CHAPTER 5: DIRECT CURRENT MOTORS

Example 5.1, Page number 182

In [1]:
#Variable declaration
l = 10.0    #Conductor length(m)
B = 0.56    #Magnetic flux density(T)
I = 2.0     #Current through conductor(A)

#Calculation
F = B*I*l   #Magnitude of force(N)

#Result
print('Magnitude of force , F = %.1f N' %F)

Magnitude of force , F = 11.2 N

Example 5.2, Page number 189

In [1]:
#Variable declaration
I = 20.0         #Total current(A)
V_t = 250.0      #Supply voltage(V) 
R_sh = 200.0     #Shunt field resistance(ohm)
R_a = 0.3        #Armature resistance(ohm)

#Calculation
I_sh = V_t/R_sh     #Shunt field current(A)
I_a = I-I_sh        #Armature current(A)
#For case(i)
E_b = V_t-R_a*I_a   #Back emf(V)
#For case(ii)
P_md = E_b*I_a      #Mechanical power developed(W) 

#Result
print('(i)  Value of back emf , E_b = %.1f V' %E_b) 
print('(ii) Mechanical power developed in the armature motor , P_md = %.1f W' %P_md)
print('\nNOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A')

(i)  Value of back emf , E_b = 244.4 V
(ii) Mechanical power developed in the armature motor , P_md = 4582.0 W

NOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A

Example 5.3, Page number 189

In [1]:
#Variable declaration
R_a = 0.7       #Armature circuit resistance(ohm)
V_t = 5.0       #Applied voltage(V)
I_anl = 5.0     #No-load armature current(A)
I_afl = 35.0    #Full-load armature current(A)

#Calculation
E_bnl = V_t-R_a*I_anl   #Back emf under no-load(V)
E_bfl = V_t-R_a*I_afl   #Back emf under full-load(V)
E_bc = E_bnl-E_bfl      #Change in back emf from no-load to full load(V)
 
#Result
print('Change in back emf from no-load to full load , E_bc = %.f V' %E_bc)

Change in back emf from no-load to full load , E_bc = 21 V

Example 5.4, Page number 191

In [1]:
#Variable declaration
I = 40.0        #Current(A)
V_t = 230.0     #Supply voltage(V)
N = 1100.0      #Speed(rpm)
R_a = 0.25      #Armature resistance(ohm)
R_sh = 230.0    #Shunt field resistance(ohm) 

#Calculation
I_sh = V_t/R_sh         #Shunt field current(A)
I_a = I-I_sh            #Armature current(A)
E_b = V_t-I_a*R_a       #Back emf(V) 
T_a = 9.55*E_b*I_a/N    #Armature torque(N-m)
 
#Result
print('Torque developed by the armature , T_a = %.2f N-m' %T_a)

Torque developed by the armature , T_a = 74.57 N-m

Example 5.5, Page number 191-192

In [1]:
#Variable declaration
P = 6.0             #Number of poles
V_t = 230.0         #Supply voltage to shunt motor(V)
Z = 450.0           #Number of conductors
R_a = 0.8           #Armature resistance(ohm)
I = 30.0            #Current drawn from supply(A)
P_0 = 5560.0        #Output power(W)
I_f = 3.0           #Current through field winding(A)
phi = 25*10**-3     #Flux per pole(Wb)

#Calculation
A = P                    #Number of parallel paths in lap winding
I_a = I-I_f              #Armature current(A)
E_b = V_t-I_a*R_a        #Back emf(V)
N = 60*A*E_b/(P*Z*phi)   #Speed(rpm)
T_sh = 9.55*P_0/N        #Shaft torque(N-m)

#Result
print('Speed , N = %.1f rpm' %N)
print('Shaft torque , T_sh = %.1f N-m' %T_sh)

Speed , N = 1111.5 rpm
Shaft torque , T_sh = 47.8 N-m

Example 5.6, Page number 193-194

In [1]:
#Variable declaration
I_Lnl = 5.0     #Current drawn at no-load(A)
V_t = 230.0     #Terminal voltage at no-load(V)
N_nl = 1000.0   #Speed at no-load(rpm)
R_a = 0.2       #Armature resistance(ohm)
R_f = 230.0     #Field resistance(ohm)
I_Lfl = 30.0    #Current drawn at full-load(A)

#Calculation
#Under No-load condition
I_sh = V_t/R_f            #Shunt field current(A)
I_a1 = I_Lnl-I_sh         #Armature current(A)
E_b1 = V_t-I_a1*R_a       #Back emf(V)
#Under Full-load condition
I_a2 = I_Lfl-I_sh         #Armature current(A)
E_b2 = V_t-I_a2*R_a       #Back emf(V)
N_2 = (E_b2/E_b1)*N_nl    #Motor speed under load condition(rpm)

#Result
print('Motor speed under load condition , N_2 = %.1f rpm' %N_2)

Motor speed under load condition , N_2 = 978.2 rpm

Example 5.7, Page number 194-195

In [1]:
#Variable declaration
I_a1 = 65.0           #Current drawn(A)
V_t = 230.0           #Supply voltage(V)
N_1 = 900.0           #Speed(rpm)
R_a = 0.2             #Armature resistance(ohm)
R_sh = 0.25           #Field resistance(ohm)
I_a2 = 15.0           #Line current(A)
phi_1 = 1.0           #Assumtion of flux(Wb)
phi_2 = 0.4*phi_1     #Flux(Wb)

#Calculation
E_b1 = V_t-I_a1*(R_a+R_sh)          #Initial back emf(V)
E_b2 = V_t-I_a2*(R_a+R_sh)          #Final back emf(V)
N_2 = N_1*E_b2*phi_1/(E_b1*phi_2)   #Speed of motor(rpm)

#Result
print('Speed of motor when line current is 15 A , N_2 = %.f rpm' %N_2)
print('\nNOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A')

Speed of motor when line current is 15 A , N_2 = 2502 rpm

NOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A

Example 5.8, Page number 197-198

In [1]:
#Variable declaration
I_Lnl = 5.0     #Current drawn at no-load(A)
V_t = 230.0     #Supply voltage at no-load(V)
R_a = 0.25      #Armature circuit resistance(ohm)
R_sh = 115      #Field circuit resistance(ohm)
I_L = 40.0      #Current under load condition(A)

#Calculation
#Under No-load condition
P_in1 = V_t*I_Lnl                          #Input power(W)
I_sh = V_t/R_sh                            #Shunt field current(A)
I_a1 = I_Lnl-I_sh                          #Armature current(A)
P_acu1 = I_a1**2*R_a                       #Armature copper loss(W)
P_shcu = I_sh**2*R_sh                      #Shunt field copper loss(W)
P_iron_friction = P_in1-(P_acu1+P_shcu)    #Iron and friction losses(W)
#Under load condition
I_a2 = I_L-I_sh                            #Armature current(A)
P_acu2 = I_a2**2*R_a                       #Armature copper loss(W)
P_loss = P_iron_friction+P_shcu+P_acu2     #Total losses(W)
P_in2 = V_t*I_L                            #Input power(W)
P_0 = P_in2-P_loss                         #Output power(W)
n = (P_0/P_in2)*100                        #Efficiency(percent)

#Result
print('Iron and friction losses , P_iron_friction = %.2f W' %P_iron_friction)
print('Efficiency , η = %.f percent' %n)

Iron and friction losses , P_iron_friction = 687.75 W
Efficiency , η = 84 percent

Example 5.9, Page number 198-199

In [1]:
#Variable declaration
I_L = 80.0          #Current drawn(A)
V_t = 220.0         #Supply voltage(V)
N = 800.0           #Speed(rpm)
R_a = 0.1           #Armature resistance(ohm)
R_sh = 50.0         #Shunt field resistance(ohm)
P_if = 1600.0       #Iron and friction losses(W)

#Calculation
I_sh = V_t/R_sh           #Shunt field current(A)
I_a = I_L-I_sh            #Armature current(A)
E_b = V_t-I_a*R_a         #Back emf(V)
#For case(i)
P_in = V_t*I_L            #Input power(W)
P_md = E_b*I_a            #Mechanical power developed in the armature(W)
P_cu = P_in-P_md          #Copper loss(W)
#For case(ii)
T_a = 9.55*E_b*I_a/N      #Armature torque(N-m)
#For case(iii)
P_0 = P_md-P_if           #Output power(W)
T_sh = 9.55*P_0/N         #Shaft torque(N-m)
#For case(iv)
n = (P_0/P_in)*100        #Efficiency(percent)

#Result
print('(i)   Copper losses , P_cu = %.2f W' %P_cu)
print('(ii)  Armature torque , T_a = %.2f N-m' %T_a)
print('(iii) Shaft torque , T_sh = %.2f N-m' %T_sh)
print('(iv)  Efficiency , η = %.f percent' %n)

(i)   Copper losses , P_cu = 1539.54 W
(ii)  Armature torque , T_a = 191.72 N-m
(iii) Shaft torque , T_sh = 172.62 N-m
(iv)  Efficiency , η = 82 percent