# Chapter 5 : Three Phase Synchronous Machines¶

## Example 5.1 Page No : 424¶

In :
import math

# Given Data;
slots = 18.;
p = 2.;             #nmber of poles
ph = 3.;            #three phase winding

# Calculations and Results
SA = (360/slots);  #slot angle
m = slots/(p*ph);  #m = nmber of slots per pole per phase
print "number of slots per pole per phase, m = %d"%(m);
print "emfs of the oils of each phase will have a time-phase difference of  %d degree mechanical "%(SA);
print "distribution factor = %f"%(k_d);

number of slots per pole per phase, m = 3
emfs of the oils of each phase will have a time-phase difference of  20 degree mechanical
distribution factor = 0.959795


## Example 5.2 Page No : 425¶

In :
import math
# Given Data
slots = 36.;    #number of slots
poles = 4.;     #number of poles
ph = 3.;        #math.single layer three phase winding

# Calculations and Results
SP = slots/ph; #number of slots per phase
print "number of slots per phase =  %d"%(SP);
m = SP/poles;  #munber of slots per pole per phase
print "number of slots per pole per phase, m = %d"%(m)
SA_m = 360/slots;        #slot angle mechanical
SA_e = (poles/2)*SA_m    #slot angle electrical
print "slot angle =  %d degree electrical"%(SA_e)
print "distribution factor =  %f"%(k_d)

number of slots per phase =  12
number of slots per pole per phase, m = 3
slot angle =  20 degree electrical
distribution factor =  0.959795


## Example 5.3 Page No : 426¶

In :
import math

# Given Data;
slots = 48.;      #number of slots
poles = 4.;       #4-pole machine
ph = 3.;          #3-phase machine

# Calculations and Results
SA = 360/slots;  #slot angle
print "total number of slots =  %d"%(slots);
print "slot angle =  %f degree mechanical"%(SA);
#coil span is 11 slot pitches
#12 slots subtend 180degress, short pitched by 1 slot
Bta = 1*180./12;
print "pitch factor = %f"%(k_p)

total number of slots =  48
slot angle =  7.500000 degree mechanical
pitch factor = 0.991445


## Example 5.4 Page No : 426¶

In :
import math

# Given Data;
slots = 72.;    #number of slots
P = 8.;         #number of poles
ph = 3.;        #3-phase machine
N = 750.;       #speed of machine in rpm

#winding is made with 36 coils having 10 turns
Fp = 0.15;     #flux per pole
fre = (P*N)/120;
NCp = 36./ph;   #nmber of coils per phase
T = NCp*10;    #number of turns per phase
k_p = 1;       #math.since full pitched pitch factor is 1

# Calculations and Results
print "flux per pole = %fWb"%(Fp)
print "number of turns per phase = %d"%(T);
print "pitch factor = %f"%(k_p);
m = slots/(P*ph); #slots per pole per phase
SA_m = 360/slots; #slot angle mechanical
SA_e = (P/2)*SA_m;
print "distribution factor = %f"%(k_d);
E = 4.44*Fp*fre*T*k_d*k_p;
print "RMS vale of emf induced per phase = %fV"%(E)

flux per pole = 0.150000Wb
number of turns per phase = 120
pitch factor = 1.000000
distribution factor = 0.959795
RMS vale of emf induced per phase = 3835.341142V


## Example 5.5 Page No : 427¶

In :
import math
# Given Data;
print ("E(line to line) =   440V");
E_l = 440.;    #line-to-line voltage
E_p = E_l/(math.sqrt(3));
N = 750.;      #speed in rpm
fre = 50.;     #frequency

# Calculations and Results
P = (120*fre)/N;
print "P =   %d"%(P);
print "Eper phase) =   %dV"%(E_p);
ph = 3;       #3-phase machine
m = 2;        #number of slots per pole per phase
slots = m*P*ph;       #total number of stator slots
SA_m = 360/slots;     #slot angle mechanical
SA_e = (P/2)*SA_m;    #slot angle electrical
k_p = 1;              #assuming full pitch
print "slot angle =   %d degree electrical"%(SA_e);
print "pitch factor = %f"%(k_p);
print "distribution factor =   %f"%(k_d);
#2 slots per pole per phase
NSp = 2*P;            #number of slots per phase
NTc = 4;              #number of turns per coil
T = 8*NTc;            #number of turns per phase
Fp = E_p/(4.44*fre*T*k_d*k_p);
print "flux per pole =   %fWb"%(Fp);

E(line to line) =   440V
P =   8
Eper phase) =   254V
slot angle =   30 degree electrical
pitch factor = 1.000000
distribution factor =   0.965926
flux per pole =   0.037021Wb


## Example 5.6 Page No : 428¶

In :
import math
#chapter 5
#example 5.6
#page 428

# Given Data;
slots = 144.;   #number of slots
ph = 3.;        #3-phase machine
P = 16.;        #number of poles
Cp = 10.;       #number of conducters per slot
Fp = 0.03;     #flux per pole
Ns = 375.;      #synchronous speed

# Calculations and Results
fre = (Ns*P)/120;     #frequency
print "frequency = %d"%(fre);
m = slots/(P*ph);     #number of slots per pole per phase
print "number of slots per pole per phase, m =   %d"%(m);
SA_m = 360/slots;     #slot angle mechanical
SA_e = (P/2)*SA_m;    #slot angle electrical
k_p = 1               #no short pitching
print "short pitch =   %d"%(k_p);
print "distribution factor =   %f"%(k_d);
T = (slots*10)/(2*ph);
print "number of turns per phase, T =   %d"%(T);
E = 4.44*Fp*fre*T*k_d*k_p;
print "RMS value of induced emf per phase, E =   %fV"%(E);
print "induced emf across the linesis %fV "%(math.sqrt(3)*E);

frequency = 50
number of slots per pole per phase, m =   3
short pitch =   1
distribution factor =   0.959795
number of turns per phase, T =   240
RMS value of induced emf per phase, E =   1534.136457V
induced emf across the linesis 2657.202289V


## Example 5.7 Page No : 428¶

In :
import math

# Given Data;
slots = 90.;   #number of slots
P = 10.;       #number of poles
ph = 3.;       #3-phase machine
fre = 50.;     #frequency
Fp = 0.16;    #flux per pole
E_l = 11000.;  #line voltage
SA_m = 360/slots;  #machanical slot angle

# Calculations and Results
SA_e = (P/2)*SA_m; #electrical slot angle
m = slots/(ph*P);
print "slot angle = %d degree elecrical"%(SA_e)
print "number of slots per pole per phase, m = %d"%(m);
k_p = 1;           #assuming full pitch
print "pitch factor = %d"%(k_p);
print "distribution factor = %f"%(k_d);
E_p = E_l/math.sqrt(3);
T = E_p/(4.44*Fp*fre*k_p*k_d);
print "total number of armature conductors, Z =  %d"%(2*T);

slot angle = 20 degree elecrical
number of slots per pole per phase, m = 3
pitch factor = 1
distribution factor = 0.959795
total number of armature conductors, Z =  372


## Example 5.8 Page No : 429¶

In :
import math

# Given Data;
print ("P = 6   , f = 50");
P = 6.;
f = 50.;
Sp = 12.;         #slots per pole
Cs = 4.;         #conductors per slot
Fp = 1.5;

# Calculations and Results
TS = Sp*P
print "total number of slots = %d"%(TS);
print "total number of slots per phase =  %d"%( TS/3);
print "total number of conductors per phase =  %d"%(( TS*Cs)/3);
T = ((TS*Cs)/3)/2;
print "total number of turns per phase = %d"%(T)
m = (TS/(P*3));
print "number of slots per pole per phase, m =  %d"%(m);
SA_m = 360/TS;             #slot angle mechanical
SA_e = (P/2)*SA_m;
print "distribution factor = %f"%(k_d);
print ("coil pitch is 5/6 of full-pitch");
bheta = 180-(5./6)*180;     #short pitch angle
print "short pitch angle =  %d degrees"%(bheta)
print "pitch factor =  %f "%(k_p);
E = 4.44*Fp*f*T*k_d*k_p;
print "induced per phase =   %fV"%(E)

P = 6   , f = 50
total number of slots = 72
total number of slots per phase =  24
total number of conductors per phase =  96
total number of turns per phase = 48
number of slots per pole per phase, m =  4
distribution factor = 0.957662
coil pitch is 5/6 of full-pitch
short pitch angle =  30 degrees
pitch factor =  0.965926
induced per phase =   14785.689892V


## Example 5.9 Page No : 439¶

In :
import math

# Given Data;
OP = 500000.;     #output power
V_l = 3300.;      #line voltage

# Calculations and Results
I_l = OP/(math.sqrt(3)*V_l);      #line current
print "line current =   %fA"%(I_l);
#for star connected alternater, line current is equal to phase current
I_a = I_l;
pf = 0.8;        #power factor
phi = math.degrees(math.acos(pf));
R_a = 0.3;       #synchronous resistance
X_s = 4;         #synchronous reactance
V_p = V_l/math.sqrt(3);
print "phase voltage =   %fV"%(V_p)
print "induced emf =   %f V/Phase"%(E  )
PR = ((E-V_p)*100)/V_p;
print "percentage regulation =   %f percent"%(PR);

line current =   87.477314A
phase voltage =   1905.255888V
induced emf =   2152.469556 V/Phase
percentage regulation =   12.975353 percent


## Example 5.10 Page No : 440¶

In :
import math

# Given Data
V = 2000.;
V_oc = 500.;         #open circuit voltage
I_sc = 100.;         #short circuit current
I_a = 100.;
R_s = 0.8;          #armature resistance

# Calculations and Results
Z_s = V_oc/I_sc;    #synchronous impedence
print "Z_s =  %d ohm"%(Z_s);
X_s = math.sqrt(Z_s**2-R_s**2);
print "X_s =  %f ohm"%(X_s);
pf = 1;
phi = math.degrees(math.acos(pf));
print ("At unity power factor");
print "";
print "induced emf =  %fV"%(E);
R = ((E-V)*100)/V;
print "regulation =  %f percent"%(R);
pf = 0.71;
phi = math.degrees(math.acos(pf));
print ("At 0.71 lagging power factor");
print "";
print "induced emf =  %fV"%(E);
R = ((E-V)*100)/V;
print "regulation =  %fpercent"%(R);
pf = 0.8;
phi = math.degrees(math.acos(pf));
print ("At 0.8 leading power factor");
print "";
print "induced emf =  %fV"%(E);
R = ((E-V)*100)/V;
print "regulation =   %fpercent"%(R);

Z_s =  5 ohm
X_s =  4.935585 ohm
At unity power factor

induced emf =  2137.755833V
regulation =  6.887792 percent
At 0.71 lagging power factor

induced emf =  2422.283821V
regulation =  21.114191percent

induced emf =  1822.487197V
regulation =   -8.875640percent


## Example 5.11 Page No : 441¶

In :
# Given Data;
print ("field exitation current = 10A");
V_oc = 900.;      #induced emf on open circuit
I_sc = 150.;      #short circuit current

# Calculations and Results
Z_s = V_oc/I_sc; #synchronous impedence
print "synchronous impedence, Z_s =  %d ohm"%(Z_s);
I_a = 60;
print "internal voltage drop when the load current is 60amp =   %d V"%(I_a*Z_s);

field exitation current = 10A
synchronous impedence, Z_s =  6 ohm
internal voltage drop when the load current is 60amp =   360 V


## Example 5.12 Page No : 441¶

In :
import math

# Given Data;
KVA = 2000.;
V = 6600.;     #rating
V_p = 6600./math.sqrt(3);
I_a = (KVA*1000)/(math.sqrt(3)*V);
R_a = 0.4;    #armature resistance
X_s = 4.5     #synchronous reactance
pf = 0.8;

# Calculations and Results
phi = math.degrees(math.acos(pf));
print "V/phase =  %dV "%(V_p)
print "E =   %f V per phase"%(E);
R = ((E-V_p)*100)/V_p;
print "percentage change in terminal voltage =  %f percent"%(R);

V/phase =  3810V
E =   4378.515597 V per phase
percentage change in terminal voltage =  14.906234 percent


## Example 5.13 Page No : 442¶

In :
import math

# Given Data;
KVA = 1200.;       #output power
print "output power = %d"%(KVA)
V_l = 3300.;       #line voltage
R_a = 0.25;       #armature resistance

# Calculations and Results
I_l = (KVA*1000)/(math.sqrt(3)*V_l);    #line current
#for star connected I_l = I_a
I_a = I_l;
V_p = V_l/math.sqrt(3);
print "V per phase =  %dV"%(V_p)
#field current of 40A produces short circuit current of 200A and open circuit emf 1100
v_l = 1100;
i_s = 200;
Z_s =  v_l/(math.sqrt(3)*i_s);     #synchronous impedence
print "Synchronous impedance, Zs = %f ohm"%(Z_s)
X_s = math.sqrt(Z_s**2-R_a**2);      #synchronous reactance
print ("(a)for 0.8 lagging power facor");
pf = 0.8;
phi = math.degrees(math.acos(pf));
print "induced emf, E = %f V"%(E);
R = ((E-V_p)*100)/V_p;
print "regulation = %f percent"%(R);
pf = 0.8;
phi = math.degrees(math.acos(pf));
print "induced emf, E =  %f V"%(E);
R = ((E-V_p)*100)/V_p;
print "regulation = %f percent"%(R);

output power = 1200
V per phase =  1905V
Synchronous impedance, Zs = 3.175426 ohm
(a)for 0.8 lagging power facor
induced emf, E = 2398.732590 V
regulation = 25.900810 percent
induced emf, E =  1647.716860 V
regulation = -13.517293 percent


## Example 5.14 Page No : 443¶

In :
import math

# Given Data;
print ("star connected alternator")
KVA = 1500.;         #rating
ph = 3.;            #3-phase
V_l = 6600.;       #voltage
Ra = 0.4         #armature resistance
Xs = 6.;         #reactance

# Calculations and Results
Ia = (KVA*1000)/(math.sqrt(3)*V_l);
print "Full-load current =  %d A"%(Ia);
V = V_l/math.sqrt(3);
print "Voltage per phase = %d V"%(V);
print ("for 0.8 lagging power facor");
pf = 0.8;           #power factor
phi = math.degrees(math.acos(pf));
print "induced emf = %f V"%(E);
print ("then at 0.8 leading power factor");
Vt = 4743; #solved manually
print "termial Voltage, line-to-line = %d V"%(math.sqrt(3)*Vt)

star connected alternator
Voltage per phase = 3810 V
for 0.8 lagging power facor
induced emf = 4366.072552 V
then at 0.8 leading power factor
termial Voltage, line-to-line = 8215 V


## Example 5.15 Page No : 450¶

In :
import math
#chapter 5
#example 5.15
#page 450

# Given Data;
La = 5000.;
pf = 0.8;

# Calculations and Results
phi = math.degrees(math.acos(pf));
print "math.tan phi =   %f"%(math.degrees(math.atan(phi)));
print ("FOR ALTERNATOR A");
pf_a = 0.9;
phi_a = math.degrees(math.acos(pf_a));
print "math.tan phi_a =   %f"%(math.degrees(math.atan(phi_a)));
print "reactive load =   %d KVAr"%(8000*math.degrees(math.atan(phi_a)));
print ("Active Load A = 5000kW");
print "Reactive load A =   %dkVAr"%(5000*math.degrees(math.atan(phi_a)));
print "Active load of B =   %dkW"%(L-La);
a = ((8000*math.degrees(math.atan(phi)))-(5000*math.degrees(math.atan(phi_a))))
print "Reactive load of B =   %dkVAr"%(a);
B = a/(L-La);
phi_b = math.degrees(math.atan(B));
print "phi_b =   %f"%(phi_b)
print "Power Factor of B =   %f"%(math.cos(math.radians(phi_b)));

math.tan phi =   88.446382
FOR ALTERNATOR A
math.tan phi_a =   87.783943
Active load of B =   3000kW
Reactive load of B =   268651kVAr
phi_b =   89.360211
Power Factor of B =   0.011166


## Example 5.16 Page No : 451¶

In :
import math

# Given Data
V = 6600.;
ph = 3.;  #3-phase alternators

# Calculations and Results
print ("Two alternators in parallel connection");
pf = 0.8;
Ia = 438;       #armature current
phi = math.degrees(math.acos(pf));
print "Active component of current =   %fA"%(Ac);
print "Reactive component of current =   %fA"%(Rc);
print "Current supplied by each alternator = %fA"%(Il/2);
print "Active component of current supplied by each alternator =   %fA"%(Ac/2);
print "Reactive component of current supplied by each alternator =   %fA"%(Rc/2);
print ("Since steam supply is same,the active component remain the same ");
RIl = math.sqrt(Ia**2-(Ac/2)**2);
print "Reactive component of Il =  %dA"%(RIl);
RI2 = (Rc-RIl);
print "reactive component of I2 =   %fA"%(RI2);
I2 = math.sqrt((Ac/2)**2+(RI2)**2);
print " I2 =   %fA"%(I2);
phi_2 = math.degrees(math.atan(RI2/(Ac/2)));
print "phi 2 =   %f degrees"%(phi_2);
print "math.cos phi 2 =  %f"%(math.cos(math.radians(phi_2)));

Two alternators in parallel connection
Active component of current =   874.773135A
Reactive component of current =   656.079851A
Current supplied by each alternator = 546.733209A
Active component of current supplied by each alternator =   437.386568A
Reactive component of current supplied by each alternator =   328.039926A
Since steam supply is same,the active component remain the same
Reactive component of Il =  23A
reactive component of I2 =   632.906796A
I2 =   769.336091A
phi 2 =   55.352588 degrees
math.cos phi 2 =  0.568525


## Example 5.17 Page No : 455¶

In :
import math
# Given Data;
print ("power factor of existing load is 0.8 lagging");
pf = 0.8; #power factor

# Calculations and Results
phi = math.degrees(math.acos(pf));
print "phi =  %d degree"%(phi);
kVAr1 = (L*math.degrees(math.atan(phi)));
print "kVAr1 =  %d "%(kVAr1);
print ("output for the synchronous motor is 200kW");
output = 200.;
efficiency = 0.9;
kW = (output/efficiency);
print "Input to the synchronous motor =  %fkW"%(kW);
TL = (L+kW); # total load
print "Total load on the system =  %fkW"%(TL);
print ("overall power factor of the load is to be raised to 0.92 lagging");
pf = 0.92;
phi = math.degrees(math.acos(pf));
kVAr2 = (TL*math.degrees(math.atan(phi)))
print "kVAr2 = %f"%(kVAr2);
kVAr = kVAr1-kVAr2;
print "lagging kVAr of synchronous codenser =  %f"%(kVAr);
print "leading kVAr supplied by the motor =  %f"%(kVAr);
phi = math.degrees(math.atan(kVAr/kW));
print "phi =  %d degree"%(phi);
print "KVA rating of the synchronous motor =  %f"%(kW/math.cos(math.radians(phi)));

power factor of existing load is 0.8 lagging
phi =  36 degree
kVAr1 =  70757
output for the synchronous motor is 200kW
Input to the synchronous motor =  222.222222kW
Total load on the system =  1022.222222kW
overall power factor of the load is to be raised to 0.92 lagging
kVAr2 = 89463.266068
lagging kVAr of synchronous codenser =  -18706.160461
leading kVAr supplied by the motor =  -18706.160461
phi =  -89 degree
Power factor of the synchronos motor =  0.011879 leading
KVA rating of the synchronous motor =  18707.480373