Question

Two parallel alternators each rated 11 kV 50 MVA and each having sub transient reactance of 15% are connected to a bus bar. What is the fault level for a fault on one of the feeders near the bus bar?
1. 666.67 MVA
2. 333.33 MVA
3. 1333.5 MVA
4. 100 MVA

Answer 1

Correct Answer - Option 1 : 666.67 MVA

Short Circuit Capacity:

• It is the measure of the electrical strength of the bus
• It is also called short-circuit MVA stated in MVA.
• It determines the dimension of the bus bar and the interrupting capacity of a circuit breaker.

 

Finding the fault level:

\({I_{SC}} = \frac{{{V_b}}}{Z} = \frac{{{V_b}/{Z_b}}}{{Z/{Z_b}}} = \frac{{{I_b}}}{{{Z_{p.u.}}}}\)

Where,

I_SC is the short circuit current

V_b is the base voltage

Z impedance of the alternator

Z_b is the base impedance of the alternator

Z_p.u. per unit impedance of the alternator

\(\sqrt 3 {V_b}{I_{SC}} = \frac{{\sqrt 3 {V_b}{I_b}}}{{{Z_{p.u.}}}}\)

\(MV{A_{SC}} = \frac{{MV{A_b}}}{{{Z_{p.u.}}}}\)

The percent value is the per-unit value multiplied by 100 (Z% = Z_p.u. × 100)

Calculation:

Given,

MVA_b = 50 MVA

As the alternators having only reactance (Z_p.u. = X_p.u.)

Z_p.u. = 15 / 100

= 0.15

As the alternators are connected in parallel

X_p.u. = 0.15 / 2

= 0.075

The fault level for a fault on one of the feeders near the bus bar

MVA_SC = 50 / 0.075

= 666.67 MVA

Therefore, the fault level for a fault on one of the feeders near the bus bar is 666.67 MVA