Impedance for motor starting
Impedance for motor starting
(OP)
How do we calculate the impedance (R + jX) at the motor terminals for the calculation of voltage drop at the time of motor starting? Once the impedance is known, we can multiply the locked rotor current with impedance and divide by nominal system voltage to find the voltage drop at motor terminals at the time of motor tarting. I'd assume that the impedance at the motor terminals can be calculated from the impedance of the utility, transformer, and cables: but this procedure is not readily available in the books. Thanks






RE: Impedance for motor starting
http://www.openelectrical.org/wiki/index.php?title...
RE: Impedance for motor starting
Also they have used the power factor of generator as 0.85. The power factor is usually for the loads. When we say a generator has a power factor of 0.85, does it mean that it can supply full power to loads which have power factor of 0.85? Thanks
RE: Impedance for motor starting
If the generator is rated in kW, use the power factor to convert to a KVA base.
Bill
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"Why not the best?"
Jimmy Carter
RE: Impedance for motor starting
RE: Impedance for motor starting
As per Network Feeders:
Zs=Zf=c*Vn^2/Sf Sf=Ssys[short-circuit apparent power of the System at service point].
As waross said:
If instead of Sg1= 4,000 kVA let’s put Sf=500 MVA Zs=1.05*11^2/500=0.254ohm
Usually Xsys/Rsys=5 Xs=Zs/sqrt(1+1/5) Xs=0.254/sqrt(1.2)=0.232 ohm Rs=0.232/5=0.064 ohm.Now substitute for the old data of the generator the new data of the network and continue the calculation in the same way. Eo=Vs=Vn*(1+Zs/Zeq) and so on.
And I agree with dpc: the motor starting power factor is too high.