Kaplan #21

[Machiavelli999]

A three-phase, delta-wye connected, 30-MVA, 33/11-kV transformer is protected by a differential relay. The CT current ratio on the primary side is 500:5 and that on the secondary side is 2000:5. The relay current setting for faults drawing up to 200% of the rated current is nearly:

a. 3.31 amps

b. 6.82 amps

c. 7.94 amps

d. 8.32 amps

So, that's the problem. The solution is essentially to figure out how much current the relay sees on the HV side and the LV side using the CT ratios given.

And you calculate,

Ihv = (30MVA) / (sqrt(3) * 33 kV) = 524.86A

Ilv = (30MVA) / (sqrt(3) * 11 kV) = 1,574.6A

And then the relay sees,

Irelayhv = Ihv / (500/5) = 5.2486 A

Irelaylv = Ilv / (2000/5) = 3.9365

And then differential relay setting should be Irelayhv - Irelaylv = 1.3121A!

But, that's not an answer. And the reason is the book says that on the secondary side, the current the relay sees should be calculated as follows:

Irelaylv = Ilv / (2000/5) * sqrt(3)???

Why would I multiply by sqrt(3). I know it's a delta-wye transformer, but I thought the differential relay taps the lines of on the primary and secondary and therefore the ratios between the two sides should just be the ratio of the transformer.

Why am I wrong?

 

[cableguy]

It's a delta-wye transformer, but the CT's get connected wye-delta respectively to account for the phase shift. You end up multiplying the low side of the CT currents by a sqrt(3).

Discussed here also:

http://engineerboards.com/index.php?showtopic=13493