# CHAPTER 4: DC DYNAMO TORQUE RELATIONS-DC MOTORS¶

## Example 4.1, Page number 100¶

In [1]:
import math

#Variable declaration
d = 0.5        #Diameter of the coil(m)
l = 0.6        #Axial length of the coil(m)
B = 0.4        #Flux density(T)
I = 25.0       #Current carried by the coil(A)
theta = 60.0   #Angle between the useful force & the interpolar reference axis(degree)

#Calculation
F = B*I*l                             #Force developed on each coil side(N)
f = F*math.sin(theta*math.pi/180)     #Useful force at the instant the coil lies at an angle of 60° w.r.t the interpolar ref axis(N)
r = d/2                               #Radius of the coil(m)
T_c = f*r                             #Torque developed(N-m)
T_c1 = T_c*0.2248*3.281               #Torque developed in lb-ft by first method
T_c2 = T_c*0.737562                   #Torque developed in lb-ft by second method

#Result
print('Case(a): Force developed on each coil side , F = %.f N' %F)
print('Case(b): Useful force at the instant the coil lies at an angle of 60° w.r.t the interpolar ref axis , f = %.1f N' %f)
print('Case(c): Torque developed , T_c = %.1f N-m' %T_c)
print('Case(d): Torque developed by first method , 1.3 N.m * 0.2248 lb/N * 3.281 ft/m = %.2f lb-ft' %T_c1)
print('         Torque developed by second method , 1.3 N.m * 0.737562 lb.ft/N.m = %.2f lb-ft' %T_c2)

Case(a): Force developed on each coil side , F = 6 N
Case(b): Useful force at the instant the coil lies at an angle of 60° w.r.t the interpolar ref axis , f = 5.2 N
Case(c): Torque developed , T_c = 1.3 N-m
Case(d): Torque developed by first method , 1.3 N.m * 0.2248 lb/N * 3.281 ft/m = 0.96 lb-ft
Torque developed by second method , 1.3 N.m * 0.737562 lb.ft/N.m = 0.96 lb-ft


## Example 4.2, Page number 100¶

In [1]:
import math

#Variable declaration
d = 18.0       #Diameter of the coil(inches)
l = 24.0       #Axial length of the coil(inches)
B = 24000.0    #Flux density(lines/sq.inches)
I = 26.0       #Current carried by the coil(A)
theta = 60.0   #Angle between the useful force & the interpolar ref axis(degree)

#Calculation
F = (B*I*l*10**-7)/1.13             #Force developed on each conductor(lb)
f = F*math.sin(theta*math.pi/180)   #Useful force at the instant the coil lies at an angle of 60° w.r.t the interpolar ref axis(lb)
r = d/2                             #Radius of the coil(inches)
T_c = f*(r*1.0/12)                  #Torque developed(lb.ft/conductor)

#Result
print('Case(a): Force developed on each conductor , F = %.3f lb' %F)
print('Case(b): Useful force , f = %.2f lb' %f)
print('Case(c): Torque developed , T_c = %.3f lb-ft/conductor' %T_c)

Case(a): Force developed on each conductor , F = 1.325 lb
Case(b): Useful force , f = 1.15 lb
Case(c): Torque developed , T_c = 0.861 lb-ft/conductor


## Example 4.3, Page number 102¶

In [1]:
#Variable declaration
Z = 700.0       #Number of conductors
d = 24.0        #Diameter of the armature of the dc motor(inches)
l = 34.0        #Axial length of the coil(inches)
B = 50000.0     #Flux density(lines/sq.inches)
I = 25.0        #Current carried by the coil(A)
per = 0.7       #Conductors lying directly under poles

#Calculation
F_av = (B*I*l*10**-7)/1.13*(Z*per)   #Average total force tending to rotate the armature(lb)
r = d/2                              #Radius of the coil(inches)
T_av = F_av*(r/12)                   #Armature torque(lb-ft)

#Result
print('Case(a): Average total force tending to rotate the armature , F_av = %.f lb' %F_av)
print('Case(b): Armature torque , T_av = %.f lb-ft' %T_av)

Case(a): Average total force tending to rotate the armature , F_av = 1843 lb
Case(b): Armature torque , T_av = 1843 lb-ft


## Example 4.4, Page number 102¶

In [1]:
#Variable declaration
slots = 120.0                #Number of armature slots
conductors_per_slot = 6.0    #Number of conductors per slot
B = 60000.0                  #Flux density(lines/sq.inches)
d = 28.0                     #Diameter of the armature(inches)
l = 14.0                     #Axial length of the coil(inches)
A = 4.0                      #Number of parallel paths
span = 0.72                  #Pole arcs span 72% of the armature surface
I = 133.5                    #Armature current(A)

#Calculation
Z_Ta = slots*conductors_per_slot*span     #Number of armature conductors
F_t = (B*I*l)/(1.13*10**7 *A)*Z_Ta        #Force developed(lb)
r = (d/2)/12                              #Radius of the armature(feet)
T = F_t*r                                 #Total torque developed(lb-ft)

#Result
print('Total developed armature torque , T = %.f lb-ft' %T)

Total developed armature torque , T = 1500 lb-ft


## Example 4.5, Page number 103¶

In [1]:
#Variable declaration
slots = 120.0                #Number of armature slots
conductors_per_slot = 6.0    #Number of conductors per slot
B = 60000.0                  #Flux density(lines/sq.inches)
d = 28.0                     #Diameter of the armature(inches)
l = 14.0                     #Axial length of the coil(inches)
A = 4.0                      #Number of parallel paths
span = 0.72                  #Pole arcs span 72% of the armature surface
T_a = 1500.0                 #Total armature torque(lb-ft)

#Calculation
Z = slots*conductors_per_slot         #Number of armature conductors
r = (d/2)/12                          #Radius of the armature(feet)
I_a = T_a*A*1.13*10**7/(B*l*Z*r*span) #Total external armature current(A)

#Result
print('Total external armature current , I_a = %.1f A' %I_a)

Total external armature current , I_a = 133.5 A


## Example 4.6, Page number 104¶

In [1]:
#Variable declaration
phi_orig = 1.0    #Original flux
Ia_orig = 1.0     #Original armature current
T_orig = 150.0    #Original torque(N-m)
phi_new = 0.9     #New flux
Ia_new = 1.5      #New armature current

#Calculation
T_new = T_orig*(phi_new/phi_orig)*(Ia_new/Ia_orig)     #New torque produced(N-m)

#Result
print('New torque produced , T = %.1f N-m' %T_new)

New torque produced , T = 202.5 N-m


## Example 4.7, Page number 105¶

In [1]:
#Variable declaration
R_a = 0.25    #Armature resistance(ohm)
BD = 3.0      #Brush contact drop(V)
V = 120.0     #Applied voltage(V)
E_ca = 110.0  #counter EMF at a given load(V)

#Calculation
I_a_a = (V-(E_ca+BD))/R_a            #Armature current(A)
I_a_b = (V-(E_cb+BD))/R_a            #Armature current(A)
del_Ec = ((E_ca-E_cb)/E_ca)*100      #Change in counter EMF(percent)
del_Ia = ((I_a_b-I_a_a)/I_a_a)*100   #Change in armature current(percent)

#Result
print('Case(a): Aramature current , I_a = %.f A' %I_a_a)
print('Case(b): Aramature current due to additional load , I_a = %.f A' %I_a_b)
print('Case(c): Change in counter EMF , δE_c = %.2f percent' %del_Ec)
print('         Change in armature current , δI_a = %.1f percent' %del_Ia)

Case(a): Aramature current , I_a = 28 A
Case(b): Aramature current due to additional load , I_a = 48 A
Case(c): Change in counter EMF , δE_c = 4.55 percent
Change in armature current , δI_a = 71.4 percent


## Example 4.8, Page number 106¶

In [1]:
#Variable declaration
V_a = 120.0       #Rated terminal voltage of the DC motor(V)
R_a = 0.2         #Armature circuit resistance(ohm)
R_sh = 60.0       #Shunt field resistance(ohm)
I_l = 40.0        #Line current at full-load(A)
BD = 3.0          #Brush voltage drop(V)
S_orig = 1800.0   #Rated full-load speed(rpm)

#Calculation
#Case(a)
I_f = V_a/R_sh                     #Field current(A)
I_a_fl = I_l-I_f                   #Armature current at full-load(A)
E_c_orig = V_a-(I_a_fl*R_a+BD)     #Back EMF at full-load(V)
I_a_nl = I_a_fl/2                  #Armature current at half-load(A)
E_c_final = V_a-(I_a_nl*R_a+BD)    #Back EMF at half load(V)
S_a = S_orig*(E_c_final/E_c_orig)  #Speed at full load(rpm)
#Case(b)
I_a_b = I_a_fl*per                 #Armature current at 125% overload(A)
E_c_b = V_a-(I_a_b*R_a+BD)         #Back EMF at 125% overload(V)
S_b = S_orig*(E_c_b/E_c_orig)      #Speed at 125% overload(rpm)

#Result
print('Case(a): Speed at half load , S = %.f rpm' %S_a)
print('Case(b): Speed at an overload of 125 perecent , S = %.f rpm' %S_b)

Case(a): Speed at half load , S = 1863 rpm
Case(b): Speed at an overload of 125 perecent , S = 1769 rpm


## Example 4.9, Page number 106¶

In [1]:
#Variable declaration
I_l_orig = 40.0         #Original line current(A)
I_l_final = 66.0        #Final line current(A)
phi_orig = 1.0          #Original flux
phi_final = 112.0/100   #Final flux
V_a = 120.0             #Rated terminal voltage of the DC motor(V)
R_sh_orig = 60.0        #Original Field circuit resistance(ohm)
R_sh_final = 50.0       #Decreased final field circuit resistance(ohm)
R_a = 0.2               #Armature circuit resistance(ohm)
BD = 3.0                #Brush voltage drop(V)
S_orig = 1800.0         #Rated full-load speed(rpm)

#Calculation
I_f_orig = V_a/R_sh_orig                              #Original Field current(A)
I_a_orig = I_l_orig-I_f_orig                          #Original Armature current at full-load(A)
E_c_orig = V_a-(I_a_orig*R_a+BD)                      #Back EMF at full load(V)
I_f_final = V_a/R_sh_final                            #Final field current(A)
I_a_final = I_l_final-I_f_final                       #Final Armature current(A)
E_c_final = V_a-(I_a_final*R_a+BD)                    #Final EMF induced(V)
S = S_orig*(E_c_final/E_c_orig)*(phi_orig/phi_final)  #Final speed of the motor(rpm)

#Result
print('Speed of the motor , S = %.f rpm' %S)

Speed of the motor , S = 1532 rpm


## Example 4.10, Page number 108¶

In [1]:
#Variable declaration
I_a_1 = 38.0      #Armature current at full-load(A) from example 4-8a
E_c_1 = 109.4     #Back EMF at full-load(V)
S_1 = 1800.0      #Speed at full-load(rpm)
I_a_2 = 19.0      #Armature current at half-load(A) from example 4-8a
E_c_2 = 113.2     #Back EMF at half-load(V)
S_2 = 1863.0      #Speed at half-load(rpm)
I_a_3 = 47.5      #Armature current at 125% overload(A) from example 4-8b
E_c_3 = 107.5     #Back EMF at 125% overload(V)
S_3 = 1769.0      #Speed at 125% overload(rpm)
I_a_4 = 63.6      #Armature current at overload(A) from example 4-9
E_c_4 = 104.3     #Back EMF at overload(V)
S_4 = 1532.0      #Speed at overload(rpm)

#Calculation
P_d_1 = E_c_1*I_a_1    #Armature power developed at full-load(W)
P_d_2 = E_c_2*I_a_2    #Armature power developed at half-load(W)
P_d_3 = E_c_3*I_a_3    #Armature power developed at 125% overload(W)
P_d_4 = E_c_4*I_a_4    #Armature power developed at overload(W)

#Result
print(' Example \t I_a \t E_c \t Speed \t P_d or (E_c*I_a)')
print(' _______________________________________________________________________')
print(' 4-8a \t\t %d \t %.1f \t %d \t %d W at full-load' %(I_a_1,E_c_1,S_1,P_d_1))
print('      \t\t %d \t %.1f \t %d \t %.f W at half-load' %(I_a_2,E_c_2,S_2,P_d_2))
print(' 4-8b \t\t %.1f \t %.1f \t %d \t %d W at 125 percent overload' %(I_a_3,E_c_3,S_3,P_d_3))
print(' 4-9  \t\t %.1f \t %.1f \t %d \t %d W at overload' %(I_a_4,E_c_4,S_4,P_d_4))
print(' _______________________________________________________________________')

 Example 	 I_a 	 E_c 	 Speed 	 P_d or (E_c*I_a)
_______________________________________________________________________
4-8a 		 38 	 109.4 	 1800 	 4157 W at full-load
19 	 113.2 	 1863 	 2151 W at half-load
4-8b 		 47.5 	 107.5 	 1769 	 5106 W at 125 percent overload
4-9  		 63.6 	 104.3 	 1532 	 6633 W at overload
_______________________________________________________________________


## Example 4.11, Page number 110¶

In [1]:
#Variable declaration
T_a = 6.5     #Torque(dyne-centimeters)
T_b = 10.6    #Torque in (gram-centimeters)
T_c = 12.2    #Torque in (ounce-inches)

#Calculation
T_a_Nm = T_a*1.416*10**-5*7.0612*10**-3    #Torque(N-m)
T_a_lbft = T_a*1.416*10**-5*5.208*10**-3   #Torque(lb-ft)
T_b_Nm = T_b*(1/72.01)*7.0612*10**-3       #Torque(N-m)
T_b_lbft = T_b*(1/72.01)*5.208*10**-3      #Torque(lb-ft)
T_c_Nm = T_c*7.0612*10**-3                 #Torque(N-m)
T_c_lbft = T_c*5.208*10**-3                #Torque(lb-ft)

#Result
print('Case(a): Torque , T = %.1e N-m' %T_a_Nm)
print('         Torque , T = %.1e lb-ft' %T_a_lbft)
print('Case(b): Torque , T = %.2e N-m' %T_b_Nm)
print('         Torque , T = %.2e lb-ft' %T_b_lbft)
print('Case(c): Torque , T = %.3e N-m' %T_c_Nm)
print('         Torque , T = %.2e lb-ft' %T_c_lbft)

Case(a): Torque , T = 6.5e-07 N-m
Torque , T = 4.8e-07 lb-ft
Case(b): Torque , T = 1.04e-03 N-m
Torque , T = 7.67e-04 lb-ft
Case(c): Torque , T = 8.615e-02 N-m
Torque , T = 6.35e-02 lb-ft


## Example 4.12, Page number 110¶

In [1]:
#Variable declaration
V_a = 120.0    #Rated terminal voltage of dc shunt motor(V)
R_a = 0.2      #Armature resistance(ohm)
BD = 2.0       #Brush drop(V)
I_a = 75.0     #Full-load armature current(A)

#Calculation
I_st = (V_a-BD)/R_a           #Current at the instant of starting(A)
percentage = I_st/I_a*100     #Percentage at full load

#Result
print('Current at the instant of starting , I_st = %.f A (Counter EMF is zero)' %I_st)
print('Percentage at full load = %d percent' %percentage)

Current at the instant of starting , I_st = 590 A (Counter EMF is zero)
Percentage at full load = 786 percent


## Example 4.13, Page number 111¶

In [1]:
#Variable declaration
V_a = 120.0          #Rated terminal voltage of dc shunt motor(V)
R_a = 0.2            #Armature resistance(ohm)
BD = 2.0             #Brush drop(V)
I_a = 75.0           #Full-load armature current(A)
I_a_new = 1.5*I_a    #Armature current at 150% rated load(A)
E_c_a = 0            #Back EMF at starting(V)
E_c_b = 0.25* V_a    #Back EMF is 25% of Va at 150% rated load(V)
E_c_c = 0.5*V_a      #Back EMF is 50% of Va at 150% rated load(V)

#Calculation
R_s_a = (V_a-E_c_a-BD)/I_a_new-R_a   #Starting resistance at starting(ohm)
R_s_b = (V_a-E_c_b-BD)/I_a_new-R_a   #Starting resistance 25% of Va(ohm)
R_s_c = (V_a-E_c_c-BD)/I_a_new-R_a   #Starting resistance 50% of Va(ohm)
E_c_d = V_a-(I_a*R_a+BD)             #Counter EMF at full-load without starting resistance(V)

#Result
print('Case(a): Starting resistance at the instant of starting , R_s = %.2f Ω' %R_s_a)
print('Case(b): Starting resistance at 25 percent of armature voltage , R_s = %.3f Ω' %R_s_b)
print('Case(c): Starting resistance at 50 percent of armature voltage , R_s = %.3f Ω' %R_s_c)
print('Case(d): Counter EMF at full-load without starting resistance , E_c = %.f V' %E_c_d)

Case(a): Starting resistance at the instant of starting , R_s = 0.85 Ω
Case(b): Starting resistance at 25 percent of armature voltage , R_s = 0.582 Ω
Case(c): Starting resistance at 50 percent of armature voltage , R_s = 0.316 Ω
Case(d): Counter EMF at full-load without starting resistance , E_c = 103 V


## Example 4.14, Page number 115¶

In [1]:
#Variable declaration
T_orig = 160.0          #Original torque developed(lb-ft)
I_a_orig = 140.0        #Original armature current(A)
phi_f_orig = 1.6*10**6  #Original field flux(lines)
T_final_a = 190.0       #Final torque developed when reconnected as a cumulative compound motor(lb-ft)
I_a_b = 154.0           #Final armature current(A)

#Calculation
phi_f = phi_f_orig*(T_final_a/T_orig)                  #Field flux(lines)
percentage = (phi_f/phi_f_orig)*100-100                #Percentage increase in flux
phi_f_final = 1.1*phi_f                                #Final field flux(lines)
T_f = T_final_a*(I_a_b/I_a_orig)*(phi_f_final/phi_f)   #Final torque developed(lb-ft)

#Result
print('Case(a): Flux increase due to series field , Φ_f = %.1f percent' %percentage)
print('Case(b): Final torque , T_f = %.f lb-ft' %T_f)

Case(a): Flux increase due to series field , Φ_f = 18.8 percent
Case(b): Final torque , T_f = 230 lb-ft


## Example 4.15, Page number 115¶

In [1]:
#Variable declaration
I_a_orig = 25.0    #Original armature current(A)
I_a_final = 30.0   #Final armature current(A)
T_orig = 90.0      #Original torque developed(lb-ft)
phi_orig = 1.0     #Original flux
phi_final = 1.1    #Final flux

#Calculation
T_a = T_orig*(I_a_final/I_a_orig)**2   #Final torque developed if field is unsaturated(lb-ft)
T_b = T_orig*(I_a_final/I_a_orig)*(phi_final/phi_orig) #Final torque developed when current rises to 30A and flux increases by 10%

#Result
print('Case(a): Torque when field is unsaturated , T = %.1f lb-ft' %T_a)
print('Case(b): Torque when current rises to 30 A and flux increases by 10 percent , T = %.1f lb-ft' %T_b)

Case(a): Torque when field is unsaturated , T = 129.6 lb-ft
Case(b): Torque when current rises to 30 A and flux increases by 10 percent , T = 118.8 lb-ft


## Example 4.16, Page number 119¶

In [1]:
#Variable declaration
V_a = 230.0   #Rated armature voltage(V)
P = 10.0      #Rated power(hp)
S = 1250.0    #Rated speed(rpm)
R_A = 0.25    #Armature resistance(ohm)
R_p = 0.25    #Interpole resistance(ohm)
BD = 5.0      #Brush voltage drop(V)
R_s = 0.15    #Series field resistance(ohm)
R_sh = 230.0  #Shunt field resistance(ohm)
I_fl = 55.0   #Line current at rated load(A)
I_ol = 4.0    #No-load line current(A)

#Calculation
#Case(a)
R_a = R_A+R_p                          #Effective armature resistance(ohm)
I_f = V_a/R_sh                         #Field current(A)
I_a = I_ol-I_f                         #Armature current at no-load(A)
E_c_o = V_a-(I_a*R_a+BD)               #No-load back EMF(V)
I_a_fl = I_fl-I_f                      #Armature current at full-load(A)
#Case(b)
hp = P_d/746.0                         #Internal horse power(hp)

#Result
print('Case(a): Speed at rated load , S_r = %.f rpm' %S_r)
print('Case(b): Internal power in watts , P_d = %.f W' %P_d)
print('         Internal horse-power developed , P_d = %.1f hp' %hp)

Case(a): Speed at rated load , S_r = 1603 rpm
Case(b): Internal power in watts , P_d = 10692 W
Internal horse-power developed , P_d = 14.3 hp


## Example 4.17, Page number 119¶

In [1]:
#Variable declaration
V_a = 230.0    #Rated armature voltage(V)
P = 10.0       #Rated power(hp)
S = 1250.0     #Rated speed(rpm)
R_A = 0.25     #Armature resistance(ohm)
R_p = 0.25     #Interpole resistance(ohm)
BD = 5.0       #Brush voltage drop in volt
R_s = 0.15     #Series field resistance in ohm
R_sh = 230.0   #Shunt field resistance in ohm
phi_1 = 1.0    #Original flux per pole
I_fl = 55.0    #Line current at rated load(A)
phi_2 = 1.25   #Flux increased by 25% due to long-shunt cumulative connection
I_ol = 4.0     #No-load line current(A)

#Calculation
R_a = R_A+R_p                          #Effective armature resistance(ohm)
I_f = V_a/R_sh                         #Field current(A)
I_a = I_ol-I_f                         #Armature current at no-load(A)
E_c_o = V_a-(I_a*R_a+BD)               #No-load back EMF(V)
I_a_fl = I_fl-I_f                      #Armature current at full-load(A)
#Case(a)
E_c_o1 = V_a-(I_a*R_a+I_a*R_s+BD)      #No-load back EMF for long shunt cumulative connection(V)
S_n1 = S_o*(E_c_o1/E_c_o)              #Speed at no-load(rpm)
#Case(b)
E_c_full_load_lsh = V_a-(I_a_fl*R_a+I_a_fl*R_s+BD )       #Back EMF at full-load for long-shunt cumulative motor(V)
#Case(c)
hp = P_d/746.0                                            #Internal horse power(hp)
T_shunt = hp*5252/S_r                                     #Internal torque at full-load for shunt motor(lb-ft)
I_a1 = I_a_fl                                             #Armature current for shunt motor(A)
I_a2 = I_a_fl                                             #Armature current for long-shunt cumulative motor(A)
T_comp = T_shunt*(phi_2/phi_1)*(I_a2/I_a1)                #Internal torque at full load for long-shunt cumulative motor(lb-ft)
#Case(d)
Horsepower = (E_c_full_load_lsh*I_a_fl)/746               #Internal horsepower of compound motor based on flux increase(hp)

#Result
print('Case(a): Speed at no-load , S_n1 = %d rpm' %S_n1)
print('Case(b): Speed at rated-load , S_r = %d rpm' %S_r_lsh)
print('Case(c): Internal torque at full-load with series field , T_comp = %.2f lb-ft' %T_comp)
print('         Internal torque at full-load without series field , T_shunt = %.2f lb-ft' %T_shunt)
print('Case(d): Internal horsepower of the compound motor , Horsepower = %.1f hp' %Horsepower)
print('Case(e): The internal hp exceeds the rated hp because the power developed in the motor must also overcome the internal mechanical rotational losses')
print('\nNOTE: The change in obtained answer from that of textbook is due to more precision i.e more number of decimal places in this case')

Case(a): Speed at no-load , S_n1 = 1806 rpm
Case(b): Speed at rated-load , S_r = 1230 rpm
Case(c): Internal torque at full-load with series field , T_comp = 58.68 lb-ft
Internal torque at full-load without series field , T_shunt = 46.94 lb-ft
Case(d): Internal horsepower of the compound motor , Horsepower = 13.7 hp
Case(e): The internal hp exceeds the rated hp because the power developed in the motor must also overcome the internal mechanical rotational losses

NOTE: The change in obtained answer from that of textbook is due to more precision i.e more number of decimal places in this case


## Example 4.18, Page number 120¶

In [1]:
#Variable declaration
P = 25.0      #Power rating of a series motor(hp)
V_a = 250.0   #Rated voltage(V)
R_a = 0.1     #Armature circuit resistance(ohm)
BD = 3.0      #Brush voltage drop(V)
R_s = 0.05    #Series field resistance(ohm)
I_a = 85.0    #Armature current(A)
I_a1 = 100.0  #Armature current(A)(case a)
I_a2 = 40.0   #Armature current(A)(case b)
S_1 = 600.0   #Speed(rpm)
R_d = 0.05    #Diverter resistance(ohm)

#Calculation
#Case(a)
E_c2 = V_a-I_a1*(R_a+R_s)-BD       #Back EMF when Ia = 100 A(V)
E_c1 = V_a-I_a*(R_a+R_s)-BD        #Back EMF when Ia = 85 A(V)
S_2 = S_1*(E_c2/E_c1)*(I_a/I_a1)   #Speed(rpm)
#Case(b)
E_c3 = V_a-I_a2*(R_a+R_s)-BD       #Back EMF when Ia = 40 A(V)
S_3 = S_1*(E_c3/E_c1)*(I_a/I_a2)   #Speed(rpm)
#Case(c)
R_sd = (R_s*R_d)/(R_s+R_d )        #Effective series field resistance(ohm)
E_c2_new = V_a-I_a1*(R_a+R_sd)-BD  #Back EMF when Ia = 100 A(V)
S_2_new = S_1*(E_c2_new/E_c1)*(I_a/(I_a1/2))  #Speed(rpm)
E_c3_new = V_a-I_a2*(R_a+R_sd)-BD  #Back EMF when Ia = 40 A(V)
S_3_new = S_1*(E_c3_new/E_c1)*(I_a/(I_a2/2))  #Speed(rpm)

#Result
print('Case(a): Speed when current is 100 A , S_2 = %.f rpm' %S_2)
print('Case(b): Speed when current is 40 A , S_3 = %d rpm' %S_3)
print('Case(c): Speed when current is 100 A and using a diverter , S_2 = %.f rpm' %S_2_new)
print('         Speed when current is 40 A and using a diverter , S_3 = %.f rpm' %S_3_new)

Case(a): Speed when current is 100 A , S_2 = 505 rpm
Case(b): Speed when current is 40 A , S_3 = 1311 rpm
Case(c): Speed when current is 100 A and using a diverter , S_2 = 1021 rpm
Speed when current is 40 A and using a diverter , S_3 = 2634 rpm


## Example 4.19, Page number 121¶

In [1]:
#Variable declaration
S_n1 = 1810.0    #No-load speed(rpm) From Ex. 4-16
S_n2 = 1806.0    #No-load speed(rpm) From Ex. 4-17
S_n3 = 1311.0    #No-load speed(rpm) From Ex. 4-18

#Calculation
SR_1 = (S_n1-S_f1)/S_f1*100   #Speed regulation for shunt motor(%)
SR_2 = (S_n2-S_f2)/S_f2*100   #Speed regulation for compound motor(%)
SR_3 = (S_n3-S_f3)/S_f3*100   #Speed regulation for series motor(%)

#Result
print('Case(a): Speed regulation , SR(shunt) = %.1f percent' %SR_1)
print('Case(b): Speed regulation , SR(compound) = %.1f percent' %SR_2)
print('Case(c): Speed regulation , SR(series) = %.1f percent' %SR_3)

Case(a): Speed regulation , SR(shunt) = 12.9 percent
Case(b): Speed regulation , SR(compound) = 46.7 percent
Case(c): Speed regulation , SR(series) = 159.6 percent


## Example 4.20, Page number 121¶

In [1]:
import math

#Variable declaration
SR = 0.1                    #Speed regulation of a shunt motor

#Calculation

#Result
print('Case(b): No-load speed , S = %.f rpm' %S)

Case(a): No-load speed , ω_n1 = 66π rad/s
Case(b): No-load speed , S = 1980 rpm


## Example 4.21, Page number 122¶

In [1]:
#Variable declaration
S_int = 1603.0    #Internal rated speed(rpm)
S_ext = 1250.0    #External rated speed(rpm)
hp_int = 14.3     #Internal horsepower
hp_ext = 10.0     #External horsepower

#Calculation
T_int = hp_int*5252/S_int   #Internal torque(lb-ft)
T_ext = hp_ext*5252/S_ext   #External torque(lb-ft)

#Result
print('Case(a): Internal torque , T_int = %.2f lb-ft' %T_int)
print('Case(b): External torque , T_ext = %.1f lb-ft' %T_ext)
print('Case(c): Internal hp developed due electromagnetic torque is used internally to overcome mechanical losses of the motor reducing the torque available at its shaft to perform work')

Case(a): Internal torque , T_int = 46.85 lb-ft
Case(b): External torque , T_ext = 42.0 lb-ft
Case(c): Internal hp developed due electromagnetic torque is used internally to overcome mechanical losses of the motor reducing the torque available at its shaft to perform work


## Example 4.22, Page number 123¶

In [1]:
#Variable declaration
P = 50.0    #Power rating of the servo motor(W)
S = 3000.0  #Full-load speed of the servo motor(rpm)

#Calculation
T_lbft = (7.04*P)/S          #Output torque(lb-ft)
T_ounceinch = T_lbft*192     #Output torque(ounce-inches)

#Result
print('Output torque available at the motor pulley , T = %.1f oz-in' %T_ounceinch)

Output torque available at the motor pulley , T = 22.5 oz-in


## Example 4.23, Page number 123¶

In [1]:
import math

#Variable declaration
P = 50.0    #Power rating of the servo motor(W)
S = 3000.0  #Full-load speed of the servo motor(rpm)

#Calculation
T_Nm = P/omega                         #Output torque(N-m)
T_ounceinch = T_Nm*1/(7.0612*10**-3)   #Output torque(oz-in)

#Result

Case(a): Motor speed in radians per second = 314.2 rad/s