A 460 V 25 hp 60 Hz fourpole Yconnected induction motor has

A 460 V. 25 hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances per phase referred to the stator circuit: R_1 = 0.641 Ohm X_1 = 1.106 Ohm R_2 = 0.332 Ohm X_2 = 0.464 Ohm X_M = 26.3 Ohm The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor\'s Speed Stator current Power Factor P_mech P_shaft T_mech and T_shaft Efficiency

Solution

given v=460

output power P=25 HP= 25*735.5 watts = 18.38 KW.

f=60 hz, Poles = 4 Y- connected

synchrounous speed (N_s)= (120*f)/p = (120*60)/4 = 1800 rpm

a ) motor speed N_r =N_s (1-S) = 1800 * (1- 0.022) = 1760.4 rpm (given s=2.2%)

b ) stator current (I_s)= phasor sum of ( rotor current(I_r) + magnetising current (I_m) )

I_m = V_ph / x_m = (460/1.732)/26.3 = 10.09 A

I_{r =} V_ph/(sqrt((r₁₊r₂/s)^2 + (x₁ + x₂)^2)

substituting the values of V_ph , r₁ , r₂ , x₁ , x ₂, s , we get

I_{r = 16.804 A}

_{so Is=} phasor sum of (I_r + I_m)

I_{s =} sqrt(I_r ² + I_m ²)

I_s = 19.32 A

C) power factor = (R₂ /Z₂ ) = ( R₂ /sqrt( R₂² + SX₂² ))

on solving power factor= 0.998

D) mechanical power

mechanical power= shaft power + mechanical losses

given output power = shaft power = 18.38KW and mechanical losses= 1100 W

therefore mechanical power (P_m ) = 18.38 KW + 1.1 KW = 19.48 KW

now shaft power (P_shaft )= output power = 18.38 KW

E) Mechanical torque

T_mech= P_mech /w( angular speed) = P_mech /(2*3.14*N_r )/60

on solving T_mech = 105.72 N-m

shaft torque

T_shaft = P_shaft /w( angular speed) = P_shaft / (2*3.14*N_r )/60

on solving T_shaft = 99.75 N-m

F) efficiency

efficiency = P_out / (P_out +losses)

given losses= 1100 W and output power = 18,380 W

on substituting we get

efficiency = 94.3 %