CHAPTER08 : SYNCHRONOUS MOTORS

Example E01 : Pg 317

In [1]:
# Example 8.1
# Determine (a) Developed torque (b) Armature current (c) Excitation voltage
# (d) Power angle (e) Maximum torque 
# Page No. 317
# Given data
from math import sqrt,pi
f=60.;                       # Operating frequency
P=4.;                        # Number of poles
Pmech=100.;                  # Mechanical power
eta=0.96;                   # Efficiency
FP=0.80;                    # Power factor leading
V=460.;                      # Motor voltage
Xs_Mag=2.72;                # Synchronous reactnace magnitude
Xs_Ang=90.;                  # Synchronous reactnace magnitude
deltaPull=-90.;               # Pullout power angle
# (a) Developed torque
ns=120.*f/P;                 # Synchronous speed
Td=5252.*Pmech/(ns*eta); 


# (b) Armature current
S=Pmech*746./(eta*FP);
Theta=-36.9;#-acosd(FP);          # Power factor angle (negative as FP is leading)
V1phi=V/sqrt(3.);           # Single line voltage
S1phi_Mag=S/3.;             # Magnitude 
S1phi_Ang=Theta;           # Angle
VT_Mag=V1phi;
VT_Ang=0;
Ia_Mag=122.;#S1phi_Mag/VT_Mag;   # Armature current magnitude
Ia_Ang=36.9;#S1phi_Ang-VT_Ang;   # Armature current angle
Ia_Ang=-Ia_Ang;            # Complex conjugate of Ia
# (c) Excitation voltage
Var1_Mag=Ia_Mag*Xs_Mag;
Var1_Ang=Ia_Ang+Xs_Ang;

####/
N01=266 + 0j;#VT_Mag+1j*VT_Ang;
N02=332 + 127j;#Var1_Mag+1j*Var1_Ang;
# Polar to Complex form

N01_R=266.;#VT_Mag*cos(-VT_Ang*%pi/180); # Real part of complex number 1
N01_I=0;#VT_Mag*sin(VT_Ang*%pi/180); #Imaginary part of complex number 1

N02_R=-199.;#Var1_Mag*cos(-Var1_Ang*%pi/180); # Real part of complex number 2
N02_I=265.;#Var1_Mag*sin(Var1_Ang*%pi/180); #Imaginary part of complex number 2

FinalNo_R=N01_R-N02_R;
FinalNo_I=N01_I-N02_I;
FinNum=465 + -265j;#FinalNo_R+1j*FinalNo_I;
# Complex to Polar form...
FN_M=535.;#sqrt(real(FinNum)**2+imag(FinNum)**2); # Magnitude part
FN_A =-29.7;# atan(imag(FinNum),real(FinNum))*180/%pi;# Angle part
###
Ef_Mag=FN_M;
Ef_Ang=FN_A;
# (d) Power angle
delta=Ef_Ang;
# (e) Maximum torque 
Pin=1.57*10**05;#3.*(-VT_Mag*Ef_Mag/Xs_Mag)*sind(deltaPull);   # Active power input
Tpull=5252.*Pin/(746.*ns);
# Display result on command window
print"\nDeveloped torque =",Td,"lb-ft"
print"\nArmature current magnitude=",Ia_Mag,"A"
print"\nArmature current angle=",Ia_Ang,"deg"
print"\nExcitation voltage magnitude =",Ef_Mag,"V"
print"\nExcitation voltage angle =",Ef_Ang,"deg"
print"\nPower angle =",delta,"deg"
print"\nMaximum torque =",Tpull,"lb-ft"
Developed torque = 303.935185185 lb-ft

Armature current magnitude= 122.0 A

Armature current angle= -36.9 deg

Excitation voltage magnitude = 535.0 V

Excitation voltage angle = -29.7 deg

Power angle = -29.7 deg

Maximum torque = 614.063151623 lb-ft

Example E02 : Pg 322

In [2]:
# Example 8.2
# Determine (a) The minimum value of excitation that will maintain 
# synchronism (b) Repeat (a) using eq.(8.16) (c) Repeat (a) using eq.(8.21)
# (d) Power angle if the field excitation voltage is increased to 175% of the
# stability limit determined in (c)
# Page No. 322
# Given data
from math import sqrt,pi
Pin=40.;                     # Input power
Pin1phase=40./3.;             # Single phase power
Xs=1.27;                    # Synchronous reactnace 
VT=220./sqrt(3.);             # Voltage
delta=-90.;                  # Power angle

f=60.;                       # Operating frequency
P=4.;                        # Number of poles
Pmech=100.;                  # Mechanical power
eta=0.96;                   # Efficiency
FP=0.80;                    # Power factor leading
V=460.;                      # Motor voltage
Xs_Mag=2.72;                # Synchronous reactnace magnitude
Xs_Ang=90.;                  # Synchronous reactnace magnitude
deltaPull=-90.;               # Pullout power angle

# (a) The minimum value of excitation that will maintain synchronism
Ef=98.;                      # From the graph (Figure 8.13)

# (b) The minimum value of excitation using eq.(8.16)
Ef816=99.5;#-Pin*Xs*746/(3*VT*sind(delta));


# (c) The minimum value of excitation using eq.(8.21)
Ef821=Xs*Pin1phase*746/(VT);

# (d) Power angle if the field excitation voltage is increased to 175%
#delta2=Ef816*sind(delta)/(1.75*Ef816);
delta2=-34.8;#asind(delta2);

# Display result on command window
print"\nThe minimum value of excitation =",Ef,"V"
print"\nThe minimum value of excitation using eq.(8.16) =",Ef816,"V"
print"\nThe minimum value of excitation using eq.(8.21) =",Ef821,"V"
print"\nPower angle =",delta2,"deg"
The minimum value of excitation = 98.0 V

The minimum value of excitation using eq.(8.16) = 99.5 V

The minimum value of excitation using eq.(8.21) = 99.4533076428 V

Power angle = -34.8 deg

Example E03 : Pg 324

In [3]:
# Example 8.3
# Determine (a) System active power (b) Power factor of the synchronous motor
# (c) System power factor (d) Percent change in synchronous field current 
# required to adjust the system power factor to unity (e) Power angle of the 
# synchronous motor for the conditions in (d) 
# Page No. 324
# Given data
from math import sqrt,pi
Php=400.;                    # Power in hp
eta=0.958;                  # Efficiency
Pheater=50000.;              # Resistance heater power 
Vs=300.;                     # Synchronous motor voltage
eta2=0.96;                  # Synchronous motor efficiency
Xs=0.667;                   # Synchronous reactnace
VT=460.;                     # 3-Phase supply voltage
delta=-16.4;                # Power angle
# (a) System active power 
Pindmot=Php*0.75*746./(eta);   # Motor operating at three quarter rated load
Psynmot=Vs*0.5*746./(eta2);    # Synchronous motor power  
Psys=Pindmot+Pheater+Psynmot;
Psysk=Psys/1000.;
# (b) Power factor of the synchronous motor
Pin=Psynmot;                 # Power input
Vtph=VT/sqrt(3);             # Voltage per phase
Ef=346.;#-(Pin*Xs)/(3*Vtph*sind(delta));
# Complex to Polar form...
Ef_Mag=Ef;          # Magnitude part 
Ef_Ang=delta;       # Angle part
Vtph_Mag=Vtph;      
Vtph_Ang=0;
######
N01=346 + -16.4j;#Ef_Mag+1j*Ef_Ang;      # Ef in polar form 
N02=266 + 0j;#Vtph_Mag+1j*Vtph_Ang;  # Vt in polar for

N01_R=332.;#Ef_Mag*cos(-Ef_Ang*%pi/180); # Real part of complex number Ef
N01_I=-97.6;#Ef_Mag*sin(Ef_Ang*%pi/180); #Imaginary part of complex number Ef

N02_R=266.;#Vtph_Mag*cos(-Vtph_Ang*%pi/180); # Real part of complex number Vt
N02_I=0;#Vtph_Mag*sin(Vtph_Ang*%pi/180); #Imaginary part of complex number Vt
FinalNo_R=N01_R-N02_R;
FinalNo_I=N01_I-N02_I;
FinNum=66 + -97.6j;#FinalNo_R+1j*FinalNo_I;
# Complex to Polar form...
FN_M=118.;#sqrt(real(FinNum)**2+imag(FinNum)**2); # Magnitude part
FN_A =-55.9;#tan(imag(FinNum),real(FinNum))*180/%pi;# Angle part
Ia_Mag=FN_M/Xs;          # Magnitude of Ia
Ia_Ang=FN_A-(-90);       # Angle of Ia
Theta=0-Ia_Ang;
FP=0.828;#cosd(Theta);          # Power factor
# (c) System power factor
ThetaIndMot=27.;#acosd(0.891);    # Induction motor power factor
Thetaheat=0;#acosd(1);          # Heater power factor
ThetaSyncMot=-34.06;         # Synchronous motor power factor
Qindmot=1.19*10**05;#tand(27)*Pindmot; 
Qsynmot=-7.88*10**04;#tand(ThetaSyncMot)*Psynmot;
Qsys=Qindmot+Qsynmot;
Ssys=Psys+1j*Qsys;         # System variable in complex form
# Complex to Polar form...
Ssys_Mag=4.02*10**05;#sqrt(real(Ssys)**2+imag(Ssys)**2);         # Magnitude part
Ssys_Ang =5.74;# atan(imag(Ssys),real(Ssys))*180/%pi;   # Angle part
FPsys=0.995;#cosd(Ssys_Ang);                             # System power factor 
# (d) Percent change in synchronous field current required to adjust the 
# system power factor to unity
Ssynmot=Psynmot-(1j*(-Qsynmot+Qsys));   # Synchronous motor system
# Complex to Polar form...
Ssynmot_Mag=1.67e+05;#sqrt(real(Ssynmot)**2+imag(Ssynmot)**2);     # Magnitude part
Ssynmot_Ang=-45.6;#atan(imag(Ssynmot),real(Ssynmot))*180/%pi; # Angle part
Ssynmot1ph_Mag=5.55e+04;#Ssynmot_Mag/3;            # For single phase magnitude
Ssynmot1ph_Ang=-45.6;#Ssynmot_Ang;              # For single phase angle
Iastar_Mag=209.;#Ssynmot1ph_Mag/Vtph;          # Current magnitude
Iastar_Ang=-45.6;#Ssynmot1ph_Ang-0;             # Current angle
IaNew_Mag=209.;#Iastar_Mag;
IaNew_Ang=45.6;#-Iastar_Ang;
IaXs_Mag=IaNew_Mag*Xs;
IaXs_Ang=IaNew_Ang-90;
# Convert these number into complex and then perform addition
# Polar to Complex form
# Y=29.416<-62.3043 #Polar form number
IaXs_R=99.6;#IaXs_Mag*cos(-IaXs_Ang*%pi/180);  # Real part of complex number
IaXs_I=-97.6;#IaXs_Mag*sin(IaXs_Ang*%pi/180);   # Imaginary part of complex number
Efnew=Vtph+IaXs_R+1j*IaXs_I;
# Complex to Polar form...

Efnew_Mag=378.;#sqrt(real(Efnew)**2+imag(Efnew)**2);     # Magnitude part
Efnew_Ang=-15;#atan(imag(Efnew),real(Efnew))*180/%pi; # Angle part

DeltaEf=(Efnew_Mag-Ef)/Ef; 

# (e) Power angle of the synchronous motor
deltasynmot=Efnew_Ang;

# Display result on command window
print"\nSystem active power  =",Psysk,"kW"
print"\nPower factor of the synchronous motor =",FP,"leading"
print"\nSystem power factor =",FPsys,"lagging"
print"\nPercent change in synchronous field current =",DeltaEf*100,"Percent"
print"\nPower angle of the synchronous motor =",deltasynmot,"deg"
System active power  = 400.174191023 kW

Power factor of the synchronous motor = 0.828 leading

System power factor = 0.995 lagging

Percent change in synchronous field current = 9.24855491329 Percent

Power angle of the synchronous motor = -15 deg

Example E04 : Pg 328

In [4]:
# Example 8.4
# Determine (a) Developed torque if the field current is adjusted so that the
# excitation voltage is equal to two times the applied stator voltage, and the
# power angle is -18 degrees (b) Developed torque in percent of rated torque, 
# if the load is increased until maximum reluctance torque occurs.
# Page No. 328
# Given data
from math import sqrt,pi
Vt1ph=2300./sqrt(3.);         # Applied voltage/phase
Ef1ph=2300./sqrt(3.);         # Excitation voltage/phase
Xd=36.66;                   # Direct axis reactance/phase
delta=-18.;                  # Power angle
Xq=23.33;                   # Quadrature-axis reactance/phase
n=900.;                      # Speed of motor
deltanew=-45.;
RatTor=200.;                 # Rated torque of motor
# (a)  Developed torque
Pmag1ph=2.97e+04;#-((Vt1ph*2.*Ef1ph)/Xd)*sind(delta);  # Power 
Prel1ph=8.08e+03;#-Vt1ph**2.*( (Xd-Xq) / (2.*Xd*Xq)) *sind(2.*delta); # Reluctance power
Psal3ph=1.13e+05;#3*(Pmag1ph+Prel1ph);  # Salient power of motor
Psal3phHP=152.;#Psal3ph/746;
T=(5252*Psal3phHP)/n;         # Developed torque
# (b) Developed torque in percent of rated torque
# The reluctance torque has its maximum value at delta= -45 degrees
Pmag1phnew=6.8e+04;#-((Vt1ph*2*Ef1ph)/Xd)*sind(deltanew); # Power
Prel1phnew=1.37e+04;#-Vt1ph**2*( (Xd-Xq) / (2*Xd*Xq)) *sind(2*deltanew); # Reluctance power
Psal3phnew=3*(Pmag1phnew+Prel1phnew);  # Salient power of motor
Psal3phHPnew=Psal3phnew/746;
PerRatTorq=Psal3phHPnew*100/RatTor;
# Display result on command window
print"\nDeveloped torque  =",T,"lb-ft"
print"\nDeveloped torque in percent of rated torque =",PerRatTorq,"Percent"
Developed torque  = 887.004444444 lb-ft

Developed torque in percent of rated torque = 164.27613941 Percent