200 V shunt motor power, torque and efficiency

Given: $V = 200$ V, $R_a = 0.12\ \Omega$, $R_{sh} = 250\ \Omega$, rotational loss = 250 W, $I_L = 10$ A, $N = 1500$ rpm.

Currents and back emf

$$\begin{aligned} I_{sh} &= \frac{200}{250} = 0.8\ \text{A} \\ I_a &= I_L - I_{sh} = 10 - 0.8 = 9.2\ \text{A} \\ E_b &= V - I_a R_a = 200 - 9.2 \times 0.12 = 198.896\ \text{V} \end{aligned}$$

(i) Mechanical power developed

$$P_m = E_b I_a = 198.896 \times 9.2 = 1829.84\ \text{W}$$

(ii) Power output

$$P_{out} = P_m - P_{rot} = 1829.84 - 250 = 1579.84\ \text{W}$$

(iii) Load (shaft) torque

$$\begin{aligned} \omega &= \frac{2\pi N}{60} = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T_L &= \frac{P_{out}}{\omega} = \frac{1579.84}{157.08} = 10.06\ \text{N·m} \end{aligned}$$

(iv) Full-load efficiency

$$\begin{aligned} P_{in} &= V I_L = 200 \times 10 = 2000\ \text{W} \\ \eta &= \frac{1579.84}{2000} \times 100 = 78.99\% \end{aligned}$$

Check of losses: armature copper $= 9.2^2 \times 0.12 = 10.16$ W, field copper $= 200 \times 0.8 = 160$ W, rotational 250 W; total 420.16 W $= 2000 - 1579.84$ W.

Answer: (i) $P_m \approx 1829.8$ W, (ii) $P_{out} \approx 1579.8$ W, (iii) $T_L \approx 10.06$ N·m, (iv) $\eta \approx 79.0\%$.