Question

Problem 8 6000 V, V2 231 V, f 50 20 kVA, V II.1. A single-phase transformer rated at S Hz, was tested at no-load and short-circuit conditions. During the open-circuit test LV side was 231V, I 8.5A supplied and the readings are as follows: V, = = Po-191W. Short-circuit test was curried out with HV side supplied and the following readings were obtained: Vsc = 220V, Isc = 2.71A, Psc = 420W. The measured LV winding resistance is equal to R2 = 0.032. a) Determine equivalent circuit parameters referred to HV side b) Find the supplied voltage that gives the rated current at short-circuited secondary c) What is the short-circuit current if the HV side is supplied with the 0.5 rated voltage and 60 Hz frequency

Also, if you happen to have any idea of what book this came out of (or one that has very similar problems) I would GREATLY appreciate that info. Thanks!!

Answer 1

Each step is done one by one. For short circuit tests, magnetizing branch parameters are neglected. You can refer Chapman or Fitzgerald Kingsley books for such problems.

(a) From open-circuit that, core loss resistance R_c = V^2_0/P_0 = 23162/191 = 279.38 ohm p.f = cos theta = P_0/v_0 I_0 = 191/231 times 8.5 = 0.097 therefore sin theta = squareroot 1 - 0.097^2 = 0.995 therefore I_m = I_0 sin theta = 8.5 times 0.995 = 8.46A therefore Magnetizing reactance X_m = v_0/I_m = 231/8.46 = 27.31 ohm But R_c and x_m are w.r.t LV side converting to HV side, R_c = 279.38 times (6000/231)^2 = 188483.72 ohm x_m = 27.31 times (6000/231)^2 = 18424.96 ohm \r\n From short circuit test, Z_eq = v_sc/I_sc = 220/2.71 = 81.18 ohm R_eq = p_sc/I^2_sc = 420/2.71^2 = 57.19 ohm therefore x_eq = squareroot Z^2_eq - R_eq = squareroot 81.18^2 - 57.19^2 = 57.62 ohm But R_2 = 0.032 times (6000/231)^2 = 25.59 ohm w.r.t HV. Therefore R_1 = 57.19 - 25.59 = 31.6 ohm X_1 = X_2 = 57.62/2 = 28.81 ohm/w.r.t. HV side. \r\n (b) Rated current = rated VA/rated voltage = 20 times 10^3/6000...