A 220-V 1.5-hp 50-Hz six-pole capacitor-start induction motor has the following main-winding impedances: R1 = 1.3 Ω R2 = 1.73 Ω XM = 105 Ω X1 = 2.01 Ω X2 = 2.01 Ω At a slip of 0.05, the motor’s rotational losses are 291 W. The rotational losses may be assumed constant over the normal operating

Question

A 220-V 1.5-hp 50-Hz six-pole capacitor-start induction motor has the following main-winding impedances:

[latex]\begin{array}{ll}R_1=1.30 \ \Omega & X_1=2.01 \ \Omega \R_2=1.73 \ \Omega & X_2=2.01 \ \Omega\end{array} \quad X_M=105 \ \Omega[/latex]

At a slip of 0.05, the motor’s rotational losses are 291 W. The rotational losses may be assumed constant over the normal operating range of the motor. Find the following quantities for this motor at 5 percent slip:

(a) Stator current
(b) Stator power factor
(c) Input power
(d) [latex] { P }_{ AG } [/latex]
(e) [latex] { P }_{ conv } [/latex]
(f) [latex] { P }_{ out } [/latex]
(g) [latex] { au }_{ ind } [/latex]
(h) [latex] { au }_{ load } [/latex]
(i) Efficiency [latex] \eta [/latex]

Accepted Answer

The equivalent circuit of the motor is shown below

The impedances [latex]Z_F ext { and } Z_B[/latex] are:

[latex]Z_F=\frac{\left(R_2 / s+j X_2 ight)\left(j X_M ight)}{R_2 / s+j X_2+j X_M}[/latex]

The slip s = 0.05 , so [latex]R_2 / s=(1.73 \ \Omega) / 0.05=20.6 \ \Omega[/latex]

[latex]\begin{aligned}& Z_F=\frac{(20.6+j 2.01)(j 105)}{20.6+j 2.01+j 105}=19.12+j 5.654 \ \Omega \& Z_B=\frac{\left[R_2 /(2-s)+j X_2 ight]\left(j X_M ight)}{R_2 /(2-s)+j X_2+j X_M}\end{aligned}[/latex]

The slip s = 0.05 , so [latex]R_2 /(2-s)=(1.73 \ \Omega) /(2-0.05)=0.887 \ \Omega[/latex]

[latex]Z_B=\frac{(0.887+j 2.01)(j 105)}{0.887+j 2.01+j 105}=0.854+j 1.979 \ \Omega[/latex]

(a) The input stator current is

[latex]\begin{aligned}& \mathrm{I}_1=\frac{\mathbf{V}}{R_1+j X_1+0.5 Z_F+0.5 Z_B} \& \mathrm{I}_1=\frac{220 \angle 0^{\circ} \mathrm{V}}{(1.30+j 2.01)+0.5(19.12+j 5.654)+0.5(0.854+j 1.979)}=17.32 \angle-27.3^{\circ} \mathrm{A}\end{aligned}[/latex]

(b) The stator power factor is

[latex]\mathrm{PF}=\cos 27.3^{\circ}=0.889 ext { lagging }[/latex]

(c) The input power is

[latex]P_{\mathrm{IN}}=V I \cos heta=(220 \mathrm{~V})(17.32 \mathrm{~A}) \cos 27.3^{\circ}=3386 \mathrm{~W}[/latex]

(d) The air-gap power is

[latex]\begin{aligned}& P_{\mathrm{AG}, F}=I_1^2\left(0.5 R_F ight)=(17.32 \mathrm{~A})^2(9.56 \ \Omega)=2868 \mathrm{~W} \& P_{\mathrm{AG}, B}=I_1^2\left(0.5 R_B ight)=(17.32 \mathrm{~A})^2(0.427 \ \Omega)=128 \mathrm{~W} \& P_{\mathrm{AG}}=P_{\mathrm{AG}, F}-P_{\mathrm{AG}, B}=2868 \mathrm{~W}-128 \mathrm{~W}=2740 \mathrm{~W}\end{aligned}[/latex]

(e) The power converted from electrical to mechanical form is

[latex]\begin{aligned}& P_{ ext {conv }, F}=(1-s) P_{\mathrm{AG}, F}=(1-0.05)(2868 \mathrm{~W})=2725 \mathrm{~W} \& P_{ ext {conv }, B}=(1-s) P_{\mathrm{AG}, B}=(1-0.05)(128 \mathrm{~W})=122 \mathrm{~W} \& P_{ ext {conv }}=P_{ ext {conv }, F}-P_{ ext {conv }, B}=2725 \mathrm{~W}-122 \mathrm{~W}=2603 \mathrm{~W}\end{aligned}[/latex]

(f) The output power is

[latex]P_{ ext {OUT }}=P_{ ext {conv }}-P_{ ext {rot }}=2603 \mathrm{~W}-291 \mathrm{~W}=2312 \mathrm{~W}[/latex]

(g) The synchronous speed for a 6 pole 50 Hz machine is 1000 r/min, so induced torque is

[latex] au_{ ext {ind }}=\frac{P_{\mathrm{AG}}}{\omega_{\mathrm{sync}}}=\frac{2740 \mathrm{~W}}{(1000 \mathrm{\ r} / \mathrm{min})\left\lgroup\frac{2 \pi \mathrm{\ rad}}{1 \mathrm{\ r}} ight group\left\lgroup\frac{1 \mathrm{~min}}{60 \mathrm{~s}} ight group}=26.17 \mathrm{~N}⋅\mathrm{m}[/latex]

(h) The load torque is

[latex] au_{ ext {load }}=\frac{P_{ ext {OUT }}}{\omega_m}=\frac{2312 \mathrm{~W}}{(0.95)(1000 \mathrm{\ r} / \mathrm{min})\left\lgroup\frac{2 \pi \mathrm{\ rad}}{1 \mathrm{\ r}} ight group\left\lgroup\frac{1 \mathrm{~min}}{60 \mathrm{~s}} ight group}=23.24 \mathrm{~N}⋅\mathrm{m}[/latex]

(i) The overall efficiency is

[latex]\eta=\frac{P_{\mathrm{OUT}}}{P_{\mathrm{IN}}} imes 100 \%=\frac{2312 \mathrm{~W}}{3386 \mathrm{~W}} imes 100 \%=68.3 \%[/latex]

Question 9.5: A 220-V 1.5-hp 50-Hz six-pole capacitor-start induction moto...

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