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Claculation of motor power as per FEM standard

Claculation of motor power as per FEM standard

Claculation of motor power as per FEM standard

(OP)
I would like to know if rotating and other inertias (GD^2) are not given then how to calculate electrical power required for acceleration. I am talking abour Container Gantry Cranes. I am an electrical engineer.

RE: Claculation of motor power as per FEM standard

Without knowing the inertias of the system, you can't calculate the power requirement.

The rotor I^2*R losses during acceleration of an unloaded motor will be exactly equal to the change in kinetic energy of the rotor (1/2*J*w^2).

The stator I^2*R losses during acceleration of the same unloaded motor will be Rs/Rr times the rotor I^2*R losses (assuming small motor... neglect rotor resistance variation from skin effect).

Total energy involved in bringing the inertia up to speed (unloaded) would be rotor heating plus stator heating plus KE rotor = 1/2*J*w^2 + Rs/Rr * 1/2*J*w^2 + 1/2*J*w^2  =
[2+ Rs/Rr] * (1/2)Jw^2.

If you add a load during the start, you have to multiply the above rotor and stator heating by a factor which represembles the average value of Te/(Te-Tm) during the start.  You will also have to integrate the total work done against that lod during the start.

Here are some equations to support the rotor heating calculation (I believe the others are more straightforward)

a) RotorHeating = RH= Int{I^2R2 dt}  =
b) RH  = Int{I^2R2 / (ds/st) ds}
c) What is ds/dt
(1) dN/dt = T/J (f=ma type relationship)
(a) N =(1-s)Nsync
(2) dN/dt =-Nsync ds/dt = T/J
(3) ds/dt=-T/J/Nsync
(4) ds/dt = -1/(JNsync)*T = 1/J*[Pelec]/[N]
(a) = -1/(JNsync)*[I^2R2*(1-s)/s]/[Nsync*(1-s)]
(b) = -1/(JNsync)*[I^2R2/s]/[Nsync]
(c) -I^2R2/[sJNsync^2]
d) RH  = Int{-I^2R2 / (I^2R2/sJNsync^2) ds}
e) RH  = Int{-sJ Nsync ds; s=0..sFinal}
(1) = [-0.5JNsync^2*s^2@sFinal-s0]
(a) sNsync=Nsync-N
(2) = [-0.5J(Nsync-N)^2@sFinal-s0]
(a) times -1
(3) = [0.5J(N-Nsync)^2@sFinal-s0]
(4) 0.5JNfinal^2 – 0.5JN0^2 = deltaKE
f) RH = deltaKE
h) Note that if we add load, then we have to use
(1) ds/dt = (Te-Tm)/J  
(2) This is equivalent to the old (unloaded) integral where the integrand is increased by factor of Te/(Te-Tm) => graphical approach suggested in EPRI PPERM V6.
(3) RH  = Int{I^2R2 / (I^2R2/sJ) / [(Te-Tm)/Te]) ds}
(4) RH  = Int{sJ*[Te/(Te-Tm)]) ds}


By the way, what does FEM stand for?

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