Power input to liquid rheostat 1

[electrifire]

We have a three stage liquid rheostat connected to a large medium voltage wound rotor motor and we are trying to determine the power (or heat) induced into the solution for the rheostar. This heat must be removed by a cooler on hot days in the south and this heat transfewr is the concern. The secondary nameplate voltage and corrent are known.

 

[busbar]

Possibly: thread237-24735

 

[Marke]

The total power dissipated in the rotor resistors will be typically similar to the full speed kinetic energy of the driven load. The exact amount can vary if the load torque curve is significant relative to the start torque curve.
Best regards, Mark Empson

http://www.lmphotonics.com

 

[electrifire]

Thanks to busbar and marke for their responses. A review of the suggested thread did not provide info needed.

The secondary motor nameplate info, 2000 volts and 1350 amps, indicates 4,671Kva will go into the liquid rheostat solution.This does not seem a reasonable value for the heat to be disipated.

Any additional help to determine the value of the heat input to the liquid rheostat solution is appreciated.

 

[electricpete]

Here is a discussion I have posted previously on the subject of heat dissipated for starting of a normal cage rotor motor (constant resistance). It indicates the rotor heating during starting of an unloaded motor is approx the same as the total kinetic energy that the load will have once started. If there is a load present than the heating is somewhat higher as indicated near the end of the message. Maybe you can modify this approach to account for varying resistance steps.
RotorHeating
RH= Int{I^2R2 dt} =
RH = Int{I^2R2 / (ds/st) ds}
What is ds/dt
dN/dt = T/J (f=ma type relationship)
N =(1-s)Nsync
dN/dt =-Nsync ds/dt = T/J
ds/dt=-T/J/Nsync
ds/dt = -1/(JNsync)*T = 1/J*[Pelec]/[N]
= -1/(JNsync)*[I^2R2*(1-s)/s]/[Nsync*(1-s)]
= -1/(JNsync)*[I^2R2/s]/[Nsync] -I^2R2/[sJNsync^2]
RH = Int{-I^2R2 / (I^2R2/sJNsync^2) ds}
RH = Int{-sJ Nsync ds; s=0..sFinal}
= [-0.5JNsync^2*s^2@sFinal-s0] sNsync=Nsync-N
= [-0.5J(Nsync-N)^2@sFinal-s0]
= [-0.5J(sFinalNsync)^2] - [-0.5J(s0*Nsync)^2]
= [-0.5J(s0*Nsync)^2] - [-0.5J(sFinalNsync)^2]
= 1/2*J*(N_sync*s_initial)^2 - 1/2*J*(N_sync*s_final)^2

IF s_0 is 1 and s_final ~0, then RH = 1/2*J*(w_sync)^2 = (1/2)*KE

Note that if we add load, then we have to use
ds/dt = (Te-Tm)/J
This is equivalent to the old (unloaded) integral where the integrand is increased by factor of Te/(Te-Tm) => can be solved graphically or numerically using the new integral approach suggested in EPRI PPERM V6.
RH = Int{I^2R2 / (I^2R2/sJ) / [(Te-Tm)/Te]) ds}
The loaded heating is approx the unloaded heating times the average of Te/(Te-Tm) over all speeds from s=0 to s=1.