# Chapter 6 : Synchronous Machines¶

## Example 6.1 Page No : 297¶

In [1]:
# GIVEN DATA

f = 50;                             # Generating Frequency in Hertz

# CALCULATIONS
# For Case(a)

Ns_a = 3000;                        # Synchronous speed in RPM
p_a = (120*f)/Ns_a;                 # Number of poles

# For Case(b)

Ns_b = 1000;                        # Synchronous speed in RPM
p_b = (120*f)/Ns_b;                 # Number of poles

# For Case(c)

Ns_c = 300;                         # Synchronous speed in RPM
p_c = (120*f)/Ns_c;                 # Number of poles

# For Case(d)

Ns_d = 40;                          # Synchronous speed in RPM
p_d = (120*f)/Ns_d;                 # Number of poles

# DISPLAY RESULTS

print ("EXAMPLE : 6.1 : SOLUTION :-") ;
print " For Casea) Ns = %.f, p = %.f  "%(Ns_a,p_a);
print " For Caseb) Ns = %.f, p = %.f  "%(Ns_b,p_b);
print " For Casec) Ns = %.f, p = %.f  "%(Ns_c,p_c);
print " For Cased) Ns = %.f, p = %.f  "%(Ns_d,p_d);

EXAMPLE : 6.1 : SOLUTION :-
For Casea) Ns = 3000, p = 2
For Caseb) Ns = 1000, p = 6
For Casec) Ns = 300, p = 20
For Cased) Ns = 40, p = 150


## Example 6.2 Page No : 301¶

In [2]:
# GIVEN DATA

f = 60.;                             # Generating Frequency in Hertz
Ns = 1200.;                          # Synchronous speed in RPM
Ns_r = 1000.;                        # Alternator running speed in RPM

# CALCULATIONS

p = (120*f)/Ns;                     # Total number of poles
f_r = (p*Ns_r)/120;                 # Alternator running frequency in Hertz

# DISPLAY RESULTS

print ("EXAMPLE : 6.2 : SOLUTION :-") ;
print " Alternator running frequency, f = %.f Hz  "%(f_r);

EXAMPLE : 6.2 : SOLUTION :-
Alternator running frequency, f = 50 Hz


## Example 6.3 Page No : 303¶

In [3]:
# GIVEN DATA

p = 2.;                                      # Total number of poles

# CALCULATIONS
# For Case(a)

f_a = 10;                                   # Frequency in Hertz
Ns_a = (120*f_a)/p;                         # Synchronous speed in RPM

# For Case(b)

f_b = 50;                                   # Frequency in Hertz
Ns_b = (120*f_b)/p;                         # Synchronous speed in RPM

# For Case(c)

f_c = 60;                                   # Frequency in Hertz
Ns_c = (120*f_c)/p;                         # Synchronous speed in RPM

# For Case(d)

f_d = 100;                                  # Frequency in Hertz
Ns_d = (120*f_d)/p;                         # Synchronous speed in RPM

# For Case(e)

f_e = 400;                                  # Frequency in Hertz
Ns_e = (120*f_e)/p;                         # Synchronous speed in RPM

# DISPLAY RESULTS

print ("EXAMPLE : 6.3 : SOLUTION :-");
print " For Case a) When f = %.f, Synchronous speed, Ns = %.f RPM "%(f_a,Ns_a)
print " For Case b) When f = %.f, Synchronous speed, Ns = %.f RPM "%(f_b,Ns_b)
print " For Case c) When f = %.f, Synchronous speed, Ns = %.f RPM "%(f_c,Ns_c)
print " For Case d) When f = %.f, Synchronous speed, Ns = %.f RPM "%(f_d,Ns_d)
print " For Case e) When f = %.f, Synchronous speed, Ns = %.f RPM "%(f_e,Ns_e)

EXAMPLE : 6.3 : SOLUTION :-
For Case a) When f = 10, Synchronous speed, Ns = 600 RPM
For Case b) When f = 50, Synchronous speed, Ns = 3000 RPM
For Case c) When f = 60, Synchronous speed, Ns = 3600 RPM
For Case d) When f = 100, Synchronous speed, Ns = 6000 RPM
For Case e) When f = 400, Synchronous speed, Ns = 24000 RPM


## Example 6.4 Page No : 304¶

In [6]:
import math
import cmath

# GIVEN DATA

print " EXAMPLE : 6.4 :           Given Data ";
print " Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560 ";
print " IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24 ";
m = 3;                         # Total Number of phase
p = 6;                         # Total number of Poles
V = 400;                       # Operating voltage in Volts
I = 13.5;                      # Operating current in Amphere
N = 1000;                      # Speed in RPM
Ia_scc = 13.5;                 # SCC test Armature current in Amphere at If = 9.5 A
If_scc = 9.5;                  # SCC test field Rated current in Amphere
Ia_zpf = 13.5;                 # ZPF test Armature current in Amphere at If = 24 A
If_zpf = 24;                   # ZPF test field Rated current in Amphere

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
Ra = 1.0;                                   # Armature resistance in Ohms

# For case (a)

BC = 120;                                   # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386
Xl = BC/(math.sqrt(3)*Ia_scc);                   # Per phase leakage reactance in Ohms

# For Case (b.1) 0.8 pf Lagging

pfa_b1 = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
R_b1 = 10; A_b1 = 9.5;                                                    #From OCC the field current required for Er_b1 (Should be in Line-line Voltage) Er_b1 = 379.12V will get R_b1 & A_b1 value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b1 = 136.35;                                                        # Angle between R_b1 & A_b1 (figure 6.16(a) & Page no:-388) = 90'+9.48'+36.87' = 136.35'
F_b1 = math.sqrt((R_b1**2)+(A_b1**2)-(2*R_b1*A_b1*math.degrees(math.degrees(math.cos(math.radians(angle_b1))))));              # From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere
Eo_b1 = 525;                                                              # Corresponding to field current F_b1 = 18.12 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
r_b1 = 100*((Eo_b1-V)/V);                                                  # Percentage regulation

# For Case (b.2) 0.8 pf Leading

pfa_b2 = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
R_b2 = 8.3; A_b2 = 9.5;                                                   #From OCC the field current required for Er_b2 (Should be in Line-line Voltage) Er_b1 = 363.71 V will get R_b2 & A_b2 value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b2 = 70.61;                                                         # Angle between R_b2 & A_b2 (figure 6.16(b) & Page no:-388) = 90'+17.48'-36.87' = 70.61'
F_b2 = cmath.sqrt((R_b2**2)+(A_b2**2)-(2*R_b2*A_b2*math.degrees(math.degrees(math.cos(math.radians(angle_b2))))));              # From phasor diagram in figure 6.16(b) & Page no:-388 the neccessary field excitation in Amphere
Eo_b2 = 338;                                                              # Corresponding to field current F_b2 = 10.36 A the open circuit EMF from OCC is 338 V (Figure6.15 & Page no:-386)
r_b2 = 100*((Eo_b2-V)/V);                                                  # Percentage regulation

# For Case (b.3) Unity pf Leading

pfa_b3 = math.acos(math.radians(1.0));                                                      # Power factor angle in degree
R_b3 = 13; A_b3 = 9.5;                                                    #From OCC the field current required for Er_b3 (Should be in Line-line Voltage) Er_b1 = 440.30 V will get R_b3 & A_b3 value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b3 = 105.81;                                                        # Angle between R_b3 & A_b3 (figure 6.16(c) & Page no:-388) = 90'+15.81' = 105.81'
F_b3 = cmath.sqrt((R_b3**2)+(A_b3**2)-(2*R_b3*A_b3*math.degrees(math.degrees(math.cos(math.radians(angle_b3))))));              # From phasor diagram in figure 6.16(c) & Page no:-388 the neccessary field excitation in Amphere
Eo_b3 = 520;                                                              # Corresponding to field current F_b2 = 18.10 A the open circuit EMF from OCC is 520 V (Figure6.15 & Page no:-386)
r_b3 = 100*((Eo_b3-V)/V);                                                  # Percentage regulation

# For Case (b.4) ZPF Lagging

pfa_b4 = math.degrees(math.acos(math.radians(0)));                                                      # Power factor angle in degree
R_b4 = 18; A_b4 = 9.5;                                                    #From OCC the field current required for Er_b4 (Should be in Line-line Voltage) Er_b4 = 521 V will get R_b4 & A_b4 value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b4 = 177.57;                                                        # Angle between R_b4 & A_b4 = 90'-2.43'+90' = 177.57'
F_b4 = cmath.sqrt((R_b4**2)+(A_b4**2)-(2*R_b4*A_b4*math.degrees(math.degrees(math.cos(math.radians(angle_b4))))));              # The neccessary field excitation in Amphere
Eo_b4 = 570;                                                              # Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
r_b4 = 100*((Eo_b4-V)/V);                                                  # Percentage regulation

# For Case (b.4) ZPF Lagging

pfa_b5 = math.degrees(math.acos(math.radians(0)));                                                      # Power factor angle in degree
R_b5 = 6.0; A_b5 = 9.50;                                                    #From OCC the field current required for Er_b5 (Should be in Line-line Voltage) Er_b5 = 280.70 V will get R_b5 & A_b5 value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b5 = 4.77;                                                        # Angle between R_b5 & A_b5 = 90'-4.77'-90' = 4.77'
F_b5 = cmath.sqrt((R_b5**2)+(A_b5**2)-(2*R_b5*A_b5*math.degrees(math.degrees(math.cos(math.radians(angle_b5))))));              # The neccessary field excitation in Amphere
Eo_b5 = 135;                                                              # Corresponding to field current F_b4 = 27.50 A the open circuit EMF from OCC is 135 V (Figure6.15 & Page no:-386)
r_b5 = 100*((Eo_b5-V)/V);                                                  # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.4 : SOLUTION :-") ;
print " a)  Per phase leakage reactance, Xl = %.2f Ohms "%(Xl)
print " For Case b.1) 0.8 pf Lagging   Open circuit EMF EMF = %.f V "%(Eo_b1)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b1)
print " For Case b.2) 0.8 pf Leading   Open circuit EMF, EMF = %.f V "%(Eo_b2)
print " Percenatge Regulation, R = %.1f Percenatge "%(r_b2)
print " For Case b.3) Unity pf Lagging   Open circuit EMF, EMF = %.f V "%(Eo_b3)
print " Percenatge Regulation, R = %.f Percenatge "%(r_b3)
print " For Case b.4) ZPF Lagging   Open circuit EMF, EMF = %.f V"%(Eo_b4)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b4)
print " For Case b.5) ZPF Leading   Open circuit EMF, EMF = %.f V "%(Eo_b5)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b5)
print (" Calculated Answer in Tabular Column")
print "   Power Factor                 0.8 Lag      0.8 Lead        1.0       ZPF Lag      ZPF Lead "
print " Open circuit EMF V)            %.f           %.f            %.f       %.f          %.f "%(Eo_b1,Eo_b2,Eo_b3,Eo_b4,Eo_b5)
print " Percenatge Regulation           %.2f         %.1f          %.f.       %.2f        %.2f "%(r_b1,r_b2,r_b3,r_b4,r_b5)

 EXAMPLE : 6.4 :           Given Data
Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560
IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24
EXAMPLE : 6.4 : SOLUTION :-
a)  Per phase leakage reactance, Xl = 5.13 Ohms
For Case b.1) 0.8 pf Lagging   Open circuit EMF EMF = 525 V
Percenatge Regulation, R = 0.00 Percenatge
For Case b.2) 0.8 pf Leading   Open circuit EMF, EMF = 338 V
Percenatge Regulation, R = -100.0 Percenatge
For Case b.3) Unity pf Lagging   Open circuit EMF, EMF = 520 V
Percenatge Regulation, R = 0 Percenatge
For Case b.4) ZPF Lagging   Open circuit EMF, EMF = 570 V
Percenatge Regulation, R = 0.00 Percenatge
For Case b.5) ZPF Leading   Open circuit EMF, EMF = 135 V
Percenatge Regulation, R = -100.00 Percenatge
Open circuit EMF V)            525           338            520       570          135
Percenatge Regulation           0.00         -100.0          0.       0.00        -100.00


## Example 6.5 Page No : 307¶

In [7]:
import math

# GIVEN DATA

print " EXAMPLE : 6.5a Data is same as Example 6.4:           Given Data ";
print " Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560 ";
print " IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24 ";
m = 3;                         # Total Number of phase in Induction Motor
p = 6;                         # Total number of Poles of Induction Motor
V = 400;                       # Operating voltage of the Induction motor in Volts
I = 13.5;                      # Operating current of the Induction motor in Amphere
N = 1000;                      # speed of the Induction motor in RPM
Ia_scc = 13.5;                 # SCC test Armature current in Amphere at If = 9.5 A
If_scc = 9.5;                  # SCC test field Rated current in Amphere
Ia_zpf = 13.5;                 # ZPF test Armature current in Amphere at If = 24 A
If_zpf = 24;                   # ZPF test field Rated current in Amphere

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
Ra = 1.0;                                   # Armature resistance in Ohms
v = V/math.sqrt(3);                              # Rated phase Voltage in Volts

# For case (a)

EMF_a1 = 345;                               # From OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386 open-circuit line-line voltage per phase is 345vVfor If = 9.50A in Volts
Zs_a1 = EMF_a1/(Ia_zpf*math.sqrt(3));            # Unsaturated synchronous impedance at If=9.50A in Ohms
Xs_a1 = math.sqrt((Zs_a1**2)-(Ra**2));             # Synchronous reactance at If =9.50A in Ohms
Ia_a2 = 15.75;                              # Current from SCC in Figure6.15 & Page no:-386 is 15.75A for correspounding to the rated Voltage in Volts
Zs_a2 = V/(Ia_a2*math.sqrt(3));                  # Unsaturated synchronous impedance at If=9.50A in Ohms
Xs_a2 = math.sqrt((Zs_a2**2)-(Ra**2));             # Synchronous reactance at If =9.50A in Ohms

# For Case (b.1) 0.8 pf Lagging

pfa_b1 = math.acos(math.radians(0.8));                                                     # Power factor angle in degree
E_b1 = math.sqrt(real_b1**2+imag_b1**2);                                        # Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for lagging pf
del_b1 = math.degrees(math.atan(imag_b1/real_b1));                                         # Power angle in degree
r_b1 = 100*(E_b1-v)/v;                                                   # Percantage regulation

# For Case (b.2) 0.8 pf Leading

pfa_b2 = math.acos(math.radians(0.8));                                                     # Power factor angle in degree
E_b2 = math.sqrt(real_b2**2+imag_b2**2);                                        # Induced Voltage pr phase in Volts from Figure6.19 (b) & Page no:-394 shows the phasor diagram for leading pf
del_b2 = math.degrees(math.atan(imag_b2/real_b2));                                         # Power angle in degree
r_b2 = 100*(E_b2-v)/v;                                                   # Percantage regulation

# For Case (b.3) Unity pf

pfa_b3 = math.acos(math.radians(1.0));                                                     # Power factor angle in degree
real_b3 = (v+Ia_zpf*Ra);
imag_b3 = (Ia_zpf*Xs_a1);
E_b3 = math.sqrt(real_b3**2+imag_b3**2);                                        # Induced Voltage pr phase in Volts from Figure6.19 (a) & Page no:-394 shows the phasor diagram for unity pf
del_b3 = math.degrees(math.atan(imag_b3/real_b3));                                         # Power angle in degree
r_b3 = 100*(E_b3-v)/v;                                                   # Percantage regulation

# For Case (b.4) ZPF pf Lagging

pfa_b4 = math.degrees(math.acos(math.radians(0)));                                                       # Power factor angle in degree
real_b4 = (v+Ia_zpf*Xs_a1);
imag_b4 = (-Ia_zpf*Ra);
E_b4 = math.sqrt(real_b4**2+imag_b4**2);                                        # Induced Voltage pr phase in Volts ZPF for lagging pf
del_b4 = math.degrees(math.atan(imag_b4/real_b4));                                         # Power angle in degree
r_b4 = 100*(E_b4-v)/v;                                                   # Percantage regulation

# For Case (b.5) ZPF pf Leading

pfa_b5 = math.degrees(math.acos(math.radians(0)));                                                       # Power factor angle in degree
real_b5 = (v-Ia_zpf*Xs_a1);
imag_b5 = (Ia_zpf*Ra);
E_b5 = math.sqrt(real_b5**2+imag_b5**2);                                        # Induced Voltage pr phase in Volts ZPF for lagging pf
del_b5 = math.degrees(math.atan(imag_b5/real_b5));                                         # Power angle in degree
r_b5 = 100*(E_b5-v)/v;                                                   # Percantage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.5a : SOLUTION :-") ;
print " a.1)  Synchronous reactance for rated current at If = %.2f, Xs = %.2f Ohms "%(If_scc,Xs_a1)
print " a.2)  Synchronous reactance for rated per phase Voltage  at v  = %.f, Xs = %.2f Ohms "%(v,Xs_a2)
print " For Case b.1) 0.8 pf Lagging   Induced EMF per phase , EMF = %.2f V "%(E_b1)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b1)
print " Power angle = %.2f degree "%(del_b1)
print " For Case b.2) 0.8 pf Leading   Induced EMF per phase, EMF = %.2f V "%(E_b2)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b2)
print " Power angle = %.2f degree "%(del_b2)
print " For Case b.3) Unity pf Lagging   Induced EMF per phase, EMF = %.2f V "%(E_b3)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b3)
print " Power angle = %.2f degree "%(del_b3)
print " For Case b.4) ZPF Lagging   Induced EMF per phase, EMF = %.2f V"%(E_b4)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b4)
print " Power angle = %.1f degree "%(del_b4)
print " For Case b.5) ZPF Leading   Induced EMF per phase, EMF = %.2f V "%(E_b5)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b5)
print " Power angle = %.2f degree "%(del_b5)
print (" Calculated Answer in Tabular Column")
print "   Power Factor                 0.8 Lag      0.8 Lead        1.0         ZPF Lag         ZPF Lead "
print " Open circuit EMF V)           %.2f        %.2f         %.2f       %.2f          %.2f "%(E_b1,E_b2,E_b3,E_b4,E_b5)
print " Percenatge Regulation           %.2f         %.2f          %.2f.       %.2f         %.2f "%(r_b1,r_b2,r_b3,r_b4,r_b5)
print " Power angle                    %.2f         %.2f          %.2f.       %.1f           %.2f "%(del_b1,del_b2,del_b3,del_b4,del_b5)

 EXAMPLE : 6.5a Data is same as Example 6.4:           Given Data
Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560
IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24
EXAMPLE : 6.5a : SOLUTION :-
a.1)  Synchronous reactance for rated current at If = 9.50, Xs = 14.72 Ohms
a.2)  Synchronous reactance for rated per phase Voltage  at v  = 231, Xs = 14.63 Ohms
For Case b.1) 0.8 pf Lagging   Induced EMF per phase , EMF = 653910.50 V
Percenatge Regulation, R = 283051.55 Percenatge
Power angle = 84.54 degree
For Case b.2) 0.8 pf Leading   Induced EMF per phase, EMF = 653897.98 V
Percenatge Regulation, R = 283046.13 Percenatge
Power angle = 87.65 degree
For Case b.3) Unity pf Lagging   Induced EMF per phase, EMF = 315.03 V
Percenatge Regulation, R = 36.41 Percenatge
Power angle = 39.11 degree
For Case b.4) ZPF Lagging   Induced EMF per phase, EMF = 429.88 V
Percenatge Regulation, R = 86.14 Percenatge
Power angle = -1.8 degree
For Case b.5) ZPF Leading   Induced EMF per phase, EMF = 34.93 V
Percenatge Regulation, R = -84.88 Percenatge
Power angle = 22.74 degree
Open circuit EMF V)           653910.50        653897.98         315.03       429.88          34.93
Percenatge Regulation           283051.55         283046.13          36.41.       86.14         -84.88
Power angle                    84.54         87.65          39.11.       -1.8           22.74


## Example 6.6 Page No : 309¶

In [8]:
import math
import cmath

# GIVEN DATA

print " EXAMPLE : 6.4  Data is same as Exaple 6.4 :           Given Data ";
print " Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560 ";
print " IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24 ";
m = 3;                         # Total Number of phase in Induction Motor
p = 6;                         # Total number of Poles of Induction Motor
V = 400;                       # Operating voltage of the Induction motor in Volts
I = 13.5;                      # Operating current of the Induction motor in Amphere
N = 1000;                      # speed of the Induction motor in RPM
Ia_scc = 13.5;                 # SCC test Armature current in Amphere at If = 9.5 A
If_scc = 9.5;                  # SCC test field Rated current in Amphere
Ia_zpf = 13.5;                 # ZPF test Armature current in Amphere at If = 24 A
If_zpf = 24;                   # ZPF test field Rated current in Amphere

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
Ra = 1.0;                                   # Armature resistance in Ohms
v = V/math.sqrt(3);                              # Rated phase voltage in Volts

# For Case (a) 0.8 pf Lagging

pfa_a = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
# E_a = v+(Ia_scc*(cosd(pfa_a)-%i*sind(pfa_a))*Ra);
R1_a = 11.8; A_a = 9.50;                                                 #From OCC the field current required for E_a (Should be in Line-line Voltage) E_a = 419.05V will get R1_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)
angle_a = 124.95;                                                        # Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'-1.92'+36.87' = 124.95'
F_a = math.sqrt((R1_a**2)+(A_a**2)-(2*R1_a*A_a*math.degrees(math.cos(math.radians(angle_a)))));                 # From phasor diagram in figure 6.21(a) & Page no:-400 the neccessary field excitation in Amphere
Eo_a = 538;                                                              # Corresponding to field current F_a = 18.94 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)
r_a = 100*((Eo_a-V)/V);                                                  # Percentage regulation

# For Case (b) 0.8 pf Leading

pfa_b = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
R1_b = 11.80; A_b = 9.50;                                                 #From OCC the field current required for E_b (Should be in Line-line Voltage) E_b = 419.10V will get R1_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b = 55.07;                                                         # Angle between R1_b & A_b (Figure6.21b & Page no:-400) = 90'-1.92'-36.87' = 55.07'
F_b = cmath.sqrt((R1_b**2)+(A_b**2)-(2*R1_b*A_b*math.degrees(math.cos(math.radians(angle_b)))));                 # From phasor diagram in figure 6.21(b) & Page no:-400 the neccessary field excitation in Amphere
Eo_b = 382;                                                              # Corresponding to field current F_b = 10.10 A the open circuit EMF from OCC is 382 V (Figure6.15 & Page no:-386)
r_b = 100*((Eo_b-V)/V);                                                  # Percentage regulation

# For Case (c) Unity pf

pfa_c = math.degrees(math.acos(math.radians(1)));                                                        # Power factor angle in degree
R1_c = 12.10; A_c = 9.50;                                                #From OCC the field current required for E_c (Should be in Line-line Voltage) E_c = 423.50V will get R1_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)
angle_c = 90;                                                            # Angle between R1_a & A_a (Figure6.21a & Page no:-400) = 90'
F_c = math.sqrt((R1_c**2)+(A_c**2)-(2*R1_c*A_c*math.degrees(math.cos(math.radians(angle_c)))));                 # From phasor diagram in figure 6.21(c) & Page no:-400 the neccessary field excitation in Amphere
Eo_c = 480;                                                              # Corresponding to field current F_c = 15.38 A the open circuit EMF from OCC is 538 V (Figure6.15 & Page no:-386)
r_c = 100*((Eo_c-V)/V);                                                  # Percentage regulation

# For Case (d) ZPF Lagging

pfa_d = math.acos(math.radians(0.0));                                                      # Power factor angle in degree
R1_d = 11.20; A_d = 9.50;                                                #From OCC the field current required for E_d (Should be in Line-line Voltage) E_d = 400.80V will get R1_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)
angle_d = 179.40;                                                        # Angle between R1_d & A_d = 90'-0.6'+90' = 179.40'
F_d = math.sqrt((R1_d**2)+(A_d**2)-(2*R1_d*A_d*math.degrees(math.cos(math.radians(angle_d)))));                 # From phasor diagram the neccessary field excitation in Amphere
Eo_d = 545;                                                              # Corresponding to field current F_d = 18.12 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)
r_d = 100*((Eo_d-V)/V);                                                  # Percentage regulation

# For Case (d) ZPF Lagging

pfa_e = math.acos(math.radians(0.0));                                                      # Power factor angle in degree
R1_e = 11.20; A_e = 9.50;                                                #From OCC the field current required for E_e (Should be in Line-line Voltage) E_d = 400.80V will get R1_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)
angle_e = 0.60;                                                          # Angle between R1_e & A_e = 90'+0.6'-90' = 0.60'
F_e = cmath.sqrt((R1_e**2)+(A_e**2)-(2*R1_e*A_e*math.degrees(math.cos(math.radians(angle_e)))));                 # From phasor diagram the neccessary field excitation in Amphere
Eo_e = 63;                                                               # Corresponding to field current F_e = 1.70 A the open circuit EMF from OCC is 545 V (Figure6.15 & Page no:-386)
r_e = 100*((Eo_e-V)/V);                                                  # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.6 : SOLUTION :-") ;
print " For Case a) 0.8 pf Lagging   Open circuit EMF, EMF = %.f V "%(Eo_a)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_a)
print " For Case b) 0.8 pf Leading   Open circuit EMF, EMF = %.f V "%(Eo_b)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b)
print " For Case c) Unity pf Lagging   Open circuit EMF, EMF = %.f V "%(Eo_c)
print " Percenatge Regulation, R = %.f Percenatge "%(r_c)
print " For Case d) ZPF Lagging   Open circuit EMF, EMF = %.f V"%(Eo_d)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_d)
print " For Case e) ZPF Leading   Open circuit EMF, EMF = %.f V "%(Eo_e)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_e)
print (" Calculated Answer in Tabular Column")
print "   Power Factor                 0.8 Lag      0.8 Lead        1.0       ZPF Lag      ZPF Lead "
print " Open circuit EMF V)            %.f           %.f          %.f         %.f          %.f "%(Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)
print " Percenatge Regulation           %.2f         %.2f          %.f       %.2f        %.2f "%(r_a,r_b,r_c,r_d,r_e)

 EXAMPLE : 6.4  Data is same as Exaple 6.4 :           Given Data
Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560
IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24
EXAMPLE : 6.6 : SOLUTION :-
For Case a) 0.8 pf Lagging   Open circuit EMF, EMF = 538 V
Percenatge Regulation, R = 0.00 Percenatge
For Case b) 0.8 pf Leading   Open circuit EMF, EMF = 382 V
Percenatge Regulation, R = -100.00 Percenatge
For Case c) Unity pf Lagging   Open circuit EMF, EMF = 480 V
Percenatge Regulation, R = 0 Percenatge
For Case d) ZPF Lagging   Open circuit EMF, EMF = 545 V
Percenatge Regulation, R = 0.00 Percenatge
For Case e) ZPF Leading   Open circuit EMF, EMF = 63 V
Percenatge Regulation, R = -100.00 Percenatge
Open circuit EMF V)            538           382          480         545          63
Percenatge Regulation           0.00         -100.00          0       0.00        -100.00


## Example 6.7 Page No : 314¶

In [9]:
import math
import cmath

# GIVEN DATA

print " EXAMPLE : 6.7  Data as same as Example 6.4  :           Given Data ";
print " Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560 ";
print " IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24 ";
m = 3;                         # Total Number of phase in Induction Motor
p = 6;                         # Total number of Poles of Induction Motor
V = 400;                       # Operating voltage of the Induction motor in Volts
I = 13.5;                      # Operating current of the Induction motor in Amphere
N = 1000;                      # speed of the Induction motor in RPM
Ia_scc = 13.5;                 # SCC test Armature current in Amphere at If = 9.5 A
If_scc = 9.5;                  # SCC test field Rated current in Amphere
Ia_zpf = 13.5;                 # ZPF test Armature current in Amphere at If = 24 A
If_zpf = 24;                   # ZPF test field Rated current in Amphere

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.15 & Page no:-386
Ra = 1.0;                                   # Armature resistance in Ohms
BC = 120;                                   # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.15 & Page no:-386
Xl = BC/(math.sqrt(3)*Ia_scc);                   # Per phase leakage reactance in Ohms for this referring to example6.4 & page no:- 386

# For Case (a) 0.8 pf Lagging

pfa_a = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
R_a = 9.8; A_a = 9.5;                                                    #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 479.60V will get R_a & A_a value Respectively from SCC (Figure6.15 & Page no:-386)
angle_a = 126.87;                                                        # Angle between R_a & A_a (Figure6.22(a) & Page no:-403) = 90'+36.87' = 126.87'
F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a)))));                   # From phasor diagram in figure 6.22(a) & Page no:-403 the neccessary field excitation in Amphere
Eo_a = 560;                                                              # Corresponding to field current ( OF'=OF+FF'), F_a = 17.28 + 6.2 = 23.46 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)
r_a = 100*((Eo_a-V)/V);                                                  # Percentage regulation

# For Case (b) 0.8 pf Leading

pfa_b = math.acos(math.radians(0.8));                                                      # Power factor angle in degree
R_b = 9.8; A_b = 9.5;                                                    #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 363.90 V will get R_b & A_b value Respectively from SCC (Figure6.15 & Page no:-386)
angle_b = 53.13;                                                         # Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'-36.87' = 53.13'
F_b = cmath.sqrt((R_b**2)+(A_b**2)-(2*R_b*A_b*math.degrees(math.cos(math.radians(angle_b)))));                   # From phasor diagram in figure 6.22(b) & Page no:-403 the neccessary field excitation in Amphere
Eo_b = 380;                                                              # Corresponding to field current ( OF'=OF+FF') F_b = 8.62+1.5=10.12 A the open circuit EMF from OCC is 380 V (Figure6.15 & Page no:-386)
r_b = 100*((Eo_b-V)/V);                                                  # Percentage regulation

# For Case (c) Unity pf Leading

pfa_c = math.acos(math.radians(1.0));                                                      # Power factor angle in degree
R_c = 9.8; A_c = 9.5;                                                    #From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 440.11 V will get R_c & A_c value Respectively from SCC (Figure6.15 & Page no:-386)
angle_c = 90;                                                            # Angle between R_c & A_c (Figure6.22c & Page no:-403) = 90' = 90'
F_c = math.sqrt((R_c**2)+(A_c**2)-(2*R_c*A_c*math.degrees(math.cos(math.radians(angle_c)))));                   # From phasor diagram in figure 6.22(c) & Page no:-403 the neccessary field excitation in Amphere
Eo_c = 510;                                                              # Corresponding to field current ( OF'=OF+FF') F_c = 13.65+3.0=16.65A the open circuit EMF from OCC is 510 V (Figure6.15 & Page no:-386)
r_c = 100*((Eo_c-V)/V);                                                  # Percentage regulation

# For Case (d) ZPF Lagging

pfa_d = math.degrees(math.acos(math.radians(0)));                                                        # Power factor angle in degree
R_d = 9.8; A_d = 9.5;                                                    #From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 570.20 V will get R_d & A_d value Respectively from SCC (Figure6.15 & Page no:-386)
angle_d = 180.0;                                                         # Angle between R_d & A_d = 90'+90' = 180'
F_d = math.sqrt((R_d**2)+(A_d**2)-(2*R_d*A_d*math.degrees(math.cos(math.radians(angle_d)))));              # The neccessary field excitation in Amphere
Eo_d = 600;                                                              # Corresponding to field current ( OF'=OF+FF') F_d = 19.3+16=35.30 A the open circuit EMF from OCC is 525 V (Figure6.15 & Page no:-386)
r_d = 100*((Eo_d-V)/V);                                                  # Percentage regulation

# For Case (e) ZPF Lagging

pfa_e = math.degrees(math.acos(math.radians(0)));                                                        # Power factor angle in degree
R_e = 9.8; A_e = 9.50;                                                   #From OCC the field current required for Er_e (Should be in Line-line Voltage) Er_e = 281.10 V will get R_e & A_e value Respectively from SCC (Figure6.15 & Page no:-386)
angle_e = 0.0;                                                           # Angle between R_e & A_e = 90'-90' = 0.0'
F_e = cmath.sqrt((R_e**2)+(A_e**2)-(2*R_e*A_e*math.degrees(math.cos(math.radians(angle_e)))));                   # The neccessary field excitation in Amphere
Eo_e = 5;                                                                # Corresponding to field current ( OF'=OF+FF') F_e = 0.0+0.30=0.30 A the open circuit EMF from OCC is 5 V (Figure6.15 & Page no:-386)
r_e = 100*((Eo_e-V)/V);                                                  # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.7 : SOLUTION :-") ;
print "  Per phase leakage reactance, Xl = %.2f Ohms "%(Xl)
print " For Case a) 0.8 pf Lagging   Open circuit EMF, EMF = %.f V "%(Eo_a)
print " Percenatge Regulation, R = %.f Percenatge "%(r_a)
print " For Case b) 0.8 pf Leading   Open circuit EMF, EMF = %.f V "%(Eo_b)
print " Percenatge Regulation, R = %.f Percenatge "%(r_b)
print " For Case c) Unity pf Lagging   Open circuit EMF, EMF = %.f V "%(Eo_c)
print " Percenatge Regulation, R = %.f Percenatge "%(r_c)
print " For Case d) ZPF Lagging   Open circuit EMF, EMF = %.f V"%(Eo_d)
print " Percenatge Regulation, R = %.f Percenatge "%(r_d)
print " For Case e) ZPF Leading   Open circuit EMF, EMF = %.f V "%(Eo_e)
print " Percenatge Regulation, R = %.2f Percenatge "%(r_e)
print (" Calculated Answer in Tabular Column")
print "   Power Factor                 0.8 Lag      0.8 Lead        1.0       ZPF Lag      ZPF Lead "
print " Open circuit EMF V)            %.f           %.f            %.f       %.f          %.f "%(Eo_a,Eo_b,Eo_c,Eo_d,Eo_e)
print " Percenatge Regulation           %.f             %.f          %.1f        %.f        %.2f "%(r_a,r_b,r_c,r_d,r_e)

 EXAMPLE : 6.7  Data as same as Example 6.4  :           Given Data
Vocv  215      284      320      380      400      422      452      472      488      508      520      532      540      552      560
IfA    6.5      8        9       10       11       12        14       15       16       17      18        19       20       22       24
EXAMPLE : 6.7 : SOLUTION :-
Per phase leakage reactance, Xl = 5.13 Ohms
For Case a) 0.8 pf Lagging   Open circuit EMF, EMF = 560 V
Percenatge Regulation, R = 0 Percenatge
For Case b) 0.8 pf Leading   Open circuit EMF, EMF = 380 V
Percenatge Regulation, R = -100 Percenatge
For Case c) Unity pf Lagging   Open circuit EMF, EMF = 510 V
Percenatge Regulation, R = 0 Percenatge
For Case d) ZPF Lagging   Open circuit EMF, EMF = 600 V
Percenatge Regulation, R = 0 Percenatge
For Case e) ZPF Leading   Open circuit EMF, EMF = 5 V
Percenatge Regulation, R = -100.00 Percenatge
Open circuit EMF V)            560           380            510       600          5
Percenatge Regulation           0             -100          0.0        0        -100.00


## Example 6.9 Page No : 315¶

In [10]:
import math
from numpy import imag,real

# GIVEN DATA

print " EXAMPLE : 6.9 :           Given Data ";
print " VockV   10     10.80     11.50     12.10       12.60       13       14       14.50     14.80 ";
print " IfA    175      200       225       250        275        300       400       450      500 ";
p = 6;                         # Total number of Poles of Alternator
V = 11*10**3;                   # Operating voltage of the Alternator in Volts
N = 1500;                      # speed of the Alternator in RPM
Ia_scc = 2099;                 # SCC test Armature current in Amphere at If = 200 A
If_scc = 200;                  # SCC test field Rated current in Amphere
Ia_pt = 2099;                  # Pottier test Armature current in Amphere at If = 450 A
If_pt = 450;                   # Pottier test field Rated current in Amphere
VA = 40*10**6;                  # VA rating of the Alternator in Volts-Amphere
f = 50;                        # Operating Frequency of the Alternator in Hertz
pf = 0.8;                      # Power factor (lagging)

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407

v = V/math.sqrt(3);                                                  # Rated phase Voltage in Volts
I = VA/(math.sqrt(3)*V);                                             # Full-load phase current in Amphere
Xl = 0.4481;                                                    # Leakage reactance in Ohms From OCC and SCC test Graph or Pottier triangle in Figure6.24 & Page no:-407

# For Case(a) General Method

pfa_a = math.degrees(math.acos(math.radians(pf)));                                                       # Power factor angle in degree
R_a = 208.4; A_a = 200;                                                    #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 11043.66 V will get R_a & A_a value Respectively from SCC (Figure6.24 & Page no:-407)
angle_a = 131.93;                                                        # Angle between R_a & A_a (Figure6.25(a) & Page no:-408) = 90'+5.06'+36.87' = 131.93'
F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a)))));                   # From phasor diagram in figure 6.25(a) & Page no:-408 the neccessary field excitation in Amphere
Eo_a = 13720;                                                            # Corresponding to field current, F_a = 373 A the open circuit EMF from OCC is 560 V (Figure6.15 & Page no:-386)
r_a = 100*((Eo_a-V)/V);                                                  # Percentage regulation

# For Case (b) ASA Method

pfa_b = math.degrees(math.acos(math.radians(pf)));                                                       # Power factor angle in degree
R_b = 160; A_b = 200;                                                    #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 11043.66 V will get R_b & A_b value Respectively from SCC (Figure6.24 & Page no:-407)
angle_b = 126.87;                                                        # Angle between R_b2 & A_b2 (Figure6.22b & Page no:-403) = 90'+36.87' = 126.87'
F_b = math.sqrt((R_b**2)+(A_b**2)-(2*R_b*A_b*math.degrees(math.cos(math.radians(angle_b)))));                   # From phasor diagram in figure 6.25(b) & Page no:-408 the neccessary field excitation in Amphere
Eo_b = 13660;                                                            # Corresponding to field current ( OF'=OF+FF') F_b = 337.88+15.38=337.88  A the open circuit EMF from OCC is 13660 V (Figure6.15 & Page no:-386)
r_b = 100*((Eo_b-V)/V);                                                  # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.9 : SOLUTION :-") ;
print " For Case a) GeneralZPF) Method   Induced EMF, EMF = %.f < %.2f V "%(abs(Er_a),math.degrees(math.atan2(Er_a.imag,Er_a.real)))
print " Percenatge Regulation, R = %.2f Percenatge "%(r_a)
print " For Case b) ASA Method   Induced EMF, EMF = %.f < %.2f V "%(abs(Er_b),math.degrees(math.atan2(Er_b.imag,Er_b.real)))
print " Percenatge Regulation, R = %.2f Percenatge "%(r_b)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- For Case a) GeneralZPF) Method a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f  "%(abs(Er_a),math.degrees(math.atan2(Er_a.imag,Er_a.real)))
print "                                     For Case b) ASA Method          a) Induced EMF = 6376<-5.07 degree instead of %.f < %.2f  "%(abs(Er_b),math.degrees(math.atan2(Er_b.imag,Er_b.real)))
print " CALCULATION OF THE POWER ANGLE IS NOT CALCULATED IN THE TEXT BOOK FOR THIS PROBLEM "
print " INDUCED EMF AND PERCENTAGE REGULATION IS APPROXIMATED VALUE BECACUSE IN THE TEXT BOOK, CALCULATED INDUCED EMF IS WRONGLY PRINTED"

 EXAMPLE : 6.9 :           Given Data
VockV   10     10.80     11.50     12.10       12.60       13       14       14.50     14.80
IfA    175      200       225       250        275        300       400       450      500
EXAMPLE : 6.9 : SOLUTION :-
For Case a) GeneralZPF) Method   Induced EMF, EMF = 54351 < -82.49 V
Percenatge Regulation, R = 0.00 Percenatge
For Case b) ASA Method   Induced EMF, EMF = 54351 < -82.49 V
Percenatge Regulation, R = 0.00 Percenatge
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- For Case a) GeneralZPF) Method a) Induced EMF = 6376<-5.07 degree instead of 54351 < -82.49
For Case b) ASA Method          a) Induced EMF = 6376<-5.07 degree instead of 54351 < -82.49
CALCULATION OF THE POWER ANGLE IS NOT CALCULATED IN THE TEXT BOOK FOR THIS PROBLEM
INDUCED EMF AND PERCENTAGE REGULATION IS APPROXIMATED VALUE BECACUSE IN THE TEXT BOOK, CALCULATED INDUCED EMF IS WRONGLY PRINTED


## Example 6.10 Page No : 318¶

In [11]:
import math
from numpy import real,imag

# GIVEN DATA

m = 3;                         # Total Number of Phase in Alternator
p = 2;                         # Total number of Poles of Alternator
V = 11*10**3;                   # Operating voltage of the Alternator in Volts
VA = 10*10**6;                  # VA rating of the Alternator in Volts-Amphere
f = 50;                        # Operating Frequency of the alternator in Hertz
pf = 0.8;                      # Power factor (lagging)
Vf = 12*10**3;                  # Operating field voltage of the Alternator in Volts
If = 160;                      # Field Current in Amphere
Ra = 0.05;                     # Armature resistance per phase in Ohms
Xs = 1.5;                      # Winding leakage reactance per phase in Ohms
A = 150;                     # The armature MMF at rated current is equivalent to Field Current in Amphere

# CALCULATIONS

Vt = V/math.sqrt(3);                                     # Rated per phase Voltage in Volts
Ia = VA/(math.sqrt(3)*V);                                # Rated Armature Current in Amphere
pfa = math.degrees(math.acos(math.radians(pf)));                                    # Power factor angle in degree
R_a = 90 + math.degrees(math.atan2(Er.imag,Er.real));                # Angle of R in Degree
R = 160 * ( 1j * (R_a) * math.pi/180)**2;               # (Line-line Voltage) Er = 11902.40V will get R from Air gap Characteristics
A_n = A * ( 1j * (-pfa) * math.pi/180)**2;
F = R - A_n;                                        # Field Current required to produce the excitation EMF in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 6.10: SOLUTION :-");
print " a) Field Current required to produce the excitation EMF, F = %.2f A "%(abs(F))

EXAMPLE : 6.10: SOLUTION :-
a) Field Current required to produce the excitation EMF, F = 22.74 A


## Example 6.11 Page No : 322¶

In [12]:
import math

# GIVEN DATA

print " EXAMPLE : 6.11 :           Given Data ";
print " Voc V   12     13       13.8      14.5        15.1  ";
print " IfA    175      200       225       250        275   ";
V = 11*10**3;                   # Operating voltage of the Synchronous generator in Volts
VA = 50*10**6;                  # VA rating of the Synchronous generator in Volts-Amphere
f = 50;                        # Operating Frequency of the Synchronous generator in Hertz
N = 1500;                      # Speed of the Synchronous generator in RPM
If_scc = 200;                  # SCC test field Rated current in Amphere at rated Short circuit current
If_zpf = 400;                  # ZPF test field Rated current in Amphere at rated voltage and rated current
pf = 0.8;                      # Power factor (lagging)

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.30 & Page no:-413

Vt = V/math.sqrt(3);                                     # Rated per phase Voltage in Volts
Ia = VA/(math.sqrt(3)*V);                                # Rated Armature Current in Amphere
pfa = math.degrees(math.acos(math.radians(pf)));                                    # Power factor angle in degree
O = 13000;                                          # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
Xs = O/(math.sqrt(3)*Ia);                                # Synchronous reactance per phase in Ohms
BC = 4000;                                          # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
Xl = BC/(math.sqrt(3)*Ia );                              # Per phase leakage reactance in Ohms

# For Case (a) General (ZPF) Method

R_a = 220; A_a = 200;                                                     #From OCC the field current required for Er_a (Should be in Line-line Voltage) Er_a = 13776V will get R_a & A_a value Respectively from SCC (Figure6.30 & Page no:-403)
angle_a = 140.3;                                                          # Angle between R_a & A_a = 90'+13.43'+36.87' = 140.3'
F_a = math.sqrt((R_a**2)+(A_a**2)-(2*R_a*A_a*math.degrees(math.cos(math.radians(angle_a)))));                    # From phasor diagram in figure 6.16(a) & Page no:-388 the neccessary field excitation in Amphere
Eo_a = 20000;                                                             # Corresponding to field current F_a = 470.90 A the open circuit EMF from OCC is 20000 V (Figure6.30 & Page no:-413)
r_a = 100*((Eo_a-V)/V);                                                   # Percentage regulation

# For Case(b) EMF Method

F_b = 500;                                                                #From OCC the field current required for Er_b (Should be in Line-line Voltage) Er_b = 21404 V will get 500A from SCC (Figure6.15 & Page no:-386)

# For Case (c) MMF Method

Er_c = Vt+Ia*(math.degrees(math.cos(math.radians(pfa)))-1j*math.degrees(math.sin(math.radians(pfa))))*0;                                  # Induced Voltage in Volts ( Zero is multipied because Armature reismath.tance is zero (not mentioned))
R_c = 160; A_c = 200;                                                     #From OCC the field current required for Er_c (Should be in Line-line Voltage) Er_c = 11000 V will get R_c & A_c value Respectively from SCC (Figure6.30 & Page no:-413)
angle_c = 126.27;                                                         # Angle between R_c & A_c  = 90'-0'+36.87' = 126.27' {can refer figure 6.21a at page no:-400}
F_c = math.sqrt((R_c**2)+(A_c**2)-(2*R_c*A_c*math.degrees(math.cos(math.radians(angle_c)))));                    # From phasor diagram {can refer figure 6.21a at page no:-400} the neccessary field excitation in Amphere

# For Case (d) ASA Method

R_d = 220; A_d = 200;                                                     #From OCC the field current required for Er_d (Should be in Line-line Voltage) Er_d = 13800 V will get R_d & A_d value Respectively from SCC (Figure6.30 & Page no:-413)
angle_d = 126.87;                                                         # Angle between R_d & A_d = 90'+36.87' = 126.87'{can refer figure 6.22a at page no:-40}
F_d1 = math.sqrt((R_d**2)+(A_d**2)-(2*R_d*A_d*math.degrees(math.cos(math.radians(angle_d)))));                   # from Phasor diagram {can refer figure 6.2a at page no:-400 The neccessary field excitation in Amphere
F_d = F_d1 + 30;                                                          # from Phasor diagram {can refer figure 6.2a at page no:-400 The Total neccessary field excitation in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 6.11 : SOLUTION :-") ;
print " a) Leakage Reactance, Xl = %.2f Ohms "%(Xl)
print " b) Synchronous Reactance, Xs = %.2f Ohms "%(Xs)
print " For Case a) General ZPF) Method  Field Current required for maintaing the rated terminal voltage\
for rated kVA rating at %.2f Lagging Power factor , F = %.2f A "%(pf,F_a)
print " For Case a) EMF Method  Field Current required for maintaing the rated terminal voltage for rated\
kVA rating at %.2f Lagging Power factor , F = %.f A "%(pf,F_b)
print " For Case a) MMF Method  Field Current required for maintaing the rated terminal voltage for rated\
kVA rating at %.2f Lagging Power factor , F = %.2f A "%(pf,F_c)
print " For Case a) ASA Method  Field Current required for maintaing the rated terminal voltage for rated kVA\
rating at %.2f Lagging Power factor , F = %.f A "%(pf,F_d)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- For Case a) General ZPF) Method  a) Field Current required for maintaining the rated terminal voltage for rated kVA rating at %.2f Lagging Power factor , F = 470.90 A instead of %.2f A "%(pf,F_a);

 EXAMPLE : 6.11 :           Given Data
Voc V   12     13       13.8      14.5        15.1
IfA    175      200       225       250        275
EXAMPLE : 6.11 : SOLUTION :-
a) Leakage Reactance, Xl = 0.88 Ohms
b) Synchronous Reactance, Xs = 2.86 Ohms
For Case a) General ZPF) Method  Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1991.92 A
For Case a) EMF Method  Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 500 A
For Case a) MMF Method  Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1494.97 A
For Case a) ASA Method  Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 1795 A
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- For Case a) General ZPF) Method  a) Field Current required for maintaining the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 470.90 A instead of 1991.92 A


## Example 6.12 Page No : 324¶

In [13]:
import math

# GIVEN DATA

print " EXAMPLE : 6.12 :           Given Data ";
print " VocV   175     250      280      300        330       350      370       380 ";
print " IfA     10       17       20       23         30       38       50        60 ";
print " VzpfV      -        -       -       0         130     210      265      280 ";
V = 433;                       # Operating voltage of the Alternator in Volts
N = 3000;                      # speed of the Alternator in RPM
VA = 20*10**3;                  # VA rating of the Alternator in Volts-Amphere
f = 50;                        # Operating Frequency of the Alternator in Hertz
pf = 0.8;                      # Power factor (lagging)

# CALCULATIONS
# Some of the data obtained from OCC and SCC test Graph or Pottier triangle in Figure6.35 & Page no:-420

Vt = V/math.sqrt(3);                                     # Rated per phase Voltage in Volts
Ia = VA/(math.sqrt(3)*V);                                # Rated Armature Current in Amphere
pfa = math.degrees(math.acos(math.radians(pf)));                                    # Power factor angle in degree
O = 298;                                            # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
Xs = O/(math.sqrt(3)*Ia);                                # Synchronous reactance per phase in Ohms
BC = 70;                                            # Open circuit Voltage in Volts obtained from OCC and SCC test Graph or Pottier triangle Figure6.30 & Page no:-413
Xl = BC/(math.sqrt(3)*Ia );                              # Per phase leakage reactance in Ohms
c = 380-60;                                         # The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation
y = 694.50;                                         # The open Voltage voltage is 694.50V (line-line) its Obatained by extrapolation
# Extrapolation Equation is y = (x*(380-370)/(60-50))*c
x = y - c;                                          # The required field current in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 6.12 : SOLUTION :-") ;
print " a) Leakage Reactance, Xl = %.2f Ohms "%(Xl)
print " b) Synchronous Reactance, Xs = %.2f Ohms "%(Xs)
print " c) Field Current required for maintaing the rated terminal voltage for rated kVA\
rating at %.2f Lagging Power factor , F = %.2f A "%(pf,x)


 EXAMPLE : 6.12 :           Given Data
VocV   175     250      280      300        330       350      370       380
IfA     10       17       20       23         30       38       50        60
VzpfV      -        -       -       0         130     210      265      280
EXAMPLE : 6.12 : SOLUTION :-
a) Leakage Reactance, Xl = 1.52 Ohms
b) Synchronous Reactance, Xs = 6.45 Ohms
c) Field Current required for maintaing the rated terminal voltage for rated kVA rating at 0.80 Lagging Power factor , F = 374.50 A


## Example 6.13 Page No : 328¶

In [14]:
import math

# GIVEN DATA

V = 400;                       # Operating voltage of the Synchronous generator in Volts
VA = 60*10**3;                  # VA rating of the Synchronous generator in Volts-Amphere
f = 50;                        # Operating Frequency of the Synchronous generator in Hertz
xd = 1.5;                      # Direct axis reactances in Ohms
xq = 0.6;                      # Quadrature axis reactances in Ohms

# CALCULATIONS

I = VA/(math.sqrt(3)*V);                             # Rated current in Amphere
v = V/math.sqrt(3);                                  # Rated Phase Votage in Volts

# For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)

pf_a = 0.8;                                                     # Power factor
pfa_a = math.degrees(math.acos(math.radians(pf_a)));                                            # Power factor angle in deg
Iq_a = I*math.cos(math.radians(pfa_a+pa_a));                                      # Current in Amphere
Id_a = I*math.sin(math.radians(pfa_a+pa_a));                                      # Current in Amphere
pr_a = ((Eo_a-v)/v)*100;                                        # Percentage regulation

# For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)

pf_b = 1.0;                                                     # Power factor
pfa_b= math.degrees(math.acos(math.radians(pf_b)));                                            # Power factor angle in deg
pr_b = ((Eo_b-v)/v)*100;                                        # Percentage regulation

# For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)

pf_c = 0.8;                                                     # Power factor
pfa_c = math.degrees(math.acos(math.radians(pf_c)));                                            # Power factor angle in deg
pr_c = ((Eo_c-v)/v)*100;                                        # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.13: SOLUTION :-");
print " For Case a) 0.80 lagging Power factor   Induced EMF, EMF = %.2f V "%(Eo_a)
print " Power angle = %.3f degree "%(pa_a)
print " Percenatge Regulation, R = %.1f Percenatge "%(pr_a)
print " For Case b) Unity Power factor   Induced EMF, EMF = %.2f V "%(Eo_b)
print " Power angle = %.2f degree "%(pa_b)
print " Percenatge Regulation, R = %.2f Percenatge "%(pr_b)
print " For Case c) 0.80 leading Power factor   Induced EMF, EMF = %.2f V "%(Eo_c)
print " Power angle = %.2f degree "%(pa_c)
print " Percenatge Regulation, R = %.2f Percenatge "%(pr_c)

EXAMPLE : 6.13: SOLUTION :-
For Case a) 0.80 lagging Power factor   Induced EMF, EMF = 7673.86 V
Power angle = 0.742 degree
Percenatge Regulation, R = 3222.9 Percenatge
For Case b) Unity Power factor   Induced EMF, EMF = 7673.84 V
Power angle = 0.93 degree
Percenatge Regulation, R = 3222.87 Percenatge
For Case c) 0.80 leading Power factor   Induced EMF, EMF = 7212.02 V
Power angle = -0.87 degree
Percenatge Regulation, R = 3022.90 Percenatge


## Example 6.14 Page No : 332¶

In [15]:
import math

# GIVEN DATA

v = 1.0;                       # Operating voltage of the Synchronous generator in pu
xd = 1.0;                      # Direct axis reactances in pu
xq = 0.5;                      # Quadrature axis reactances in pu
I = 1.0;                       # Rated current in pu

# CALCULATIONS

# For Case (a) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)

pf_a = 0.8;                                                     # Power factor
pfa_a = math.degrees(math.acos(math.radians(pf_a)));                                            # Power factor angle in deg
pr_a = ((Eo_a-v)/v)*100;                                             # Percentage regulation

# For Case (b) Unity Power factor (Refer figure 6.37 page no. 422)

pf_b = 1.0;                                                     # Power factor
pfa_b= math.degrees(math.acos(math.radians(pf_b)));                                            # Power factor angle in deg
pr_b = ((Eo_b-v)/v)*100;                                             # Percentage regulation

# For Case (c) 0.80 lagging Power factor (Refer figure 6.36 page no. 421)

pf_c = 0.8;                                                     # Power factor
pfa_c = math.degrees(math.acos(math.radians(pf_c)));                                            # Power factor angle in deg
pr_c = ((Eo_c-v)/v)*100;                                              # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.14: SOLUTION :-");
print " For Case a) 0.80 lagging Power factor   Induced EMF, EMF = %.4f V "%(Eo_a)
print " Power angle = %.1f degree "%(pa_a)
print " Percenatge Regulation, R = %.2f Percenatge "%(pr_a)
print " For Case b) Unity Power factor   Induced EMF, EMF = %.2f V "%(Eo_b)
print " Power angle = %.2f degree "%(pa_b)
print " Percenatge Regulation, R = %.2f Percenatge "%(pr_b)
print " For Case c) 0.80 leading Power factor   Induced EMF, EMF = %.4f V "%(Eo_c)
print " Power angle = %.2f degree "%(pa_c)
print " Percenatge Regulation, R = %.2f Percenatge "%(pr_c)

EXAMPLE : 6.14: SOLUTION :-
For Case a) 0.80 lagging Power factor   Induced EMF, EMF = 58.2957 V
Power angle = 0.8 degree
Percenatge Regulation, R = 5729.57 Percenatge
For Case b) Unity Power factor   Induced EMF, EMF = 58.30 V
Power angle = 0.97 degree
Percenatge Regulation, R = 5729.56 Percenatge
For Case c) 0.80 leading Power factor   Induced EMF, EMF = 56.2959 V
Power angle = -0.83 degree
Percenatge Regulation, R = 5529.59 Percenatge


## Example 6.15 Page No : 337¶

In [16]:
import math
from numpy import real,imag

# GIVEN DATA

If = 1.25;                     # Given that rated voltage at air gap line for this field current in pu
IF = 0.75;                     # Rated current in SC test for this field current in pu
Ia = 1.0;                      # Rated current in Per unit
pf = 0.8;                      # Power factor
V = 1.0;                       # Rated Volatge in pu

# CALCULATIONS

pfa = math.degrees(math.acos(math.radians(pf)));                                        # Power factor angle in deg
Voc = (V*IF)/If;                                        # Open circuit volatge in pu
xs = Voc/Ia;                                            # Syncronous reactance in pu
a = abs(E)*If;

# DISPLAY RESULTS

print ("EXAMPLE : 6.15: SOLUTION :-");
print " Induced EMF, E = %.2f < %.2f pu  "%(abs(E),math.degrees(math.atan2(E.imag,E.real)))
print " The field current required for %.2f pu voltage on air gap line %.1f pu "%(abs(E),a)

EXAMPLE : 6.15: SOLUTION :-
Induced EMF, E = 35.38 < 0.78 pu
The field current required for 35.38 pu voltage on air gap line 44.2 pu


## Example 6.16 Page No : 340¶

In [17]:
import math

# GIVEN DATA

V = 440;                       # Operating voltage of the alternator in Volts
VA = 20*10**3;                  # VA rating of the alternator in Volts-Amphere
f = 50;                        # Operating Frequency of the alternator in Hertz
N = 3000;                      # Rotation of the alternator in RPM
Ra = 0.0;                      # Armature reismath.tance in Ohms
xl = 0.6;                      # Armature reactances in Ohms
pf = 0.8;                      # Power factor lagging
pfa = math.degrees(math.acos(math.radians(pf)));                # ower factor angle in deg
p = (120*f)/N;                 # Number of poles
w = (2*math.pi*f);                 #  Rotation speed in Radians per second
v = V/math.sqrt(3);                 # Rated phase voltage in Volts
I = VA/(math.sqrt(3)*V);            # Rated curent in Amphere
If = I;                        # Given field current = armature current from SCC test in Amphere
E = 16*If;                     # Open-circuit EMF at field current in Volts given from Equation E = 16If refer page no. 431
xs = E/(If*math.sqrt(3));           # Synchronous reactance in Ohms
pa = math.degrees(math.atan(193.98/399.49));                                      # From above equation Eo
pr = ((Eo-v)/v)*100;                                            # Percent regulation
P = (3*v*Eo*math.degrees(math.sin(math.radians(pa))))/(xs*1000);                                # Power inKilo-Watts
T = (P*1000)/w;                                                 # Torque devolped in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.16: SOLUTION :-");
print " Induced EMF, EMF = %.f V "%(Eo)
print " Power angle = %.2f degree "%(pa)
print " Power, P = %.3f kW "%(P)
print " Counter Torque, T = %.2f N-m "%(T)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :-  a) Induced EMF, EMF = 471 V instead of %.f V "%(Eo)
print " b) Power angle = 18.05 degree instead of %.2f degree "%(pa)
print " c) Power, P = 12.003 kV instead of %.3f kW "%(P)
print " d) Counter Torque, T = 38.23 N-m instead of %.2f N-m "%(T)
print " From Calculation of the Induced EMFE), rest all the Calculated values in the TEXT BOOK is WRONG because\
of the Induced EMFE) value is WRONGLY calculated and the same used for the further Calculation part "

EXAMPLE : 6.16: SOLUTION :-
Induced EMF, EMF = 14144 V
Power angle = 25.90 degree
Power, P = 29202.939 kW
Counter Torque, T = 92955.84 N-m
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :-  a) Induced EMF, EMF = 471 V instead of 14144 V
b) Power angle = 18.05 degree instead of 25.90 degree
c) Power, P = 12.003 kV instead of 29202.939 kW
d) Counter Torque, T = 38.23 N-m instead of 92955.84 N-m
From Calculation of the Induced EMFE), rest all the Calculated values in the TEXT BOOK is WRONG because of the Induced EMFE) value is WRONGLY calculated and the same used for the further Calculation part


## Example 6.17 Page No : 342¶

In [18]:
import math

# GIVEN DATA
V = 440.;                       # Operating voltage of the Synchronous Motor in pu
E = 200.;                       # Induced voltage in Volts
xs = 8.0;                      # Synchronous reactance in Ohms
f = 50.;                        # Frequency in Hertz
pa = 36.;                       # Power angle in degree

# CALCULATIONS

v = V/math.sqrt(3);                 # Rated phase voltage in Volts
ws = 2*math.pi*f;                  # Synchronous speed in Radians per second
# To calculate the power factor angle refer page no 438 n figure 6.50
# Since E*math.cos(delta) < v so Power factor is lagging, let power factor angle be theta from ohasor diagram figure 6.50:- page no. 438
# v = E*math.cos(delta) + I*xs*math.sin(theta), I*math.sin(theta) = (254-0.809*200)/8 = 11.525
# Similarly, E*math.sin(delta) = I*xs*math.cos(theta), I*math.cos(theta) = (200*0.59)8 = 14.70
# From above two equations, math.tan(theta) = 0.784
theta = -38.1;                                  # Power factor angle in degree (minus sign because of lagging)
pf = math.degrees(math.cos(math.radians(theta)));                               # Power factor lagging
I = 14.7/math.degrees(math.cos(math.radians(theta)));                           # Line current in Amphere (I*math.cos(theta) = 14.7)
p = 3 * v * 14.7;                               # Input to motor in watts ( p = 3*V*I*math.cos(theta), I*math.cos(theta) = 14.7)
P = (3*E*v*math.degrees(math.sin(math.radians(pa))))/(xs*1000);                 # Power in Kilo-watts
T = (P*1000)/ws;                                # Torque in Newton-meter
# For Power factor unity
# let the current will be I2, thus 3*v*I2 = 3*v*I*math.cos(theta) , I2 = I*math.cos(theta) = 14.10 A
# let ecitation will be E2, thus v = E2*math.cos(delta2) and E2*math.sin(delta2) = I2*xs, E2*math.cos(delta2) = 254 and E2*math.sin(delta2) = 117.60, by solving these two equations we get E2 = math.sqrt(254**2+117.6**2) = 279.90 V and delta2 = math.degrees(math.atan(117.6/254) = 24.84 degree
E2 = 279.90;
delta2 = 24.84;
P_1 = (3*v*E2*math.degrees(math.sin(math.radians(delta2))))/(xs*1000);                     # Power in kilo-watts
T_1 = (P_1*1000)/ws;                                       # Torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.17: SOLUTION :-");
print " a) Line current, I = %.2f A "%(I)
print " b) Power factor angle = %.1f degree "%(theta)
print " c) Power , P = %.3f kW "%(P)
print " d) Torque , T = %.2f N-m "%(T)
print " e) Power factor = %.2f lagging "%(pf)
print " To make the Power factor to UNITY requirements are:-  a) Excitation EMF, E = %.2f V "%(E2)
print " b) Power angle = %.2f degree "%(delta2)
print " c) Power , P = %.3f kW "%(P_1)
print " d) Torque , T = %.2f N-m "%(T_1)

EXAMPLE : 6.17: SOLUTION :-
a) Line current, I = 0.33 A
b) Power factor angle = -38.1 degree
c) Power , P = 641.645 kW
d) Torque , T = 2042.42 N-m
e) Power factor = 45.09 lagging
To make the Power factor to UNITY requirements are:-  a) Excitation EMF, E = 279.90 V
b) Power angle = 24.84 degree
c) Power , P = 641.781 kW
d) Torque , T = 2042.85 N-m


## Example 6.18 Page No : 347¶

In [19]:
import math

# GIVEN DATA

v = 1100;                       # Operating voltage of the Synchronous Motor in Volts
p = 4;                         # Total number of Poles
m = 3;                         # number of phase
xs = 5.0;                      # Synchrouons reactances in Ohms
f = 50;                        # Frequency in Hertz
delta = 9;                     # Power angle in degree
p_hp = 150;                    # Motor delivering power in HP
eta = 89/100;                  # Efficiency of motor

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
ws = (4*math.pi*f)/p;                               # Synchronous speed in Radians per second
# We have (746*150)/0.89) = 125730.34 W = math.sqrt(3)*1100*I*math.cos(theta) refer page no. 440, thus we get I*math.cos(theta) = 12530.34/(1100*math.sqrt(3)) = 65.99 and E*math.sin(delta) = I*xs*math.cos(theta)
E = (xs*65.99)/math.degrees(math.sin(math.radians(delta)));                     # Exitation EMF per phase in Volts
# math.since E*math.cos(delta) > V, therefore the machine is over excited and power factor is leading, thus we get V = E*math.cos(delta) + I*xs*math.sin(theta), I*math.sin(theta) = (635.1-2109.2*math.cos(9)/5 = -289.586 and we have I*math.cos(theta) = 65.99 thus by solving these two equations we get theta = math.degrees(math.atan(-286.586/65.99) = 77.16 degre
theta = 77.16;                                 # Power factor angle in degree
I = 65.99/math.degrees(math.cos(math.radians(theta)));                         # Current in Amphere
P = (3*V*E*math.degrees(math.sin(math.radians(delta))))/(xs*1000);             # Power in kilo-Watts
T = (P*1000)/ws;                               # torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.18: SOLUTION :-");
print " a) Excitation EMF, E = %.1f V "%(E)
print " b) Line current, I = %.2f A "%(I)
print " c) Power factor = %.3f leading "%(pf)
print " d) Power , P = %.4f kW "%(P)
print " e) Torque , T = %.2f N-m "%(T)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :-  a) Power , P = 13.0667 kW instaed of %.4f kW "%(P)
print " b) Torque , T = 83.22 N-m instaed of %.2f N-m "%(T)
print " From Calculation of the PowerP), rest all the Calculated values in the TEXT BOOK is WRONG because of the\
PowerP) value is WRONGLY calculated and the same used for the further Calculation part "

EXAMPLE : 6.18: SOLUTION :-
a) Excitation EMF, E = 36.8 V
b) Line current, I = 5.18 A
c) Power factor = 12.733 leading
d) Power , P = 125.7278 kW
e) Torque , T = 800.41 N-m
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :-  a) Power , P = 13.0667 kW instaed of 125.7278 kW
b) Torque , T = 83.22 N-m instaed of 800.41 N-m
From Calculation of the PowerP), rest all the Calculated values in the TEXT BOOK is WRONG because of the PowerP) value is WRONGLY calculated and the same used for the further Calculation part


## Example 6.19 Page No : 351¶

In [20]:
import math

# GIVEN DATA

v = 440;                       # Operating voltage of the Synchronous Motor in Volts
p = 6;                         # Total number of Poles
m = 3;                         # Number of phase
xs = 5;                        # Synchrouons reactances per phase in Ohms
f = 50;                        # Frequency in Hertz
p_hp = 10;                     # Motor delivering power in HP
loss = 1000;                   # Total iron,copper and friction losses in Watts
pf = 0.8;                      # Power factor lagging
I = 10;                        # Motor drawing current in Amphere at 0.8 PF lagging

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
ws = (4*math.pi*f)/p;                               # Synchronous speed in Radians per second
theta = math.degrees(math.acos(math.radians(pf)));                              # Power factor angle in degree
Po = p_hp*746;                                  # Output power in Watts
Pi = Po+loss;                                   # Input power in Watts
eta = (Po/Pi)*100;                              # Efficiency
# we have V = E*math.cos(delta) - I*xs*math.sin(theta), 254 = E*math.cos(delta) - 5*10*0.6, so E*math.cos(delta) = 254 + 30 = 284 and E*math.sin(delta) = I*xs*math.cos(theta) = 5*10*0.8 = 40 by solving these two equations we get delta = math.degrees(math.atan(40/284) = 8.01 degree
delta = 8.01;                                    # Power angle in degree
E = 40/math.degrees(math.sin(math.radians(delta)));                              # Induced EMF per phase in Volts
P = (3*V*E*math.degrees(math.sin(math.radians(delta))))/(xs*1000);               # Power in Kilo-watts
T = (P*1000)/ws;                                 # Torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.19: SOLUTION :-");
print " a) Efficiency, eta = %.2f Percent "%(eta)
print " b) Induced EMF, E = %.f V per phase and      Power Torque) angle = %.2f degree "%(E,delta)
print " c) Power , P = %.4f kW "%(P)
print " d) Torque , T = %.2f N-m "%(T)

EXAMPLE : 6.19: SOLUTION :-
a) Efficiency, eta = 0.00 Percent
b) Induced EMF, E = 5 V per phase and      Power Torque) angle = 8.01 degree
c) Power , P = 6.0968 kW
d) Torque , T = 58.22 N-m


## Example 6.20 Page No : 352¶

In [21]:
import math

# GIVEN DATA
v = 11*10**3;                   # Operating voltage of the Synchronous Motor in Volts
p = 4;                         # Total number of Poles
m = 3;                         # number of phase
xs = 7;                        # Synchrouons reactances per phase in Ohms
f = 50;                        # Frequency in Hertz
KVA = 1500;                    # KVA rating (whole)
kva = 500;                     # Each case KVA rating

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
ws = (4*math.pi*f)/p;                               # Synchronous speed in Radians per second
I = (math.sqrt(3)*kva)/v;                            # Phase Current in Amphere

# For Case (a) 0.8 pf lagging

pf_a = 0.8;                                     # Power factor lagging
pfa_a = math.degrees(math.acos(math.radians(pf_a)));                            # Power factor angle in degree
# we have E*math.cos(delta) = V - I*xs*math.sin(theta) = 6351-78.73*7*0.6 = 6020.334 and E*math.sin(delta) = I*xs*math.cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6036.46 V and delta = math.degrees(math.atan(440.888/6020.334) = 4.19 degree
E_a = 6036.46;                                 # Induced Voltage in Volts
delta_a = 4.19;                                # Power angle in degree
P_a = (3*V*E_a*math.degrees(math.sin(math.radians(delta_a))))/(xs*10**6);       # Power in Mega-Watts
T_a = (P_a*10**6)/ws;                           # Torque in Newton-meter

# For Case (b) 0.8 pf leading

pf_b = 0.8;                                     # Power factor lagging
pfa_b = math.degrees(math.acos(math.radians(pf_b)));                            # Power factor angle in degree
# we have E*math.cos(delta) = V + I*xs*math.sin(theta) = 6351+78.73*7*0.6 = 6681.666 and E*math.sin(delta) = I*xs*math.cos(theta) = 78.73*7*0.8 = 440.888 thus we get by sloving these two equatins E = 6696.2 V and delta = math.degrees(math.atan(440.888/6681.666) = 3.78 degree
E_b = 6696.2;                                  # Induced Voltage in Volts
delta_b = 3.78;                                # Power angle in degree
P_b = (3*V*E_b*math.degrees(math.sin(math.radians(delta_b))))/(xs*10**6);       # Power in Mega-Watts
T_b = (P_b*10**6)/ws;                           # Torque in Newton-meter

# For Case (c) UPf

pf_c = 1.0;                                     # Power factor lagging
pfa_c = math.degrees(math.acos(math.radians(pf_c)));                            # Power factor angle in degree
# we have E*math.cos(delta) = V = 6351 and E*math.sin(delta) = I*xs = 78.73*7 = 551.11 thus we get by sloving these two equatins E = 6374.9 V and delta = math.degrees(math.atan(551.11/6351) = 4.96 degree
E_c = 6374.9;                                  # Induced Voltage in Volts
delta_c = 4.96;                                # Power angle in degree
P_c = (3*V*E_c*math.degrees(math.sin(math.radians(delta_c))))/(xs*10**6);       # Power in Mega-Watts
T_c = (P_c*10**6)/ws;                           # Torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.20: SOLUTION :-");
print " For Case a) 0.80 pf lagging :-  a) Induced EMF, E = %.2f V "%(E_a)
print " b) Power , P = %.1f MW "%(P_a)
print " c) Torque , T = %.2f N-m "%(T_a)
print " For Case b) 0.80 pf leading :-  a) Induced EMF, E = %.1f V "%(E_b)
print " b) Power , P = %.3f MW "%(P_b)
print " c) Torque , T = %.2f N-m "%(T_b)
print " For Case a) UPf :-  a) Induced EMF, E = %.1f V "%(E_c)
print " b) Power , P = %.2f MW "%(P_c)
print " c) Torque , T = %.f N-m "%(T_c)

EXAMPLE : 6.20: SOLUTION :-
For Case a) 0.80 pf lagging :-  a) Induced EMF, E = 6036.46 V
b) Power , P = 68.8 MW
c) Torque , T = 437869.39 N-m
For Case b) 0.80 pf leading :-  a) Induced EMF, E = 6696.2 V
b) Power , P = 68.843 MW
c) Torque , T = 438268.77 N-m
For Case a) UPf :-  a) Induced EMF, E = 6374.9 V
b) Power , P = 85.95 MW
c) Torque , T = 547202 N-m


## Example 6.21 Page No : 355¶

In [22]:
import math

# GIVEN DATA

v = 440;                       # Operating voltage of the Synchronous Motor in Volts
f = 50;                        # Operating Frequency of the Synchronous Motor in Hertz
xd = 10;                       # Direct axis reactances in Ohms
xq = 7.0;                      # Quadrature axis reactances in Ohms
p = 6;                         # Total number of Poles
pf = 0.8;                      # Power factor lagging
i = 10;                        # Motor drawing current in Amphere

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
ws = (4*math.pi*f)/p;                               # Synchronous speed in Radians per second
theta = math.degrees(math.acos(math.radians(pf)));                              # Power factor angle in degree
delta = math.degrees(math.atan( (i*xq*math.degrees(math.cos(math.radians(theta)))) / (V+i*xq*math.degrees(math.sin(math.radians(theta)))) ));             # Power angle for motoring mode in degree
# delta = atand((i*xq*cosd(theta))/(V+i*xq*sind(theta)));             // Power angle for motoring mode in degree
Iq = i*math.cos(math.radians(theta+delta));                                           # Current in Amphere
Id = i*math.sin(math.radians(theta+delta));                                           # Current in Amphere
Eo = V*math.degrees(math.cos(math.radians(delta))) + Id*xd;                                         # Induced EMF in Volts

# when the machine is running as alternator, the magnitude of induced EMF = 323.38V. Let the new current will be Inew at lagging power factor thetanew. Now torque angle is 10.71 deg from phasor diagram Figure 6.51 and page no. 444 we get V+Id*xd*math.cos(delta)-Iq*xq*math.sin(delta) = Eo*math.cos(delta), 254+9.825*Id-1.3Iq = 317.75, 9.825*Id-1.3*Iq = 63.75, 7.56*Id-Iq = 49 and we have Id*xd*math.sin(delta)+Iq*xq*math.cos(math.radianselta) = Eo*math.sin(delta), 1.85*Id+6.88*Iq = 60.1, 0.27*Id+Iq = 8.74 by solving these two equations we get Idnew = 123.85/10.095 = 12.27A and Iqnew = 5.43A
Iqnew = 5.43;                                   # New current in Amphere
Idnew = 12.27;                                  # New current in Amphere
Inew = math.sqrt(Idnew**2 + Iqnew**2);                 # New total Current in Amphere
# We know that torque angle, math.tan(delta) = (I*xd*math.cos(theta))/(V+I*Xq*math.sin(theta)) so by calutaion for new power factor angle thetanew we get, math.tan(10.17) = (13.42*7*math.cos(thetanew))/(254+13.42*7*math.sin(thetanew)), 0.189(254+13.42*7*math.sin(thetanew) = 13.42*7*math.cos(thetanew), 48-93.94math.cos(thetanew)+17.75*math.sin(thetanew) = 0 by solving this equatuon we gwt thetanew = 49.5 lagging
thetanew = 49.5;                                    # New power factor angle in degree
pfnew = math.degrees(math.cos(math.radians(thetanew)));                             # Power factor lagging

# DISPLAY RESULTS

print ("EXAMPLE : 6.21: SOLUTION :-");
print " a) Induced EMF, E = %.2f V "%(Eo)
print " b) Power Torque) angle = %.2f degree "%(delta)
print "     Power , P = %.2f W "%(P)
print "     Torque , T = %.2f N-m "%(T)
print " c) when the machine is running as alternator requirements are:-      New Current = %.2f A"%(Inew)
print "     Power factor = %.3f lagging "%(pfnew)

EXAMPLE : 6.21: SOLUTION :-
a) Induced EMF, E = 14653.83 V
b) Power Torque) angle = 0.75 degree
Power , P = 840567.62 W
Torque , T = 8026.83 N-m
c) when the machine is running as alternator requirements are:-      New Current = 13.42 A
Power factor = 37.211 lagging


## Example 6.22 Page No : 359¶

In [23]:
import math
from numpy import imag,real

# GIVEN DATA

E1 = 1100 + (1j*0);                         # EMFs of two identicel synchronous Generators in Volts per phase
Zl = 1.0 + (1j*1.0);                        # Load impedance in Ohms per phase
Zs1 = 0.15 + (1j*2.1);                      # Synchronous impedance in Ohms per phase
Zs2 = 0.2 + (1j*3.3);                       # Synchronous impedance in Ohms per phase
f = 50;                                     # Frequency in Hertz

# CALCULATONS

Ys1 = 1/Zs1;                                        # Synchronous Admitmath.tance in Ohms per phase
Ys2 = 1/Zs2;                                        # Synchronous Admitmath.tance in Ohms per ohase
V = ((E1*Ys1)+(E2*Ys2))/(Yl+Ys2+Ys1);               # Terminal Voltage in Volts per phase (From Millman's Theorem)
I1 = (E1-V)/Zs1;                                    # Individual current in Amphere per phase
I2 = (E2-V)/Zs2;                                    # Individual current in Amphere per phase
P1 = abs(V)*abs(I1)*math.cos(math.radians(math.degrees(math.atan2(V.imag,V.real)))-math.degrees(math.atan2(I1.imag,I1.real)));                                  # Per phase actice power in Watts
P2 = abs(V)*abs(I2)*math.cos(math.radians(math.degrees(math.atan2(V.imag,V.real)))-math.degrees(math.atan2(I2.imag,I2.real)));                                  # Per phase actice power in Watts
Ic = (E2-E1)/(Zs1+Zs2);                             # No-load circulating current in Amphere per phase

# DISPLAY RESULTS

print ("EXAMPLE : 6.22 : SOLUTION :-");
print " a) Terminal Voltage per phase, V = %.2f < %.1f V "%(abs(V),math.degrees(math.atan2(V.imag,V.real)))
print " b) Individual currents per phase, I1 = %.f < %.1f A      I2 = %.1f < %.1f A "%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)),abs(I2),math.degrees(math.atan2(I2.imag,I2.real)))
print " c) Per phase Active Power , P1 = %.f W      P2 = %.1f W  "%(P1,P2)
print " d) No-load current per phase, Ic = %.2f < %.2f A "%(abs(Ic),math.degrees(math.atan2(Ic.imag,Ic.real)))

EXAMPLE : 6.22 : SOLUTION :-
a) Terminal Voltage per phase, V = 14088.39 < -24.8 V
b) Individual currents per phase, I1 = 6221 < 67.3 A      I2 = 15122.8 < -86.1 A
c) Per phase Active Power , P1 = 14868984 W      P2 = -148034989.2 W
d) No-load current per phase, Ic = 11444.43 < -91.38 A


## Example 6.23 Page No : 364¶

In [24]:
import math

# GIVEN DATA

p = 4;                          # Number of the poles in the Alternator
f = 50;                         # Frequency in Hertz
pkw = 500;                      # Alternator delivering load in kilo-watts
pkwinc = 1000;                  # Generator increases its share of the common elictrical in kilo-watts
Kj = 1.5;                       # Inertia acceleration coefficient for the combined prime mover-alternator in N-m/elec deg/second square
Kd = 12;                        # Damping torque coefficient in N-m/elec deg/second
delta1 = 9;                     # Initial value of the Power angle in degree

# CALCULATIONS

delta2 = (pkwinc/pkw)*delta1;                               # Final value (maximum value) of the Power angle in degree (considering Linear variation)
ws = (4*math.pi*f)/p;                                           # Rotational speed in Radians per second
Ts = (pkw*1000)/ws;                                         # Synchornizing torque at 500kW in N-m
Ks = Ts/delta1;                                             # Synchornizing torque cofficient at 500kW in N-m/elec-deg
# Laplace transform of the swing Equation can be written as :- s**2 + ((Kd/Kj)*s) + (Ks/Kj) = 0, s**2 + (12/1.5)s + (353.86/1.5) = 0 and compring with the smath.degrees(math.atanard equation s**2 + s(2*zeta*Wn) + Wn**2 = 0 we get:- mentined below  (refer page no. 454 and 455)
Wn = math.sqrt(Ks/Kj);                                           # Natural frequency of oscillations in Radians per second
fn = Wn/(2*math.pi);                                            # Frequency of natural oscillations in Hertz
zeta = (1*Kd)/(2*Wn*Kj);                                    # Damping ratio
Wd = Wn*(math.sqrt(1-zeta**2));                                   # Frequency of damped oscillations in radians/s
fd = Wd/(2*math.pi);                                            # Frequency of damped oscillations in Hertz
ts = 5/(zeta*Wn);                                           # Settling time in second
deltamax = delta1 + 1.42*(delta2-delta1);                   # The maximum overshoot for damping ratio of 0.2604 is about 42% the maximum appoximate value of the overshoot in terms of 1% tolearance band in Electrical degree

# DISPLAY RESULTS

print ("EXAMPLE : 6.23: SOLUTION :-");
print " a.1) Final value maximum value) of the Power angle considering Linear variation), delta2 = %.f degree "%(delta2)
print " a.2) Natural frequency of oscillations, Ns = %.2f radians/s "%(Wn)
print " a.3) Damping ratio, zeta = %.4f "%(zeta)
print " a.4) Frequency of damped oscillations, Wd = %.2f radians/s "%(Wd)
print " a.5) Settling time, ts = %.2f seconds "%(ts)
print "  b)  The maximum overshoot for damping ratio of 0.2604 is about 42 percent the maximum appoximate value of the overshoot in terms of 1 percent tolearance band is, deltamax = %.2f degree "%(deltamax)
print " FOR CASE C CANNOT BE DO IT IN THIS BECAUSE AS IT REQUIRES MATLAB SIMULINK "

EXAMPLE : 6.23: SOLUTION :-
a.1) Final value maximum value) of the Power angle considering Linear variation), delta2 = 18 degree
a.2) Natural frequency of oscillations, Ns = 15.36 radians/s
a.3) Damping ratio, zeta = 0.2605
a.4) Frequency of damped oscillations, Wd = 14.83 radians/s
a.5) Settling time, ts = 1.25 seconds
b)  The maximum overshoot for damping ratio of 0.2604 is about 42 percent the maximum appoximate value of the overshoot in terms of 1 percent tolearance band is, deltamax = 21.78 degree
FOR CASE C CANNOT BE DO IT IN THIS BECAUSE AS IT REQUIRES MATLAB SIMULINK


## Example 6.25 Page No : 365¶

In [25]:
import math

# GIVEN DATA

v = 440;                       # Operating voltage of the Synchronous generator in Volts
f = 50;                        # Operating Frequency of the Synchronous generator in Hertz
m = 3;                         # Total number of Phase
pf = 0.8;                      # Power factor lagging
Il = 100;                      # Motor drawing current in Amphere
xs = 2;                        # Synchronous reactances in Ohms
delta = 20;                    # Power angle in degree
P = 50*10**3;                   # Total Power developed by the motor in Watts
Ppp = (50*10**3)/3;             # Power developed by the motor per phase in Watts

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
Eo = (Ppp*xs)/(3*V*math.degrees(math.sin(math.radians(delta))));                # Per phase Induced voltage in Volts
# Let us assume thetam is Power factor angle in degree and Im is the Motor current now, from phasor diagram figure 6.67 page no. 465 we get, Eo*math.degrees(math.cos(math.radians(delta)) = V+Im*xs*math.degrees(math.sin(math.radians(thetam)), Im*math.degrees(math.sin(math.radians(thetam)) = ((383.84*math.degrees(math.cos(math.radians(20)))-254.03)/2 = 53.35 and Im*xs*math.degrees(math.cos(math.radians(thetam)) = Eo*math.sin(delta), Im*math.degrees(math.cos(math.radians(theta)) = ((383.84*math.degrees(math.sin(math.radians(20)))/2 = 65.60 by sloving these two equations we get Im = math.sqrt(65.60**2 + 53.35**2) = 84.56 A and thetam = math.degrees(math.atan(53.35/65.60) = 39.13 degree
Im = math.sqrt(65.60**2 + 53.35**2);                           # Motor current in Amphere
thetam = math.degrees(math.atan(53.35/65.60));                            # Power factor angle in degree
kVA = (math.sqrt(3)*V*Im*math.degrees(math.sin(math.radians(thetam))))/1000;                 # Rective kVA of the motor in kVAR
ovpf = math.degrees(math.cos(math.radians(thetaR)));                                    # Overall Power factor lagging

# DISPLAY RESULTS

print ("EXAMPLE : 6.25: SOLUTION :-");
print " a) Rective kVA of the motor = %.3f kVAR "%(kVA)
print " b) Overall Power factor of the load and motor = %.4f lagging and "%(ovpf)
print "     resultant magnitude) current = %.2f A "%(IR)

EXAMPLE : 6.25: SOLUTION :-
a) Rective kVA of the motor = 1344.961 kVAR
b) Overall Power factor of the load and motor = 47.0232 lagging and
resultant magnitude) current = 4677.18 A


## Example 6.27 Page No : 367¶

In [26]:
import math
from numpy import real,imag

# GIVEN DATA

v = 440;                       # Operating voltage of the Synchronous generator in Volts
f = 50;                        # Operating Frequency of the Synchronous generator in Hertz
m = 3;                         # Total number of Phase
xs = 5;                        # Synchronous reactances in Ohms
Eo = 500;                      # Indduced Voltage in Volts per phase
R1 = 0.1;                      # Circuit Parameter in Ohms
R2 = 0.1;                      # Circuit Parameter in Ohms
X1 = 1.55;                     # Circuit Parameter in Ohms
X2 = 1.55;                     # Circuit Parameter in Ohms
s = 0.03;                      # Slip
P = 30*10**3;                   # Total Power developed by the motor in Watts

# CALCULATIONS

V = v/math.sqrt(3);                                  # Phase voltage in Volts
Ii = V/math.sqrt((R1+R2/s)**2 + (X1+X2)**2);           # Per phase induction motor current in Amphere
thetal = math.degrees(math.atan((X1+X2)/(R1+R2/s)));              # Power factor angle of the induction motor in degree
pf = math.degrees(math.cos(math.radians(thetal)));                              # Power factor of the induction motor lagging
# Let us assume thetam is leading Power factor angle in degree and Im is the synchronous Motor current now, from phasor diagram figure 6.70 page no. 469
delta = math.asin(math.radians((xs*P)/(3*V*Eo)));                 # Power angle in degree
# From phasor diagram figure 6.70 page no. 469 we have, Im*xs*math.cos(thetam) = Eo*math.sin(delta), Im*math.cos(delta) = ((500*math.sin(math.radians(23.18))/5 = 39.37 and Eo*math.degrees(math.cos(math.radians(delta)) = V+Im*xs*math.degrees(math.sin(math.radians(thetam)), Im*math.degrees(math.sin(math.radians(thetam)) = ((500*math.cos(math.radians(23.18))-254.03)/5 = 41.12 by sloving these two equations we get Im = math.sqrt(39.37**2 + 41.12**2) = 56.93 A and thetam = math.degrees(math.atan(41.12/39.37) = 46.25 degree
Im = math.sqrt(39.37**2 + 41.12**2);                           # Motor current in Amphere
thetam = math.degrees(math.atan(41.12/39.37));                            # Power factor angle in degree
kVA = (math.sqrt(3)*V*Im*math.degrees(math.sin(math.radians(thetam))))/1000;                 # Rective kVA of the motor in kVAR
II = Ii * ( 1j * (-thetal) * math.pi/180)**2;               # Induction Motor current in Amphere
Im = Im * ( 1j * thetam * math.pi/180)**2;                  # Synchronous Motor current in Amphere
It = II + Im;                                           # Total per phase current in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 6.27: SOLUTION :-");
print " a) Reactive kVA of the motor = %.3f kVAR "%(kVA)
print " b) Individual currents:-      Induction Motor current, II = %.2f + i%.2f) A      Synchronous Motor current,\
Im = %.2f + i%.2f) A "%(II.real,II.imag,Im.real,Im.imag)
print " c) resultant overall) current = %.2f < %.2f A "%(abs(It),math.degrees(math.atan2(It.imag,It.real)))
print " d) Overall Power factor = %.4f leading "%(ovpf)

EXAMPLE : 6.27: SOLUTION :-
a) Reactive kVA of the motor = 1036.641 kVAR
b) Individual currents:-      Induction Motor current, II = -29.62 + i-0.00) A      Synchronous Motor current, Im = -37.09 + i0.00) A
c) resultant overall) current = 66.71 < 180.00 A
d) Overall Power factor = -57.2958 leading


## Example 6.28 Page No : 371¶

In [27]:
import math

# GIVEN DATA

V = 400;                       # Operating voltage of the Synchronous generator in Volts
f = 50;                        # Operating Frequency of the Synchronous generator in Hertz
xd = 12;                      # Direct axis reactances in Ohms
xq = 5;                       # Quadrature axis reactances in Ohms
delta = 15;                   # Power(Torque) angle in degree
p = 2;                        # Number of the poles
m = 3;                        # Number of the phase

# CALCULATIONS

v = V/math.sqrt(3);                                          # Rated Phase Votage in Volts
Ns = (120*f)/p;                                         # Operating speed in RPM
Ws = (2*math.pi*f)/(p/2);                                   # Synchronous speed in radians/s
T = (3*v**2*math.sin(math.radians(2*delta))/(2*Ws))*((1/xq)-(1/xd));       # Developed Torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.28: SOLUTION :-");
print " a) Operating speed, Ns = %.f RPM "%(Ns)
print " b) Developed Torque , T = %.2f N-m "%(T)

EXAMPLE : 6.28: SOLUTION :-
a) Operating speed, Ns = 3000 RPM
b) Developed Torque , T = 0.00 N-m


## Example 6.29 Page No : 373¶

In [28]:
import math

# GIVEN DATA

f = 400;                             # Operating Frequency of the Synchronous generator in Hertz
Ld = 50*10**-3;                       # Direct axis reactances in Henry
Lq = 15*10**-3;                       # Quadrature axis reactances in Henry
delta = 15;                          # Power(Torque) angle in degree
p = 2;                               # Number of the poles
m = 3;                               # Number of the phase
I = 10;                              # Operating current in Amphere

# CALCULATIONS

Ns = (120*f)/p;                                               # Operating speed in RPM
Ws = (2*math.pi*f)/(p/2);                                         # Synchronous speed in radians/s
xd = 2*math.pi*f*Ld;                                              # Direct axis reactances in reactance
xq = 2*math.pi*f*Lq;                                              # Quadrature axis reactances in reactance
E1 = 0;                                                       # Induced EMF in Volts (Its ZERO beacuse when field winding current is zero)
v = xq*I;                                                    # Applied voltage in Volts
T = (3*v**2*math.sin(math.radians(2*delta))/(2*Ws))*((1/xq)-(1/xd));             # Developed Torque in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.29: SOLUTION :-");
print " a) Operating speed, Ns = %.f RPM "%(Ns)
print " b) Developed Torque , T = %.5f N-m "%(T)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) xd = 12.56 instead of %.2f Ohms  "%(xd);
print "                                     b) xq = 3.768 instead of %.2f Ohms  "%(xq);
print "                                     c) v = 36.68 instead of %.2f V  "%(v);
print "                                     d) T = 0.07875 instead of %.4f N-m  "%(T);
print " From Calculation of the d-axis and q-axis reactance xd and xq respectively),\
rest all the Calculated values in the TEXT BOOK is WRONG because of the d-axis and q-axis reactance xd and\
xq respectively) value is WRONGLY calculated and the same used for the further Calculation part "

EXAMPLE : 6.29: SOLUTION :-
a) Operating speed, Ns = 24000 RPM
b) Developed Torque , T = 0.78750 N-m
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) xd = 12.56 instead of 125.66 Ohms
b) xq = 3.768 instead of 37.70 Ohms
c) v = 36.68 instead of 376.99 V
d) T = 0.07875 instead of 0.7875 N-m
From Calculation of the d-axis and q-axis reactance xd and xq respectively), rest all the Calculated values in the TEXT BOOK is WRONG because of the d-axis and q-axis reactance xd and  xq respectively) value is WRONGLY calculated and the same used for the further Calculation part


## Example 6.30 Page No : 375¶

In [29]:
import math

# GIVEN DATA

f = 50;                       # Operating Frequency of the Synchronous generator in Hertz
p = 2;                        # Number of the poles
Pt = 800;                     # Total loss in Watts
Pr = 10;                      # Rotational loss in Watts

# CALCULATIONS

Ws = (4*math.pi*f)/p;                                     # Synchronous speed in radians/s
Ph = Pt-Pr;                                           # Hysteresis loss refered to stator in Watts
Th = Ph/Ws;                                           # Torque at the shaft in Newton-meter

# DISPLAY RESULTS

print ("EXAMPLE : 6.30: SOLUTION :-");
print " a) Power at the shaft, Ph = %.f W "%(Ph)
print " )b) Torque at the shaft , Th = %.2f N-m "%(Th)

EXAMPLE : 6.30: SOLUTION :-
a) Power at the shaft, Ph = 790 W
)b) Torque at the shaft , Th = 2.51 N-m


## Example 6.31 Page No : 379¶

In [30]:
import math

# GIVEN DATA

Pi = 2*10**6;                                # Power input in Volt-Amphere
v = 6.6*10**3;                               # Operating voltage in Volts

# CALCULATIONS

I = Pi/(v*math.sqrt(3));                         # Rated current in Amphere
V = v/math.sqrt(3);                              # Phase voltage in Volts
xs = v/(I*math.sqrt(3));                         # Synchronous reactance in Ohms

# For case (a) 0.8 pf lagging

pf_a = 0.8;                                 # Power factor
pfa_a = math.degrees(math.acos(math.radians(pf_a)));                        # Power factor angle in degree
E_a = math.sqrt(a_a**2 + b_a**2);                  # Induced EMF in Volts
delta_a = math.degrees(math.atan(b_a/a_a));                   # Torque (power) angle in degree
P_a = (3*V*E_a*math.degrees(math.sin(math.radians(delta_a))))/(xs*10**6);    # Power developed in MVA

# For case (b) 0.8 pf leading

pf_b = 0.8;                                 # Power factor
pfa_b = math.degrees(math.acos(math.radians(pf_b)));                        # Power factor angle in degree
E_b = math.sqrt(a_b**2 + b_b**2);                  # Induced EMF in Volts
delta_b = math.degrees(math.atan(b_b/a_b));                   # Torque (power) angle in degree
P_b = (3*V*E_b*math.degrees(math.sin(math.radians(delta_b))))/(xs*10**6);    # Power developed in MVA

# For case (c) UPF

pf_c = 1.0;                                 # Power factor
pfa_c = math.degrees(math.acos(math.radians(pf_c)));                        # Power factor angle in degree
a_c = V;
b_c = I*xs;
E_c = math.sqrt(a_c**2 + b_c**2);                  # Induced EMF in Volts
delta_c = math.degrees(math.atan(b_c/a_c));                   # Torque (power) angle in degree
P_c = (3*V*E_c*math.degrees(math.sin(math.radians(delta_c))))/(xs*10**6);    # Power developed in MVA

print ("EXAMPLE : 6.31: SOLUTION :-");
print " For Case a) 0.80 lagging Power factor   Induced EMF, EMF = %.2f V "%(E_a)
print " Power Torque) angle = %.2f degree "%(delta_a)
print " Power developed, P = %.1f MVA "%(P_a)
print " For Case b) 0.80 leading Power factor   Induced EMF, EMF = %.f V "%(E_b)
print " Power Torque) angle = %.2f degree "%(delta_b)
print " Power developed, P = %.3f MVA "%(P_b)
print " For Case c) Unity Power Factor   Induced EMF, EMF = %.1f V "%(E_c)
print " Power Torque) angle = %.2f degree "%(delta_c)
print " Power developed, P = %.1f MVA "%(P_c)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :-  xs = 20.14 instead of %.2f Ohms  "%(xs);
print "  For Case a) 0.80 lagging Pf a.1)  E = 6561.42 instead of %.2f V  "%(E_a);
print "                               a.2)  delta = 25.45 instead of %.2f degree  "%(delta_a);
print "  For Case b) 0.80 leading Pf b.1)  E = 3290 instead of %.1f V  "%(E_b);
print "                               b.2)  delta = 58.98 instead of %.2f degree  "%(delta_b);
print "                               b.3)  Power developed = 1.617 instead of %.3f MVA  "%(P_b);
print "  For Case c) UPF  c.1)  E = 5190.2 instead of %.2f V  "%(E_c);
print "                    c.2)  delta = 42.77 instead of %.2f degree  "%(delta_c);
print " In all the three cases from Calculation of the Synchronous reactance xs), \
rest all the Calculated values in the TEXT BOOK is WRONG because of the Synchronous reactance xs) value is WRONGLY calculated and the same used for the further Calculation part "

EXAMPLE : 6.31: SOLUTION :-
For Case a) 0.80 lagging Power factor   Induced EMF, EMF = 222136.39 V
Power Torque) angle = 0.79 degree
Power developed, P = 91.7 MVA
For Case b) 0.80 leading Power factor   Induced EMF, EMF = 214516 V
Power Torque) angle = -0.81 degree
Power developed, P = -91.673 MVA
For Case c) Unity Power Factor   Induced EMF, EMF = 5388.9 V
Power Torque) angle = 45.00 degree
Power developed, P = 114.6 MVA
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :-  xs = 20.14 instead of 21.78 Ohms
For Case a) 0.80 lagging Pf a.1)  E = 6561.42 instead of 222136.39 V
a.2)  delta = 25.45 instead of 0.79 degree
For Case b) 0.80 leading Pf b.1)  E = 3290 instead of 214516.1 V
b.2)  delta = 58.98 instead of -0.81 degree
b.3)  Power developed = 1.617 instead of -91.673 MVA
For Case c) UPF  c.1)  E = 5190.2 instead of 5388.88 V
c.2)  delta = 42.77 instead of 45.00 degree
In all the three cases from Calculation of the Synchronous reactance xs),  rest all the Calculated values in the TEXT BOOK is WRONG because of the Synchronous reactance xs) value is WRONGLY calculated and the same used for the further Calculation part


## Example 6.32 Page No : 382¶

In [31]:
import math

# GIVEN DATA
# Refer phasor diagram figure 6.76 and page no. 476

pf = 0.8;                                               # Power factor lagging
pa = math.degrees(math.acos(math.radians(pf)));                                         # Power factor angle in degree
v = 1.0 * (1j * pa * math.pi/180)**2;                      # Operating voltage of the alternator in pu
xd = 0.8;                                               # Direct axis reactances in pu
xq = 0.4;                                               # Quadrature axis reactances in pu
I = 1.0;                                                # Current in pu taking this as reference

# CALCULATIONS

A = v + (1j*xq*I);
delta = math.degrees(math.atan2(A.imag,A.real))-pa;                      # Power angle in degree
Iq = I * math.degrees(math.cos(math.radians(math.degrees(math.atan2(A.imag,A.real)))));                  # d-axis currents in Amphere

Id = I * math.degrees(math.sin(math.radians(math.degrees(math.atan2(A.imag,A.real)))));                  # q-axis currents in Amphere
E = abs(v)*math.degrees(math.cos(math.radians(delta))) + Id*xd;                         # Induced EMF per phase in Per unit
pr = ((abs(E)-abs(v))/abs(v))*100;                      # Percentage regulation

# DISPLAY RESULTS

print ("EXAMPLE : 6.32: SOLUTION :-");
print " a) Induced EMF per phase, E = %.4f < %.2f pu "%(E,delta)
print " b) Power angle = %.2f degree "%(delta)
print " C) Percenatge Regulation, R = %.2f Percent "%(pr)
print " IN THIS PROBLEM PERCENTAGE REGULATION IS NOT CALCULATED IN THE TEXT BOOK"

EXAMPLE : 6.32: SOLUTION :-
a) Induced EMF per phase, E = 28.1608 < 81.43 pu
b) Power angle = 81.43 degree
C) Percenatge Regulation, R = 1061.88 Percent
IN THIS PROBLEM PERCENTAGE REGULATION IS NOT CALCULATED IN THE TEXT BOOK


## Example 6.33 Page No : 386¶

In [32]:
import math

# GIVEN DATA

v = 6.6*10**3;                  # Operating voltage of the Synchronous motor in Volts
P = 5*10**6;                    # Operating power of the Synchronous motor in Watts
pf = 1.0;                      # Power factor
xd = 3.0;                      # Direct axis reactances in Ohms
xq = 1.0;                      # Quadrature axis reactances in Ohms
eta = 0.98;                    # OPerating efficiency

# CALCULATIONS

V = v/math.sqrt(3);                                  # Per phase voltage in Volts
I = P/(eta*v*math.sqrt(3));                         # Line current in Amphere
delta = math.degrees(math.atan((xq*I)/v));                       # power angle in degree
# Actual electromagnetic torque in N-m
Ratio = Tmax/T;                                # Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque

# DISPLAY RESULTS

print ("EXAMPLE : 6.33: SOLUTION :-");
print " a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is %.2f "%(Ratio)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) delta = 2.41 instead of %.2f degree  "%(delta);
print "                                     b) E = 6379 instead of %.2f V  "%(E);
print "                                     c) Ratio = 10.84 instead of %.2f  "%(Ratio);
print " From Calculation of the Power angle delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle delta) value is WRONGLY calculated and the same used for the further Calculation part "

EXAMPLE : 6.33: SOLUTION :-
a) Ratio of the Maximum electromagnetic torque to the actual electromagnetic torque is 14.82
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) delta = 2.41 instead of 3.87 degree
b) E = 6379 instead of 1978904207.10 V
c) Ratio = 10.84 instead of 14.82
From Calculation of the Power angle delta), rest all the Calculated values in the TEXT BOOK is WRONG because of the Power angle delta) value is WRONGLY calculated and the same used for the further Calculation part