Plugging resistance for 30 kW DC shunt motor

Data: $V = 220$ V, $N = 535$ rpm, $R_a = 0.086\ \Omega$, $I_a = 150$ A (full load), limit $I_b = 200$ A.

Back emf at full-load speed

$$\begin{aligned} E_b &= V - I_a R_a = 220 - 150\times0.086 = 207.1\ \text{V} \\ \omega &= \frac{2\pi\times535}{60} = 56.03\ \text{rad/s} \\ K\phi &= \frac{207.1}{56.03} = 3.6966\ \text{V s/rad} \end{aligned}$$

Series resistance for plugging

At the instant of plugging, the reversed supply voltage and the back emf act in the same direction round the armature circuit:

$$\begin{aligned} R_a + R &= \frac{V + E_b}{I_b} = \frac{220 + 207.1}{200} = 2.1355\ \Omega \\ R &= 2.1355 - 0.086 = 2.05\ \Omega \end{aligned}$$

Initial braking torque

$$T_{b0} = K\phi\, I_b = 3.6966 \times 200 = 739.3\ \text{N·m}$$

Torque at half speed (267.5 rpm)

$$\begin{aligned} E &= \frac{207.1}{2} = 103.55\ \text{V} \\ I &= \frac{220 + 103.55}{2.1355} = 151.51\ \text{A} \\ T &= 3.6966 \times 151.51 = 560.1\ \text{N·m} \end{aligned}$$

Answer: series resistance = 2.05 Ω; initial braking torque = 739.3 N·m; torque at half speed = 560.1 N·m.

(Full-load torque for comparison $= 3.6966 \times 150 = 554.5$ N·m, so braking starts at 1.33 × full-load torque.)