# Chapter 2 : D.C. Motors¶

## Example 2.1 page no : 5¶

In :
# Variables
V = 220.
I_a = 30. 			#armature currnet
R_a = 0.75 			#Armature resistance

# Calculations
E_b = V - I_a*R_a   			# Since V =  E_b+ I_a*R_a

# Results
print 'Induced EMF or back EMF in the motor is %.1f V'%(E_b)

Induced EMF or back EMF in the motor is 197.5 V


## Example 2.2 page no : 6¶

In :
# Variables
Pole = 4.
A = Pole 			#for lap winding
V = 230.
Z = 250. 			#number of armature conductors
phi = 30.*10**-3 			#flux per pole in weber
I_a = 40.
R_a = 0.6 			#Armature resistance

# Calculations
E_b = V - I_a*R_a   			# Since V =  E_b+ I_a*R_a
N = E_b * 60*A/(phi*Pole*Z)   			#because E_b  =  phi*P*N*Z/(60*A)

# Results
print 'Back emf is %.0f V and running speed is %.0f rpm'%(E_b,N)

Back emf is 206 V and running speed is 1648 rpm


## Example 2.3 page no : 9¶

In :
# Variables
Pole = 4.
A = Pole 			#for lap winding
Z = 480.			#number of armature conductors
phi = 20.*10**-3 			#flux per pole in weber
I_a = 50. 			#Armature current

# Calculations
T_a  =  0.159*phi*I_a*Pole*Z/A  			#Gross torque developed by armature

# Results
print 'Gross torque developed by armature is %.3f N-m'%(T_a)

Gross torque developed by armature is 76.320 N-m


## Example 2.4 page no : 10¶

In :
import math

# Variables
Pole = 4.
A = Pole 			#for lap winding
V = 230.
R_a = 0.8 			#Armature resistance
N_0 = 1000. 			#no load speed in rpm
Z = 540. 			#number of armature conductors
phi = 25.*10**-3 			#flux per pole in weber

# Calculations and Results
E_b0  =  phi*Pole*N_0*Z/(60*A)  			#induced emf

#part(i)
print 'i)Induced e.m.f  =  %.0f V'%(E_b0)

#part(ii)
I_a0  =  (V- E_b0)/R_a  			#because V =  E_b0+ I_a0*R_a
print 'ii)Armature current  =  %.2f A'%(I_a0)

#part(iii)
stray_losses  =  E_b0*I_a0  			#on no load ,power developed is fully power required to overcome strya losses
print 'iii)Stray loss  =  %.2f W'%(stray_losses)

#part(iv)
T_f  =  E_b0*I_a0/(2*math.pi*N_0/60)  			#lost torque
print 'iv)Lost torque  =  %.3f N-m'%(T_f)

i)Induced e.m.f  =  225 V
ii)Armature current  =  6.25 A
iii)Stray loss  =  1406.25 W
iv)Lost torque  =  13.429 N-m


## Example 2.5 page no : 21¶

In :
# Variables
Pole = 4.
Z = 200. 			    #No of armature conductors
A = 2.   			    #wave connected armature
V = 250.
phi = 25.*10**-3 			#flux per pole in weber
I_a  = 60.
I_L  = I_a  	    		#armature current
R_a = 0.15
R_se = 0.2       			#resistances of armature and series field winding

# Calculations
E_b =  V - I_a*(R_a+R_se)  			#induced emf
N = E_b * 60*A/(phi*Pole*Z)   			#because E_b  =  phi*P*N*Z/(60*A)

# Results
print 'Required speed is %.0f r.p.m'%(N)

Required speed is 1374 r.p.m


## Example 2.6 page no : 22¶

In :
# Variables
V = 250.
R_a = 0.3
R_sh = 200. 			#Armature and shunt field winding

# Calculations
I_sh = V/R_sh 			#shunt current
I_a = I_L-I_sh 			#armature current
E_b =  V - I_a*R_a 			#emf generated

# Results
print 'Armature current is %.2f A'%(I_a)
print 'Back e.m.f is %.3f V'%(E_b)

Armature current is 18.75 A
Back e.m.f is 244.375 V


## Example 2.7 page no : 22¶

In :
# Variables
V = 220.
R_a = 0.3
R_sh = 110. 			#resistance of armature and shunt field winding

N_0 = 1000. 			#no load speed in r.p.m
I_L0  = 6. 			#line current on no load
I_sh =  V/R_sh 			#no load shnt current
I_a0  =  I_L0 - I_sh 			#no load armature current
E_b0 =  V - I_a0*R_a 			#no load induced emf

# Calculations
I_sh_FL =  V/R_sh
I_L_FL = 50 			#line current at full load
I_a_FL =  I_L_FL - I_sh_FL			#full load armature current
E_b_FL =  V - I_a_FL * R_a 			#full load induced emf
#using speed equation as treating phi as constant
N_FL = N_0 * (E_b_FL/E_b0)

# Results
print 'Speed on full load is %.2f r.p.m'%(N_FL)

Speed on full load is 939.67 r.p.m


## Example 2.8 page no : 23¶

In :
# Variables
R_a = 0.2
R_se  = 0.3 			#resistance of armature and series field winding
#following variables correspond to load 1
V = 250.
N_1 = 800.
I_1 = 20.
I_a1 = I_1
I_se1 =  I_a1

# Calculations
E_b1 =  V - I_a1*(R_a+R_se)
#following variables correspond to load 2
I_2 = 50.
I_a2 = I_2
E_b2 =  V - I_a2*(R_a+R_se)

#from speed equation it can be derived that

N_2  =  N_1 * (E_b2/E_b1) * (I_a1/I_a2)

# Results
print 'Speed on motor on no load  is %.0f r.p.m'%(N_2)

Speed on motor on no load  is 300 r.p.m


## Example 2.9 page no : 31¶

In :
# Variables
V = 250.
R_a = 0.3
R_sh = 200.  			#resistance of armature and shunt field winding
I_L1 = 22.
I_sh1 = V/R_sh 			#initial shunt current before adding 150 ohms resistance
I_a1  =  I_L1 - I_sh1 			#initial armature current before adding 150 ohms resistance
N_1 = 1500. 			#initial speed before adding 150 ohms resistance

# Calculations
#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying further
I_sh2 = V/(R_sh + R_x) 			#new shunt current
I_a2 =  I_sh1*I_a1/I_sh2  			#New armature current

E_b1 = V - I_a1*R_a   			#induced emf before adding 150 ohms resistance
E_b2 = V - I_a2*R_a   			#new emf

N_2  =  N_1 * (E_b2/E_b1) * (I_sh1/I_sh2) 			#new speed

# Results
print 'New armature current and speed are %.4f A and %.f r.p.m respectively'%(I_a2,N_2)

New armature current and speed are 36.3125 A and 2575 r.p.m respectively


## Example 2.10 page no : 36¶

In :
import math

# Variables
V = 250.
R_a = 0.15
R_se = 0.1
R_x = 0.1 			#Resimath.tance of armature , series field winding and extra resistance
N_1  =  800. 			#initial speed before load torque is increased
I_1 =  30.
I_a1 = I_1
I_se1  =  I_1  			#initial currents

# Calculations
T_2_by_T_1  =  1 + (50./100)  			#50 percent increase as mentioned in question
I_se2_by_I_a2  =  R_x/(R_x + R_se)  			#from the figure

#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying ,solving further
I_a2 = math.sqrt(I_a1*I_se1*T_2_by_T_1/I_se2_by_I_a2) 			#new armature current
I_se2  =  I_se2_by_I_a2 *I_a2 			#new series field current

E_b1  =  V - I_a1*R_a - I_se1*R_se 			#indiced emf initially
E_b2  =  V - I_a2*R_a - I_se2*R_se 			#new induced emf
N_2  =  N_1 * (E_b2/E_b1) * (I_se1/I_se2) 			#required speed

# Results
print 'The required running speed of motor is %.3f r.p.m'%(N_2)

The required running speed of motor is 912.743 r.p.m


## Example 2.11 page no : 38¶

In :
import math

# Variables
V = 220.
I_1 = 50.
I_a1 = I_1  			#Currents before adding extra resistance
T_2_by_T_1  = 0.5
R_t = 0.15 			#R_e + R_se  = 0.15

# Calculations
I_a2  = I_a1 * math.sqrt(T_2_by_T_1)  			#Because T (prop.) I_a**2
E_b1 = V-I_a1*(R_t) 			#induced emf before adding extra resistance
N_1 = 500.
N_2 = 300. 			#speeds before and adding extra resistance

#N (prop.) E_b/phi (prop.) E_b/I_a
E_b2 = E_b1 *(I_a2/I_a1)*(N_2/N_1) 			#induced emf after adding resistance
R_x =  (V-E_b2)/I_a2 -R_t 			#because E_b2 = V - I_a2*(R_a + R_se + R_x)

# Results
print 'Desired extrea resistance =  %.4f ohms '%(R_x)

Desired extrea resistance =  3.5225 ohms


## Example 2.12 page no : 43¶

In :
import math

# Variables
R_a =  1.
I_a = 1.2
V = 50.

# Calculations and Results
#part(i)
E_b  =  V - I_a*R_a
rot_loss_NL  = E_b*I_a 			#no load rotational loss
print 'i)No load rotational losses  =  %.2f W'%(rot_loss_NL)

#part(ii)
omega_2000 = 2*math.pi*2000/60 			#angular velocity when speed of motor  = 2000 rpm
K_m = E_b/omega_2000   			#to determine K_m
V = 48.
omega_1800 = 2*math.pi*1800/60 			#angular velocity when speed of motor  = 1800 rpm
E_b = K_m*omega_1800
I_a  =  (V-E_b)/R_a  			#armature current
P_dev  =  E_b*I_a			#power developed
motor_output  =   P_dev - rot_loss_NL
print 'ii)Motor output  =  %.f W'%(motor_output)

#part(iii)
E_b = 0. 			#when motor stalls
V_stall = 20. 			#voltage during stalling
I_a = V_stall/R_a 			#armature current during stalling
T_stall  =  K_m*I_a 			#stalling torque
print 'iii)Stalling torque  =  %.2f N-m'%(T_stall)
print 'partii answer is slightly different due to inaccurate calculation of Power developed'

i)No load rotational losses  =  58.56 W
ii)Motor output  =  121 W
iii)Stalling torque  =  4.66 N-m
partii answer is slightly different due to inaccurate calculation of Power developed


## Example 2.13 page no : 49¶

In :
# Variables
V = 120.
R_a = 0.2
R_sh = 60. 			#armature and field resistance
I_L1 = 40.
N_1 = 1800.

# Calculations
I_sh =  V/R_sh

I_a1 = I_L1 - I_sh
E_b1  =  V -I_a1*R_a  			#Induced emf at half load
T2_by_T1  = 1./2
I_a2 = I_a1*(T2_by_T1)   			#T (prop.)I_a
E_b2 = V- I_a2*R_a			#induced emf at half load
N_2  =  N_1 *(E_b2/E_b1)  			#N (prop.) E_b as phi is constant

# Results
print 'Speed on half load condition is %.2f r.p.m'%(N_2)

Speed on half load condition is 1860.85 r.p.m


## Example 2.14 page no : 50¶

In :
# Variables
R_a = 0.08
E_b1 = 242.
V = 250.

# Calculations and Results
#part(i)
I_a1 =  (V-E_b1)/R_a
print 'i)Armature current  =  %.0f A'%(I_a1)

#part(ii)
N = 0.
E_b = 0. 			#because N = 0
I_a_start = V/R_a
print 'ii)Starting armature current  =  %.0f A'%(I_a_start)

#part(iii)
I_a2 = 120.
E_b2 = V-I_a2*R_a
print 'iii)Back emf if armature current is changed to 120 A =    %.1f V'%(E_b2)

#part(iv)
I_a = 87.
N_m = 1500.
E_g = V + I_a*R_a 			#induced emf
N_g = N_m*(E_g/E_b1)			#as E (prop.) N
print 'iv)Generator speed to deliver 87 A at 250 V  =  %.1f rpm'%(N_g)

i)Armature current  =  100 A
ii)Starting armature current  =  3125 A
iii)Back emf if armature current is changed to 120 A =    240.4 V
iv)Generator speed to deliver 87 A at 250 V  =  1592.7 rpm


## Example 2.15 page no : 51¶

In :
import math
from numpy import roots

# Variables
shaft_output  =  80.*746 			#coverted to watts
eta =  80./100           			#efficiency
V = 250.
N_1 = 1200.
R_a = 0.04
R_sh  =  250. 			            #armature and shunt field resistance

# Calculations and Results
power_input  =  shaft_output/eta
I_L =  power_input /V
I_sh =  V / R_sh
I_a  =  I_L - I_sh
E_b1  =  V - I_a*R_a

gross_mechanical_power =  E_b1*I_a 			#electrical equivalent of mechanical power developed
stray_losses  =   gross_mechanical_power - shaft_output
print 'Mechanical power developed on full load  =  %.3f kW'%(gross_mechanical_power/1000)

#on no load shaft_output = 0 and entire gross power is used to overcome stray losses
Eb0_Ia0 =  stray_losses
#E_b0  =  V - I_a0*R_a  ... solving for I_0
p = [R_a, -V, Eb0_Ia0]
ans = roots(p)
I_a0 = ans 			#first root is ignored math.since its too large
I_L0  = I_sh+I_a0 			#current drawn from supply
E_b0  =  V - I_a0*R_a

#From speed equation N (prop.) E_b
N_0  =  N_1*(E_b0/E_b1)
print 'No load speed and current are %.4f rpm and %.2f A respectively'%(N_0,I_L0)

Mechanical power developed on full load  =  70.812 kW
No load speed and current are 1250.9121 rpm and 45.85 A respectively


## Example 2.16 page no : 53¶

In :
# Variables
V = 250.
P = 4.
R_a = 0.1
R_sh  = 124. 			#armature and shunt field resistance
I_L0 = 4.
N_0 = 1200.
I_L_1 = 61.

# Calculations
I_sh = V/R_sh
I_a0 = I_L0-I_sh
V_brush =  2 			#voltage loss due to brush
E_b0 =  V - I_a0*R_a- V_brush

I_a1 = I_L_1 - I_sh
E_b1 = V - I_a1*R_a -V_brush

phi1_by_phi0 = 1-(5./100)   			#weakened by 5 %
N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0

# Results
print 'Full load speed is %.3f r.p.m'%(N_1)

Full load speed is 1234.102 r.p.m


## Example 2.17 page no : 54¶

In :
# Variables
V = 250.
R_a = 0.15
R_sh = 167.67 			#armature and shunt field resistance
N_0 = 1280. 			#speed at no load

I_L1  =  67.  			#current drawn on full load
I_sh  =  V / R_sh  			#as shunt motor
I_a1 =  I_L1- I_sh
E_b1 =  V - I_a1*R_a

I_L0 = 6.5
I_a0  =  I_L0 - I_sh
E_b0  =  V - I_a0*R_a

# Calculations and Results
#part(i) USING SPEED EQUATION
#N (prop.) E_b/phi (prop.)E_b   			#as phi is constant
N_1  =  N_0 * (E_b1 / E_b0)
print 'i)Full load speed  =  %.3f r.p.m'%(N_1)

#part(ii)
speed_regulation  =  100* ((N_0-N_1)/N_1)
print 'ii)Speed regulation  =  %.2f percent '%(speed_regulation )

#part(iii)
shaft_output_FL  =  E_b1*I_a1 - E_b0*I_a0  			#full load power developed - stray losses
hp_rating  =  shaft_output_FL /746
print 'iii)HP rating of machine  =  %.2f h.p'%(hp_rating)

#part(iv)
power_input =  V*I_L1
eta =  100*(shaft_output_FL/power_input)  			#full load efficiency
print 'iv)Full load efficiency  =  %.2f percent'%(eta)

i)Full load speed  =  1233.396 r.p.m
ii)Speed regulation  =  3.78 percent
iii)HP rating of machine  =  19.42 h.p
iv)Full load efficiency  =  86.48 percent


## Example 2.18 page no : 55¶

In :
# Variables
V = 200.
R_a = 0.5
R_se = 0.2
R_x = 0.2 			#armature and series field resistance; extra resistance
I_a1 = 20.
I_1 = I_a1
I_se1 = I_a1
I_a2 = 20.
I_2 = I_a2
I_se2 =  I_2 *(R_x/(R_se+R_x))

# Calculations
E_b1  =  V -I_a1*R_a - I_a1*R_se
E_b2  =  V -I_a2*R_a - I_se2*R_se

phi2_by_phi1 = 70./100
N_1 = 1000
N_2 = N_1*(E_b2/E_b1)  /phi2_by_phi1        			#N (prop.) E_b/phi

# Results
print 'Required speed is %.2f r.p.m'%(N_2)

Required speed is 1443.93 r.p.m


## Example 2.19 page no : 57¶

In :
import math

# Variables
V = 110.
P = 4.
R_a  =  0.1
R = 0.01  			#A resistance of 0.01 ohms
R_se = R+R

# Calculations
#case(i)
I_1 = 50.
I_a1 = I_1
N_1 = 700.
E_b1  =  V -I_a1*(R_a + R_se)

#T (prop) phi*I_a   from torque equation                                     (1)

#phi_1 (prop.) I_a1                                                           (2)
#case(ii) when I_a2 gets divided to half
#phi_2 (prop.) I_a2/2                                                         (3)

#combining (1)(2)(3) and T1 = T2
I_a2  =  math.sqrt(2*I_a1**2)
R_se_eqvt = (R*R)/(R+R)   			#Equavalent of parallel combination
E_b2  =  V - I_a2*R_a - I_a2* R_se_eqvt

#using speed equation N (prop.) E_b / phi  and using (2) and (3)
N_2  =  N_1 *( E_b2/E_b1) *(I_a1/(I_a2/2))

# Results
print 'Speed after reconnection  =  %.3f r.p.m'%(N_2)

Speed after reconnection  =  976.389 r.p.m


## Example 2.20 page no : 58¶

In :
import math

# Variables
P = 4.
I_a1 =  50.
N_1 = 2000.
V = 230.

# Calculations and Results
#phi_1 is proportioanl to total ampere-turns produced by field coils
#phi_1 (prop.) I_a1*P*n (prop.) 200*n                                         (1)

#After reconnection ,phi_2 proportional to ampere turns divided as follows
#phi_2 (prop.) [I_a2/2*2*n + I_a2/2*2*n]  (prop.) 2*n*I_a2                    (2)

# Dividing (1) and (2) ,(phi_1/phi_2) = 100 / I_a2                             (3)

#T (prop.) phi*I_a AND T (prop.) N**2                                          (4)(5)
#therefore N**2 (prop.) phi*I_a                                                (6)

#N (prop.) E_b/phi (prop.) 1/phi   ..
#Because drops across windings can be neglected , E_b1 = E_b2
#therefore N (prop.) 1/phi                                                    (7)

#using (7) and (6) phi**3 (prop.) 1/I_a                                        (8)

#combining (3) and (8)
I_a2  =   (50.*100**3)**(1./4)    			#new armature current
print 'New armature current =  %.3f A'%(I_a2)
#combining (6) and (7)   N**3 (prop.) I_a1
N_2 = N_1 *(I_a2/I_a1)**(1./3)
print 'New motor speed  = %.3f r.p.m'%(N_2)

New armature current =  84.090 A
New motor speed  = 2378.414 r.p.m


## Example 2.22 page no : 61¶

In :
import math

# Variables
V = 200.
I_a1 = 30.
R_t = 1.5  			#R_a + R_se
E_b1 =  V - I_a1*R_t
N2_by_N1 = (60./100)

# Calculations
#T (prop.) I_a**2  and T (prop.) N_3....therefore I_a**2 (prop.) N**3
I_a2   =  I_a1*math.sqrt(N2_by_N1**3)

#N (prop.) E_b/I_a
N2_by_N1
E_b2  =  E_b1 *(I_a2/I_a1)*N2_by_N1
R_x =  (V- E_b2)/I_a2 - R_t        			#because E_b2 =  V - I_a2*(R_x+R_t)

# Results
print 'Additional resistance to be added in series with motor circuit  =  %.3f ohms'%(R_x)

Additional resistance to be added in series with motor circuit  =  9.744 ohms


## Example 2.23 page no : 63¶

In :
import math
from numpy import array,roots

# Variables
V = 250.
R_a = 0.4
R_sh = 100. 			#armature and shunt field resistance
I_sh1 = V / R_sh
P_out_FL  =  10 * 735.5
eta = 85./100 			#efficiency
P_in =  P_out_FL/eta
I_L1 =  P_in /V
I_a1 =  I_L1 - I_sh1

# T (prop.) phi*I_a (prop.) I_sh*I_a    because phi (prop.) I_sh
#Bu torque is constant..
Ia2_Ish2 =  I_a1*I_sh1
E_b1 =  V - I_a1*R_a

#N (prop.) E_b/I_sh
#put E_b2 =  V - I_a2*R_a  and solving further for I_sh2 we get ,I_sh2**2 - 1.8824 I_sh2 +0.2417 = 0
p = array([1, -1.8824, 0.2417])
ans = roots(p)
I_sh2 = ans
#root 1 was considered because its always easier to attain root(1) because less resistacne is needeed
#R_x in series with field
R_x  =  (V/I_sh2) -R_sh   			#because I_sh2  =  V/(R_sh + R_x)
print 'Extra resistance to be added   =  %.2f ohms'%(R_x)

Extra resistance to be added   =  43.37 ohms


## Example 2.24 page no : 64¶

In :
# Variables
R_t = 1. 			#R_t  =  R_se + R_a
V_1 =  230.
N_1 = 300.
N_2 = 375.
I_1 = 15.
I_a1 = I_1

# Calculations
#T (prop.) I_a**2  and T (prop.) N_2....therefore I_a**2 (prop.) N**2
I_a2 = I_a1 *(N_2/N_1)
E_b1  =  V_1 - I_a1*(R_t)

#N (prop.) E_b/I_a
E_b2 =  E_b1*(I_a2/I_a1)*(N_2/N_1)
V_2 = E_b2 + I_a2* (R_t)  			#because E_b2  =  V_2 - I_a2*(R_a+R_se)

# Results
print 'Voltage supply needed  =  %.4f V'%(V_2)

Voltage supply needed  =  354.6875 V


## Example 2.25 page no : 66¶

In :
# Variables
I_L1 = 30.
V = 230.
R_sh = 230.
R_a = 1.
I_sh =  V / R_sh
I_a1 =  I_L1 - I_sh
E_b1  =  V - I_a1*R_a

# Calculations
#T (prop.) phi*I_a (prop.) I_a   as phi is constant
#and torque is constant
I_a2  =  I_a1
N2_by_N1 =  1./2
#N (prop.) E_b/phi (prop.) E_b
E_b2 =  E_b1 *(N2_by_N1)
R_x =  (V- E_b2)/I_a2 - R_a     			#Because E_b2  =  V - I_a2*(R_a + R_x)

# Results
print 'resistance to be inserted in series  =  %.4f ohms '%(R_x)

resistance to be inserted in series  =  3.4655 ohms


## Example 2.26 page no : 67¶

In :
# Variables
T_1 = 40.  			#initial torque
#phi_1 is initial flux
#phi_2 is new flux
#T_2 is new torque
#I_a1 is initial current
#I_a2 is new current
phi2_by_phi1  =  1- (30./100)  			#decrease by 30 percent
Ia2_by_Ia1 = 1+(15./100)       			#increase by 15 percent

# Calculations
#T (prop.)phi*I_a
T_2 = T_1*(phi2_by_phi1)*(Ia2_by_Ia1)

# Results
print 'New torque is %.1f N-m'%(T_2)

New torque is 32.2 N-m


## Example 2.27 page no : 67¶

In :
# Variables
V = 230.
N_1 = 1000.
N_2 = 950.
R_a = 0.5
R_sh = 230. 			#armature and shunt field resistance
I_L1 = 10.

# Calculations
I_sh  =  V/R_sh
I_a1  =  I_L1 - I_sh

#T (prop.) phi*I_a  (prop.) I_a with phi constant and T is constant due to full-load
I_a2 = I_a1

E_b1  =  V - I_a1*R_a
E_b2 = E_b1*(N_2/N_1) 			#N (prop.) E_b /phi (prop.) E_b   as phi is constant

R_x  =  (V-E_b2)/I_a2  -R_a

# Results
print 'resistance to be inserted in series with armature  =  %.4f ohms'%(R_x)

resistance to be inserted in series with armature  =  1.2528 ohms


## Example 2.28 page no : 68¶

In :
# Variables
V = 250.
N_0 = 1000.
I_0 = 5.
R_a = 0.2
R_sh = 250. 			#armature and shunt field resistance
I_L = 50. 			#on no load
I_sh = V / R_sh
I_a0  =  I_0 - I_sh
I_a   =  I_L - I_sh
E_b0  =  V- I_a0*R_a
E_b1  =  V- I_a *R_a

# Calculations
phi1_by_phi0  = 1-(3./100) 			#weakens by 3 percent
#N (prop.) E_b/phi
N_1  =  N_0 *(E_b1/E_b0) /phi1_by_phi0

# Results
print 'Speed when loaded and drawing 50A current is %.3f r.p.m'%(N_1)

Speed when loaded and drawing 50A current is 993.695 r.p.m


## Example 2.29 page no : 69¶

In :
import math

# Variables
V = 230.
I_a0 = 3.3
R_a = 0.3
R_sh = 160. 			#armature and shunt field resistance
I_L1 = 40.
N_0 = 1000.
E_b0  =  V - I_a0*R_a
I_sh = V/ R_sh
I_a1  =  I_L1 - I_sh
E_b1  =  V - I_a1*R_a
phi1_by_phi0 =  1- (4./100) 			#weakening by 4 percent

# Calculations and Results
N_1  =  N_0 *(E_b1/E_b0)/(phi1_by_phi0)  			#because N (prop.) E_b/phi
print 'Full load speed is %.4f rpm'%(N_1)
T_0  =  E_b0*I_a0/(2*math.pi*N_0/60)
T_1  =  T_0*(I_a1/I_a0)*phi1_by_phi0   			# because T (prop.) phi*I_a
print 'Full load developed torque is %.4f N-m'%(T_1)

Full load speed is 993.5485 rpm
Full load developed torque is 80.9585 N-m


## Example 2.30 page no : 70¶

In :
# Variables
V = 220.
I_L = 52.
N_1 = 750.
N_2 = 600.
R_a = 0.2
R_sh  =  110. 			#armature and shunt field resistance

# Calculations and Results
I_sh = V/ R_sh
I_a1 =  I_L - I_sh
I_a2 = I_a1			#T (prop.) I_a  and T is constant
E_b1  =  V - I_a1*R_a

#N (prop.) E_b/phi (prop.) E_b
E_b2  =  E_b1*(N_2/N_1)
R_x   =  (V- E_b2)/I_a2 -R_a  			#Because E_b2  =  V - I_a2*(R_a+R_x)
print 'resistance to be connected in series  =  %.2f ohms'%(R_x)

#After R_x gets connected in series with armature and 110 ohms in series with field winding
N_1 = 600.
I_sh2 = V /(R_sh+110)
I_a1 = 50.
I_sh1 = 2.
I_sh2 = 1.
#T (prop.) I_a*I_sh and T doesn't vary
I_a2  =  I_a1*(I_sh1/I_sh2)
E_b1  =  V - I_a1*(R_a+R_x)
E_b2  =  V - I_a2*(R_a+R_x)
N_2  =  N_1*(E_b2/E_b1)*(I_sh1/I_sh2) 			#Because N (prop.) E_b/I_sh
print 'New speed =  %.3f rpm'%(N_2)

resistance to be connected in series  =  0.84 ohms
New speed =  828.571 rpm


## Example 2.31 page no : 72¶

In :
# Variables
V = 230.
R_a = 0.15
R_sh = 250. 			#armature and shunt field resistance
I_a1 = 50.
I_a2 =  80.
N_1 = 800.
N_2 = 1000.
I_sh1 =  V / R_sh

# Calculations
E_b1  =  V - I_a1*R_a
E_b2  =  V - I_a2*R_a

I_sh2 = I_sh1*(E_b2/E_b1)*(N_1/N_2) 			#Because N (prop.) E_b/ I_sh
R_x =  (V/I_sh2 ) - R_sh    			#because I_sh2  =  V /(R_x+ R_sh)

# Results
print 'resistance to be added is R_x = %.0f ohms'%(R_x)

resistance to be added is R_x = 69 ohms


## Example 2.32 page no : 74¶

In :
# Variables
V = 230.
R_a = 0.5
N_1 = 800.
N_2 = 600.
I_a2  = 20.
I_a1 = I_a2
E_b1  =  V - I_a1*R_a

# Calculations
#N (prop.) E_b/phi (prop.) E_b         as phi is constant
E_b2 = E_b1 *(N_2/N_1)
R_x  =  (V -E_b2)/I_a2  - R_a   			#because E_b2  =  V - I_a2*(R_a+R_x)

# Results
print 'Additional resistance required  =  %.2f ohms '%(R_x)

Additional resistance required  =  2.75 ohms


## Example 2.33 page no : 74¶

In :
# Variables
V = 220.
R_a = 0.5
R_x = 5. 			#armature resistacne and extra resistance
I_1 = 15.
I_se1 = I_1
I_se2 = I_se1
I_2 = I_se2
N_1 = 800.

# Calculations
E_b1  =  V - I_1*R_a
E_b2  =  V - I_2*(R_a+R_x)

N_2 =  N_1*(E_b2/E_b1)*(I_se1/I_se2) 			#because N (prop.) E_b/I_se

# Results
print 'New speed of rotor  =  %.3f r.p.m'%(N_2)

New speed of rotor  =  517.647 r.p.m


## Example 2.34 page no : 75¶

In :
# Variables
V = 250.
I_a1 = 20.
R_a = 0.5
N_1 = 1000.
N_2 = 500.

# Calculations and Results
#T (prop.) I_a and T_1 = T_2
I_a2 = I_a1
E_b1  =  V - I_a1*R_a

#N (prop.) E_b
E_b2 =  E_b1 *(N_2/N_1)
R_x =  (V-E_b2)/I_a2 - R_a   			#because E_b2  =  V - I_a2*(R_a+R_x)
print 'Additional resistance  =  %.0f ohms'%(R_x)
T3_by_T2 = 0.5   			#torque is halved
I_a3 =  I_a2 *(T3_by_T2)   			#new armature current
E_b3  =  V - I_a3*(R_x + R_a)
N_3 = E_b3*N_2 / E_b2 			#N (prop.) E_b
print 'New speed  =  %.3f rpm'%(N_3)

Additional resistance  =  6 ohms
New speed  =  770.833 rpm


## Example 2.35 page no : 76¶

In :
# Variables
P_out =  100*735.5
V = 500.
P = 4.
A = 2.			# due to wave winding
Z = 492. 			#no of conductors
phi = 50.*10**-3 			#flux per pole
eta = 92./100 			#efficiency
P_in =  P_out/eta
R_a = 0.1
R_sh = 250. 			#amature and shunt field resistance

# Calculations
I_L = P_in/V
I_sh  =  V/ R_sh
I_a  =  I_L - I_sh
E_b  =  V - I_a*R_a
N = E_b*60*A/(phi*P*Z)  			#because E_b =   phi*P*N*Z/(60*A)

T_sh =  P_out/(2*math.pi*N/60) 			#Useful torque

# Results
print 'i)Speed at full load  =  %.4f rpm'%(N)
print 'ii)Useful torque  =  %.2f N-m'%(T_sh)
print 'Answer mismatches due to improper approximation'

i)Speed at full load  =  590.5011 rpm
ii)Useful torque  =  1189.41 N-m
Answer mismatches due to improper approximation


## Example 2.36 page no : 77¶

In :
from numpy import roots
import math

# Variables
N_1 = 1000.
I_1 = 50.
I_a1 = I_1
V = 250.
R_x = 4.4
R_t = 0.6   			#R_t  =  R_a+R_se
E_b1 = V - I_a1*(R_t)

# Calculations
#T (prop.)I_a**2   T (prop.) N**2  .... hence N (prop.) I_a
#N (prop.) E_b /I_a
#combining both  E_b (prop.) I_a**2
#using E_b2  =  V - I_a2*(R_a + R_se + R_x) and solving for I_a2 we get 0.088 I_a2**2 +5 I_a2 -250 = 0
p = [0.088 ,5, -250]
ans = roots(p)
I_a2 = ans 			#root(1) is ignored as it is -ve
E_b2  =  V - I_a2*(R_t + R_x)
N_2 = N_1*(E_b2/E_b1)*(I_a1/I_a2)

# Results
print 'Motor speed  =  %.2f r.p.m'%(N_2)

Motor speed  =  639.79 r.p.m


## Example 2.37 page no : 78¶

In :
from numpy import roots
import math

# Variables
V = 250.
I_a1 = 20.
R_sh =  250.
R_a = 0.5 			#shunt field and armature resistance
I_sh1 =  V / R_sh
E_b1  =  V - I_a1*R_a

#T (prop.) phi*I_a (prop.) I_sh*I_a
#math.since T_1  =  T_2
I_sh2_I_a2  =  I_sh1*I_a1
I_sh2_I_a2  =  I_sh1*I_a1  			# = 20

# Calculations
#N (prop.) E_b/I_sh
#E_b1  =  V - I_a1*R_a
#Solving further for I_a2 we get I_a2**2 -500 I_a2 + 12800
p = [1, -500, 12800]
ans = roots(p)
I_a2 = ans 			#higher root is neglected
I_sh2 =  I_sh2_I_a2 / I_a2
R_x =  (V / I_sh2) - R_sh 			#resistance to be inserted in shunt field

# Results
print 'resistance to be inserted  =  %.4f ohms '%(R_x)

resistance to be inserted  =  88.3129 ohms


## Example 2.38 page no : 79¶

In :
import math

# Variables
V = 250.
N_1 = 1000.
I_L1 = 25.
R_a = 0.2
R_sh = 250. 			#armature and shunt field resistance
V_brush =  1. 			#voltage drop due to brushes

# Calculations and Results
I_sh1  =  V/R_sh
I_a1 =  I_L1 - I_sh1
E_b1 =  V- I_a1*R_a - 2 *V_brush

I_L2 = 50.
I_sh2 = I_sh1   			#as flux weakensby armature reaction shunt field current remains same
I_a2 =  I_L2 - I_sh2
E_b2 =  V- I_a2*R_a - 2 *V_brush

phi2_by_phi1 =  1- (3./100)  			#weakens by 3 percent
N_2 =  N_1*(E_b2/E_b1)/ phi2_by_phi1   			#N (prop.) E_b/phi
print 'New speed  =  %.3f rpm'%(N_2)
T_1 =  E_b1*I_a1/(2*math.pi*N_1/60)
T_2 =  E_b2*I_a2/(2*math.pi*N_2/60)
print 'Torque before field weakening  =  %.4f N-m'%(T_1)
print 'Torque after  field weakening  =  %.4f N-m'%(T_2)

New speed  =  1009.733 rpm
Torque before field weakening  =  55.7373 N-m
Torque after  field weakening  =  110.3831 N-m


## Example 2.39 page no : 80¶

In :
import math

# Variables
V = 220.
R_a = 0.5
R_x = 1. 			#armature resistance and extra resistance
N_FL = 500. 			#full load speed in r.p.m
I_a_FL = 30.

# Calculations and Results
E_b_FL =  V- I_a_FL * R_a
#T (prop.) I_a... T is constant
I_a_dash_FL  =  I_a_FL
E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)
#N (prop.) E_b/phi   (prop.) E_b
N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)
print 'i)Speed at full load torque  = %.4f r.p.m'%(N_dash_FL)

#part(ii)
T2_by_T1  =  2
I_a_dash_FL   =   I_a_FL  *(T2_by_T1)
E_b_dash_FL  =   V- I_a_dash_FL * (R_a+R_x)
N_dash_FL  =  N_FL*(E_b_dash_FL/E_b_FL)
print 'ii)Speed at double full load torque  = %.3f r.p.m'%(N_dash_FL)

#part(iii) ...stalling
E_b = 0 			#as speed is zero in case of stalling torque
I_a_stall = (V-E_b)/(R_a+R_x)
T_FL  =  E_b_FL * I_a_FL/(2*math.pi*N_FL/60)
T_stall  =  T_FL *(I_a_stall/ I_a_FL)
print 'iii)Stalling torque  =  %.3f Nm'%(T_stall)

i)Speed at full load torque  = 426.8293 r.p.m
ii)Speed at double full load torque  = 317.073 r.p.m
iii)Stalling torque  =  574.231 Nm


## Example 2.40 page no : 81¶

In :
import math

# Variables
V = 230.
I_a1 = 30.
R_a = 0.4
R_x = 1.1			#armature resistance and extra resistance
N_1 = 500.

# Calculations and Results
#part(i)
E_b1 =  V - I_a1*R_a
I_a2  =  I_a1 			#I_a is constant as T , phi are constant
E_b2 =  V - I_a2*(R_a+R_x)
N_2  =  N_1 *(E_b2/E_b1) 			#Because N (prop.) E_b/phi (prop.) E_b
print 'i)Speed at full load torque  =  %.3f r.p.m'%(N_2)

#part(ii)
T2_by_T1 = 1.5
I_a2 =  I_a1 * T2_by_T1
E_b2 =  V - I_a2*(R_a+R_x)
N_2  =  N_1 *(E_b2/E_b1) 			#Because N (prop.) E_b/phi (prop.) E_b
print 'ii)Speed at 1.5 times full load torque  =  %.3f r.p.m'%(N_2)

i)Speed at full load torque  =  424.312 r.p.m
ii)Speed at 1.5 times full load torque  =  372.706 r.p.m