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Solved Problems on Single-Phase Induction Motors


Problem-1

A 6-pole, 50-Hz, 1-phase induction motor runs at a speed of 900 rpm. Determine the frequency of currents in the cage rotor?

Solution

Ns=120fP=120×506=1000 rpmSlip, s=10009001000=0.1=10%

The two rotor current frequencies are :

sf=0.1×50=5 Hz(2s)f=1.9×50=95 Hz

Problem-2

For a single-phase motor of 2 HP rating, supply voltage is 240 V ac. If the efficiency is 70% and power factor is 0.8, find the input current.

Solution

1 HP=746 WPout=2 HP=2×746=1492 WInput Power (P)=Output powerη=14920.7=2131.43 W 2131.43=240×1×0.8 I=11.1 A

Problem-3

A 4-pole single-phase induction motor is rotating in a clockwise direction at a speed of 1000 rpm having a voltage of 100 V with a frequency of 50 Hz. At a standstill, if the rotor resistance is 1.7 Ω, then in the backward branch what will be the effective resistance?


Solution

  • The effective rotor resistance in the forward branch (Rf)

    Rf=R22s
  • The effective rotor resistance in the backward branch (Rb)

    Rb=R22(2s)

    Where R2= Standstill rotor resistance referred to as the main stator winding.

Ns=120×504=1500 rpms=150010001500=5001500=0.33Rb=1.72(20.33)=1.72×1.670.5  Ω

Problem-4

A 6 pole, 50 Hz induction motor has an equivalent rotor resistance of 0.01 Ω/phase. If its stalling speed is 900 rpm, determine the resistance that must be inserted in rotor windings per phase to obtain maximum torque at starting?


Solution

Synchronous speed, Ns=120×50/6=1000 rpmStalling speed=900 rpmSlip at stalling speed,s=1000900/1000=0.1
 Slip at maximum torque; SmT=R2/X2=0.01/X2
  • To obtain maximum torque at starting

  • Rotor resistance =\mathrm{R}^{\prime}_2 at starting \mathrm{s}=1

    \begin{aligned} \mathrm{S}_m&=\mathrm{R}^{\prime} 2 / \mathrm{X}_2 \\ \Rightarrow 1 & =R_2^{\prime} / 0.1\\ \Rightarrow R^{\prime}{ }_2 & =0.1 ~\Omega / \text{phase} \\ R_{ext} &=0.1-0.01\\ &=0.09 ~\Omega / \text{phase} \end{aligned}

Problem-5

A single phase, 230 \mathrm{~V}, 50 \mathrm{~Hz}, 4 pole, capacitor start induction motor has the following standstill impedance.

\begin{aligned} \text{Main winding}: \mathrm{Zm} & =(8.0+j 5.0)~ ~\Omega \\ \text{auxiliary winding}: \mathrm{Za} & =(9.0+\mathrm{j} 6.0) ~\Omega \end{aligned}

Determine the value of the starting capacitor required to produce 90^{\circ} phase difference between the current in the main and auxiliary windings.

Solution

\begin{aligned} \text{Main winding}~Z_m&=(8.0+j5.0)\Omega =9.434\angle32^{\circ}~\Omega \\ \text{Auxiliary winding}~Z_a & =(9.0+j6.0)~\Omega=10.82\angle33.69^\circ~\Omega \end{aligned}
image
  • \overrightarrow{I_m} lags behind voltage \vec{V} by angle \theta_m = 32^{\circ}

  • \begin{aligned}&\overrightarrow{Z_{\mathrm{aw}}}=\overrightarrow{Z_{\mathrm{a}}}+\frac{1}{\mathrm{j\omega C}}\\&=9-\mathrm{j} (\frac{1}{\omega C}-6)=Z_{\mathrm{aw}}\angle-\theta_{a} \\ \theta_{a} & =\tan^{-1}\frac{(\frac{1}{\omega C}-6)}{9} \end{aligned}
    Total impedance of auxiliary winding :
  • Phase difference :

    \begin{aligned} &\theta_{m} +\theta_{a} =90^{\circ}\\ &\tan^{-1}\left(\frac{\frac1{\omega C}-6}{9}\right)+32^{\circ}=90^{\circ} \Rightarrow& \frac{1}{\omega C}=20.4 \end{aligned}
    C=\frac{1}{2\pi\times50\times20.4}=156.03~ \mu F

Problem-6

  • A 120 V, 60 Hz, single-phase induction motor of the capacitor-start type has a main winding with 180 effective turns and an auxiliary starting winding with 250 effective turns. With the rotor stationary, the input impedance of the main winding is 5+j10~\Omega and that of the auxiliary winding is 13.89 + j15.50~\Omega.

    Determine at standstill:

    • The magnitude of the forward and backward-rotating fields

    • The size of the starting capacitor necessary to yield a single rotating magnetic field at starting.


Solution

image
  • The positive ({F}_{m_1}) and negative ({F}_{m_2}) sequence components of the motor’s mmf:

\begin{bmatrix}{F}_{m_1}\\{F}_{m_2}\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -j \\ 1 & j \end{bmatrix} \begin{bmatrix} N_m I_m \\ N_a I_a \end{bmatrix}
\begin{aligned} {F}_{m_1} & =\frac{1}{2}\left(N_m I_m-j N_a I_a\right) \\ & =\frac{1}{2}(120)\left(\frac{180}{5+j 10}-j \frac{250}{13.89+j 15.50}\right) \\ & =1349.10 \angle-94.5^{\circ} \mathrm{A} \mathrm{rms} \\ {F}_{m_2} & =\frac{1}{2}\left(N_m I_m+j N_a I_a\right) \\ & =\frac{1}{2}(120)\left(\frac{180}{5+j 10}+j \frac{250}{13.89+j 15.50}\right) \\ & =1041.7 \angle-21.6^{\circ} \mathrm{A} \mathrm{rms} \end{aligned}
  • {F}_{a_1}=j{F}_{m_1}=1349.10\angle{-4.5^\circ}\text{A rms}
    The positive-sequence component of the auxiliary winding’s mmf
  • {F}_{a_2}=-j{F}_{m_2}=1041.7\angle{-111.6^\circ}\text{A rms}
    the negative-sequence component
  • Thus, the net forward-rotating field is

    \begin{aligned} {F}_1={F}_{a_1}+j {F}_{m_1} & =1349.1\left(1 \angle-4.5^{\circ}+1 \angle-4.5^{\circ}\right) \\ & =2,698.2 \angle-4.5^{\circ} \mathrm{A} \mathrm{rms} \end{aligned}
  • The net backward-rotating field is

    \begin{aligned} {F}_2={F}_{a_2}+j {F}_{m_2} & =1041.7\left(1 \angle-111.6^{\circ}+1 \angle-21.6^{\circ}\right) \\ & =90.9~ \angle 155.9^{\circ} \mathrm{A} \mathrm{rms} \end{aligned}
  • To have a single rotating field, the backward-rotating mmf must be equal to zero. That is,

    \frac{1}{2}(120)\left(\frac{180}{5+j 10}+j \frac{250}{Z_a}\right)=0
  • The impedance of the auxiliary winding is

    Z_a=13.89+j\left(15.5-X_c\right)

    where X_c is the reactance of the capacitor.

  • From the last two relationships,

    \begin{aligned} Z_a & =-j\left(\frac{250(5+j 10)}{180}\right)=13.89+j\left(15.5-X_c\right) \\ X_c &=22.44 ~\Omega\\ \Rightarrow~C & =\frac{1}{\omega X_c} =\frac{1}{2 \pi(60)(22.44)} =118.2 \mu \mathrm{F} \end{aligned}