The circuit model parameters in Ω /phase (referred to stator) of a two-phase, 1 kW, 220 V, 4-pole, 50 Hz squirrel-cage motor are given below: R1 = 3 Ω R2 = 2.6 Ω X1 = X2 = 2.7 Ω X = 110 Ω The windage, friction and core losses equal 200 W. The applied voltages are adjusted such that Va = 110 ∠ 90°

Question

The circuit model parameters in [latex] \Omega ext { phase } [/latex] (referred to stator) of a two-phase, 1 kW, 220 V, 4-pole, 50 Hz squirrel-cage motor are given below:

[latex] R_{1}=3 \Omega \quad R_{2}=2.6 \Omega[/latex]

[latex] X_{l}=X_{2}=2.7 \Omega \quad X=110 \Omega[/latex]

The windage, friction and core losses equal 200 W.

The applied voltages are adjusted such that

[latex] \bar{V}_{a}=110 \angle 90^{\circ}[/latex] and [latex] \bar{V}_{m}=220 \angle 0^{\circ}[/latex]

(a) Calculate the starting torque and starting current (in each phase).

(b) Calculate the motor performance at s = 0.04.

(c) With the motor running at s = 0.04, the phase a gets open-circuited. What voltage will be developed across this phase?

Accepted Answer

From Eqs (10.38a) [latex] \bar{Z}_{11}=\left(\frac{\bar{Z}_{1 m}}{2}+\frac{\bar{Z}_{1 a}}{2 a^{2}}+\bar{Z}_{f} ight)[/latex] and (10.38b) [latex] \bar{Z}_{12}=\frac{1}{2}\left(\frac{\bar{Z}_{1 a}}{a^{2}}-\bar{Z}_{1 m} ight)[/latex] (a = 1)

[latex] \bar{V}_{f}=\frac{1}{2}\left(220-j 110 \angle 90^{\circ} ight)=165 V[/latex]

[latex] \bar{V}_{b}=\frac{1}{2}\left(220+j 110 \angle 90^{\circ} ight)=55 V[/latex]

(a) s = 1

[latex] \bar{Z}_{f}=\bar{Z}_{b}=j 110 \|(2.6+j 2.7)[/latex]

[latex] =3.66 \angle 47.4^{\circ}=2.48+j 2.69[/latex]

[latex] \bar{Z}_{f} ext { (total) }=\bar{Z}_{b} ext { (total) }=(3+j 2.7)+(2.48+j 2.69)[/latex]

[latex] =5.48+j 5.39=7.69 \angle 44.5^{\circ}[/latex]

[latex] \bar{I}_{f}=\frac{165}{7.69 \angle 44.5^{\circ}}=21.45 \angle-44.5^{\circ}=15.3-j 15.03[/latex]

[latex] \bar{I}_{b}=\frac{55}{7.69 \angle 44.5^{\circ}}=7.15 \angle-44.5^{\circ}=5.1-j 5.01[/latex]

[latex] \bar{I}_{s}=\frac{2}{157} imes 2.48\left[(21.45)^{2}-(7.15)^{2} ight][/latex]

= 12.9 Nm

[latex] \bar{I}_{m}=\bar{I}_{f}+\bar{I}_{b}=20.4-j 20.04=28.6 \angle-44.4^{\circ}[/latex]

[latex] I_{m}=28.6 A[/latex]

[latex] \bar{I}_{a}=j\left(\bar{I}_{f}-\bar{I}_{b} ight)=10.2+j 10.02=14.3 \angle 44.5^{\circ}[/latex]

[latex] I_{a}=14.3 A[/latex]

(b) [latex] \bar{Z}_{f}=j 110 \|\left(\frac{2.6}{0.04}+j 2.7 ight)[/latex]

[latex] =55.1 \angle 32.4^{\circ}=46.5+j 29.5[/latex]

[latex] \bar{Z}_{b}=j 110 \|\left(\frac{2.6}{2-0.04}+j 2.7 ight)[/latex]

[latex] =2.93 \angle 64.5^{\circ}=1.26+j 2.64[/latex]

[latex] \bar{Z}_{f}( ext { total })=(3+j 2.7)+(46.5+j 29.5)[/latex]

[latex] =49.5+j 32.2=59 \angle 33^{\circ}[/latex]

[latex] \bar{Z}_{b}( ext { total })=(3+j 2.7)+(1.26+j 2.64)[/latex]

[latex] =4.26+j 5.34=6.83 \angle 51.4^{\circ}[/latex]

[latex] \bar{I}_{f}=\frac{165}{59 \angle 33^{\circ}}=2.79 \angle-33^{\circ}=2.34-j 1.52[/latex]

[latex] \bar{I}_{b}=\frac{55}{6.83 \angle 51.4^{\circ}}=8.05 \angle-51.4^{\circ}=5.02-j 6.29[/latex]

[latex] n_{s}=1500 rpm[/latex] or [latex] \omega_{s}=157.1 rad / s[/latex]

[latex] T_{s}=\frac{2}{\omega_{s}}\left(I_{f}^{2} R_{f}-I_{f}^{2} R_{b} ight)[/latex]

[latex] =\frac{2}{157.1}\left[(2.79)^{2} imes 46.5-(8.05)^{2} imes 1.26 ight][/latex]

= 3.57 N m

[latex] \bar{I}_{m}=\bar{I}_{f}+\bar{I}_{b}=(2.34-j 1.52)+(5.02-j 6.29)[/latex]

[latex] =7.36-j 7.81=10.73 \angle-46.7^{\circ}[/latex]

[latex] \bar{I}_{a}=j\left(\bar{I}_{f}-\bar{I}_{b} ight)=j[(2.34-j 1.52)-(5.02-j 6.29)][/latex]

[latex] =-3.5+j 8.63=-9.31 \angle 112.1^{\circ}[/latex]

[latex] I_{m}=10.73 A , l_{a}=9.31 A[/latex]

[latex] P_{m}=\omega_{s}(1-s) T=157.1(1-0.04) imes 3.57=538.4 W[/latex]

[latex] P_{ ext {out }}=538.4-200=338.4 W[/latex]

[latex] P_{i n}^{m}=220 imes 10.73 imes \cos 46.7^{\circ}=1619 W[/latex]

[latex] P_{i n}^{a}=220 imes 5.47 imes \cos 29.3^{\circ}=1049 W[/latex]

[latex] P_{i n}=1619+1049=2668 W[/latex]

[latex] \eta=\frac{338.4}{2668}=12.7 \% [/latex] (low because of unbalanced operation)

(c) [latex] \bar{I}_{a}=j\left(\bar{I}_{f}-\bar{I}_{b} ight)=0[/latex]

Or [latex] \bar{I}_{f}=\bar{I}_{b}[/latex] (i)

[latex] \bar{V}_{m}=\bar{V}_{m f}+\bar{V}_{m b}=220 V[/latex] (ii)

[latex] \bar{V}_{a}=j\left(\bar{V}_{m f}-\bar{V}_{m b} ight)=? [/latex] (iii)

[latex] \bar{I}_{f}=\frac{\bar{V}_{m f}}{\bar{Z}_{f} ext { (total) }}=\frac{\bar{V}_{m b}}{\bar{Z}_{b} ext { (total) }}=\bar{I}_{b}[/latex]

Or [latex] \frac{\bar{V}_{m f}}{\bar{V}_{m b}}=\frac{\bar{Z}_{f} ext { (total) }}{\bar{Z}_{b} ext { (total) }}[/latex] (iv)

[latex] =\frac{59 \angle 33^{\circ}}{6.83 \angle 51.4^{\circ}}=8.63 \angle-18.4^{\circ}[/latex] (v)

Solving Eqs (ii) and (v)

[latex] \bar{V}_{m f}=198.2 \angle-1.9^{\circ}=198.1-j 6.6[/latex]

[latex] \bar{V}_{m b}=22.96 \angle 16.5^{\circ}=22.01+j 6.5[/latex]

[latex] \bar{V}_{a}=j(176.1-j 13.1)[/latex]

[latex] =13.1+j 176.1=176.6 \angle 85.7^{\circ}[/latex]

[latex] V_{a}=176.6 V[/latex]

It may be seen that the angle of [latex] V_{a}[/latex] is slightly short of 90°.

Question 10.6: The circuit model parameters in Ω /phase (referred to stator...

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