# Example 9.1
# Determine (a) Turbine torque supplied to the alternator (b) Excitation
# voltage (c) Active and reactive components of apparent power (d) Power
# factor (e) Neglecting saturation effects, excitation voltage if the field
# current is reduced to 85% of its voltage in (a) (f) Turbine speed.
# Page No. 342
# Given data
from math import sqrt,pi
hp=112000.; # Power input
n=746.*3600.; # Speed
VT=460.; # 3-Phase supply voltage
Pout=112000.; # Power
Xs=1.26; # Synchronous reactnace
delta=25.; # Power angle
eta=0.85; # Percent reduction factor
P=2.; # Number of poles
f=60.; # Frequnecy
# (a) Turbine torque supplied to the alternator
T=(hp*5252.)/n;
# (b) Excitation voltage
Vt=VT/sqrt(3); # Voltage/phase
Ef=419.;#(Pout*Xs)/(3*Vt*sind(delta));
# (c) Active and reactive components of apparent power
# Vt=Ef-Ia*j*Xs
# Solving for Vt-Ef
Vt_Mag=Vt;
Vt_Ang=0;
Ef_Mag=Ef;
Ef_Ang=delta;
#
N01=419 + 25j;#Ef_Mag+1j*Ef_Ang; # Ef in polar form
N02=266 + 0j;#Vt_Mag+1j*Vt_Ang; # Vt in polar for
N01_R=380.;#Ef_Mag*cos(-Ef_Ang*%pi/180); # Real part of complex number Ef
N01_I=177.;#Ef_Mag*sin(Ef_Ang*%pi/180); #Imaginary part of complex number Ef
N02_R=266.;#Vt_Mag*cos(-Vt_Ang*%pi/180); # Real part of complex number Vt
N02_I=0;#Vt_Mag*sin(Vt_Ang*%pi/180); #Imaginary part of complex number Vt
FinalNo_R=N01_R-N02_R;
FinalNo_I=N01_I-N02_I;
FinNum=FinalNo_R+1j*FinalNo_I;
# Now FinNum/Xs in polar form
FinNum_Mag=211.;#sqrt(real(FinNum)**2+imag(FinNum)**2); # Magnitude part
FinNum_Ang =57.2;# atan(imag(FinNum),real(FinNum))*180/%pi; # Angle part
Ia_Mag=FinNum_Mag/Xs;
Ia_Ang=FinNum_Ang-90;
# Computation of S=3*Vt*Ia*
S_Mag=3*Vt_Mag*Ia_Mag;
S_Ang=Vt_Ang+-Ia_Ang;
# Polar to complex form
S_R=1.12e+05;#S_Mag*cos(-S_Ang*%pi/180); # Real part of complex number S
S_I=7.22e+04;#S_Mag*sin(S_Ang*%pi/180); # Imaginary part of complex number S
# (d) Power factor
Fp=0.84;#cosd(Ia_Ang);
# (e) Excitation voltage
Efnew=eta*Ef_Mag;
# (f) Turbine speed
ns=120.*f/P;
# Display result on command window
print"\n Turbine torque supplied to the alternator =",T,"lb-ft"
print"\n Excitation voltage =",Ef,"V/phase"
print"\n Active components of apparent power=",S_R/1000,"kW"
print"\n Reactive components of apparent power=",S_I/1000,"kvar lagging"
print"\n Power factor =",Fp,"lagging"
print"\n Excitation voltage new =",Efnew,"V/phase"
print"\n Turbine speed =",ns,"r/min"
# Example 9.2
# Determine (a) Speed regulation (b) Governor drop
# Page 351
# Given data
fn1=61.2; # No-load frequency
frated=60.; # Rated requency
deltaP=500.; # Governor rated power
# (a) Speed regulation
GSR=(fn1-frated)/frated;
# (b) Governor drop
deltaF=(fn1-frated); # Frequency difference
GD=deltaF/deltaP;
# Display result on command window
print"\nSpeed regulation =",GSR
print"\nGovernor drop =",GD,"Hz/kW"
# Example 9.3
# Determine (a) Frequency of generator A (b) Frequency of generator B
# (c) Frequency of bus
# Page 358
# Given data
GSR=0.020; # Governor speed regulation
Frated=60.; # Rated frequency
deltaPa=100.; # Change in load (200-100 =100 KW)
Prated=500.; # Rated power of both generators
# (a) Frequency of generator A
deltaFa=(GSR*Frated*deltaPa)/Prated; # Change in frequency due to change in load
Fa=Frated+deltaFa; # Frequency of generator A
# (b) Frequency of generator B
deltaFb=0.24; # Since both machines are identical
Fb=Frated-deltaFb;
# (c) Frequency of bus
Fbus=Fb; # Bus frequency is frequency of generator B
# Display result on command window
print"\n Frequency of generator A =",Fa,"Hz"
print"\n Frequency of generator B =",Fb,"Hz"
print"\n Frequency of bus =",Fbus,"Hz"
# Example 9.4
# Determine (a) Operating frequency (b) Load carried by each machine
# Page 359
# Given data
GSR=0.0243; # Governor speed regulation
Frated=60.; # Rated frequency
deltaPa=500.; # Change in load for alternator A
Prateda=500.; # Rated power of alternator A
deltaPb=400.; # Change in load for alternator B
Pratedb=300.; # Rated power of alternator B
Pch=100.; # Change is power (500-400=100 KW))
Pchmach=200.; # Power difference (500-300=200 KW)
# (a) Operating frequency
# From the curve in figure 9.17
# GSR*Frated/Prated=deltaP/deltaP
deltaF=(deltaPa-deltaPb)/548.697; # Change in frequency
Fbus=60.5-deltaF;
# (b) Load carried by each machine
deltaPa=(deltaF*Prateda)/(GSR*Frated); # Change in power for machine A
deltaPb=Pch-deltaPa; # Change in power for machine B
Pa=Pchmach+deltaPa;
Pb=Pchmach+deltaPb;
# Display result on command window
print"\nOperating frequency =",Fbus,"Hz"
print"\nLoad carried by machine A =",Pa,"kW"
print"\nLoad carried by machine B =",Pb,"kW"
# Example 9.5
# Determine (a) Bus frequency (b) Load on each machine
# Page 360
# Given data
Padd=720; # Additional load connected
GD=0.0008; # Governor droop
f=60.2; # Frequency of machine
Pbus=900; # Bus load
# (a) Bus frequency
deltaPa=Padd/2;
deltaPb=deltaPa; # Since both machines have identical governor drops
deltaF=GD*deltaPa; # Change in frequency
Fbus=f-deltaF;
# (b) Load on each machine
Pa=(2/3)*Pbus+deltaPa; # Load on machine A
Pb=(1/3)*Pbus+deltaPb; # Load on machine B
# Display result on command window
print"\n Bus frequency =",Fbus,"Hz"
print"\n Load on machine A =",Pa,"kW"
print"\n Load on machine B =",Pb,"kW"
# Example 9.6
# Determine (a) System kilowatts (b) System frequency (c) kilowatt loads
# carried by each machine
# Page 361
# Given data
Pres=440.; # Resistive load
PF=0.8; # Power factor
Pind=200.; # Induction motor power
Palt=210.; # Alternator bus load
deltaPa=70.; # Change in load for machine A
f=60.; # Frequency
deltaPb=70.; # Change in load for machine B
deltaPc=70.; # Change in load for machine C
# (a) System kilowatts
deltaPbus=Pres+PF*Pind; # Increase in bus load
Psys=Palt+deltaPbus;
# (b) System frequency
GDa=(60.2-f)/deltaPa; # Governor droop for machine A
GDb=(60.4-f)/deltaPb; # Governor droop for machine B
GDc=(60.6-f)/deltaPc; # Governor droop for machine C
# From the figure 9.18(b)
deltaF=600./(350.+175.+116.6667) ;
f2=f-deltaF;
# (c) Kilowatt loads carried by each machine
Pa2=deltaPa+350.*deltaF;
Pb2=deltaPb+175.*deltaF;
Pc2=deltaPc+116.6667*deltaF;
# Display result on command window
print"\nSystem kilowatts =",Psys,"kW"
print"\nSystem frequency =",f2,"Hz"
print"\nKilowatt loads carried by machine A =",Pa2,"kW"
print"\nKilowatt loads carried by machine B =",Pb2,"kW"
print"\nKilowatt loads carried by machine C =",Pc2,"kW"
# Example 9.7
# Determine (a) Active and reactive components of the bus load (b) If the
# power factor of generator A is 0.94 lagging, determine the reactive power
# supplied by each machine.
# Page 366
# Given data
Pbuspower=500.; # Power supplied
Pind=200.; # Induction motor power
PF=0.852; # Percent power factor
NA=2.; # Number of alternators
LPF=0.94; # Lagging power factor
# (a) Active and reactive components of the bus load
Pbus=Pbuspower+Pind*PF; # Active component of the bus load
ThetaMot=31.6;#acosd(PF); # Power angle of motor
Qbus=105.#Pind*sind(ThetaMot); # Reactive component the bus load
# (b) Reactive power supplied by each machine
Pa=Pbus/NA; # Alternator A power
ThetaA=19.9;#acosd(LPF); # Alternator A angle
Qa=122.;#tand(ThetaA)*Pa; # Reactive power supplied by machine A
Qb=Qbus-Qa; # Reactive power supplied by machine B
# Display result on command window
print"\nActive component of the bus load =",Pbus,"kW"
print"\nReactive component of the bus load =",Qbus,"kvar"
print"\nReactive power supplied by machine A =",Qa,"kvar"
print"\nReactive power supplied by machine B =",Qb,"kvar"
# Example 9.8
# Computation of per-unit impedance of a generator
# Page 368
# Given data
from math import sqrt,pi
P=100000.; # Power of synchronous generator
V=480.; # Voltage of synchronous generator
Ra=0.0800; # Resistive component
Xs=2.3; # Reactive component
# Computation of per-unit impedance of a generator
Sbase=P/3.; # Rated apparent power per phase
Vbase=V/sqrt(3.); # Rated voltage per phase
Zbase=Vbase**2./Sbase; # Rated impedance
Rpu=Ra/Zbase; # Per unit resistance
Xpu=Xs/Zbase; # Per unit reactance
Zpu=0.0347 + 0.998j;#Rpu+1j*Xpu; # Per unit impedance
# Complex to Polar form...
Zpu_Mag=0.999;#sqrt(real(Zpu)**2+imag(Zpu)**2); # Magnitude part
Zpu_Ang =88.;# atan(imag(Zpu),real(Zpu))*180/pi; # Angle part
# Display result on command window
print"\nPer-unit impedance magnitude =",Zpu_Mag,"Ohm"
print"\nPer-unit impedance angle =",Zpu_Ang,"deg\n"
# Example 9.9
# Determine (a) Excitation voltage (b) Power angle (c) No load voltage,
# assuming the field current is not changed (d) Voltage regulation (e) No load
# voltage if the field current is reduced to 80% of its value at rated load.
# Page 369
# Given data
from math import sqrt,pi,sin,cos
V=4800.; # Voltage of synchronous generator
PF=0.900; # Lagging power factor
S_Mag=1000000./3.;
Xa_Mag=13.80; # Synchronous reactance
Xa_Ang=90.;
Vt_Ang=0;
# (a) Excitation voltage
Vt=V/sqrt(3);
Theta=25.8;#acosd(PF); # Angle
Ia_Magstar=S_Mag/Vt; # Magnitude of curent
Ia_Angstar=Theta-0; # Angle of current
Ia_Mag=Ia_Magstar;
Ia_Ang=-Ia_Angstar;
# Ef=Vt+Ia*j*Xa
# First compute Ia*Xa
IaXa_Mag=Ia_Mag*Xa_Mag;
IaXa_Ang=Ia_Ang+Xa_Ang;
# Polar to Complex form for IaXa
IaXa_R=IaXa_Mag*cos(-IaXa_Ang*pi/180); # Real part of complex number
IaXa_I=IaXa_Mag*sin(IaXa_Ang*pi/180); # Imaginary part of complex number
# Vt term in polar form
Vt_Mag=Vt;
Vt_Ang=Vt_Ang;
# Polar to Complex form for Vt
Vt_R=Vt_Mag*cos(-Vt_Ang*pi/180); # Real part of complex number
Vt_I=Vt_Mag*sin(Vt_Ang*pi/180); # Imaginary part of complex number
# Ef in complex form
Ef_R=IaXa_R+Vt_R;
Ef_I=IaXa_I+Vt_I;
Ef=3.49e+03 + 1.49e+03j;#Ef_R+%i*Ef_I;
# Complex to Polar form for Ef
Ef_Mag=3.8e+03;#sqrt(real(Ef)**2+imag(Ef)**2); # Magnitude part
Ef_Ang=23.1;# atan(imag(Ef),real(Ef))*180/%pi; # Angle part
# (b) Power angle
PA=Ef_Ang;
# (c) No load voltage, assuming the field current is not changed
# From figure 9.23 (b)
VolAxis=Vt_Mag/30; # The scale at the given voltage axis
Ef_loc=Ef_Mag/VolAxis; # Location of Ef voltage
Vnl=33.4*VolAxis; # No load voltage
# (d) Voltage regulation
VR=(Vnl-Vt)/Vt*100;
# (e) No load voltage if the field current is reduced to 80%
Vnlnew=31*VolAxis;
# Display result on command window
print"\nExcitation voltage =",Ef_Mag,"V"
print"\nPower angle =",PA,"deg"
print"\nNo load voltage =",Vnl,"V"
print"\nVoltage regulation =",VR,"Percent"
print"\nNo load voltage when field current is reduced to 80 percent =",Vnlnew,"V "
# Example 9.10
# Repeat the example 9.9 assuming 90 % leading power factor
# Determine (a) Excitation voltage (b) Power angle (c) No load voltage,
# assuming the field current is not changed (d) Voltage regulation (e) No load
# voltage if the field current is reduced to 80% of its value at rated load.
# Page 372
# Given data
from math import sqrt,pi,sin,cos
V=4800.; # Voltage of synchronous generator
PF=0.900; # Lagging power factor
S_Mag=1000000./3.;
Xa_Mag=13.80; # Synchronous reactance
Xa_Ang=90.;
Vt_Ang=0;
# (a) Excitation voltage
Vt=V/sqrt(3.);
Theta=25.8;#acosd(PF); # Angle
Ia_Magstar=S_Mag/Vt; # Magnitude of curent
Ia_Angstar=Theta-0; # Angle of current
Ia_Mag=Ia_Magstar;
Ia_Ang=Ia_Angstar;
# Ef=Vt+Ia*j*Xa
# First compute Ia*Xa
IaXa_Mag=Ia_Mag*Xa_Mag;
IaXa_Ang=Ia_Ang+Xa_Ang;
# Polar to Complex form for IaXa
IaXa_R=IaXa_Mag*cos(-IaXa_Ang*pi/180); # Real part of complex number
IaXa_I=IaXa_Mag*sin(IaXa_Ang*pi/180); # Imaginary part of complex number
# Vt term in polar form
Vt_Mag=Vt;
Vt_Ang=Vt_Ang;
# Polar to Complex form for Vt
Vt_R=Vt_Mag*cos(-Vt_Ang*pi/180); # Real part of complex number
Vt_I=Vt_Mag*sin(Vt_Ang*pi/180); # Imaginary part of complex number
# Ef in complex form
Ef_R=IaXa_R+Vt_R;
Ef_I=IaXa_I+Vt_I;
Ef=2.05e+03 + 1.49e+03j;#Ef_R+1j*Ef_I;
# Complex to Polar form for Ef
Ef_Mag=2.53e+03;#sqrt(real(Ef)**2+imag(Ef)**2); # Magnitude part
Ef_Ang=36.1;#atan(imag(Ef),real(Ef))*180/%pi; # Angle part
# (b) Power angle
PA=Ef_Ang;
# (c) No load voltage, assuming the field current is not changed
# From figure 9.23 (b)
VolAxis=Vt_Mag/30.; # The scale at the given voltage axis
Ef_loc=Ef_Mag/VolAxis; # Location of Ef voltage
Vnl=29.*VolAxis; # No load voltage
# (d) Voltage regulation
VR=(Vnl-Vt)/Vt*100.;
# Display result on command window
print"\nExcitation voltage =",Ef_Mag,"V"
print"\nPower angle =",PA,"deg"
print"\nNo load voltage =",Vnl,"V"
print"\nVoltage regulation =",VR,"Percent"
print'The leading power factor resulted in a negativr voltage regulation'
# Example 9.11
# Determine (a) Equivalent armature resistance (b) Synchronous reactance
# (c) Short-circuit ratio
# Page 377
# Given data
from math import sqrt,pi
Vdc=10.35; # DC voltage
Idc=52.80; # DC current
VOCph=240./sqrt(3.); # Open-circuit phase voltage
ISCph=115.65; # Short-circuit phase current
P=50000.;
V=240.; # Supply voltage
# (a) Equivalent armature resistance
Rdc=Vdc/Idc; # DC resistance
Rgamma=Rdc/2.;
Ra=1.2*Rgamma; # Armature resistance
# (b) Synchronous reactance
Zs= VOCph/ISCph; # Synchronous impedance/phase
Xs=sqrt(Zs**2-Ra**2.);
# (c) Short-circuit ratio
Sbase=P/3; # Power/phase
Vbase=V/sqrt(3.); # Voltage/phase
Zbase=Vbase**2./Sbase;
Xpu=Xs/Zbase; # Per unit synchronous reactance
SCR=1./Xpu; # Short-circuit ratio
# Display result on command window
print"\nEquivalent armature resistance =",Ra,"Ohm"
print"\nSynchronous reactance =",Xs,"Ohm"
print"\nShort-circuit ratio =",SCR