CHAPTER 4 - Characteristics of AC motor¶

EXAMPLE 4.1 - PG NO.65¶

In [1]:
#Example 4.1, Page 65
import numpy
import math
l=([1, -3, 1])
a=numpy.asarray(l[0])
b=numpy.asarray(l[1])
c=numpy.asarray(l[2])
d = (b**2) - (4*a*c)
sol1 = (-b-math.sqrt(d))/(2*a)
sol2 = (-b+math.sqrt(d))/(2*a)
print('Part a')
print('roots of the equation when slip at max torque')
print'%.3f %s %.3f' %(sol1,'or',sol2)

l1=([1, -1.719,0.146])
a1=numpy.asarray(l1[0])
b1=numpy.asarray(l1[1])
c1=numpy.asarray(l1[2])
d1 = (b1**2) - (4*a1*c1)
solution1 = (-b1-math.sqrt(d1))/(2*a1)
solution2 = (-b1+math.sqrt(d1))/(2*a1)
print('\n')
print('Part b')
print('roots of the equation when slip at max load')
print'%.2f %s %.2f' %(solution1,'or',solution2)

Part a
roots of the equation when slip at max torque
0.382 or 2.618

Part b
roots of the equation when slip at max load
0.09 or 1.63


EXAMPLE 4.3 - PG NO.70¶

In [2]:
#Example 4.3, Page 70
import numpy
import math
l=([0.04, -0.0266,.0016])
a=numpy.asarray(l[0])
b=numpy.asarray(l[1])
c=numpy.asarray(l[2])
d = (b**2.) - (4.*a*c)
sol1 = (-b-math.sqrt(d))/(2.*a)
sol2 = (-b+math.sqrt(d))/(2.*a)
print('Part a')
print('roots of the equation that slip will run is')
print'%.3f %s %.3f' %(sol1,'or',sol2)

Part a
roots of the equation that slip will run is
0.067 or 0.598


EXAMPLE 4.4.a - PG NO. 71¶

In [3]:
#example 4.4.a
#page no. 71
import math
r1=0.02#
x=0.04#
TmaxatVby2=0.02+math.sqrt(0.0004+0.04)
TmaxatV=2*(0.02+math.sqrt(0.0004+0.01))
Tmax=TmaxatVby2/TmaxatV
print'%s %.5f' %('Tmax at rated voltage and frequency= ',Tmax)

Tmax at rated voltage and frequency=  0.90587


EXAMPLE 4.4.b - PG NO. 71¶

In [4]:
#example 4.4.b
#page no. 71
import math
r1=0.02#
x=0.04#
TstatVby2=(0.04*0.04)+(0.2*0.2)
TstatV=2*((0.04*0.04)+(0.01))
Tst=TstatVby2/TstatV
print'%s %.3f' %('Tst at rated voltage and frequency =',Tst)

Tst at rated voltage and frequency = 1.793


EXAMPLE 4.5 - PG NO.74¶

In [3]:
#Example 4.5, page 74
import math
pole=24.
Ns=245.#in rpm
N=(120.*50.)/pole#synchronous speed in rpm
f=(N-Ns)/N
p=110.#in kw
T=(p*1000.*60.)/(2.*math.pi*Ns)
v1=440./math.sqrt(3)#in v
ws=(2*math.pi*250)/60
s=0.02
R=0.03125#in ohm
x=math.sqrt(((3*R*v1**2)/(T*ws*s))-(R/s)**2)#by rearranging formula
print'%s %.5f %s' %('Stator resistance per phase is=',x,'ohm')
#calculating original resistance

#Example 4.1, Page 65
import numpy
import math
l=([3190,-3235,72.78])
a=numpy.asarray(l[0])
b=numpy.asarray(l[1])
c=numpy.asarray(l[2])
d = (b**2.) - (4.*a*c)
sol1 = (-b-math.sqrt(d))/(2.*a)
sol2 = (-b+math.sqrt(d))/(2.*a)
print('Part a')
print('The value of original resistance is')
print'%.4f %s %.5f' %(sol1,'or',sol2)

#Taking r=0.99108
r=(0.99108-R)/1.25**2

print'%s %.4f %s' %('The value of resistance to be added is =',r,' ohm')

Stator resistance per phase is= 0.50358 ohm
Part a
The value of original resistance is
0.0230 or 0.99109
The value of resistance to be added is = 0.6143  ohm


EXAMPLE 4.6 - PG NO.85¶

In [4]:
#Example 4.6, Page no 85
import math
print("Part ii")
new_sin_delta=math.sin(math.pi/4)/.95
delta=math.asin(new_sin_delta)
x=math.degrees(delta)
print'%s %.2f %s' %('The value of delta is =',x,'degree')

Part ii
The value of delta is = 48.10 degree