Question

Q3. A single-phase, 120 V, 60 Hz, four-pole, split-phase induction motor has the following equivalent circuit parameters: Stator parameters: The Main winding (R =1.512, X=2.522), and the auxiliary winding (R=2.512, X=2.52), Rotor parameters referred to the stator side: (R=1N, X=1.592). The magnetizing reactance is equal to 4082. (a)Determine the standstill impedances of both windings. (b)Determine the starting torque and the starting current of the motor if it is started from rated voltage mains as a split phase induction motor. (c)Determine the value of the capacitor to be connected in series with the auxiliary winding to produce maximum starting torque per ampere of starting current. Determine the value of starting torque and starting current. (d)Compare the starting torque per ampere of starting current for cases (b) and (c) |(25M)

Answer 1

Main winding :

R = 1.5 ohm

X =2.5 ohm

Auxillary winding:

R =2.5 ohm

X = 2.5 ohm

Rotor parameters :

R2=1 ohm

X2=1.5 ohm

Magnetizing recatance ,Xm=40 ohm

a) Standstill impedance of both winding:

The impedance of forward running rotor :

Z jum( 3 + jx2) 2 + j(Im + 22

At start s=1 so,

Z; jam (r2 + j.cz) r2 + 2m +.02)

Z; j40(1+j1.5) 1+ (40 + 1.5)

Zf = 0.92 +j1.46

$Z_{f}=1.73 \angle 57.6^{\circ}$

The impedance of backward running rotor:

jxml 792 2-3 Zo -jx2) (2-3) + j(Im +.cz)

$Z_{b}=\frac{jx_{m}(\frac{r_{2}}{2}+{j x_{2}})}{\frac{r_{2}}{2}+{j(x_{m}+x_{2}})}$

At starting we have

Zf =Zb

main winding :Z1 =R1+X1+Zf +Zb =1.5+j2.5+2*1.73$\angle$57.6^o=3.35+j5.42

$Z_{1}=6.37\angle 58.25^{\circ}$

Auxillary winding :Z1 = R1+X1+Zf +Zb=2.5+j2.5+2*1.73$\angle$57.6^o=3.42+j3.96

$Z_{1}=5.23\angle 49.13^{\circ}$

b)

Starting current $I=\frac{V}{x_{m }+X_{f }+X_{b }}$

Im 120 20° 240 +2(1.73257.6°)

$Im=0.12-j2.79$

$Im=2.79\angle -87.52^{\circ}$

Starting torque =$T_{s}=\frac{2*I_{m}(Z_{f}+Z_{b})sin\Theta _{m})}{\omega syn}$

$N_{s}=120 \frac{f}{p}$

$N_{s}=120 \frac{60}{4}$

${\omega syn}=2\pi *N_{s}$

ufis 27 * 1800
${\omega syn}=3600\pi$

$T_{s}=\frac{2*I_{m}(Z_{f}+Z_{b})sin\Theta _{m})}{\omega syn}$

$T_{s}=\frac{2*2.79*2(1.73\angle 57.6^{\circ})sin87.52}{3600\pi }$

$T_{s}=(1.35+j2.14)*10^{-4}Nm$

$T_{s}=2.53*10^{-4}\angle57.75^{\circ} Nm$

c)  $X_{c}=3.96-\frac{3.96*3.42-6.37\sqrt{(3.42+3.35)*3.42}}{3.35}$

From the real and imaginary values of the main and the auxillary winding

C= $C=\frac{1}{\omega X_{c}}$

=1.733*10^-5 F

Starting current at this condition :

$Ia=\frac{V}{x_{m }+X_{f }+X_{b }-jX_{c}}$

$Ia=\frac{120 \angle 0^{\circ}}{j40 +2*(1.73\angle 57.6^{\circ})-j5.1}$

=0.15 -j3.16

=$Ia=3.16\angle -87.19^{\circ}$

starting torque at this condition:

$T_{r}=\frac{2*I_{m}I_{a}(Z_{f}+Z_{b})sin(\Theta _{a}-\Theta _{m}))}{\omega syn}$

$T_{r}=\frac{2*2.79*2(1.73\angle 57.6^{\circ})sin(87.52-87.19)}{3600\pi }$

$T_{r}=(5.26+j8.3)*10^{-6}$

$T_{r}=9.83*10^{-6} \angle 57.6^{\circ}$

d)

$\frac{Im}{Ts}=$$Im=2.79\angle -87.52^{\circ}$/$T_{s}=2.53*10^{-4}\angle57.75^{\circ} Nm$=$11027.66\angle -145.27^{\circ}$

$\frac{Ia}{Tr}=$$Ia=3.16\angle -87.19^{\circ}$/$T_{r}=9.83*10^{-6} \angle 57.6^{\circ}$Nm=$321464.9\angle -144.79^{\circ}$

Hence a higher value of current per torque value is obtained.