Suppose that the motor in Problem 9-1 is started and the auxiliary winding fails open while the rotor is accelerating through 400 r/min. How much induced torque will the motor be able to produce on its main winding alone? Assuming that the rotational losses are still 51 W, will this motor continue

Question

Suppose that the motor in Problem 9-1 is started and the auxiliary winding fails open while the rotor is accelerating through 400 r/min. How much induced torque will the motor be able to produce on its main winding alone? Assuming that the rotational losses are still 51 W, will this motor continue accelerating or will it slow down again? Prove your answer.

Accepted Answer

At a speed of 400 r/min, the slip is

[latex]s=\frac{1800 \ r / min -400 \ r / min }{1800 \ r / min }=0.778[/latex]

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 = 0.778 , so [latex]R_2 / s=(2.80 \ \Omega) / 0.778=3.60 \ \Omega[/latex]

[latex]\begin{aligned}& Z_F=\frac{(3.60+j 2.56)(j 60.5)}{3.60+j 2.56+j 60.5}=3.303+j 2.645 \ \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 = 0.778 , so [latex]R_2 /(2-s)=(2.80 \ \Omega) /(2-0.778)=2.291 \ \Omega[/latex]

[latex]Z_B=\frac{(2.291+j 2.56)(j 60.5)}{2.291+j 2.56+j 60.5}=2.106+j 2.533 \ \Omega[/latex]

The input current is

[latex]\begin{aligned}& I _1=\frac{ \mathbf{V} }{R_1+j X_1+0.5 Z_F+0.5 Z_B} \& I _1=\frac{120 \angle 0^{\circ} V }{(2.00+j 2.56)+0.5(3.303+j 2.645)+0.5(2.106+j 2.533)}=17.21 \angle-47.6^{\circ} A\end{aligned}[/latex]

The air-gap power is

[latex]\begin{aligned}& P_{ AG , F}=I_1^2\left(0.5 \ R_F ight)=(17.21 \ A )^2(3.303 \ \Omega)=978.3 \ W \& P_{ AG , B}=I_1^2\left(0.5 \ R_B ight)=(17.21 \ A )^2(2.106 \ \Omega)=623.8 \ W \& P_{ AG }=P_{ AG , F}-P_{ AG , B}=978.3 \ W -623.8 \ W =354.5 \ W\end{aligned}[/latex]

The power converted from electrical to mechanical form is

[latex]\begin{aligned}& P_{ ext {conv }, F}=(1-s) P_{ AG , F}=(1-0.778)(978.3 \ W )=217.2 \ W \& P_{ ext {conv }, B}=(1-s) P_{ AG , B}=(1-0.778)(623.8 \ W )=138.5 \ W \& P_{ conv }=P_{ conv , F}-P_{ conv , B}=217.2 \ W -138.5 \ W =78.7 \ W\end{aligned}[/latex]

The output power is

[latex]P_{ OUT }=P_{ ext {conv }}-P_{ ext {rot }}=78.7 \ W -51 \ W =27.7 \ W[/latex]

The induced torque is

[latex] au_{ ext {ind }}=\frac{P_{ AG }}{\omega_{ sync }}=\frac{354.5 \ W }{(1800 \ r / min )\left\lgroup\frac{2 \pi \ rad }{1 \ r } ight group\left\lgroup\frac{1 \ min }{60 \ s } ight group}=1.88 \ N⋅m[/latex]

Assuming that the rotational losses are still 51 W, this motor will still be able to speed up because [latex]P_{ ext {conv }}[/latex] is 78.7 W, while the rotational losses are 51 W, so there is more power than it required to cover the rotational losses. The motor will continue to speed up.

Question 9.3: Suppose that the motor in Problem 9-1 is started and the aux...

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