Practice 13

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The short-circuit test of a transformer is to connect the terminals of the secondary winding shorted, and this provision applied to a low voltage primary Ucc (shunt voltage) so that flow through this winding rated current.


Fundamentals of Practice

This test provides the value of the nominal copper losses as short-circuit iron losses are negligible due to applying a low voltage Ucc, induction is reduced in the same proportion as the iron losses are roughly proportional to the square of the induction, it is clear who will be negligible.
If Ucc = 5% U 1n => Bcc ~ 0.05 B

where B = maximal induction with the nominal
Bcc = maximal induction with short-circuit voltage

then: P FECCAS = (0.05) 2 · P Fe = 1 / 400 · P Fe

where: P Fe = iron losses nominal
P FECCAS = iron losses in the short circuit test

The power absorbed by the transformer short-circuit will be the nominal copper losses.
The test also allows to calculate the short-circuit power factor, voltage drops in resistive and reactive relative and absolute value and the value of resistance and short-circuit reactance.
If Pcc is the power input, UCC is the reduced stress or applied and R 1n is the current flow equal to the rated current, it will have the short-circuit power factor is cos öcc = Pcc / Ucc · Iin
Resistive voltage drop in absolute value: U Rcc = Ucc · cos cc
Relative voltage drop in absolute value: U Xcc = Ucc · sin ö cc
The short circuit voltage and voltage drops in resistive and reactive relative values will be: cc% = Ucc · 100 / U 1n
to CCR% = URCC · 100 / U 1n
å Uxcc · xcc% = 100 / U 1n

The value of shunt resistance and reactance are:
Rcc Rcc = U / I 1n
Xcc = Uxcc / I 1n

The measures should be made quickly to prevent the overheating of the windings.

Equipment needed

A single-phase transformer, an AC power source, a voltmeter to ac cos a True or reduced and an ammeter to ca

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