Chapter 19: Three phase systems

Example 1, page no. 299

In [1]:
from __future__ import division
import math
#initializing  the  variables:
Vl  =  415;#  in  Volts
Rp  =  30;#  in  ohms

#calculation:
Vp  =  Vl/(3**0.5)
Ip  =  Vp/Rp
Il  =  Ip

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  system  phase  voltage  is  ",round(Vp,2),"  V"
print  "\n  (b)phase  current  is  ",round(Ip,2),"  A"
print  "\n  (c)line  current  is  ",round(Il,2),"  A"


  Result  



  (a)the  system  phase  voltage  is   239.6   V

  (b)phase  current  is   7.99   A

  (c)line  current  is   7.99   A

Example 2, page no. 299

In [2]:
from __future__ import division
import math
#initializing  the  variables:
R  =  30;#  in  ohms
L  =  0.1273;#  in  Henry
Ip  =  5.08;#  in  Amperes
f  =  50;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
Zp  =  (R*R  +  XL*XL)**0.5
Il  =  Ip
Vp  =  Ip*Zp
Vl  =  Vp*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)line  voltage  is  ",round(Vl,2),"  V"


  Result  



  (a)line  voltage  is   439.89   V

Example 4, page no. 301

In [1]:
from __future__ import division
import math
#initializing  the  variables:
V  =  415;#  in  Volts
PR  =  24000;#  in  Watt
Py  =  18000;#  in  Watt
Pb  =  12000;#  in  Watt
VR  =  240;#  in  Volts
Vy  =  240;#  in  Volts
Vb  =  240;#  in  Volts

#calculation:
 #For  a  star-connected  system  VL  =  Vp*(3**0.5)
Vp  =  V/(3**0.5)
phir  =  90*math.pi/180
phiy  =  330*math.pi/180
phib  =  210*math.pi/180
 #  I  =  P/V    for  a  resistive  load
IR  =  PR/VR
Iy  =  Py/Vy
Ib  =  Pb/Vb
Inh  =  IR*math.cos(phir)  +  Ib*math.cos(phib)  +  Iy*math.cos(phiy)
Inv  =  IR*math.sin(phir)  +  Ib*math.sin(phib)  +  Iy*math.sin(phiy)
In  =  (Inh**2  +  Inv**2)**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)cuurnt  in  R  line  is  ",round(IR,2),"  A,  cuurnt  in  Y  line  is  ",round(Iy,2),"  A  "
print   "and  cuurnt  in  B  line  is  ",round(Ib,2),"  A"
print  "\n  (b)cuurnt  in  neutral  line  is  ",round(In,2),"  A"


  Result  



  (a)cuurnt  in  R  line  is   100.0   A,  cuurnt  in  Y  line  is   75.0   A  and  cuurnt  in  B  line  is   50.0   A

  (b)cuurnt  in  neutral  line  is   43.3   A

Example 5, page no. 302

In [2]:
from __future__ import division
import math
#initializing  the  variables:
R  =  30;#  in  ohms
L  =  0.1273;#  in  Henry
VL  =  440;#  in  Volts
f  =  50;#  in  Hz

#calculation:
XL  =  2*math.pi*f*L
Zp  =  (R*R  +  XL*XL)**0.5
Vp  =  VL
 #Phase  current
Ip  =  Vp/Zp
 #For  a  delta  connection,
IL  =  Ip*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  phase  current  ",round(Ip,2),"  A"
print  "\n  (b)line  current  ",round(IL,2),"  A"


  Result  



  (a)the  phase  current   8.8   A

  (b)line  current   15.24   A

Example 6, page no. 302

In [3]:
from __future__ import division
import math
#initializing  the  variables:
IL  =  15;#  in  Amperes
VL  =  415;#  in  Volts
f  =  50;#  in  Hz

#calculation:
 #For  a  delta  connection
Ip  =  IL/(3**0.5)#phase  current
Vp  =  VL
 #Capacitive  reactance  per  phase
Xc  =  Vp/Ip
C  =  1/(2*math.pi*f*Xc)


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  is  ",round(C*1E6,2),"uF"


  Result  



  capacitance  is   66.43 uF

Example 7, page no. 303

In [1]:
from __future__ import division
import math
#initializing  the  variables:
R  =  3;#  in  ohms
XL  =  4;#  in  ohms
VL  =  415;#  in  Volts

#calculation:
 #For  a  star  connection:
 #IL  =  Ip
 #VL  =  Vp*(3**0.5)
VLs  =  VL
Vps  =  VLs/(3**0.5)
 #Impedance  per  phase,
Zp  =  (R*R  +  XL*XL)**0.5
Ips  =  Vps/Zp
ILs  =  Ips
 #For  a  delta  connection:
 #VL  =  Vp
 #IL  =  Ip*(3**0.5)
VLd  =  VL
Vpd  =  VLd
Ipd  =  Vpd/Zp
ILd  =  Ipd*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  line  voltage  for  star  connection  is  ",round(VLs,2),"  V  "
print   "and the  phase  voltage  for  star  connection  is  ",round(Vps,2),"  V  "
print   "and the  line  voltage  for  delta  connection  is  ",round(VLd,2),"  V  "
print   "and the  phase  voltage  for  delta  connection  is  ",round(Vpd,2),"  V"
print  "\n  (b)the  line  current  for  star  connection  is  ",round(ILs,2),"  A  "
print   "and  the  phase  current  for  star  connection  is  ",round(Ips,2),"  A  "
print   "and the  line  current  for  delta  connection  is  ",round(ILd,2),"  A  "
print   "and the  phase  current  for  delta  connection  is  ",round(Ipd,2),"  A"


  Result  



  (a)the  line  voltage  for  star  connection  is   415.0   V  
and the  phase  voltage  for  star  connection  is   239.6   V  
and the  line  voltage  for  delta  connection  is   415.0   V  
and the  phase  voltage  for  delta  connection  is   415.0   V

  (b)the  line  current  for  star  connection  is   47.92   A  
and  the  phase  current  for  star  connection  is   47.92   A  
and the  line  current  for  delta  connection  is   143.76   A  
and the  phase  current  for  delta  connection  is   83.0   A

Example 8, page no. 304

In [5]:
from __future__ import division
import math
#initializing  the  variables:
Rp  =  12;#  in  ohms
VL  =  415;#  in  Volts

#calculation:
 #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
Vp  =  VL/(3**0.5)#  since  the  resistors  are  star-connected
 #Phase  current,  Ip
Zp  =  Rp
Ip  =  Vp/Zp
 #For  a  star  connection
IL  =  Ip
 #  For  a  purely  resistive  load,  the  power  factor  cos(phi)  =  1
pf  =  1
P  =  VL*IL*(3**0.5)*pf


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)total  power  dissipated  by  the  resistors  is  ",round(P,2),"  W"


  Result  



  (a)total  power  dissipated  by  the  resistors  is   14352.08   W

Example 9, page no. 304

In [8]:
from __future__ import division
import math
#initializing  the  variables:
P  =  5000;#  in  Watts
IL  =  8.6;#  in  amperes
VL  =  400;#  in  Volts

#calculation:
 #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
pf  =  P/(VL*IL*(3**0.5))


#Results
print  "\n\n  Result  \n\n"
print  "\n  power  factor  is  ",round(pf,3)


  Result  



  power  factor  is   0.839

Example 10, page no. 304

In [9]:
from __future__ import division
import math
#initializing  the  variables:
R  =  10;#  in  ohms
L  =  0.042;#  in  Henry
VL  =  415;#  in  Volts
f  =  50;#  in  Hz

#calculation:
 #For  a  star  connection:
 #IL  =  Ip
 #VL  =  Vp*(3**0.5)
XL  =  2*math.pi*f*L
Zp  =  (R*R  +  XL*XL)**0.5
VLs  =  VL
Vps  =  VLs/(3**0.5)
 #Impedance  per  phase,
Ips  =  Vps/Zp
ILs  =  Ips
 #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
pfs  =  R/Zp
Ps  =  VLs*ILs*(3**0.5)*pfs

 #For  a  delta  connection:
 #VL  =  Vp
 #IL  =  Ip*(3**0.5)
VLd  =  VL
Vpd  =  VLd
Ipd  =  Vpd/Zp
ILd  =  Ipd*(3**0.5)
 #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
pfd  =  R/Zp
Pd  =  VLd*ILd*(3**0.5)*pfd


#Results
print  "\n\n  Result  \n\n"
print  "\n  total  power  dissipated  in  star  is  ",round(Ps,2)," W  and  in  delta  is  ",round(Pd,2)," W"


  Result  



  total  power  dissipated  in  star  is   6283.29  W  and  in  delta  is   18849.88  W

Example 11, page no. 305

In [10]:
from __future__ import division
import math
#initializing  the  variables:
Po  =  12750;#  in  Watts
pf  =  0.77;#  power  factor
eff  =  0.85;
VL  =  415;#  in  Volts

#calculation:
 #eff  =  power_out/power_in
Pi  =  Po/eff
 #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
IL  =  Pi/(VL*(3**0.5)*pf)#  line  current
 #For  a  delta  connection:
 #IL  =  Ip*(3**0.5)
Ip  =  IL/(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)power  input  is  ",round(Pi,2)," W"
print  "\n  (b)line  current  is  ",round(IL,2),"  A"
print  "\n  (c)phase  current  is  ",round(Ip,2),"  A"


  Result  



  (a)power  input  is   15000.0  W

  (b)line  current  is   27.1   A

  (c)phase  current  is   15.65   A

Example 13, page no. 308

In [1]:
from __future__ import division
import math
#initializing  the  variables:
R  =  30;#  in  ohms
XL  =  40;#  in  ohms
VL  =  400;#  in  Volts

#calculation:
Zp  =  (R*R  +  XL*XL)**0.5
 #a  delta-connected  load
Vp  =  VL
 #Phase  current
Ip  =  Vp/Zp
IL  =  Ip*(3**0.5)
 #Alternator  output  power  is  equal  to  the  power  dissipated  by  the  load.
 #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
pf  =  R/Zp
P  =  VL*IL*(3**0.5)*pf
 #Alternator  output  kVA,
S  =  VL*IL*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  current  supplied  by  the  alternator  is  ",round(IL,2),"  A"
print  "\n  (b)output  power  is  ",round(P/1000,2),"KW  and  kVA  of  the  alternator  is  ",round(S/1000,2),"kVA"


  Result  



  (a)the  current  supplied  by  the  alternator  is   13.86   A

  (b)output  power  is   5.76 KW  and  kVA  of  the  alternator  is   9.6 kVA

Example 14, page no. 308

In [2]:
from __future__ import division
import math
#initializing  the  variables:
R  =  30;#  in  ohms
C  =  80E-6;#  in  Farads
f  =  50;#  in  Hz
VL  =  400;#  in  Volts

#calculation:
 #Capacitive  reactance
Xc  =  1/(2*math.pi*f*C)
Zp  =  (R*R  +  Xc*Xc)**0.5
pf  =  R/Zp
 #a  delta-connected  load
Vp  =  VL
 #Phase  current
Ip  =  Vp/Zp
IL  =  Ip*(3**0.5)
 #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)
P  =  VL*IL*(3**0.5)*pf
 #Alternator  output  kVA,
S  =  VL*IL*(3**0.5)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  phase  current    is  ",round(Ip,2),"  A"
print  "\n  (b)the  line  current    is  ",round(IL,2),"  A"
print  "\n  (c) power  is  ",round(P/1000,2),"kW"
print  "\n  (d)kVA  of  the  alternator  is  ", round(S/1000,2),"kVA"


  Result  



  (a)the  phase  current    is   8.03   A

  (b)the  line  current    is   13.9   A

  (c) power  is   5.8 kW

  (d)kVA  of  the  alternator  is   9.63 kVA

Example 15, page no. 309

In [14]:
from __future__ import division
import math
#initializing  the  variables:
Pi1  =  8000;#  in  Watts
Pi2  =  4000;#  in  Watts

#calculation:
 #Total  input  power
Pi  =  Pi1  +  Pi2
phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))
 #Power  factor
pf  =  math.cos(phi)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)power  input  is  ",round(Pi,2),"W"
print  "\n  (b)power  factor  is  ",round(pf,2)


  Result  



  (a)power  input  is   12000.0 W

  (b)power  factor  is   0.87

Example 16, page no. 310

In [15]:
from __future__ import division
import math
#initializing  the  variables:
Pi  =  12000;#  in  Watts
pf  =  0.6;#  power  factor

#calculation:
 #If  the  two  wattmeters  indicate  Pi1  and  Pi2  respectively
 #  Pit  =  Pi1  +  Pi2
Pit  =  Pi
 #  Pid  =  Pi1  -  Pi2
 #power  factor  =  0.6  =  cos(phi)
phi  =  math.acos(pf)
Pid  =  Pit*math.tan(phi)/(3**0.5)
 #Hence  wattmeter  1  reads
Pi1  =  (Pid  +  Pit)/2
 #wattmeter  2  reads
Pi2  =  Pit  -  Pi1


#Results
print  "\n\n  Result  \n\n"
print  "\n  reading  in  each  wattameter  are  ",round(Pi1,2),"W  and  ",round(Pi2,2),"W"


  Result  



  reading  in  each  wattameter  are   10618.8 W  and   1381.2 W

Example 17, page no. 310

In [16]:
from __future__ import division
import math
#initializing  the  variables:
Pi1  =  10000;#  in  Watts
Pi2  =  -3000;#  in  Watts

#calculation:
 #Total  input  power
Pi  =  Pi1  +  Pi2
phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))
 #Power  factor
pf  =  math.cos(phi)


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)power  input  is  ",round(Pi,2),"W"
print  "\n  (b)power  factor  is  ",round(pf,2)


  Result  



  (a)power  input  is   7000.0 W

  (b)power  factor  is   0.3

Example 18, page no. 311

In [17]:
from __future__ import division
import math
#initializing the variables:
R = 8; # in ohms
XL = 8; # in ohms
VL = 415; # in Volts

#calculation:
#For a star connection:
#IL = Ip
#VL = Vp*(3**0.5)
VLs = VL
Vps = VLs/(3**0.5)
#Impedance per phase,
Zp = (R*R + XL*XL)**0.5
Ips = Vps/Zp
ILs = Ips
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)
pf = R/Zp
Ps = VLs*ILs*(3**0.5)*pf
#If wattmeter readings are P1 and P2 then P1 + P2 = Pst
Pst = Ps
# Pid = Pi1 - Pi2
phi = math.acos(pf)
Psd = Pst*math.tan(phi)/(3**0.5)
#Hence wattmeter 1 reads
Ps1 = (Psd + Pst)/2
#wattmeter 2 reads
Ps2 = Pst - Ps1

#For a delta connection:
#VL = Vp
#IL = Ip*(3**0.5)
VLd = VL
Vpd = VLd
Ipd = Vpd/Zp
ILd = Ipd*(3**0.5)
#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)
Pd = VLd*ILd*(3**0.5)*pf
#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt
Pdt = Pd
# Pid = Pi1 - Pi2
Pdd = Pdt*math.tan(phi)/(3**0.5)
#Hence wattmeter 1 reads
Pd1 = (Pdd + Pdt)/2
#wattmeter 2 reads
Pd2 = Pdt - Pd1

#results
print "\n\n Result \n\n"
print "(a)When the coils are star-connected the wattmeter readings are", round(Ps1,2),"W and ",round(Ps2,2),"W"
print "(b)When the coils are delta-connected the wattmeter readings are are", round(Pd1,2),"W and", round(Pd2,2),"W"


 Result 


(a)When the coils are star-connected the wattmeter readings are 8489.35 W and  2274.71 W
(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W