Control power transformer - Secondary's view of primary's short circuit

[brainsalad]

Hello,

Suppose that a 480/120V, 350VA transformer is fused on the secondary, and that there are 10,000A fault current available on the primary. Is it correct that the secondary fuse should be rated to interrupt the available 10,000A? If so, has anyone witnessed such an event in practice? Thanks for your thoughts.

 

[dpc]

The available fault current on the secondary will be much less than 10,000 A due to the impedance of the transformer.

 

[waross]

I would expect 40 to 60 amps available short circuit current on the secondary. Possibly a little more, most likely less than 40 Amps.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[brainsalad]

Thanks. So in general, secondary fuses will never see primary fault amps, even through a shorted out transformer?

 

[alon13]

I don't think you will see 10KA fault in the secondary.
On another note, if you are bound to NEC rules, you should put a fuse on the primary side to protect the transformer in case there was an internal short.

 

[LionelHutz]

I ~suppose~ you could argue that it's possible for a transfrormer failure to cause one of the high side terminals to be shorted to a low side terminal and then the control power shorting to ground would cause a fault where the secondary fuse sees the line to ground fault current. But, that's not something that is considered in practice when selecting the secondary fuse.

 

[dpc]

Even if that happened, the primary protection somewhere upstream should operate for a 10 kA fault.

 

[ScottyUK]

It's not hard to find a standard industrial fuselink of a size suitable for the control circuit which is capable of interrupting at least 80kA. Most common MCBs can clear 10kA too.


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If we learn from our mistakes I'm getting a great education!

 

[brainsalad]

Thanks for your thoughts. To show that the actual secondary fault current (as "seen" through the fuse) through a 480/120V, 150VA, Z = 3.5%, control power transformer, while considering the 10,000A available fault amps (at the primary), does the following make sense?

Using secondary as base,

I_base = 150VA / 120V = 1.25 A

Z_base = 120V / 1.25A = 96 Ohm

I_fault = 10000A / 1.25A = 8000 A pu

Z_fault = 1.0V / 8000 Ohm = 1.25e-4 Ohm

Then fault current seen by secondary fuse is I = 1.0 / (0.035 + 1.25e-4) = 28.47A ?

 

[mgtrp]

Short answer... no, you've messed up your per unit, both conversion to and conversion back from.

Longer answer... ignore the system impedance and assume only the transformer impedance, giving you:
If(pu) = 1 / 0.035 = 28.57 pu
If(A) = 28.57 x (150VA/120V) = 35.71 A

You could add in the system impedance, but you'll probably find the change is negligible.