CHAPTER04 : PRINCIPLES OF THREE PHASE INDUCTION MOTORS

Example E01 : Pg 140

In [1]:
#  Example 4.1
#  Computation of synchronous speed of a six pole induction motor
#  Page No. 140
#  Given data
f=60.;                #  Frequency
p=6.;                 #  Number of poles
fs=f*0.85;           #  Frequency is 85% of its rated value
ns=120.*fs/p;         #  Synchronous speed

#  Display result on command window
print"Synchronous speed of a six pole induction motor =",ns,"r/min"
Synchronous speed of a six pole induction motor = 1020.0 r/min

Example E02 : Pg 143

In [2]:
#  Example 4.2
#  Computation of (a) Frequency (b) Induced voltage of six pole induction motor
#  Page No. 143
#  Given data
f=60.;                #  Frequency
p=6.;                 #  Number of poles
nr=1100.;             #  Rotor speed
Ebr=100.;             #  Blocked rotor voltage

#  (a) Synchronous speed
ns=120.*f/p;              #  Synchronous speed

#  (b) Slip
s=(ns-nr)/ns;              #  Slip

#  (c) Rotor frequency
fr=s*f;              #  Rotor frequency

#  (d) Rotor voltage
Er=s*Ebr;              #  Rotor voltage


#  Display result on command window
print"Synchronous speed =",ns,"r/min"
print"Slip =",s
print"Rotor frequency =",fr,"Hz"
print"Rotor voltage =",Er,"V"
Synchronous speed = 1200.0 r/min
Slip = 0.0833333333333
Rotor frequency = 5.0 Hz
Rotor voltage = 8.33333333333 V

Example E03 : Pg 146

In [3]:
#  Example 4.3
#  Determine (a) Synchronous speed (b) Slip (c) Rotor impedance (d) Rotor current
#  (e) Rotor current if changing the shaft load resulted in 1.24 percenr slip 
#  (f) Speed for the condition in (e) 
#  Page No. 146
#  Given data
fs=60.;                  #  Frequency
p=6.;                    #  Number of poles
nr=1164.;                #  Rotor speed
Rr=0.10;                #  Equivalent resistance
Xbr=0.54;               #  Equivalent reactance
Ebr=150.;                #  Blocked rotor voltage per phase
s1=0.0124;              #  Percent slip

#  (a) Synchronous speed
ns=120.*fs/p;               #  Speed 

#  (b) Slip
s=(ns-nr)/ns; 

#  (c) Rotor impedance  
Zr=3.33+0.54j;#(Rr/s)+%i*Xbr;
#  Complex to Polar form...
Zr_Mag=3.38;#sqrt(real(Zr)**2+imag(Zr)**2);         #  Magnitude part
Zr_Ang=9.2;#atan(imag(Zr),real(Zr))*180/%pi;     #  Angle part

#  (d) Rotor current
Ir_Mag=Ebr/Zr_Mag;                          #  Magnitude
Ir_Ang=0-Zr_Ang;                            #  Angle

#  (e) Rotor current if changing the shaft load resulted in 1.24 percent slip 
Zrnew=8.06+0.54j;#Rr/s1+%i*Xbr;
#  Complex to Polar form...
Zrnew_Mag=8.08;#sqrt(real(Zrnew)**2+imag(Zrnew)**2);         #  Magnitude part
Zrnew_Ang=3.83;#atan(imag(Zrnew),real(Zrnew))*180/%pi;     #  Angle part

Irnew_Mag=Ebr/Zrnew_Mag;                             #  Magnitude
Irnew_Ang=0-Zrnew_Ang;                               #  Angle

#  (f) Speed for the condition in (e) 
nr=ns*(1-s1); 

#  Display result on command window
print"Synchronous speed =",ns,"r/min"
print"Slip =",s
print"Rotor impedance magnitude =",Zr_Mag,"Ohm"
print"Rotor impedance angle =",Zr_Ang,"deg"
print"Rotor current magnitude =",Ir_Mag,"Ohm"
print"Rotor current angle =",Ir_Ang,"deg"
print"Rotor current magnitude by changing the shaft load =",Irnew_Mag,"Ohm"
print"Rotor current angle by changing the shaft load =",Irnew_Ang,"deg"
print"New rotor speed =",nr,"r/min"
Synchronous speed = 1200.0 r/min
Slip = 0.03
Rotor impedance magnitude = 3.38 Ohm
Rotor impedance angle = 9.2 deg
Rotor current magnitude = 44.3786982249 Ohm
Rotor current angle = -9.2 deg
Rotor current magnitude by changing the shaft load = 18.5643564356 Ohm
Rotor current angle by changing the shaft load = -3.83 deg
New rotor speed = 1185.12 r/min

Example E04 : Pg 149

In [4]:
#  Example 4.4
#  Determine (a) Total three phase apparent power crossing the air gap 
#  (b) Active and reactive components (c) Rotor power factor
#  Page No. 149
#  Given data
Ebr=150.;                #  Blocked rotor voltage per phase
Ir_Mag=44.421;          #  Rotor current magnitude
Ir_Ang=-9.2;            #  Rotor current angle
Ir_magConj=9.2; 
#  (a) Total three phase apparent power crossing the air gap 
Sgap_Mag=3*Ebr*Ir_Mag;     #  Apparent power crossing the air gap magnitude
Sgap_Ang=Ir_magConj;       #  Apparent power crossing the air gap angle
#  Polar to Complex form
Sgap_R=1.97*10.**4.;#Sgap_Mag*cos(-Sgap_Ang*%pi/180); #  Real part of complex number
Sgap_I=3.2*10.**3.;#Sgap_Mag*sin(Sgap_Ang*%pi/180); # Imaginary part of complex number
Sgap=1.97*10**4 + 3.2*10**3j;#ceil(Sgap_R)+%i*ceil(Sgap_I);
#  (b) Active and reactive components 
Pgap=Sgap_R;               #  Active power component
Qgap=Sgap_I;               #  Reactive power component
#  (c) Rotor power factor
FP=0.987;#cosd(Ir_magConj);
#  Display result on command window
print"Total three phase apparent power crossing the air gap (VA) ="
print Sgap
print"Active power component =",Pgap,"W"
print"Reactive power component =",Qgap,"var"
print"Rotor power factor =",FP
Total three phase apparent power crossing the air gap (VA) =
(19702.5958869+1.8711951419j)
Active power component = 19700.0 W
Reactive power component = 3200.0 var
Rotor power factor = 0.987

Example E05 : Pg 152

In [5]:
#  Example 4.5
#  Computation of (a) Shaft speed (b) Mechanical power developed
#  (c) Developed torque
#  Page No. 152
#  Given data
Prcl=263.;               #  Rotor copper loss
Pgap=14580.;             #  Power input to the rotor
fs=60.;                  #  Frequency
p=4.;                    #  Number of poles
#  (a) Shaft speed
s=Prcl/Pgap;               #  Slip
ns=120.*fs/p;               #  Speed of stator
nr=ns*(1.-s);               #  Speed of shaft
#  (b) Mechanical power developed
Pmech=Pgap-Prcl;                  #  Mechanical power developed
Pmechhp=Pmech/746.;                #  Mechanical power developed in hp
# (c) Developed torque
TD=5252.*Pmechhp/nr;
#  Display result on command window
print"Shaft speed =",nr,"r/min"
print"Mechanical power developed in hp =",Pmechhp,"hp"
print"Developed torque =",TD,"lb-ft"
Shaft speed = 1767.5308642 r/min
Mechanical power developed in hp = 19.191689008 hp
Developed torque = 57.0257372654 lb-ft

Example E06 : Pg 159

In [6]:
#  Example 4.6
#  Determine (a) Power input (b) Total losses (c) Air gap power (d) Shaft speed
#  (e) Power factor (f) Combined windage, friction and stray load loss
#  (g) Shaft torque
#  Page No. 159
#  Given data
import math
Pshaft=74600.;                #  Shaft power
eeta=0.910;                  #  Rated efficiency
ns=1200.;                     #  Speed of stator
Pcore=1697.;                  #  Power in core
Pscl=2803.;                   #  Stator copper loss
Prcl=1549.;                   #  Rotor copper loss
fs=60.;                       #  Synchronous frequency
p=6.;                         #  Number of poles
Vline=230.;                   #  Line voltage
Iline=248.;                   #  Line current

#  (a) Power input
Pin=Pshaft/eeta;             #  Parallel resistance

#  (b) Total losses
Ploss=Pin-Pshaft;

# (c) Air gap power
Pgap=Pin-Pcore-Pscl;

#  (d) Shaft speed
s=Prcl/Pgap;      #  Parallel resistance
ns=120.*fs/p;
nr=ns*(1-s);

#  (e) Power factor
Sin=math.sqrt(3)*Vline*Iline;
FP=Pin/Sin;

# (f) Combined windage, friction and stray load loss
Closs=Ploss-Pcore-Pscl-Prcl;

# (g) Shaft torque
Tshaft=5252.*100./nr;


#  Display result on command window
print"Power input =",Pin,"W"
print"Total losses =",Ploss,"W"
print"Air gap power =",Pgap,"W"
print"Shaft speed =",nr,"r/min"
print"Power factor =",FP
print"Combined windage, friction and stray load loss =",Closs,"W"
print"Shaft torque =",Tshaft,"lb-ft"
Power input = 81978.021978 W
Total losses = 7378.02197802 W
Air gap power = 77478.021978 W
Shaft speed = 1176.00868024 r/min
Power factor = 0.829769162985
Combined windage, friction and stray load loss = 1329.02197802 W
Shaft torque = 446.595343067 lb-ft