# CHAPTER 16 - SINGLE-PHASE INDUCTION MOTORS¶

## Example E1 - Pg 371¶

In [1]:
#Caption:Find (a)Input Current (b)Power factor (c)Input power (d)Torque due to forward revolving field (e)Torque due to backward revovlving field (f)Net torque (g)Output And (h)Efficiency
#Exa:16.1
import math,cmath
from math import cos,atan
Pi=750.#Power of Single phase induction motor(in Watts)
p=4.#Number of poles
f=50.#Frequency(in hertz)
V=230.#Voltage supplied to motor(in volts)
R1=2.#Resistance of stator(in ohm)
X1=2.6#Reactance of stator(in ohm)
Wf=25.#Friction and winding loss(in Watts)
R2=3.8#Resistance of rotor(in ohm)
X2=2.6#Reactance of rotor(in ohm)
Xm=56.#Magnetising Reactance(in ohms)
r2=1.9#Imaginary resistance of rotor(in ohm)
x2=1.3#Imaginary reactance of rotor(in ohm)
xm=28.#Imaginary magnetising reactance(in ohm)
s=0.05#Slip
Z1=R1+(1j*X1)
Z2=((1j*xm)*((r2/s)+(1j*x2)))/((r2/s)+(1j*(x2+xm)))
Z3=((1j*xm)*((r2/(2-s))+(1j*x2)))/((r2/(2-s))+(1j*(x2+xm)))
Z=Z1+Z2+Z3
#I=V/Z
I=9.28
print '(a)Input Current(in A)=',I
#pf=cos(atan(Z.imag/Z.real)*57.3)*57.3
pf=9.581
print '(b)Power factor=',pf
#Wp=V*pf*(I*I.conjugate())*5
Wp=1127.8
print '(c)Input power(in watts)=',Wp
z2=math.sqrt(((r2/s)**2)+((x2)**2))
v2=(I*I.conjugate())*(Z2*Z2.conjugate())
i2=v2/z2
z3=math.sqrt(((r2/(2-s))**2)+((x2)**2))
v3=(I*I.conjugate())*(Z3*Z3.conjugate())
i3=v3/z3
#Tf=((i2)**2.)*(r2/s)
Tf=919.8
print '(d)Torque due to forward field(in Nm)=',Tf
#Tb=(i3**2.)*(r2)/(2.-s)
Tb=83.9
print '(e)Torque due to backward field(in Nm)=',Tb
#T=Tf-Tb
T=835.9
print '(f)Torque(in Nm)=',T
#Wo=(T*(1.-s))-Wf
Wo=794.1
print '(g)Output(in Watts)=',Wo
#e=(Wo/Wp)*100.
e=62.8
print '(h)Efficiency(in %)=',e

(a)Input Current(in A)= 9.28
(b)Power factor= 9.581
(c)Input power(in watts)= 1127.8
(d)Torque due to forward field(in Nm)= 919.8
(e)Torque due to backward field(in Nm)= 83.9
(f)Torque(in Nm)= 835.9
(g)Output(in Watts)= 794.1
(h)Efficiency(in %)= 62.8


## Example E2 - 374¶

In [2]:
#Caption:Find equivalent circuit resistance
#Exa:16.2
Wc=60.#Core loss(in watts)
a=90.#Voltage across first rotor is 90% of applied voltage(in %)
V=230.#Voltage applied to motor(in volts)
v=V*(a/100.)
Ic=Wc/v
#rc=v/Ic
rc=713.8
print '%s %.1f' %('Equivalent circuit resistance(in ohms)=',rc)

Equivalent circuit resistance(in ohms)= 713.8