CHAPTER05 : CLASSIFICATION PERFORMANCE APPLICATIONS AND OPERATION OF THREE PHASE INDUCTION MACHINES

Example E01 : Pg 173

In [1]:
#  Example 5.1
#  Computation of minimum value of (a) Locked rotor torque (b) Breakdown torque
#  (c) Pull up torque
#  Page No. 173
#  Given data
f=60.;               #  Frequency in Hz
p=6.;                #  Number of poles
hp=10.;              #  Horsepower
n=1150.;             #  Rated speed of machine
ns=120.*f/p;
#  (a) Locked rotor torque
Trated=hp*5252./n;   #  Rated torque 
Tlockedrotor=2.25*Trated;
#  (b) Breakdown torque
Tbreakdown=1.90*Trated;
#  (c) Pull up torque
Tpullup=1.65*Trated;
#  Display result on command window
print"Locked rotor torque =",Tlockedrotor,"lb-ft"
print"Breakdown torque =",Tbreakdown,"lb-ft"
print"Pull up torque =",Tpullup,"lb-ft"
Locked rotor torque = 102.756521739 lb-ft
Breakdown torque = 86.772173913 lb-ft
Pull up torque = 75.3547826087 lb-ft

Example E02 : Pg 180

In [2]:
# Example 5.2
# Determine (a) Slip (b) Line current (c) Apparent power, active power, 
# reactive power and power factor of the motor (d) Equivalent rotor curret
# (e) Stator copper loss (f) Rotor copper loss (g) Core loss (h) Air-gap
# power (i) Mechanical power developed (j) Developed torque (k) Shaft 
# horsepower (l) Shaft torque (m) Effiency 
# Page No. 180
# Given data
from math import sqrt,pi,sin,cos 
f=60.;                   # Frequency
P=6.;                    # Number of poles
nr=1185.;
R1=0.200;               # Motor resistance
R2=0.250;
X1=1.20;                # Motor reactance
X2=1.29;
Rfe=317.;                # Field resistance
XM=42.;                  # Motor reactance
V=460.;                  # Voltage rating
PFPS=166.;               # Stray loss

# (a) Slip 
ns=(120.*f)/P;
s=(ns-nr)/ns;            # Speed difference

# (b) Line current
Z2=20 + 1.29j;#(R2/s)+%i*X2;
# Complex to Polar form...
Z2_Mag=20.;#sqrt(real(Z2)**2+imag(Z2)**2);       # Magnitude part
Z2_Ang =3.69;#atan(imag(Z2),real(Z2))*180/%pi; # Angle part

Z0_Num_Mag=Rfe*XM;   # Z0 numerator
Z0_Num_Ang=0+90;  
   
Z0_Den_R=Rfe;       # Z0 denominator
Z0_Den_I=XM;
Z0_Den=317 + 42j;#Z0_Den_R+%i*Z0_Den_I;
# Complex to Polar form...
Z0_Den_Mag=320.;#sqrt(real(Z0_Den)**2+imag(Z0_Den)**2);       # Magnitude part
Z0_Den_Ang =7.55;#atan(imag(Z0_Den),real(Z0_Den))*180/%pi; # Angle part

Z0_Mag=Z0_Num_Mag/Z0_Den_Mag;            # Magnitude of Z0
Z0_Ang=Z0_Num_Ang-Z0_Den_Ang;            # Angle of Z0

# Polar to Complex form
Z0_R=Z0_Mag*cos(-Z0_Ang*pi/180);       # Real part of complex number
Z0_I=Z0_Mag*sin(Z0_Ang*pi/180);        # Imaginary part of complex number

# ZP computation
ZP_Num_Mag=Z2_Mag*Z0_Mag;              # ZP numerator magnitude
ZP_Num_Ang=Z2_Ang+Z0_Ang;              # ZP numerator angle

ZP_Den_R=25.5;#real(Z2)+Z0_R;                # Real part of ZP denominator
ZP_Den_I=42.6;#imag(Z2)+Z0_I; 
ZP_Den=25.5 + 42.6j;#lZP_Den_R+%i*ZP_Den_I;           # ZP in complex form

# Complex to Polar form...
ZP_Den_Mag=49.6;#sqrt(real(ZP_Den)**2+imag(ZP_Den)**2);       # Magnitude part
ZP_Den_Ang =59.1;#atan(imag(ZP_Den),real(ZP_Den))*180/%pi; # Angle part

ZP_Mag=ZP_Num_Mag/ZP_Den_Mag;           # Final vlaue of ZP in polar form
ZP_Ang=ZP_Num_Ang-ZP_Den_Ang;
# Polar to Complex form
ZP_R=ZP_Mag*cos(-ZP_Ang*pi/180);      # Real part of complex number
ZP_I=ZP_Mag*sin(ZP_Ang*pi/180);       # Imaginary part of complex number

# Zin computation
ZP=15 + 7.65j;#ZP_R+%i*ZP_I;                       # Parallel impedance
Z1=0.2 + 1.2j;#R1+%i*X1;
Zin=Z1+ZP;                             # Input impedance
# Complex to Polar form...
Zin_Mag=17.6;#sqrt(real(Zin)**2+imag(Zin)**2);       # Magnitude part
Zin_Ang =30.2;#atan(imag(Zin),real(Zin))*180/%pi; # Angle part

# I1 computation
I1_Mag=(V/sqrt(3.))/Zin_Mag;           # I1 magnitude
I1_Ang=0-Zin_Ang;                     # I1 angle

# (c) Apparent power, active power, reactive power and power factor of the motor
S_Mag=3.*(V/sqrt(3.))*I1_Mag;          # S magnitude
S_Ang=0-(-Zin_Ang);                  # S angle

# Polar to Complex form
S_R=S_Mag*cos(-S_Ang*pi/180);       # Real part of complex number
S_I=S_Mag*sin(S_Ang*pi/180);        # Imaginary part of complex number

FP=0.864;#cosd(S_Ang);                      # Power factor

# (d) Equivalent rotor curret
E2_Mag=I1_Mag*ZP_Mag;                # E2 magnitude
E2_Ang=I1_Ang+ZP_Ang;                # E2 angle

I2_Mag=E2_Mag/Z2_Mag;                # I2 magnitude
I2_Ang=E2_Ang-Z2_Ang;                # I2 angle

# (e) Stator copper loss 
Pscl=3.*I1_Mag**2.*R1;

# (f) Rotor copper loss
Prel=3.*I2_Mag**2.*R2;

# (g) Core loss 
Pcore=3.*(E2_Mag**2./Rfe);

# (h) Air-gap power
Pgap=Prel/s;

# (i) Mechanical power developed
Pmech=Prel*(1.-s)/s;

# (j) Developed torque 
TD=(21.12*I2_Mag**2*R2)/(s*ns);

# (k) Shaft horsepower
LOSS=Pscl+Prel+Pcore+PFPS;
Pshaft=(S_R-LOSS)/746.;

# (l) Shaft torque
T=5252.*Pshaft/nr;

# (m) Effiency 
eta=Pshaft/S_R*746.;

# Display result on command window
print"\nSlip =",s
print"\nLine current magnitude =",I1_Mag,"A"
print"\nLine current angle =",I1_Ang,"deg"
print"\nApparent power =",S_R,"W"
print"\nActive power =",S_I,"var"
print"\nReactive power =",S_Mag,"VA"
print"\nPower factor of the motor =",FP
print"\nEquivalent rotor curret magnitude =",I2_Mag,"A"
print"\nEquivalent rotor curret angle =",I2_Ang,"deg"
print"\nStator copper loss =",Pscl,"W"
print"\nRotor copper loss =",Prel,"W"
print"\nCore loss =",Pcore,"W"
print"\nAir-gap power =",Pgap,"W"
print"\nMechanical power developed =",Pmech,"W"
print"\nDeveloped torque =",TD,"lb-ft"
print"\nShaft horsepower =",Pshaft,"hp"
print"\nShaft torque =",T,"lb-ft"
print"\nEffiency =",eta
Slip = 0.0125

Line current magnitude = 15.0898365811 A

Line current angle = -30.2 deg

Apparent power = 10390.9402326 W

Active power = 6047.67163108 var

Reactive power = 12022.7272727 VA

Power factor of the motor = 0.864

Equivalent rotor curret magnitude = 12.6578934123 A

Equivalent rotor curret angle = -6.85 deg

Stator copper loss = 136.621900826 W

Rotor copper loss = 120.166699229 W

Core loss = 606.519617558 W

Air-gap power = 9613.33593829 W

Mechanical power developed = 9493.16923906 W

Developed torque = 56.3982375046 lb-ft

Shaft horsepower = 12.5491045777 hp

Shaft torque = 55.6184786853 lb-ft

Effiency = 0.90094176325

Example E03 : Pg 184

In [3]:
# Example 5.3
# Computation of (a) Speed at which maximum torque is developed (b) Maximum 
# torque that the machine can develop (c) Rated shaft torque (d) Which NEMA 
# design fits this motor?
# Page No. 184
# Given data
from math import sqrt
f=60.;                  # Frequency in Hz
p=4.;                   # Number of poles
hp=40.;                 # Horsepower
n=1751.;                # Rated speed of machine
v=460./sqrt(3.);       # Voltage
s=0.1490;              # Slip
R2=0.153;              # Rotor resistance 
R1=0.102;
X1=0.409;              # Rotor reactance
X2=0.613;

# (a) Speed at which maximum torque is developed 
STDmax=R2/(sqrt(R1**2.+(X1+X2)**2.));
ns=120.*f/p;                          #stator spped
nr=ns*(1.-s);

# (b) Maximum torque that the machine can develop
TDmax=(21.12*v**2.)/(2.*ns*(sqrt(R1**2.+(X1+X2)**2.)+R1));

# (c) Rated shaft torque
TDshaft=hp*5252./n;

# Display result on command window
print"\nSpeed at which maximum torque is developed =",nr,"r/min"
print"\nMaximum torque that the machine can develop =",TDmax,"lb-ft"
print"\nRated shaft torque =",TDshaft,"lb-ft"
print"\nMaximum torque is developed at slip of 0.1490 and \nhence machine is placed in design A category"
Speed at which maximum torque is developed = 1531.8 r/min

Maximum torque that the machine can develop = 366.489979327 lb-ft

Rated shaft torque = 119.977155911 lb-ft

Maximum torque is developed at slip of 0.1490 and 
hence machine is placed in design A category

Example E04 : Pg 185

In [4]:
# Example 5.4
# Computation of (a) Amount of torque that must be removed from the motor 
# shaft to maintain 1760r/min (b) Expected minimum startimg torque for the 
# lower voltage (c) Percent change in developed torque caused by 10% drop in 
# system voltage.
# Page No. 185
# Given data
hp=50.;              # Horsepower
n=1760.;             # Rated speed of machine
v1=460.;
# (a) Amount of torque that must be removed from the motor shaft to maintain 
# 1760r/min
v2=v1*0.90;
Trated=hp*5252./n;       #Rated torque 
TD2=Trated*(v2/v1)**2.;
Treduction=Trated-TD2;
# (b) Expected minimum startimg torque for the lower voltage
Tlr=1.40*Trated;
Tlr2=Tlr*(v2/v1)**2;
# (c) Percent change in developed torque caused by 10% drop in system voltage
Tchange=(TD2-Trated)/Trated;
Tchanger=(Tlr2-Tlr)/Tlr;
# Display result on command window
print"\n Amount of torque that must be removed from the motor shaft =",Treduction,"lb-ft"
print"\n Expected minimum starting torque for the lower voltage =",Tlr2,"lb-ft"
print"\n Percent change in developed torque =",Tchanger*100,"Percent"
 Amount of torque that must be removed from the motor shaft = 28.3488636364 lb-ft

 Expected minimum starting torque for the lower voltage = 169.197954545 lb-ft

 Percent change in developed torque = -19.0 Percent

Example E05 : Pg 187

In [5]:
# Example 5.5
# Computation of minimum value of (a) Shaft speed (b) Rotor current referred 
# to the stator
# Page No. 187
# Given data
from math import sqrt
f=60.;               # Frequency in Hz
p=12.;               # Number of poles
nr=591.1;           # Rated speed of machine
v=575.;              # Voltage rating of the machine
R2=0.055;

# (a) Shaft speed
ns=120.*f/p;        # Speed (r/min)
s1=(ns-nr)/ns;     # Slip 1
s2=1.25*s1;        # Slip 2
nr1=ns*(1.-s2);

# (b) Rotor current referred to the stator
V=v/sqrt(3.);
I2=V*s2/R2;

# Display result on command window
print"\nShaft speed =",nr1,"r/min"
print"\nRotor current referred to the stator =",I2,"A"
Shaft speed = 588.875 r/min

Rotor current referred to the stator = 111.916287976 A

Example E06 : Pg 190

In [6]:
# Example 5.6
# Determine (a) New operating speed if a system disturbance causes a 10% drop
# in voltage and 6% drop in frequency (b) New shaft horsepower.
# Page No. 190
# Given data
etaV=0.90;               # Efficiency related to voltage
V=230.;                   # Voltage
etaF=0.94;               # Efficiency related to voltage
f=60.;                    # Frequency
N=6.;                     # Number of poles
nr1=1175.;                # Speed of motor
P=20.;                    # Horsepower of motor

# (a)  New operating speed if a system disturbance causes a 10% drop in 
# voltage and 6% drop in frequency
V2=etaV*V;              # New voltage after 10% drop
f2=etaF*f;              # New frequency after 6% drop
ns1=120.*f/N;
ns2=120.*0.94*f/N;
s1=(ns1-nr1)/ns1;       # Speed difference

s2=s1*((V/V2)**2.)*(f2/f);  
nr2=ns2*(1.-s2);         # New speed

# (b) New shaft horsepower
P2=P*(nr2/nr1);         # With a constant torque load T2=T1

# Display result on command window
print"\nNew operating speed in case of voltage and frequency drop =",nr2,"r/min"
print"\nNew shaft horsepower =",P2,"hp"
New operating speed in case of voltage and frequency drop = 1100.72839506 r/min

New shaft horsepower = 18.7358024691 hp

Example E07 : Pg 192

In [7]:
# Example 5.7
# Determine expected locked-rotor line current
# Page No. 192
# Given data
Ir1=151.;                   # Rated current
V1=230.;                    # Rated voltage
V2=220.;                    # Motor starting voltage
F1=60.;                     # Rated frequency
F2=50.;                     # Motor starting frequency
# Expected locked-rotor line current
Ir2=Ir1*((V2/F2)/(V1/F1));
# Display result on command window
print"\n Expected locked-rotor line current =",Ir2,"A"
 Expected locked-rotor line current = 173.32173913 A

Example E08 : Pg 193

In [8]:
# Example 5.8
# Determine (a) Expected minimum locked-rotor torque (b) Repeat (a) when 
# voltage and frequency dropped to 230V and 58Hz  
# Page No. 193
# Given data
HPrated=75.;            # Rated horsepower
nrated=1750.;           # Rated speed
V1=240.;                # Rated voltage
V2=230.;                # Voltage after drop
F1=60.;                 # Rated frequency
F2=58.;                 # Frequency after drop

# (a) Expected minimum locked-rotor torque
Trated=5252.*HPrated/nrated;   # Rated torque
Tlr=Trated*1.2;               # Minimum locked-rotor torque is 120% rated 

# (b) Expected minimum locked-rotor torque when voltage and frequency dropped 
# to 230V and 58Hz  
Tlr2=Tlr*((V2/F2)**2.)*((F1/V1)**2.);

# Display result on command window
print"\nExpected minimum locked-rotor torque =",Tlr,"lb-ft"
print"\nExpected minimum locked-rotor torque after drop =",Tlr2,"lb-ft"
Expected minimum locked-rotor torque = 270.102857143 lb-ft

Expected minimum locked-rotor torque after drop = 265.465984372 lb-ft

Example E09 : Pg 194

In [9]:
# Example 5.9
# Determine (a) Shaft r/min (b) Slip 
# Page No. 194
# Given data
from math import sqrt
F1=60.;             # Rated frequency
N=4.;               # Number of poles
F2=50.;             # New frequency
ns=1770.;           # Rated speed

# (a) Shaft r/min
ns60=120.*F1/N;       # Speed at rated ferquency 
ns50=120.*F2/N;       # Speed at 50 Hz frequency
s60=(ns60-ns)/ns60;  # Slip at 60 Hz frequency

# Using eq. (5.16) and by solving..s50=29.251/nr50
# Using eq. (4.3) and solving for nr50 we get the quadratic equation..
# Using various values of quadratic equations, we have
a=1.;
b=-1500.;
c=43876.5;
r1=(-b+sqrt(b**2-4*a*c))/(2.*a); # Root 1

r2=(-b-sqrt(b**2-4*a*c))/(2.*a); # Root 2
# Answer 'r2' is not valid

# (b) Slip 
s50=(ns50-r1)/ns50;

# Display result on command window
print"\nShaft speed =",r1,"r/min"
print"\nSlip =",s50
Shaft speed = 1470.15519161 r/min

Slip = 0.019896538928

Example E10 : Pg 198

In [10]:
# Example 5.10
# Determine (a) Range of rotor speed (b) Required rheostat resistance
# Page No. 198
# Given data
from math import sqrt
F=60.;                  # Frequency of motor
P=14.;                  # Number of poles
SL=0.395;              # Low speed point
SH=0.02;               # High speed point
Stdmax=0.74;           # Value at which TD is maximum (from curve B)
R1=0.403;              # Motor resistance
R2=0.317;
X1=1.32;               # Motor reactance
X2=1.32;
a=3.8;                 # Ratio of stator turns/phase to rotor turns/phase

# (a) Range of rotor speed
ns=120.*F/P;           # Speed
nrl=ns*(1.-SL);        # Rotor low speed
nrh=ns*(1.-SH);        # Rotor high speed

# (b) Required rheostat resistance
Rrhe=Stdmax*(sqrt(R1**2.+(X1+X2)**2.))-R2;
Rehereq=Rrhe/a**2.;

# Display result on command window
print"\n Low range of rotor speed =",nrl,"r/min"
print"\n High range of rotor speed =",nrh,"r/min"
print"\n Required rheostat resistance =",Rehereq,"Ohm/phase"
 Low range of rotor speed = 311.142857143 r/min

 High range of rotor speed = 504.0 r/min

 Required rheostat resistance = 0.114905179241 Ohm/phase

Example E11 : Pg 201

In [11]:
# Example 5.11
# Determine (a) Rotor frequency (b) Slip at which TDmax occurs (c) Rotor speed
# at 1/2 rated torque load (d) Required rheostat resistance (e) Rated torque
# Page No. 201
# Given data
from math import sqrt
S=0.0159;                # Slip
Fbr=50.;                  # Rated frequency
R1=0.00536;              # Motor resistance
R2=0.00613;
X1=0.0383;               # Motor reactance
X2=0.0383;
Rrhe=0;                  # Initial rheostat resistance
P=4.;                     # Number of poles
NR=1000.;                 # Rated speed
s1=0.0159;               # Slip of rheostat
a=2.;                     # Stator to rotor turns ratio
hp=400.;                  # Motor horsepower

# (a) Rotor frequency
fr=S*Fbr;

# (b) Slip at which TDmax occurs
Stdmax=(R2+Rrhe)/(sqrt(R1**2.+(X1+X2)**2.));

# (c) Rotor speed at 1/2 rated torque load 
s=S*(0.5)*(R2/R2);  # Rotor speed at 1/2 rated torque
ns=120.*Fbr/P;        
nr=ns*(1.-s);         # Rotor speed

# (d) Required rheostat resistance
s2=(ns-NR)/ns;
Rrhe2=((s2/s1)*(1./0.5)*(R2+Rrhe))-R2;  # rheostat resistance
Rrheostat=Rrhe2/a**2.;

# (e) Rated torque
nr1=ns*(1.-s1);       # Rated speed
T=hp*5252./nr1;



# Display result on command window
print"\n Rotor frequency =",fr,"Hz"
print"\n Slip at which TDmax occurs =",Stdmax
print"\n Rotor speed at 1/2 rated torque =",nr,"r/min"
print"\n Required rheostat resistance =",Rrheostat,"Ohm/phase"
print"\n Rated torque =",T,"lb-ft"
 Rotor frequency = 0.795 Hz

 Slip at which TDmax occurs = 0.079830908464

 Rotor speed at 1/2 rated torque = 1488.075 r/min

 Required rheostat resistance = 0.0627232651992 Ohm/phase

 Rated torque = 1423.16160282 lb-ft

Example E12 : Pg 202

In [12]:
# Example 5.12
# Determine the percent increase or decrease in rotor circuit resistance
# Page No. 202
# Given data
Stdmax1=0.45;       # Maximum torque condition 1
Stdmax2=0.80;       # Maximum torque condition 2
# Percent increase or decrease in rotor circuit resistance
PerCh=1/(Stdmax1/Stdmax2);
PerCh=PerCh-1;
# Display result on command window
print"\nPercent change in rotor circuit resistance =",PerCh*100,"Percent increase"
Percent change in rotor circuit resistance = 77.7777777778 Percent increase

Example E13 : Pg 208

In [13]:
# Example 5.13
# Determine the expected in-rush current
# Page No. 208
# Given data
from math import sqrt
kva1=5.6;             # KVA/hp lower limit from table 5.9
hp=150.;               # Motor horsepower
Vline=460.;            # Line voltage
kva2=6.3;             # KVA/hp upper limit from table 5.9
# Expected in-rush current
# Lower limit of expected range of in-rush current is
Ilrss=(kva1*hp*1000)/(sqrt(3)*Vline);  
# Upper limit of expected range of in-rush current is
Iulss=(kva2*hp*1000)/(sqrt(3)*Vline);  
# Display result on command window
print"\n Lower limit of expected range of in-rush current =",Ilrss,"A"
print"\n Upper limit of expected range of in-rush current =",Iulss,"A"
 Lower limit of expected range of in-rush current = 1054.29179591 A

 Upper limit of expected range of in-rush current = 1186.0782704 A

Example E14 : Pg 211

In [14]:
# Example 5.14
# Determine (a) Percent voltage unbalance (b) Expected approximate temp. rise
# if operating at rated load in a 40 deg ambient (c) Expected insulation life 
# (d) Required derating of motor to prevent shortening isulation life.
# Page No. 211
# Given data
from math import sqrt
VL1=460.;                 # Line voltage 1
VL2=455.;                 # Line voltage 2
VL3=440.;                 # Line voltage 3 
Trated=110.;              # Rated temp. (from table 5.8)
hp=30.;                   # Motor horsepower

# (a) Percent voltage unbalance 
Vavg=(VL1+VL2+VL3)/3.;        # Average line voltage

#VD1=abs(VL1-Vavg);           # Voltage deviation from the average 
#VD2=abs(VL2-Vavg); 
#VD3=abs(VL3-Vavg); 
#VD=[VD1 VD2 VD3];          
#VDMax=max(VD);               # Choose maximum value of voltage deviation
PerUBV=2.58;#(VDMax/Vavg)*100;

# (b) Expected approximate temp. rise if operating at rated load in a 40 deg
PerDeltaT=2.*PerUBV**2.;        # Percent change in temp.
Tubv=Trated*(1.+(PerDeltaT/100.));

# (c) Expected insulation life 
DeltaT=Tubv-Trated;         # Percent increase in motor temp.
RL=1./(2.**(DeltaT/10.));       # Relative life on insulation
EL=RL*20;

# (d) Required derating of motor to prevent shortening isulation life
ReqDer=hp*0.92;

# Display result on command window
print"\nPercent voltage unbalance  =",PerUBV
print"\nExpected approximate temperature rise  =",Tubv,"deg C"
print"\n Expected insulation life =",EL,"years"
print"\n Required derating of motor =",ReqDer,"hp"
Percent voltage unbalance  = 2.58

Expected approximate temperature rise  = 124.64408 deg C

 Expected insulation life = 7.2476842001 years

 Required derating of motor = 27.6 hp

Example E15 : Pg 213

In [15]:
# Example 5.15
# Determine the machine parameters in ohms
# Page No. 213
# Given data
from math import sqrt 
V=460.;             # Motor voltage
hp=50.;             # Motor horsepower
r1=0.021;          # Resistance
r2=0.020;
x1=0.100;          # Reactance
x2=0.0178;
rfe=20.;           
Xm=3.68;           # Motor reactance

# Machine parameters in ohms
Vbase=V/sqrt(3.);         # Base voltage
Pbase=hp*746./3.;          # Base power
Zbase=Vbase**2./Pbase;     # Base impedance

R1=r1*Zbase;
X1=x1*Zbase;
R2=r2*Zbase;
X2=x2*Zbase;
Rfe=rfe*Zbase;
XM=Xm*Zbase;

# Display result on command window
print"\n Motor resistance 1 =",R1,"Ohm"
print"\n Motor reactance 1 =",X1,"Ohm"
print"\n Motor resistance 2 =",R2,"Ohm"
print"\n Motor reactance 2 =",X2,"Ohm"
print"\n Field resistance =",Rfe,"Ohm"
print"\n Reactance of motor =",XM,"Ohm"
 Motor resistance 1 = 0.119131367292 Ohm

 Motor reactance 1 = 0.567292225201 Ohm

 Motor resistance 2 = 0.11345844504 Ohm

 Motor reactance 2 = 0.100978016086 Ohm

 Field resistance = 113.45844504 Ohm

 Reactance of motor = 20.8763538874 Ohm

Example E16 : Pg 218

In [16]:
# Example 5.16
# Determine (a) R1, R2, X1, X2, XM and the combined core, friction and windage 
# loss (b) Express the no-load current as a percent of rated current
# Page No. 218
# Given data
from math import sqrt
P3ph=2573.4;                 # 3-ph power of induction motor
Vline=36.2;                  # Line voltage
Iline=58;                    # Line current
P3phnl=4664.4;               # No load power
Vlinenl=460.;                 # No load line volatge
Ilinenl=32.7;                # No load line current
Vdc=12.;                      # DC voltage
Idc=59.;                      # DC current
F1=60.;                       # Rated frequency
F2=15.;                       # Test frequency
Irated=57.8;                 # Rated current
 
# (a) R1, R2, X1, X2, XM and the combined core, friction and windage loss
Pbr15=P3ph/3.;                # Power/phase
Vbr15=Vline/sqrt(3.);         # Voltage/phase
Ibr15=Iline;
PNL=P3phnl/3.;                # No load power/phase
VNL=Vlinenl/sqrt(3.);         # No load voltage/phase
INL=Ilinenl;                 # No load current/phase

# Determination of R1
Rdc=Vdc/Idc;
R1=Rdc/2.;

# Determination of R2
Zbr15=Vbr15/Ibr15;          # Impedance
Rbr15=Pbr15/Ibr15**2.;
R2=Rbr15-R1;

# Determination of X1 and X2
Xbr15=sqrt(Zbr15**2.-Rbr15**2.);
Xbr60=Xbr15*(F1/F2);
X1=0.4*Xbr60;               # From Table 5.10
X2=0.6*Xbr60;

# Determination of XM
SNL=VNL*INL;
QNL=sqrt(SNL**2.-PNL**2.);
XNL=QNL/INL**2.;
XM=XNL-X1;

# Determination of combined friction, windage and core loss
Ploss=PNL-(INL**2.*R1);

# (b) No-load current as a percent of rated current
PerINL=INL*100./Irated;

# Display result on command window
print"\n Motor resistance 1 =",R1,"Ohm/phase"
print"\n Motor reactance 1 =",X1,"Ohm/phase"
print"\n Motor resistance 2 = ",R2,"Ohm/phase"
print"\n Motor reactance 2 =",X2,"Ohm/phase"
print"\n Reactance of motor =",XM,"Ohm/phase"
print"\n Combined friction, windage and core loss =",Ploss,"W/phase"
print"\n No-load current as a percent of rated current =",PerINL,"Percent"
 Motor resistance 1 = 0.101694915254 Ohm/phase

 Motor reactance 1 = 0.407379662436 Ohm/phase

 Motor resistance 2 =  0.153299139443 Ohm/phase

 Motor reactance 2 = 0.611069493654 Ohm/phase

 Reactance of motor = 7.58314688225 Ohm/phase

 Combined friction, windage and core loss = 1446.05864407 W/phase

 No-load current as a percent of rated current = 56.5743944637 Percent

Example E17 : Pg 223

In [17]:
# Example 5.17
# Determine the active power that the motor, driven as an induction generator
# delivers to the system.
# Page No. 223
# Given data
from math import sqrt,pi,sin,cos
ns=1200.;                # Speed
nr=1215.;
R1=0.200;               # Motor resistance
R2=0.250;
X1=1.20;                # Motor reactance
X2=1.29;
Rfe=317.;                # Field resistance
XM=42.;                  # Motor reactance
V=460.;                  # Voltage rating

# Active power of the motor computation
s=(ns-nr)/ns;            # Speed difference
Z2=-20 + 1.29j;#(R2/s)+%i*X2;

# Complex to Polar form...
Z2_Mag=20.;#sqrt(real(Z2)**2+imag(Z2)**2);       # Magnitude part
Z2_Ang =176.;#atan(imag(Z2),real(Z2))*180/%pi; # Angle part

Z0_Num_Mag=Rfe*XM;   # Z0 numerator
Z0_Num_Ang=0+90;  
   
Z0_Den_R=Rfe;       # Z0 denominator
Z0_Den_I=XM;
Z0_Den=317 + 42j;#Z0_Den_R+%i*Z0_Den_I;
# Complex to Polar form...
Z0_Den_Mag=320.;#sqrt(real(Z0_Den)**2+imag(Z0_Den)**2);       # Magnitude part
Z0_Den_Ang =7.55;#atan(imag(Z0_Den),real(Z0_Den))*180/%pi; # Angle part

Z0_Mag=Z0_Num_Mag/Z0_Den_Mag;            # Magnitude of Z0
Z0_Ang=Z0_Num_Ang-Z0_Den_Ang;            # Angle of Z0

# Polar to Complex form
Z0_R=Z0_Mag*cos(-Z0_Ang*pi/180);       # Real part of complex number
Z0_I=Z0_Mag*sin(Z0_Ang*pi/180);        # Imaginary part of complex number

# ZP computation
ZP_Num_Mag=Z2_Mag*Z0_Mag;              # ZP numerator magnitude
ZP_Num_Ang=Z2_Ang+Z0_Ang;              # ZP numerator angle

ZP_Den_R=-14.5;#real(Z2)+Z0_R;                # Real part of ZP denominator
ZP_Den_I=42.6;#imag(Z2)+Z0_I; 
ZP_Den=-14.5 + 42.6j;#ZP_Den_R+%i*ZP_Den_I;           # ZP in complex form

# Complex to Polar form...
ZP_Den_Mag=45.;#sqrt(real(ZP_Den)**2+imag(ZP_Den)**2);       # Magnitude part
ZP_Den_Ang =109.;# atan(imag(ZP_Den),real(ZP_Den))*180/%pi; # Angle part

ZP_Mag=ZP_Num_Mag/ZP_Den_Mag;           # Final vlaue of ZP in polar form
ZP_Ang=ZP_Num_Ang-ZP_Den_Ang;
# Polar to Complex form
ZP_R=ZP_Mag*cos(-ZP_Ang*pi/180);      # Real part of complex number
ZP_I=ZP_Mag*sin(ZP_Ang*pi/180);       # Imaginary part of complex number

# Zin computation
ZP=-16.1 + 9.3j;#ZP_R+%i*ZP_I;                       # Parallel impedance
Z1=0.2 + 1.2j;#R1+%i*X1;
Zin=Z1+ZP;                             # Input impedance
# Complex to Polar form...
Zin_Mag=19.;#sqrt(real(Zin)**2+imag(Zin)**2);       # Magnitude part
Zin_Ang =146.;#atan(imag(Zin),real(Zin))*180/%pi; # Angle part

# I1 computation
I1_Mag=(V/sqrt(3))/Zin_Mag;           # I1 magnitude
I1_Ang=0-Zin_Ang;                     # I1 angle

# S computation
S_Mag=3*(V/sqrt(3))*I1_Mag;          # S magnitude
S_Ang=0-(-Zin_Ang);                  # S angle

# Polar to Complex form
S_R=S_Mag*cos(-S_Ang*pi/180);       # Real part of complex number
S_I=S_Mag*sin(S_Ang*pi/180);        # Imaginary part of complex number

# Display result on command window
print"Active power of the motor =",S_R,"W"
Active power of the motor = -9232.86054488 W

Example E18 : Pg 231

In [18]:
# Example 5.18
# Computation of (a) Locked rotor torque and the expected average in rush 
# current (b) Repeat part (a) assuming motor is started at reduced voltage 
# with 65% tap (c) In rush line current line current when starting at reduced 
# voltage
# Page No. 231
# Given data
from math import sqrt
P=125.;                 # Rated Voltage
n=1141.;                # Speed of machine
hp=125.;                # Horsepower rating of device 
Vline=460.;             # Line voltage
ns=1200.;               # Stator speed
s=0.125;               # Slip
ILS=683.;               # Current at low side
# (a) Locked rotor torque and the expected average in rush current
Trated=P*5252./(n);                 # Rated torque
Tlr=1.25*Trated;                   # Locked rotor torque
kVA=(6.3+7.1)/2.;
Ilr=(kVA*1000.*hp)/(Vline*sqrt(3.)); # In-rush current
# (b) Locked rotor torque and the expected average in rush current when motor 
# is started at reduced voltage
V2=0.65*Vline;                    # Voltage impressed across the stator
I=Ilr*0.65;                       # Average in-rush current
T2=Tlr*(V2/Vline)**2.;              # Locked rotor toreque
nr=ns*(1.-s);
# (c) In rush line current line current when starting at reduced voltage
a=1./0.65;                         # Bank ratio of autotransformer
IHS=ILS/a;
# Display result on command window
print"\nLocked rotor torque =",Tlr,"lb-ft"
print"\nExpected average in-rush current =",Ilr,"A"
print"\nLocked rotor torque when motor is started at reduced voltage =",T2,"lb-ft"
print"\nIn-rush line current =",IHS,"A"
Locked rotor torque = 719.215600351 lb-ft

Expected average in-rush current = 1051.15402271 A

Locked rotor torque when motor is started at reduced voltage = 303.868591148 lb-ft

In-rush line current = 443.95 A

Example E19 : Pg 233

In [19]:
# Example 5.19
# Computation of (a) Locked rotor current per phase and minimum locked rotor 
# torque when starting (b) Locked rotor current per phase when motor is delta 
# connected (c) Code letter 
# Page No.233
# Given data
from math import sqrt
V=460.;                 # Rated Voltage
Z=0.547;               # Locked rotor impedance
n=1750.;                # Speed of machine
hp=60.;                 # Horsepower rating of device
f=60.;                  # Frequency of motor 
# (a) Locked rotor current per phase and minimum locked rotor torque 
Vphase=V/sqrt(3.);        # Voltage/phase
Ilr1=Vphase/Z;            # Locked rotor current/phase
Trated=hp*5252./(n);      # Rated torque
Tlr=1.4*Trated;          # Locked rotor torque
T2=Tlr*(Vphase/V)**2.;
# (b) Locked rotor current per phase when motor is delta connected 
Ilr=V/Z;                 # Locked rotor current/phase
Il=Ilr*sqrt(3.);          # Line current
# (c) Code letter
Slr=sqrt(3.)*V*Il/1000.;   # Code letter at rated voltage
kVA=Slr/f;
# Display result on command window
print"\n Locked rotor current per phase =",Ilr1,"A"
print"\n Minimum locked rotor torque =",T2,"lb-ft"
print"\n Locked rotor current per phase when motor is delta connected =",Il,"A"
print"\n Code letter =",kVA
 Locked rotor current per phase = 485.523078295 A

 Minimum locked rotor torque = 84.032 lb-ft

 Locked rotor current per phase when motor is delta connected = 1456.56923488 A

 Code letter = 19.3418647166

Example E20 : Pg 235

In [20]:
# Example 5.20
# Computation of (a) Resistance of the resistors required to limit the locked 
# rotor current to 3 times rated current (b) Stator voltage per phase at 
# locked rotor (c) Expected minimum locked rotor torque when starting as a 
# percent of rated torque
# Page No. 235
# Given data
from math import sqrt
Ilr=3.*78.;                 # Locked rotor current
Vbranch=132.79;           # Branch voltage
Rlr=0.2549;               #Locked rotor resistance
Xlr=0.0978;               #Locked rotor impedance
f=60.;                     #Frequency of motor 
Zlr=0.273;

# (a) Resistance of the resistors required to limit the locked rotor current 
# to 3 times rated current
Rex=sqrt((Vbranch**2./Ilr**2.)-(Rlr**2.))-Xlr;

# (b) Stator voltage per phase at locked rotor 
IZlr=Ilr*Zlr;
VT1_N=IZlr;

# (c) Expected minimum locked rotor torque when starting as a percent of 
# rated torque
# From table 5.1 --> Minimum locked rotor torque = 150% rated torque

# Display result on command window

print"\nResistance of the resistors required =",Rex,"Ohm"
print"\nStator voltage per phase at locked rotor =",VT1_N,"V"
print'\nExpected minimum locked rotor torque = 1.5 Trated'
Resistance of the resistors required = 0.409208864143 Ohm

Stator voltage per phase at locked rotor = 63.882 V

Expected minimum locked rotor torque = 1.5 Trated

Example E21 : Pg 236

In [21]:
# Example 5.21
# Computation of Inductance and voltage rating of each series connected 
# inductor required to limit the starting current to approximately 2*Irated. 
# Page No. 236
# Given data
from math import sqrt,pi
KVA=6.7;                    # Average locked rotor KVA/hp
hp=7.5;                     # Motor horsepower
Vline=208.;                  # Line voltage
I=48.;                       # Total current
Rlr=0.294;                  # Locked rotor resistance
Xlr=0.809;                  # Locked rotor impedance
f=60.;                       # Frequency of motor

# Corresponding approximate load current
Ilr=KVA*1000.*hp/(sqrt(3.)*Vline);      
Vphase=Vline/sqrt(3.);            # Voltage/phase

# Applying ohm's law to one phase
Zlr=Vphase/Ilr;               # Impedance
Xex=sqrt((Vphase**2./I**2.)-(Rlr**2.))-Xlr;
L=Xex/(2.*pi*f);
L=L*10.**03;
VXl=I*Xex;

# Display result on command window
print"\nThe inductance of each series connected inductor =",L,"mH"
print"\nThe voltage rating of each series connected inductor =",VXl,"V"
The inductance of each series connected inductor = 4.44444609821 mH

The voltage rating of each series connected inductor = 80.4248018577 V