# Chapter16 - Polyphase Systems¶

## Exa 16.1 page 344¶

In [15]:
print 'For star connected load'
Il=50000/((3**0.5)*440*0.85)
print "Line current=%.2f A"%(Il)
Iph=Il
print "Phase current=%.2f A"%(Iph)
Il=50000/((3**0.5)*440*0.85)
print "Line current=%.2f A"%(Il)
Iph=Il/(3**0.5)
print "Phase current=%.2f A"%(Iph)

For star connected load
Line current=77.19 A
Phase current=77.19 A
Line current=77.19 A
Phase current=44.56 A


## Exa 16.2 page 345¶

In [16]:
print 'For star connection'
Zph=(12**2+5**2)**0.5
Eph=440/(3**0.5)
Iph=Eph/Zph
Il=Iph
print "Line current=%.2f A"%(Il)
P_total=(3**0.5)*440*Il*12/(Zph*1000)
print "Total Power=%.2f kW"%(P_total)

print 'For Delta connection'
Zph=(12**2+5**2)**0.5
Eph=440
Iph=Eph/Zph
Il=Iph*(3**0.5)
print "Line current=%.2f A"%(Il)
P_total=(3**0.5)*440*Il*12/(Zph*1000)
print "Total Power=%.2f kW"%(P_total)

For star connection
Line current=19.54 A
Total Power=13.75 kW
For Delta connection
Line current=58.62 A
Total Power=41.24 kW


## Exa 16.3 page 346¶

In [17]:
pf=(1.8*1000)/(1100*(3**0.5))
Z=1100/100
R=Z*pf
print "Resistance of the load=%.2f ohm"%(R)
Xl=(121-108)**0.5
L=Xl/314
print "Inductive reactance of the load=%.2f H"%(L)

Resistance of the load=10.39 ohm
Inductive reactance of the load=0.01 H


## Exa 16.4 page 346¶

In [18]:
from math import cos, pi
Eph=400/(3**0.5)
print "\nPhase voltage=%.2f V"%(Eph)
P_total=(3**0.5)*400*30*cos(30*pi/180)/1000
print "Total power=%.2f kW"%(P_total)

Phase voltage=230.94 V
Total power=18.00 kW


## Exa 16.5 page 347¶

In [19]:
Out_motor=80*735.5
Input_motor=Out_motor/0.8
I_alternator_phase=120.64
I_motor_phase= I_alternator_phase/(3**0.5)
print "Current in each motor phase=%.2f A"%(I_motor_phase)
print "Current in each generator phase=%.2f A"%(I_alternator_phase)

Current in each motor phase=69.65 A
Current in each generator phase=120.64 A


## Exa 16.6 page 348¶

In [20]:
El=400
Eph=El
Impedance_per_phase= (10**2+15**2)**0.5
Iph= 400/Impedance_per_phase
print "Phase current=%.2f A"%(Iph)
Il=Iph*3**0.5
print "Line current=%.2f A"%(Il)
pf=10/Impedance_per_phase
print "Power factor=%.2f "%(pf)
P_total=(3**0.5)*El*Il*pf/1000
print "Total Power=%.2f kW"%(P_total)
VAR=(3**0.5)*El*Il*15/(Impedance_per_phase*1000)
print "Reactive volt ampers=%.2f KVAR"%(VAR)
VA=(3**0.5)*El*Il/1000
print "Total Volt ampers=%.2f kVA"%(VA)

Phase current=22.19 A
Line current=38.43 A
Power factor=0.55
Total Power=14.77 kW
Reactive volt ampers=22.15 KVAR
Total Volt ampers=26.63 kVA


## Exa 16.7 page 349¶

In [21]:
print 'Star connections'
R=20
Iph=440/(3**0.5*R)
P_total=3*Iph**2*R
print 'when one of the resistor get disconnected'
Iph=440/(2*20)
P_total_new=2*Iph**2*R
P_reduction=(P_total-P_total_new)*100/P_total
print "Reduction in Power=%.2f percent"%(P_reduction)
print 'Delta connections'
R=20
Iph=440/(R)
P_total=3*Iph**2*R
print 'when one of the resistor get disconnected'
Iph=440/(20)
P_total_new=2*Iph**2*R
P_reduction=(P_total-P_total_new)*100/P_total
print "Reduction in Power=%.2f percent"%(P_reduction)

Star connections
when one of the resistor get disconnected
Reduction in Power=50.00 percent
Delta connections
when one of the resistor get disconnected
Reduction in Power=33.33 percent


## Exa 16.8 page 350¶

In [22]:
from __future__ import division
R=3
XL=4
Z=(R**2+XL**2)**0.5
Iph1=440/(3**0.5*Z)
IL1=Iph1
print "Line current=%.1f A"%(IL1)
P=3*Iph1**2*R
print "Power=%.0f W"%(P)
pf1=R/Z
print "power factor=%.2f (lag)"%(pf1)
IL2=IL1*(4/5)
Iph2=IL2/3**0.5
XL2=440/Iph2
C2=1*10**6/(2*50*28.755)
print "Capacitance=%.1f uF"%(C2)

Line current=50.8 A
Power=23232 W
power factor=0.60 (lag)
Capacitance=347.8 uF


## Exa 16.9 page 350¶

In [23]:
IL=11000
Eph=IL/3**0.5
print "\nLine to neutral voltage=%.2f V"%(Eph)
E_Each_phase=Eph
print "\nVoltage induced in Each phase winding=%.2f V"%(E_Each_phase)
T=(242/360)*(1/50)*1000
print "\nTime interval=%.2f ms"%(T)
IL_peak=(2**0.5)*IL
print "\nPeak line voltage=%.2f V"%(IL_peak)

Line to neutral voltage=6350.85 V

Voltage induced in Each phase winding=6350.85 V

Time interval=13.44 ms

Peak line voltage=15556.35 V


## 16.10 page 351¶

In [24]:
P_consumed=3000/3
E_per_phase=440/(3**0.5)
IL=P_consumed/E_per_phase
print "\nCurrent in each line=%.2f A"%(IL)
R=E_per_phase/IL
print "\nResistance of resistor=%.2f ohm"%(R)

Current in each line=3.94 A

Resistance of resistor=64.53 ohm


## Exa 16.11 page 351¶

In [25]:
VL=1100
IL=100
pf=150*1000/(3**0.5*VL*IL)
E_per_phase=VL/3**0.5
Zph=E_per_phase/100
Rph=pf*Zph
Xc=(Zph**2-Rph**2)**0.5
C=10**6/(2*pi*50*Xc)
print 'Circuit Constants are'
print "\nR=%.2f ohm"%(Rph)
print "\nC=%.2f uF"%(C)

Circuit Constants are

R=5.00 ohm

C=812.89 uF


## Exa 16.12 page 352¶

In [26]:
from math import degrees, acos, tan
#P_input=W1+W2=15000........(i)
pf=0.4
phi=degrees(acos(0.4) )
a=tan(phi*pi/180)
#tand(phi)=(3**0.5)*(W1-W2)/(W1+W2)
#on solving  W1-W2=3464.2 ..............(ii)
#From (i) and (ii) we can calculate
W1=9.232
W2=5.768
print "\nW1=%.2f kW"%(W1)
print "\nW2=%.2fkW "%(W2)

W1=9.23 kW

W2=5.77kW


## Exa 16.13 page 352¶

In [27]:
from math import atan
W1=10
W2=-1.2
P_absorbed=W1+W2
print "\nPower=%.2f kW"%(P_absorbed)
phi=degrees(atan((3**0.5)*(W1-W2)/(W1+W2)) )
pf=cos(phi*pi/180)
print "\nPower Factor=%.2f "%(pf)

Power=8.80 kW

Power Factor=0.41


## Exa 16.14 page 353¶

In [28]:
P_input=10*735.5/0.82
#P_input=W1+W2=8974........(i)
pf=0.4
phi=degrees(acos(0.83))
a=tan(phi*pi/180)
#tand(phi)=(3**0.5)*(W1-W2)/(W1+W2)
#on solving  W1-W2=3482 ..............(ii)
#From (i) and (ii) we can calculate
W1=6.228
W2=2.746
print "\nW1=%.2f kW"%(W1)
print "\nW2=%.2fkW "%(W2)

W1=6.23 kW

W2=2.75kW