25 kW separately excited DC machine as motor or generator

Given: open-circuit (generated) EMF $E = 125$ V (constant, since speed and field current are constant), $R_a = 0.02\ \Omega$, speed $N = 3000$ rpm.

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 3000}{60} = 314.16\ \text{rad/s}$$

Take armature current positive when it flows out of the machine (generator action): $I_a = \frac{E - V_t}{R_a}$.

(i) Terminal voltage 128 V

$$I_a = \frac{125 - 128}{0.02} = -150\ \text{A}$$

The negative sign means current flows into the armature: the machine runs as a motor, taking 150 A from the 128 V supply.

$$\begin{aligned} \text{Terminal power} &= V_tI_a = 128 \times 150 = 19200\ \text{W} = 19.2\ \text{kW (input)} \\ \text{Electromagnetic power} &= EI_a = 125 \times 150 = 18750\ \text{W} = 18.75\ \text{kW} \\ T &= \frac{EI_a}{\omega} = \frac{18750}{314.16} = 59.68\ \text{N·m} \end{aligned}$$

The torque is a motoring torque (in the direction of rotation). The difference $19200 - 18750 = 450$ W $= 150^2 \times 0.02$ is the armature copper loss.

(ii) Terminal voltage 124 V

$$I_a = \frac{125 - 124}{0.02} = 50\ \text{A}$$

Positive, so the machine runs as a generator, delivering 50 A.

$$\begin{aligned} \text{Terminal power} &= 124 \times 50 = 6200\ \text{W} = 6.2\ \text{kW (output)} \\ \text{Electromagnetic power} &= 125 \times 50 = 6250\ \text{W} = 6.25\ \text{kW} \\ T &= \frac{6250}{314.16} = 19.89\ \text{N·m} \end{aligned}$$

This torque opposes rotation and must be supplied by the prime mover. Loss $= 6250 - 6200 = 50$ W $= 50^2 \times 0.02$.

Answer: (i) $I_a = 150$ A (motoring), $P_t = 19.2$ kW, $P_{em} = 18.75$ kW, $T = 59.68$ N·m; (ii) $I_a = 50$ A (generating), $P_t = 6.2$ kW, $P_{em} = 6.25$ kW, $T = 19.89$ N·m.

This shows that a DC machine changes from generator to motor simply when the terminal voltage exceeds the generated EMF.