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thermo electronics

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 Author Topic: thermo electronics  (Read 1027 times)
jt
Newbie

Posts: 2

 « on: November 30, 2009, 06:58:43 AM »

hi, im doing a college project on a thermoelectric pwoer source using a peltier thermoelectric unit (TEU). i have my heat source and am trying to calculate my power output. i have calculated the heat in Watts (joules/second) but i am having difficulty in converting this into Watts (electrical).

any help would be appreciated!!!

cheers,
john.
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Hero999
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Posts: 2471

 « Reply #1 on: November 30, 2009, 09:11:41 AM »

What do you mean Watts are J/s?
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jt
Newbie

Posts: 2

 « Reply #2 on: November 30, 2009, 10:25:37 AM »

theres a difference between Watts electrical and Watts heat loss but i cant figure out how to convert heat loss to electrical.
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Hero999
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 « Reply #3 on: November 30, 2009, 10:48:58 AM »

Heat loss is power which is not converted to electrical energy.

Calculate the thermal power in, calculate the electrical power out and work out the efficiency.
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awright
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Posts: 17

 « Reply #4 on: December 14, 2009, 04:45:10 AM »

Hero999, the joule is a unit of energy and the watt is a rate of energy flow, i.e., power.  One watt is equal to a rate of energy flow of one joule per second, as jt stated.  A watt is a watt is a watt in any consistent system of units, be they thermal, mechanical, or electrical.  Mathmatically, they are directly convertable.

John's problem is not a mathmatical conversion of units, but rather figuring out how effectively the thermoelectric device will convert the heat power in into electrical power out.  I believe this can be quite a complicated function of the source and sink temperatures, the area and thermal conductivity of the device, the Seeback Coefficient of the device, and the electrical resistances of the device and the load.  Seeback devices (not Peltier devices that convert electrical energy into thermal energy, although they are essentially identical except for the application and how they are optimized for the application) are very inefficient at converting heat energy into electrical energy.  I believe one would require either a manufacturer's rating of the performance of the device at John's specific operating conditions or data derived from experimantation on the device to predict the output of the device.  Alternatively, just assume an efficiency of about 10% and apply that factor to the input thermal power.

awright
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Hero999
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Electronics God

Posts: 2471

 « Reply #5 on: December 14, 2009, 06:33:30 AM »

It's actually pretty simple.

He already knows the heat input.

All he needs to do is calculate the power output which is as simple as connecting a resistor, measuring the voltage and current, and applying Ohm's law.

Then he'll have the power in and power out, enabling him to calculate the efficiency.

If this is a college project, he could look at the hot and cold temperatures, calculate the efficiency of an ideal Carnot heat engine and see how his thermoelectric heat engine measures up to it.
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