Power input to liquid rheostat
Power input to liquid rheostat
(OP)
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.





RE: Power input to liquid rheostat
Possibly: Thread237-24735
RE: Power input to liquid rheostat
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: Power input to liquid rheostat
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.
RE: Power input to liquid rheostat
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.