A 3-phase, 25 kW, 400 V, 50 Hz, 8-pole induction motor has rotor resistance of 0.08 \omega and standstill reactance of 0.4 \omega . The effective stator/rotor turn ratio is 2.5/1. The motor is to drive a constant-torque load of 250 Nm. Neglect stator impedance.
(a) Calculate the minimum resistance to be added in rotor circuit for the motor to start up on load.
(b) At what speed would the motor run, if the added rotor resistance is (i) left in the circuit, and (ii) subsequently short circuited.
Motor circuit seen on rotor side is shown in Fig. 9.45; stator impedance having been neglected.
I_{2}=\frac{V_{1} / a}{\sqrt{\left(R_{2} / s\right)^{2}+X_{2}^{2}}}T=\frac{3}{\omega_{s}} \cdot I_{2}^{2} R_{2} / s
=\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2}\left(R_{2} / s\right)}{\left(R_{2} / s\right)^{2}+X_{2}^{2}} (i)
T(\text { start })=\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2} R_{2}}{R_{2}^{2}+X_{2}^{2}} ; s=1 (ii)
Let external resistance added to rotor circuit be R_{2} (ext). Then
R_{2}(\text { total })=R_{2 t}=R_{2}+R_{2}(\text { ext }) (iii)
Then T( start )=\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2} R_{2 t}}{R_{2 t}^{2}+X_{2}^{2}} (iv)
(a) a=2.5, \quad X_{2}=0.4 \Omega, \quad R_{2}=0.08 \Omega
T(start) = T(load) = 250 Nm; This is minimum starting torque.
Actual starting must be sufficiently more than this.
V=400 / \sqrt{3}=231 V
n_{s}=750 rpm or \omega_{s}=78.54 rad / s
Substituting values in Eq. (ii)
250=\left(\frac{3}{78.54}\right) \cdot \frac{(231 / 2.5)^{2} R_{2 t}}{R_{2 t}^{2}+(0.4)^{2}} (iv)
Or R_{2 t}^{2}-1.304 R_{2 t}+0.16=0
Or R_{2 t}=0.137 \Omega, 1.167 \Omega
The T-s characteristics with these two values of R_{2t} are drawn in Fig. 9.46. It is easy to see that with R_{2 l}=1.167 \Omega the motor will not start as motor torque reduces \left(T_{\text {motor }}< T_{\text {load }}\right) ) for s < 1.
So we select R_{2 t}=0.137 \Omega
\Rightarrow R_{2}( ext )=0.137-0.08=0.057 \Omega
(b)(i) R_{2 t}=0.137 \Omega ; External resistance included in circuit.
Substituting values in Eq. (i)
250=\left(\frac{3}{78.54}\right) \frac{(231 / 2.5)^{2}\left(R_{2 t} / s\right)}{\left(R_{2 t} / s\right)^{2}+(0.4)^{2}}This equation has the same solution as Eq. (v). Thus
R_{2 t} / s=0.137,0.167With R_{2 t}=0.137 \Omega , we get
s = 1,0.137/1.167
= 0.117 (as shown in Fig. 9.42)
Motor speed, n = 750 (l – 0.117) = 662 rpm
(ii) With external resistance cut out
250=\left(\frac{3}{78.54}\right) \frac{(231 / 2.5)^{2}\left(R_{2} / s\right)}{\left(R_{2} / s\right)^{2}+(0.4)^{2}}The solution would yield as before
R_{2} / s=0.137, R_{2} / s=1.167The solution points are indicated in the T-s characteristic drawn in Fig. 9.47. The motor will run at
s = 0.067 \Rightarrow 700 rpm
The point s = 0.584 on T-s characteristic is unstable as
the torque-speed slope here is positive and the motor will speed up beyond this.
