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I read very informative stuff (dated Sept 07) regarding Make and Break Duty of breakers. I would appreciate if someone can clarify why single phase make duty is greater than three phase duty, and is this always the case? Is that due to maximum asymmetry during single phase fault?? Thanks.

Responding to Shaikh's 06-Nov-08 (06:11) query... there are two types of electrical faults; symmetrical and unsymmetrical. Solution of the former is simple because it is a balanced 3-ph system that can be represented as a single-phase series circuit.

A solution of the latter is not as simple because it is unbalanced. Thus, a mathematical technique called Symmetrical Components is used. For more detail refer to Control List Topic:

http://control.com/thread/1026222892

How, then, does the above relate to your question?

Any unsymmetrical-fault can be represented by 3 "balanced" sets of impedances: the first is referred to as the positive-sequence impedance, Z1; the second, the negative-sequence impedance, Z2; and the third, the zero-sequence impedance, Zo.

For the three-phase short circuit, current is determined by the formula:

I(3) = Epn/Z1, where Epn is the phase-to-neutral voltage.

For the phase-to-ground fault, current is:

I(1) = (3xEpn)/(Z1+Z2+Zo).

It is obvious if the 3 impedances were equal, then the phase-to-ground fault current would equal the 3-phase case. However, with some electrical equipment, (e.g. a generator) the positive and negative sequence impedances are equal, but the zero-sequence impedance is smaller. Thus, the phase-to-ground fault current is larger than that of the 3-phase case.

If you require additional info, let me know.

Regards, Phil Corso (cepsicon@aol.com)

Thanks Phil. I was thinking on the lines that when fault occurs at maximum current (zero voltage), initial current is zero due to equal and opposite DC offset. Current reaches its maximum (negative) value after half a cycle and that specifies Make Duty. If this argument is valid for both single and three phase faults and based on what you said that due to smaller ZPS impedance, single phase fault current can be higher than 3 phase one, is this the reason we see higher single phase duty?

Regards

Adding to my 06-Nov-08 (22:43) response:

1) Phase-to-ground fault-current is not always greater than the 3-phase case.

2) Fault-current asymmetry has more to do with the L/R ratio of the total fault path impedance between the source and the point of fault. This affects closing or making duty.

3) Z1 and Z2 are not always equal.

FYI, the formula for a phase-to-phase, fault-current formula is

I(2)= (sqrt(3)xEpn)/(Z1+Z2)

If a neutral impedance (e.g. NGR, or NGT), Zn, is involved, then the fault-current formula is

I(1)= 3xEpn/(Z1+Z2+Zo+3xZn)

Regards, Phil Corso (cepsicon@aol.com)