Chapter 22: Three-phase induction motors

Example 1, page no. 389

In [1]:
from __future__ import division
import math
#initializing  the  variables:
f  =  50;#  in  Hz
p  =  2/2;#  number  of  pairs  of  poles

#calculation:
 #ns  is  the  synchronous  speed,  
    #f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and 
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
nsrpm  =  ns*60


#Results
print  "\n\n  Result  \n\n"
print  "\nsynchronous  speed  of  the  motor  is  ",nsrpm,"  rev/min"


  Result  



synchronous  speed  of  the  motor  is   3000.0   rev/min

Example 2, page no. 389

In [2]:
from __future__ import division
import math
#initializing  the  variables:
f  =  60;#  in  Hz
ns  =  900/60;#  in  rev/sec

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  
    #p  is  the  number  of  pairs  of  poles.
p  =  f/ns
np  =  p*2


#Results
print  "\n\n  Result  \n\n"
print  "\nnumber  of  poles  is  ", round(np,2)


  Result  



number  of  poles  is   8.0

Example 3, page no. 390

In [3]:
from __future__ import division
import math
#initializing  the  variables:
p  =  2/2;#  number  of  pairs  of  poles
ns  =  6000/60;#  in  rev/sec

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and 
    #


#Results
print  "\n\n  Result  \n\n"
print  "frequency  is  ",f,"  Hz"


  Result  


frequency  is   100.0   Hz

Example 4, page no. 391

In [4]:
from __future__ import division
import math
#initializing  the  variables:
p  =  4/2;#  number  of  pairs  of  poles
f  =  50;#  in  Hz
nr  =  1455/60;#  in  rev/sec

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and 
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
 #The  slip,  s
s  =  ((ns  -  nr)/ns)*100#  in  percent


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)  synchronous  speed  is  ",ns,"  rev/sec"
print  "\n(b)  slip  is  ",s,"  percent"


  Result  



(a)  synchronous  speed  is   25.0   rev/sec

(b)  slip  is   3.0   percent

Example 5, page no. 392

In [5]:
from __future__ import division
import math
#initializing  the  variables:
p  =  2/2;#  number  of  pairs  of  poles
f  =  60;#  in  Hz
s  =  0.02;#  slip

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and 
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
 #The  the  rotor  runs  at
nr  =  ns*(1  -  s)
 #frequency  of  the  e.m.f.  induced  in  the  rotor  bars  is
fr  =  ns  -  nr

#Results
print  "\n\n  Result  \n\n"
print  "\n(a)  synchronous  speed  is  ",ns,"  rev/sec"
print  "\n(b)  rotor  speed  is  ",nr,"  rev/sec"
print  "\n(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is  ",fr,"  Hz"


  Result  



(a)  synchronous  speed  is   60.0   rev/sec

(b)  rotor  speed  is   58.8   rev/sec

(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is   1.2   Hz

Example 6, page no. 392

In [6]:
from __future__ import division
import math
#initializing  the  variables:
f  =  50;#  in  Hz
nr  =  1200/60;#  in  rev/min
s  =  0.04;#  slip

#calculation:
 #the  synchronous  speed.
ns  =  nr/(1  -  s)
nsrpm  =  ns*60

#Results
print  "\n\n  Result  \n\n"
print  "\n  synchronous  speed  is  ",nsrpm,"  rev/min"


  Result  



  synchronous  speed  is   1250.0   rev/min

Example 7, page no. 394

In [7]:
from __future__ import division
import math
#initializing  the  variables:
p  =  8/2;#  number  of  pairs  of  poles
f  =  50;#  in  Hz
fr  =  3;#  in  Hz

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
 #fr  =  s*f
s  =  (fr/f)
 #the  rotor  speed.
nr  =  ns*(1  -  s)
nrrpm  =  nr*60


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)  slip  is  ",s*100,"  percent"
print  "\n  (b)  rotor  speed  is  ",nrrpm,"  rev/min"


  Result  



(a)  slip  is   6.0   percent

  (b)  rotor  speed  is   705.0   rev/min

Example 8, page no. 396

In [8]:
from __future__ import division
import math
#initializing  the  variables:
Psi  =  32000;#  in  Watts
Psl  =  1200;#  in  Watts
s  =  0.05;#  slip
Pfl  =  750;#  in  Watts

#calculation:
 #Input  power  to  rotor  =  stator  input  power  -  stator  losses
Pi  =    Psi  -  Psl
 #slip  =  rotor  copper  loss/rotor  input
Pl  =  s*Pi
 #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -  rotor  losses
Pr  =  Pi  -  Pl
 #Output  power  of  motor  =  power  developed  by  the  rotor  -  friction  and  windage  losses
Po  =  Pr  -  Pfl
 #Efficiency  of  induction  motor  =  (output  power/input  power)*100
eff  =  (Po/Psi)*100#  in  percent


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)  rotor  copper  loss  is  ",Pl,"  Watt"
print  "\n(b)  Total  mechanical  power  developed  by  the  rotor  is  ",Pr,"  W"
print  "\n(c)  Output  power  of  motor  is  ",Po,"  Watt"
print  "\n(d)  efficiency  of  induction  motor  is  ",round(eff,2),"  percent"


  Result  



(a)  rotor  copper  loss  is   1540.0   Watt

(b)  Total  mechanical  power  developed  by  the  rotor  is   29260.0   W

(c)  Output  power  of  motor  is   28510.0   Watt

(d)  efficiency  of  induction  motor  is   89.09   percent

Example 9, page no. 397

In [9]:
from __future__ import division
import math
#initializing  the  variables:
Psi  =  32000;#  in  Watts
Psl  =  1200;#  in  Watts
Pfl  =  750;#  in  Watts
x  =  0.35;

#calculation:
 #The  slip,  s
s  =  1-x
 #Input  power  to  rotor  =  stator  input  power  -  stator  losses
Pi  =    Psi  -  Psl
 #slip  =  rotor  copper  loss/rotor  input
Pl  =  s*Pi
 #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -  rotor  losses
Pr  =  Pi  -  Pl
 #Output  power  of  motor  =  power  developed  by  the  rotor  -  friction  and  windage  losses
Po  =  Pr  -  Pfl
 #Efficiency  of  induction  motor  =  (output  power/input  power)*100
eff  =  (Po/Psi)*100#  in  percent


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)  rotor  copper  loss  is  ",Pl,"  Watt"
print  "\n(b)  efficiency  of  induction  motor  is  ",round(eff,2),"  percent"


  Result  



(a)  rotor  copper  loss  is   20020.0   Watt

(b)  efficiency  of  induction  motor  is   31.34   percent

Example 10, page no. 398

In [10]:
from __future__ import division
import math
#initializing  the  variables:
V  =  415;#  in  Volts
f  =  50  ;#  in  Hz
nr  =  24;#  in  rev/sec
p  =  4/2;#  no.  of  pole  pairs
R2  =  0.35;#  in  Ohms
X2  =  3.5;#  in  Ohms
tr  =  0.85;#  turn  ratio  N2/N1
Pl  =  770;#  in  Watt
m  =  3;#  no.  of  phases

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
 #The  slip,  s
s  =  ((ns  -  nr)/ns)*100#  in  percent
 #Phase  voltage,  E1  =  V/(3**0.5)
E1  =  V/(3**0.5)
 #Full  load  torque
T  =  (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2  +  (X2*(s/100))**2))
 #Output  power,  including  friction  losses
Pm  =  2*math.pi*nr*T
 #power  output
Po  =  Pm  -  Pl
 #Maximum  torque  occurs  when  R2  =  Xr  =  0.35  ohm
 #Slip  
sm  =  R2/X2
 #maximum  torque,  Tm  
Tm  =  (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2  +  (X2*sm)**2))
 #speed  at  which  maximum  torque  occurs
nrm  =  ns*(1  -  sm)
nrmrpm  =  nrm*60
 #At  the  start,  i.e.,  at  standstill,  slip,  s=1
ss  =  1
 #starting  torque
Ts  =  (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2  +  (X2*ss)**2))


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)Synchronous  speed  is  ",round(ns,2),"  rev/sec"
print  "\n(b)Slip  is  ",round(s,2),"  percent"
print  "\n(c)Full  load  torque  is  ",round(T,2),"  Nm"
print  "\n(d)power  output  is  ",round(Po,2),"W"
print  "\n(e)maximum  torque  is  ",round(Tm,2),"  Nm"
print  "\n(f)speed  at  which  maximum  torque  occurs  is  ",round(nrmrpm,2),"rev/min"
print  "\n(g)starting  torque  is  ",round(Ts,2),"  Nm"


  Result  



(a)Synchronous  speed  is   25.0   rev/sec

(b)Slip  is   4.0   percent

(c)Full  load  torque  is   78.05   Nm

(d)power  output  is   10998.99 W

(e)maximum  torque  is   113.17   Nm

(f)speed  at  which  maximum  torque  occurs  is   1350.0 rev/min

(g)starting  torque  is   22.41   Nm

Example 11, page no. 400

In [11]:
from __future__ import division
import math
#initializing  the  variables:
V  =  415;#  in  Volts
f  =  50  ;#  in  Hz
nr  =  24;#  in  rev/sec
p  =  4/2;#  no.  of  pole  pairs
R2  =  0.35;#  in  Ohms
X2  =  3.5;#  in  Ohms
tr  =  0.85;#  turn  ratio  N2/N1
m  =  3;#  no.  of  phases

#calculation:
 #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and 
    #p  is  the  number  of  pairs  of  poles.
ns  =  f/p
 #The  slip,  s
s  =  ((ns  -  nr)/ns)*100#  in  percent
 #Phase  voltage,  E1  =  V/(3**0.5)
E1  =  V/(3**0.5)
 #rotor  current,
Ir  =  (s/100)*E1*tr/((R2**2  +  (X2*(s/100))**2)**0.5)
 #Rotor  copper  loss  
Pcl  =  m*R2*(Ir**2)
 #starting  current,
ss  =1
I2  =  ss*tr*E1/((R2**2  +  (X2*ss)**2)**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)rotor  current  is  ",round(Ir,2),"  A"
print  "\n(b)Total  copper  loss  is  ",round(Pcl,2),"  W"
print  "\n(c)starting  current  is  ",round(I2,2),"  A"


  Result  



(a)rotor  current  is   21.61   A

(b)Total  copper  loss  is   490.37   W

(c)starting  current  is   57.9   A

Example 12, page no. 401

In [12]:
from __future__ import division
import math
#initializing  the  variables:
V  =  415;#  in  Volts
Psl  =  650;#  in  Watt
pf  =  0.87;#  power  factor

#calculation:
Pm  =  11770;#  watts  from  part  (d),  Problem  22.10
Pcl  =  490.35;#  watts,  Rotor  copper  loss,  from  part  (b),  Problem  22.11
 #Stator  input  power
P1  =  Pm  +  Pcl  +  Psl
Po  =  11000#  watts,  Net  power  output,  from  part  (d),  Problem  22.10
 #efficiency  =  (output/input)  *100
eff  =  (Po/P1)*100#  in  percent
 #Power  input,  P1  =  (3**0.5)*VL*IL*cos(phi)
 #  pf  =  cos(phi)
 #supply  current,  IL
I  =  P1/((3**0.5)*V*pf)


#Results
print  "\n\n  Result  \n\n"
print  "\n(aStator  input  power  is  ",round(P1,2)," W"
print  "\n(b)efficiency  is  ",round(eff,2),"  percent"
print  "\n(c)supply  current  is  ",round(I,2),"  A"


  Result  



(aStator  input  power  is   12910.35  W

(b)efficiency  is   85.2   percent

(c)supply  current  is   20.64   A

Example 13, page no. 401

In [13]:
from __future__ import division
import math
#initializing  the  variables:
V  =  415;#  in  Volts
f  =  50  ;#  in  Hz
nr  =  24;#  in  rev/sec
p  =  4/2;#  no.  of  pole  pairs
R2  =  0.35;#  in  Ohms
X2  =  3.5;#  in  Ohms

#calculation:
 #At  the  moment  of  starting,  slip,  
s  =  1
 #Maximum  torque  occurs  when  rotor  reactance  equals  rotor  resistance
 #for  maximum  torque
R2  =  s*X2


#Results
print  "\n\n  Result  \n\n"
print  "\nresistance  of  the  rotor  is  ",R2,"  Ohm"


  Result  



resistance  of  the  rotor  is   3.5   Ohm