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Trans. short circuit current with lower voltage

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bdf5526

Electrical
Nov 26, 2007
51
Hi, I have trans with 20.2mva, 21kv/6.3kv, z=9.5%. If we apply 9.5% of primary rated voltage(21kv) to the primary with sec. short circuit we should have full load current at secondary winding.

If I apply 21volt(3ph) instead of 21kv to primary with secondary short circuited the current at secondary will be,

First we find the leakage reactance:-

1)Trans per unit Z= 0.095
2)Trans base impedence refer to secondary, (kv)*(kv)/base mva = (6.3)*(6.3)/20.2 =1.96 ohm
3) transformer resistance refer to secondary, Z(real)
=Z(pu)*Z(base)= 0.095 x 1.96 = 0.1862 ohm

for 3ph Z(real)= square root 3*0.1862 = 1.732*0.1863 = 0.322 ohm

Now if I apply 21 volt 3ph at primary, my sec. voltage is 6.3volt,

With secondary short circuited my SC current will be Isc=V/Z=6.3v/0.322ohm = 19.53 amp

Am I correct?

I have to fully understand this before put my hand on the real trans. soon, where I shall use 400V 3ph at 21kv winding.

Thanks in advance.
 
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Hi BDF.
You can apply 400V on the 20kV side w/o any problem, it's standard procedure, btw, on the 6.3kV too.


I=S/(Ux1.73) of trafo, Isc=I/Uk(0.095).
Instead 22kV you apply 0.4kV it's all, that meant SC current will be Isc/55.
Good Luck.
Slava
 
HI Slavag,

Thanks for the info. Voltage injection at HV winding with secondary winding short circuited will be used to check the all the CT loop at short circuited side, this is what we are going to do.

Thank again.
 
Your answer is correct but I think your calculation is a bit overcomplicated or not easy to follow.

Rated secondary current equals 1851 A.

Should be applying 9.5% of 21 kV (or 1995 V) to get rated current.

If you apply 21 volts, you should expect to get

V/1995 * 1851 or

21/1995 * 1851 = 19.49 A

So it just amounts to a voltage divider times rated secondary current.
 
Hi everyone,

Looks like I have complecated the principle and the calculation. Thanks magoo2 and slavag.

This is the reason why I like this site.

Thanks.
 
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