Full-load torque, rotor current and copper loss of 440 V motor

Given: $V_L = 440$ V (Y), $f = 50$ Hz, $P = 4$, $N = 1425$ rpm, rotor impedance at standstill $= 0.4 + j4\ \Omega$/phase, rotor/stator turns ratio $K = 0.8$. Stator impedance is neglected.

Slip and rotor emf

$$\begin{aligned} N_s &= \frac{120 \times 50}{4} = 1500\ \text{rpm}, \quad s = \frac{1500 - 1425}{1500} = 0.05 \\ E_1 &= \frac{440}{\sqrt{3}} = 254.03\ \text{V/phase} \\ E_2 &= K E_1 = 0.8 \times 254.03 = 203.23\ \text{V/phase (standstill)} \end{aligned}$$

(ii) Rotor current

$$\begin{aligned} I_2 &= \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}} = \frac{0.05 \times 203.23}{\sqrt{0.4^2 + (0.05 \times 4)^2}} \\ &= \frac{10.161}{\sqrt{0.16 + 0.04}} = \frac{10.161}{0.4472} = 22.72\ \text{A} \end{aligned}$$

(iii) Full-load rotor copper loss

$$P_{cu2} = 3 I_2^2 R_2 = 3 \times 22.72^2 \times 0.4 = 619.5\ \text{W}$$

(i) Full-load torque

$$\begin{aligned} P_{ag} &= \frac{P_{cu2}}{s} = \frac{619.5}{0.05} = 12\,390\ \text{W} \\ \omega_s &= \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T &= \frac{P_{ag}}{\omega_s} = \frac{12\,390}{157.08} = 78.88\ \text{N·m} \end{aligned}$$

Check: $P_m = (1-s)P_{ag} = 11\,771$ W and $T = P_m/\omega = 11\,771/149.23 = 78.88$ N·m, which agrees. (This is the gross developed torque; mechanical losses are not given.)

Answer: (i) Full-load torque = 78.88 N·m; (ii) rotor current = 22.72 A; (iii) rotor copper loss = 619.5 W.