solve this question please The percent impedance of a transf

solve this question please.

The percent impedance of a transformer is typically determined by a short circuit test. In such a test, the secondary of the transformer is shorted and the voltage on the primary is increased until rated current flows in the transformer windings. The applied voltage that produces rated current divided by the rated voltage of the transformer is equal to the per-unit impedance of the transformer. A short circuit test on a 150 KVA, 7200-240 V transformer provides the following results: Primary voltage at 20.8 primary amperes = 208.8 V Determine the %Z of the transformer. Calculate the ohmic impedance of the transformer in primary and secondary terms. How much current would flow in the transformer if its secondary would become shorted during normal operating conditions? (Consider source impedance to be zero.)

Solution

1)

% Z can Be determinded by formula as below :

Z% =  Impedance Voltage /   Rated Voltage   x 100

   = 208.8 / 7200 * 100

   = 2.9 %   

Z% = 2.9 %   

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2)

KVA =150 kva

Primary V= 7200

secondary V = 240 volt

20.8 primary Amp caused 208.8 Volt in primary

NOw,

Assume 7200 V voltage and 150 KVA.
Base amps = 150 /( 7200x1.732)= 12.04 A. Base ohms = 7200 x 7200/150 = 345.6 ohms

ohmic impedance at 150 KVA = 2.9% x 345.6 = 10.02 ohms.

Ohmic impedance =10.02 ohms.

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3)   

as per the formula of I_fl

I_{fl =} (KVA x 1000 / sec voltage ) / 3

= ( 150 X 1000 / 240  ) / 3

= 361.27

I_fl = 361.27