Question

i want it step by step. Thank u.

1. A 24 kV, 200 MVA round rotor synchnronous generator (Xs 1.0, r -0) is synchronized to a bus whose voltage is 1.0 @0 deg. At synchronization iç 1000A (actual). The mechanical power is increased until PG 0.8pu. Now if is adjusted from 1000A to 1600A (with PG and Va held constant) (a) Plot a curve of Ial versus if (b) Plot a curve of QG versus if (c) When Ia is minimum, what is La(magnitude and angle)? What is the power factor? (d) Find Xs in ohms

Answer 1

Given, V_rated = 24 kv = V_b S_rated = 200 m v a = s_b X_s = 1 p.u I_b = 200/squareroot 3 times 24 I_b = 4.811 K Amps Z_b = V_b^2/S_b = 24 times 24 /200 Z_b = 2.88 ohm rightarrowX_s = 1 p.u times 2.88 ohm rightarrowX_s = 2.88 ohm E_f = V 0 degree + j times X_s I_a 0 degree = 24 0 degree + j (2.88)(4.811) (Assuming units pf rightarrowphi = 0 degree) = (24 0 degree + j 13.856) k volts E_f_1 s = 27.713 30 degree @ unity pf s_1 = 30 degree P_G = Efv/X_s sin delta = 27.713 times 24/2.88 sin 30 degree P_G_1 = 115.471 m watts P_G = 115.471/200 rightarrowP_G_1 = 0.5774 p.4 \r\n According to the problem, mechanical i/p is increased so that P_G = 0.8 pu rightarrowP_G_2 = 0.8 = Efv /X_s sin delta { E_f_2 = ? V = 1 pu, X}