# Chapter 1: Synchronous Machines¶

## Example 1.1, Page 24¶

In [52]:
#given data
slots=24;#no. of slotes
NoOfPhase=3;#no of phase
MotorSpeed=1450;#in rpm
N=1500;#Synchonous Speed in rpm
f=50;#in Hz

#Calculations&Results
print "As the winding is in single layer, each slot contains one coil slide only.";
CoilSlidePerSlot=1;#coil slide per slot
CoilSlidePerCoil=2;#coil slide per Coil
TotalCoils=slots*CoilSlidePerSlot/CoilSlidePerCoil;#no. of coils
print "Total no. of Coils : ",TotalCoils;
P=120*f/N;#no. of poles
print "No. of poles : ",P;
PolesPitch=slots/P;#unitless
print "Poles Pitch = ",PolesPitch;
print "In case of single layer winding, the pole ppitch is generally taken in odd numbers only"
print "let pole pitch = 5(for short pitch winding)";
PolesPitch=5;#for short pitch winding
print "Coil Span or coil through = 1-6";
CoilsPerPolePerPhase=TotalCoils/(P*NoOfPhase);#No. of Coils/Pole/Phase
print "No. of Coils/Pole/Phase = ",CoilsPerPolePerPhase;
pair_of_poles=2;#no. of pair of poles
TotalElectricalDegree=360*pair_of_poles;#in degree
ElectricalDegreesPerSlot=TotalElectricalDegree/slots;#in degree electrical
print "Electrical Degrees/Slot = ",ElectricalDegreesPerSlot;
Slots_required=120/ElectricalDegreesPerSlot;#No. of slotes required for proper phase displacement
print "No. of slotes required for proper phase displacement = ",Slots_required;
print "Winding Table is as follows :";
print "Coil No.     Connection Lead from     Coil Span     Phase and Group No.";
print "   1                   A1               1-6                  A1";
print "   2                                    3-8                  C4";
print "   3                   B1               5-10                 B1";
print "   4                                    7-12                 A2";
print "   5                   C1               9-14                 C1";
print "   6                                   11-16                 B2";
print "   7                                   13-18                 A3";
print "   8                                   15-20                 C2";
print "   9                                   17-22                 B3";
print "  10                                   19-24                 A4";
print "  11                                   21-2                  C3";
print "  12                                   23-4                  B4";

As the winding is in single layer, each slot contains one coil slide only.
Total no. of Coils :  12
No. of poles :  4
Poles Pitch =  6
In case of single layer winding, the pole ppitch is generally taken in odd numbers only
let pole pitch = 5(for short pitch winding)
Coil Span or coil through = 1-6
No. of Coils/Pole/Phase =  1
Electrical Degrees/Slot =  30
No. of slotes required for proper phase displacement =  4
Winding Table is as follows :
Coil No.     Connection Lead from     Coil Span     Phase and Group No.
1                   A1               1-6                  A1
2                                    3-8                  C4
3                   B1               5-10                 B1
4                                    7-12                 A2
5                   C1               9-14                 C1
6                                   11-16                 B2
7                                   13-18                 A3
8                                   15-20                 C2
9                                   17-22                 B3
10                                   19-24                 A4
11                                   21-2                  C3
12                                   23-4                  B4


## Example 1.2, Page 28¶

In [53]:
#given data
slots=24;#no. of slotes
P=4;#no. of poles
CoilPitch=5;#unitless
NoOfPhase=3;#no of phase

#Calculations&Results
print "As the winding is in double layer, each slot contains 2 coils per slide.";
CoilSlidePerSlot=2;#coil slide per slot
CoilSlidePerCoil=2;#coil slide per Coil
TotalCoils=slots*CoilSlidePerSlot/CoilSlidePerCoil;#no. of coils
print "Total no. of Coils : ",TotalCoils;
PolesPitch=slots/P;#unitless
print "Pole Pitch = ",PolesPitch;
CoilsPerPolePerPhase=TotalCoils/(P*NoOfPhase);#No. of Coils/Pole/Phase
print "No. of Coils/Pole/Phase = ",CoilsPerPolePerPhase;
pair_of_poles=2;#no. of pair of poles
TotalElectricalDegree=360*pair_of_poles;#in degree
ElectricalDegreesPerSlot=TotalElectricalDegree/slots;#in degree electrical
print "Electrical Degrees/Slot = ",ElectricalDegreesPerSlot;
Slots_required=120/ElectricalDegreesPerSlot;#No. of slotes required for proper phase displacement
print "No. of slots required for proper phase displacement = ",Slots_required;
print "ie. Phase A1 is brought out from slot no. = 1";
print "Phase B1 at slot no. = 1+4 = 5";
print "Phase C1 at slot no. = 5+4 = 9";
print "Col Connection - end to start to start";
print "Winding Table is as follows :";
print "Coil No.     Connection Lead from     Coil Span     Phase and Group No.";
print "   1                   A1               1-6                  A1";
print "   2                                    2-7                    ";
print "   3                                    3-8                  C4";
print "   4                                    4-9                    ";
print "   5                   B1               5-10                 B1";
print "   6                                    6-11                   ";
print "   7                                    7-12                 A2";
print "   8                                    8-13                   ";
print "   9                   C1               9-14                 C1";
print "  10                                   10-15                   ";
print "  11                                   11-16                 B2";
print "  12                                   12-17                   ";
print "  13                                   13-18                 A3";
print "  14                                   14-19                   ";
print "  15                                   15-20                 C2";
print "  16                                   16-21                   ";
print "  17                                   17-22                 B3";
print "  18                                   18-23                   ";
print "  19                                   19-24                 A4";
print "  20                                   20-1                    ";
print "  21                                   21-2                  C3";
print "  22                                   22-3                    ";
print "  23                                   23-4                    ";
print "  24                                   24-5                  B4";

As the winding is in double layer, each slot contains 2 coils per slide.
Total no. of Coils :  24
Pole Pitch =  6
No. of Coils/Pole/Phase =  2
Electrical Degrees/Slot =  30
No. of slots required for proper phase displacement =  4
ie. Phase A1 is brought out from slot no. = 1
Phase B1 at slot no. = 1+4 = 5
Phase C1 at slot no. = 5+4 = 9
Col Connection - end to start to start
Winding Table is as follows :
Coil No.     Connection Lead from     Coil Span     Phase and Group No.
1                   A1               1-6                  A1
2                                    2-7
3                                    3-8                  C4
4                                    4-9
5                   B1               5-10                 B1
6                                    6-11
7                                    7-12                 A2
8                                    8-13
9                   C1               9-14                 C1
10                                   10-15
11                                   11-16                 B2
12                                   12-17
13                                   13-18                 A3
14                                   14-19
15                                   15-20                 C2
16                                   16-21
17                                   17-22                 B3
18                                   18-23
19                                   19-24                 A4
20                                   20-1
21                                   21-2                  C3
22                                   22-3
23                                   23-4
24                                   24-5                  B4


## Example 1.3, Page 35¶

In [2]:
from math import pi, sqrt, sin, cos

#given data
StatorSlots=36;#No. of stator slots
Poles=4;#No. of poles
#coilSpan=1:8;#unitless

#Calculations&Results
SlotsPerPole=StatorSlots/Poles;# no. of slots per pole
print "Slots/Pole = ",SlotsPerPole;
print "Here the coil span falls short by, ",((2./9)*180)," Degree";
alfa=40*pi/180;#short pitch angle in degree
Kp=cos(alfa/2);#Coil span Factor
print "Pitch Factor or coil span factor : ",round(Kp,1);

Slots/Pole =  9
Here the coil span falls short by,  40.0  Degree
Pitch Factor or coil span factor :  0.9


## Example 1.4, Page 38¶

In [6]:
from math import pi, sqrt, sin, cos

#given data
NoOfPhase=3;#no of phase
P=16;#No. of pole alternator
Slots=144;#No. of slots
Conductors=10;#per slot
fi=0.03;#in Weber
N=375;#machine speed in rpm

#Calculations
f=P*N/120;#in Hz
SlotsPerPole=Slots/P;#unitless
m=Slots/(P*NoOfPhase);#unitless
Beta=180/(SlotsPerPole);#in Degree
Kd=sin(m*Beta/2*pi/180)/(m*sin(Beta/2*pi/180));#unitless
TotalConductors=Conductors*Slots;#no. of conductors
TotalConductorsPerPhase=Conductors*Slots/NoOfPhase;#no. of conductors/phase
TurnsPerPhase=TotalConductorsPerPhase/2;#No. of turns per phase
EMFPerPhase=4.44*Kd*fi*f*TurnsPerPhase;#in Volt
LineVoltage=sqrt(3)*EMFPerPhase;#in Volt

#Results
print "Frequency in Hz : ",f;
print "Phase Electromotive force in Volt : ",round(EMFPerPhase,1);
print "Line Electromotive force in Volt : ",round(LineVoltage,3);

Frequency in Hz :  50
Phase Electromotive force in Volt :  1534.1
Line Electromotive force in Volt :  2657.202


## Example 1.5, Page 39¶

In [7]:
from math import pi, sqrt, sin, cos

#given data
NoOfPhase=3;#no of phase
SlotsPerPhase=3;#o. of slots

#Calculations&Results
m=SlotsPerPhase;#no. of slots
SlotsPerPolePerPhase=SlotsPerPhase*NoOfPhase;#unitless
Beta=180./SlotsPerPolePerPhase;#in degree
print "The phase difference between the induced emf in two coils lying in adjacent slots = ",(Beta)," degree";
Kb=sin((m*Beta*pi)/(2*180))/(m*sin((Beta*pi)/(2*180)));#unitless
print "The breadth factor :  ",round(Kb,3);

The phase difference between the induced emf in two coils lying in adjacent slots =  20.0  degree


## Example 1.6, Page 39¶

In [9]:
from math import pi, sqrt, sin, cos

#given data
StatorSlots=24#No. of stator slots
Poles=4;#No. of poles
SlotsPerPole=StatorSlots/Poles;# no. of slots per pole
#coilSpan=1:6;#unitless

#Calculations&Results
print "If the sides of the coil are placed in slots 1 and 7, then it is full pitched, \nIf the coil slides are placed" \
+ " in 1 and 6 then it is short pitched and the distance equal to 5/6th of pole-pitch.";
print "Since it falls short by 1/6th of the pole-pitch, hence it is short by :"
theta=180/SlotsPerPole;#in Degree
print (theta)," Degree."
Kp=cos(theta/2*pi/180);#unitless
print "Pitch-factor Kp : ",round(Kp,4);

If the sides of the coil are placed in slots 1 and 7, then it is full pitched,
If the coil slides are placed in 1 and 6 then it is short pitched and the distance equal to 5/6th of pole-pitch.
Since it falls short by 1/6th of the pole-pitch, hence it is short by :
30  Degree.
Pitch-factor Kp :  0.9659


## Example 1.7, Page 40¶

In [10]:
from math import pi, sqrt, sin, cos
#given data
NoOfPhase=3;#no of phase
Eph=3300/sqrt(3);#in Volts
f=50;#in Hz
Poles=12;#No. of poles
StatorSlots=144#No. of stator slots

#Calculations
SlotsPerPhase=StatorSlots/NoOfPhase;#no. of slots/phase
Conductors=5;#per slot
ConductorsPerphase=SlotsPerPhase*Conductors;#Conductors/Phase
S=ConductorsPerphase;#Conductors/phase
SlotsPerPolePerPhase=SlotsPerPhase/Poles;#no. of slots/phase
Kf=1.11;#Form Factor
Kp=1;#For concentric winding
fi=Eph/(2*Kf*Kb*Kp*S*f);#in weber

#Result
print "The Flux per pole in weber : ",round(fi,4);

The Flux per pole in weber :  0.0745


## Example 1.8, Page 41¶

In [12]:
from math import pi, sqrt, sin, cos

#given data
Phase=3;#no. of phase
f=50;#in Hz
P=16;#No. of pole alternator
Slots=144.;#No. of slots
conductors=10;#conductors per slot

#Calculations&Results
fi=2.48*10**-2;#in weber
n=Slots/P;#No. of slots/pole
Zr=Slots*conductors/Phase;#No. of conductors/Phase
T=Zr/2;#N. of turns/phase
Beta=180/n;#Angular displacement between slots in degree
m=n/Phase;#No. of slots/pole/Phase
Kd=sin(m*Beta/2*pi/180)/(m*sin(Beta/2*pi/180));#Distribution factor :unitless
print "The coil span falls short of 2 slots i.e. ",(2*180/9)," degree.";
alfa=40;#short pitch angle in degree
Kp=(cos(alfa/2*pi/180));#Unitless
#Formula : f=P*N/120;#in Hz
N=120*f/P;#in rpm
Ep=4.44*Kd*Kp*fi*f*T;#in Volts
LineVoltage=sqrt(3)*Ep;#in Volts
print "The speed is ",(N)," in rpm";
print "The line emf is ",round(LineVoltage,2)," Volts";
#Note : Answer in the book is not accurate for last part due to rounding off errors

The coil span falls short of 2 slots i.e.  40  degree.
The speed is  375  in rpm
The line emf is  2064.15  Volts


## Example 1.9, Page 58¶

In [14]:
from math import pi, sqrt, sin, cos

#given data
RatedPower=100;#in KVA
RatedPower=100*1000;#in VA
VL=1040;#in Volt
Phase=3;#Machine phase
If=40;#in Ampere
Isc=200;#in Ampere
EL=1040;#in Volt

#Calculations&Results
Eph=EL/sqrt(3);#in Volt
Zs=Eph/Isc;#in Ohm
Rs=0.2;#in Ohm
Xs=sqrt(Zs**2-Rs**2);#in Ohm
IL=19.25;#in Ampere
V=3000/sqrt(3);#in Volt
#At 0.8 power factor lagging
IRa=IL*0.2;#in Volt
IXs=IL*Xs;#in Volt
Vsin_fi=V*0.6;#in Volt
Vcos_fi=V*0.8;#in Volt
Eo=sqrt((Vcos_fi+IRa)**2+(Vsin_fi+IXs)**2);#in Volts
Regulation=((Eo-V)/V)*100;#in %
print "Full load percentage regulation at a power factor of 0.8 lagging : ",round(Regulation,1);

Eo=sqrt((Vcos_fi+IRa)**2+(Vsin_fi-IXs)**2);#in Volts
Regulation=((Eo-V)/V)*100;#in %
print "Full load percentage regulation at a power factor of 0.8 leading : ",round(Regulation,2);
print "Negative regulation due to leading power factor.";

Full load percentage regulation at a power factor of 0.8 lagging :  2.2
Full load percentage regulation at a power factor of 0.8 leading :  -1.78
Negative regulation due to leading power factor.


## Example 1.10, Page 60¶

In [16]:
from math import pi, sqrt, sin, cos

#given data
RatedPower=50;#in KVA
RatedPower=50*1000;#in VA
VL=173;#in Volts
Ra=0.1;#in Ohm

#Calculations&Results
VP=VL/sqrt(3);#in Volts
print "Some exciting curent on short circuit produces a current of 100 A.";
OC_PhaseVoltage=100;#in Volt
SC_Current=100;#in Ampere
Zs=OC_PhaseVoltage/SC_Current;#n ohm
Xs=sqrt(Zs**2-Ra**2);#in Ohm
print "Impedence of the alternator in Ohm : ",round(Xs,2);
V=400;#in Volts
I_FL=RatedPower/(sqrt(3)*V);#in Ampere
V=400/sqrt(3);#in Volts
Eo=sqrt((V+I_FL*Ra)**2+(I_FL*Xs)**2);#in Volts
Regulation=(Eo-V)*100/V;#in %
print "Regulation at U.P.F. in % :",round(Regulation,2);

Some exciting curent on short circuit produces a current of 100 A.
Impedence of the alternator in Ohm :  0.99
Regulation at U.P.F. in % : 7.71


## Example 1.11, Page 61¶

In [20]:
from math import pi, sqrt, cos, sin

#given data
OutputPower=500;#in KVA
OutputPower=500*1000;#in VA
VL=3300;#in Volts
Ra=0.3;#in Ohm
Xs=4;#in Ohm
PF=0.8;#Lagging Power factor

#Calculations&Results
#Formula : outputPower=sqrt(3)*VL*IL
IL=OutputPower/(sqrt(3)*VL);#in Ampere
print "For a star connected alternator, line current is equal to phase current. Therefore Ia=IL";
Ia=IL;#in Ampere
#PF=cosd(fi)=0.8 and sind(fi)=0.6
cos_fi=0.8;#Power factor
sin_fi=0.6;#Unitless
VPerPhase=VL/sqrt(3);#in Volts
E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in Volts/Phase
Regulation=(E-VPerPhase)*100/VPerPhase;#in %
print "Voltage Regulation at Full Load in % :",round(Regulation,2);

For a star connected alternator, line current is equal to phase current. Therefore Ia=IL
Voltage Regulation at Full Load in % : 12.98


## Example 1.12, Page 62¶

In [21]:
from math import pi, sqrt, sin, cos

#given data
V=2000;#in Volt
Ia=100;#in Ampere
OC_Voltage=500;#in Volt
SC_Current=100;#in Ampere

#Calculations&Results
Zs=OC_Voltage/SC_Current;#in ohm
Ra=0.8;#in Ohm
Xs=sqrt(Zs**2-Ra**2);#in Ohm
#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)
#Part (a)  : at unity pf
cos_fi=1;#Unitless
sin_fi=0;#Unitless
E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi+Ia*Xs)**2)
Regulation=(E-V)*100/V;#in %
print "Regulation at U.P.F. in % :",round(Regulation,2);

#Part (b)  : at 0.71 pf lagging
cos_fi=0.71;#Unitless
sin_fi=0.704;#Unitless
E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi+Ia*Xs)**2)
Regulation=(E-V)*100/V;#in %
print "Regulation at 0.71 pf lagging in % :",round(Regulation,2);

#Part (c)  : at 0.8 pf leading
cos_fi=0.8;#Unitless
sin_fi=0.6;#Unitless
E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi-Ia*Xs)**2)
Regulation=(E-V)*100/V;#in %
print "Regulation at 0.8 pf leading in % :",round(Regulation,2);

Regulation at U.P.F. in % : 6.89
Regulation at 0.71 pf lagging in % : 21.1
Regulation at 0.8 pf leading in % : -8.88


## Example 1.13, Page 63¶

In [118]:
#given data
Ia=60;#in Ampere

#Calculations&Results
print "The value of synchronous impedence at this excitation :" ;
OC_Voltage=900;#in Volt
SC_Current=150;#in Ampere
Zs=OC_Voltage/SC_Current;#in ohm
print "Zs equals to ",(Zs)," Ohm";
print "Internal Voltage drop when tthe load current is 60A=Ia*Zs=",(Ia*Zs)," Volts";

The value of synchronous impedence at this excitation :
Zs equals to  6  Ohm
Internal Voltage drop when tthe load current is 60A=Ia*Zs= 360  Volts


## Example 1.14, Page 63¶

In [119]:
from math import pi, sqrt, sin, cos

#given data
V=6600;#in Volts
OutputPower=2000;#in KVA

#Calculations
OutputPower=2000*1000;#in VA
#Formula : outputPower=sqrt(3)*VL*IL
IL=OutputPower/(sqrt(3)*V);#in Ampere
Ia=IL;#in Ampere
Ra=0.4;#in Ohm
Xs=4.5;#in Ohm
#PF=cosd(fi)=0.8 and sind(fi)=0.6
cos_fi=0.8;#Power factor
sin_fi=0.6;#Unitless
VPerPhase=V/sqrt(3);#in Volts
E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in Volts/Phase
Regulation=(E-VPerPhase)*100/VPerPhase;#in %

#Result
print "Percentage Change in Terminal Voltage :",round(Regulation,1);

Percentage Change in Terminal Voltage : 14.9


## Example 1.15, Page 63¶

In [120]:
from math import pi, sqrt, sin, cos

#given data
OutputPower=1200;#in KVA
OutputPower=1200*1000;#in VA
V=3300;#in Volt
Ra=0.25;#in Ohm

#Calculations&Results
#Formula : outputPower=sqrt(3)*VL*IL
IL=OutputPower/(sqrt(3)*V);#in Ampere
Ia=IL;#in Ampere
VPerPhase=V/sqrt(3);#in Volts
OC_Voltage=1100;#in Volt
SC_Current=200;#in Ampere
Zs=OC_Voltage/(sqrt(3)*SC_Current);#in ohmRa
Xs=sqrt(Zs**2-Ra**2);#in Ohm
#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)

#Part (a)  : For lagging pf load
cos_fi=0.8;#Unitless
sin_fi=0.6;#Unitless
E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);
Regulation=(E-VPerPhase)*100/VPerPhase;#in %
print "Regulation at U.P.F. in % :",round(Regulation,2);

cos_fi=0.8;#Unitless
sin_fi=0.6;#Unitless
E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi-Ia*Xs)**2)
Regulation=(E-VPerPhase)*100/VPerPhase;#in %
print "Regulation at 0.71 pf lagging in % :",round(Regulation,2);

Regulation at U.P.F. in % : 25.9
Regulation at 0.71 pf lagging in % : -13.52


## Example 1.16, Page 64¶

In [123]:
from math import pi, sqrt, sin, cos

#given data
OutputPower=1500;#in KVA
OutputPower=1500*1000;#in VA
V=6600;#in Volt
Ra=0.4;#in Ohm
Xs=6;#in Ohm per phase
pf=0.8;#lagging power factor

#Calculations&Results
#Formula : outputPower=sqrt(3)*VL*IL
Ia=OutputPower/(sqrt(3)*V);#in Ampere
VPerPhase=V/sqrt(3);#in Volts
#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)
cos_fi=0.8;#Unitless
sin_fi=0.6;#Unitless
E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in volt
print "Induced emf in volt : ",round(E,2);
print "As excitation remains constant, E at 4364 volt remains constant.";
E=4364;#in Volt
V=4743;#in Volts
TerminalVoltage=sqrt(3)*V;#in Volts
print "Terminal voltage line to line in Volts : ",round(TerminalVoltage);
#Note ans of 1st part is wrong in the books

Induced emf in volt :  4366.07
As excitation remains constant, E at 4364 volt remains constant.
Terminal voltage line to line in Volts :  8215.0


## Example 1.17, Page 81¶

In [32]:
from math import pi, sqrt, sin, cos, atan, tan

#given data
OutputPower=2500;#in KVA
OutputPower=2500*1000;#in VA
V=6600;#in Volt

#Calculations
KW1=2500;#in KWatts
KVAR1=0;#Kwatts
KW2=5000;#in KWatts
cos_fi=0.707;#unitless
sin_fi=0.707;#unitless
KVAR2=KW2*sin_fi/cos_fi;#Kwatts
TotalKW=KW1+KW2;#in KWatts
TotalKVAR=KVAR1+KVAR2;#Kwatts
#For first Machine
KWm=4000;#in KWatts
cos_fi=0.8;#unitless
sin_fi=0.6;#unitless
KVARm=KWm*sin_fi/cos_fi;#Kwatts
#so, second machine will be supplying
KW=TotalKW-KWm;#in Kwatts
print "KW output of second machine : ",KW;
KVAR=TotalKVAR-KVARm;#in KWatts
tan_fi=KVAR/KW;#unitless
fi=atan(tan_fi);#ib degree
#Power factor of other machine
pf=cos(fi);#unitless

#Result
print "Power factor of other machine : ",round(pf,4)," lagging";

KW output of second machine :  3500
Power factor of other machine :  0.8682  lagging


## Example 1.18, Page 82¶

In [26]:
from math import pi, sqrt, sin, cos

#given data

KW1=500;#in KWatts
KVAR1=0;#Kwatts

KW2=1000;#in KWatts
pf=0.9;#lagging
cos_fi=0.9;#unitless
fi=acos(pf);
sin_fi=sin(fi);#unitless
KVAR2=KW2*sin_fi/cos_fi;#Kwatts

KW3=800;#in KWatts
pf=0.8;#lagging
cos_fi=0.8;#unitless
fi=acos(pf);
sin_fi=sin(fi);#unitless
KVAR3=KW3*sin_fi/cos_fi;#Kwatts

KW4=500;#in KWatts
pf=0.9;#lagging
cos_fi=0.9;#unitless
fi=acos(pf);
sin_fi=sin(fi);#unitless
KVAR4=-KW4*sin_fi/cos_fi;#Kwatts
#TOtalKW and TotalKVAR
TotalKW=KW1+KW2+KW3+KW4;#in KWatts
TotalKVAR=KVAR1+KVAR2+KVAR3+KVAR4;#in KWAtts
#For the first Machine :
KW=1500;#n Kwatts
cos_fi=0.95;#unitless
sin_fi=0.3123;#unitless
KVAR=KW*sin_fi/cos_fi;#Kwatts

KW1=TotalKW-KW;#in KWatts
KVAR1=TotalKVAR-KVAR;#in Volts
print "KW supplied by other machine : ",(TotalKW-KW);
print "KVAR supplied by other machine : ",round(TotalKVAR-KVAR);  #answer differs due to rounding off the digits
tan_fi=KVAR1/KW1;#unitless
#fi=atand(tan_fi);#in degree
cos_fi=cos(atan(tan_fi));#unitless

#Result
print "Power factor of the other machine : ",round(cos_fi,3);

KW supplied by other machine :  1300
KVAR supplied by other machine :  349.0
Power factor of the other machine :  0.966


## Example 1.19, Page 82¶

In [25]:
from math import pi, sqrt, sin, cos, acos, tan, atan
#given data

KW1=MW1*1000;#in KWatts
KVAR1=0;#Kwatts

KW2=MW2*1000;#in KWatts
pf=0.8;#unitless
cos_fi=0.8
fi=acos(pf);
sin_fi=sin(fi);#unitless
KVAR2=KW2*sin_fi/cos_fi;#Kwatts

TotalKVAR=KVAR1+KVAR2;#in KWatts
#For first machine :
KWm=MWm*1000;#in KWatts
cos_fi=0.866;#unitless
fi=acos(cos_fi);
tan_fi=tan(fi);#unitless
KVARm=KWm*tan_fi;#in KWatts
#so, load supplied by the second machine
KW2=TotalKW-KWm;#in Kwatts
print "Load of other machine,KW : ",(KW2);
KVAR2=TotalKVAR-KVARm;#in Kwatts
tan_fi=KVAR2/KW2;#unitless
fi=atan(tan_fi);#in degree
cos_fi=cos(atan(tan_fi));#unitless

#Result
print "Power factor of the other machine : ",round(cos_fi,3);

Load of other machine,KW :  28000
Power factor of the other machine :  0.925


## Example 1.20, Page 84¶

In [29]:
from math import pi, sqrt, sin, cos, atan, tan, acos

#given data
cos_fi=0.8;#unitless
fi=acos(cos_fi);
tan_fi=tan(fi);#unitless

#For Alternator A :
cos_fi_A=0.9;#unitless
fi_A=acos(cos_fi_A);
tan_fi_A=tan(fi_A);#unitless

#For A:

#For B :
fi_B=atan(tan_fi_B);#in degree
cos_fi=cos(atan(tan_fi_B));#unitless

#Result
print "Power factor of the other machine : ",round(cos_fi,3);

Power factor of the other machine :  0.642


## Example 1.21, Page 97¶

In [30]:
from math import pi, sqrt, sin, cos

#given data
V=6600;#in Volts
KW=6000.;#in KWatts
pf=0.8;#unitless
cos_fi=pf;#unitless
Eff=90.;#in %

#Calculations&Results
#Part (a) :
KVA=KW/cos_fi;#in KVAR
print "KVA rating of the alternator : ",(KVA)," KVA";
#Part (b) :
TotalRating=KVA;#in KVA
VA=TotalRating*1000;#in VA
I=VA/(sqrt(3)*KW);#in Ampere
print "Current Rating in Ampere : ",round(I,2);
#Part (c) :
Input=KW/(Eff/100);#in KW
print "Power Input(in KW) :",round(Input,2);
Input=Input*1000/735.5;#in hp
print "Power Input(in hp) :",round(Input,2);

KVA rating of the alternator :  7500.0  KVA
Current Rating in Ampere :  721.69
Power Input(in KW) : 6666.67
Power Input(in hp) : 9064.13


## Example 1.22, Page 97¶

In [31]:
from math import pi, sqrt, sin, cos

#given data
Ecoil=8000;#in Volts
Icoil=418;#in Ampere
pf=80.;#in % lgging

#Calculations&Results
pf=pf/100;#in fraction
cos_fi=pf;#unitless
#Part (i) :
EL=sqrt(3)*Ecoil;#in volt
print "Line volts(in V): ",round(EL);
#Part (ii) :
IL=Icoil;#in Ampere
print "Line Current in Ampere : ",IL;
#Part (iii) :
Rating=sqrt(3)*EL*IL/1000;#in KVA
print "Rating (in KVA) :",Rating;
#Part (iv) :

Line volts(in V):  13856.0