Problem 92-4
Source: Problem developed by the site author.
A single-phase 25 KVA transformer, with a primary voltage of 13,200 V and a secondary voltage of 220 V,
it has a primary winding resistance of 0.3 Ω and a secondary winding resistance of 0.005 Ω. A load of 20 KVA with a FP = 0.75 is connected to the secondary. Iron losses are 120 W.
a) In this condition, find the efficiency of the transformer.
b) If you change the load to 25 KVA with FP = 0.9, what is the new efficiency?
Solution of the Problem 92-4
Item a
Initially we will determine the value of the electrical current in the secondary, I2, of the transformer.
I2 = Scarga / V2 = 20,000 / 220 = 90.91 A
The transform ratio is a = 13,200 / 220 = 60. We can calculate the current in the transformer primary, that is,
I1. So:
I1 = I2 / a = 90.91 / 60 = 1.52 A
With the data of the problem we can calculate R'eq, which is the equivalent resistance of the
transformer windings referenced to the primary. So:
R'eq = R1 + R2 a2 = 54 + 0.015 x 602
Performing the calculation, we have:
R'eq = 108 Ω
Now we have the values needed to calculate copper losses, or:
Pcu = R'eq I12 = 108 x (1.52)2
Performing the calculation, we have:
Pcu = 249.52 W
Now let's use the equation eq. 92-21 to calculate the transformer efficiency, remembering that
R''eq I22 = R'eq I12. So we have:
η = 13,200 x 1.52 x 0.75 / (13,200 x 1.52 x 0.75 + 220 + 249.52 )
Performing the calculation, we have:
η = 0.9697 ou η = 96.97%
Item b
Initially we will determine the value of the electric current in the primary, I1, from the transformer to the power nominal.
I1 = Scarga / V1 = 25,000 / 13,200 = 1.894 A
Thus, the new value of copper losses is:
Pcu = R'eq I12 = 108 x (1.894)2
Performing the calculation, we have:
Pcu = 387.42 W
Calculating V1 I1 FP = 22,500.72 W. Therefore, the value of the transformer efficiency, using eq.92-21 is:
