# Chapter 2 : Direct Current Machines¶

## Example 2.4 Page No : 92¶

In [1]:
# Given Data
P = 2.         #number of poles
Z = 400.        #number of conducters
n = 300.        #speed in rpm
E = 200.         #voltage of generator
A = 2.           #number of parallel paths
N = 1200.        #number of turns in each field coil

# Calculations and Results
phi = (E*60*A)/(Z*n*P)   #flux at the end of 0.15sec
t = 0.15                     #time
print "magnitude of flux at the end of 15sec is %f wb"%(phi)
e = N*(phi/t)
print "induced emf in the field coil =  %d volts"%(e)

magnitude of flux at the end of 15sec is 0.100000 wb
induced emf in the field coil =  800 volts


## Example 2.6 Page No : 93¶

In [2]:
# Given Data
P = 8.            #number of poles
A = 8.            #number of parallel paths in the armature
Z = 960.          #number of conductors
N = 400.          #speed in rpm
phi = 0.04       #flux per pole

# Calculations
E = (phi*Z*N*P)/(60*A)        #emf generated onopen circuit condition

# Results
print "emf generated on open circuit condition, E = %d volts"%(E)

emf generated on open circuit condition, E = 256 volts


## Example 2.7 Page No : 97¶

In [3]:
# Given Data
E = 180.;                #induced emf at 500rpm
N = 500.;                #speed in rpm

# Calculations and Results
K1 = (E/N)
print "K1 = %f"%(K1)
E1 = (K1*600)           #induced emf at 600rpm
print " induced emf at 600rpm is = %d V"%(E1)

K1 = 0.360000
induced emf at 600rpm is = 216 V


## Example 2.8 Page No : 97¶

In [4]:
# Given Data
E1 = 220.;           #induced emf at N1 speed in volts
N1 = 750.;          # speed
K1 = (E1/N1)
E2 = 250.;          #induced emf at speed N2
N2 = E2/K1
print "speed at induced emf of 250V  = %d rpm"%(N2)
print ("when induced emf is 250V and speed 700 rpm")
E3 = 250.;            #induced emf at N3 speed
N3 = 700.;            #speed
ratio = (E3*N1)/(E1*N3)
Pi = (ratio-1)*100
print "percentage increase in flux is %f percent"%(Pi)

speed at induced emf of 250V  = 852 rpm
when induced emf is 250V and speed 700 rpm
percentage increase in flux is 21.753247 percent


## Example 2.9 Page No : 98¶

In [5]:
# Given Data
E = 200.      #emf induced
I = 15.        #armature current
n = 1200.          #speed in rpm

# Calculations and Results
omega = (2*3.14*n)/60;
print "omega = %f "%(omega)
T = (E*I)/omega;
print "electromagnetic torque = %f Nm"%(T)

omega = 125.600000
electromagnetic torque = 23.885350 Nm


## Example 2.10 Page No : 98¶

In [6]:
# Given Data
n = 10.;             #number of turns in 1 coil
l = 0.2;
d = 0.2;          #diameter in metres
B = 1.;              #uniform magnetic field density in weber per m**2
N = 1500.;            #speed in rpm

# Calculations and Results
r = (d/2);           #radius in metres
E = (B*l*((2*3.14*N)/60)*r*2*n);
print "total induced emf = %f V"%(E)
R = 4;                #total resistance in ohms
I = E/R;
print "The current through the armature coil when connected to the load, I = %f A"%(I)
T = (E*I)/((2*3.14*N)/60)
print "torque = %f Nm"%(T)

total induced emf = 62.800000 V
The current through the armature coil when connected to the load, I = 15.700000 A
torque = 6.280000 Nm


## Example 2.11 Page No : 99¶

In [7]:
# Given Data
V = 230.;            #armature voltage supply in volts
Ia = 12.;            #armature current in amperes
Ra = 0.8;            #armature resistance in ohms
N = 100.;               #speed in radian per second

# Calculations and Results
E = (V-(Ia*Ra))
print "induced emf, E = %fV"%(E)
Te = (E*Ia)/N
print "the electromagnetic torque = %fNm"%(Te)
Pi = V*Ia
print "electrical input to the armature, Pinput =  %dW"%(Pi)
Pd = Te*N
print "mechanical developed = %fW"%(Pd)
loss = (Ia**2*Ra)
print "armature copper loss = %fW"%(loss)

induced emf, E = 220.400000V
the electromagnetic torque = 26.448000Nm
electrical input to the armature, Pinput =  2760W
mechanical developed = 2644.800000W
armature copper loss = 115.200000W


## Example 2.13 Page No : 101¶

In [10]:
# Given Data
P = 50000.;            #power delivered in watts
V = 250.;              #voltage in volts
Ra = 0.02;            #armature resistance in ohms
Rf = 50.;              #field resistance in ohms

# Calculations and Results
If = V/Rf             #field current in amperes
Ng = 400.;             #speed in generating condition in rpm
print "field current, If = %dA"%(If)
Il = P/V               #load current in amperes
print "Load current, If = %dA"%(Il)
Ia = If+Il             #armature current in amperes
print "Aramture current, If = %dA"%(Ia)
Eg = (V+(Ia*Ra))
print ("At motor condition")
Ia = (Il-If)
print "Aramture current, If = %dA"%(Ia)
Em = (V-(Ia*Ra))
print "Em = %fV"%(Em)
Nm = (Ng*Em)/Eg
print "Speed of the motor = %drpm"%(Nm)

field current, If = 5A
Aramture current, If = 205A
At motor condition
Aramture current, If = 195A
Em = 246.100000V
Speed of the motor = 387rpm


## Example 2.14 Page No : 101¶

In [11]:
# Given Data
V = 250.;                #voltage supply in volts
Ra = 0.12;              #armature resistance in ohms
Rf = 100.;                #field resistance in ohms
Il = 80.;                #load current in amperes

# Calculations and Results
If = V/Rf
print "Field current, If = %f"%(If)
print ("When machine is generating")
Ia = Il+If
Eg = (V+(Ia*Ra))
print "Ia = %fA"%(Ia)
print "Eg = %fV"%(Eg)
print ("When machine is motoring")
Ia = Il-If
Em = (V-(Ia*Ra))
print "Ia = %fA"%(Ia)
print "Eg = %fV"%(Em)
ratio = Eg/Em
print "Ratio of speeds = %f"%(ratio)

Field current, If = 2.500000
When machine is generating
Ia = 82.500000A
Eg = 259.900000V
When machine is motoring
Ia = 77.500000A
Eg = 240.700000V
Ratio of speeds = 1.079767


## Example 2.15 Page No : 102¶

In [12]:
# Given Data
V = 550.;            #voltage supply in volts
P = 16.;            #number of poles
N = 150.;             #speed in rpm
Z = 2500.;            #number of armature conductors
A = 16.;
Power = 1500000.;        #power in watt
Cl = 25000.;              #full-load copper loss
B = 0.9;                #flux density in the pole

# Calculations and Results
Ia = Power/V
print "Full load current = %fA"%(Ia)
Ra = Cl/(Ia**2)
print "Ra = %fohms"%(Ra)
E = V+(Ia*Ra)
print "Induced emf = %fvolts"%(E)
phi = (E*60*A)/(Z*N*P)
print "flux density = %fWb/m**2"%(B)
print "flux = %fWb"%(phi)
area = (phi/B)
print " Area of pole shoe = %fcm**2"%(area*10000)

Full load current = 2727.272727A
Ra = 0.003361ohms
Induced emf = 559.166667volts
flux density = 0.900000Wb/m**2
flux = 0.089467Wb
Area of pole shoe = 994.074074cm**2


## Example 2.16 Page No : 103¶

In [13]:
# Given Data
Cd = 0.76;            #commutator diameter in metres
Cr = .38;              #commutator radius in metres
bw = 1.5*10**(-2);        #brush width in metres
N = 600.;                #speed in rpm
n = 10.;                #speed in rps

# Calculations and Results
V = Cr*(2*3.14*n);
print "peripheral speed of commutator, V = %fm/sec"%(V);
Tc = bw/V;
print "Time of commutation = %fseconds"%(Tc)

peripheral speed of commutator, V = 23.864000m/sec
Time of commutation = 0.000629seconds


## Example 2.17 Page No : 123¶

In [14]:
# Given Data
V = 240.;            #supply voltage in volts
N = 800.;            #speed in rpm
Ia = 2.;              #armeture current in amperes
Ra = 0.4;            #armature resistance in ohms
Rf = 160.;            #field resistance in ohms
Il1 = 30.;            #line current in amperes

# Calculations and Results
E = V-(Ia*Ra);       #induced emf in volts
print "E = %fV"%(E)
If = V/Rf;             #field current in amperes
print "If = %fA"%(If)
K1 = E/(If*N);
print "K1 = %f"%(K1)
print ("At a load of 30A")
Ia1 = (Il1-If);
E1 = V-(Ia1*Ra);
N1 = 950;               #speed in rpm
If1 = E1/(K1*N1);
print "If1 = %fA"%(If1);
Rr = V/If1;
R = (Rr-Rf);
print "Extra resistance required in the field circuit, R = %fohms"%(R)

At no-load
E = 239.200000V
If = 1.500000A
K1 = 0.199333
If1 = 1.207182A
Extra resistance required in the field circuit, R = 38.810149ohms


## Example 2.18 Page No : 124¶

In [15]:
# Given Data
V = 230.;                    #voltage supply in volts
Ia = 20.;                    #armature current in amperes
Ra = 0.5;                    #armature resistance in ohms

# Calculations and Results
E = V-(Ia*Ra);
print "E = %dV"%(E)
print ("when extra resistance is added in the armature circuit,the speed is halved")
E2 = E/2;
R = ((V-E2)/Ia)-Ra;
print ("The load torque is conmath.atant")
print "extra resistance in the armature circui, R = %fohms"%(R)
print ("The load torque directly proportional to square of speed")
print ("if N is halfed, Iais one-fourthed")
Ia2 = Ia/4;
R = ((V-E2)/Ia2)-Ra;
print "extra resistance in the armature circui, R = %fohms"%(R)

E = 220V
when extra resistance is added in the armature circuit,the speed is halved
extra resistance in the armature circui, R = 5.500000ohms
The load torque directly proportional to square of speed
if N is halfed, Iais one-fourthed
extra resistance in the armature circui, R = 23.500000ohms


## Example 2.19 Page No : 125¶

In [17]:
# Given Data
V = 250.;                    #voltage supply in volts
Ia = 50.;                    #armature current in amperes
Ra = 0.3;                    #armature resistance in ohms
N = 1000.;

# Calculations and Results
E = V-(Ia*Ra);
print "E = %dV"%(E)
print ("when extra resistance is added in the armature circuit when the speed is 800rpm")
N2 = 800.;
E2 = (E*N2)/N;
print "E at 800rpm = %dV"%(E2)
R = ((V-E2)/Ia)-Ra;
print "extra resistance in the armature circui, R = %fohms"%(R)
print ("if load is halfed,Ia will be halfed")
Ia2 = Ia/2;
E1 = V-(Ia2*(Ra+R));
print "E1 = %dV"%(E1)
N1 = (N2*E1)/E2;
print "N1 = %frpm"%(N1)

E = 235V
when extra resistance is added in the armature circuit when the speed is 800rpm
E at 800rpm = 188V
extra resistance in the armature circui, R = 0.940000ohms
if load is halfed,Ia will be halfed
E1 = 219V
N1 = 931.914894rpm


## Example 2.20 Page No : 125¶

In [18]:
# Given Data
Il = 5.;            #current in amperes al no-load
V = 250.;            #voltage in volts
Rf = 250.;            #field resistance in ohms

# Calculations and Results
If1 = V/Rf;            #field current in amperes
Ia1 = Il-If1;            #armature current
Ra = 0.2;                    #armature resistance in ohms
print ("at a load current of 50A")
Il2 = 50;                #load current in amperes
#armature reaction weakens by 3percent
If2 = 0.97;                    #current in amperes
Ia2 = Il2-If2;
N1 = 1000;
E1 = (V-(Ia1*Ra));
E2 = (V-(Ia2*Ra));
N2 = (N1*E2)/(0.97*E1);
print "N2 = %frpm"%(N2)

at a load current of 50A
N2 = 993.670467rpm


## Example 2.21 Page No : 126¶

In [19]:
# Given Data
P = 4;                  #pole
V = 500;                #shunt motor  in volts
Ia = 60;                    #armature current in amperes
Ra = 0.2;                   #armature resistance in ohms

# Calculations and Results
E = V-(Ia*Ra)-2;
print "voltage drop across each brush = %fV"%(E)
phi = 0.03;                 #flux per pole in Wb
Z = 720.;                    #total armature current in volts
A = 2;
N = (E*60*A)/(phi*Z*P)
print "full load speed of the motor = %frpm"%(N)

voltage drop across each brush = 486.000000V
full load speed of the motor = 675.000000rpm


## Example 2.22 Page No : 126¶

In [21]:
import math
# Given Data
V = 440;                #primary voltage in volts
Ia = 50;                #armature current in amperes
Ra = 0.2;                #armature resistance in ohms
N = 600;                  #speed in rpm
E = V-(Ia*Ra);           #emf induced  in volts before adding extra resistance
#E = K*phi*N = K1*Ia*N
K1 = E/(Ia*N);

# Calculations and Results
#we have the relation T = Kt1*Ia**2, T1 = Kt1*Ia1**2
#when torque is half, say torque be T1
#T1 = T/2. r = T/T1
r = 2;
Ia1 = math.sqrt(Ia**2/r);
print "Ia1 = %fA"%(Ia1);
#extra resistance R is introduced in the circuit
N1 = 400;
E1 = (K1*Ia1*N1);
R = ((V-E1)/Ia1)-Ra;
print "value of extra resistance added = %fohms"%(R)

Ia1 = 35.355339A
value of extra resistance added = 6.511746ohms


## Example 2.23 Page No : 127¶

In [22]:
# Given Data
V = 200.;                        #voltage in volts
Ia = 20.;                        #armature current in amperes
Ra = 0.5;                        #armature resistance in ohms
Rse = 0.2;                        #field winding resistance in ohms

# Calculations and Results
E = V-(Ia*(Ra+Rse));
print "In first case, E = %fV"%(E)
#E = k*phi*N
N = 1000;                        #speed in rpm
Kphi = E/N;
#a resistance R is connected in parallel with the series field which is called diverter
print ("when resistace R is added and new conditions")
I = 20;                        #total current flowing
#current is equally devided between series field and diverter
Ise2 = I/2;
#flux at 10A current is 20percent of flux at 20A current
p = 0.70;                #percentage of flux
Kpih1 = p*Kphi;
E1 = (V-((Ia*Ra)+(Ise2*Rse)));
print "Induced emf = %fV"%(E1)
#new speed is N1
N1 = E1/(p*Kphi)
print "N1 = %frpm"%(N1)

In first case, E = 186.000000V
when resistace R is added and new conditions
Induced emf = 188.000000V
N1 = 1443.932412rpm


## Example 2.24 Page No : 128¶

In [23]:
# Given Data
V = 200.;                             #motor runs in volts
Ia = 15.;                                  #current taken in amperes
Ra = 1.;                                #motor resistance in ohms

# Calculations and Results
E1 = V-(Ia*Ra);
print "resistance when 1ohm = %fV"%(E1)
R = 5;                                   #resistance
E2 = V-(Ia*(Ra+R))
print "resistance when 5ohms connected in series = %fV"%(E2)
N1 = 800;                                #speed of motor in rpm
N2 = N1*(E2/E1);
print "speed at which motor will run when resistance is 5ohms = %frpm"%(N2)

resistance when 1ohm = 185.000000V
resistance when 5ohms connected in series = 110.000000V
speed at which motor will run when resistance is 5ohms = 475.675676rpm


## Example 2.25 Page No : 135¶

In [24]:
# Given Data
P = 8.;                          #pole
Z = 107.;                        #generator with slots
Ia = 1000.;                    #current containing in amperes
Bag = 0.32;                      #gap flux density in Wb/m**2
lg = 0.012;                          #interpole air gap in meters
pi = 3.14;

# Calculations
Mu = (4*pi*10**-7)
AT = (((Ia*Z)/(2*P))+((Bag*lg)/Mu));

# Results
print "current for each commutating pole = %f"%(AT)

current for each commutating pole = 9744.824841


## Example 2.26 Page No : 135¶

In [25]:
# Given Data
Bag = 0.3;                                  #flux density in the interpole air gap in Wb/m**2

# Calculations and Results
Ia = 200000./200;                        #armature current in amperes
print "Armature current = %f"%(Ia)
Z = 540.;                          #Number of armature conductors
Zt = 540./2;                            #Number armature winding turns
print "Number armature winding turns = %f"%(Zt)
A = 6.;                             #the winding lap
Ap = Zt/A;                        #Number of armature turns per parallel path
print "Number of armature turns per parallel path = %f"%(Ap)
P = 6;                              #pole
Np = ((Ia*Ap)/P);
print "Number of armature ampere turns per pole = %f"%(Np)
lg = 0.01;                              #inter pole air gap in meters
pi = 3.14;
Mu = (4*pi*10**-7)
Nipg = ((Bag*lg)/Mu);                          #Air gap
print "ampere turns for the air gap = %f"%(Nipg)
NipI = (Np+Nipg);                                #total interpole ampere
print "Total interpole ampere turns = %f"%(NipI)
Nip = (NipI/Ia);
print "Number of turns needed on each commutating pole = %f"%(Nip)

Armature current = 1000.000000
Number armature winding turns = 270.000000
Number of armature turns per parallel path = 45.000000
Number of armature ampere turns per pole = 7500.000000
ampere turns for the air gap = 2388.535032
Total interpole ampere turns = 9888.535032
Number of turns needed on each commutating pole = 9.888535


## Example 2.27 Page No : 128¶

In [26]:
# Given Data
N = 960.;                           #speed in rpm
F = 23.;                            #effictive load in kgf

# Calculations and Results
r = 45./2;                               #radius of the drum
print "radius of the drum = %fcm"%(r)
pi = 3.14;
OP = (2*pi*N*F*r*9.81)/(60*100);
print "output power = %fW"%(OP)

Vi = 230.;                  #motor input in volts
Ci = 28.;                       #input current in amperes
IP = (Vi*Ci);
print "input power  = %fW"%(IP)
Effi = (OP/IP)*100;
print "Efficiency of the motor = %fpercent"%(Effi)

radius of the drum = 22.500000cm
output power = 5101.043040W
input power  = 6440.000000W
Efficiency of the motor = 79.208743percent


## Example 2.29 Page No : 145¶

In [27]:
# Given Data
I = 440.;                      #input at no-load in watt
V = 220.;                        #voltage in volts
Ic = I/V;                      #input current at no-load in amperes
i = 1;                    #input current in amperes
A = 2;                      #current in amperes
C = A-i;                    #armature current at no-load in amperes
L = I-((((C)**2)*0.5)+(V*C));                #iron,friction and windage losses in watt
a = 40;                         #motor current in amperes
OP = (V*a);
Ra = 0.5;

# Calculations and Results
Effi = (OP*100)/(OP+(((a+i)**2)*Ra)+(V*i)+L)
print "Efficiency as a generator when delivering 40A at 220V = %fpercent"%(Effi)
Eff = ((OP-(((a-i)**2)*Ra)-(V*C)-L)/OP)*100;
print "Efficiency as a motor when taking 40A from at 220V = %fpercent"%(Eff)

Efficiency as a generator when delivering 40A at 220V = 87.301587percent
Efficiency as a motor when taking 40A from at 220V = 86.363636percent


## Example 2.30 Page No : 147¶

In [29]:
# Given Data
V = 400.;                            #motor in volts
Rf = 200.;                            #field resistance in ohms
If = V/Rf;                          #current in amperes
i = 5;                             #current at no load in amperes
IP = V*i;                            #motor input at no load
Ia = 3;                             #aramture current in amperes
Ra = 0.5;                           #armature resistance in ohms

# Calculations and Results
L = IP-(((Ia)**2)*Ra)-(V*If);                     #iron,friction and windage in losses in watt
print "iron, friction and windage in losses = %fW"%(L)
At = 50.;                                        #armature total current in amperes
A = At-2;                                       #armature current in amperes
Ls = (((A)**2)*Ra)+(V*If)+L;                      #Losses
Eff = (((V*At)-Ls)/(V*At))*100;
print "Efficiency of full load = %fpercent"%(Eff)
#flux is consmath.tant
E1 = V-(Ia*Ra);                           #induced emf in the armature at no load
E2 = V-(A*Ra);                             #induced emf in the armature at full load
# math.since N1/N2 = E1/E2

iron, friction and windage in losses = 1195.500000W
Efficiency of full load = 84.262500percent
Percentage change in speed from no load to full load = 5.646173percent


## Example 2.31 Page No : 148¶

In [30]:
# Given Data
Ra = 0.5;                #armature resistance in ohms
Rf = 750.;               #field circuit resistance in ohms
V = 500.;                       #voltage in volts

# Calculations
If = V/Rf;                          #current in amperes
l = 3.;                          #line current in amperes
i = 2.33;                          #current in motor in amperes
I = 0.67;                         #current i amperes
L = (V*l)-(((i)**2)*Ra)-(V*I);                         #Iron,friction and windage losses
O = 20.;                               #generator
OP = (O*1000)/V;                #output current of the generator under loaded condition in amperes
Ia = I+OP;            #output in amperes
Effi = (O*1000*100)/((O*1000)+(((Ia)**2)*Ra)+(V*I)+L);

# Results
print "efficiency of the machine = %fpercent"%(Effi)

efficiency of the machine = 89.588435percent


## Example 2.32 Page No : 149¶

In [31]:
# Given Data
Ig = 25.;               #current of generator in amperes
I = 30.;                   #current in motor in amperes
Il = I-Ig;              #current in amperes
Ra = 0.25;                #resistance in ohms
Gl = ((Ig)**2)*Ra;                #loss in generator in watt
M = ((I)**2)*Ra;                    #loss in motor in watt
T = Gl+M;                   #total loss in watt
V = 100.;             #voltage in volts
P = V*Il;               #power supplied from mains in watt
L = P-T;                  #iron,friction and windages losses in the two machines in ohms
l = L/2;                   #iron,friction and windages losses in each machines in ohms
IP = I*V;                    #input

# Calculations and Results
Eff = ((IP-M-l)/IP)*100;
print "Efficiency of the motor = %fpercent"%(Eff)
OP = Ig*V;                 #output
Effi = ((OP)/(OP+Gl+l))*100;
print "Efficiency of the generator = %fpercent"%(Effi)

Efficiency of the motor = 90.520833percent
Efficiency of the generator = 92.059839percent


## Example 2.33 Page No : 150¶

In [32]:
# Given Data
V = 440.;                    #voltage in volts
P = 200.*1000;               #power in watt
Ig = P/V;              #rated current of each machine in amperes

# Calculations
#assume losses to be equal
I = 90;              #addition currnet supply
Effi = math.sqrt(Ig/(Ig+I))*100;

# Results
print "approximate efficiency = %fpercent"%(Effi)

approximate efficiency = 91.363261percent


## Example 2.34 Page No : 150¶

In [33]:
# Given Data
Ig = 2000.;                             #output current of generator in amperes
I = 380.;                               #Input current from supply mains in amperes

# Calculations and Results
Effi = math.sqrt(Ig/(Ig+I))*100;                  #Efficiency of generator assuming equal efficiencies of the two machines
print "Efficiences of the generator at full load assuming equal efficiencies = %fpercent"%(Effi)
S = 22.;                            #Shunt field current of generator
G = Ig+S;                        #Armature current of generator in amperes
R = 0.01;                               #resistance of the armature circuit of each machine in ohms
Gc = ((G)**2)*R;                          #copper loss in arrmature circuit of generator in W
V = 500.;                                #Voltage in volts
L = V*S;                              #loss in the field circuit of the generator in W
T = Ig+I;                            #total current suuply in amperes
Sf = 17.;                                        #shunt field current of motor in amperes
A = T-Sf;                              #armature current in motor in amperes
Lc = ((A)**2)*R;                        #loss in armature circuit of motor in amperes
Lf = V*Sf;                                 #loss in the shunt field circuit of motor in W
Tin = V*I;                      #total input to motor and generator in W
Ml = Tin-(Gc+L+Lc+Lf);                     #iron,friction and windage loss in both machines in W
Me = Ml/2;                                   #iron,friction and windage loss in each machine in W
p = 1000.;                     #power in kW
OP = (Ig*V)/p;                        #full load output of the generator
Eff = (p*100)/(p+((Gc+L+Me)/1000));
print "Efficiency of the generator at full load = %fpercent"%(Eff)

Efficiences of the generator at full load assuming equal efficiencies = 91.669850percent
Efficiency of the generator at full load = 91.846461percent