Hey guys. 
Just studying for my exam and I'm a little stuck on this question. mainly part b)
A three phase, four pole, 60Hz motor is rated at 10hp, 208V and 1755rpm. The approximate equivalent circuit referred to the stator is as follows;
R1 = 0.15, R2 = 0.15, X1 = 0.4, X2 = 0.25, Xm = 30
The combined rotational losses(including frictional, winding and core losses) are 500W. The motor operates at rated speed when connected to a 208V 60Hz source. Assuming the shunt component (Xm) is connected at the front of the equivalent circuit, calculate
a)Synchronous speed and slip. draw the equivalent circuit.
b)Line current and input power factor
c)induced (airgap) power and torque
d)Output horsepower and torque
e)Efficiency of the motor
Solutions.
S = 0.025, ILine = 20.25A, PF = 0.95 lagging, Pg = 6787W, Tg = 36Nm, Hp = 8.2Hp, Tout = 33.28Nm, efficiency = 88%
my attempt
a) Nsyc = (120*60)/4 = 1800
s = (1800 - 1755)/1800 = 0.025
b)This is where I'm a bit lost for what formula to use
is IL = (Vphase/(R1 + (R2/s)) ?
= (208/(√3))/(0.15 + (0.15/0.025)) = 19.52A.... Not quite right
Thanks guys
Just studying for my exam and I'm a little stuck on this question. mainly part b)
A three phase, four pole, 60Hz motor is rated at 10hp, 208V and 1755rpm. The approximate equivalent circuit referred to the stator is as follows;
R1 = 0.15, R2 = 0.15, X1 = 0.4, X2 = 0.25, Xm = 30
The combined rotational losses(including frictional, winding and core losses) are 500W. The motor operates at rated speed when connected to a 208V 60Hz source. Assuming the shunt component (Xm) is connected at the front of the equivalent circuit, calculate
a)Synchronous speed and slip. draw the equivalent circuit.
b)Line current and input power factor
c)induced (airgap) power and torque
d)Output horsepower and torque
e)Efficiency of the motor
Solutions.
S = 0.025, ILine = 20.25A, PF = 0.95 lagging, Pg = 6787W, Tg = 36Nm, Hp = 8.2Hp, Tout = 33.28Nm, efficiency = 88%
my attempt
a) Nsyc = (120*60)/4 = 1800
s = (1800 - 1755)/1800 = 0.025
b)This is where I'm a bit lost for what formula to use
is IL = (Vphase/(R1 + (R2/s)) ?
= (208/(√3))/(0.15 + (0.15/0.025)) = 19.52A.... Not quite right
Thanks guys