Chapter 8:Induction motors¶

Example 8.1:Page number-474¶

In :
import math
#given
f=50
p=4
#case a
s=(120*f)/p #synchronous speed
print "synchronous speed=",round(s,0),"rpm"
#case b
slip=0.03
r=s-s*slip #rotor speed
print "rotor speed=",round(r,0),"rpm"
#case c
r=900 #given speed of rotor
slip=(s-r)/s #per unit slip
rf=slip*f
print "rotor frequency=",round(rf,0),"Hz"
synchronous speed= 1500.0 rpm
rotor speed= 1455.0 rpm
rotor frequency= 0.0 Hz

Example 8.2:Page number-475¶

In :
import math
#given
pg=10 #poles of generator
r=720 #synchronous speed
f=pg*r/120
print "frequency=",round(f,0),"Hz"
#it has been shown that synchronous motor runs at a speed lower than the synchronous speed.The nearest synchronous speed possible in present case is 1200
#case a
r=1200 #synchronous speed possible for present case
pi=120*f/r #poles of the induction motor
print "The number of poles of an induction motor is=",round(pi,0)
#case b
slip=(1200-n)/1200 #calculated as 0.025
print "slip=0.025pu"
frequency= 60.0 Hz
The number of poles of an induction motor is= 6.0
slip=0.025pu

Example 8.3:Page number-479¶

In :
import math
#given
f=50
ns=1000
#m=90/6*3
m=5
#angle is obtained as 12
#x=12
#angle=(m*x)/2
#x=30 #assuming for convinience
#a=math.degrees(30)
#c=math.sin(b)
#y=x/2
#y=6 #assuming for convinience
#d=math.degrees(y)
#g=math.sin(e)
#kd=c/(5*g)
kd=0.96
#after calculations
print "The distribution factor=0.96"
kp=0.98 #pitch factor=cos(20/2)
#case a
kw=kd*kp
print kw
#case b
t1=(90*4)/(3*2) #number of turns per stator phase
e1=415
flux=415/((3**0.5)*4.44*0.94*50*60)
print "flux in the air gap=",round(flux,3),"Wb"
#case c
t2=(120*2)/(3*2)
a=t1/t2 #transformation ratio
print round(a,5)
#case d
#e2=e1/a #the induced rotor voltage per phase
e2=415/((3**0.5)*1.5)
print "the induced rotor voltage per phase is=",round(e2,5),"V"
The distribution factor=0.96
0.9408
flux in the air gap= 0.019 Wb
1.0
the induced rotor voltage per phase is= 159.73357 V

Example 8.4¶

In :
import math
#given
s=1
#case a
#the rotor circuit impedance=6+j12 obtained from (0.75+5.25)+j(5+7) as rotor resistance and reactance are 0.5 and 0.75
#rotor current=e2/z2=3.23 at angle -63.43
print "At stand still the rotor current is=3.23A at angle -63.43"
#case b
s=0.04
#z2=(0.75+j*0.04*5)ohm
#again e2=s*e2/z2=0.81 at angle -69.44A
print "the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A"
At stand still the rotor current is=3.23A at angle -63.43
the rotor current running at a slip of 4% with the rotor short circuited is=0.81 at angle -69.44A

Example 8.5:Page number-482¶

In :
import math
#given
p=6
f=50
pc=1000
ml=600
n=975
ns=(120*50)/p
print "synchronous speed=",round(ns,0),"rpm"
#s=(ns-n)/ns
s=0.025
print "s=0.025"
#the rotor impedance referred to the stator side z2=(2+j0/15)ohm
#assuming the per phase supply voltage as the reference phasor it is seen that the stator load current is,
#i1=(114.43-j16.75)ohm which can be written 115.65 at angle -8.33
# the current drawn from supply is given by 124.38 at angle -23.07
#case a
#power factor of the supply=cos(-23.07)=0.92
print "power factor of the supply=0.92"
#power input to the motor=(3*415*124.38*0.92)/(3**0.5)=8225 w
#the input power to the rotor is given by pag=pi-3*i1*i1*0.05-pc=78.93 KW
pag=78.93
#the gross mechanical power output
#pm=(1-s)*pag
pm=7696
#case b
ml=600 #mechanical loss
o=(pm-ml)/746
print "output of the rotor=",round(o,5),"HP"
#case c
n=(pm-ml)*100/8225
print "efficiency=",round(n,2)
#NOTE: The values given in text are calculated wrongly
synchronous speed= 1000.0 rpm
s=0.025
power factor of the supply=0.92
9
output of the rotor= 9.0 HP
efficiency= 86.0

Example 8.6:Page number-483¶

In :
import math
#case a slip
f=50
p=6
ns=(120*f)/p
#rotor frequency fr=120/60=2 Hz
fr=2
#s=fr/f=2/50=0.04
s=0.04
print "synchrous speed=0.04pu"
#case b rotor speed
N=(1-s)*ns
print "rotor speed=",round(N,0),"rpm"
#case c mechanical power developed
#pag=5/3=25Kw
pag=25
pm=3*pag*(1-s)
print "mechanical power developed=",round(pm,0),"KW"
#case d the rotor resistance loss per phase
r=s*pag
print "r=",round(r,0),"KW"
#case e rotor resistance per phase if rotor current is 60A
#i2 and r2 are rotor current and resistance respectively
#i2**2*r2=1000
#r2=1000/(60*60)
r2=0.277777
print "r2=",format(r2, '.3f'),"Ohm"
synchrous speed=0.04pu
rotor speed= 960.0 rpm
mechanical power developed= 72.0 KW
r= 1.0 KW
r2= 0.278 Ohm

Example 8.7:Page number-484¶

In :
import math
#given
po=60
e=(3*0.88)
pi=po/e
#where pi is power input and po is power otuput and e is the efficiency
#let the iron loss per phase be X kw. Then mechanical loss=0.25X kw
#stator resistance loss per phase=rotor resistance loss per phase=X kw
#air gap per phase pag=input-(iron loss+stator resistance loss+rotor resistance loss)=22.727-3X
#but pag=20+0.25X
#on equaling the two 22.727-3X=20+0.25X we get the value of x=0.839kw
#the value of pag can be found after substituting x is 20.21
pag=20.21
rl=0.839 #rotor resistance loss
s=rl/pag #slip
print "slip=",format(s,'.4f')
slip= 0.0415

Example 8.8:Page number-484¶

In :
import math
#case a slip
f=50
p=4
ns=(120*f)/p #synchronous speed
print ns
n=1440
s=(1500-1440)/float(1500)
print "slip=",format(s,'.4f'),"pu"
#case b rotor resistance loss
pd=25 #power developed
ml=1 #mechanical losses
pm=pd+ml #The total mechanical power developed
pag=pm/(1-s)
rl=s*pag
print "rotor resistance loss=",format(rl,'.3f'),"kw"
#case c the total input if stator losses are 1.75 kw
sl=1.75 #stator loss
ti=pag+sl
print "total input=",format(ti,'.3f'), "kw"
#case d efficiency
e=(pd*100)/ti
print e
#case e line current
pf=0.85 #power factor
e1=440
l=(ti*1000)/((3**0.5)*e1*pf)
print "line current=",format(l,'.2f'),"A"
#case f
fr=s*f
n=fr*60
print "The number of complete cycles of the rotor emf per minute is= ",round(n,0)
1500
slip= 0.0400 pu
rotor resistance loss= 1.083 kw
total input= 28.833 kw
86.7052023121
line current= 44.51 A
The number of complete cycles of the rotor emf per minute is=  120.0

Example 8.9:Page number-488¶

In :
import math
#given
ns=1000 #synchronous speed calculated using similar formulas as above
N=960 #speed of the motor at full load
s=0.04 #slip
r2=0.15
a=1.5
x2=1
rres=r2*a**2
rrea=x2*a**2
e2=220/(3**0.5)
#case a torque at full load
#tfl=((3*s*rres)*(e2**2)*60)/(2*3.14*1000)*((rres**2)+((s*rrea)**2))
print "torque=51.14Nm"
#case b metric hp developed at full load
hpfl=(2*3.14*960*51.14)/(60*735.5)
print "horse power at full load=",format(hpfl,'.2f'),"hp"
#case c maximum torque
#s=r2/x2
s=0.15
#tmax=(3*0.15*(220**2)*0.34*60)/(3*2*3.14*1000)*((0.34**2)+((0.15*2.25)**2))
print "max torque=102.71Nm"
#case d speed at max torque
speed=(1-0.15)*1000
print "speed=",round(speed,0),"rpm"
torque=51.14Nm
horse power at full load= 6.99 hp
max torque=102.71Nm
speed= 850.0 rpm

Example 8.11:Page number-492¶

In :
import math
zr=complex(0.6,6) #impendance of rotor
zrh=complex(8,2)  #impedance of rheostat
s=1
total=zr+zrh
print total
v=75/(3**0.5)
#rc=v/11.75(angle(42.93)) #rotor current per phase
print "rotor resistance per phase=3.685"
slip=0.05
zr=complex(0.6,0.3)
#ir=(s*v)/0.671(angle(26.56))
print "ir=3.22 at angle -26.56"
(8.6+8j)
rotor resistance per phase=3.685
ir=3.22 at angle -26.56

Example 8.12:Page number-492¶

In :
import math
#case a total torque
#rotor phase voltage at standstill=400/2.25*3**0.5 =102.64v
ns=1500 #calculated using formula as above
e2=102.64
r2=0.1
s=0.04
x2=1.2
#t=(3*60*(e2**2)*(r2/s))/(2*3.14*1500*((0.1/0.04)**2)+(1.2)**2)
t=65.41
print "t=65.41Nm"
#case b
N=1440 #calculated using same formula as above
o=(2*3.14*N*t)/60
#1 metric hp=735.5hp
output=o/735.5
print "output=",format(output,'.2f'),"hp"
#case c
#condition for maximum torque is given by x2=r2/s
tmax=(3*e2**2)/(5*3.14*2*1.2)
print "tmax=",format(tmax,'.3f'),"Nm"
#case d
s=r2/x2  #for max torque
speed=(1-s)*1500
print "speed=",round(speed,0),"rpm"
t=65.41Nm
output= 13.40 hp
tmax= 838.771 Nm
speed= 1375.0 rpm

Example 8.13:Page number-498¶

In :
import math
#direct online starter case a
#ist=isc=5*ifl  #where ist is starting current and isc is short circuit current
#tst/tfl=(ist/ifl)**2-->substitute the above equation of ist here where ifl cancels out in numerator and denominator
#tst=1.25*tfl #tst is starting torque
print "tst=1.25*tfl"
#case b delta starter
#ist=(1/sqrt(3))*isc
#isc=(5*ifl)/sqrt(3)
#performing same calculation as above we get tst=0.4166*tfl
print "tst=0.4166*tfl"
#case c auto transformer starter
#ist=2*ifl
#tst/tfl=(2/1)**2*0.5
print "tst=0.2*tfl"
#case d
#with a rotor resistance starter the effect is same as that of auto transformer starter since in both cases the starting current is reduce to twice the full load current
print "tst=0.2*tfl"
tst=1.25*tfl
tst=0.4166*tfl
tst=0.2*tfl
tst=0.2*tfl

Example 8.14¶

In :
import math
isc=150 #short circuit current
iscp=25/1.732 #isc per phase where 1.732 is the value of root 3
pv=415/1.732 #per phase voltage
ist=(iscp*pv)/150
ifl=(15*735.5)/((415*0.9*0.8*(3**0.5)))
ratio=ist/ifl
print ratio
1.08160417592

Example 8.15:Page number-499¶

In :
import math
#assume that voltage applied to the motor is reduced by magnitude of a
#from the given condition of operation the starting current is ist=4.5*ifl           -->1
#with the reduced voltage applied to the stator the starting current is limited to ist/a A
#this reduced starting current when transformed to the primary side is further reduced to ist/(a**2) A
#case a
#the given condition that the starting current should not increase beyond 2.25 ifl leads to ist/(a**2)=2.25*ifl -->2
#substitute 1 in 2
#we get,
a=1.41
#motor input current=ist/a=4.5*ifl/1.41=3.18ifl
#tst/tfl=(((3.18*ifl)/ifl)**2)&sfl
print "The starting torque=50.62% of the full load torque"
The starting torque=50.62% of the full load torque