A 3-phase, 25 kW, 400 V, 50 Hz, 8-pole induction motor has rotor resistance of 0.08 Ωand standstill reactance of 0.4 Ω . 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

Question

A 3-phase, 25 kW, 400 V, 50 Hz, 8-pole induction motor has rotor resistance of 0.08 [latex] \omaga[/latex] and standstill reactance of 0.4 [latex] \omaga[/latex] . 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.

Accepted Answer

Motor circuit seen on rotor side is shown in Fig. 9.45; stator impedance having been neglected.

[latex] I_{2}=\frac{V_{1} / a}{\sqrt{\left(R_{2} / s ight)^{2}+X_{2}^{2}}}[/latex]

[latex] T=\frac{3}{\omega_{s}} \cdot I_{2}^{2} R_{2} / s[/latex]

[latex] =\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2}\left(R_{2} / s ight)}{\left(R_{2} / s ight)^{2}+X_{2}^{2}}[/latex] (i)

[latex] T( ext { start })=\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2} R_{2}}{R_{2}^{2}+X_{2}^{2}} ; s=1[/latex] (ii)

Let external resistance added to rotor circuit be [latex] R_{2}[/latex] (ext). Then

[latex] R_{2}( ext { total })=R_{2 t}=R_{2}+R_{2}( ext { ext })[/latex] (iii)

Then [latex] T( start )=\frac{3}{\omega_{s}} \cdot \frac{(V / a)^{2} R_{2 t}}{R_{2 t}^{2}+X_{2}^{2}}[/latex] (iv)

(a) [latex] a=2.5, \quad X_{2}=0.4 \Omega, \quad R_{2}=0.08 \Omega[/latex]

T(start) = T(load) = 250 Nm; This is minimum starting torque.

Actual starting must be sufficiently more than this.

[latex] V=400 / \sqrt{3}=231 V[/latex]

[latex] n_{s}=750 rpm[/latex] or [latex] \omega_{s}=78.54 rad / s[/latex]

Substituting values in Eq. (ii)

[latex] 250=\left(\frac{3}{78.54} ight) \cdot \frac{(231 / 2.5)^{2} R_{2 t}}{R_{2 t}^{2}+(0.4)^{2}}[/latex] (iv)

Or [latex] R_{2 t}^{2}-1.304 R_{2 t}+0.16=0[/latex]

Or [latex] R_{2 t}=0.137 \Omega, 1.167 \Omega[/latex]

The T-s characteristics with these two values of [latex] R_{2t}[/latex] are drawn in Fig. 9.46. It is easy to see that with [latex] R_{2 l}=1.167 \Omega[/latex] the motor will not start as motor torque reduces [latex] \left(T_{ ext {motor }}< T_{ ext {load }} ight) [/latex] ) for s < 1.

So we select [latex] R_{2 t}=0.137 \Omega[/latex]

[latex] \Rightarrow[/latex] [latex] R_{2}( ext )=0.137-0.08=0.057 \Omega[/latex]

(b)(i) [latex] R_{2 t}=0.137 \Omega[/latex] ; External resistance included in circuit.

Substituting values in Eq. (i)

[latex] 250=\left(\frac{3}{78.54} ight) \frac{(231 / 2.5)^{2}\left(R_{2 t} / s ight)}{\left(R_{2 t} / s ight)^{2}+(0.4)^{2}}[/latex]

This equation has the same solution as Eq. (v). Thus

[latex] R_{2 t} / s=0.137,0.167[/latex]

With [latex] R_{2 t}=0.137 \Omega[/latex] , 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

[latex] 250=\left(\frac{3}{78.54} ight) \frac{(231 / 2.5)^{2}\left(R_{2} / s ight)}{\left(R_{2} / s ight)^{2}+(0.4)^{2}}[/latex]

The solution would yield as before

[latex] R_{2} / s=0.137, R_{2} / s=1.167[/latex]

The solution points are indicated in the T-s characteristic drawn in Fig. 9.47. The motor will run at

s = 0.067 [latex] \Rightarrow[/latex] 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.

Question 9.15: A 3-phase, 25 kW, 400 V, 50 Hz, 8-pole induction motor has r...

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