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.
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.