# Example 11.1
# Computation of (a) The armature current when operating at rated conditions
# (b) The resistance and power rating of an external resistance required in
# series with the shunt field circuit to operate at 125 percent rated speed
# Page No. 448
# Given data
HP=40.; # hp rating of the device
Perratedload=0.902; # Percentage rated load
VT=240.; # Voltage value of motor
RF=99.5; # Resistance of shunt motor
Nf=1231.; # Turns per pole of the shunt motor
Ra=0.0680; # Armature resistance
RIP=0.0198; # Interpole winding resistance
Rs=0.00911; # Resistance of series field winding
Bp1=0.70; # Flux density for a net mmf
n1=1150.; # Speed of shunt motor
# (a) The armature current when operating at rated conditions
P=HP*746./Perratedload;
IT=P/VT; # Total current
IF=VT/RF; # Field current
Ia=IT-IF;
# (b) The resistance and power rating of an external resistance required in
# series with the shunt field circuit to operate at 125 percent rated speed
Fnet=Nf*IF; # Corresponding mmf from magnetization curve
Racir=Ra+RIP+Rs;
n2=n1*1.25; # 125 percent rated speed
# Shaft load is adjusted to value that limits the armature current to 115%
# of rated current
Bp2=Bp1*(n1/n2)*((VT-Ia*Racir*1.15)/(VT-Ia*Racir))
FF=2.3*1000.;
IF1=FF/Nf;
Rx=(VT/IF1)-RF;
PRx=(IF1**2.)*Rx;
# Display result on command window
print"The armature current =",Ia,"A"
print"The resistance rating =",Rx,"Ohm"
print"The power rating =",PRx,"W"
# Note: Answer varies due to round-off errors
# Example 11.2
# Computation of (a) Shunt field current (b) Armature current (c) Developed
# torque (d) Armature current if a resistor inserted in series with the shunt
# field circuit caused the speed to increase to 900 r/min (e) External
# resistance required in series with the shunt field circuit to operate
# at 900 r/min
# Page No. 450
# Given data
HP=125.;
perratedload=0.854; # Percentage rated load
VT=240.; # Voltage value of motor
RF=49.2; # Resistance of shunt motor
Nf=577.; # Turns per pole of the shunt motor
Ns=4.5;
Ra=0.0172; # Armature resistance
RIP=0.005; # Interpole winding resistance
Rs=0.0023; # Resistance of series field winding
n1=850.; # Speed of shunt motor
n2=900.;
F2=4000.;
# (a) Shunt field current
IF=VT/RF; # Field current
# (b) Armature current
Pin=HP*746./perratedload; # Input power
IT=Pin/VT; # Total current
Ia1=IT-IF;
# (c) Developed torque
Racir=Ra+RIP+Rs;
Ea=VT-Ia1*Racir; # Armature emf
Pmech=Ea*Ia1; # Mechanical power
TD=Pmech*5252./n1/746.; # Torque developed
# (d) Armature current if a resistor inserted in series with the shunt field
# circuit caused the speed to increase to 900 r/min
Ia2=Ia1*n2/n1;
# (e) External resistance required in series with the shunt field circuit to
# operate at 900 r/min
IF2=(F2-0.90*Ns*Ia2)/Nf;
Rx=(VT/IF2)-RF;
# Display result on command window
print"\n Shunt field current =",IF,"A"
print"\n Armature current =",Ia1,"A"
print"\n Developed torque =",TD,"lb-ft"
print"\n Armature current if a resistor inserted in series =",Ia2,"A"
print"\n External resistance required =",Rx,"Ohm"
# Example 11.3
# Computation of Speed if the load is reduced to a value that causes the
# armature current to be 30 percent of the rated current
# Page No.453
# Given data
HP=100.;
perratedload=0.896; # Percentage rated load
VT=240.; # Voltage value of motor
Ns=14.; # Number of turns/pole in series field
Ra=0.0202; # Armature resistance
RIP=0.00588; # Interpole winding resistance
Rs=0.00272; # Resistance of series field winding
n1=650.; # Speed of shunt motor
Bp2=0.34; # Air gap flux density from magnetization curve
Bp1=0.87; # Air gap flux density from magnetization curve
# Computation of Speed if the load is reduced to a value that causes the
# armature current to be 30 percent of the rated current
Pin=HP*746./perratedload; # Input power
IT=Pin/VT; # Total current
Ia=IT; # Armature current
Racir=Ra+RIP+Rs; # Resistance of armature circuit
Fnet1=Ns*Ia*(1.-0.080); # Net mmf
Fnet2=0.30*Fnet1; # Net mmf from magnetization curve
n2=n1/((VT-(Ia*Racir))/Bp1 * Bp2/(VT-(0.30*Ia*Racir)));
# Display result on command window
print"Speed of the motor =",round(n2,2),"r/min"
# Example 11.4
# Computation of resistance using linear approximation and values are
# compared with results obtained in example 11.1
# Page No. 456
# Given data
HP=40.; # hp rating of the device
percentratedload=0.902; # Percentage rated load
VT=240; # Voltage value of motor
RF=99.5; # Resistance of shunt motor
Nf=1231.; # Turns per pole of the shunt motor
Ra=0.0680; # Armature resistance
RIP=0.0198; # Interpole winding resistance
Rs=0.00911; # Resistance of series field winding
Bp1=0.70; # Flux density for a net mmf
n1=1150.; # Speed of shunt motor
n2=1.25*n1;
IT=137.84;
# Computation of resistance using linear approximation and values are
# compared with results obtained in example 11.1
IF=VT/RF; # Field current
Ia1=IT-IF; # Armature current
Fnet1=Nf*IF; # Net mmf
Racir=Ra+RIP+Rs; # Armature circuit resistance
Fnet2=Fnet1*(n1/n2)*((VT-Ia1*Racir*1.15)/(VT-Ia1*Racir));
IF1=Fnet2/Nf; # Field current
Rx=(VT/IF1)-RF; # External resistance required
# Display result on command window
print"The resistance rating of an external resistance =",Rx,"Ohm"
# Example 11.5
# Computation using linear approximation to show the gross error that occurs
# when a linear assumption is applied to compound motors operating at overload
# conditions
# Page No. 456
# Given data
Nf=577.; # Turns per pole of the shunt motor
IF=4.88; # Field current
Ns=4.5;
IA=450.09; # Armature current
F2=4367.8; # mmf
VT=240.; # Voltage value of motor
RF=49.2; # Resistance of shunt motor
HP=125.;
perratedload=0.854; # Percentage rated load
Rx1=17.8; # Value of resistance in Example 11.2
Fnet1=(Nf*IF)+ (0.90 * Ns*IA);
Ia2=Fnet1*IA/F2; # Armature current
If2=(F2 - Ns*Ia2*0.90)/Nf;
Rx=(VT/If2)-RF; # External resistance required
# Error introduced by linear approximation
PE=(17.8-Rx)/17.8*100;
# Display result on command window
print"External resistance required in series =",Rx,"Ohm"
print"Error introduced by linear approximation =",PE,"Percent"
# Example 11.6
# Determine (a) Torque developed when operating at rated speed (b) Developed
# torque required at half rated speed (c) Armature voltage required for half
# rated speed
# Page No. 460
# Given data
VT=750.; # Voltage value of motor
Nf=1231.; # Turns per pole of the shunt motor
Ra=0.00540; # Armature resistance
RIPcw=0.00420; # Interpole winding resistance
N=955.; # Speed of shunt motor
Ia1=1675.; # Armature current
# (a) Torque developed when operating at rated speed
Racir=Ra+RIPcw;
Ea=VT-Ia1*Racir;
Pmech=Ea*Ia1;
TD=Pmech*5252./N/746.;
# (b) Developed torque required at half rated speed
T2=TD*(0.5*N/N)**2.;
# (c) Armature voltage required for half rated speed
Ia2=T2*Ia1/TD;
V2=(0.5*N/N)*(VT-Ia1*Racir) + Ia2*Racir ;
# Shaft load is adjusted to value that limits the armature current to 115 % of rated current
# Display result on command window
print"Torque developed when operating at rated speed =",TD,"lb-ft"
print"Developed torque required at half rated speed =",T2,"lb-ft"
print"Armature voltage required for half rated speed =",V2,"V"
# Example 11.7
# Computation of the resistance of a dynamic braking resistor that will be
# capable of developing 500 lb-ft of braking torque at a speed of 1000 r/min.
# Page No. 464
# Given data
T1=910.; # Torque load
Pshaft=199.257*746.; # Power of shaft
eeta=0.940; # Efficiency
VT=240.; # Rated voltage
T2=500.; # Braking torque
n1=1000.; # Windage and friction speed
n2=1150.; # Speed of motor
Rf=52.6; # Field resistance
Racir=0.00707; # Combined armature,compensating winding and # interpolar resistance
# Resistance of a dynamic braking resistor
Pshaft=T1*n2/5252.; # Shaft power
Pin=Pshaft*746./eeta; # Input power
IT=Pin/VT; # Total current
If=VT/Rf; # Field current
Ia1=IT-If; # Armature current
Ea1=VT-Ia1*Racir; # Armature emf
Ia2=Ia1*T2/T1; # Armature current
Ea2=Ea1*n1/n2;
RDB=(Ea2-Ia2*Racir)/Ia2; # Resistance
# Display result on command window
print"Resistance of the dynamic braking resistor =",RDB,"Ohm"