples I will need in my thermopile.
Ignoring the fact I won't be sitting down to hand make fourteen hundred thermocouples unless I start smoking a lot more meth, I went on to figure out my total resistance in the circuit.
Each thermocouple will be made of a one centimeter cube of copper, one of nickel, and connected via a bar of copper, two centimeters long, one centimeter wide and .3 centimeters thick.
Now adding each of these resistances together we get the total resistance for one thermocouple: 20.067/10^6 ohms
Now if we multiply those ohms by the number of thermocouples, g, we should get the total circuit, which turns out to be .0280938 ohms
The relation of amperes to ohms and volts is given by E=IR, and we know two of those already, E and R.
I=320.355 amperes, which seems silly. I just recently fried an amp-meter trying to see how many amperes came from a car battery, much more than ten.
If the radio circuit had a resistance of 25 ohms this would work, and that may be the case, but something seems to be going wrong I here.
|Jul13-12, 02:40 PM||#2|
Your radio will probably draw less than 1 A, regardless if you device could supply 300A. Your calculated result is probably correct, thermocouples are low impedance devices and their current is mainly limited by the internal resistance. But their Carnot efficiency is not very good. If you would short-wire the device you would for some amount of time get 320 Amps out of it. But that would mean a power of about 2880 Watts if the efficiency of the device would be 5% you would have to supply 57600 Watts as waste heat from somewhere. Keeping that 300 degree difference between that much copper will be the real challenge, so I don't know your power source but it would probably have to be burning pretty hot. Just calculate the thermal conductivity of your 1400 copper bolts (and maybe the area).
|Jul23-12, 11:30 PM||#3|
Thanks deadbeef. I was hoping to get some work done on the project before replying, but life keeps happening.
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