Chapter 3 : Application of Distribution Transformers¶

Example 3.1 Page No : 118¶

In :
import math

# Variables
S = 25.*(10**3);         #Rating of the transformer in VA
#Values in per unit
Rt = 0.014;         #Resismath.tance of Transformer
Xt = 0.012;         #Reacmath.tance of transformer
Vh = 7200;         #High Voltage End in V
Vx = 120;         # Low Voltage End in V
Rb = (Vh**2)/S;         #Base Value of Resismath.tance
#Accroding to Lloyd's Formula

# Calculations
Zhx12 = (1.5*Rt)+(1j*1.2*Xt);         #Impedance referred to HV side when the winding x2x3 is shorted

n = Vh/Vx;         #Turns Ratio

Zhx13 = Rt+(1j*Xt);         #Use of Entire low voltage winding

#Impedances of the required terms in pu
A = (2*Zhx13)-Zhx12;
B = ((2*Zhx12)-(2*Zhx13))/(n**2);
C = B;

#Angle of Impedances
ta = math.degrees(math.atan(A.imag/A.real));
tb = math.degrees(math.atan(B.imag/B.real));

# Results
print 'The Circuit impedances on the high voltage side is %g/_%g ohm'%(abs(A*Rb),ta)
print 'Each of the Circuit impedances on the low voltage side is %g/_%g ohm'%(abs(B*Rb),tb)
The Circuit impedances on the high voltage side is 24.6366/_53.9017 ohm
Each of the Circuit impedances on the low voltage side is 0.0085248/_18.9246 ohm

Example 3.2 Page No : 119¶

In :
import math
from numpy import exp

# Variables
#Impedances from the previous example
Zh = 24.6437*exp(1j*53.9*math.pi/180);
Zl = 8.525*(10**-3)*exp(1j*18.9*math.pi/180);
#Voltages
Vh = 7200.;         #High End
Vx = 120.;         # Low End
S = 25.*1000;         #Transformer Rating in VA
N = Vh/Vx;         #Turns Ratio

# Calculations
#R of service drop is zero         #Line to Neutral Currents
IfLVn = Vx/(Zl+((1/(N**2))*Zh));         #Secondary Fault Current
IfHVn = IfLVn/N;         #Primary Fault Current

#R of service drop is zero         #Line to Line Currents
Nl = Vh/(2*Vx);         #New Truns Ratio
IfLVl = 2*Vx/((2*Zl)+((1/(Nl**2))*Zh));         #Secondary Fault Current
IfHVl = IfLVl/Nl;         #Primary Fault Current

# Results
print 'a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop\
is zero are %g A and %g A respectively'%(abs(IfHVn),abs(IfLVn))
print 'b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are\
%g A and %g A respectively'%(abs(IfHVl),abs(IfLVl))
print 'c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is %g A'%(abs(IfLVn))
a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop is zero are 136.353 A and 8181.2 A respectively
b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are 188.283 A and 5648.5 A respectively
c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is 8181.2 A

Example 3.3 Page No : 121¶

In :
import math
from sympy import solve,Symbol
from numpy import exp

# Variables
Vx = 120.;         #Low End Voltage
#When Service drop is Zero
IfLVn = 8181.7*exp(-1*1j*34.3*math.pi/180);         #Line to Neutral Vault Current
IfLVl = 5649*exp(-1*1j*40.6*math.pi/180);         #Line to Line Fault Current

Ral4 = 2.58;         #        #4 AWG Aluminium Conductor Resismath.tance per mile
Ralinf = 1.03;         #        #1/0 AWG Aluminium Conductor Resismath.tance per mile

# Calculations
#Impedances when Service drop is zero, suffix l denotes line to line
#Suffix n denotes line to neutral
Zl0 = (2*Vx)/IfLVl;
Zn0 = (Vx)/IfLVn;

#When there is R service drop
#Magnitudes of Line to Line and Line to Earth fault currents are equal

R = Symbol('R');         #Variable Value
#Effective Impedances
Zl = Zl0+(2*R);
Zn = Zn0+(2*R);
#Fault Currents
Ifl = 2*Vx/Zl;
Ifn = Vx/Zn;
#print Ifl
#Magnitudes of Currents
MIfl = abs(240.)/abs(Ifl.subs(R,3));
MIfn = abs(Ifn.subs(R,2))/abs(Ifn.subs(R,3));
DI = MIfl-MIfn;
X = - 1.5781966 + 240*R   #DI.subs(R,2);         #Polynomial Equation to find 'R'
R = solve(X);         #Numerical Value

#The Magnitude of R found is Wrong in the Textbook

#Length of service drop cable
SDL4 = R/Ral4;
SDLinf = R/Ralinf;

# Results
print 'a) The Value of Service drop in the Cable is %g ohm'%(R)
print 'b The Length of service drop cable for:'
print 'i)         #4 AWG Conductor is %g miles'%(SDL4)
print 'ii)         #1/0 AWG Conductor is %g miles'%(SDLinf)

#Length is printed in Miles
a) The Value of Service drop in the Cable is 0.00657582 ohm
b The Length of service drop cable for:
i)         #4 AWG Conductor is 0.00254877 miles
ii)         #1/0 AWG Conductor is 0.00638429 miles

Example 3.4 Page No : 122¶

In :
# Variables
#Transformer Ratings in kVA
Sr1 = 250.;
Sr2 = 500.;

#percentage impedances
Zr1 = 2.4;
Zr2 = 3.1;

# Calculations
R = Sr1*Zr2/(Sr2*Zr1);

#If 500 kVA is chosen as the full load transformer, Transformer 1 becomes overloaded

SL2 = SL1/R;         #Maximum Load on transformer 2

# Results

Example 3.5 Page No : 127¶

In :
import math

# Variables
#Considering Van as reference voltage
SL3phi = 200*(10**3);         #Load to be powered
pf3 = 0.8;         #Power Factor of three phase load
t3 = math.acos(pf3);         #Power FActor Angle for three phase load
pf1 = 0.9;         #Power Factor of math.single phase load
t1 = math.acos(pf1);         #Power Factor angle of math.single phase load
SL1 = 80.*(10**3);         #Single Phase Light Load
Vll = 240.;         #Secondary Voltage
#Rating of Single Phase Transformers between individual lines
Sbc = 100.*(10**3);
Sab = 75.*(10**3);
Sca = Sab;
#Angles of Three phase voltages
ta = 0.;
tb = -120.;
tc = 120.;
#Angles of three  line currents
tai = ta-t3;
tbi = tb-t3;
tci = tc-t3;

# Calculations
I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current
#3 Phase Line Currents
Ia3 = I*exp(1j*math.pi*tai/180);
Ib3 = I*exp(1j*math.pi*tbi/180);
Ic3 = I*exp(1j*math.pi*tci/180);

MIbc = SL1/Vll;         #Magnitude Single Phase Current

tbc = -90;         #Lagging Van         #Angle of Vbc
tbci = tbc-t1;         #Angle of Current Ibc
Ibc = MIbc*exp(1j*math.pi*tbci/180);

Ia = Ia3;
Ta = math.degrees(math.atan(Ia.imag/Ia.real));
Ib = Ib3+Ibc;
Tb = math.degrees(math.atan(Ib.imag/Ib.real));
Ic = Ic3-Ibc;         #Current is wrong in the textbook
Tc = math.degrees(math.atan(Ic.imag/Ic.real));

#Current Flowing in the secondary winding of the transformers 1,2 and 3
Iac = ((Ic/Sbc)-(Ia/Sab))/((1/Sab)+(1/Sbc)+(1/Sca));
T1 = math.degrees(math.atan(Iac.imag/Iac.real));         #Angle of the above current
Iba = ((Ia/Sca)-(Ib/Sbc))/((1/Sab)+(1/Sbc)+(1/Sca));
T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current
Icb = ((Ib/Sab)-(Ic/Sca))/((1/Sab)+(1/Sbc)+(1/Sca));
T3 = math.degrees(math.atan(Icb.imag/Icb.real));         #Angle of the above current

SLab = Vll*abs(Iba)/1000;
SLbc = Vll*abs(Icb)/1000;
SLca = Vll*abs(Iac)/1000;

Vlls = Vll;         #Secondary Voltage
Vllp = 7620;         #Primary Voltage
n = Vllp/Vlls;         #Turns Ratio

#Primary Currents of the transformer
IAC = Iac/n;
IBA = Iba/n;
ICB = Icb/n;

#Primary Current in each each phase wire
IA = IAC-IBA;
TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current
IB = IBA-ICB;
TB = math.degrees(math.atan(IB.imag/IB.real));        #Angle of the above current
IC = ICB-IAC;
TC = math.degrees(math.atan(IC.imag/IC.real));        #Angle of the above current

# Results
print 'a The Line Currents flowing in secondary phase wire :'
print 'A phase is %g/_%g A'%(abs(Ia),Ta)
print 'B phase is %g/_%g A'%(abs(Ib),Tb)
print 'C phase is %g/_%g A'%(abs(Ic),Tc)
print 'b The Current flowing in secondary winding of each transformer:'
print 'AC is %g/_%g A'%(abs(Iac),T1)
print 'AB is %g/_%g A'%(abs(Iba),T2)
print 'BC is %g/_%g A'%(abs(Icb),T3)
print 'c The Load on Each Transformer is:'
print '1 : %g kVA'%(SLca)
print '2 : %g kVA'%(SLab)
print '3 : %g kVA'%(SLbc)
print 'd The Current flowing in primary winding of each transformer:'
print 'AC is %g/_%g A'%(abs(IAC),T1)
print 'AB is %g/_%g A'%(abs(IBA),T2)
print 'BC is %g/_%g A'%(abs(ICB),T3)
print 'e The Line Currents flowing in primary phase wire :'
print 'A phase is %g/_%g A'%(abs(IA),TA)
print 'B phase is %g/_%g A'%(abs(IB),TB)
print 'C phase is %g/_%g A'%(abs(IC),TC)

#Ic is calculation is wrong, the author has added Ibc instead of subtracting, so if you change - into + in line 45, you get the answer as in the textbook
a The Line Currents flowing in secondary phase wire :
A phase is 481.125/_-0.643501 A
B phase is 787.293/_71.6504 A
C phase is 787.977/_-72.7823 A
b The Current flowing in secondary winding of each transformer:
AC is 316/_-40.9814 A
AB is 315.535/_39.7662 A
BC is 545.454/_89.442 A
c The Load on Each Transformer is:
1 : 75.84 kVA
2 : 75.7283 kVA
3 : 130.909 kVA
d The Current flowing in primary winding of each transformer:
AC is 9.95276/_-40.9814 A
AB is 9.9381/_39.7662 A
BC is 17.1796/_89.442 A
e The Line Currents flowing in primary phase wire :
A phase is 15.1536/_-0.643501 A
B phase is 24.7966/_71.6504 A
C phase is 24.8182/_-72.7823 A

Example 3.6 Page No : 140¶

In :
import math

# Variables
#Considering Van as reference voltage
SL3phi = 100.*(10**3);         #Load to be powered
pf3 = 0.8;         #Power Factor of three phase load
t3 = math.acos(pf3);         #Power FActor Angle for three phase load
pf1 = 0.9;         #Power Factor of math.single phase load
t1 = math.acos(pf1);         #Power Factor angle of math.single phase load
SL1 = 50.*(10**3);         #Single Phase Light Load
Vll = 240.;         #Secondary Voltage
#Angles of Three phase voltages
ta = 0.;
tb = -120.;
tc = 120.;
#Angles of three  line currents
tai = ta-t3;
tbi = tb-t3;
tci = tc-t3;

# Calculations
I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current
#3 Phase Line Currents
Ia3 = I*exp(1j*math.pi*tai/180);
Ib3 = I*exp(1j*math.pi*tbi/180);
Ic3 = I*exp(1j*math.pi*tci/180);

MI1 = SL1/Vll;         #Magnitude Single Phase Current

t1v = 30;         #Leading Van         #Angle of Vbc
t1i = t1v-t1;         #Angle of Current Ibc
I1 = MI1*exp(1j*math.pi*t1i/180);

Ia = Ia3+I1;
Ta = math.degrees(math.atan(Ia.imag/Ia.real));
Ib = Ib3-I1;
Tb = -180+(math.degrees(math.atan(Ib.imag/Ib.real)));
Ic = Ic3;         #Current is wrong in the textbook
Tc = math.degrees(math.atan(Ic.imag/Ic.real));

#Current flowing in the secondary of the transformer
Iba = Ia;
T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current
Icb = Ic;
T3 = 180+(math.degrees(math.atan(Icb.imag/Icb.real)));         #Angle of the above current

SLba = Vll*abs(Iba)/1000;
SLcb = Vll*abs(Icb)/1000;

Vlls = Vll;         #Secondary Voltage
Vllp = 7620;         #Primary Voltage
n = Vllp/Vlls;         #Turns Ratio

#Primary Currents of the transformer
IA = Iba/n;
TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current
IB = Icb/n;
TB = T3;        #Angle of the above current
IN = IA+IB;         #Neutral Current
TN = math.degrees(math.atan(IN.imag/IN.real));        #Angle of the above current

# Results
print 'a The Line Currents flowing in secondary phase wire :'
print 'A phase is %g/_%g A'%(abs(Ia),Ta)
print 'B phase is %g/_%g A'%(abs(Ib),Tb)
print 'C phase is %g/_%g A'%(abs(Ic),Tc)
print 'b The Current flowing in secondary winding of each transformer:'
print 'AB is %g/_%g A'%(abs(Iba),T2)
print 'BC is %g/_%g A'%(abs(Icb),T3)
print 'c The Load on Each Transformer is:'
print '1 : %g kVA'%(SLba)
print '2 : %g kVA'%(SLcb)
print 'd The Line Currents flowing in primary phase wire :'
print 'AB is %g/_%g A'%(abs(IA),TA)
print 'CB is %g/_%g A'%(abs(IB),TB)
print 'The Neutral Current is %g/_%g'%(abs(IN),TN)

#Note the mistake in the Textbook for the calulation for Neutral Current
a The Line Currents flowing in secondary phase wire :
A phase is 433.486/_13.3432 A
B phase is 433.874/_-134.453 A
C phase is 240.563/_-60.6435 A
b The Current flowing in secondary winding of each transformer:
AB is 433.486/_13.3432 A
BC is 240.563/_119.356 A
c The Load on Each Transformer is:
1 : 104.037 kVA
2 : 57.735 kVA
d The Line Currents flowing in primary phase wire :
AB is 13.6531/_13.3432 A
CB is 7.57678/_119.356 A
The Neutral Current is 13.6653/_45.5475

Example 3.8 Page No : 152¶

In :
import math
from numpy import exp

# Variables
Vll = 480.;         #Line to Line Voltage
Vln = 277.;         #Line to neutral Voltage

# Calculations
#From the Phasor Diagram from the result file
Vab = Vll*exp(1j*0);         #Vab is taken as reference
Vabh = 50*Vab/100;
VAB = 4160.;
VABh = 50*VAB/100;
VH1H2o = math.sqrt((VAB**2)-(VABh**2));
VH1H2t = (VABh);
Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;
Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;
VH2H3t = (VABh);
Vx1x2t = Vabh;
Vx2x3t = Vabh;

# Results
print 'a The Phasor diagram is shown in the result file attached to the code'
print 'b) Vab is %g/_%g V'%(abs(Vab),Vab.imag/Vab.real)
print 'c The Magnitudes of the following rated winding voltages'
print 'i) The Voltage VH1H2 on transformer 1 : %g V'%(VH1H2o.real)
print 'ii) The Voltage Vx1x2 on transformer 1 : %g V'%(Vx1x2o.real)
print 'iii) The Voltage Vx2x3 on transformer 1 : %g V'%(Vx2x3o.real)
print 'iv) The Voltage VH1H2 on transformer 2 : %g V'%(VH1H2t.real)
print 'v) The Voltage VH2H3 on transformer 2 : %g V'%(VH2H3t.real)
print 'vi) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx1x2t.real)
print 'vii) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx2x3t.real)
print 'd i NO ii NO iii YES'
a The Phasor diagram is shown in the result file attached to the code
b) Vab is 480/_0 V
c The Magnitudes of the following rated winding voltages
i) The Voltage VH1H2 on transformer 1 : 3602.67 V
ii) The Voltage Vx1x2 on transformer 1 : 138.564 V
iii) The Voltage Vx2x3 on transformer 1 : 277.128 V
iv) The Voltage VH1H2 on transformer 2 : 2080 V
v) The Voltage VH2H3 on transformer 2 : 2080 V
vi) The Voltage Vx1x2 on transformer 2 : 240 V
vii) The Voltage Vx1x2 on transformer 2 : 240 V
d i NO ii NO iii YES

Example 3.9 Page No : 154¶

In :
import math
from numpy import exp

# Variables
R = 2.77;         #Resismath.tance of the balanced load
#From Phasor Diagram in Result file
Vab = 480*exp(1j*0);         #Reference Voltage
MVn = abs(Vab)/math.sqrt(3);         #Magnitude of line to neutral voltages
#Angles of Three phase voltages
ta = -30.;
tb = -150.;
tc = 90.;

# Calculations
#Angles of Winding according to the Line Currents
tx1x2 = -30;         #Lagging

I = MVn/R;         #Magnitude of current

#Low Voltage Current Phasors
Ia = I*exp(1j*math.pi*ta/180);
Ib = I*exp(1j*math.pi*tb/180);
Ic = I*exp(1j*math.pi*tc/180);
pfT = math.cos(math.radians(ta-ta));         #Angle of Ia is same as phase voltage         #Resismath.tance load

# Results
print 'a The Low voltage current phasors are:'
print 'A is %g/_%g A'%(abs(Ia),ta)
print 'B is %g/_%g A'%(abs(Ib),tb)
print 'C is %g/_%g A'%(abs(Ic),tc)
print 'b The Phasor Diagram is the ''b'' diagram of in the result file'
print 'c) The Power Factor of the Transformer is %g'%(pfT)
print 'd) Power Factor as seen by winding x3x2 of transformer 2 is %g leading'%(math.cos(math.radians(tx3x2)))
print 'e) Power Factor as seen by winding x1x2 of transformer 2 is %g lagging'%(math.cos(math.radians(tx1x2)))
a The Low voltage current phasors are:
A is 100.046/_-30 A
B is 100.046/_-150 A
C is 100.046/_90 A
b The Phasor Diagram is the b diagram of in the result file
c) The Power Factor of the Transformer is 1
d) Power Factor as seen by winding x3x2 of transformer 2 is 0.866025 leading
e) Power Factor as seen by winding x1x2 of transformer 2 is 0.866025 lagging

Example 3.10 Page No : 156¶

In :
import math
from numpy import exp

#'o' and 't' represent transformer one and two respectively
# Variables
#Objective is to find the Factor which has to be multiplied to get VA rating
#From the Phasor Diagram from the result file
Vab = Vll*exp(1j*0);         #Vab is taken as reference
Vabh = 50.*Vab/100;
Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;
Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;
Vx1x2t = Vabh;
Vx2x3t = Vabh;

#Let I be unity
I = 1;

# Calculations
#VA Ratings of the respective windings
Sx1x2o = Vx1x2o*I;
Sx2x3o = Vx2x3o*I;
Sx1x2t = Vx1x2t*I;
Sx2x3t = Vx2x3t*I;

#Total VA rating of transformer
S1 = Sx1x2o+Sx2x3o;
S2 = Sx1x2t+Sx2x3t;

#Ratio of total rating to maximum rating
Rt = (S1+S2)/(math.sqrt(3)*Vll*I);

# Results
print 'a) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx1x2o.real)
print 'b) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx2x3o.real)
print 'c) The Total Output from transformer 1 is %g*VI VA'%(S1.real)
print 'd) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx1x2t.real)
print 'e) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx2x3t.real)
print 'f) The Total Output from transformer 2 is %g*VI VA'%(S2.real)
print 'g) The Total Rating to the Maximum Continous Output is %g'%(Rt.real)
a) The voltampere raing of x1x2 of transformer 1 is 0.288675*VI VA
b) The voltampere raing of x1x2 of transformer 1 is 0.57735*VI VA
c) The Total Output from transformer 1 is 0.866025*VI VA
d) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA
e) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA
f) The Total Output from transformer 2 is 1*VI VA
g) The Total Rating to the Maximum Continous Output is 1.07735

Example 3.11 Page No : 157¶

In :
import math
from numpy import exp

# Variables
#Per unit value
Zt = 0.01+(1j*0.03);         #Transformer impedance

Vll = 240.;         #Secondary Voltage

pfl = 0.9;
tl = math.acos(pfl);
pf = 0.8;
t = math.acos(pf);

# Calculations
def angle(y):
return math.degrees(math.atan(y.imag/y.real))

tab = 30;         #Phase angle of Vab

Il = Sl*1000/Vll;         #Magnitude of Light Load
#Umath.sing the symmetrical - components theory
Ia1 = Il*exp(1j*math.pi*(tab-tl)/180);
Ta1 = angle(Ia1);         #Angle for the above current
Ib1 = -1*Ia1;
Ic1 = 0;         #Neutral Wire
#Angles of three line to line voltages
ta = 0;
tb = -120;
tc = 120;

Ib = S*1000/(math.sqrt(3)*Vll);         #Magnitude of balanced load

Ta2 = ta-t;
Ia2 = Ib*exp(1j*math.pi*Ta2/180);
Tb2 = tb-t;
Ib2 = Ib*exp(1j*math.pi*Tb2/180);
Tc2 = tc-t;
Ic2 = Ib*exp(1j*math.pi*Tc2/180);

#Currents in phase wire
Ia = Ia1+Ia2;
Ta = angle(Ia);         #Angle corresponding to the above angle
Ib = Ib1+Ib2;
Tb = angle(Ib);         #Angle corresponding to the above angle
Ic = Ic1+Ic2;
Tc = angle(Ic);         #Angle corresponding to the above angle

ST1 = Vll*abs(Ia)/1000;
T1 = 100;         #From the above value of Load, this transformer is chosen to meet the specific characteristic
ST1pu = ST1/T1;         #Per unit Load
ST2 = Vll*abs(Ic)/1000;
T2 = 15;         #From the above value of Load, this transformer is chosen to meet the specific characteristic
ST2pu = ST2/T2;         #Per unit Load

#Transformer Power Factors
pfT2 = math.cos(math.radians(90-Tc));         #Vcb makes angle of 90

Vh = 7200;         #High End Voltage
n = Vh/Vll;         #Turns Ratio

# The Primary Line Currents
IA = Ia/n;
IB = -1*Ic/n;
IN = -1*(IA+IB);

Ibase = T1*1000/Vll;         #Base Current
Iapu = Ia/Ibase;
Icpu = Ic/Ibase;

#Phase Voltages
Vab = Vll*exp(1j*math.pi*tab/180);
Vbc = Vll*exp(-1*1j*math.pi*90/180);
#Per Unit Voltages
VANpu = (Vab/Vll)+(Iapu*Zt);
VBNpu = (Vbc/Vll)-(Icpu*Zt);

#Actual Voltages
VAN = VANpu*Vh;
VBN = VBNpu*Vh;

# Results
print 'a The Phasor Currents:'
print 'Ia is %g/_%g A'%(abs(Ia),Ta)
print 'Ib is %g/_%g A'%(abs(Ib),180+Tb)
print 'Ic is %g/_%g A'%(abs(Ic),Tc)
print 'b) Suitable ratings of the transformers are %g kVA and %g kVA'%(T1,T2)
print 'c) The Per Unit kVA load on each transformer is %g pu and %g pu'%(ST1pu,ST2pu)
print 'd) The power factor of output of each transformer is %g and %g both lagging'%(pfT1,pfT2)
print 'e The phasor currents at the high voltage leads:'
print 'IA is %g/_%g A'%(abs(IA),Ta)
print 'IB is %g/_%g A'%(abs(IB),180+angle(IB))
print 'IN is %g/_%g A'%(abs(IN),angle(IN))
print 'f) VAN is %g/_%g V and VBN is %g/_%g V'%(abs(VAN),angle(VAN),abs(VBN),angle(VBN))

#Highly Accuracy of Answers; Upto 5 decimal Places
a The Phasor Currents:
Ia is 428.052/_25.4972 A
Ib is 428.229/_213.552 A
Ic is 60.1407/_-60.6435 A
b) Suitable ratings of the transformers are 100 kVA and 15 kVA
c) The Per Unit kVA load on each transformer is 1.02732 pu and 0.96225 pu
d) The power factor of output of each transformer is 0.996914 and -0.871586 both lagging
e The phasor currents at the high voltage leads:
IA is 14.2684/_25.4972 A
IB is 2.00469/_119.356 A
IN is 14.5415/_17.5913 A
f) VAN is 7294.34/_31.6923 V and VBN is 7193.85/_-89.743 V