Problem 92-8   Source:   Example 2-3 - page 66 -    FITZGERALD, A. E. , KINGSLEY, Charles Jr. , UMANS, Stephen D. -    Book: Electric Machinery - 5th ed. - McGraw-Hill - 1990.

The 50 kVA, 2,400:240 V transformer whose constants are given in the previous problem is used to step down the voltage at the load end of a feeder whose impedance is 0.30 + j1.60  Ω. The voltage V₁ at the sending end of the feeder is 2,400 V.

Find the voltage at the secondary terminals of the transformer when the load connected to its secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops in the transformer and feeder caused by the exciting current.

Solution of the Problem 92-8

In the Figura 92-8.1 we see the representation of the transformer with all the values referenced to the high voltage side, where the equivalent impedance of the transformer is represented according to the eq. 92-08. Thus, the value of equivalent impedance is equal to Z_eq = 1.42 + j1.82  Ω. On the other hand, the combined series impedance (Z_total) of the transformer and feeder is equal to Z_total = 1.72 + j3.42  Ω.

From the rated primary power and voltage, we can calculate the current in the load referenced to the high voltage side, or:

To facilitate the solution of the problem, we will base ourselves on the phasor diagram shown in Figura 92-8.2 and remembering that:

As we know the value of I_L, R and X we can find the values of V_R and V_X, or:

Now, based on Figura 92-8.2 and using a basic trigonometry we can write:

In addition, we know that:

Substituting for the numerical values, we find:

Thus, we conclude that:

And therefore, reflecting this value for the secondary, remembering that a = 10, we have: