Converting to Per-Unit System

[JatTiw]

Dear Members,
I had confusion regarding converting impedance in PU system. A 3-phase transfomer(5MVA, 13.8/4.16kV & 7% impedance) connected to source (Fault level 255MVA). The load connected on 4.16kV is 2000hp (suppose impdance-Xpu). The base MVA-100 & base kV-13.8. For calculating the PU impedance of load at base voltage, the impedance at 4.16kV is to be multiplied with voltage factor.

 

[cuky2000]

Utility: Zu = 100MVA/225MVA=0.44pu

Transformer: Zt = (100MVA/5MVA).(7% /100%) = 1.4 pu

Motor (assumed all Induction): Zm = (100MVA)/(4x2.0MVA) = 12.5 pu
(1MVA~1/1000hp @ 0.8 PF).

Cable Impedance?? (No Data) Zc = Length(Impedance per length )* 100MVA/(4.16^2*1000).pu


One of the simplifications of the PU calculation is that the equivalent circuit impedance will not require voltage transformation.

 

[jlazucena]

jlazucena

The data that you have is
System 255 MVA scc (Short circuit capacity)
XFMR 5MVA, 13.8/4.16 and 7% impedance
Motors 2000hp (Here you should know the average size of these motors that will affect the X/R ratio of the impedance of the motors)A conservative value for the motor impedance is 16.7% but you can use 20% if you neglect the impedance of the motor-feeders.

Know to solve your system you need to define two variables, usually these variables are one voltage and system capacity.
You can use 13.8kV and 100MVA
Then the impedance of the system refered to 100MVA-base would be 100/225=.444 pu
The impedance of the XFMR refered to 100MVA-base would be
0.07pu*100MVA/5MVA=1.4 pu (Note: 7% means 0.07 pu)
If you neglect the X/R ratio of the impedances the short circuit current (SCC)available at the secondary of the transformer from the system would be
Isc = 1/(1.4+.444) = .5423 pu

Now you need to find the base current at the secondary side of the transformer
I base = 100/(4.16*1.732) = 13.879 kAmps
And the SCC would be
Isc = 13.879 * .5423 = 7.526 kAmps (Or 7526 Amps)

The contribution of the motors can be calculated in the same way (Note added in parellel pls), that means you have to refered the impedance of the motor to the bases of the system, as follows

Xnew pu = X given pu *((base kVgiven/kVnew)^2)/(base MVAnew/MVAgiven)

Please note that you can solve using the capacity of the XFMR as a base, then:
MVA base is 5 MVA and Voltage base is again 13.8 kV

Then the impedance of the system refered to 5MVA-base would be 5/225=..0222 pu
The impedance of the XFMR refered to its own 5MVA-base would be 0.07 pu (Note: 7% means 0.07 pu)
If you neglect the X/R ratio of the impedances the short circuit current(SCC)available at the secondary of the transformer from the system would be
Isc = 1/(.07+.0222) = 10.846 pu

Now, the base current at the secondary side of the transformer
I base = 5.0/(4.16*1.732) = 0.694 kAmps
And the SCC would be
Isc = 0.694 * 10.846 = 7.526 kAmps (Or 7526 Amps)