# CHAPTER 7: PARALLEL OPERATION¶

## Example 7.1, Page number 182¶

In :
#Variable declaration
R_sh = 120.0   #Shunt field resistance(ohm)
R_a = 0.1      #Armature resistance(ohm)
V_L = 120.0    #Line voltage(V)
E_g1 = 125.0   #Generated voltage by dynamo A(V)
E_g2 = 120.0   #Generated voltage by dynamo B(V)
E_g3 = 114.0   #Generated voltage by dynamo C(V)

#Calculation
#Case(a) #1
I_gA = (E_g1-V_L)/R_a  #Current in the generating dynamo A(A)
I_f = V_L/R_sh         #Shunt field current(A)
I_a1 = I_gA+I_f        #Armature current for dynamo A(A)
I_L1 = I_gA            #Current delivered by dynamo A to the bus(A)
#2
I_gB = (E_g2-V_L)/R_a #Current in the generating dynamo B(A)
I_a2 = I_gB+I_f       #Armature current for dynamo B(A)
I_L2 = I_gB           #Current delivered by dynamo B to the bus(A)
#3
I_gC = (V_L-E_g3)/R_a #Current in the generating dynamo C(A)
I_a3 = I_gC           #Armature current for dynamo C(A)
I_L3 = I_gC+I_f       #Current received by dynamo C from the bus(A)
#Case(b) #1
P_LA = V_L*I_L1       #Power delivered to the bus by dynamo A(W)
P_gA = E_g1*I_a1      #Power generated by dynamo A(W)
#2
P_LB = V_L*I_L2       #Power delivered to the bus by dynamo B(W)
P_gB = E_g2*I_a2      #Power generated by dynamo B(W)
#3
P_LC = V_L*I_L3       #Power delivered by the bus to dynamo C(W)
P_gC = E_g3*I_a3      #Power generated by dynamo C(W)

#Result
print('Case(a) 1: Line current delivered by dynamo A to the bus , I_LA = %.f A' %I_L1)
print('           Armature current , I_a = %.f A' %I_a1)
print('        2: Line current delivered by dynamo B to the bus , I_LB = %.f A. Thus dynamo B is floating' %I_L2)
print('           Armature current , I_a = %.f A' %I_a2)
print('        3: Line current received by dynamo C from the bus , I_LC = %.f A' %I_L3)
print('           Armature current , I_a = %.f A' %I_a3)
print('\nCase(b) 1: Power delivered to the bus by dynamo A , P_LA = %.f W' %P_LA)
print('           Power generated by dynamo A , P_gA = %.f W' %P_gA)
print('        2: Dynamo B neither receives or delivers power , P_LB = %.f W' %P_LB)
print('           Power generated by dynamo B to excite its field , P_gB = %.f W' %P_gB)
print('        3: Power delivered by the bus to dynamo C , P_LC = %.f W' %P_LC)
print('           Internal power delivered in the direction of rotation of its prime mover , P_gC = %.f W' %P_gC)

Case(a) 1: Line current delivered by dynamo A to the bus , I_LA = 50 A
Armature current , I_a = 51 A
2: Line current delivered by dynamo B to the bus , I_LB = 0 A. Thus dynamo B is floating
Armature current , I_a = 1 A
3: Line current received by dynamo C from the bus , I_LC = 61 A
Armature current , I_a = 60 A

Case(b) 1: Power delivered to the bus by dynamo A , P_LA = 6000 W
Power generated by dynamo A , P_gA = 6375 W
2: Dynamo B neither receives or delivers power , P_LB = 0 W
Power generated by dynamo B to excite its field , P_gB = 120 W
3: Power delivered by the bus to dynamo C , P_LC = 7320 W
Internal power delivered in the direction of rotation of its prime mover , P_gC = 6840 W


## Example 7.2, Page number 182¶

In :
#Variable declaration
R_a = 0.1       #Armature resistance(ohm)
R_f = 100.0     #Field circuit resistance(ohm)
V_L_b = 120.0   #Bus voltage(V)
V_L_a = 140.0   #Generated voltage of the generator(V)

#Calculation
#Case(a)
V_f = V_L_a                  #Voltage across the field(V)
I_f_a = V_f/R_f              #Field current(A)
I_a_a = I_f_a                #Armature current(A)
E_g_a = V_L_a+I_a_a*R_a      #Generated EMF(V)
P_g_a = E_g_a*I_a_a          #Generated power(W)
#Case(b)
I_a_b = (E_g_a-V_L_b)/R_a    #Armature current(A)
I_f_b = V_L_b/R_f            #Field current(A)
I_Lg = I_a_b-I_f_b           #Generated line current(A)
P_L = V_L_b*I_Lg             #Power generated across the lines(W)
E_g_b = V_L_a
P_g_b = E_g_b*I_a_b          #Generated power(W)

#Result
print('Case(a): Generated current before it is connected to the bus , I_a = %.1f A' %I_a_a)
print('         Generated power before it is connected to the bus , P_g = %.1f W' %P_g_a)
print('Case(b): Generated armature current after it is connected to the bus , I_a = %.1f A' %I_a_b)
print('         Generated line current after it is connected to the bus , I_Lg = %.1f A' %I_Lg)
print('         Generated power across the line after it is connected to the bus , P_g = %.f W' %P_L)
print('         Generated power after it is connected to the bus , P_g = %.f W' %P_g_b)

Case(a): Generated current before it is connected to the bus , I_a = 1.4 A
Generated power before it is connected to the bus , P_g = 196.2 W
Case(b): Generated armature current after it is connected to the bus , I_a = 201.4 A
Generated line current after it is connected to the bus , I_Lg = 200.2 A
Generated power across the line after it is connected to the bus , P_g = 24024 W
Generated power after it is connected to the bus , P_g = 28196 W


## Example 7.3, Page number 183¶

In :
#Variable declaration
R_a1 = 0.1      #Armature resistance of shunt generator 1(ohm)
R_a2 = 0.1      #Armature resistance of shunt generator 2(ohm)
R_a3 = 0.1      #Armature resistance of shunt generator 3(ohm)
R_L = 2.0       #Load resistance(ohm)
E_g1 = 127.0    #Voltage generated by shunt generator 1(V)
E_g2 = 120.0    #Voltage generated by shunt generator 2(V)
E_g3 = 119.0    #Voltage generated by shunt generator 3(V)

#Calculation
#Case(a)
V_L = ((E_g1/R_a1)+(E_g2/R_a2)+(E_g3/R_a3))/((1/R_a1)+(1/R_a2)+(1/R_a3)+(1/R_L))
#Case(b)
I_L1 = (E_g1-V_L)/R_a1   #Current delivered/received by generator 1(A)
I_L2 = (E_g2-V_L)/R_a2   #Current delivered/received by generator 2(A)
I_L3 = (E_g3-V_L)/R_a3   #Current delivered/received by generator 3(A)
I_L = -V_L/R_L           #Current received by 2 ohm load(A)
#Case(c)
I_a1 = I_L1              #Armature current for generator 1(A)
I_a2 = I_L2              #Armature current for generator 2(A)
I_a3 = I_L3              #Armature current for generator 3(A)
P_g1 = E_g1*I_a1         #Power generated by generator 1(W)
P_g2 = E_g2*I_a2         #Power generated by generator 2(W)
P_g3 = E_g3*I_a3         #Power generated by generator 3(W)
#Case(d)
P_L1 = V_L*I_L1          #Power delivered/received from generator 1(W)
P_L2 = V_L*I_L2          #Power delivered/received from generator 2(W)
P_L3 = V_L*I_L3          #Power delivered/received from generator 3(W)
P_L = V_L*I_L            #Power delivered/received from 2 ohm load(W)

#Result
print('Case(a): Terminal bus voltage , V_L = %.f V' %V_L)
print('Case(b): Current delivered to the bus by generator 1 , I_L1 = %.f A (to bus)' %I_L1)
print('         Current delivered to the bus by generator 2 , I_L2 = %.f A' %I_L2)
print('         Current received by the generator 3 from the bus , I_L3 = %.f A (from bus)' %I_L3)
print('         Current received from the bus by load , I_L3 = %.f A (from bus)' %I_L)
print('Case(c): Power generated by generator 1 , P_g1 = %.f W' %P_g1)
print('         Power generated by generator 2 , P_g2 = %.f W (floating)' %P_g2)
print('         Power generated by generator 3 , P_g3 = %.f W' %P_g3)
print('Case(d): Power delivered to the bus from generator 1 , P_L1 = %.f W' %P_L1)
print('         Power delivered to the bus from generator 2 , P_L2 = %.f W' %P_L2)
print('         Power received from the bus by generator 3 , P_L2 = %.f W' %P_L3)
print('         Power received from the bus by load , P_L = %.f W' %P_L)

Case(a): Terminal bus voltage , V_L = 120 V
Case(b): Current delivered to the bus by generator 1 , I_L1 = 70 A (to bus)
Current delivered to the bus by generator 2 , I_L2 = 0 A
Current received by the generator 3 from the bus , I_L3 = -10 A (from bus)
Current received from the bus by load , I_L3 = -60 A (from bus)
Case(c): Power generated by generator 1 , P_g1 = 8890 W
Power generated by generator 2 , P_g2 = 0 W (floating)
Power generated by generator 3 , P_g3 = -1190 W
Case(d): Power delivered to the bus from generator 1 , P_L1 = 8400 W
Power delivered to the bus from generator 2 , P_L2 = 0 W
Power received from the bus by generator 3 , P_L2 = -1200 W
Power received from the bus by load , P_L = -7200 W


## Example 7.4, Page number 184¶

In :
#Variable declaration
P1 = 300.0          #Power rating of generator 1(kW)
P2 = 600.0          #Power rating of generator 2(kW)
V = 220.0           #Voltage rating of generator 1 and 2(V)
V_o = 250.0         #No-load voltage applied to both the generators(V)
V_1 = 230.0         #Terminal voltage(V)
V_2 = 240.0         #Terminal voltage(V)

#Calculation
#Case(a)
kW1_a = (V_o-V_1)/(V_o-V)*P1   #Load carried by generator 1(kW)
kW2_a = (V_o-V_1)/(V_o-V)*P2   #Load carried by generator 2(kW)
#Case(b)
kW1_b = (V_o-V_2)/(V_o-V)*P1   #Load carried by generator 1(kW)
kW2_b = (V_o-V_2)/(V_o-V)*P2   #Load carried by generator 2(kW)
#Case(c)
frac_a = (V_o-V_1)/(V_o-V)     #Fraction of rated kW carried by each generator
frac_b = (V_o-V_2)/(V_o-V)     #Fraction of rated kW carried by each generator

#Result
print('Case(a): When the terminal voltage is 230 V Generator 1 carries = %.f kW' %kW1_a)
print('         When the terminal voltage is 230 V Generator 2 carries = %.f kW' %kW2_a)
print('Case(b): When the terminal voltage is 240 V Generator 1 carries = %.f kW' %kW1_b)
print('         When the terminal voltage is 240 V Generator 2 carries = %.f kW' %kW2_b)
print('Case(c): Both generators carry no-load at 250 V ; %.3f rated load at %d V ; %.3f rated load at %d V ; and rated load at %d V ;' %(frac_b,V_2,frac_a,V_1,V))

Case(a): When the terminal voltage is 230 V Generator 1 carries = 200 kW
When the terminal voltage is 230 V Generator 2 carries = 400 kW
Case(b): When the terminal voltage is 240 V Generator 1 carries = 100 kW
When the terminal voltage is 240 V Generator 2 carries = 200 kW
Case(c): Both generators carry no-load at 250 V ; 0.333 rated load at 240 V ; 0.667 rated load at 230 V ; and rated load at 220 V ;


## Example 7.5, Page number 191¶

In :
#Variable declaration
E_1 = 220.0    #Terminal voltage of alternator 1(V)
E_2 = 222.0    #Terminal voltage of alternator 2(V)
f_1 = 60.0     #Frequency of alternator 1(Hz)
f_2 = 59.5     #Frequency of alternator 2(Hz)

#Calculation
E_max = (E_1+E_2)/2    #Maximum effective voltage across each lamp(V)
E_min = (E_2-E_1)/2    #Minimum effective voltage across each lamp(V)
f = f_1-f_2            #Frequency of the voltage across the lamps(Hz)
E_peak = E_max/0.7071  #Peak value of the voltage across each lamp(V)
n = (1.0/2)*f_1        #Number of maximum light pulsations per minute

#Result
print('Case(a): Maximum effective voltage across each lamp , E_max/lamp = %.f V (rms)' %E_max)
print('         Minimum effective voltage across each lamp , E_min/lamp = %.f V' %E_min)
print('Case(b): Frequency of the voltage across the lamps , f = %.1f Hz' %f)
print('Case(c): Peak value of the voltage across each lamp , E_peak = %.f V' %E_peak)
print('Case(d): Number of maximum light pulsations per minute , n = %.f pulsations/min' %n)

Case(a): Maximum effective voltage across each lamp , E_max/lamp = 221 V (rms)
Minimum effective voltage across each lamp , E_min/lamp = 1 V
Case(b): Frequency of the voltage across the lamps , f = 0.5 Hz
Case(c): Peak value of the voltage across each lamp , E_peak = 313 V
Case(d): Number of maximum light pulsations per minute , n = 30 pulsations/min


## Example 7.6, Page number 191¶

In :
#Variable declaration
E_1 = 220.0   #Voltage generated by alternator 1(V)
E_2 = 220.0   #Voltage generated by alternator 2(V)
f_1 = 60.0    #Frequency of alternator 1(Hz)
f_2 = 58.0    #Frequency of alternator 2(Hz)

#Calculation
E_max = (E_1+E_2)/2    #Maximum effective voltage across each lamp(V)
f = f_1-f_2            #Frequency of the voltage across the lamp(Hz)
E_min = (E_2-E_1)/2    #Minimum effective voltage across each lamp(V)

#Result
print('Case(a): Maximum effective voltage across each lamp , E_max/lamp = %.f V' %E_max)
print('         Frequency of the voltage across each lamp , f = %.f Hz' %f)
print('Case(b): The voltages are equal and opposite in the local circuit')
print('Case(c): Minimum effective voltage across each lamp , E_min/lamp = %.f V' %E_min)
print('         Frequency of the voltage across each lamp , f = 0 Hz')
print('Case(d): The voltages are in phase in the local circuit')

Case(a): Maximum effective voltage across each lamp , E_max/lamp = 220 V
Frequency of the voltage across each lamp , f = 2 Hz
Case(b): The voltages are equal and opposite in the local circuit
Case(c): Minimum effective voltage across each lamp , E_min/lamp = 0 V
Frequency of the voltage across each lamp , f = 0 Hz
Case(d): The voltages are in phase in the local circuit


## Example 7.7, Page number 193¶

In :
import math
import cmath

#Variable declaration
E_1 = 220.0    #Terminal voltage of alternator 1(V) From Ex. 7-5
E_2 = 222.0    #Terminal voltage of alternator 2(V)
f_1 = 60.0     #Frequency of alternator 1(Hz)
f_2 = 59.5     #Frequency of alternator 2(Hz)
E1 = 220.0     #Voltage generated by alternator 1(V) From Ex. 7-6
E2 = 220.0     #Voltage generated by alternator 2(V)
f1 = 60.0      #Frequency of alternator 1(Hz)
f2 = 58.0      #Frequency of alternator 2(Hz)
R_a1 = 0.1     #Armature resistance of alternator 1(ohm)
R_a2 = 0.1     #Armature resistance of alternator 2(ohm)
X_a1 = 0.9     #Armature reactance of alternator 1(ohm)
X_a2 = 0.9     #Armature reactance of alternator 2(ohm)

#Calculation
Z_1 = complex(R_a1,X_a1)   #Effective impedance of alternator 1(ohm)
Z_2 = complex(R_a2,X_a2)   #Effective impedance of alternator 2(ohm)
#In Ex.7-5
E_r = E_2-E_1              #Effective voltage generated(V)
I_s = E_r/(Z_1+Z_2)        #Synchronizing current in the armature(A)
#In Ex.7-6
Er = E2 -E1                #Effective voltage generated(V)
Is = Er/(Z_1+Z_2)          #Synchronizing current in the armature(A)

#Result
print('In Ex.7-5 the synchronizing current in the armatures of both alternators , I_s = %.3f∠%.2f° A' %(abs(I_s),cmath.phase(I_s)*180/math.pi))
print('In Ex.7-6 the synchronizing current in the armatures of both alternators , I_s = %.f∠%.f° A' %(abs(Is),cmath.phase(Is)*180/math.pi))

In Ex.7-5 the synchronizing current in the armatures of both alternators , I_s = 1.104∠-83.66° A
In Ex.7-6 the synchronizing current in the armatures of both alternators , I_s = 0∠0° A


## Example 7.8, Page number 195¶

In :
import math
import cmath

#Variable declaration
E_gp1 = 200.0    #Terminal voltage of alternator 1(V)
E_gp2 = 220.0    #Terminal voltage of alternator 2(V)
R_a1 = 0.2       #Armature resistance of alternator 1(ohm)
R_a2 = 0.2       #Armature resistance of alternator 2(ohm)
X_a1 = 2.0       #Armature reactance of alternator 1(ohm)
X_a2 = 2.0       #Armature reactance of alternator 1(ohm)

#Calculation
#Case(a)
Z_p1 = complex(R_a1,X_a1)   #Effective impedance of alternator 1(ohm)
Z_p2 = complex(R_a2,X_a2)   #Effective impedance of alternator 2(ohm)
E_r = (E_gp2-E_gp1)         #Effective voltage generated(V)
I_s = E_r/(Z_p1+Z_p2)       #Synchronizing current in the armature(A)
Is = abs(I_s)               #Magnitude of Synchronizing current(A)
theta = cmath.phase(I_s)*180/math.pi         #Angle of Synchronizing current(degree)
P_2 = E_gp2*Is*math.cos(theta*math.pi/180)   #Generator action developed by alternator 2(W)
#Case(b)
P_1 = -E_gp1*Is*math.cos(theta*math.pi/180)  #Synchronizing power received by alternator 1(W)
#Case(c)
P1 = abs(P_1)                                #Magnitude of P1(W)
losses = P_2-P1                              #Power loss in both armatures(W)
check = E_r*Is*math.cos(theta*math.pi/180)   #Verifying losses(W) by Eq.7-7
double_check = Is**2*(R_a1+R_a2)             #Verifying losses(W) by Eq.7-7
#Case(d)
V_p2  = E_gp2-Is*abs(Z_p2)                   #Generator action(V)
V_p1  = E_gp1+Is*abs(Z_p1)                   #Motor action(V)

#Result
print('Case(a): Generator action developed by alternator 2 , P_2 = %.1f W' %P_2)
print('Case(b): Synchronizing power received by alternator 1 , P_1 = %.1f W' %P_1)
print('Case(c): Power loss in both armature , Losses = %.f W' %losses)
print('Case(d): Terminal voltage of alternator 2 , V_p2 = %.f V (generator action)' %V_p2)
print('         Terminal voltage of alternator 1 , V_p2 = %.f V (motor action)' %V_p1)
print('Case(e): Phasor diagram is shown in Fig 7-14 in textbook page no 195')

Case(a): Generator action developed by alternator 2 , P_2 = 108.9 W
Case(b): Synchronizing power received by alternator 1 , P_1 = -99.0 W
Case(c): Power loss in both armature , Losses = 10 W
Case(d): Terminal voltage of alternator 2 , V_p2 = 210 V (generator action)
Terminal voltage of alternator 1 , V_p2 = 210 V (motor action)
Case(e): Phasor diagram is shown in Fig 7-14 in textbook page no 195


## Example 7.9, Page number 199¶

In :
import math
import cmath

#Variable declaration
E_2 = 230.0*cmath.exp(1j*180.0*math.pi/180)    #Voltage generated by alternator 2(V)
E_1 = 230.0*cmath.exp(1j*20.0*math.pi/180)     #Voltage generated by alternator 1(V)
Z = 2.01*cmath.exp(1j*84.3*math.pi/180)        #Impedance(ohm)

#Calculation
E_r = E_2+E_1                                  #Total voltage generated by Alternator 1 and 2(V)
Z_1 = Z
Z_2 = Z
#Case(a)
I_s = E_r/(Z_1+Z_2)                            #Synchronizing current(A)
Is = abs(I_s)                                  #Magnitude of Synchronizing current(A)
I_s_a = cmath.phase(I_s)*180/math.pi           #Phase angle of Synchronizing current(degrees)
#Case(b)
E_gp1 = abs(E_1)
E_gp1_Is = (cmath.phase(E_1)*180/math.pi)-I_s_a     #Angle(degree)
P_1 = E_gp1*Is*math.cos(E_gp1_Is*math.pi/180)       #Synchronizing power developed by alternator 1(W)
#Case(c)
E_gp2 = abs(E_2)
E_gp2_Is = (cmath.phase(E_2)*180/math.pi)-I_s_a     #Angle(degree)
P_2 = E_gp2*Is*math.cos(E_gp2_Is*math.pi/180)       #Synchronizing power developed by alternator 2(W)
#Case(d)
P2 = abs(P_2);
losses = P_1-P2                                     #Losses in the armature(W)
theta = cmath.phase(Z)*180/math.pi                  #Angle(degree)
check = abs(E_r)*Is*math.cos(theta*math.pi/180)     #Verifying losses(W) by Eq.7-7
R_aT = 2*Z.real                                     #Total armature resistance of alternator 1 and 2(ohm)
double_check = Is**2*R_aT                           #Verifying losses(W) by Eq.7-7

#Result
print('Case(a): Synchronizing current , I_s = %.2f∠%.1f° A' %(Is,I_s_a))
print('Case(b): Synchronizing power developed by alternator 1 , P_1 = %.f W (power delivered to bus)' %P_1)
print('Case(c): Synchronizing power developed by alternator 2 , P_2 = %.f W (power received from bus)' %P_2)
print('Case(d): Losses in the armature , Losses = %.f W' %losses)
print('\nNOTE: Changes in obtained answer from that of textbook is due to more precision i.e more number of decimal places')

Case(a): Synchronizing current , I_s = 19.87∠15.7° A
Case(b): Synchronizing power developed by alternator 1 , P_1 = 4557 W (power delivered to bus)
Case(c): Synchronizing power developed by alternator 2 , P_2 = -4400 W (power received from bus)
Case(d): Losses in the armature , Losses = 158 W

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


## Example 7.10, Page number 200¶

In :
import math
import cmath

#Variable declaration
E_2 = 230.0*cmath.exp(1j*180.0*math.pi/180)    #Voltage generated by alternator 2(V)
E_1 = 230.0*cmath.exp(1j*20.0*math.pi/180)     #Voltage generated by alternator 1(V)
Z = 6.0*cmath.exp(1j*50.0*math.pi/180)         #Impedance(ohm)

#Calculation
E_r = E_2+E_1                                  #Total voltage generated by Alternator 1 and 2(V)
Z_1 = Z
Z_2 = Z
#Case(a)
I_s = E_r/(Z_1+Z_2)                            #Synchronizing current(A)
Is = abs(I_s)                                  #Magnitude of Synchronizing current(A)
I_s_a = cmath.phase(I_s)*180/math.pi           #Phase angle of Synchronizing current(degrees)
#Case(b)
E_gp1 = abs(E_1)
E_gp1_Is = (cmath.phase(E_1)*180/math.pi)-I_s_a     #Angle(degree)
P_1 = E_gp1*Is*math.cos(E_gp1_Is*math.pi/180)       #Synchronizing power developed by alternator 1(W)
#Case(c)
E_gp2 = abs(E_2)
E_gp2_Is = (cmath.phase(E_2)*180/math.pi)-I_s_a     #Angle(degree)
P_2 = E_gp2*Is*math.cos(E_gp2_Is*math.pi/180)       #Synchronizing power developed by alternator 2(W)
#Case(d)
P2 = abs(P_2);
losses = P_1-P2                                     #Losses in the armature(W)
theta = cmath.phase(Z)*180/math.pi                  #Angle(degree)
check = abs(E_r)*Is*math.cos(theta*math.pi/180)     #Verifying losses(W) by Eq.7-7
R_aT = 2*Z.real                                     #Total armature resistance of alternator 1 and 2(ohm)
double_check = Is**2*R_aT                           #Verifying losses(W) by Eq.7-7

#Result
print('Case(a): Synchronizing current , I_s = %.2f∠%.1f° A' %(Is,I_s_a))
print('Case(b): Synchronizing power developed by alternator 1 , P_1 = %.f W (power delivered to bus)' %P_1)
print('Case(c): Synchronizing power developed by alternator 2 , P_2 = %.f W (power received from bus)' %P_2)
print('Case(d): Losses in the armature , Losses = %.f W' %losses)
print('\nNOTE: Changes in obtained answer from that of textbook is due to more precision i.e more number of decimal places')

Case(a): Synchronizing current , I_s = 6.66∠50.0° A
Case(b): Synchronizing power developed by alternator 1 , P_1 = 1326 W (power delivered to bus)
Case(c): Synchronizing power developed by alternator 2 , P_2 = -984 W (power received from bus)
Case(d): Losses in the armature , Losses = 342 W

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


## Example 7.11, Page number 207¶

In :
import math
import cmath
import numpy

#Variable declaration
V_AB = 100.0*cmath.exp(1j*0.0*math.pi/180)      #Voltage supplied across A & B(V)
V_BC = 100.0*cmath.exp(1j*-120.0*math.pi/180)   #Voltage supplied across B & C(V)
V_CA = 100.0*cmath.exp(1j*120.0*math.pi/180)    #Voltage supplied across C & A(V)
I1_1 = complex(6,0)
I1_2 = complex(-3,0)
I2_1 = complex(-3,0)
I2_2 = complex(3,-4)
V_1 = complex(100,0)
V_2 = complex(-50,-86.6)

#Calculation
A = [[I1_1,I2_1],[I1_2,I2_2]]          #Matrix containing mesh equations array
delta = numpy.linalg.det(A)            #Determinant of A
#Case(a)
I_1 = numpy.linalg.det([[V_1,I2_1],[V_2,I2_2]])/delta   #Mesh current I_1(A)
I_2 = numpy.linalg.det([[I1_1,V_1],[I1_2,V_2]])/delta    #Mesh current I_2(A)
#Case(b)
I_A = I_1                      #Line current(A)
I_B = I_2-I_1                  #Line current(A)
I_C = -I_2                     #Line current(A)
#Case(c)
Z_A = complex(3,0)             #Impedance(ohm)
Z_B = complex(3,0)             #Impedance(ohm)
Z_C = complex(0,-4)            #Impedance(ohm)
V_AO = I_A*Z_A                 #Phase voltage(V)
V_BO = I_B*Z_B                 #Phase voltage(V)
V_CO = I_C*Z_C                 #Phase voltage(V)

#Result
print('Case(a): Mesh current , I_1 = (%.3f%.4fj) A  = %.1f∠%.2f° A' %(I_1.real,I_1.imag,abs(I_1),cmath.phase(I_1)*180/math.pi))
print('         Mesh current , I_2 = (%.2f%.3fj) A  = %.2f∠%.2f° A' %(I_2.real,I_2.imag,abs(I_2),cmath.phase(I_2)*180/math.pi))
print('Case(a): Line current , I_A = (%.2f%.3fj) A  = %.1f∠%.2f° A' %(I_A.real,I_A.imag,abs(I_A),cmath.phase(I_A)*180/math.pi))
print('         Line current , I_B = (%.3f%.3fj) A  = %.3f∠%.2f° A' %(I_B.real,I_B.imag,abs(I_B),cmath.phase(I_B)*180/math.pi))
print('         Line current , I_C = (%.2f+%.3fj) A  = %.2f∠%.2f° A' %(I_C.real,I_C.imag,abs(I_C),cmath.phase(I_C)*180/math.pi))
print('Case(c): Phase voltage , V_AO = %.1f∠%.2f° V' %(abs(V_AO),cmath.phase(V_AO)*180/math.pi))
print('         Phase voltage , V_BO = %.2f∠%.1f° V' %(abs(V_BO),cmath.phase(V_BO)*180/math.pi))
print('         Phase voltage , V_CO = %.2f∠%.2f° V' %(abs(V_CO),cmath.phase(V_CO)*180/math.pi))
print('Case(d): The phasor diagram is shown in fig 7-23b in textbook page no.208')

Case(a): Mesh current , I_1 = (26.157-3.5589j) A  = 26.4∠-7.75° A
Mesh current , I_2 = (18.98-7.118j) A  = 20.27∠-20.56° A
Case(a): Line current , I_A = (26.16-3.559j) A  = 26.4∠-7.75° A
Line current , I_B = (-7.176-3.559j) A  = 8.010∠-153.62° A
Line current , I_C = (-18.98+7.118j) A  = 20.27∠159.44° A
Case(c): Phase voltage , V_AO = 79.2∠-7.75° V
Phase voltage , V_BO = 24.03∠-153.6° V
Phase voltage , V_CO = 81.09∠69.44° V
Case(d): The phasor diagram is shown in fig 7-23b in textbook page no.208