Answer / mbao richie

simply means the bus rated at that fault level can
withstand the electrical pressure(voltage) for which it is
designated for that period of time

Answer / srikanth

Dear friends,

in the above calculations people consider the % impedence
but they are not calculate the fault impedence to get MVA
of transformer.

Answer / jigar shah

suppose transformer (11 KV/433) is of 1000 KVA & %
impedance is 4 .LT side Fault level can be calculated as
below :

KVA = 1.732 X V X I X Cos Phi - PF assumed : 0.95

LT Rated current (I) = (1000 * 1000)/(1.732*433*0.95)
= 1481 A

Fault current on LT side = FLC * 100/(% impedance)
= 1481 * 100/5
= 29.62 KA

Answer / laxminarayan

suppose transformer (11 KV/433) is of 1000 KVA & %
impedance is 4 .LT side Fault level can be calculated as
below :

KVA = 1.732 X V X I X Cos Phi - PF assumed : 0.95

LT Rated current (I) = (1000 * 1000)/(1.732*433*0.95)
= 1481 A

Fault current on LT side = FLC * 100/(% impedance)
= 1481 * 100/5
= 29.62 KA

In case of legging current of 20°, pf is .94 and leading of 20°,pf is .94 , in mathematicaly both are same but in realy how it is differ ?

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