CHAPTER06 : SINGLE PHASE INDUCTION MOTORS

Example E01 : Pg 257

In [1]:
# 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"
 Main winding current magnitude = 29.8 A

 Main winding current angle = -60.3 deg

 Auxillary winding current magnitude = 9.66 A

 Auxillary winding current angle = -42.6 deg

 Phase displacement angle = 17.7 deg

 Locked rotor torque in terms of the machine constant = 87.4 .Ksp

 External resistance required = 5.25 Ohm

 Locked rotor torque = 22.5 .Ksp

 Percent increase in locked rotor torque = -74.2562929062 Percent increase

Example E02 : Pg 265

In [2]:
# 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
 Required capacitance = 1281.76980266 microF

 Percent increase in locked rotor torque = 216.526610644 Percent

Example E03 : Pg 271

In [3]:
# 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"
 NEMA standard horsepower rating of machine = 52.5 hp

 Required running capacitance = 1590.0 microF

 Additional capacitance required for starting = 12210.0 microF

Example E04 : Pg 274

In [4]:
# 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"
 Motor line current = 41.4829962669 A

 Motor  phase current = 41.4829962669 A

 Motor line current if one line opens = 71.8506571845 A

 Motor phase current if one line opens = 71.8506571845 A

 Line  current if the power factor is 82.0 percent = 73.9536032484 A

 Phase current if the power factor is 82.0 percent = 73.9536032484 A