# Example 6.1
# Determine (a) Locked rotor current in each winding (b) Phase displacement
# angle between the two currents (c) Locked rotor torque in terms of the
# machine constant (d) External resistance required in series with the auxillary
# winding in order to obtain a 30 degree phase displacement between the currents
# in the two windings (e) Locked rotor torque for the conditions in (d)
# (f) Percent increase in locked rotor torque due to the addition of external
# resistance
# Page No. 257
# Given data
Zmw=2.00+1j*3.50 # Main winding impedance
Zaw=9.15+1j*8.40 # Auxillary winding impedance
VT=120.; # Transformer voltage
Xaw=8.40; # Auxillary winding reactance
Raw=9.15; # Auxillary winding resistance
# (a) Locked rotor current in each winding
# Main winding impedance in polar form
# Complex to Polar form...
Zmw_Mag=4.03;#sqrt(real(Zmw)**2+imag(Zmw)**2); # Magnitude part
Zmw_Ang=60.3;#atan(imag(Zmw),real(Zmw))*180/%pi; # Angle part
# Auxillary winding impedance in polar form
# Complex to Polar form...
Zaw_Mag=12.4;#sqrt(real(Zaw)**2+imag(Zaw)**2); # Magnitude part
Zaw_Ang=42.6;#atan(imag(Zaw),real(Zaw))*180/%pi; # Angle part
# Main winding current
Imw_Mag=29.8;#VT/Zmw_Mag; # Main winding current magnitude
Imw_Ang=-60.3;#0-Zmw_Ang; # Main winding current angle
# Auxillary winding current
Iaw_Mag=9.66;#VT/Zaw_Mag; # Auxillary winding current magnitude
Iaw_Ang=-42.6;#0-Zaw_Ang; # Auxillary winding current angle
# (b) Phase displacement angle between the two currents
Alpha=17.7;#abs(Imw_Ang-Iaw_Ang);
# (c) Locked rotor torque in terms of the machine constant
Tlr=87.4;#Imw_Mag*Iaw_Mag*sind(Alpha);
# (d) External resistance required in seris with the auxillary winding in
# order to obtain a 30 degree phase displacement between the currents in the
# two windings
Theta_awi=-30.3;#Imw_Ang+30; # Required phase angle
Theta_awz=30.3;#-Theta_awi;
Rx=5.25;#(Xaw/tand(Theta_awz))-Raw;
# (e) Locked rotor torque for the conditions in (d)
Zawnew=14.4 + 8.4j;#Raw+Rx+1j*Xaw; # Auxillary winding impedance
# Complex to Polar form...
Zmwnew_Mag=16.7;#sqrt(real(Zawnew)**2+imag(Zawnew)**2); # Magnitude part
Zmwnew_Ang=30.3;#atan(imag(Zawnew),real(Zawnew))*180/%pi; # Angle part
Iawnew_Mag=7.2;#VT/Zmwnew_Mag; # Auxillary winding current magnitude
Iawnew_Ang=-30.3;#0-Zmwnew_Ang; # Auxillary winding current magnitude
Tlenew=22.5;#107;#Imw_Mag*Iawnew_Mag*sind(30);
# (f) Percent increase in locked rotor torque due to the addition of external
# resistance
PI=(Tlenew-Tlr)/Tlr*100.;
# Display result on command window
print"\n Main winding current magnitude =",Imw_Mag,"A"
print"\n Main winding current angle =",Imw_Ang,"deg"
print"\n Auxillary winding current magnitude =",Iaw_Mag,"A"
print"\n Auxillary winding current angle =",Iaw_Ang,"deg"
print"\n Phase displacement angle =",Alpha,"deg"
print"\n Locked rotor torque in terms of the machine constant =",Tlr,".Ksp"
print"\n External resistance required =",Rx,"Ohm"
print"\n Locked rotor torque =",Tlenew,".Ksp"
print"\n Percent increase in locked rotor torque =",PI,"Percent increase"
# Example 6.2
# Determine (a) Capacitance required in series with the auxillary winding
# in order to obtain a 90 degree phase displacement between the current in
# the main winding and the current in the auxillary winding at locked rotor
# (b) Locked rotor torque in terms of the machine constant
# Page No. 265
# Given data
from math import sqrt,pi
Zmw=2.00+1j*3.50 # Main winding impedance
Zaw=9.15+1j*8.40 # Auxillary winding impedance
VT=120.; # Transformer voltage
Xaw=8.40; # Auxillary winding reactance
Raw=9.15; # Auxillary winding resistance
f=60.; # Frequency
Tlr=107.1; # Original torque
# (a) Capacitance required in series with the auxillary winding
# Main winding impedance in polar form
# Complex to Polar form...
Zmw_Mag=4.03;#sqrt(real(Zmw)**2.+imag(Zmw)**2.); # Magnitude part
Zmw_Ang=60.3;#atan(imag(Zmw),real(Zmw))*180./pi; # Angle part
# Auxillary winding impedance in polar form
# Complex to Polar form...
Zaw_Mag=12.4;#sqrt(real(Zaw)**2.+imag(Zaw)**2.); # Magnitude part
Zaw_Ang=42.6;#atan(imag(Zaw),real(Zaw))*180/pi; # Angle part
# Main winding current
Imw_Mag=29.8;#VT/Zmw_Mag; # Main winding current magnitude
Imw_Ang=-60.3;#0-Zmw_Ang; # Main winding current angle
# Auxillary winding current
Iaw_Mag=9.66;#VT/Zaw_Mag; # Auxillary winding current magnitude
Iaw_Ang=-42.6;#0-Zaw_Ang; # Auxillary winding current angle
Theta_awi=90-60.26; # Required phase angle
Theta_awz=-Theta_awi;
Xc=13.6;#Xaw-Raw*tand(Theta_awz); # Capacitive reactance
C=1./2.*pi*f*Xc; # Required capacitance
# (b) Locked rotor torque in terms of the machine constant
Zawnew=9.15 + -5.23j;#Raw+1j*Xaw-1j*Xc; # Auxillary winding impedance
# Complex to Polar form...
Zawnew_Mag=10.5;#sqrt(real(Zawnew)**2+imag(Zawnew)**2); # Magnitude part
Zawnew_Ang=-29.7;#atan(imag(Zawnew),real(Zawnew))*180/%pi; # Angle part
Iawnew_Mag=11.4;#VT/Zawnew_Mag; # Auxillary winding current magnitude
Iawnew_Ang=29.7;#0-Zawnew_Ang; # Auxillary winding current magnitude
Tlenew=339.;#Imw_Mag*Iawnew_Mag*sind(90);
# Percent change increase in locked rotor torque
PI=(Tlenew-Tlr)/Tlr*100;
# Display result on command window
print"\n Required capacitance =",C,"microF"
print"\n Percent increase in locked rotor torque =",PI,"Percent"
#Note: Capacitor computation is wrong in the book
# Example 6.3
# Determine (a) NEMA standard horsepower rating of machine (b) Required
# running capacitance (c) Additional capacitance required for starting
# Page No. 271
# Given data
hp=35.; # Power in hp
p=3.; # Number of phase
f=60.; # Frequency
# (a) NEMA standard horsepower rating of machine
Prated3ph=hp*p/2.;
# (b)Required running capacitance
C1=26.5*f;
# (c) Additional capacitance required for starting.
C2=230.*f-C1;
# Display result on command window
print"\n NEMA standard horsepower rating of machine =",Prated3ph,"hp"
print"\n Required running capacitance =",C1,"microF"
print"\n Additional capacitance required for starting =",C2,"microF"
# Example 6.4
# Computation of (a) Motor line current and motor phase current (b) Motor line
# current and motor phase current if one line opens (c) Line and phase
# currents if the power factor when single phasing is 82.0 percent.
# Page No. 274
# Given data
from math import sqrt,pi
Vline=2300.; # Line voltage
Fp3ph=3.; # Frequency of three phase
PF=0.844; # Power factor
PF1=0.820; # 82.2 percent power factor
Pin=350.*746./(0.936*2); # Input power
# (a) Motor line current and motor phase current
Iline3ph=Pin/(sqrt(3)*Vline*PF);
Iphase3ph=Iline3ph;
#(b) Motor line current and motor phase current if one line opens
Iline1ph=(sqrt(3)*Iline3ph*PF)/PF;
Iphase1ph=Iline1ph;
# (c) Line and phase currents if the power factoe when single phasing is 82.0 percent.
Iline=(Iline1ph*PF)/PF1;
Iphase=Iline;
# Display result on command window
print"\n Motor line current =",Iline3ph,"A"
print"\n Motor phase current =",Iphase3ph,"A"
print"\n Motor line current if one line opens =",Iline1ph,"A"
print"\n Motor phase current if one line opens =",Iphase1ph,"A"
print"\n Line current if the power factor is 82.0 percent =",Iline,"A"
print"\n Phase current if the power factor is 82.0 percent =",Iphase,"A"