CHAPTER 6: CONTROL AND STARTING OF A DC MOTOR

Example 6.1, Page number 212

In [1]:
#Variable declaration
V_t = 220.0      #Supply voltage(V)
I_a1 = 10.0      #Armature current(A)
N_1 = 900.0      #Speed(rpm)
R_a = 1.0        #Armature resistance(ohm)
N_2 = 500.0      #Reduced speed(rpm)

#Calculation
E_b1 = V_t-I_a1*R_a           #Initial back emf(V)
R = E_b1/I_a1*(1-(N_2/N_1))   #Value of additional resistance(ohm)

#Result
print('Value of additional resistance , R = %.1f ohm' %R)

Value of additional resistance , R = 9.3 ohm

Example 6.2, Page number 212-213

In [1]:
#Variable declaration
V_t = 230.0    #Supply voltage(V)
I_a1 = 15.0    #Armature current(A)
N_1 = 650.0    #Speed(rpm)
R_a = 0.4      #Armature resistance(ohm)
R = 1.0        #Variable resistance in series with the armature(ohm)

#Calculation 
E_b1 = V_t-I_a1*R_a          #Initial back emf(V)
#At full load torque
E_b2 = V_t-I_a1*(R+R_a)      #Final back emf(V)
N_2 = N_1*(E_b2/E_b1)        #Speed at full load torque(rpm)

#At half load torque
I_a2hl = I_a1/2              #Armature current(A)
E_b2hl = V_t-I_a2hl*(R+R_a)  #Back emf(V)
N_2hl = N_1*(E_b2hl/E_b1)    #Speed at half load torque(rpm)

#Result
print('Speed at full load torque , N_2 = %.1f rpm' %N_2)
print('Speed at half load torque , N_2 = %.1f rpm' %N_2hl)

Speed at full load torque , N_2 = 606.5 rpm
Speed at half load torque , N_2 = 636.9 rpm

Example 6.3, Page number 215-216

In [1]:
#Variable declaration
V_t = 230.0          #Supply voltage(V)
R_af = 0.2           #Total resistance of armature & field(ohm)
I_a1 = 10.0          #Armature current for certain load(A)
N = 1000.0           #Motor speed for certain load(rpm)
R_1 = 0              #Variable resistance for certain load(ohm)
I_a2 = 8.0           #Armature current for other load(A)
R_2 = 2.0            #Variable resistance for other load(ohm)
phi_1 = 1.0          #Assuming flux in certain load(Wb)
phi_2 = 0.8*phi_1    #Flux in other load(Wb)

#Calculation
#For certain load
R_a1 = R_af+R_1                     #New armature resistance(ohm)
E_b1 = V_t-I_a1*R_a1                #Back emf(V)
#For other load
R_a2 = R_af+R_2                     #New armature resistance(ohm)
E_b2 = V_t-I_a2*R_a2                #Back emf(V)
N_2 = (E_b2/E_b1)*(phi_1/phi_2)*N   #New speed(rpm) 

#Result
print('New speed , N_2 = %.1f rpm' %N_2)

New speed , N_2 = 1164.5 rpm

Example 6.4, Page number 216-218

In [1]:
#Variable declaration
N = 1100.0           #Speed of dc series motor(rpm)
P = 4.0              #Number of poles
path = 4.0           #Number of parallel paths
I_a1 = 15.0          #Supply current to dc series motor(A)
V_t = 220.0          #Supply voltage(V)
R_a = 0.9            #Series armature resistance(ohm)
R_se = 0.6           #Series field resistance(ohm)
I_a2 = 25            #Supply current to dc series motor(A)
phi_1 = 1.0          #Assuming flux for 15 A case(Wb)
phi_2 = 0.8*phi_1    #Flux in 25 A case(Wb)

#Calculation
#First case
R_se1 = R_se                         #Total series field resistance(ohm)
E_b1 = V_t-I_a1*(R_a+R_se1)          #Back emf(V)
#Second case
R_se2 = R_se1/path                   #Total series field resistance(ohm)
E_b2 = V_t-I_a2*(R_a+R_se2)          #Back emf(V)
N_2 = (E_b2/E_b1)*(phi_1/phi_2)*N    #New speed(rpm) 

#Result
print('Speed for second case , N_2 = %.1f rpm = %.f rpm' %(N_2,N_2))

Speed for second case , N_2 = 1348.9 rpm = 1349 rpm

Example 6.5, Page number 233-234

In [1]:
#Variable declaration
V_t = 230.0         #Shunt motor supply voltage(V)
R_a = 0.4           #Armature resistance(ohm)
I_a = 30.0          #Armature current(A)
n = 3.0             #Number of steps

#Calculation
R_1 = V_t/I_a              #Maximum resistance(ohm)
k = (R_1/R_a)**(1.0/3)     #Constant
R_2 = R_1/k                #Resistance(ohm)
R_3 = R_2/k                #Resistance(ohm)
R_4 = R_3/k                #Resistance(ohm)
R_1step = R_1-R_2          #Resistance of the first step(ohm)
R_2step = R_2-R_3          #Resistance of the second step(ohm)
R_3step = R_3-R_4          #Resistance of the third step(ohm)

#Result
print('Resistance of the first step , R_1step = %.1f ohm' %R_1step)
print('Resistance of the second step , R_2step = %.1f ohm' %R_2step)
print('Resistance of the third step , R_3step = %.2f ohm' %R_3step)

Resistance of the first step , R_1step = 4.8 ohm
Resistance of the second step , R_2step = 1.8 ohm
Resistance of the third step , R_3step = 0.67 ohm

Example 6.6, Page number 234-236

In [1]:
#Variable declaration
V_t = 220.0      #Shunt motor supply voltage(V)
P_0 = 3550.0     #Output power(W) 
n = 0.85         #Efficiency

#Calculation
P_in = P_0/n             #Input power(W)
P_tloss = P_in-P_0       #Total loss(W)
I_a = P_in/V_t           #Armature current(A)
P_cu = P_tloss/2         #Copper loss(W)
R_a = P_cu/I_a**2        #Armature resistance(ohm)
I_1 = 2*I_a              #Maximum starting current(A)
R_1 = V_t/I_1            #Maximum resistance(ohm)
k = (R_1/R_a)**(1.0/4)   #Constant
R_2 = R_1/k              #Resistance(ohm)
R_3 = R_2/k              #Resistance(ohm)
R_4 = R_3/k              #Resistance(ohm)
R_5 = R_4/k              #Resistance(ohm)
R_1step = R_1-R_2        #Resistance of the first step(ohm)
R_2step = R_2-R_3        #Resistance of the second step(ohm)
R_3step = R_3-R_4        #Resistance of the third step(ohm)
R_4step = R_4-R_5        #Resistance of the fourth step(ohm)

#Result
print('Resistance of the first step , R_1step = %.1f ohm' %R_1step)
print('Resistance of the second step , R_2step = %.2f ohm' %R_2step)
print('Resistance of the third step , R_3step = %.2f ohm' %R_3step)
print('Resistance of the fourth step , R_4step = %.2f ohm' %R_4step)

Resistance of the first step , R_1step = 2.2 ohm
Resistance of the second step , R_2step = 1.36 ohm
Resistance of the third step , R_3step = 0.85 ohm
Resistance of the fourth step , R_4step = 0.53 ohm