Saturday, September 12, 2015

Short circuit test on Transformers;Why Short circuit test performed on HT side

As we have seen in open circuit test that it is done on low voltage winding. Short circuit test is done of High voltage winding. In this case Ammeter, Voltmeter, Wattmeter are connected on High voltage winding of the transformer. This test is done to calculate copper losses in Transformer at full load. This test is performed on HT side not on LT side as Short circuit test in Transformer is conducted at its rated current which can be easily obtained by applying 5-6% of normal voltage.





If we conduct Short circuit test on LT side by short circuiting HT side then Voltage of HT winding falls to zero which leads to very high current on High voltage winding leading to burning out of winding.

For no load test on Transformers visit link:-
http://electrialstandards.blogspot.in/2015/09/no-load-test-on-transformers-open.html


Now lets take a Example if we have Transformer 2 MVA, 415V/11000V then 
if we perform Short circuit in LT side the current will be:-

2     X      10^6 =   2782A
1.732 X 415

Now if short circuit test is done on HT side then rated current will be:-
2     X      10^6 =   104A
1.732 X 11000

Now you will see how much the difference in rated currents.



Also Since short circuit test is done at rated current so it can be easily obtained in HT side then on LT side as rated current value is low on HT side instead of LT side as there the difference equal to Transformation ratio.

Connection diagram for performing Short circuit test on Transformer in as Shown in figure below:-


Short circuit test on Transformer

With the help of Variac voltage is applied and increased slowly until the ammeter on HT side read the rated current. Now readings of all the connected instruments are recorded. The ammeter reads the full load current of primary side which will be If. As voltage applied for achieving this full load current will be very small core losses can be ignored. Now wattmeter connected on HT side will give copper losses. Now let’s assume wattmeter reading is Ps.c. and Voltmeter reading as Vs.c. then


Ps.c.= If2 R

Where R is equivalent resistance.