Electrical Engineering - Transformer Excercises, Study notes of Electrical and Electronics Engineering

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Transformer Exercises

Question: 1

• (^) The O.C and S.C test data are given below for a single phase, 5 kVA, 200V/400V, 50Hz transformer.
• (^) Draw the equivalent circuit of the transformer
• (^) (i) referred to LV side and
• (^) (ii) referred to HV side inserting all the parameter values.

Solution

• (^) After knowing the value of cosθo and sinθo and referring to the no load phasor diagram, Im 1 and Icl 1 can be easily calculated as follows.

Solution

Solution

• (^) It may be noted that from the O.C test data we

can get the parallel branch impedances namely

the magnetizing reactance and the resistance

representing the core loss referred to the side

where measurements have been taken.

Computation with S.C test data

• (^) Since the test has been carried out from the HV side with LV side shorted, we draw the equivalent circuit referred to the HV side as shown in figure
• (^) Parameter values are denoted by using suffix 2. Important point to note here is the absence of the parallel branch.

Solution

• (^) Calculation of series parameters is rather simple

and as follows.

Solution

Equivalent circuit referred LV

side

• (^) The parallel branch parameters Rcl 1 = 266.67Ω and Xm 1 = 200Ω have already been calculated wrt LV side. Naturally no further transformations are necessary.
• (^) However, series parameters re 2 and xe 2 have been calculated above from test data.
• So we need to calculate re 1 and xe 1 in order to correctly represent the equivalent circuit referred to primary side.

Solution

Equivalent circuit referred HV

side

• (^) Here we note that series parameters referred to

HV side are already known to be re 2 = 1.12Ω and

xe 2 = 1.14Ω. However, the parallel branch

parameters are to be transformed as follows.

Solution

• (^) We are now in a position to draw the equivalent

circuit of the same transformer referred to the HV

side as shown in figure

Solution

Solution

ii. We know maximum efficiency occurs when x^2 Pcu = Pcore, where Pcu is the full load copper loss and Pcore is the iron loss. Now Pcu = 175 W and Pcore = 120 W.