A 220-V, 6-pole, 50-Hz, single-winding single-phase induction motor has the following equivalent circuit parameters as referred to the stator. R1m = 3.0 Ω, X1m = 5.0 Ω R2 = 1.5 Ω , X2 = 2.0 Ω Neglect the magnetizing current. When the motor runs at 97% of the synchronous speed, compute the

Question

A 220-V, 6-pole, 50-Hz, single-winding single-phase induction motor has the following equivalent circuit parameters as referred to the stator.

[latex] R_{1m} = 3.0 \Omega[/latex], [latex] X_{1m} = 5.0 \Omega[/latex]

[latex] R_{2} = 1.5 \Omega[/latex], [latex] X_{2} = 2.0 \Omega[/latex]

Neglect the magnetizing current. When the motor runs at 97% of the synchronous speed, compute the following:

(a) The ratio [latex] E_{mf} /E_{mb}[/latex]

(b) The ratio [latex] V_{f} /V_{b}[/latex]

(c) The ratio [latex] T_{f} /T_{b}[/latex]

(d) The gross total torque.

(e) The ratios [latex] T_{f}[/latex] /(Total torque) and [latex] T_{b}[/latex]/(Total torque)

Accepted Answer

Slip = s = 1 – 0.97 = 0.03

(a) From Fig. 10.5(c),

[latex] \frac{E_{m f}}{E_{m b}}=\frac{Z_{f}}{Z_{b}}=\frac{\left|j X \|\left(\frac{R_{2}}{s}+j X_{2} ight) ight|}{\left|j X \|\left(\frac{R_{2}}{2-s}+j X_{2} ight) ight|}[/latex]

Since magnetizing current is neglected, [latex] X=\infty [/latex]

[latex]\frac{E_{m f}}{E_{m b}}=\frac{\left|R_{2} / s+j X_{2} ight|}{\left|R_{2}(2-s)+j X_{2} ight|}[/latex]

[latex] =\frac{|1.5 / 0.03+j 2|}{|1.5 /(2-0.03)+j 2|}=23.38[/latex]

(b) [latex] V_{f}[/latex] and [latex] V_{b}[/latex] are components of stator voltage [latex] V_{m}[/latex] , i.e

[latex] \bar{V}_{m}=\bar{V}_{f}+\bar{V}_{b}[/latex]

These components are defined by redrawing the circuit model of Fig. 10.5(c) in the symmetrical form of Fig. 10.6. For the purpose of this problem [latex] X=\infty[/latex] , therefore

Impedance offered to [latex] V_{f}[/latex] component [latex] =\frac{1}{2}\left[\left(3+\frac{1.5}{0.03} ight)+j(5+2) ight]=\frac{1}{2}(53+j 7)[/latex]

Impedance offered to [latex] V_{b}[/latex] component [latex] =\frac{1}{2}\left[\left(3+\frac{1.5}{1.97} ight)+j(5+2) ight]=\frac{1}{2}(3.76+j 7)[/latex]

Hence [latex] \frac{V_{f}}{V_{b}}=\frac{|53+j 7|}{|3.76+j 7|}=6.73[/latex]

(c) From Eqs (10.8a) [latex] T_{f}=\frac{1}{\omega_{s}} P_{g f}[/latex] and (10.8b) [latex] T_{b}=\frac{1}{\omega_{s}} P_{g b}[/latex]

[latex] \frac{T_{f}}{T_{b}}=\frac{P_{g f}}{P_{g b}}=\frac{\frac{1}{2} I_{m}^{2} R_{2} / s}{\frac{1}{2} I_{m}^{2} R_{2} /(2-s)}=\frac{2-s}{s}=\frac{2-0.03}{0.03}=65.7[/latex]

(d) Total impedance as seen from stator terminal is

[latex] \bar{Z}( ext { Total })=\frac{1}{2}[(53+j 7)+(3.76+j 7)]=28.38+j 7=29.2 \angle 13.9^{\circ}[/latex]

[latex] I_{m}=\frac{220}{29.2}=753 A[/latex]

[latex] n_{s}=\frac{120 imes 50}{6}=1000 rpm[/latex]

[latex] \omega_{s}=\frac{2 \pi imes 1000}{60}=104.72 rad / s[/latex]

[latex] T_{f}=\frac{1}{\omega_{s}} I_{m}^{2} \frac{R_{2}}{2 s}[/latex] (i)

[latex] T_{b}=\frac{1}{\omega_{s}} I_{m}^{2} \frac{R_{2}}{2(2-s)}[/latex] (ii)

[latex] T_{ ext {total }}=T_{f}-T_{b}[/latex]

[latex] T_{ ext {total }}=\frac{I_{m}^{2} R_{2}}{2 \omega_{s}}\left(\frac{1}{s}-\frac{1}{2-s} ight)[/latex] (iii)

[latex] =\frac{(7.53)^{2} imes 1.5}{2 imes 104.72}\left(\frac{1}{0.03}-\frac{1}{1.97} ight)=13.31 Nm[/latex]

(e) From Eqs (i), (ii) and (iii)

[latex] \frac{T_{f}}{T_{ ext {total }}}=\frac{1 / s}{1 s-1 /(2-s)}=\frac{1}{1-\frac{s}{2-s}}=1.015[/latex]

[latex] \frac{T_{b}}{T_{ ext {total }}}=\frac{1 /(2-s)}{1 / s-1 /(2-s)}=\frac{1}{\frac{2-s}{s}-1}=0.015[/latex]

Question 10.1: A 220-V, 6-pole, 50-Hz, single-winding single-phase inductio...

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