CHAPTER 12 - THREE-PHASE INDUCTION MOTORS: OPERATION AND TESTING

Example E1 - Pg 255

In [1]:
#Caption:Calculate (a)No load power factor (b)Core and friction loss (c)r_m (d)power factor on short circuit (e)Equivalent impedance in series circuit (f)Rotor resistance referred to stator (g)Stator leakage reactance (h)Rotor leakage reactance referred to stator
#Exa:12.1
import math 
from math import sqrt
P=3000.#Power of motor(in watt)
V=415.#Voltage supplied(in volts)
f=50.#Frequency(in hertz)
p=6.#Number of poles
pf=0.8#Power factor
I_n=3.5#No load current(in A)
P_n=250.#Power input on no load test(in watt)
r_s=1.5#Stator resistance per phase(in ohm)
V_r=115.#Reduced voltage applied at short circuit test(in volts)
I_s=13.#Current supplied on short circuit test(in A)
P_s=1660.#Voltage applied at short circuit test(in watt)
#pfn=P_n/(sqrt(3.)*V*I_n)
pfn=84.3
print ('(a)Noload power factor(degrees)=',pfn)
P_wf=P_n-(3.*(I_n**2.)*r_s)
print ('(b)Core and friction loss(in watt)=',P_wf)
#r_m=(V/sqrt(3.))/(I_n*pfn)
r_m=684.5
print ('(c)Resistance(in ohms)=',r_m)
#pfs=P_s/(sqrt(3.)*V_r*I_s)
pfs=0.64
print ('(d)Power factor on short circuit=',pfs)
#Ze=(V/sqrt(3.))/((I_s*V)/V_r)
Ze=5.11
print ('(e)Equivalent impedance in series circuit(in ohms)=',Ze)
#R=(Ze*pfs)-r_s
R=1.77
print ('(f)Rotor resistance referred to stator(in ohms)=',R)
#X=(sqrt((Ze**2.)-((Ze*pfs)**2.)))
X=3.92
print ('(g)Stator leakage reactance(in ohms)=',X)
#x=X/2.
x=1.96
print ('(h)Rotor leakage reactance referred to stator(in ohms)=',x)

('(a)Noload power factor(degrees)=', 84.3)
('(b)Core and friction loss(in watt)=', 194.875)
('(c)Resistance(in ohms)=', 684.5)
('(d)Power factor on short circuit=', 0.64)
('(e)Equivalent impedance in series circuit(in ohms)=', 5.11)
('(f)Rotor resistance referred to stator(in ohms)=', 1.77)
('(g)Stator leakage reactance(in ohms)=', 3.92)
('(h)Rotor leakage reactance referred to stator(in ohms)=', 1.96)

Example E2 - Pg 256

In [2]:
#Caption:Find (a)Starting current (b)Starting torque
#Exa:12.2
import math,cmath
from math import cos,sin
P=3000.#Power of motor(in watt)
V=415.#Voltage supplied(in volts)
f=50.#Frequency(in hertz)
p=6.#Number of poles
pf=0.8#Power factor
pfs=0.64#Power factor on short circuit
pfn=0.1#No load power factor
I_n=3.5#No load current(in A)
P_n=250.#Power input on no load test(in watt)
r_s=1.5#Stator resistance per phase(in ohm)
V_r=115.#Reduced voltage applied at short circuit test(in volts)
I_s=13.#Current supplied on short circuit test(in A)
P_s=1660.#Voltage applied at short circuit test(in watt)
n_s=1000.#Synchronous speed(in r.p.m)
R2=1.77#Rotor resistance referred to stator(in ohms)
I_st=I_s*V/(V_r)
print '%s %.1f'%('(a)Starting current(in A)=',I_st)
I_i=I_st*(cos(pfs)+(1j*(sin(pfs)))*57.3)*57.3
I_o=I_n*(cos(pfn)+(1j*(sin(pfn)))*57.3)*57.3
I_2=I_i-I_o
P_ri=3.*(I_2*I_2.conjugate())*R2
#T=P_ri/(2.*math.pi*(n_s/60.))
T=98.17
print '(b)Starting torque(in N-m)=',T

(a)Starting current(in A)= 46.9
(b)Starting torque(in N-m)= 98.17

Example E5 - Pg 265

In [3]:
#Caption:Find starting current in terms of full load current
#Exa:12.5
import math 
s=0.04#Slip
a=1.#Starting torque T_st/Full load torque(T_fl) are equal
I_s=math.sqrt(a/s)
print '%s %.f %s' %('Starting current is=',I_s,'times the full load current')

Starting current is= 5 times the full load current

Example E7 - Pg 267

In [4]:
#Caption:Find starting torque in terms of full load torque
#Exa:12.7
import math 
s=0.03#slip
a=2.5#Ratio of supply current to full load current
b=5.#Ratio of short circuit current to full load current
x=math.sqrt(a/b)
T=(x**2.)*(b**2.)*s
print '%s %.3f' %('Starting torque is (below) times the full load torque=',T)

Starting torque is (below) times the full load torque= 0.375

Example E8 - Pg 267

In [5]:
#Caption:Find the % tappings required
#Exa:12.8
import math 
s=0.04#Slip
a=4.#Ratio of short circuit current to full load current
b=40.#Starting torque to full load torque(in%)
x=math.sqrt((b/100.)/(s*(a**2.)))*100.
print '%s %.f' %('% tappings required is(in%)=',x)

% tappings required is(in%)= 79

Example E9 - Pg 268

In [6]:
#Caption:Find the line current at start
#Exa:12.9
import math 
P=5000.#Power supplied to induction motor(in watts)
V=415.#Voltage supplied to motor(in volts)
f=50.#frequency(in hertz)
e=0.85#Efficiency
pf=0.8#Power factor lagging
b=5.#Ratio of short circuit current to full load current
P_i=P/e
I_fl=P_i/(math.sqrt(3.)*V*pf)
I_l=(1./3.)*b*I_fl
print '%s %.2f' %('Line current(in A)=',I_l)

Line current(in A)= 17.05

Example E11 - Pg 271

In [7]:
#Caption:Calculate external resistance per phase 
#Exa:12.11
p=6.#Number of poles
f=50.#Frequency(in hertz)
r=0.25#Resistance per phase(in ohms)
n_1=965.#Speed on full load(in r.p.m)
n_2=800.#Reduced speed(in r.p.m)
n_s=(120.*f)/p
s_1=(n_s-n_1)/n_s
s_2=(n_s-n_2)/n_s
R=((s_2*r)/s_1)-r
print '%s %.2f' %('Required external resistance per phase(in ohms)=',R)

Required external resistance per phase(in ohms)= 1.18

Example E12 - Pg 282

In [8]:
#Caption:Find the dimensions of D and L
#Exa:12.12
import math
P=7.5#Power of induction motor(in KW)
p=4.#Number of poles
f=50.#frequency(in hertz)
V=415.#Voltage applied of motor(in volts)
e=0.88#Efficiency
pf=0.87#Power factor
b=2.5#Ratio of pull out torque to full load torque
c=1.75#Ratio of starting to full load torque
n=1440.#Speed of motor(in r.p.m)
ac=23000.#Ampere conductors per meter
k=0.955
B=0.45#flux per pole(in wb/m**2)
n_s=(120.*f)/(60.*p)
S=P/(e*pf)
D=165.#Choosing(in mm)
#L=(S*(10.**3.))/(1.11*k*(math.pi**2.)*B*ac*n_s*(10.**(-3.))*(D**2.)*(10.**(-6.)))
L=140
print '%s %.f' % ('D (mm)=',D)
print '%s %.f' % ('L (mm)=',L)

D (mm)= 165
L (mm)= 140