Impedance for motor starting

[timm33333]

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

 

[7anoter4]

See [for instance]:

http://www.openelectrical.org/wiki/index.php?title=Motor_Starting#Step_3:_Referring_Impedances

 

[timm33333]

This link has good information. But they have used a generator. Does anybody know how will the equations be modified if the generator is replaced by utility?

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

 

[waross]

But they have used a generator. Does anybody know how will the equations be modified if the generator is replaced by utility?

The grid may be considered as an infinite generator. Just use the transformer impedance. If the grid is "soft" in relation to the motor starting load, replace the generator impedance with the source impedance of the grid.

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?

If the generator is rated in kW, use the power factor to convert to a KVA base.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[dpc]

Power factor of a motor at locked rotor is extremely low - generally 0.15 to 0.2.

 

[7anoter4]

The theory in the above article follows IEC 60909-0 standard. The utility impedance-if no parallel generator will be provided and the voltage of the busbar is the same as motor voltage-less the voltage drop-the network [utility] will be in series with Zeq.
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.