CHAPTER 7: THREE-PHASE INDUCTION MOTOR

Example 7.1, Page number 246-247

In [1]:
#Variable declaration
V = 230.0        #Supply voltage(V)
P = 4.0          #Number of poles
f = 50.0         #Frequency(Hz)
N_l = 1445.0     #Full load speed(rpm)

#Calculation
#For case(i)
N_s = 120*f/P       #Synchronous speed(rpm)
#For case(ii)
s = (N_s-N_l)/N_s   #Slip
#For case(iii)
f_r = s*f           #Rotor frequency(Hz)

#Result
print('(i)   Synchronous speed , N_s = %.f rpm' %N_s)
print('(ii)  Slip , s = %.4f ' %s)
print('(iii) Rotor frequency , f_r = %.1f Hz' %f_r)

(i)   Synchronous speed , N_s = 1500 rpm
(ii)  Slip , s = 0.0367 
(iii) Rotor frequency , f_r = 1.8 Hz

Example 7.2, Page number 247-248

In [1]:
#Variable declaration
E_BR = 120.0     #Voltage under blocked condition(V)
P = 4.0          #Number of poles
f = 50.0         #Frequency(Hz)
N_l = 1450.0     #Speed(rpm)

#Calculation
N_s = 120*f/P        #Synchronous speed(rpm)
s = (N_s-N_l)/N_s    #Slip
f_r = s*f            #Rotor frequency(Hz)
E_r = s*E_BR         #Rotor voltage(V)

#Result
print('Synchronous speed , N_s = %.f rpm' %N_s)
print('Rotor frequency , f_r = %.2f Hz' %f_r)
print('Rotor voltage , E_r = %.2f V' %E_r)
print('\nNOTE : Changes in answer is due to precision i.e more number of decimal places')

Synchronous speed , N_s = 1500 rpm
Rotor frequency , f_r = 1.67 Hz
Rotor voltage , E_r = 4.00 V

NOTE : Changes in answer is due to precision i.e more number of decimal places

Example 7.3, Page number 250

In [1]:
#Variable declaration
V_0 = 230.0    #Supply voltage(V)
P = 4.0        #Number of poles
T_0 = 230.0    #Original torque(N-m)
V_s = 150.0    #Stator voltage(V)
I_0 = 560.0    #Starting current(A)

#Calculation
T_st = (V_s/V_0)**2*T_0    #Starting torque(N-m)
I_st = I_0*(V_s/V_0)       #Starting current(A)

#Result
print('Starting torque , T_st = %.1f N-m' %T_st)
print('Starting current , I_st = %.1f A' %I_st)

Starting torque , T_st = 97.8 N-m
Starting current , I_st = 365.2 A

Example 7.4, Page number 254

In [1]:
#Variable declaration
f = 50.0        #Frequency(Hz)
P = 8.0         #Number of poles
a = 0.03        #Full load slip
R_2 = 0.01      #Rotor resistance per phase(ohm)
X_2 = 0.1       #Standstill reactance per phase(ohm)

#Calculation
#For case(i)
N_s = 120*f/P               #Synchronous speed(rpm)
s = R_2/X_2                 #Slip at maximum torque
N_l = (1-s)*N_s             #Rotor speed at maximum torque(rpm)
#For case(ii)
T = (s**2+a**2)/(2*a*s)     #Ratio of maximum torque to full load torque

#Result
print('(i)  Speed at which maximum torque occurs , N_l = %.f rpm' %N_l)
print('(ii) Ratio of the maximum torque to full load torque , T_max/T_f = %.2f ' %T)

(i)  Speed at which maximum torque occurs , N_l = 675 rpm
(ii) Ratio of the maximum torque to full load torque , T_max/T_f = 1.82

Example 7.5, Page number 260

In [1]:
#Variable declaration
V = 440.0       #Supply voltage(V)
P = 4.0         #Number of poles
P_ag = 1500.0   #Rotor input(W)
P_rcu = 250.0   #Copper loss(W)
f = 50.0        #Frequency(Hz)

#Calculation
#For case(i)
s = P_rcu/P_ag         #Slip
#For case(ii)
N_s = 120*f/P          #Synchronous speed(rpm)
#For case(iii)
N_l = (1-s)*N_s        #Shaft speed(rpm)
#For case(iv)
P_mech = (1-s)*P_ag    #Mechanical power developed(W)

#Result
print('(i)   Slip , s = %.2f' %s)
print('(ii)  Synchronous speed , N_s = %.f rpm' %N_s)
print('(iii) Shaft speed , N_l = %.f rpm' %N_l)
print('(iv)  Mechanical power developed , P_mech = %.f W' %P_mech)
print('\nNOTE : Changes in answer is due to precision i.e more number of decimal places')

(i)   Slip , s = 0.17
(ii)  Synchronous speed , N_s = 1500 rpm
(iii) Shaft speed , N_l = 1250 rpm
(iv)  Mechanical power developed , P_mech = 1250 W

NOTE : Changes in answer is due to precision i.e more number of decimal places

Example 7.6, Page number 264

In [1]:
import math

#Variable declaration
V_1 = 150.0              #Supply voltage(V)
P = 4.0                  #Number of poles
f = 50.0                 #Frequency(Hz)
Z_1 = complex(0.12,0.16) #Per phase standstill stator impedance(ohm)
Z_2 = complex(0.22,0.28) #Per phase standstill rotor impedance(ohm)

#Calculation
Z_eq = Z_1+Z_2                                  #Equivalent impedance(ohm)
R_eq = Z_eq.real
P_mech =  3*V_1**2/(2*(R_eq+abs(Z_eq)))*10**-3  #Maximum mechanical power developed(kW)
R_2 = Z_2.real
s_mp = R_2/(abs(Z_eq)+R_2)                      #Slip
W_s = 2*math.pi*2*f/P                           #Synchronous speed(rad/s)
W = (1-s_mp)*W_s                                #Speed of rotor(rad/s)
T_mxm = P_mech*1000/W                           #Maximum torque(N-m) 

#Result
print('Maximum mechanical power , P_mech = %.2f kW' %P_mech)
print('Maximum torque , T_mxm = %.2f N-m' %T_mxm)
print('Slip , s_mp = %.2f ' %s_mp)
print('\nNOTE : Changes in answer is due to precision i.e more number of decimal places')

Maximum mechanical power , P_mech = 37.66 kW
Maximum torque , T_mxm = 334.65 N-m
Slip , s_mp = 0.28 

NOTE : Changes in answer is due to precision i.e more number of decimal places

Example 7.7, Page number 265-266

In [1]:
#Variable declaration
V = 440.0           #Supply voltage(V)
P = 6.0             #Number of poles
f = 50.0            #Frequency(Hz)
P_a = 45000.0       #Input power(W)
N_l = 900.0         #Speed(rpm)
P_tloss = 2000.0    #Total stator losses(W)
P_fw = 1000.0       #Friction and windage losses(W)

#Calculation
N_s = 120*f/P                 #Synchronous speed(rpm)
s = (N_s-N_l)/N_s             #Slip
P_ag = (P_a-P_tloss)          #Air gap power(W)
P_rcu = s*P_ag                #Rotor copper loss(W)
P_mech = P_ag-P_rcu           #Mechanical power(W)
P_0 = P_mech-(P_tloss+P_fw)   #Output power(W)
n = (P_0/P_ag)*100            #Efficiency(percent)

#Result
print('(i)   Slip , s = %.1f ' %s)
print('(ii)  Rotor copper loss , P_rcu = %.f W' %P_rcu)
print('(iii) Shaft or Output power , P_0 = %.f W' %P_0)
print('(iv)  Efficiency , η = %.f percent' %n)
print('\nNOTE : ERROR : Friction & windage losses are 1 kW not 1.5 kW as given in textbook question')

(i)   Slip , s = 0.1 
(ii)  Rotor copper loss , P_rcu = 4300 W
(iii) Shaft or Output power , P_0 = 35700 W
(iv)  Efficiency , η = 83 percent

NOTE : ERROR : Friction & windage losses are 1 kW not 1.5 kW as given in textbook question

Example 7.8, Page number 268-269

In [1]:
#Variable declaration
v_s = 120.0      #Train speed(km/h)
f = 50.0         #Stator frequency(Hz)

#Calculation
v_s1 = v_s*1000/(60*60)    #Train speed(m/s)
w = v_s1/(2*f)             #Length of the pole-pitch(m)

#Result
print('Length of the pole-pitch of linear induction motor , w = %.2f m' %w)

Length of the pole-pitch of linear induction motor , w = 0.33 m