# Chapter 16: DC MACHINES¶

## Example 16.1,Page number: 518¶

In [1]:
#Question:
"""Finding the terminal voltage,output current and total power generated by a dc generator."""

#Variable Declaration:
e=2.1                 #Average emf generated in each conductor(in Volts)
Z=480.0               #Number of conductors in armature
P=8.0                 #Number of poles

#Calculations:
A_1=P
E_1=e*(Z/A_1)
Po_1=E_1*I_L_1
A_2=2
E_2=e*(Z/A_2)
Po_2=E_2*I_L_2

#Result:
print("(a) When the armature is lap wound \n ")
print "The terminal voltage on no load is %.2f V" %(E_1)
print "The output current on full load is %.2f A" %(I_L_1)
print "The total power generated on full load is %e W" %(Po_1)
print("\n(b)When the armature is wave wound \n ")
print "The terminal voltage on no load is %.2f V" %(E_2)
print "The output current on full load is %.2f A" %(I_L_2)
print "The total power generated on full load is %e W" %(Po_2)

(a) When the armature is lap wound

The terminal voltage on no load is 126.00 V
The output current on full load is 1600.00 A
The total power generated on full load is 2.016000e+05 W

(b)When the armature is wave wound

The terminal voltage on no load is 504.00 V
The output current on full load is 400.00 A
The total power generated on full load is 2.016000e+05 W


## Example 16.2,Page number: 520¶

In [10]:
#Question:
"""Finding the emf induced in the armature of a dc generator. """

#Variable Declaration:
slots=65.0            #Number of slots
cond_per_slot=12.0    #Number of conductors per slot
A=4.0                 #Number of parallel paths
P=4.0                 #Number of poles
flux=0.02             #Flux per pole(in Webers)
N=1200.0              #Speed of operation of the dc generator(in rpm)

#Calculations:
Z=slots*cond_per_slot
E=(flux*Z*N*P)/(60*A)

#Result:
print "The emf induced in the armature is %.2f V" %(E)

The emf induced in the armature is 312.00 V


## Example 16.3,Page number: 520¶

In [11]:
#Question:
""" Finding the induced emf in a dc machine."""

#Variable Declaration:
N1=500.0              #Initial speed of operation of dc machine(in rpm)
E1=180.0              #Induced emf at 500rpm(in Volts)
N2=600.0              #New speed of operation(in rpm)

#Calculations:
E2=(N2/N1)*E1

#Result:
print "The induced emf when the machine runs at 600 rpm is %.2f V" %(E2)

The induced emf when the machine runs at 600 rpm is 216.00 V


## Example 16.4,Page number: 520¶

In [2]:
#Question:
""""Finding the percentage increase in the field flux in a dc generator."""

#Variable Declaration:
N1=750.0              #Initial speed of operation of dc machine(in rpm)
E1=220.0              #Induced emf at 750 rpm(in Volts)
E2_a=250.0            #New emf(in Volts)

#Calculations and Result:
N2_a=(E2_a/E1)*N1
E2_b=250.0
N2_b=600.0
flux_ratio=(E2_b/E1)*(N1/N2_b)

#Result:
print "(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is %d rpm " %(round(N2_a,0))
print "(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is %d percent" %(round(((flux_ratio-1)*100),0))

(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is 852 rpm
(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is 42 percent


## Example 16.5,Page number: 525¶

In [15]:
#Question:
""" Finding the emf induced in the armature."""

#Variable Declaration:
Rsh=110.0             #Resistance of shunt field coil(in Ohms)
Ra=0.02               #Armature resistance(in Ohms)

#Calculations:
Ish=V/Rsh
Ia=I_L+Ish
Eg=V+(Ia*Ra)

#Result:
print "The emf induced in the armature is %.2f V" %(Eg)

The emf induced in the armature is 450.00 V


## Example 16.6,Page number: 525¶

In [16]:
#Question:
""" Finding the total armature current and the generated emf."""

#Variable Declaration:
no_of_lamps=100.0         #Number of lamps
P=60.0                    #Power rating of each lamp(in Watts)
V=200                     #Voltage rating of each lamp(in Volts)
Ra=0.2                    #Armature resistance(in Ohms)
Rsh=50                    #Shunt field resistance(in Ohms)
Poles=4.0                 #Number of poles
no_of_brushes=2.0         #Number of brushes
brush_drop_per_brush=1.0  #Brush drop at each brush(in Volts)

#Calculations:
I1=P/V
I_L=no_of_lamps*I1
Ish=V/Rsh
Ia=Ish+I_L
A=Poles
Ic=Ia/A
brush_drop=no_of_brushes*brush_drop_per_brush
Eg=V+(Ia*Ra)+brush_drop

#Result:
print "The total armature current is %.2f A" %(Ia)
print "The current per path is %.2f A" %(Ic)
print "The generated emf is %.2f V" %(Eg)

The total armature current is 34.00 A
The current per path is 8.50 A
The generated emf is 208.80 V


## Example 16.7,Page number: 525¶

In [3]:
#Question:
"""Finding the emf generated in a compound-wound dc generator."""

#Variable Declaration:
Rsh=130.0                #Shunt field resistance(in Ohms)
Ra=0.1                   #Armature resistance(in Ohms)
Rse=0.1                  #Series field resistance(in Ohms)
no_of_brushes=2          #Number of brushes
brush_drop_per_brush=1.0 #Brush drop at each brush(in Volts)

#Calculations:
Ise=I_L
Vse=Ise*Rse
Vsh=V+Vse
Ish=Vsh/Rsh
Ia=I_L+Ish
brush_drop=no_of_brushes*brush_drop_per_brush
Eg=V+Vse+(Ia*Ra)+brush_drop

#Result:
print "The emf generated is %.2f V." %(Eg)

The emf generated is 272.20 V.


## Example 16.8,Page number: 528¶

In [18]:
#Question:
"""Finding the emf generated,the copper losses,and efficiency of a shunt generator."""

#Variable Declaration:
V=200.0               #Terminal voltage(in Volts)
Ra=0.05               #Armature resistance(in Ohms)
Rsh=50.0              #Shunt field resistance(in Ohms)
loss=1000.0           #Friction losses(in Watts)

#Calculations:
I_L=Po/V
Ish=V/Rsh
Ia=Ish+I_L
Eg=V+(Ia*Ra)
copper_loss=(pow(Ish,2)*Rsh)+(pow(Ia,2)*Ra)
effi=Po/(Po+copper_loss+loss)

#Result:
print "(a)The emf generated is %.2f V" %(Eg)
print "(b)The copper loss is %.2f W" %(copper_loss)
print "(c)The efficiency is %.2f percent" %(effi*100)

(a)The emf generated is 207.70 V
(b)The copper loss is 1985.80 W
(c)The efficiency is 90.95 percent


## Example 16.9,Page number:529¶

In [19]:
#Question:
"""Finding armature resistance and the load-current corresponding to maximum efficiency of dc shunt generator."""

from math import sqrt,pow

#Variable Declaration:
Rsh=52.5               #Shunt field resistance(in Ohms)
stray_loss=710.0       #Stray losses(in Watts)

#Calculations:
Po=V*I_L
Pin=Po/effi
total_loss=Pin-Po
Ish=V/Rsh
Ia=I_L+Ish
sh_copp_loss=pow(Ish,2)*Rsh
const_loss=sh_copp_loss+stray_loss
arma_copp_loss=total_loss-const_loss
Ra=arma_copp_loss/(pow(Ia,2))
Ia_max_effi=sqrt(const_loss/Ra)
I_L_max_effi=Ia_max_effi-Ish

#Result:
print "The armature resistance is %.5f ohm " %(Ra)
print "The load current corresponding to maximum efficiency is %.2f A" %(I_L_max_effi)

The armature resistance is 0.07576 ohm
The load current corresponding to maximum efficiency is 139.04 A


## Example 16.10,Page number: 534¶

In [20]:
#Question:
"""Finding the number of series turns required per pole for a level-compounded generator."""

#Variable Declaration:
sh_turns=1500.0       #Number of turns in the shunt winding

#Calculations:

#Result:
print "The number of series turns required per pole is %d " %(Nse)

The number of series turns required per pole is 30


## Example 16.11,Page number: 536¶

In [21]:
#Question:
"""Finding the back emf generated in a dc shunt motor."""

#Variable Declaration:
Ra=0.1                #Armature resistance(in Ohms)
Rsh=250.0             #Shunt field resistance(in Ohms)

#Calculations:
Ish=V/Rsh
Ia=I_L-Ish
Eb=V-(Ia*Ra)

#Result:
print "The back emf generated in the motor is %.2f V" %(Eb)

The back emf generated in the motor is 246.00 V


## Example 16.12,Page number: 536¶

In [23]:
#Question:
"""Finding the speed of a dc motor."""

#Variable Declaration:
A=2.0                 #Number of parallel paths
P=4.0                 #Number of poles
Z=888.0               #Number of conductors
flux=23e-03           #Flux per pole(in Webers)
Ia=50.0               #Armature current(in Amperes)
Ra=0.28               #Armature resistance(in Ohms)
V=440.0               #Rated voltage(in Volts)

#Calculations:
Eb=V-(Ia*Ra)
N=(60*A*Eb)/(flux*Z*P)

#Result:
print "The speed of the motor is %d rpm" %(round(N,0))

The speed of the motor is 626 rpm


## Example 16.13,Page number: 536¶

In [24]:
#Question:
"""Finding the speed of a dc motor."""

#Variable Declaration:
V1=460.0              #Initial supply voltage(in Volts)
N1=900.0              #Speed of motor at 460-V(in rpm)
V2=200.0              #Final supply voltage(in Volts)

#Calculations:
kflux=V1/N1
N2=V2/(0.7*kflux)

#Result:
ans_N2="The approximate speed of the motor when the motor is connected across a 200V supply is %d rpm" %(round(N2,0))
print(ans_N2)

The approximate speed of the motor when the motor is connected across a 200V supply is 559 rpm


## Example 16.14,Page number: 537¶

In [25]:
#Question:
"""Finding the speed and the gross torque developed by the armature of a dc motor."""

from math import pi

#Variable Declaration:
V=480.0                #Rated voltage(in Volts)
Ia=110.0               #Armature current at rated voltage(in Amperes)
Ra=0.2                 #Armature resistance(in Ohms)
flux=50e-03            #Flux per pole(in Webers)
A=6.0                  #Number of parallel paths
P=6.0                  #Number of poles
Z=864.0                #Number of conductors

#Calculations:
Eb=V-(Ia*Ra)
N=(60*A*Eb)/(flux*Z*P)
torque=((flux*Z)/(2*pi))*(P/A)*Ia

#Result:
print "(a)The speed of the motor is %d rpm " %(round(N,0))
print "(b)The gross torque developed by the armature is %.2f Nm" %(torque)

(a)The speed of the motor is 636 rpm
(b)The gross torque developed by the armature is 756.30 Nm


## Example 16.15,Page number: 538¶

In [26]:
#Question:
"""Finding the power generated in the armature winding of a dc generator."""

from math import pi

#Variable Declaration:
N=900.0               #Operating speed of generator(in rpm)
torque=2e03           #Torque(in N-metre)
P_losses=8e03         #Power losses(in Watts)

#Calculations:
Pin=(2*pi*torque*N)/60.0
Pd=Pin-P_losses

#Result:
print "The power generated in the armature winding is %e W" %(Pd)

The power generated in the armature winding is 1.804956e+05 W


## Example 16.16,Page number: 540¶

In [27]:
#Question:
"""Finding the speed of a series motor when the current changes."""

#Variable Declaration:
V=230.0               #Supply voltage(in Volts)
Ra=0.12               #Armature resistance(in Ohms)
Rse=0.03              #Series field resistance(in Ohms)
Ia1=110.0             #Current at 230 V(in Amperes)
flux1=24e-03          #Flux per pole at 110 A(in Webers)
N1=600.0              #Speed at 230 V(in rpm)
Ia2=50.0              #Armature current(in Amperes)
flux2=16e-03          #Flux per pole at 50 A(in Webers)

#Calculations:
Eb1=V-Ia1*(Ra+Rse)
k=Eb1/(N1*flux1)
Eb2=V-Ia2*(Ra+Rse)
N2=Eb2/(k*flux2)

#Result:
print "The speed of the motor when the currenthas fallen to 50 A is %d rpm" %(round(N2,0))

The speed of the motor when the currenthas fallen to 50 A is 938 rpm


## Example 16.17,Page number: 540¶

In [28]:
#Question:
"""Finding the current drawn by the machine."""

#Variable Declaration:
V=250.0                #Supply voltage(in Volts)
Ra=0.2                 #Armature resistance(in Ohms)

#Calculations:
Eb_1=0
Ia_1=(V-Eb_1)/Ra
Eb_2=200
Ia_2=(V-Eb_2)/Ra
Eb_3=250
Ia_3=(V-Eb_3)/Ra
Eb_4=-250
Ia_4=(V-Eb_4)/Ra

#Result:
print "(a)When the machine is at rest,"
print "The current drawn by the machine is %.2f A" %(Ia_1)
print "(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,"
print "The current drawn by the machine is %.2f A" %(Ia_2)
print "(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,"
print "The current drawn by the machine is %.2f A" %(Ia_3)
print "(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,"
print "The current drawn by the machine is %.2f A" %(Ia_4)

(a)When the machine is at rest,
The current drawn by the machine is 1250.00 A
(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,
The current drawn by the machine is 250.00 A
(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,
The current drawn by the machine is 0.00 A
(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,
The current drawn by the machine is 2500.00 A


## Example 16.18,Page number: 541¶

In [29]:
#Question:
"""Finding the speed and the gross torque developed by the armature of a dc series motor."""

from math import pi

#Variable Declaration:
P=6.0                 #Number of poles
A=6.0                 #Number of parallel paths
Z=864.0               #Number of conductors
flux=50e-03           #Flux per pole(in Webers)
Ia=110.0              #Armature current(in Amperes)
Ra=0.18               #Armature resistance(in Ohms)
Rse=0.02              #Series field resistance(in Ohms)

#Calculations:
Eb=V-Ia*(Ra+Rse)
N=(60*A*Eb)/(flux*Z*P)
torque=(60*Eb*Ia)/(2*pi*N)

#Result:
print "(a)The speed of the motor is %d rpm" %(round(N,0))
print "(b)The gross torque developed by the armature is %.2f Nm " %(torque)

(a)The speed of the motor is 636 rpm
(b)The gross torque developed by the armature is 756.30 Nm


## Example 16.19,Page number: 541¶

In [30]:
#Question:
"""Finding the series resistance to reduce the speed of a shunt motor."""

#Variable Declaration:
V=220.0               #Rated voltage of the motor(in Volts)
Ia=22.0               #Armature current(in Amperes)
Ra=0.45               #Armature resistance(in Ohms)
N1=700.0              #Initial speed of motor(in rpm)
N2=450.0              #Final speed of motor(in rpm)

#Calculations:
E1=V-(Ia*Ra)
E2=(N2/N1)*E1
R=((V-E2)/Ia)-Ra

#Result:
print "The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is %.3f ohm " %(R)

The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is 3.411 ohm


## Example 16.20,Page number: 541¶

In [31]:
#Question:
"""Finding the speed of a dc series motor."""

#Variable Declaration:
V=230.0               #Rated voltage of the dc series motor(in Volts)
Ra=0.2                #Armature resistance(in Ohms)
Rse=0.1               #Series field resistance(in Ohms)
Ia1=40.0              #Line current at rated voltage(in Amperes)
N1=1000.0             #Speed of motor at rated voltage(in rpm)
Ia2=20.0              #Line current at 230 V(in Amperes)

#Calculations:
Eb1=V-Ia1*(Ra+Rse)
Eb2=V-Ia2*(Ra+Rse)
""" Eb=k*flux*N """
N2=(Eb2*N1)/(Eb1*0.6)

#Result:
print "The speed of the motor for a line current of 20A at 230V is %d rpm" %(round(N2,0))

The speed of the motor for a line current of 20A at 230V is 1713 rpm


## Example 16.21,Page number: 543¶

In [36]:
#Question:
"""Finding the terminal voltage of a dc shunt generator."""

#Variable Declaration:
P=4                    #Number of poles
turns=260              #Number of turns in the armature winding
R=0.006                #Resistance of each turn of armature(in Ohms)
flux=0.08              #Useful flux per pole(in Webers)
N=1000                 #Speed of the generator(in rpm)

#Calculations:
Z=2*turns
A=P
Eg=(flux*Z*N*P)/(60.0*A)
Rw=turns*R
R1=Rw/4.0
Ra=R1/4
V=Eg-(I_L*Ra)

#Result:
print "The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is %.3f V." %(V)

The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is 687.971 V.


## Example 16.22,Page number: 544¶

In [43]:
#Question:
"""Finding the armature current,the emf induced and the flux per pole for a dc shunt generator."""

#Variable Declaration:
P=8.0                 #Number of poles
A=2.0                 #Number of parallel paths
Z=778                 #Number of conductors
Ra=0.24               #Armature resistance(in Ohms)
Rsh=250.0             #Shunt field resistance(in Ohms)
N=500                 #Speed of the dc shunt generator(in rpm)

#Calculations:
I_L=V/R_L
Ish=V/Rsh
Ia=I_L+Ish
Eg=V+(Ia*Ra)
flux=(60.0*A*Eg)/(Z*N*P)

#Result:
print "The armature current is %.2f A." %(Ia)
print "The emf induced is %.2f V." %(Eg)
print "The flux per pole is %e Wb." %(flux)

The armature current is 21.00 A.
The emf induced is 255.04 V.
The flux per pole is 9.834447e-03 Wb.


## Example 16.23,Page number: 544¶

In [1]:
#Question:
"""Finding the percentage reduction in speed of dynamo."""

#Variable Declaration:
Po_1=500e03             #Initial power output(in Watts)
Po_2=250e03             #Final power output(in Watts)
V=500.0                   #Constant excitation voltage(in Volts)
Ra=0.015                #Resistance between the terminals of dynamo(in Ohms)

#Calculations:
Ia_1=Po_1/V
E1=V+(Ia_1*Ra)
Ia_2=Po_2/V
E2=V+(Ia_2*Ra)
""" Since excitation emf remains constant in the two cases,we have

E=(flux*Z*N*P)/(60*A) where E=emf generated;Z=number of conductors;N=speed of motor(in rpm);P=number of poles;A=number of parallel paths;
flux=useful flux per pole(in Wb).

N=KE, where K is a constant.

Hence,fractional reduction in speed is given as,

(N1-N2)/N1=((K*(E1-E2))/(K*E1)). """

fract=((E1-E2)/E1)*100

#Result:
print "The percentage reduction in speed of the dynamo is %.3f percent." %(fract)

The percentage reduction in speed of the dynamo is 1.456 percent.


## Example 16.24,Page number: 545¶

In [2]:
#Question:
"""Finding the voltage between the far end of the feeder and the bus-bar of a dc series generator."""

#Variable Declaration:
Rt=0.3                #Resistance of transmission line(in Ohms)
I_L_1=160.0           #Load current in first case(in Amperes)
I_L_2=50.0           #Load current in second case(in Amperes)

#Calculations:
Vt_1=I_L_1*Rt
Vb_1=(50.0/200.0)*I_L_1
Vd_1=Vt_1-Vb_1
Vt_2=I_L_2*Rt
Vb_2=(50.0/200.0)*I_L_2
Vd_2=Vt_2-Vb_2

#Result:
print "(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is %.2f V." %(Vd_1)
print "(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is %.2f V." %(Vd_2)

(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is 8.00 V.
(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is 2.50 V.


## Example 16.25,Page number: 545¶

In [5]:
#Question:
""" Finding the emf generated and the armature current in a dc long-shunt compound generator. """

#Variable Declaration:
V=500                 #Terminal voltage(in Volts)
Ra=0.05               #Armature resistance(in Ohms)
Rse=0.03              #Series field resistance(in Ohms)
Rsh=250               #Shunt field resistance(in Ohms)
brush_drop=1.0        #Brush contact drop(in Volts)

#Calculations:
Ish=V/Rsh
Ia=Ish+I_L
Eg=V+(Ia*(Ra+Rse))+brush_drop

#Result:
print "The armature current is %.2f A." %(Ia)
print "The emf generated is %.2f V" %(Eg)

The armature current is 52.00 A.
The emf generated is 505.16 V


## Example 16.26,Page number: 545¶

In [8]:
#Question:
""" Finding the voltage and the power generated by a dc generator."""

#Variable Declaration:
P=8                   #Number of poles
Z=500                 #Number of conductors on the armature
flux =0.02            #Magnetic flux per pole(in Webers)
N=1800                #Speed of the generator(in rpm)
I=5.0                 #Allowable current per path(in Amperes)

#Calculations:
A_1=2
Eg_1=(flux*Z*N*P)/(60*A_1)
A_2=P
Eg_2=(flux*Z*N*P)/(60*A_2)
Ia_1=A_1*I
Pd_1=Eg_1*Ia_1
Ia_2=A_2*I
Pd_2=Eg_2*Ia_2

#Result:
print "When the armature is wave wound:"
print "(a)The generated voltage is %.2f V." %(Eg_1)
print "(b)The kW generated by the machine is %.2f kW." %(Pd_1/1000.0)
print "\nWhen the armature is lap wound:"
print "(a)The generated voltage is %.2f V." %(Eg_2)
print "(b)The kW generated by the machine is %.2f kW." %(Pd_2/1000.0)

When the armature is wave wound:
(a)The generated voltage is 1200.00 V.
(b)The kW generated by the machine is 12.00 kW.

When the armature is lap wound:
(a)The generated voltage is 300.00 V.
(b)The kW generated by the machine is 12.00 kW.


## Example 16.27,Page number: 546¶

In [15]:
#Question:
"""Finding the emf generated and the copper losses in a dc shunt generator."""

#Variable Declaration:
V=250.0               #Terminal voltage(in Volts)
Ra=0.02               #Armature resistance(in Ohms)
Rsh=50.0              #Shunt-field resistance(in Ohms)
loss=950.0            #Iron and frictional losses(in Watts)

#Calculations:
Ish=V/Rsh
Ia=I_L+Ish
Eg=V+(Ia*Ra)
copp=(Ia*Ia*Ra)+(V*Ish)
Po=V*I_L
tot_loss=copp+loss
Pin=Po+tot_loss
Pe=Pin-loss
mech_effi=Pe/Pin
ele_effi=Po/Pe
comm_effi=Po/Pin

#Result:
print "(a)The emf generated is %.2f V." %(Eg)
print "(b)The copper losses is %.2f W." %(copp)
print "(c)The output of the prime mover is %.3f kW." %(Pin/1000.0)
print "(d)The commercial efficiency is %.2f.\n   The mechanical efficiency is %.2f." %((comm_effi*100),(mech_effi*100))
print "   The electrical efficiency is %.2f." %(ele_effi*100)

(a)The emf generated is 254.00 V.
(b)The copper losses is 2050.00 W.
(c)The output of the prime mover is 51.750 kW.
(d)The commercial efficiency is 94.20.
The mechanical efficiency is 98.16.
The electrical efficiency is 95.96.


## Example 16.28,Page number: 547¶

In [13]:
#Question:
"""Finding the back emf generated by a dc shunt motor."""

#Variable Declaration:
V=250.0               #Terminal Voltage(in Volts)
Ra=0.2                #Armature resistance(in Ohms)
Rsh=250.0             #Field resistance(in Ohms)

#Calculations:
Ish=V/Rsh
Ia1=I_L1-Ish
E1=V-(Ia1*Ra)
Ia2=I_L2-Ish
E2=V-(Ia2*Ra)
"""As the motor is shunt-wound,the flux remains constant.The emf generated is directly proportional to the speed."""
N2=(E2/E1)*N1
speed_drop=(N1-N2)/N1

#Result:
print "(a)The back emf generated at no-load is %.3f V." %(E1)
print "(b)On loading,\n   The back emf generated is %.2f V.\n   The speed of the motor is %d rpm." %(E2,round(N2,0))
print "   The percentage speed drop is %.3f percent." %(speed_drop*100)

(a)The back emf generated at no-load is 249.800 V.
The back emf generated is 240.00 V.
The speed of the motor is 961 rpm.
The percentage speed drop is 3.923 percent.


## Example 16.29,Page number: 547¶

In [6]:
#Question:
"""Finding the value of starting resistance for a shunt motor."""

#Variable Declaration:
Po=14920.0            #Output power(in Watts)
V=240.0               #Supply voltage(in Volts)
Ra=0.25               #Armature resistance(in Ohms)

#Calculations:
Pin=Po/effi
I_L=Pin/V
Ist=1.5*I_L
Rt=V/Ist
Rst=Rt-Ra

#Result:
print "The starting resistance for the shunt motor is %.3f Ohms." %(Rst)

The starting resistance for the shunt motor is 1.963 Ohms.


## Example 16.30,Page number: 548¶

In [9]:
#Question:
"""Finding the speed and efficiency of a dc shunt motor."""

#Variable Declaration:
V=500.0               #Voltage rating of the dc shunt motor(in Volts)
Ra=0.2                #Armature resistance(in Ohms)
Ish=1.0               #Field current(in Amperes)

#Calculations:
Ia1=I_L1-Ish
E1=V-(Ia1*Ra)
Ia2=I_L2-Ish
E2=V-(Ia2*Ra)
""" For a shunt motor,the flux remains constant and hence E is directly proportional to speed of the motor(N).

E=kN where k is a constant. """
N2=(E2/E1)*N1
"""At no-load,the power taken by the motor mainly meets the constant losses(iron and frictional losses)."""
Pc=V*I_L1
"""On loading,the copper loss in shunt field winding is negligible compared to the copper loss in armature winding."""
Pv=Ia2*Ia2*Ra
Pin=V*I_L2
effi=(Pin-(Pv+Pc))/Pin

#Result:
print "The speed of the dc shunt motor on loading is %d rpm." %(round(N2,0))
print "The efficiency of the motor is %.2f percent." %(effi*100)

The speed of the dc shunt motor on loading is 962 rpm.
The efficiency of the motor is 92.08 percent.


## Example 16.31,Page number: 548¶

In [8]:
#Question:
"""Finding the speed of a dc generator running as a shunt motor."""

#Variable Declaration:
Ra=0.02               #Armature resistance(in Ohms)
Rsh=50.0              #Shunt-field resistance(in Ohms)
V=250.0               #Terminal voltage(in Volts)
Po=50e03              #Output power(in Watts)
N1=500.0              #Speed of the dc generator(in rpm)

#Calculations:
Ish=V/Rsh
""" When working as a generator,the machine supplies a load of 50 kW at 250 V. """
I_L=Po/V
Ia1=I_L+Ish
E1=V+(Ia1*Ra)
"""When working as a motor,the machine takes a power of 50 kW at 250 V. """
Ia2=I_L-Ish
E2=V-(Ia2*Ra)
N2=(E2/E1)*N1
""" NOTE: The field current and the flux per pole is same in both cases."""

#Result:
print "The speed of the machine running as a shunt motor is %d rpm." %(round(N2,0))

The speed of the machine running as a shunt motor is 484 rpm.


## Example 16.32,Page number: 548¶

In [16]:
#Question:
"""Finding the applied voltage and the current to run the motor."""

#Variable Declaration:
Ra=0.6                #Resistance of the armature(in Ohms)
Rse=0.4               #Series field resistance(in Ohms)
Ia1=20.0              #Initial armature current(in Amperes)
V1=400.0              #Initial terminal voltage(in Volts)
N1=250.0              #Initial speed of the motor(in rpm)
N2=350.0              #Final speed of the motor(in rpm)

#Calculations:
""" In a series motor,torque is directly proprtional to the square of the armature current(Ia).

Given: Torque is directly proprtional to the square of the speed(N).

Therefore, Ia is directly proportional to N. Ia=kN where k is a constant. """
Ia2=(N2/N1)*Ia1
E1=V1-(Ia1*(Ra+Rse))
"""In a series motor,as the flux is directly proportional to Ia,the back emf is proportional to (Ia*N). """
E2=E1*((Ia2*N2)/(Ia1*N1))
V2=E2+(Ia2*(Ra+Rse))

#Result:
print "The applied voltage is %.2f V and the current is %.2f A to run the motor at 350 rpm." %(V2,Ia2)

The applied voltage is 772.80 V and the current is 28.00 A to run the motor at 350 rpm.