Heat to Power Equations

[keithc]

Does anybody have some heat to electrical power equations??<br>We are moving from shafted printing presses to shaftless, so the printing cylinders need to be cooled. <br>

 

[azzmi]

Hi,
I don't know if is it useful for your case but,
1 kilowatt-hour is equal to 859 kilocalories or 3413 BTUs.
But your case is not easy because you have to know the thermal capacitance of your physical system.
Regards,

 

[jbartos]

One may use the following at the beginning:
P = A/t = Q x c x (T2 - T1)/t
where
P - Power in Watts
A - Energy in Wattseconds
t - Time in seconds
Q - Mass in kg (i.e. kilograms), or l (liters for water)
1 liter of water = 1 kg
c - constant (specific heat) = 1 kcal/(degree C x kg)for water
kcal - kilocalorie
T2 - Higher Temperature in degree C (centigrade or Celsia)
T1 - Lower Temperature in degree C
1 kcal = 1 x 4186 Joule/second = 4160 Watts

 

[jbartos]

Small correction of the previous posting and small clarification.
1 kcal (International Table calorie) = 4186.8 Joules =
4186.8 Wattseconds
1 kcal (thermochemical calorie) = 4184.0 Joules =
4184.0 Wattseconds
Reference:
Fink D. G., Beaty H. W., &quot;Standard Handbook for Electrical Engineers,&quot; 13-th Edition, McGraw-Hill, Inc., 1993, page 1-13

 

[azzmi]

Hello Jbartos,
Your second post is correct. But if you make some basical mathematic operations on your result, you will see that;
a kilowatt-hour is equal to 859 kilocalories. And even this form of equation may be more useful for which cases are electrical power is known.

The equation that you send may be useful if constant c for steel (or any material)can be obtained.

(P = A/t = Q x c x (T2 - T1)/t)(from Jbartos)
Regards

 

[jbartos]

Suggestion:
Reference:
Baumeister T., et. al., &quot;Marks' Standard Handbook for Mechanical Engineers,&quot; 8th Edition, McGraw-Hill Book Company, 1978, page 4-9, Table 17. Mean Specific Heats of Various Solids and Liquids between 32 and 212 degree F, in Btu/(Lb degree F)
Matter Specific Heat Cp [Btu/(Lb deg F)
Nickel Steel 0.109
Rose's Metal 0.05
Red Brass 0.09
German Silver 0.095
Solders 0.04-0.045
Asbestos 0.2
Brick 0.22
Coal 0.3
Normal Glass 0.199
Gypsum 0.259
Ice at 32degF 0.487
India rubber 0.48
Marble 0.21
Sand 0.195
Wood-Fir 0.65
Machine oil 0.4
Sea water 0.94
Normal water at 32degF 1.007
Olive oil 0.4

 

I am trying to calculate how hot a lead battery terminal will get under a 2000 A load for 3 minutes. I have been trying to find an equation that will give me the heat generated, in a wire as a fucntion of resistance, current etc..

I will assume no heat transferred from the terminal and see if the temperature would be expected to reach the melting temperature.

Any thoughts on this?

Thanks Simon

 

[azzmi]

Hi Simon,
I think, first of all, we have to calculate the total electrical power loss which is dissipated in your battery at its operation conditions.
For this calculation, the internal resistance of the battery must be determined. Simply you can make two measurements to determine roughly the resistance value. first of these measurements is at no load condition. (measure the output terminal voltage). Then one more measurement under a known load conditions. now you can calculate internal resistance of your battery roughly by using measured voltage and current values.
Now we can calculate internal electrical power loss of the battery. (square of current times internal resistance)
Then a transformation can be made between electrical power value to its thermal equivalent.
And then (i am not sure anymore)
you have to calculate total mass of lead, (I think mass of the others -liquid and plastic- can be ignored). (But if I were you, I don't)
Now the question turns to how much X(calculated)X kcals of thermal energy heats X(calculated)X kilograms of lead.
I think there is enough additional information in other posts which are sent before.
I hope, it will be helpful..
[sig]<p>Azmi Demirel<br><a href=mailto:azzmi@elk.itu.edu.tr>azzmi@elk.itu.edu.tr</a><br><a href=

http://triton.elk.itu.edu.tr/~azzmi>Welcome</a><br>I

am handsome but married [/sig]

 

[jbartos]

References:
1. IEEE Std 946-1992 &quot;Recommended Practice for the Design of DC Auxiliary Power Systems for Generating Stations&quot;
2. IEEE Std 450-1987 &quot;Recommended Practice for Maintenance, Testing, and Replacement of Large Lead Storage Batteries for Generating Stations and Substations&quot;
There will be different heat sources.
A. One is due to battery internal resistance. Considering the battery volume and 2000A for 3 minutes, the battery will warm up somewhat depending on its internal resistance.
The internal resistance may be 0.00018 Ohms, which may lead to
P = 0.00018 x 2000 x 2000 = 720 Watts (about hot iron heat for three minutes)
See Reference 1 page 20.
B. The second hot spots are the battery intercell connections. One inercell connection may have the resistance anywhere from 0.00001 ohms for the large battery to 0.0001 ohms for the smaller battery. This means that the power developed there may be
P = 0.0001 x 2000 x 2000 = 400 Watts which is plenty for two small battery posts and the intertie. See Reference 2 page 15. [sig][/sig]