Chapter 3 : Transformers¶

Example 3.1 Page No : 26¶

In :
import math

# GIVEN DATA

Z = (0.05 + 0.05 * 1j) * 100;          # Transmission line parameters (impedance) in Ohms (multiplied by 100 because dismath.tance of the Transmission line is 100km)
R = 0.05 * 100;                        # Transmission line resistance in Ohms (multiplied by 100 because dismath.tance of the Transmission line is 100km)
V1 = 220;                             # Terminal voltage in Volts
V2 = 1 * 10 ** 3;                      # Terminal volatge from Generator side in Volts
P = 20 * 10 ** 3;                      # Power in Watts

# CACULATIONS

I1 = P/V1;                          # Line current for 220V in Amphere
I2 = P/V2;                          # Line current for 1kV in Amphere
I1Z = Z*I1;                         # Voltage drop due to I1 in Volts
I2Z = Z*I2;                         # Voltage drop due to I2 in Volts
Loss1 = (I1 ** 2) * R * 10 ** -3;     # Line loss for I1 in kW
Loss2 = (I2 ** 2) * R * 10 ** -3;     # Line loss for I2 in kW
Vg1 = V1 + I1Z;                     # Input Voltages on Generator Terminal in Volts
Vg2 = V2 + I2Z;                     # Input Voltages on Generator Terminal in Volts

# DISPLAY RESULTS

print ("EXAMPLE : 3.1 : SOLUTION :-") ;
print " a.1) Voltage drop due to I1 ,I1Z = % .2f+j%.2f  V  "%(I1Z.real,I1Z.imag);
print " a.2) Voltage drop due to I2 , I2Z = % .f+j%.f  V "%(I2Z.real,I2Z.imag);
print " b.1) Line loss for I1 , Loss1 = %.2f kW  "%(Loss1);
print " b.2) Line loss for I2 , Loss2 = % .2f kW "%(Loss2);
print " c.1) Input Voltages on Generator Terminal from a load terminal , Vg1 = %.2f+j%.2f = %.2f V  "%(Vg1.real,Vg1.imag,abs(Vg1));
print " c.2) Input Voltages on Generator Terminal from a Generating Station , Vg2 = % .f+j%.f = %.2f V "%(Vg2.real,Vg2.imag,abs(Vg2));
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a I1Z = 450.45+j450.45V instead of 454.55+j454.55 V" ;
print " b) Vg1 = 670.45)+j450.45) = 807.72 V  instead of % .2f+j%.2f = %.2f V "%(Vg1.real,Vg1.imag,abs(Vg1) );
EXAMPLE : 3.1 : SOLUTION :-
a.1) Voltage drop due to I1 ,I1Z =  450.00+j450.00  V
a.2) Voltage drop due to I2 , I2Z =  100+j100  V
b.1) Line loss for I1 , Loss1 = 40.50 kW
b.2) Line loss for I2 , Loss2 =  2.00 kW
c.1) Input Voltages on Generator Terminal from a load terminal , Vg1 = 670.00+j450.00 = 807.09 V
c.2) Input Voltages on Generator Terminal from a Generating Station , Vg2 =  1100+j100 = 1104.54 V
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a I1Z = 450.45+j450.45V instead of 454.55+j454.55 V
b) Vg1 = 670.45)+j450.45) = 807.72 V  instead of  670.00+j450.00 = 807.09 V

Example 3.2 Page No : 28¶

In :
import math

# GIVEN DATA
E1 = 6.6 * 10 ** 3;              # Primary voltage in Volts
E2 = 220;                       # Secondary Voltage in volts
f = 50;                         # Frequency in Hertz
phi_m = 0.06;                   # Flux in Weber
S = 50 * 10**6;                  # Rating of the math.single-phase transformer in VA

# CALCULATIONS
N1 = E1/(4.44*f*phi_m);         # Number of turns in Primary
vpn1 = E1/N1;                    # Voltage per turns in Primary in Volts/turn
N2 = E2/(4.44*f*phi_m);         # Number of turns in Secondary
vpn2 = E2/N2;                    # Voltage per turns in Secondary in Volts/turn

# DISPLAY RESULTS
print ("EXAMPLE : 3.2 : SOLUTION :-") ;
print " a.1) Number of turns in Primary , N1 = %.1f Turns nearly 496 Turns  "%(N1);
print " a.2) Number of turns in Secondary , N2 = %.1f Turns nearly 16 Turns "%(N2);
print " b.1) Voltage per turns in Primary , vpn1 = %.1f Volts/turns  "%(vpn1);
print " b.2) Voltage per turns in Secondary , vpn2 = %.2f Volts/turns  "%(vpn2);
EXAMPLE : 3.2 : SOLUTION :-
a.1) Number of turns in Primary , N1 = 495.5 Turns nearly 496 Turns
a.2) Number of turns in Secondary , N2 = 16.5 Turns nearly 16 Turns
b.1) Voltage per turns in Primary , vpn1 = 13.3 Volts/turns
b.2) Voltage per turns in Secondary , vpn2 = 13.32 Volts/turns

Example 3.3 Page No : 30¶

In :
# GIVEN DATA
f = 50;                 # Frequency in Hertz
N = 50;                 # Number of turns in Secondary
E =220;                 # Induced voltage in Volts

# CALCULATIONs
phi_m = E/(4.44*f*N);        # Maximum value of the Flux in Weber

# DISPLAY RESULTS

print ("EXAMPLE : 3.3 : SOLUTION :-") ;
print " a) Maximum value of the Flux , phi_m = % .7f Wb  "%(phi_m);
EXAMPLE : 3.3 : SOLUTION :-
a) Maximum value of the Flux , phi_m =  0.0198198 Wb

Example 3.4 Page No : 32¶

In :
# GIVEN DATA
S = 1.5;               # Transformer Rating in KVA
E1 = 220.;              # HV side voltage in volts
E2 = 40.;               # LV side voltage in volts

# CALCULATION
Ihv = (S * 10 ** 3)/E1;             # Rated HV side Curent in Amphere
Ilv = (S * 10 ** 3)/E2;             # Rated lV side Curent in Amphere

# DISPLAY RESULTS
print ("EXAMPLE : 3.4 : SOLUTION :-") ;
print " a) Rated HV side Curent , Ihv = %.2f A  "%(Ihv);
print " b) Rated LV side Curent , Ilv = %.1f A "%(Ilv);
EXAMPLE : 3.4 : SOLUTION :-
a) Rated HV side Curent , Ihv = 6.82 A
b) Rated LV side Curent , Ilv = 37.5 A

Example 3.5 Page No : 36¶

In :
import math

# GIVEN DATA
Ai = 2.3 * 10 ** -3;                 # Cross-Sectional area of the core in Meter-Square
mue_0 = 4*math.pi*10** -7;               # Permeability of the air in Henry/Meter
Fe_loss = 2.6;                      # Iron loss at the working Flux density Watts/kg
Fe_den = 7.8 * 10 ** 3;              # Density of the Iron in kg/Meter-Cube
N1 = 800.;                           # Number of Turns of the Primary winding
L = 2.5;                            # Length of the Flux path in Meter
mue_r = 1000.;                       # Relative Permeability
E = 400.;                            # Primary Volatge of the Transformer in Volts
f = 50.;                             # Frequency in Hertz

# CALCULATIONS
Bm = E/(4.44*f*Ai*800);             # Flux Density in Weber/Meter-Square
phi_m = (Bm*Ai)*10**3;               # Maximum Flux in the core in milli-Weber
F = (L*Bm)/(mue_r*mue_0);           # Magnetizing MMF in Amphere-turns
Im = F/(N1*math.sqrt(2));                # Magnetizing Current in Amphere
Vol = L*Ai;                         # Volume of the Core in Meter-Cube
W = Vol * Fe_den;                   # Weight of the Core in kg
Total_Fe_loss = Fe_loss * W;        # Total Iron loss in Watt
Ic = Total_Fe_loss/E;               # Loss component of Current in Amphere
Io= math.sqrt((Ic ** 2)+(Im ** 2));        # No load Current in Amphere
pf_angle = math.degrees(math.atan(Io/Ic));            # No load Power factor angle in degree

# DISPLAY RESULTS
print ("EXAMPLE : 3.5 : SOLUTION :-") ;print " a) Maximum Flux in the core , phi_m = %.6f mWb  "%(phi_m);
print " b) No load Current , I0 = %.3f A "%(Io);
print " c) No load Power factor angle  =  %.3f degree  "%(pf_angle);
print " d) No load Power factor =  %.4f  "%(pf);
EXAMPLE : 3.5 : SOLUTION :-
a) Maximum Flux in the core , phi_m = 2.252252 mWb
b) No load Current , I0 = 1.746 A
c) No load Power factor angle  =  80.523 degree
d) No load Power factor =  9.4336

Example 3.6 Page No : 39¶

In :
# GIVEN DATA
S = 5.;                  # Transformer Rating in kVA
V1 = 220.;               # HV side voltage in volts
V2 = 110.;               # LV side voltage in Volts
P = 4. * 10 ** 2;         # Load of the Transformer
pf = 0.8;               # Power Factor (lagging)
f = 50.;                 # Frequency in Hertz

# CALCULATIONS
a = V1/V2;              # Turn Ratio of the Transformer

# case (a) At full load
I1 = (S * 10 ** 3)/V1;       # Primary current at full load in Amphere
I2 = (S * 10 ** 3)/V2;       # Secondary Current at full Load in Amphere

# Case (b) At 4kW, 0.8 lagging pf load
I11 = (4 * 10 ** 3 * 0.8)/V1;       # Primary current At 4kW, 0.8 lagging pf load in Amphere
I22 = (4 * 10 ** 3 * 0.8)/V2;       # Secondary Current At 4kW, 0.8 lagging pf load in Amphere

# DISPLAY RESULTS
print ("EXAMPLE : 3.6 : SOLUTION :-") ;
print "   a)   Turn Ratio of the Transformer , a = %.f  "%(a);
print " b.1.1) Primary current at full load , I1 = %.2f A "%(I1);
print " b.1.2) Secondary current at full load , I2 = %.2f A  "%(I2);
print " b.2.1) Primary current at 4kW, 0.8 lagging pf load , I1 = %.3f A "%(I11);
print " b.2.1) Secondary current at 4kW, 0.8 lagging pf load , I2 = %.3f A "%(I22);
EXAMPLE : 3.6 : SOLUTION :-
a)   Turn Ratio of the Transformer , a = 2
b.1.1) Primary current at full load , I1 = 22.73 A
b.1.2) Secondary current at full load , I2 = 45.45 A
b.2.1) Primary current at 4kW, 0.8 lagging pf load , I1 = 14.545 A
b.2.1) Secondary current at 4kW, 0.8 lagging pf load , I2 = 29.091 A

Example 3.7 Page No : 41¶

In :
# GIVEN DATA
a = 100./200;                        # Turn ratio of the Ideal transformer
R = 1.0;                            # resistance across the secondary side having 200 turns in Ohms

# CALCULATIONS
R1 = (a ** 2)*R;                     # Referred value of the resistance from Primary side having 100 turns in Ohms

# DISPLAY RESULTS
print ("EXAMPLE : 3.7 : SOLUTION :-") ;
print " a) Referred value of the %.f Ohm resistance from Primary side having 100 turns , R1 = %.2f ohms  "%(R,R1);
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) Referred value of the resistance from Primary side having\
\n 100 turns = 0.025 Ohms instead of %.2f Ohms  "%(R1);
EXAMPLE : 3.7 : SOLUTION :-
a) Referred value of the 1 Ohm resistance from Primary side having 100 turns , R1 = 0.25 ohms
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) Referred value of the resistance from Primary side having
100 turns = 0.025 Ohms instead of 0.25 Ohms

Example 3.8 Page No : 44¶

In :
# GIVEN DATA
S = 60.;                # Tranformer Rating in kVA
V1 = 6600.;             # HV Side Voltage Rating of the Transformer in Volts
V2 = 220.;              # LV Side Voltage Rating of the Transformer in Volts
R1 = 7.8;              # primary resistances of the Transformer in Ohms
R2 = 0.0085;           # Secondary resistances of the Transformer in Ohms

# CALCULATIONS
a = V1/V2;             # Transformation Ratio
Rp = R1+(a**2)*R2;      # resistance referred to Primary side in Ohms
Rs = (R1/(a**2))+R2;    # resistance referred to Secondary side in Ohms
Ip = (S*1000)/V1       #  Current in Primary Side in Ampheres
Cu_loss = Rp*(Ip**2);   # Copper loss in Transformer in Watts

# DISPLAY RESULTS
print ("EXAMPLE : 3.8 : SOLUTION :-") ;
print " a) Equlivalent resistance as Referred to Primary Side, Rp = % .2f ohms "%(Rp)
print " b) Equlivalent resistance as Referred to Secondary Side, Rs = % .5f ohms "%(Rs)
print " c) Total Copper Loss, Cu_loss = % .2f W "%(Cu_loss)
EXAMPLE : 3.8 : SOLUTION :-
a) Equlivalent resistance as Referred to Primary Side, Rp =  15.45 ohms
b) Equlivalent resistance as Referred to Secondary Side, Rs =  0.01717 ohms
c) Total Copper Loss, Cu_loss =  1276.86 W

Example 3.9 Page No : 45¶

In :
# GIVEN DATA
V1 = 11000.;            # HV Side Voltage Rating of the Transformer in Volts
V2 = 440.;              # LV Side Voltage Rating of the Transformer in Volts
R = 1.0;               # resistance across the secondary side having 11kV in Ohms

# CALCULATIONS
a = V1/V2;                          # Turns ratio of the ideal transformers
R2 = (a ** 2)*R;                     # Referred value of the resistance from Primary side having 440V in Ohms

# DISPLAY RESULTS

print ("EXAMPLE : 3.9 : SOLUTION :-") ;
print " a) Referred value of the resistance from Primary side having 440V , R2 = %.f Ohms  "%(R2);
EXAMPLE : 3.9 : SOLUTION :-
a) Referred value of the resistance from Primary side having 440V , R2 = 625 Ohms

Example 3.10 Page No : 48¶

In :
import cmath
import math

# GIVEN DATA
V1 = 440.;                                       # HV Side Voltage Rating of the Transformer in Volts
V2 = 220.;                                       # LV Side Voltage Rating of the Transformer in Volts
pf_o = 0.2;                                     # No-load Power factor lagging
pf_l = 0.8;                                     # Load Power factor lagging
I_o = 5.;                                        # No-load current in Amphere
I_2 = 120.;                                      # Load current in Amphere

# CALCULATIONS
a = V1/V2;                                      # Turns ratio of the two winding Transformers
theta_o =math.degrees(math.acos(math.radians(pf_o)));                          # No load power factor of the two winding Transformers in Degrees
Io = I_o * cmath.exp(-(1j*theta_o*math.pi/180));          # No load current of the two winding Transformers (minus because lagging) in Amphere
theta =math.degrees(math.acos(math.radians(pf_l)));                            # load power factor of the two winding Transformers in Degrees
I2 = I_2 * cmath.exp(-(1j*theta*math.pi/180));            # secondary load current of the two winding Transformers (minus because lagging) in Amphere
I21 = I2/a;                                     # Secondary referred to the primaryin Amphere
I1 = Io + I21;                                  # Primary current in Amphere
I1_mag = abs(I1);                               # Primary current magnitude inj Amphere
theta_1 = (math.atan2(I1.imag,I1.real));            # Primary current angle in Degree

# DISPLAY RESULTS

print ("EXAMPLE : 3.10 : SOLUTION :-") ;
print " a) Primary current , I1 = %.2f < %.1f A  "%(I1_mag,theta_1);
EXAMPLE : 3.10 : SOLUTION :-
a) Primary current , I1 = 65.00 < -1.6 A

Example 3.11 Page No : 50¶

In :
import math
import cmath

# GIVEN DATA

S = 50.;             # kVA Rating of the Transformer
f = 50.;             # Frequency in Hertz
Wo = 190.;           # Meter Readings when HV Winding kept open in Watt
Vo = 230.;           # Meter Readings when HV Winding kept open in Volts
Io = 6.5;           # Meter Readings when HV Winding kept open in Amphere
R2 = 0.06;          # resistance of the LV Winding in Ohms
V1 = 2300.;          # Voltage across the HV Side in Volts
V2 = 230.;           # Voltage across the LV Side in Volts
AC = 230.;           # Tranformer connected to AC mains in Volts

# CALCULATIONS

a = V1/V2;                                    # Trasformation ratio of the Transformer
Wc = Wo - ((Io ** 2) * R2);                    # Core loss in Watts
Po = Wc;                                      # Core loss in Watts
Pc = Wc;                                      # Core loss in Watts
cos_theta_o = Po/(Vo*Io);                     # No load power factor
E = V1 - Io * cmath.exp(1j*(theta_o)*math.pi/180);
Rc = Pc/(Ic ** 2 );                             # Core loss resistance in Ohms
Im = Io *math.degrees(math.sin(math.radians(theta_o)));                       # Current through the Magnetizing branch in Amphre
Xm = E/Im;                                     # Magnetizing  reactance in Ohms
Ift = (S * 10 ** 3)/V2;                         # Full Load current in Amphere
Ie = (Io/Ift)*100;                             # Percentage of the Exicting Current in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 3.11 : SOLUTION :-") ;
print "  a)  Core loss , Wc = %.2f W  "%(Wc);
print " b.1) No load power factor angle , theta_o = % .2f Degree "%(theta_o);
print " b.2) No load power factor , (costheta_o) = % .6f "%(cos_theta_o );
print " c.1) Curent through Core loss Component , Ic = %.4f A  "%(Ic);
print " c.2) Core loss resistance , Rc = %.2f Ohms  "%(Rc);
print "  d)  Current through the Magnetizing Component Xm , Im = % .2f A "%(Im);
print "  e)  Percentage of the Exicting Current  = % .2f Percent "%(Ie);
EXAMPLE : 3.11 : SOLUTION :-
a)  Core loss , Wc = 187.47 W
b.1) No load power factor angle , theta_o =  89.87 Degree
b.2) No load power factor , (costheta_o) =  0.125395
c.1) Curent through Core loss Component , Ic = 0.8151 A
c.2) Core loss resistance , Rc = 282.19 Ohms
d)  Current through the Magnetizing Component Xm , Im =  372.42 A
e)  Percentage of the Exicting Current  =  2.99 Percent

Example 3.12 Page No : 54¶

In :
import math
from numpy.linalg import solve

# GIVEN DATA
N1 = 1000.;               # 1st Test at No-load condition f1 Frequency, Speed in RPM
Vo1 = 250.;               # 1st Test at No-load condition f1 Frequency, Voltage in Volts
Io1 = 0.5;               # 1st Test at No-load condition f1 Frequency, Current in Amphere
Wo1= 230.;                # 1st Test at No-load condition f1 Frequency, Power in Watts

N2 = 900.;                # 2nd Test at No-load condition f2 Frequency, Speed in RPM
Vo2 = 225.;               # 2nd Test at No-load condition f2 Frequency, Voltage in Volts
Io2= 0.5;                # 2nd Test at No-load condition f2 Frequency, Current in Amphere
Wo2 = 200.;               # 2nd Test at No-load condition f2 Frequency, Power in Watts
p = 6.;                  # Number of poles of math.single phase alternator
N = 220.;                # Number of the turns of math.single phase alternator
R = 0.66;               # resistance of the math.single phase alternator in Ohms

# CALCULATIONS
f1 = (N1*p)/120;              # 1st case Supply Frequency in Hertz
Ratio1 = Vo1/f1;              # 1st case Ratio of the Volatge and Frequency in Volts/Hertz
f2 = (N2*p)/120;              # 2nd case Supply Frequency in Hertz
Ratio2 = Vo2/f2;              # 2nd case Ratio of the Volatge and Frequency in Volts/Hertz

c = (Wo1-(Io1**2)*R)/f1;         # No-load corrected losses Eq 1 in Watts
d = (Wo2-(Io2**2)*R)/f2;         # No-load corrected losses Eq 2 in watts

x = [[ 1, f1] ,[ 1, f2] ];            # No-load corrected losses Eq 1 in watts
y = [ [c] , [d] ];                  # No-load corrected losses Eq 2 in watts

E = solve(x,y);                        # Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
A = E;                     # Solution of constant A in Watts/Hertz
B = E;                     # Solution of constant B in watts/Hertz-Sqare
Ph = f1*A;                      # Hysteresis loss at 50 Hertz in Watts
Pe = (f1**2)*B;                  # Eddy current loss at 50 Hertz in Watts

# DISPLAY RESULTS

print ("EXAMPLE : 3.12 : SOLUTION :-") ;
print " a) Hysteresis loss at %.f Hertz , Ph = %.3f W  "%(f1,Ph);
print " b) Eddy current loss at %.f Hertz , Pe = % .2f W "%(f1,Pe);
EXAMPLE : 3.12 : SOLUTION :-
a) Hysteresis loss at 50 Hertz , Ph = 151.874 W
b) Eddy current loss at 50 Hertz , Pe =  77.96 W

Example 3.13 Page No : 59¶

In :
import math
from numpy.linalg import solve

# GIVEN DATA

N1 = 1500.;               # 1st Test on Transformer at f1 Frequency and Vo1 voltage, Speed in RPM
Vo1 = 250.;               # 1st Test on Transformer at f1 Frequency and Vo1 voltage, Voltage in Volts
Wo1= 55.;                 # 1st Test on Transformer at f1 Frequency and Vo1 voltage, Power in Watts
N2 = 1200.;               # 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Speed in RPM
Vo2 = 200.;               # 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Voltage in Volts
Wo2 = 40.;                # 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Power in Watts
p = 4.;                   # Number of poles of math.single phase alternator

# CALCULATIONS
f1 = (N1*p)/120;              # 1st case Supply Frequency in Hertz
Ratio1 = Vo1/f1;              # 1st case Ratio of the Volatge and Frequency in Volts/Hertz
f2 = (N2*p)/120;              # 2nd case Supply Frequency in Hertz
Ratio2 = Vo2/f2;              # 2nd case Ratio of the Volatge and Frequency in Volts/Hertz

c = Wo1/f1;         # No-load corrected losses Eq 1 in Watts
d = Wo2/f2;         # No-load corrected losses Eq 2 in watts

x = [ [1, f1] , [1, f2] ];             # No-load corrected losses Eq 1 in watts
y = [ [c] , [d] ];                   # No-load corrected losses Eq 2 in watts

E = solve(x,y);                        # Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
A = E;                      # Solution of constant A in Watts/Hertz
B = E;                      # Solution of constant B in watts/Hertz-Sqare
Ph1 = f1*A;                      # Hysteresis loss at 50 Hertz in Watts
Pe1 = (f1**2)*B;                  # Eddy current loss at 50 Hertz in Watts
Ph2 = f2*A;                      # Hysteresis loss at 40 Hertz in Watts
Pe2 = (f2**2)*B;                  # Eddy current loss at 40 Hertz in Watts

# DISPLAY RESULTS

print ("EXAMPLE : 3.13 : SOLUTION :-") ;
print " a.1) Hysteresis loss at %.f Hertz , Ph = %.f W  "%(f1,Ph1);
print " a.2) Eddy current loss at %.f Hertz , Pe = %.f W "%(f1,Pe1);
print " b.1) Hysteresis loss at %.f Hertz , Ph = %.f W  "%(f2,Ph2);
print " b.2) Eddy current loss at %.f Hertz , Pe = %.f W "%(f2,Pe2);
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) Hysteresis loss at %.f Hertz , Ph = 25 W instead of %.f W  "%(f2,Ph2);
EXAMPLE : 3.13 : SOLUTION :-
a.1) Hysteresis loss at 50 Hertz , Ph = 30 W
a.2) Eddy current loss at 50 Hertz , Pe = 25 W
b.1) Hysteresis loss at 40 Hertz , Ph = 24 W
b.2) Eddy current loss at 40 Hertz , Pe = 16 W
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) Hysteresis loss at 40 Hertz , Ph = 25 W instead of 24 W

Example 3.14 Page No : 62¶

In :
import math

# GIVEN DATA

S = 10 * 10 ** 3;        # Rating of the Single Transformer in VA
f = 50;                 # Frequency in Hertz
Pc = 110;               # Required input no-load at normal voltage in Watts (Core loss)
Psc = 120;              # Required input Short-circuit at full-load current in Watts (copper loss or short circuit loss)

# CALCUATIONS
# case (a) for Unity power factor

cos_theta1 = 1;                                                                  # Unity Power factor
K1 = 1.0;                                                                        # Full load
K2 = 0.5;                                                                        # Half load
eta_11 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+( K1 ** 2 )*Psc);          # Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage
eta_12 = 100 * (K2*S*cos_theta1)/((K2*S*cos_theta1)+Pc+( K2 ** 2 )*Psc);          # Efficiency at unity factor and half load ( beacuse taken k2 = 0.5 ) in percentage

# case (b) for 0.8 power factor lagging

cos_theta2 = 0.8;                                                                # 0.8 power factor lagging
eta_21 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+( K1 ** 2 )*Psc);          # Efficiency at 0.8 power factor lagging and full load ( beacuse taken k1 = 1 ) in percentage
eta_22 = 100 * (K2*S*cos_theta2)/((K2*S*cos_theta2)+Pc+( K2 ** 2 )*Psc);          # Efficiency at 0.8 power factor lagging and half load ( beacuse taken k2 = 0.5 ) in percentage

# Case (c) for 0.8 poer factor leading

eta_31 = eta_21;                         # Efficiency at 0.8 power factor leading and full load will be same as the Efficiency at 0.8 power factor lagging and full load in percentage
eta_32 = eta_22;                         # Efficiency at 0.8 power factor leading and half load will be same as the Efficiency at 0.8 power factor lagging and half load in percentage

# Case (d) Maximum Efficiency assumed that unity power factor
# Psc = Pc At Maximum Efficiency

eta_41 = 100 * (K1*S*cos_theta1)/((K1*S*cos_theta1)+Pc+Pc);          # Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage

# Case (e) Maximum Efficiency assumed that 0.8 power factor lagging
# Psc = Pc At Maximum Efficiency

eta_51 = 100 * (K1*S*cos_theta2)/((K1*S*cos_theta2)+Pc+Pc);          # Maximum Efficiency at unity factor and full load ( beacuse taken k1 = 1 ) in percentage

# Case (f) Maximum Efficiency assumed that 0.8 power factor leading
# Psc = Pc At Maximum Efficiency

eta_61 = eta_51;                                                     # Maximum Efficiency at 0.8 power factor leading and full load will be same as the Maximum Efficiency at 0.8 power factor lagging and full load in percentage
out1 = K1*S*cos_theta1;                                             # Output at which maximum efficiency occurs at unity power factor at full load in Watts
out2 = K1*S*cos_theta2;                                             # Output at which maximum efficiency occurs at 0.8 power factor lagging at full load in Watts
out3 = K1*S*cos_theta2;                                             # Output at which maximum efficiency occurs at unity power factor leading at full load in Watts

# DISPLAY RESULTS

print ("EXAMPLE : 3.14 : SOLUTION :-") ;
print " a.1) Efficiency at unity power factor and full load , eta = %.2f Percent  "%(eta_11);
print " a.2) Efficiency at unity power factor and half load , eta= % .2f Percent "%(eta_12);
print " b.1) Efficiency at 0.8 power factor lagging and full load  , eta = %.2f Percent  "%(eta_21);
print " b.2) Efficiency at 0.8 power factor lagging and half load  , eta= % .2f Percent "%(eta_22);
print " c.1) Efficiency at 0.8 power factor leading and full load  , eta = %.2f Percent  "%(eta_31);
print " c.2) Efficiency at 0.8 power factor leading and half load  , eta= % .2f Percent "%(eta_32);
print "  d)  Maximum Efficiency at unity power factor and full load , eta = %.2f Percent  "%(eta_41);
print "  e)  Maximum Efficiency at 0.8 power factor lagging and full load , eta = %.2f Percent  "%(eta_51);
print "  f)  Maximum Efficiency at 0.8 power factor leading and full load , eta = %.2f Percent  "%(eta_61);
print "  g)  Output at which maximum efficiency occurs at unity power factor at full load = %.f W  "%(out1);
print "  h)  Output at which maximum efficiency occurs at 0.8 power factor lagging at full load = %.f W  "%(out2);
print "  i)  Output at which maximum efficiency occurs at 0.8 power factor leading at full load = %.f W  "%(out3);
print " IN THE ABOVE PROBLEM MAXIMUM EFFICIENCY AND THE OUTPUT AT WHICH THE MAXIMUM EFFICIENCY OCCURS IS\
\n NOT CALCULATED IN THE TEXT BOOK "
EXAMPLE : 3.14 : SOLUTION :-
a.1) Efficiency at unity power factor and full load , eta = 97.75 Percent
a.2) Efficiency at unity power factor and half load , eta=  97.28 Percent
b.1) Efficiency at 0.8 power factor lagging and full load  , eta = 97.21 Percent
b.2) Efficiency at 0.8 power factor lagging and half load  , eta=  96.62 Percent
c.1) Efficiency at 0.8 power factor leading and full load  , eta = 97.21 Percent
c.2) Efficiency at 0.8 power factor leading and half load  , eta=  96.62 Percent
d)  Maximum Efficiency at unity power factor and full load , eta = 97.85 Percent
e)  Maximum Efficiency at 0.8 power factor lagging and full load , eta = 97.32 Percent
f)  Maximum Efficiency at 0.8 power factor leading and full load , eta = 97.32 Percent
g)  Output at which maximum efficiency occurs at unity power factor at full load = 10000 W
h)  Output at which maximum efficiency occurs at 0.8 power factor lagging at full load = 8000 W
i)  Output at which maximum efficiency occurs at 0.8 power factor leading at full load = 8000 W
IN THE ABOVE PROBLEM MAXIMUM EFFICIENCY AND THE OUTPUT AT WHICH THE MAXIMUM EFFICIENCY OCCURS IS
NOT CALCULATED IN THE TEXT BOOK

Example 3.15 Page No : 65¶

In :
import math

# GIVEN DATA
# Refer figure 3.17 page no. 101

S = 500. * 10 ** 6;                   # Rating of power transformer in VA
V1 = 400. * 10**3;                    # HV side rating of the power transformer in Volts
V2 = 131. * 10**3;                    # LV side rating of the power transformer in Volts
pcu = 5.;                            # Rated Copper loss in Percentage
pi = 1.;                             # Rated Core loss in Percentage

# CALCULATIONS

Pcu = S*(pcu/100);                              # Rated Copper loss in Watts
Pi = S*(pi/100);                                # Rated Core loss in Watts
kt = 0.25*3 + 0.75*3 + 1*3 + 0.5*3 + 1.0*3 + 0.25*6 + 1.0*3;            # From graph figure 3.17 page no. 101
out = S*kt;                                     # Output energy in kilo-watt-hour
kt2 = 0.54375;                                   # From graph figure 3.17 page no. 101
eloss = 24*Pi + S*kt2;                          # Energy required in losses in kilo-watt-hour { Energy required in losses = 24*Pi + sigma(copper loss * duration)}
eta = 100*(out/(out+eloss));                    # All day efficiency

# DISPLAY RESULTS

print ("EXAMPLE : 3.15: SOLUTION :-");
print " a) All day efficiency = %.2f percent "%(eta)
EXAMPLE : 3.15: SOLUTION :-
a) All day efficiency = 95.03 percent

Example 3.16 Page No : 68¶

In :
import math

# GIVEN DATA

S = 20. * 10 ** 3;                    # Rating of the Step-down Transformer in VA
f = 50.;                             # Frequency in Hertz
V = 200.;                            # Normally supplied Voltage of Step-down Transformer in Volts
Vsc = 100.;                          # Potential difference when Secondary being Short- Circuited in Volts
Isc = 10.;                           # Primary Current when Secondary being Short- Circuited in Amphere
Cos_theta_sc = 0.28;                # Power factor when Secondary being Short- Circuited

# CALCULATIONS

I = S/V;                                    # Rated primary current in Amphere
Wsc = Vsc * Isc * Cos_theta_sc;             # Power loss when Secondary being Short- Circuited in Watts
R = Wsc/(Isc ** 2);                          # resistance of Transformer referred to primary side in Ohms
Z = Vsc/Isc;                                # Referred Impedence in Ohms
X = math.sqrt((Z**2)-(R**2));                      # Leakage reactance referred to primary side in Ohms
Er = (I*R)/V;                               # Per unit resistance in Ohms
Ex = (I*X)/V;                               # Per unit reactance in Ohms
Cos_theta1 = 1.0;                           # Unity Power factor
Cos_theta2 = 0.6;                           # 0.6 Power factor Lagging
Cos_theta3 = 0.6;                           # 0.6 Power factor Leading
Sin_theta1 = 0.0;                           # Unity Power factor
Sin_theta2 = 0.8;                           # 0.6 Power factor Lagging
Sin_theta3 = 0.8;                           # 0.6 Power factor Leading
E1 = (Er*Cos_theta1)+(Ex*Sin_theta1);       # pu Regulation at Unity Power factor
E2 = (Er*Cos_theta2)+(Ex*Sin_theta2);       # pu Regulation at 0.6 Power factor Lagging
E3 = (Er*Cos_theta3)-(Ex*Sin_theta3);       # pu Regulation at 0.6 Power factor Leading

# DISPLAY RESULTS

print ("EXAMPLE : 3.16 : SOLUTION :-") ;
print " a) pu Regulation at Unity Power factor , E = %.1f  "%(E1);
print " b) pu Regulation at 0.6 Power factor Lagging , E= % .2f "%(E2);
print " c) pu Regulation at 0.6 Power factor Leading , E= % .2f "%(E3);
EXAMPLE : 3.16 : SOLUTION :-
a) pu Regulation at Unity Power factor , E = 1.4
b) pu Regulation at 0.6 Power factor Lagging , E=  4.68
c) pu Regulation at 0.6 Power factor Leading , E= -3.00

Example 3.17 Page No : 73¶

In :
import math

# GIVEN DATA

S = 500.;                            # Rating of the 3-Phase transformer in kVA
V1 = 11. * 10 ** 3;                   # Votage rating of the 3-Phase transformer on HV side in Volts
V2 = 400.;                           # Votage rating of the 3-Phase transformer on LV side in Volts
f = 60.;                             # Frequencyin Hertz
eta = 98.;                           # Maximum Efficency of the Transformer in Percentage Operating at 80% full load and Unity Power factor
K = 0.8;                            # Beacuse 80% Full load
x = 1.0;                            # Unity Power factor
Ex = 4.5;                           # Percentage impedance

# CALCULATIONS

Out = S * K * x;                      # Output in KiloWatts at 80% full load and Unity Power factor
Inp = Out/(eta/100);                  # Input in KiloWatts at full load and Unity Power factor
Total_loss = Inp - Out;               # Total loss at full load  in KiloWatts
Cu_loss = Total_loss/2;               # Copper loss in KiloWatts at 80% full load and Unity Power factor
Pcu = Cu_loss/(K **2 );                # Full load Copper loss in KiloWatts
Er = Pcu/S;                           # Per unit resistance
theta = math.degrees(math.atan((Ex/100)/Er));           # Power factor angle at secondary terminal voltage is minimum in Degrees
Pf =math.degrees(math.cos(math.radians(theta)));                     # Load power factor for minimum volatge of the secondary terminal

# DISPLAY RESULTS

print ("EXAMPLE : 3.17 : SOLUTION :-") ;
print " a) Load power factor for minimum volatge of the secondary terminal ( math.costheta) = %.4f lagging  "%(Pf);
EXAMPLE : 3.17 : SOLUTION :-
a) Load power factor for minimum volatge of the secondary terminal ( math.costheta) = 15.6248 lagging

Example 3.19 Page No : 74¶

In :
import math

# GIVEN DATA

Sa = 200;               # Rating of the TWO 1-Phase Transformer in kVA
Z1 = 0.005 + 0.08 * 1j  # Equivalent Impedance of the Transformer-1 in Per-Unit
Z2 = 0.0075 + 0.04 * 1j # Equivalent Impedance of the Transformer-2 in Per-Unit
P = 400;                # Total load in kiloWatts
Cos_theta = 1.0;        # Unity power factor

# CALCULATIONS

kVA = P/Cos_theta;       # kVA rating of the Transformer
S = kVA;                 # kVA rating of the Transformer
S1 = ( Z2/(Z1+Z2) )*S;   # Load shared by Transformer-1 in kVA
S2 = S - S1;             # Load shared by Transformer-2 in kVA

# DISPLAY RESULTS

print ("EXAMPLE : 3.19 : SOLUTION :-") ;
print " a) Load shared by Transformer-1 , S1 = %.2f+j%.2f) kVA  "%(S1.real,S1.imag);
print " b )Load shared by Transformer-2 , S2 = %.2f+j%.2f kVA  "%(S2.real,S2.imag);
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) S1 = -131.90)+j38.47)kVA instead of %.2f+j%.2f) kVA  "%(S1.real,S1.imag);
print "                                     b) S2 = 268.1)+j.38047)kVA  instead of %.2f+j%.2f kVA  "%(S2.real,S2.imag);
EXAMPLE : 3.19 : SOLUTION :-
a) Load shared by Transformer-1 , S1 = 134.48+j-10.99) kVA
b )Load shared by Transformer-2 , S2 = 265.52+j10.99 kVA
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) S1 = -131.90)+j38.47)kVA instead of 134.48+j-10.99) kVA
b) S2 = 268.1)+j.38047)kVA  instead of 265.52+j10.99 kVA

Example 3.20 Page No : 78¶

In :
import math

# GIVEN DATA

V1 = 110;               # Primary voltage of the Two Transformers the two primaries are connected in parallel in Volts
I1 = 2.0;               # Primary Current in Amphere
P1 = 40;                # Primary power intake in Watts
V2 = 28;                # secondary voltage of the Two Transformers the two secondary are connected in phase opposition in Volts
I2 = 6.8;               # secondary Current in Amphere
P2 = 180.;               # secondary power intake in Watts
a = 110./220;            # Turn ratio of the Transformer

# CALCULATIONS

theta_o = math.acos(math.radians((a*P1)/(a*I1*V1)));                      # Primary Power factor angle in Degrees
theta_sc = math.acos(math.radians((a*P2)/(a*I2*V2)));                    # Secondary Power factor angle in Degrees
i_sc = I2 * ( math.degrees(math.cos(math.radians(theta_sc)))-math.degrees(math.sin(math.radians(theta_sc)))* 1j);      # Secondary current in Amphere
I_sc = (1/a)*i_sc;                                     # referred Secondary current in each of the primary side in Amphere
It1 = Io + I_sc;                                       # LT winding current in the 1st Transformer in Amphere
It2 = Io - I_sc;                                       # LT winding current in the 2nd Transformer in Amphere
In1 = It1 + It2;                                       # The current flowing on paralel connected LT winding (This is same as total no-load current in the two Transforemer) in Amphere

# DISPLAY RESULTS

print ("EXAMPLE : 3.20 : SOLUTION :-") ;
print " a) LT  Primary ) winding current in the 1st Transformer  It1 = %.3f+j%.4f) A  "%(It1.real,It1.imag);
print " b) LT  Primary ) winding current in the 2nd Transformer  It2= %.3f+j%.5f A "%(It2.real,It2.imag);
print " c) LT winding are connected in parallel,  the current flowing on paralel connected LT winding\
\n In1 = %.3f+j%.5f) A "%(In1.real,In1.imag);
EXAMPLE : 3.20 : SOLUTION :-
a) LT  Primary ) winding current in the 1st Transformer  It1 = 836.210+j-22.7033) A
b) LT  Primary ) winding current in the 2nd Transformer  It2= -721.662+j19.56840 A
c) LT winding are connected in parallel,  the current flowing on paralel connected LT winding
In1 = 114.549+j-3.13485) A

Example 3.21 Page No : 83¶

In :
import math

# GIVEN DATA
# Refer figures 3.31(a), 3.31(b) and 3.31(c):- Page no. 121

VaH = 220.;                   # HV side Voltage of the two winding Transformer in volts for case(a)
VaL = 110.;                   # LV side Voltage of the two winding Transformer in volts for case(a)
VbH = 330.;                   # HV side Voltage of the two winding Transformer in volts for case(b)
VbL = 220.;                    # LV side Voltage of the two winding Transformer in volts for case(b)
VcH = 330;                    # HV side Voltage of the two winding Transformer in volts for case(c)
VcL = 110.;                    # LV side Voltage of the two winding Transformer in volts for case(c)
S = 1.5;                      # Ratings of the the two winding Transformer in kVA
I1 = 6.8;                     # Rated current in HV side in Amphere
I2 = 13.6;                    # Rated current in LV side in Amphere

# CALCULATIONS
# for case(a):- figure 3.31(b) page no. 121

IbH = I2;                        # Current of Auto-Transformer in HV side in Amphere
IbL = I1 + I2;                   # Current of Auto-Transformer in LV side in Amphere
KVA_b_L = (VbL*IbL)/1000;        # LV side kVA rating of the Auto-Transformer in kVA
KVA_b_H = (VbH*IbH)/1000;        # HV side kVA rating of the Auto-Transformer in kVA

# for case(b):- figure 3.31(c) page no. 121

IcH = I1;                         # Current of Auto-Transformer in HV side in Amphere
IcL = I1 + I2;                    # Current of Auto-Transformer in LV side in Amphere
KVA_c_L = (VcL*IcL)/1000;         # LV side kVA rating of the Auto-Transformer in kVA
KVA_c_H = (VcH*IcH)/1000;         # HV side kVA rating of the Auto-Transformer in kVA

# DISPLAY RESULTS

print ("EXAMPLE : 3.21 : SOLUTION :-") ;
print " a.1) Current of Auto-Transformer in HV side for case b) , IH = %.1f A  "%(IbH);
print "       Current of Auto-Transformer in LV side for case b), IL= % .1f A "%(IbL);
print " a.2) LV side kVA rating of the Auto-Transformer for case b), KVAL = % .3f kVA  "%(KVA_b_L);
print "       HV side kVA rating of the Auto-Transformer for case b), KVAH= % .3f kVA "%(KVA_b_H);
print " b.1) Current of Auto-Transformer in HV side for case c) , IH = %.1f A  "%(IcH);
print "       Current of Auto-Transformer in LV side for case c) , IL= % .1f A "%(IcL);
print " b.2) LV side kVA rating of the Auto-Transformer for case c), KVAL = % .3f kVA  "%(KVA_c_L);
print "       HV side kVA rating of the Auto-Transformer for case c)  KVAH= % .3f kVA "%(KVA_c_H);
EXAMPLE : 3.21 : SOLUTION :-
a.1) Current of Auto-Transformer in HV side for case b) , IH = 13.6 A
Current of Auto-Transformer in LV side for case b), IL=  20.4 A
a.2) LV side kVA rating of the Auto-Transformer for case b), KVAL =  4.488 kVA
HV side kVA rating of the Auto-Transformer for case b), KVAH=  4.488 kVA
b.1) Current of Auto-Transformer in HV side for case c) , IH = 6.8 A
Current of Auto-Transformer in LV side for case c) , IL=  20.4 A
b.2) LV side kVA rating of the Auto-Transformer for case c), KVAL =  2.244 kVA
HV side kVA rating of the Auto-Transformer for case c)  KVAH=  2.244 kVA

Example 3.22 Page No : 84¶

In :
import math

# GIVEN DATA

S = 10. * 10 ** 3;                  # Rating of the Two-winding Transformer in VA
V1 = 2000.;                        # HV side voltage of the Two-winding Transformer in Volts
V2 = 200.;                         # LV side voltage of the Two-winding Transformer in Volts
V_A_H = 2200.;                     # Two-winding Transformer is connected to auto transformer HV side in Volts
V_A_L = 200.;                      # Two-winding Transformer is connected to auto transformer LV side in Volts
f = 50.;                           # Frequency in Hertz

# CALCULATIONS
# for finding (a)

I2 = S/V2;                          # Rated LV side current of winding for Step-up Auto transformer in Amphere
I1 = S/V1;                          # Rated HV side current of winding for Step-up Auto transformer in Amphere
IaH = I2;                           # The HV side current in the Auto-Transformer for Full-load in Amphere
IaL = I2 + I1 ;                     # The LV side current in the Auto-Transformer for Full-load in Amphere
VL = V1;                            # LV side voltage in Volts
VH = V1 + V2;                       # HV side voltage in Volts
KVA_a_L = (VL*IaL)/1000;            # kVA rating of LV SIDE
KVA_a_H = (VH*IaH)/1000;            # kVA rating of HV SIDE

# For finding (b)

IbH = I1;                           # HV side Rated current through the Auto-Transformer in Amphere
IbL = I1 + I2;                      # LV side Rated current through the Auto-Transformer in Amphere
KVA_b_L = (V_A_L*IbL)/1000;         # kVA rating of LV SIDE as output Auto-Transformer
KVA_b_H = (V_A_H*IbH)/1000;         # kVA rating of HV SIDE as output Auto-Transformer

# case (c)

V = V1;                             # Voltage on the Secondary, if the Commom windings are open

# DISPLAY RESULTS

print ("EXAMPLE : 3.22 : SOLUTION :-") ;
print " a.1)  HV side Curent supplied by the common windings , IH = %.f A  "%(IaH);
print " a.2)  LV side Curent supplied by the common windings , IL= %.f A "%(IaL);
print " b.1)  KVA rating of LV SIDE as output Auto-Transformer , KVAL = %.f kVA  "%(KVA_b_L);
print " b.2)  KVA rating of HV SIDE as output Auto-Transformer , KVAH= %.f kVA "%(KVA_b_H);
print "  c)   Voltage on the Secondary, if the Commom windings are open , V = %.f V  "%(V);
EXAMPLE : 3.22 : SOLUTION :-
a.1)  HV side Curent supplied by the common windings , IH = 50 A
a.2)  LV side Curent supplied by the common windings , IL= 55 A
b.1)  KVA rating of LV SIDE as output Auto-Transformer , KVAL = 11 kVA
b.2)  KVA rating of HV SIDE as output Auto-Transformer , KVAH= 11 kVA
c)   Voltage on the Secondary, if the Commom windings are open , V = 2000 V

Example 3.23 Page No : 86¶

In :
import math
import cmath

# GIVEN DATA

S = 100.;                                  # Rating of the 3-Phase Transformer in kVA
VH = 11.;                                  # HV side voltage in kilo-Volts
VL = 440.;                                 # LV side voltage in Volts
Vl = 400.;                                 # Line voltage in Volts
ZA = 0.6;                                 # Line impedance in line A in Ohms
ZB = 0.6*(0.8 + 0.6 * 1j);                # Line impedance in line B in Ohms
ZC = 0.6*(0.5 - 0.866 * 1j);              # Line impedance in line C in Ohms

# CALCULATIONS

Vp = Vl/math.sqrt(3);                                                  # Phase voltage in Volts
VAB = Vl * cmath.exp( 1j * 0 * math.pi/180);                                # Line Voltage across line A and B in Volts
VBC = Vl * cmath.exp( 1j * (-120) * math.pi/180);                           # Line Voltage across line B and C in Volts
VCA = Vl * cmath.exp( 1j * 120 * math.pi/180);                              # Line Voltage across line C and A in Volts
VAN = (Vl/math.sqrt(3)) * cmath.exp( 1j * (-30) * math.pi/180);                  # Phase Voltage across line A and Neutral in Volts
VBN = (Vl/math.sqrt(3)) * cmath.exp( 1j * (-150) * math.pi/180);                 # Phase Voltage across line B and Neutral in Volts
VCN = (Vl/math.sqrt(3)) * cmath.exp( 1j * (90) * math.pi/180);                   # Phase Voltage across line C and Neutral in Volts
IA = VAN/ZA;                                                      # Line current in line A in Amphere
IB = VBN/ZB;                                                      # Line current in line B in Amphere
IC = VCN/ZC;                                                      # Line current in line C in Amphere
IN = IA + IB + IC ;                                               # Current in the Neutral in Amphere
Y = (1/ZA)+(1/ZB)+(1/ZC);                                         # Net Admitmath.tance in mho
VN = IN/Y;                                                        # Neutral Potential in Volts
VDA = VAN - VN;                                                   # Voltage drops across the ZA in Volts
VDB = VBN - VN;                                                   # Voltage drops across the ZB in Volts
VDC = VCN - VN;                                                   # Voltage drops across the ZC in Volts

# DISPLAY RESULTS

print ("EXAMPLE : 3.23 : SOLUTION :-") ;
print " a.1) Line current in line A  , IA = %.f<%.f A  "%(abs(IA),math.degrees(math.atan2(IA.imag,IA.real)));
print " a.2) Line current in line B  , IB = %.f<%.2f A "%(abs(IB),math.degrees(math.atan2(IB.imag,IB.real)));
print " a.3) Line current in line C  , IC = %.f<%.f A  "%(abs(IC),math.degrees(math.atan2(IC.imag,IC.real)));
print " b.1) Phase Voltage across line A and Neutral , VAN = %.f<%.f V "%(abs(VAN),math.degrees(math.atan2(VAN.imag,VAN.real)));
print " b.2) Phase Voltage across line B and Neutral , VBN = %.f<%.f V  "%(abs(VBN),math.degrees(math.atan2(VBN.imag,VBN.real)));
print " b.3) Phase Voltage across line C and Neutral , VCN = %.f<%.f V "%(abs(VCN),math.degrees(math.atan2(VCN.imag,VCN.real)));
print "  c)   Neutral Potential , VN = %.1f<%.2f V  "%(abs(VN),math.degrees(math.atan2(VN.imag,VN.real)));
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) IC = 385<-90.1 V instead of %.f<%.f A  "%(abs(IC),math.degrees(math.atan2(IC.imag,IC.real)));
print "                                     b) VN = 230.5<78.17 V  instead of %.1f<%.2f V  "%(abs(VN),math.degrees(math.atan2(VN.imag,VN.real)) );
print " From Calculation of the IC, rest all the Calculated values in the TEXT BOOK is WRONG because of the IC value is WRONGLY calculated and the same used for the further Calculation part ";
EXAMPLE : 3.23 : SOLUTION :-
a.1) Line current in line A  , IA = 385<-30 A
a.2) Line current in line B  , IB = 385<173.13 A
a.3) Line current in line C  , IC = 385<150 A
b.1) Phase Voltage across line A and Neutral , VAN = 231<-30 V
b.2) Phase Voltage across line B and Neutral , VBN = 231<-150 V
b.3) Phase Voltage across line C and Neutral , VCN = 231<90 V
c)   Neutral Potential , VN = 99.7<166.53 V
[ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
WRONGLY PRINTED ANSWERS ARE :- a) IC = 385<-90.1 V instead of 385<150 A
b) VN = 230.5<78.17 V  instead of 99.7<166.53 V
From Calculation of the IC, rest all the Calculated values in the TEXT BOOK is WRONG because of the IC value is WRONGLY calculated and the same used for the further Calculation part

Example 3.24 Page No : 91¶

In :
import math

# GIVEN DATA
VL = 11000.;                     # Line-line voltage of the 3 identical 1-phase Transformer in Volts
IL = 10.;                        # Line current of the 3 identical 1-phase Transformer in Amphere
a = 10.;                         # Ratio of trun per phase of the 3 identical 1-phase Transformer

# CALCULATIONS
# For case (a) STAR-STAR

VPp_a = VL/math.sqrt(3);               # Primary phase volatge in Volts
IPp_a = IL;                       # Primary phase current in Amphere
VSp_a = VPp_a/a;                  # Secondary phase voltage in Volts
ISp_a = a*IPp_a;                  # Secondary phase current in Amphere
ISl_a = ISp_a;                    # Secondary line current in Amphere
VSl_a = VSp_a*math.sqrt(3);            # Secondary line voltage in Volts
Out_a = math.sqrt(3)*VSl_a*ISl_a/1000; # Output in kVA

# For case (b) STAR-DELTA

VPp_b = VL/math.sqrt(3);               # Primary phase volatge in Volts
IPp_b = IL;                       # Primary phase current in Amphere
VSp_b = VPp_a/a;                  # Secondary phase voltage in Volts
ISp_b = a*IPp_b;                  # Secondary phase current in Amphere
ISl_b = math.sqrt(3)*ISp_b;            # Secondary line current in Amphere
VSl_b = VSp_b;                    # Secondary line voltage in Volts
Out_b = math.sqrt(3)*VSl_b*ISl_b/1000; # Output in kVA

# For case (c) DELTA-DELTA

VPp_c = VL;                       # Primary phase volatge in Volts
IPp_c = IL/math.sqrt(3);               # Primary phase current in Amphere
VSp_c = VPp_c/a;                  # Secondary phase voltage in Volts
ISp_c = a*IPp_c;                  # Secondary phase current in Amphere
ISl_c = math.sqrt(3)*ISp_c;            # Secondary line current in Amphere
VSl_c = VSp_c;                    # Secondary line voltage in Volts
Out_c = math.sqrt(3)*VSl_c*ISl_c/1000; # Output in kVA

# For case (d) DALTA-STAR

VPp_d = VL;                       # Primary phase volatge in Volts
IPp_d = IL/math.sqrt(3);               # Primary phase current in Amphere
VSp_d = VPp_d/a;                  # Secondary phase voltage in Volts
ISp_d = a*IPp_d;                  # Secondary phase current in Amphere
ISl_d = ISp_d;                    # Secondary line current in Amphere
VSl_d = math.sqrt(3)*VSp_d;            # Secondary line voltage in Volts
Out_d = math.sqrt(3)*VSl_d*ISl_d/1000; #Output in kVA

# DISPLAY RESULTS

print ("EXAMPLE : 3.24 : SOLUTION :-") ;
print " For STAR-STAR Connection  a.1) Secondary line voltage = %.f V  "%(VSl_a);
print " a.2) Secondary line current = % .f A "%(ISl_a);
print " a.3) Primary phase current = %.f A  "%(IPp_a);
print " a.4) Secondary phase current = %.f A "%(ISp_a);
print " a.5) Output = %.2f kVA  "%(Out_a);
print " For STAR-DELTA Connection   b.1) Secondary line voltage = % .f V "%(VSl_b);
print " b.2) Secondary line current = %.f A  "%(ISl_b);
print " b.3) Primary phase current = %.f A "%(IPp_b);
print " b.4) Secondary phase current = %.f A  "%(ISp_b);
print " b.5) Output = % .2f kVA "%(Out_b);
print " For DELTA-DELTA Connection  c.1) Secondary line voltage = %.f V  "%(VSl_c);
print " c.2) Secondary line current = %.2f A "%(ISl_c);
print " c.3) Primary phase current = %.2f A  "%(IPp_c);
print " c.4) Secondary phase current = %.1f A "%(ISp_c);
print " c.5) Output = %.1f kVA  "%(Out_c);
print " For DELTA-STAR Connection  d.1) Secondary line voltage = % .2f V "%(VSl_d);
print " d.2) Secondary line current = %.1f A  "%(ISl_d);
print " d.3) Primary phase current = %.2f A "%(IPp_d);
print " d.4) Secondary phase current  = %.1f A  "%(ISp_d);
print " d.5) Output = % .1f kVA "%(Out_d);
EXAMPLE : 3.24 : SOLUTION :-
For STAR-STAR Connection  a.1) Secondary line voltage = 1100 V
a.2) Secondary line current =  100 A
a.3) Primary phase current = 10 A
a.4) Secondary phase current = 100 A
a.5) Output = 190.53 kVA
For STAR-DELTA Connection   b.1) Secondary line voltage =  635 V
b.2) Secondary line current = 173 A
b.3) Primary phase current = 10 A
b.4) Secondary phase current = 100 A
b.5) Output =  190.53 kVA
For DELTA-DELTA Connection  c.1) Secondary line voltage = 1100 V
c.2) Secondary line current = 100.00 A
c.3) Primary phase current = 5.77 A
c.4) Secondary phase current = 57.7 A
c.5) Output = 190.5 kVA
For DELTA-STAR Connection  d.1) Secondary line voltage =  1905.26 V
d.2) Secondary line current = 57.7 A
d.3) Primary phase current = 5.77 A
d.4) Secondary phase current  = 57.7 A
d.5) Output =  190.5 kVA