Reverse current (plugging) braking of 20 kW DC shunt motor

Data: $V = 220$ V, $P = 20$ kW, $N = 1200$ rpm, $R_a = 0.1\ \Omega$, $\eta = 88\ \%$. Shunt field current is not given, so it is neglected ($I_a \approx I_L$).

Rated current and back emf

$$\begin{aligned} I_a &= \frac{P}{\eta V} = \frac{20000}{0.88\times220} = 103.31\ \text{A} \\ E_b &= V - I_a R_a = 220 - 10.33 = 209.67\ \text{V} \\ \omega &= \frac{2\pi\times1200}{60} = 125.66\ \text{rad/s} \\ K\phi &= \frac{E_b}{\omega} = \frac{209.67}{125.66} = 1.6685\ \text{V s/rad} \end{aligned}$$

i) Series resistance for $I_b = 2I_a$

In reverse current braking (plugging) the armature is reversed, so the supply voltage and back emf add:

$$\begin{aligned} R_a + R &= \frac{V + E_b}{2I_a} = \frac{220 + 209.67}{206.61} = 2.0796\ \Omega \\ R &= 2.0796 - 0.1 = 1.98\ \Omega \end{aligned}$$

ii) Initial braking torque

$$T_b = K\phi\, I_b = 1.6685 \times 206.61 = 344.7\ \text{N·m}$$

iii) Torque at 400 rpm

$$\begin{aligned} E &= 209.67\times\frac{400}{1200} = 69.89\ \text{V} \\ I &= \frac{220 + 69.89}{2.0796} = 139.40\ \text{A} \\ T &= 1.6685 \times 139.40 = 232.6\ \text{N·m} \end{aligned}$$

Answer: (i) 1.98 Ω, (ii) 344.7 N·m, (iii) 232.6 N·m.

The braking torque falls as the motor slows, but does not become zero at standstill ($I = 220/2.0796 = 105.8$ A), so the supply must be switched off at zero speed or the motor will start in reverse.