# Fault calculation using per unit



## cbinla (Oct 13, 2010)

I am panicking a little bit since I thought I had this down and now I'm confused on some basic concepts.

On the link shown in the example attached, a generator feeds a motor and then there is a fault at the motor terminals.

The solution shows the impedance of the motor, 0.2 pu, being converted to base of 50MVA and 20kV. I understand when converting a per unit impedance to a base with a different base MVA and base V the following formula is used

per unit Z new = per unit Z given X (base Vgiven / base Vnew)^2 X (base VA new)/(base VA given).

If that is the case, why don't they show the motor synchronous reactance as follows:

Xm" = 0.2 (18kV/20kv)^2 X (50 MVA /25 MVA) = 0.324 per unit

I understand the rest of the problem, which is just solving a simple circuit using the impedance diagram. I am just stuck on the motor's reactance calculation to the new base of 50MVA and 20kV. I'm assuming the motor's per unit reactance is determined from the motor's base values of 25MVA and 18kV.

Thanks so much for any help.

calc_of_fault_current_example.pdf


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## DK PE (Oct 13, 2010)

cbinla said:


> The solution shows the impedance of the motor, 0.2 pu, being converted to base of 50MVA and 20kV. I understand when converting a per unit impedance to a base with a different base MVA and base V the following formula is used
> per unit Z new = per unit Z given X (base Vgiven / base Vnew)^2 X (base VA new)/(base VA given).
> 
> If that is the case, why don't they show the motor synchronous reactance as follows:
> ...


Once the base in the GENERATOR is chosen as 20kV, 50MVA, you need to write down the base voltages everywhere else in the system. Due to the right side transformer being 66kV/18kV, the base voltage in the motor circuit is 18kV. Hence you only need to change it from 25MVA to 50 MVA and you're all set.


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## cbinla (Oct 16, 2010)

Thanks, I was making it more difficult that it is. The base impedance and the base VA define the other bases for each part of the circuit. The voltages can certainly be different for each portion.


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