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Why does the temperature increase in a liquid when pumped?

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MAT73

Chemical
Feb 22, 2017
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Hello again, and thanks in advance for clarifying the following question. Thanks in advance for your help and guidance.

Is the reason that the temperature in a liquid that is being pump increases due to the difference between the Break Horse Power of the Pump and the Actual Horsepower supplied by the motor and that amount of energy being "lost" as heat going into the fluid ?

Secondly, how could I calculate the expected rise in temperature (assuming that the liquid is water) if I know the flow rate coming out of the pump, motor HP and motor efficiency ?

 
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Hallo!

Fluid in pump is heated due to:
- heat Transfer from pump Motor into fluid - this amount is ussually provided by pump supplier
- as fluid is pumped from inflow to outflow the friction appears (friction of fluid to pump area)-btw it can be assuemd by changes of pump inner energy, entropy und so on...
hope thats helps

dust123
 
MAT73,

A lack of details here which doesn't help, but in essence follow the energy.

Actual energy entering an electric pump (kW) has some losses (the motor efficiency) which often gets lost into the air, but if your motor is cooled by the process fluid ( you don't say) then yes, it can raise the temperature.

The main heat gain is through the efficiency of the pump, i.e. the difference between Brake HP of the shaft and the hydraulic power of the fluid. Often in the region 60 to 80% efficient.

Most of this loss turns into heat energy which is absorbed into the fluid.

SO unless your motor is in the process fluid, you can't do it as you've described, what you need is the pump efficiency and the shaft power at your flowrate and head difference.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
One way to get a quick estimate of the liquid heat up by pumping is by calculating the work done on the fluid:

∆t=∆P/(ρ×Cp)

Using consistent units.


 
Ultimately yes, all work turns to heat, but in the immediate term it's the efficiency losses that appear at the pump outlet.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Thanks a lot, I will gather more of the data and come back with a more detailed explanation of the system O am trying to evaluate. Thanks again.
 
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