short circuit estimation

E

Thread Starter

engineer

talking about a hypothetical situation,

if the load side of a transformer gets short (not considering breakers right now) momentarily, what happens to the primary side voltage and current?
 
C

Curt Wuollet

First pass estimation is that it acts shorted as well as zero impedance is reflected into the primary.

Actual results depend on leakage reactance, mutual coupling, wire resistance, etc. But for most practical purposes. it's a short compared to the external circuit constants.

Regards
cww
 
Engineer...

if the transformer is fed from an infinite source, that is, there is no impedance between the source and the Xfmr's primary terminals, then input voltage will remain constant.

Input current = kVA/[Sqrt(3)x(primary kV]

If there is a primary impedance, then see the example calculation in Control.Com thread:

o http://www.control.com/thread/11352793747

Regards,
Phil Corso
 
S

Sebastian Rebord

It depends of the type of transformer (ej: Dyn11), the X% of transformer and the type of shortcircuit (ej: 1ph to earth). Depending this data you get the different sequence circuits (direct, inverse and homopolar) and are able to solve it out. For example, if you have a transformer Dyn and a single phase sort circuit in the D side, then in this side you have an unbalance in the voltages), but no currents appears if you don't have an earth path on this side.

The hypothetical situation could be very very wide.
 
Engineer… I directed you to the wrong reference, so here is the example solution:

Using my Gi.Fi.E.S. method (pronounced Jiffys) for the problem solution, where Gi is for "Given"; Fi for "Find"; E for "Equation(s)"; and S for "Solution", then the answers to your two questions are:<pre>
Q1/A1
Given:
3-Ph,Xfmr with design ratngs: kVA = 15.0; kVp = 0.460; kVs = 0.208Y / 0.120kV; Imp, Z% = 5; connected to an infinite source.

Find:
3-Ph short-circuit current duty on secondary terminals, Isc(s).

Equation(s):
Infinite means zero impedance between the transformer primary and the source of supply, then the 3-Ph short-circuit current level is the Xfmr’s rated kVA, divided by its impedance expressed as a decimal Ccommonly referred to as per-unit):
o kVAsc(s) = kVA / [Z% / 100], and,
o Isc(s) = kVAsc(s) / [Sqrt(3) x kVs]

Solution:
o kVAsc(s) = 15/0.05 =300 kVA
o Isc(s) = 300 / [1.732 x 0.208] = 833 A

Q2/A2
Given:
Smame parameters as above except that 3-Ph fault-current level at Xfmr primary, Isc(p) = 22,000 A.

Find:
3-Ph short-circuit current duty on secondary, Isc(s).

Equation(s):
Typically, resistance are very small so they will be ignored! Also, I now introduce a short-cut procedure called the kVA-Value Method, or MVA-Value method.

Step 1:
Convert the primary fault-current to kVAsc(p),
o Primary kVAsc(p) = Sqrt(3) x Isc(p) x kVp.
o Xfmr short-ckt kVAsc(s) = 300 kVA, from Q1/A1, above.

Step 2:
Combine the two impedances together as if they were two resistances in parallel, that is,
o kVAsc(s) = [kVAsc(p) x kVAsc(s)] / [kVAsc(p) + kVAsc(s)].
o Isc(s) = [kVAsc(s)] / [Sqrt(3) x kVs].

Solution:
o kVAsc(p) = Sqrt(3) x 22,000 x 0.460 = 17,528 kVA.
o kVAsc(s) = [17,525 x 300] / [17525+300 ] = 295 kVA.
o Isc(s) = 295 / [1.732 x 0.208] = 818 A.</pre>
Engineer, you could have converted all impedances to ohms, percent, or per-unit, but there is quite a bit of work involved! If you are unable to process the above, please let me know!

Regards,
Phil Corso
 
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