Answer to Question #212794 in Electrical Engineering for joe

Question

A 2 MVA, 2.3 – kV three – phase, Y – connected alternator operates at rated kVA at a power factor of 80%. The dc armature winding resistance between terminals is 0.08 Ω. The field takes 70 A at 125 V from an exciter equipment. Friction and windage loss is 20 kW, iron losses is 36 kW and stray load losses are 2.0 kW. Calculate the efficiency of the alternator. Assume the effective armature – winding resistance is 120% of the DC resistance.

Answer 1

$P_{out}=s * cos \phi = 2*0.8 =1.6 MW$

$s= \sqrt{3} *v_2* I_2 \implies I_2 = \frac{s}{\sqrt{3}*v_2}= \frac{2*10^6}{\sqrt{3}*2.3*10^6}=349.9 A$

$P_{CU}=3*I_C^2*R_a = 3*349.9^2 (120 \% *0.08) = 35.26 kW$

$P_{in}=P_{out}+P_{cu}+P_{stray}+P_{core}+P_w$

$P_{in}=1.6*10^6+35.26*10^3+2*10^3+36*10^3+20*10^3= 1693.26kW$

$\eta= \frac{P_{out}}{P_{in}}= \frac{1.6*10^6}{1693.26*10^3}*100 = 94.4 \%$