Answer to Question #230229 in Electrical Engineering for KHANYA

Question

A 1.5 MVA ,3.3 kV, 40 pole, three-phase star connected asynchronous alternator has reactance of Xd = 4.01Ω/phase and Xq = 2.88Ω/phase.For a lagging power factor of 0.8 when delivering full load. Calculate a developed power: 

Answer 1

$V_t= \frac{3300}{\sqrt3}=1905.25 \\ \delta = \phi -\theta = 14.904\\ E_f= V_t \cos \delta +I_dX_d= 1905.25 \cos(14.9) +215.925*4.0=2707.0472 V\\$

$Developed \space power = \frac{E_f V_t}{X_d} \sin \delta + \frac{V_t^2}{2}(\frac{1}{X_2}-\frac{1}{X_d})\sin2 \delta\\ = \frac{4.688*3.3}{4.01} \sin 14.9 + \frac{3.3^2}{2}(\frac{1}{2.88}-\frac{1}{4.01})\sin2*14.9\\ =1.256 MW$