import math
# GIVEN DATA
N = 600; # Speed of the driven Machine in RPM
D = 2; # Diameter of the Machine in Meter
L = 0.3; # Length of the Machine in Meter
Bm = 1.0; # Flux Density in Weber per Meter-Square
# CALCULATIONS
n = N/60; # Revolution per second
v = math.pi * D * n; # Peripheral velocity in Meter per second
E = Bm * v * L; # Maximum EMF induced in the Conducter in Volts
# DISPLAY RESULTS
print ("EXAMPLE : 4.1 : SOLUTION :-") ;
print " a) Maximum EMF induced in the Conducter , E = %.3f V "%(E);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- a)Induced EMF, E = 2.826 V instead of %.3f A "%(E);
print " From Calculation of the peripheral velocity(v), rest all the Calculated values in the TEXT BOOK is WRONG because of the peripheral velocity(v) value is WRONGLY calculated and the same used for the further Calculation part "
import math
# GIVEN DATA
p = 8; # Number of the poles in Dc machine
a = 8; # Number of the Parallel path
N = 500; # Rotation per minute in RPM
phi = 0.095; # Average flux in air gap in Weber per meter
Za = 1000; # Total number of the Conductor in Armature
# CALCUALTIONS
n = N/60; # Rotation (Revolution) per Second
E = (p/a)*n*phi*Za; # EMF induced in Volts
# DISPLAY RESULTS
print ("EXAMPLE : 4.2 : SOLUTION :-") ;
print " a) EMF induced , E = %.1f A "%(E);
import math
# GIVEN DATA
E = 420; # EMF induced in Volts
N = 900; # Rotation speed in RPM
phi = 0.06; # Flux per pole in Weber per pole
Two_p = 4; # Total number of poles
# CALCULATIONS
n = N/60; # Revolution Per second
Zc = E/(Two_p*phi*n); # Number of the Conductor in Parallel Path
# DISPLAY RESULTS
print ("EXAMPLE : 4.3 : SOLUTION :-") ;
print " a) Number of the Conductor in Parallel Path , Zc = %.2f Conductors nearly 117 conductors "%(Zc);
import math
# GIVEN DATA
L = 0.3; # Length of the Machine in Meter
Ia = 10; # Current through The Conductors in Ampheres
N = 10; # Number of the Conductors in each Slot
Za = 24; # Number of the Slots
Bav = 0.6; # Average Flux Density in Telsa
D = 0.1; # Machine Daimeter in Meter
# CALCULATIONS
F = N*Ia*Bav*L; # Force due to the Single Slot in Newton
T = (Bav*L*Ia*N*D*Za)/2 # Torque produced in the Machine in Newton-Meter
# DISPLAY RESULTS
print ("EXAMPLE : 4.4 : SOLUTION :-") ;
print " a) Torque produced in the Machine, T = %.1f N-m "%(T);
import math
# GIVEN DATA
p = 4.; # Number of the Poles in the DC machine
Nt = 100.; # Number of the turns in the Dc machine
N = 600.; # Rotation speed of the DC machine in RPM
E = 220.; # EMF generated in open circuit in Volts
Z = 200.; # Total number of the Conductor in armature
# CALCUALTIONS
# For case (a) Lap Connected
a = 4.; # Number of the Poles in the DC machine
n = N/60; # Revolution per second
phi_a = (E*a)/(p*Z*n); # Useful flux per pole when Armature is Lap connected in Weber
# For case (b) Wave Connected
a = 2; # Number of the Poles in the DC machine
phi_b = (E*a)/(p*Z*n); # Useful flux per pole when Armature is Wave connected in Weber
# DISPLAY RESULTS
print ("EXAMPLE : 4.5 : SOLUTION :-") ;
print " a) Useful flux per pole when Armature is Lap connected , phi = %.1f Wb "%(phi_a);
print " B) Useful flux per pole when Armature is Lap connected , phi = %.3f Wb "%(phi_b);
import math
# GIVEN DATA
p = 6; # Number of the pole in DC Motor
Ia = 20; # Armature Current in Amphere
Z = 1000; # Number of the Conductors
a = 6; # Number of the Parallel paths
phi = 25 * 10 ** -3; # Flux per pole in Weber
# CALCULATIONS
T = (p/a)*((Z*Ia*phi)/(2*math.pi)); # Deleloped Torque in Newton-Meter
# DISPLAY RESULTS
print ("EXAMPLE : 4.6 : SOLUTION :-") ;
print " a) Developed Torque in an Six-pole DC Motor , T = %.1f N-m "%(T);
import math
# GIVEN DATA
p = 2; # Number of the Pole
N = 1000; # Rotation speed of the Armature in RPM
Ia = 20; # Armature Current in Amphere
CS = 36; # Commutator Segments
BW = 1.4; # Brush width is 1.4 times of the Commutator Segments
L = 0.09 * 10 ** -3; # Inducatnce of the each Armature Coil
# CALULATIONS
a = p; # Number of the Parallel paths (Equal to number of poles because Lap Connected Armature
n = N/60; # Revoultion per second
I = Ia/2; # Current Through the each Conductor in Amphere
v = n * CS; # Peripheral Velocity of Commutator in Commutator segments per Seconds
Tc = BW/v; # Time of the Commutation in Seconds
Er = (L*2*I)/Tc; # reactance voltage in Volts
# DISPLAY RESULTS
print ("EXAMPLE : 4.7 : SOLUTION :-") ;
print " a) reactance voltage assuming Linear Commutation , Er = %.4f V "%(Er);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- a) Tc = 0.014 s instead of %.4f s "%(Tc);
print " b) Er = 1.2857 V instead of %.4f V"%(Er);
print " From Calculation of the Time of commutation Tc), rest all the Calculated values in the TEXT BOOK is WRONG because of the Time of commutation Tc) value is WRONGLY calculated and the same used for the further Calculation part "
import math
# GIVEN DATA
N = 800; # Rotation speed of the Commutator in RPM
D = 50; # Diameter in Centimeter
BW = 1.5; # Brush Width in Centimeter
# CALCULATIONS
r = D/2; # Radius in Centimeter
n = N/60; # Revoultion per second
w = (2 * math.pi)*n; # Angular velocity
v = w*r; # Peripheral Speed in centimeter per second
Tc = (BW/v)*1000; # Time of the Commutation in Second
# DISPLAY RESULTS
print ("EXAMPLE : 4.8 : SOLUTION :-") ;
print " a) Time of the Commutation , Tc = %.4f ms "%(Tc);
import math
# GIVEN DATA
p = 4; # Number of the pole in the Generator
Ia = 100; # supplying Current by the Generator in Amphere
Za = 500; # Armature conductor
beta = 8; # Brush shift in degrees
If = 5; # Current in the Seperately exicted field winding
ratio = 0.7; # Ratio of Pole arc to Pole pitch
# CALCULATIONS
# For case (a) Lap winding
a_a = p; # Number of the Parallel Paths
AT_a = (Za*Ia)/(2*a_a*p); # Amphere turns per pole
ATd_a = (beta*Za*Ia)/(360*a_a); # Demagnetizing Armature Amphere turns per pole
ATc_a = ((1/p)-(beta/180))*((Za*Ia)/(2*a_a)); # CrossMagnetizing Armature Amphere turns per pole
ATw_a = ratio*AT_a; # Amphere turns of Compensating winding
# For case (b) Wave winding
a_b = p/2; # Number of the Parallel Paths
AT_b = (Za*Ia)/(2*a_b*p); # Amphere turns per pole
ATd_b = (beta*Za*Ia)/(360*a_b); # Demagnetizing Armature Amphere turns per pole
ATc_b = ((1/p)-(beta/180))*((Za*Ia)/(2*a_b)); # CrossMagnetizing Armature Amphere turns per pole
ATw_b = ratio*AT_b; # Amphere turns of Compensating winding
# DISPLAY RESULTS
print ("EXAMPLE : 4.9 : SOLUTION :-") ;
print " For LAP winding a.1) Amphere turns per pole, AT = %.1f AT "%(AT_a);
print " a.2) Demagnetizing Armature Amphere turns per pole, ATd = %.1f AT "%(ATd_a);
print " a.3) Cross-Magnetizing Armature Amphere turns per pole, ATc = %.1f AT "%(ATc_a);
print " a.4) Amphere turns of Compensating winding, ATw = %.1f AT "%(ATw_a);
print " For WAVE winding b.1) Amphere turns per pole, AT = %.f AT "%(AT_b);
print " b.2) Demagnetizing Armature Amphere turns per pole, ATd = %.2f AT "%(ATd_b);
print " b.3) Cross-Magnetizing Armature Amphere turns per pole, ATc = %.f AT "%(ATc_b);
print " b.4) Amphere turns of Compensating winding, ATw = %.1f AT "%(ATw_b);
import math
# GIVEN DATA
p = 6; # Number of the Poles
P = 100 * 10 ** 3; # Power rating of the DC machine in KiloWatts
V = 440; # Voltage rating of the DC machine in Volts
Z = 500; # Total number of the Armature Conductor
Ig = 1.0 * 10 ** -2; # Interpolar Air gap in Meter
Bi = 0.28; # Interpolar Flux Densist in Weber per Meter-Square
mue_0 = 4*math.pi*10** -7; # Permeability of the air in Henry/Meter
# CALCULATIONS
Ia = P/V; # Full load current in Amphere
a = p; # Number of the Parallel path (Equal to p because LAP WINDING
ATi = (Z*Ia)/(2*a*p)+((Bi*Ig)/mue_0); # Amphere turns for each Interpole
Nc = ATi/Ia; # Number of turns per pole of interpole
# DISPLAY RESULTS
print ("EXAMPLE : 4.10 : SOLUTION :-") ;
print " a) Amphere turns for each Interpol, ATi = %.2f AT "%(ATi);
print " b) Number of turns per pole of interpole, Nc = %.2f turns per pole nearly %.f turns per pole "%(Nc,Nc);
import math
# GIVEN DATA
print " EXAMPLE : 4.11 : Given Data between the Field current and Open-Circuit EMF generated by DC shunt wound Generator ";
print " IfA 0 1 2 3 4 5 6 ";
print " Vocv 10 90 170 217.5 251 272.5 281 ";
N = 1000.; # Speed of an DC Shunt wound generator on open circuit in RPM
Rf = 50.; # Shunt field resistance in Ohms
# CALCULATIONS
# Refer Figure 4.20:- Page no. 180
Vt = 277.17; # Terminal Voltage in Volts from Figure 4.20 (The slope of the resistance line Rf cuts the OCC at this Voltage [point]
Voc_r = 90; # Critical Open circuit voltage in Volts from Figure 4.20 page no. 180
If_r = 1.0; # Critical Field current in Amphere from Figure 4.20 page no. 180
Rc = Voc_r/If_r; # Crictical field resistance in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.11 : SOLUTION :-") ;
print " a) Crictical field resistance, Rc = %.f Ohms "%(Rc);
import math
# GIVEN DATA
print " EXAMPLE : 4.12 : Given Data between the Field current and Open-Circuit EMF generated by DC Machine ";
print " IfA 0 0.25 0.5 1.0 1.5 2.0 2.5 3.0 ";
print " Vocv 8 43 77 151 198 229 253 269";
N = 600; # Speed of an DC Shunt wound generator on open circuit in RPM
Rf1 = 100; # Shunt field resistance in Ohms
Rf2 = 125; # Shunt field resistance in Ohms
# CALCULATIONS
# Refer Figure 4.21:- Page no. 181
Vt1 = 253.33; # Terminal Voltage in Volts correspounding to field resistance of 100 Ohms from Figure4.21 Page no. 181 (The slope of the resistance line Rf cuts the OCC at this Voltage [point]
Vt2 = 213.33; # Terminal Voltage in Volts correspounding to field resistance of 125 Ohms from Figure 4.21 Page no. 181 (The slope of the resistance line Rf cuts the OCC at this Voltage [point]
Voc_r = 151; # Critical Open circuit voltage in Volts from Figure 4.20
If_r = 1.0; # Critical Field current in Amphere from Figure 4.20
Rc = Voc_r/If_r; # Crictical field resistance in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.12 : SOLUTION :-") ;
print " a) Crictical field resistance, Rc = %.f Ohms "%(Rc);
print " b.1) Terminal Voltage correspounding to field resistance of 100 Ohms is %.2f V "%( Vt1);
print " b.1) Terminal Voltage correspounding to field resistance of 125 Ohms is %.2f V "%( Vt2);
import math
# GIVEN DATA
N1 = 1200; # Rotation speed of the Separately excited Generator in RPM at case (1
Ia1 = 100; # Current supplied by the Generator in Amphere
V1 = 220; # Opearting Volatge of the Generator in Volts
Ra = 0.08; # Armature resistance in Ohms
N2 = 1000; # Rotation speed of the Separately excited Generator in RPM at case (2
Vb = 2.0; # Total Brush drop in Volts
# CALCULATIONS
RL = V1/Ia1; # Load resistance in Ohms
E1 = V1 + Vb + (Ra * Ia1); # Back EMF at case (1) in Volts
E2 = (N2/N1)*E1; # Back EMF at case (2) in Volts (Excitation is constant
Ia2 = (E2 - Vb)/(RL + Ra); # New load current in Amphere for case (2
# DISPLAY RESULTS
print ("EXAMPLE : 4.13 : SOLUTION :-") ;
print " a) New load current at %.f RPM , Ia2 = %.2f A "%(N2,Ia2);
import math
# GIVEN DATA
I = 50; # Curent supplied by the Separately Excitated Generator in Amphere
V = 250; # Dc bus bar in Volts
phi_1 = 0.03; # Useful Flux in Weber
Ra = 0.5; # Armature resistance in Ohms
phi_2 = 0.029; # New(Changed) Flux in Weber
# CALCULATIONS
Vd = I * Ra; # Voltage drop in the Armature in Volts
E1 = V + Vd; # EMF Generated in Volts
E2 = (phi_2/phi_1)*E1; # EMF Generated in Volts immediately after flux changes but speed will remains same
Ia = (E2 - V)/Ra; # Armature Current in Amphere immediately after flux changes
perct = 100 * (( phi_1 - phi_2)/phi_2); # Percenatge change in the speed of the machine that is required to restore the original Armature current but EMF raised to the original value and its Proportional to the speed and flux
# DISPLAY RESULTS
print ("EXAMPLE : 4.14 : SOLUTION :-") ;
print " a) Armature Current immediately after flux changes, Ia = %.1f A "%(Ia);
print " b) Percenatge change in the speed of the machine that is required to restore the original Armature current) is %.2f Percenatge "%(perct);
import math
from numpy import roots
# GIVEN DATA
S = 500 * 10 ** 3; # Rating of the Generator-1 and Generator-2
VI = 800 * 10 ** 3 # Actual load
# CALCULATIONS
#For Case (a
Voc_a = 500; # Open-circuit EMF Generator-1 and Generator-2 in Volts
I = 1000; # Full load current in Amphere
perct_1a = 2./100; # Percenatge fall of the Voltage in Generator-1
perct_2a = 3./100; # Percenatge fall of the Voltage in Generator-2
V1a = Voc_a - (perct_1a * Voc_a); # Voltage in the Generator-1 in Volts when it falls to 2% at fully loaded
V2a = Voc_a - (perct_2a * Voc_a); # Voltage in the Generator-2 in Volts when it falls to 3% at fully loaded
# From Chacteristics can be assumed linear as, for Generator 1 is V = 500 + ((500-490)*I1)/(0-1000), V = -0.01*I1+500 and for Generator 2 is V = 500 + ((500-485)*I2)/(0-1000), V = 0.015*I2+500
# When sharing load of 800KVA at voltage, the load current will be I = I1+I2 = (800*1000)/V
# From above equations we get I1 = 1.5*I2 thus, 2.5*I2 = (800*1000)/V
# Putting the above equations in the Generator 2 equation we get V = -0.015*((800*1000)/(2.5*V))+500 solving we get, 25*V**2 - 12500V + 120000 = 0
V_a = [120000, -12500, 25] #poly ([120000 -12500 25],'x','coeff'); # Expression for the load Voltage in Quadratic form
r_a = roots (V_a); # Value of the load Voltage in Volts (neglecting lower value
I_a = VI/r_a[0];
I2_a = I_a/2.5;
I1_a = 1.5*I2_a;
# For Case (b
perct = 2./100; # Percenatge fall of the Voltage in Generator-1and Generator-2
Voc_1b = 500.; # Open-circuit EMF Generator-1 in Volts
Voc_2b = 505; # Open-circuit EMF Generator-2 in Volts
I = 1000; # Full load current in Amphere
V1 = Voc_1b - (perct * Voc_1b); # Voltage in the Generator-1 in Volts when it falls to 2% at fully loaded
V2 = Voc_2b - (perct * Voc_2b); # Voltage in the Generator-2 in Volts when it falls to 2% at fully loaded
# From Chacteristics can be assumed linear as, for Generator 1 is V = 500 + ((500-490)*I1)/(0-1000), V = -0.01*I1+500 and for Generator 2 is V = 505 + ((505-494.5)*I2)/(0-1000), V = -0.0101*I2+505
# When sharing load of 800KVA at voltage, the load current will be I = I1+I2 = (800*1000)/V
# From above equations we get V = -0.01*I1 + 500, I1 = -V/0.01 + 500/0.01 = 50000 - 100*V, V = -0.0101*I2 + 505 and I2 = 505/0.0101 - V/.0101 = 50000-99.0099*V
# Putting the above equations in the Current I equation we get I = I1+I2 = (800*1000)/V = 2*50000-199.0099*V solving we get, 199.0099*V**2 - 100000V + 800000 = 0
V_b = [800000, -100000, 199.0099] #poly ([800000 -100000 199.0099],'x','coeff'); # Expression for the load Voltage in Quadratic form
r_b = roots (V_b); # Value of the load Voltage in Volts (neglecting lower value
I_b = VI/r_b[0];
I1_b = 50000-100*r_b[0]
I2_b = 50000-99.0099*r_b[0]
# DISPLAY RESULTS
print ("EXAMPLE : 4.15: SOLUTION :-");
print " For case a) Having open-circuit EMfs of 500V but their voltage falls to 2 percent and 3 percent when fully loaded Load Voltage, Load Voltage = %.2f V Load current = %.2f A Individual currents are %.2f A and %.2f A "%(r_a[0],I_a,I1_a,I2_a)
print " For case b) Having open-circuit EMfs of 500V and 505V but their governors have identical speed regulation of 2 percent when fully loaded Load Voltage, Load Voltage = %.2f V Load current = %.2f A Individual currents are %.2f A and %.2f A "%(r_b[0],I_b,I1_b,I2_b)
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- For caseb) Load voltage = 493.35 V A instead of %.2f V "%(r_b[0]);
print " Load current = 1634.73 A instead of %.2f A "%(I_b)
print " Individual currents 665 A and 1153.5 A instead of %.2f A and %.2f "%(I1_b,I2_b)
print " For Case b):- From Calculation of the Load Voltage V), rest of all the Calculated values in the TEXT BOOK is WRONG because of the value Load Voltage V) is WRONGLY calculated and the same used for the further Calculation part "
import math
# GIVEN DATA
Out_hp = 20; # Output of the Motor in HP
eta = 90./100; # Full load efficiency of the Motor
V = 220.; # Motor voltage in Volts
ns = 5; # Number of the step of Starter
Rf = 220; # Field resistance in Ohms
cr = 1.8; # Lowest Current rating is 1.8 times of the Full load current
Cu = 5./100; # Total Copper loss is 5% of the Input
# CALCULATIONS
Out = 20 * 746; # Output of the Motor in Watt
Inp = (Out/eta); # Input of the Motor in KiloWatt
I = Inp/Rf; # Full-Load Current in Amphere
Cu_l = Inp*Cu; # Total Copper loss in Watts
olf = (V ** 2)/Rf; # Ohmic loss in the Fiels in the Watts
Acu = Cu_l - olf; # Armature Copper loss in Watts
Ra = Acu/(I * I); # Armature resistance in Ohms
I2 = I * cr; # Lower Current in Amphere
n = ns - 1; # Number of the resistance
gama = ( (I2 * Ra)/Rf ) ** (1/(n + 1)); # Current Ratio
I1 = I2/gama; # Initial Current in amphere
R1 = V/I1; # Initial resistance in Ohms
R2 = gama * R1; # Initial resistance in Ohms
r1 = R1 - R2; # Graded resistance in Ohms
R3 = gama * R2; # Initial resistance in Ohms
r2 = gama * r1; # Graded resistance in Ohms
r3 = gama ** 2 * r1; # Graded resistance in Ohms
r4 = gama ** 3 * r1; # Graded resistance in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.16 : SOLUTION :-") ;
print " a) Graded resistances are %.4f Ohms, %.4f Ohms, %.4f Ohms and %.4f Ohms "%(r1,r2,r3,r4);
import math
# GIVEN DATA
I = 30.; # Initial starting Current in Amphere
ns = 5.; # Number of Steps of the starter
V = 500.; # Operating Voltage of the DC Shunt Motor in Volts
I1 = 50.; # Peak(Upper) Current limit in Amphere
Ra = 1.0; # Armature Circuit resistance in Ohms
# CALCULATIONS
R1 = V / I; # Initial resistance in Ohms
gama = ( Ra/R1) ** (1/(ns-1)); # Current Ratio
I2 = gama * I1; # Lower Current limit in Amphere
r1 = R1 * (1-gama); # Graded resistances in Ohms
r2 = gama * r1; # Graded resistances in Ohms
r3 = gama * r2; # Graded resistances in Ohms
r4 = gama * r3; # Graded resistances in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.17 : SOLUTION :-") ;
print " a)Graded resistances are %.2f Ohms, %.4f Ohms, %.4f Ohms and %.4f Ohms "%(r1,r2,r3,r4);
import math
# GIVEN DATA
V = 500; # Operating voltage of the DC series motor in Volts
P_hp = 10; # Operating Power in HP
Il = 40; # Lower currents limit in Amphere
Ih = 60; # Higher currents limit in Amphere
f = 0.5/100; # Motor flux rises by 0.5% per amphere
Rt = 0.8; # Motor terminal resistance in Ohms
eta = 90./100; # Motor efficiency
# CALCULATIONS
E1 = V-Il*Rt; # Induced EMF E1 in Volts
# Induced EMF, E2 = 500-60(0.8+r4) = 500 - 60*R4 where r4 is the fourth-step resistance, and R4 = 0.8+r4 and E1 = 1.1*E2 , 500 - 40*0.8 = 1.1*(500-60(0.8+r4)), 500-32 = 550-66*R4 thus we get, R4 = (550-500+32)/66 refer page no. 197
R4 = (V-(E1/1.1))/Ih;
r4 = R4 - Rt; # Fourth-step resistance in ohms
R3 = (V-((V-Il*R4)/1.1))/Ih;
r3 = R3 - R4; # Third-step resistance in ohms
R2 = (V-((V-Il*R3)/1.1))/Ih;
r2 = R2 - R3; # Second-step resistance in ohms
R1 = (V-((V-Il*R2)/1.1))/Ih;
r1 = R1 - R2; # First-step resistance in ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.18: SOLUTION :-");
print " a) The resistance steps in series motor stater are %.3f Ohms,%.4f Ohms, %.3f Ohms and %.2f Ohms "%(r1,r2,r3,r4)
import math
# GIVEN DATA
Vac = 250; # Operating AC Voltage in Volts
V = 220; # Operating Voltage of the separately excited DC motor in Volts
fa = 30; # Firing Angle in Degree
Out_hp = 20; # DC Motor Output in HP
La = 20 * 10 ** -3; # Armature Inducatnce in Henry
Ra = 0.15; # Armature resistance in Ohms
E_cons = 0.2; # EMF constant in Volts/RPM
eta = 90./100; # Motor Operating Efficiency
N = 1000.; # Rotational Speed of the Motor in RPM
# CALCULATIONS
out = 20 * 746; # DC Motor Output in Watt
Vt = ((Vac*2*math.sqrt(2))/math.pi)*math.degrees(math.cos(math.radians(fa))); # Average Terminal volatge in Volts
Ia = out/(V*eta); # Rated Current in Amphere
E = Vt - ( Ia * Ra ); # Back EMF in Volts
n = E/E_cons; # Speed of the Motor in RPM
e_cons = (E_cons*60)/ ( 2 * math.pi); # EMF constant in Volts-Second per radians
T = e_cons * Ia; # Devolped Torque in Newton-Meter
pi = (E*Ia)+(Ia**2*Ra); # Power intake in Watts
pi_v = Vt * Ia; # Power intake in Watts (Verification
# DISPLAY RESULTS
print ("EXAMPLE : 4.19 : SOLUTION :-") ;
print " a) Speed of the Motor, N = %.2f RPM "%(n);
print " b) Devolped Torque, T = %.2f N-m "%(T);
print " b) Power intake at Rated current and Firing angle of %.f deg, VI = %.1f W "%(fa,pi);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS IS :- a T = 114.07 N-m instead of 143.91 N-m " ;
import math
# GIVEN DATA
N = 1500; # Speed of the separately excited DC Motor in RPM
Out_hp = 100; # Output of the DC Motor in HP
V = 500; # Motor operating Volatge in Volts
VL = 440; # 3-phase Line-line Voltage in Volts
f = 50; # Frequency in Hertz
Ra = 0.0835; # Armature resistance in Ohms
La = 5.7 * 10 ** -3; # Armature Inducmath.tance in Henry
eta = 89./100; # Operating Efficiency of the Motor
E_cons = 0.35; # EMF constant in Volts per RPM
# CALCULATIONS
# For case (a
Out = Out_hp * 746; # Output of the DC Motor in Watts
Ia = Out/(V*eta); # Rated Current in Amphere
Vph = VL/math.sqrt(3); # Phase Voltage in Volts
a = (3*Vph*math.sqrt(6)) / math.pi; # constant
E = N * E_cons; # Back EMF at Rated speed
V = E + (Ia * Ra); # Terminal Volatge in Volts
alpa = math.acos(math.radians(V/a)); # Firing Angle
# For case (b
# Assumed that No load current is about 12% of full load current
Io = ( 0.12 * Ia ); # No load current in Amphere
V_b1 = a * math.degrees(math.cos(math.radians(0))); # Terminal Voltage at Firing Angle 0 deg
E_b1 = V_b1 + (Io * Ra); # Back EMF at Firing Angle 0 deg
N_b1 = E_b1/E_cons; # No load speed at Firing Angle 0 deg
V_b2 = a * math.degrees(math.cos(math.radians(30))); # Terminal Voltage at Firing Angle 30 deg
E_b2 = V_b2 + (Io * Ra); # Back EMF at Firing Angle 30 deg
N_b2 = E_b2/E_cons; # No load speed at Firing Angle 30 deg
# DISPLAY RESULTS
print ("EXAMPLE : 4.20 : SOLUTION :-") ;
print " a) Firing Angle at Rated speed and Rated Motor Current , alpa = %.2f deg "%(alpa);
print " b.1) No load speed at Firing Angle 0 deg, N = %.1f RPM "%(N_b1);
print " b.2) No load speed at Firing Angle 30 deg, N = %.1f RPM "%(N_b2);
import math
# GIVEN DATA
# Given that Back EMF is Zero Because Motor is at Smath.degrees(math.atanstill
V = 250; # DC supply Voltage to separately excited DC Motor in Volts
Ra = 1.0; # Armature resistance in Ohms
La = 30 * 10 ** -3; # Armature Inducmath.tance in Henry
E_cons = 0.19; # Motor (EMF) constant in Volts per RPM
Ia = 25; # Average Armature Current in Amphere
# CALCULATIONS
V1 = Ia * Ra; # Minimum Terminal Volatge in Volts
alpa_mini = Ia/V; # Minimum Duty Cycle
alpa_max = 1.0; # Maximum Duty Cycle
V2 = V; # Maximum Terminal Volatge in Volts when Duty cycle (alpa) is 1.0
E2 = V2 - (V1 * alpa_max); # Back EMF at Maximum Duty cycle ( i.e alpa = 1.0) in Volts
N = E2/E_cons; # Speed of the Motor
# DISPLAY RESULTS
print ("EXAMPLE : 4.21 : SOLUTION :-") ;
print " a) Range of the Speed is from 0 RPM to %.2f RPM and Range of the Duty Cycle is %.1f to %.1f "%(N,alpa_mini,alpa_max);
import math
# GIVEN DATA
N1 = 1000; # Speed of the DC shunt Motor in RPM
Out_hp = 20; # Output of the DC shunt Motor in HP
V = 220; # Motor operating Volatge in Volts
Ra = 0.9; # Armature resistance in Ohms
Rf = 200; # Field resistance in Ohms
eta = 89./100; # Operating Efficiency of the Motor
Ra_a = 0.2; # resistance inserted to the armature circuit
# CALCULATIONS
out = Out_hp * 746; # Output of the DC Motor in watts
I = out/(V * eta); # Rated current in Amphere
If = V/Rf; # Field current in Amphere
Ia1 = I - If; # Armature current in Amphere
E1 = V - (Ia1 * Ra); # Back EMF in Volts
# Assuming that Torque and Armature current is constant
E2 = V - ( Ra + Ra_a ) * Ia1; # New Back EMF in Volts
N2 = N1*(E2/E1); # New speed in RPM
# DISPLAY RESULTS
print ("EXAMPLE : 4.22: SOLUTION :-") ;
print " a) New Speed of the Motor , N2 = %.2f RPM "%(N2);
import math
from numpy import roots
# GIVEN DATA
N1 = 600; # Speed of the DC shunt Motor in RPM
Out_hp = 10; # Output of the DC shunt Motor in HP
V = 220; # Motor operating Volatge in Volts
Ra = 1.5; # Armature resistance in Ohms
Rf = 250; # Field resistance in Ohms
eta = 88./100; # Operating Efficiency of the Motor
Rf_a = 50; # resistance inserted to the field circuit
# CALCULATIONS
out = Out_hp * 746 # Output of the DC Motor in watts
I = out/(V * eta); # Rated current in Amphere
If1 = V/Rf; # Field current in Amphere
Ia1 = I - If1; # Aramature current in Amphere
E1 = V - Ra*Ia1; # Back EMF in Volts
If2 = V/(Rf+Rf_a); # New Field current in Amphere after 50 Ohms resistance inserted to the field circuit
# Refer page no. 217 we have T1 = K*If1*Ia1 proportional to 1/W1**2 and T1 = K*If2*Ia2 proportional to 1/W2**2 thus T1/T2 = (If1*Ia1)/(If2*Ia2) = (W2**2)/(W1**2) = (N2**2)/(N1**2), Ia2 = (If1*Ia1*W1**2)/(If1*W1**2) = (0.88*37.65*N2**2)/(0.733*600*600)
# Now New EMF E2 is E2 = V - Ia2*Ra, E1/E2 = (k*If1*N1)/(k*If2*N2), E2 = (0.733*N2)/(0.88*600) = 220 - (0.88*37.65*1.5*N2**2)/(0.733*600*600) Thus we have 0.001388*N2**2 = 220 - 1.833*10**-4*N2
N2 = [-220, 0.001388, 1.833*10**-4] #poly ([-220 0.001388 1.833*10**-4],'x','coeff'); # Expression for the new speed of the motor in Quadratic form
r = roots (N2); # Value of the New speed of the motor in RPM
# DISPLAY RESULTS
print ("EXAMPLE : 4.23 : SOLUTION :-") ;
print " a) New speed of the motor, N2 = %.2f RPM nearly %.f RPM "%(r[0],r[1]);
import math
# GIVEN DATA
N1 = 1000; # Speed of the DC shunt Motor in RPM
Out_hp = 10; # Output of the DC shunt Motor in HP
V = 220; # Motor operating Volatge in Volts
Ra = 0.5; # Armature resistance in Ohms
Rf = 100; # Field resistance in Ohms
eta = 90./100; # Operating Efficiency of the Motor
# CALCULATIONS
out = Out_hp * 746; # Output of the DC Motor in watts
I = out/(V * eta); # Rated current in Amphere
If = V/Rf; # Field current in Amphere
Ia = I-If; # Armature current in Amphere
E = V - (Ia*Ra); # Back EMF of the Motor in Volts
Rd = E/I; # resistance at Dynamic Braking in Ohms
Rc = (V+E)/I; # resistance at Counter Current Braking in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.24 : SOLUTION :-") ;
print " a) resistance at Dynamic Braking, Rd = %.2f Ohms "%(Rd);
print " b) resistance at Counter Current Braking, Rc = %.1f Ohms "%(Rc);
import math
from numpy import roots
# GIVEN DATA
V = 220; # Motor operating Volatge in Volts
Ra = 1.0; # Armature resistance in Ohms
Rf = 220; # Field resistance in Ohms
Ia1 = 20; # Armature Current in Amphere
N1 = 800; # Motor drving speed in RPM
N2 = 1000; # To be obtained speed in RPM
# CALCULATIONS
If = V/Rf; # Field Current in Amphere
E1 = V - ( Ia1 - If ) * Ra; # Back EMF E1 at N1 Speed in Volts
# Now we have Back EMF E2 at N2 Speed, E2 = 220-Ia2*1.0 = 220-Ia2 and the field flux be proportional to the field current, math.since torque is constant we get, If2*Ia2 = If1*Ia1 = 20
# Thus (220-Ia2)/201 = (If2*N2)/(If1*N1) = If2*(1000/(800*1.0)), 220-Ia2 = 201*(10/8)*(20/Ia2) = 5000/Ia2 solving this we get Ia2**2 - 220Ia2 + 2000 = 0
Ia2 = [5000, -220, 1] #poly ([5000 -220 1],'x','coeff'); # Expression for the new Armature current in Quadratic form
r = roots (Ia2); # Value of the New Armature current in Amphere
If2 = If*(Ia1/r[1]); # New field current in Amphere when New Armature current is 39.29A
Rfn = V/If2; # New field resistance in ohms
ERf = Rfn - Rf; # Extra resistance in Ohms
# DISPLAY RESULTS
print ("EXAMPLE : 4.25 : SOLUTION :-") ;
print " a) Extra resistance should be added in the field circuit for raimath.sing the speed to %.f RPM is = %.2f Ohms "%(N2,ERf);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- a) Ia2 = 39.29 A and 180.71 A instead of %.2f A and %.2f A "%(r[0],r[1]);
print " b) Extra resistance required is 212.22 Ohms instead of %.2f Ohms "%(ERf);
print " From Calculation of the New armature current Ia2)%( rest all the Calculated values in the TEXT BOOK\
is WRONG because of the New armature current Ia2) value is WRONGLY calculated and the same used for the further Calculation part "
import math
# GIVEN DATA
R = 2.0; # DC series Motor resistance between the terminals in Ohms
V1 = 220; # Motor Operating voltage in Volts
N1 = 500; # Rotation Sped of the DC series Motor in RPM
I1 = 22; # Current in Motor in Amphere
N2 = 600; # New Rotation Sped of the DC series Motor in RPM
# CALCULATIONS
# For case (a) when magnetic circuit is Unsaturated
E1 = V1 - (I1 * R); # Back EmF at N1 Speed in Volts
I2_a = (N2/N1)*I1; # Current in Motor at N2 speed in Amphere
E2_a = (E1*I2_a*N2)/(I1*N1); # Back EmF at N2 Speed in Volts
V2_a = E2_a + (I2_a * R); # Applied Voltage at N2 Speed in Volts
# For case (b) when magnetic circuit is Saturated
I2_b = ((N2/N1)**2)*I1; # Current in Motor at N2 speed in Amphere
E2_b = (N2/N1)*E1; # Back EmF at N2 Speed in Volts
V2_b = E2_b + (I2_b * R); # Applied Voltage at N2 Speed in Volts
# DISPLAY RESULTS
print ("EXAMPLE : 4.26 : SOLUTION :-") ;
print " a.1) Applied Voltage when magnetic circuit is Unsaturated, V2 = %.2f V "%(V2_a);
print " a.2) Current in Motor when magnetic circuit is Unsaturated, I2 = %.1f A "%(I2_a);
print " b.1) Applied Voltage when magnetic circuit is Saturated, V2 = %.2f V "%(V2_b);
print " b.2) Current in Motor when magnetic circuit is Saturated, I2 = %.2f A "%(I2_b);
import math
# GIVEN DATA
V = 220; # DC series Motor operating Volatge in Volts
Ra = 1.0; # Armature resistance in Ohms
Rf = 1.0; # Field resistance in Ohms
I1 = 20; # Armature Current in Amphere
N1 = 1800; # Motor drving speed in RPM
If = 20; # Armature Current in Amphere
Rd = 0.5; # Diverter resistance in Ohms
# CALCULATIONS
E1 = V - ( Ra + Rf ) * I1; # Back EMF in Volts
I2 = math.sqrt(3)*I1; # New Armature current in Amphere
If2 = ( Rd * I2 )/(Ra + Rd); # New field Current in Amphere
E2 = V - ( Ra + (1./3))*I1; # New BAck EMF in Volts
N2 = (N1*E2*If)/(E1*If2); # New Rotation speed of the Motor in RPM
# DISPLAY RESULTS
print ("EXAMPLE : 4.27 : SOLUTION :-") ;
print " a) New Rotation speed of the Motor at torque remains constant, N2 = %.f RPM "%(N2);
print " b.1) New Armature Current at torque remains constant, I2 = %.2f A "%(I2);
print " b.2) New Field Current at torque remains constant, If2 = %.2f A "%(If2);
import math
# GIVEN DATA
V = 220; # DC shunt Motor operating Volatge in Volts
Ra = 1.0; # Armature resistance in Ohms
Rf = 220; # Field resistance in Ohms
In1 = 5; # No-Load Current in Amphere
N1 = 1000; # Motor drving speed in RPM
inp = 10 * 10 ** 3; # Motor input in Watts
# CALCULATIONS
If = V/Rf; # Field Current in Amphere
Ian1 = In1 - If; # No load Armature Current in Amphere
E1 = V - (Ian1 * Ra); # Back EMF in Volts
Iin = inp/V; # Motor Input Current in Amphere
Ia = Iin - If; # Armature current in Amphere
E2 = V - (Ia * Ra); # New Back EMF in Volts
N2 = (N1*E2)/E1; # New Rotation speed of the Motor in RPM
Pa = E2 * Ia; # Developed Armature Power in Watts
T = Pa/((2*math.pi*N2)/60); # Developed Torque in Newton-Meter
Pi = V * In1; # No-Load input Power in Watts
Pa_cu = Ian1 ** 2 * Ra; # No-Load Armature Copper loss in Watts
F_loss = Pi - Pa_cu; # Fixed losses in Watts
Pa_cu_load = Ia ** 2 * Ra; # Loaded Armature Copper loss in Watts
Total_loss = F_loss + Pa_cu_load; # Total losses in loaded conditions in Watts
out = inp - Total_loss; # Shaft output in Watts
Ts = out/((2*math.pi*N2)/60); # Shaft torque in Newton-Meter
eta = (out/inp)*100;
# Efficiency in Percentage
# DISPLAY RESULTS
print ("EXAMPLE : 4.28 : SOLUTION :-") ;
print " a) New Rotation speed of the Motor , N2 = %.f RPM "%(N2);
print " b.1) Developed Torque, T = %.1f N-m A "%(T);
print " b.2) Shaft torque, Ts = %.2f N-m "%(Ts);
print " c) Efficiency in Percentage, eta = %.2f percent "%(eta);
import math
# GIVEN DATA
V = 220; # Shunt Motor operating Line Volatge in Volts
Ra = 0.2; # Armature resistance in Ohms
Iam = 72; # Motor Armature current in Amphere
I = 12; # Line Current in Amphere
Ifm = 1; # Motor field Current in Amphere
Ifg = 1.5; # Generator field Current in Amphere
# CALCULATIONS
Iag = Iam - I; # Geneartor Armature current in Amphere
Pfm = V * Ifm; # Loss in Motor Field winding in Watts
Pfg = V * Ifg; # Loss in Geneartor Field winding in Watts
loss_ma = Iam ** 2 * Ra; # Loss in Motor Armature circuit in Watts
loss_ga = Iag ** 2 * Ra; # Loss in Generator Armature circuit in Watts
Em = V - Iam * Ra; # Motor EMF in Volts
Eg = V + Iag * Ra; # Generator EMF in Volts
T_loss = (V*I) - (Ra*Iam**2 + Ra*Iag**2); # Total Iron and Rotational Loss in Watts
Pim = (V*Iam)+(V*Ifm); # Motor input in Watts
Wc = 0.5 * T_loss; # Total Iron and Rotational Loss in each Machine in Watts
Wm = Wc+(Ra*Iam**2)+V*Ifm; # Motor losses in Watts
Pom = Pim - Wm; # Motor output in Watts
eta_m = (1-(Wm/Pom))*100; # Motor Efficiency in Percentage
Pog = V*Iag; # Generator output in Watts
Wg = Wc+(Ra*Iag**2)+V*Ifg; # Generator losses in Watts
Pin = Pog + Wg; # Generator input power in Watts
eta_g = (1-(Wg/Pin))*100; # Generator Efficiency in Percentage
# DISPLAY RESULTS
print ("EXAMPLE : 4.29 : SOLUTION :-") ;
print " a) Motor Efficiency , eta = %.2f Percentage "%(eta_m);
print " b) Generator Efficiency , eta = %.2f Percantage "%(eta_g);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- a) Total Iron and Rotational Loss = 720 W instead of %.1f W "%(T_loss);
print " b) Pim = 15912 W instead of %.f W "%(Pim);
print " c) Wm = 1371.4 Winstead of %.1f W "%(Wm);
print " d) Pom = 14540.6 W instead of %.1f W "%(Pom);
print " e) eta_m = 90.54 Percentage instead of %.2f Percentage "%(eta_m);
print " f) eta_g = 93.22 Percentage instead of %.2f Percentage "%(eta_g);
print " From Calculation of the Total Iron and Rotational Loss in each Machine Wc), rest all the Calculated values in the TEXT BOOK is WRONG because of the Total Iron and Rotational Loss in each Machine Wc) value is WRONGLY calculated and the same used for the further Calculation part "
import math
# GIVEN DATA
Vg = 110; # Generator operating Volatge in Volts
Vm = 102; # Motor operating Volatge in Volts
Vs = 274; # Supply Volatge in Volts
Ra = 1.0; # Armature resistance in Ohms for both the Machines
Rf = 0.82; # Field resistance in Ohms for both the Machines
N = 1440; # Speed of the Set in RPM
Ig = 17.5; # Generator current in Amphere
Im = 9.5; # Motor current in Amphere
# CALCULATIONS
Pi = Vs * Im; # Input power in Watts
Pg = Vg * Ig; # Output power in Watts
Pim = Vm * Im; # Power Input to the Motor in Watts
Pl = Pi - Pg; # Losses in the entire set in Watts
Pcu = Im**2*(Ra+2*Rf) + Ig**2*Ra; # Total Copper loss for both the Machines in Watts
P_l = Pi - Pg - Pcu; # Frictional, Windage and core losses of the both Machines in Watts
Po = P_l/2; # Frictional, Windage and core loss of each Machines in Watts
eta_m = (1 - ((Po + Im**2*(Ra+Rf))/Pim))*100; # Motor Effiicency in Percentage
Pig = Pg + Po + Ig**2*Ra + Im**2*Rf; # Generator input in Watts
eta_g = (Pg / Pig)*100; # Generator Effiicency in Percentage
T = (Vg*Ig *60)/(2*math.pi*N); # Torque in Newton-Meter
# DISPLAY RESULTS
print ("EXAMPLE : 4.30 : SOLUTION :-") ;
print " a) Motor Efficiency , eta_m = %.2f percentage "%(eta_m);
print " b) Generator Efficiency , eta_g = %.2f Percentage "%(eta_g);
print " c) Torque , T = %.2f N-m "%(T);
print " [ TEXT BOOK SOLUTION IS PRINTED WRONGLY I verified by manual calculation ]" ;
print " WRONGLY PRINTED ANSWERS ARE :- a) Generator input = 2307.5 W instead of %.f W "%(Pig);
print " b) eta_g = 83.42 Percenatge instead of %.2f Percentage "%(eta_g);
print " From Calculation of the Generator input, rest all the Calculated values in the TEXT BOOK is WRONG because of the Generator input value is WRONGLY calculated and the same used for the further Calculation part "