Induction motor plugging: braking current and torque

Data: $V_{ph} = 400/\sqrt3 = 230.94$ V, $r_1 = r_2' = 1\ \Omega$, $x_1 = x_2' = 2\ \Omega$, 6 poles, 50 Hz. Magnetising branch neglected (approximate circuit).

$$\begin{aligned} N_s &= \frac{120 f}{P} = \frac{120\times50}{6} = 1000\ \text{rpm} \\ s_{fl} &= \frac{1000-950}{1000} = 0.05 \\ s_b &= \frac{N_s + N}{N_s} = \frac{1000+950}{1000} = 1.95 \ \ (\text{plugging}) \end{aligned}$$

i) Initial braking current and torque

$$I = \frac{V_{ph}}{\sqrt{(r_1 + r_2'/s)^2 + (x_1+x_2')^2}}$$ $$\begin{aligned} I_{fl} &= \frac{230.94}{\sqrt{(1+20)^2 + 4^2}} = \frac{230.94}{21.378} = 10.80\ \text{A} \\ I_b &= \frac{230.94}{\sqrt{(1+0.5128)^2 + 4^2}} = \frac{230.94}{4.276} = 54.00\ \text{A} \\ \frac{I_b}{I_{fl}} &= 5.00 \end{aligned}$$

Torque $T = \dfrac{3 I^2 r_2'}{\omega_s s}$, so

$$\begin{aligned} \frac{T_b}{T_{fl}} &= \left(\frac{I_b}{I_{fl}}\right)^2 \frac{s_{fl}}{s_b} = 5.00^2 \times \frac{0.05}{1.95} = 0.641 \end{aligned}$$

Answer (i): $I_b = 5.0\, I_{fl}$ (54.0 A), $T_b = 0.641\, T_{fl}$.

ii) Rotor resistance for $I_b = 1.5\, I_{fl}$

The braking current is largest at the start ($s = 1.95$). Required impedance:

$$\begin{aligned} Z &= \frac{230.94}{1.5\times10.80} = 14.25\ \Omega \\ 1 + \frac{R_2'}{1.95} &= \sqrt{14.25^2 - 4^2} = 13.68 \\ R_2' &= 24.72\ \Omega \ \ (\text{total, referred to stator}) \end{aligned}$$

External resistance referred to stator $= 24.72 - 1 = 23.72\ \Omega$.

Referring to the rotor (stator : rotor turns = 2 : 3, so $R_{rotor} = R' \times (3/2)^2$):

$$R_{ext} = 23.72 \times 2.25 = 53.38\ \Omega \ \text{per phase}$$

New initial braking torque:

$$\begin{aligned} \frac{T_b'}{T_{fl}} &= \frac{(1.5 I_{fl})^2 (24.72/1.95)}{I_{fl}^2 (1/0.05)} = \frac{2.25 \times 12.68}{20} = 1.43 \end{aligned}$$

Answer (ii): insert 53.4 Ω per phase in the rotor (23.7 Ω referred to stator); initial braking torque = 1.43 × full-load torque.

Adding rotor resistance reduces the current but raises the torque, because the rotor power factor improves at high slip.