CHAPTER02 : TRANSFORMER PRINCIPLES

Example E01 : Pg 42

In [1]:
#  Example 2.1
#  Computation of peak value of sinusoidal flux in a transformer
#  Page No. 42
#  Given data
Ep=240.;                #  Voltage in primary coil
Np=200.;                #  Number of turns in primary coil of transformer
f=60.;                  #  Frequency of source
#  Peak value of sinusoidal flux in a transformer
phimax=Ep/(4.44*Np*f);      
# Display result on command window
# print"\n Peak value of sinusoidal flux in a transformer = %0.4f WB ",phimax);
print'Peak value of sinusoidal flux in a transformer =',phimax,'Wb'
Peak value of sinusoidal flux in a transformer = 0.0045045045045 Wb

Example E02 : Pg 42

In [2]:
#  Example 2.2
#  Computation of (a) Turns ratio (b) Number of turns in each winding
#  (c) Magnetizing current
#  Page No. 42
Ep=2400.;               #  Induced emf in primary winding
Es=240.;                #  Induced emf in primary winding
Bmax=1.5;              #  Maximum flux density
A=50.*10.**-4.;            #  Cross section area
f=60.;                  #  Frequency
l=0.667;               #  Mean length of core
H=450.;                 #  Magnetic field intensity
#  (a) Turns ratio
Ts=Ep/Es;      
#  (b) Number of turns in each winding
phimax=Bmax*A;
Np=Ep/(4.44*f*phimax);                #  Number of primary windings
Ns=Np/Ts;                             #  Number of secondary windings
# (c) Magnetizing current
Im=H*l/Np;
# Display result on command window
print"Turns ratio =",Ts
print"Number of primary windings =",Np,"turns"
print"Number of secondary windings =",Ns,"turns"
print"Magnetizing current =",Im,"A"
Turns ratio = 10.0
Number of primary windings = 1201.2012012 turns
Number of secondary windings = 120.12012012 turns
Magnetizing current = 0.249874875 A

Example E03 : Pg 44

In [3]:
#  Example 2.3
#  Computation of (a) Exciting current and its quadrature components 
#  (b) Equalizing magnetic reactance and equivalent core loss resistance
#  (c) Magnetizing current (d)repeat (a) and (b) for the transformer in the 
#  step up mode
# Page No. 44
Fp=0.210;               #  Power factor
Pcore=138.;              #  Active power
VT=2400.;                #  Voltage applied to primary
VT1=240.;                #  240-V primary voltage -- Second case
#  (a)Exciting current and its quadrature components
Theta=77.9;#acosd(Fp);         #  Angle
Thetai=-Theta;           #  As phase angle of applied voltage is zero
Ife=Pcore/VT;            #  Exciting current
I0=Ife/Fp;               #  Quadrature component
Im=0.268;#tand(Thetai)*Ife;     #  Quadrature component
Im=Im*-1.;
#  (b) Equalizing magnetic reactance and equivalent core loss resistance
XM=VT/Im;                  #  Magnetic reactance
Rfe=VT/Ife;                #  Core-loss resistance
XM=XM/1000.;
Rfe=Rfe/1000.;
# (c) Magnetizing current
Ife1=Pcore/VT1;            #  Exciting current
I01=2.74;#Ife1/cosd(Thetai);
IM1=2.68;#tand(Thetai)*Ife1;     #  Quadrature component
IM1=IM1*-1.;
# (d) repeat (a) and (b) for the transformer in the step up mode
XM1=VT1/IM1;              #  Magnetizing reactance
Rfe1=VT1/Ife1;            #  Core-loss resistance
# Display result on command window
print"Exciting current =",Ife,"A"
print"Exciting current quadrature component 1 =",I0,"A"
print"Exciting current quadrature component 2 =",Im,"A"
print"Equivalent magnetic reactance =",XM,"kOhm"
print"Equivalent core loss resistance =",Rfe,"kOhm"
print"Exciting current in step-up mode =",Ife1,"A"
print"Exciting current in step-up mode quadrature component 1 =",I01,"A"
print"Exciting current in step-up mode quadrature component 2 =",IM1,"A"
print"Equivalent  magnetic reactance in the step up mode  =",XM1,"Ohm"
print"Equivalent core loss resistance in the step up mode =",Rfe1,"Ohm"
Exciting current = 0.0575 A
Exciting current quadrature component 1 = 0.27380952381 A
Exciting current quadrature component 2 = -0.268 A
Equivalent magnetic reactance = -8.9552238806 kOhm
Equivalent core loss resistance = 41.7391304348 kOhm
Exciting current in step-up mode = 0.575 A
Exciting current in step-up mode quadrature component 1 = 2.74 A
Exciting current in step-up mode quadrature component 2 = -2.68 A
Equivalent  magnetic reactance in the step up mode  = -89.552238806 Ohm
Equivalent core loss resistance in the step up mode = 417.391304348 Ohm

Example E04 : Pg 51

In [4]:
#  Example 2.4
#  Computation of (a) Secondary voltage (b) Load current
#  (c) Input current to the primary (d) Input impedance looking into the primary terminals
#  Page No. 51
NHS=200.;               #  Number of turns in primary
NLS=20.;                #  Number of turns in secondary
E=120.;                 #  Primary voltage magnitude
ES_Mag=12.;             #  Secondary voltage magnitude
ES_Ang=0.;              #  Secondary voltage angle
Zload_Mag=100.;         #  Load magnitude
Zload_Ang=30.;          #  Load angle 
f=60.;                  #  Frequency

#  (a)  Secondary voltage
a=NHS/NLS;
ELS=E/a;      

#  (b) Load current
IS_Mag=ES_Mag/Zload_Mag;         #  Load current magnitude
IS_Ang=ES_Ang - Zload_Ang;       #  Load current angle

# (c) Input current to the primary
Ip_Mag=IS_Mag/a;                 #  Input current to the primary magnitude
Ip_Ang=IS_Ang;                   #  Input current to the primary angle

# (d) Input impedance looking into the primary terminals
Zin_Mag=a**2.*Zload_Mag;           #  Input impedance magnitude 
Zin_Ang=Zload_Ang;               #  Input impedance angle
Zin_Mag=Zin_Mag/1000.;

#  Display result on command window
print"Turns ratio =",a
print"Secondary voltage =",ELS,"V"
print"Load current magnitude =",IS_Mag,"A"
print"Load current angle =",IS_Ang,"deg"
print"Input current to the primary magnitude =",Ip_Mag,"A"
print"Input current to the primary angle =",Ip_Ang,"deg"
print"Input impedance magnitude =",Zin_Mag,"KOhm"
print"Input impedance angle =",Zin_Ang,"deg"
Turns ratio = 10.0
Secondary voltage = 12.0 V
Load current magnitude = 0.12 A
Load current angle = -30.0 deg
Input current to the primary magnitude = 0.012 A
Input current to the primary angle = -30.0 deg
Input impedance magnitude = 10.0 KOhm
Input impedance angle = 30.0 deg

Example E05 : Pg 60

In [5]:
#  Example 2.5
#  Computation of (a) Equivalent impedance of the transformer referred to the 
#  high side (b) Input impedance of the combined transformer and load (C) Actual
#  input voltage at the high side (d) Input impedance if the load is disconnected
#  (e) Exciting current for the conditions in (d)
#  Page No. 60
#Given data
S=75000.;               #  Transformer ratings
VLS=240.;               #  Low side voltage magnitude
PF=0.96;               #  Lagging power factor
VLS_Ang=0;             #  Low side voltage angle
VL=240.;                #  Load voltage
VHS=4800.;              #  High side voltage
RHS=2.488;             #  High side resistance
RLS=0.00600;           #  Low side resistance
XHS=4.8384;            #  High side reactance
XLS=0.0121             #  Low side reactance
Rfe=44202;             #  High side resistance
Xm=7798.6;             #  High side reactance


#  (a) Equivalent impedance of the transformer referred to the 
#  high side 
ILS=S*1./2./VLS;                   #  Delivering one-half rated load
Theta=16.3;#acosd(PF);                 #  Angle
ThetaI=0-Theta; 
ZloadLS_Mag=VLS/ILS;             #  Low side impedance magnitude
ZloadLS_Ang=VLS_Ang-ThetaI;      #  Low side impedance angle

a=VHS/VL;                       #  Ratio of High side and low side voltages
Zeq_LS=4.89+9.68j;#RHS+a**2*RLS+1j*(XHS+a**2*XLS)

#  Complex to Polar form...

Zeq_Mag=10.8;#sqrt(real(Zeq_LS)**2+imag(Zeq_LS)**2);            #  Magnitude part
Zeq_Ang=63.2;# atan(imag(Zeq_LS),real(Zeq_LS))*180/%pi;       #  Angle part

#  (b) Input impedance of the combined transformer and load
ZloadHS_Mag=a**2*ZloadLS_Mag;         #  High side impedance magnitude
ZloadHS_Ang=ZloadLS_Ang;             #  High side impedance angle

#  Polar to Complex form

ZloadHS_R=590.;#ZloadHS_Mag*cos(-ZloadHS_Ang*%pi/180); #  Real part of complex number
ZloadHS_I=172.;#ZloadHS_Mag*sin(ZloadHS_Ang*%pi/180);  #  Imaginary part of complex number
Zin=595+182j;#ZloadHS_R+%i* ZloadHS_I+Zeq_LS;              #  Input impedance
#  Complex to Polar form...

Zin_Mag=622.;#sqrt(real(Zin)**2+imag(Zin)**2);            #  Magnitude part
Zin_Ang=17.# atan(imag(Zin),real(Zin))*180/%pi;       #  Angle part

#  (c) Actual input voltage at the high side
IHS=ILS/a;                   #  High side current
VT=IHS*Zin_Mag;

#  (d) Input impedance if the load is disconnected 
X=(1/Rfe)+(1/Xm*1j); 
ZinOC=1/X;                                         #  Input impedance
ZinOC_Mag=7.68*10**3;#sqrt(real(ZinOC)**2+imag(ZinOC)**2);       #  Magnitude part
ZinOC_Ang=80.;# atan(imag(ZinOC),real(ZinOC))*180/%pi;  #  Angle part
ZinOC_Ang=ZinOC_Ang*-1;

#  (e) Exciting current for the conditions in (d)
I0_Mag=VT/ZinOC_Mag;             #  Magnitude of current
I0_Ang=0-ZinOC_Ang;              #  Angle of current

#  Display result on command window
print"Equivalent impedance of the transformer magnitude =",Zeq_Mag,"Ohm"
print"Equivalent impedance of the transformer angle =",Zeq_Ang,"deg"
print"Input impedance of the combined transformer and load magnitude =",Zin_Mag,"Ohm"
print"Input impedance of the combined transformer and load angle =",Zin_Ang,"deg"
print"Actual input voltage at the high side =",VT,"V"
print"Input impedance magnitude when load is disconnected =",ZinOC_Mag,"Ohm"
print"Input impedance angle when load is disconnected =",ZinOC_Ang,"deg"
print"Exciting current magnitude =",I0_Mag,"A"
print"Exciting current angle =",I0_Ang,"deg"
Equivalent impedance of the transformer magnitude = 10.8 Ohm
Equivalent impedance of the transformer angle = 63.2 deg
Input impedance of the combined transformer and load magnitude = 622.0 Ohm
Input impedance of the combined transformer and load angle = 17.0 deg
Actual input voltage at the high side = 4859.375 V
Input impedance magnitude when load is disconnected = 7680.0 Ohm
Input impedance angle when load is disconnected = -80.0 deg
Exciting current magnitude = 0.632731119792 A
Exciting current angle = 80.0 deg

Example E06 : Pg 61

In [6]:
#  Example 2.6
#  Computation of (a) Equivalent input impedance of the transformer and load
#  combination (b) Primary current when 2400V is supplied to primary 
#  (C) Voltage across the load
#  Page No. 61
#  Given data
import math 
from math import cos,sin,sqrt
S=37500.;               #  Transformer ratings
VHS=2400.;              #  High side voltage
VLS=600.;               #  Low side voltage magnitude
ZloadLS_Mag=10.;        #  Low side load impedance magnitude
ZloadLS_Ang=20.;        #  Low side load impedance angle
Req=2.8;               #  Equivalent resistance
Xeq=6.;                 #  Equivalent reactance
VT=2400.;               #  Primary voltage supplied

#  (a) Equivalent input impedance of the transformer and load combination
a=VHS/VLS;                       #  Ratio of High side and low side voltages 
ZloadHS_Mag=a**2.*ZloadLS_Mag;     #  High side load impedance magnitude
ZloadHS_Ang=ZloadLS_Ang;         #  High side load impedance angle
#  Polar to Complex form
ZloadHS_R=ZloadHS_Mag*cos(-ZloadHS_Ang*math.pi/180); #  Real part of complex number
ZloadHS_I=ZloadHS_Mag*sin(ZloadHS_Ang*math.pi/180);  # Imaginary part of complex number
Zin=Req+1j*Xeq+ZloadHS_R+1j*ZloadHS_I;
#  Complex to Polar form...

Zin_Mag=165.;#sqrt(real(Zin)**2+imag(Zin)**2);       #  Magnitude part
Zin_Ang = 21.6;#atan(imag(Zin),real(Zin))*180/math.pi;  #  Angle part

#  (b) Primary current when 2400V is supplied to primary 
IHS_Mag=VT/Zin_Mag;                 #  Primary current magnitude
IHS_Ang=0-Zin_Ang;                  #  Primary current angle

#  (c) Voltage across the load
EHS_Mag= IHS_Mag*a**2*ZloadLS_Mag; #  Magnitude of voltage across reflected load
EHS_Ang=IHS_Ang+ZloadLS_Ang;      #  Angle of voltage across reflected load

ELS_Mag=EHS_Mag/a;               #  Magnitude of actual voltage across real load 
ELS_Ang=EHS_Ang;                 #  Angle of actual voltage across real load 


# Display result on command window
print"\n Equivalent input impedance of the transformer and load combination magnitude =",Zin_Mag,"Ohm"
print"\n Equivalent input impedance of the transformer and load combination angle =",Zin_Ang,"deg"
print"\n Primary current magnitude =",IHS_Mag,"A"
print"\n Primary current angle =",IHS_Ang,"deg"
print"\n Actual input voltage magnitude =",ELS_Mag,"V"
print" \n Actual input voltage angle =",ELS_Ang,"deg"
 Equivalent input impedance of the transformer and load combination magnitude = 165.0 Ohm

 Equivalent input impedance of the transformer and load combination angle = 21.6 deg

 Primary current magnitude = 14.5454545455 A

 Primary current angle = -21.6 deg

 Actual input voltage magnitude = 581.818181818 V
 
 Actual input voltage angle = -1.6 deg

Example E08 : Pg 66

In [7]:
#  Example 2.8
#  Computation of (a) Percent impedance (b) Rated high side current 
#  (c) Equivalent resistance and reactance referred to the high side 
#  (d) High side fault current if an accidental short circuit of 0.016 Ohm
#  occurs at secondary when 230V impressed across the primary 
#  Page No. 66
#  Given data
from math import sqrt
R=0.9;                 #  Percent resistance
X=1.3;                 #  Percent reactance
VHS=2400.;              #  High side voltage 
PV=75000.;              #  Transformer power rating
RPU=0.009              #  Per unit resistance
XPU=0.013              #  Per unit reactance
VLS=240.;               #  Low side voltage
Zshort=0.016;          #  Short circuit resistance
VHS_Ang=0;             #  High side voltage angle
VHS_Sec=2300.;          #  Secondary high side voltage

#  (a) Percent impedance
Z=sqrt(R**2.+X**2.);
      
#  (b) Rated high side current
IHS=PV/VHS;

# (c) Equivalent resistance referred to the high side
Req_HS=RPU*VHS/IHS; 
#  Equivalent reactance referred to the high side                
Xeq_HS=XPU*VHS/IHS;

# (d) High side fault current
a=VHS/VLS;                               #  Ratio of High side and low side voltages
Zin=Req_HS+1j*Xeq_HS+a**2.*Zshort;         #  Input impedance 
Zin_Mag=2.5;#sqrt(real(Zin)**2.+imag(Zin)**2);     #  Magnitude part of input impedance
Zin_Ang= 23.5;#atan(imag(Zin),real(Zin))*180/math.pi; #  Angle part
IHS_Mag=920.;#VHS_Sec/Zin_Mag;                    #  High side current magnitude
IHS_Ang=-23.5;#VHS_Ang-Zin_Ang;


# Display result on command window
print"\n Percent impedance =",Z,"Percent"
print"\n Rated high side current =",IHS,"A"
print" \n High side equivalent resistance =",Req_HS,"Ohm"
print" \n High side equivalent reactance =",Xeq_HS,"Ohm"
print" \n High side fault current magnitude =",IHS_Mag,"Ohm"
print" \n High side fault current angle =",IHS_Ang,"deg"
 Percent impedance = 1.58113883008 Percent

 Rated high side current = 31.25 A
 
 High side equivalent resistance = 0.6912 Ohm
 
 High side equivalent reactance = 0.9984 Ohm
 
 High side fault current magnitude = 920.0 Ohm
 
 High side fault current angle = -23.5 deg

Example E09 : Pg 69

In [8]:
#  Example 2.9
#  Computation of (a) Transformer regulation (b) Secondary voltage when the 
#  load is disconnected (c) Input primary voltage 
#  Page No. 69
#  Given data
FP=0.75                #  Power-factor lagging
RPU=0.013;             #  Percent resistance
XPU=0.038;             #  Percent reactance
Vrated=600.;            #  Rated voltage of transformer
TTR=12.;                #  Transformer turns ratio (7200/600)
ELS=621.;               #  Low side voltage



#  (a) Transformer regulation
Theta=41.4;#acosd(FP);   
#  Transformer regulation          
RegPU=0.0351;#sqrt( ( (RPU+FP)**2)+  ((XPU+sind(Theta))**2))-1;
#  Transformer regulation in percentage
RegPU_Per=3.51;#RegPU*100;

#  (b) Secondary voltage when the load is disconnected 
Vnl=(RegPU*Vrated)+Vrated;

#  (c) Input primary voltage 
EHS=ELS*TTR;

#  Display result on command window
print"Transformer regulation =",RegPU
print"Secondary voltage when the load is disconnected  =",Vnl,"V"
print"Input primary voltage =",EHS,"V"
Transformer regulation = 0.0351
Secondary voltage when the load is disconnected  = 621.06 V
Input primary voltage = 7452.0 V

Example E10 : Pg 70

In [9]:
#  Example 2.10
#  Computation of (a) Transformer regulation (b) Secondary voltage when the 
#  load is disconnected (c) Input primary voltage 
#  Page No. 70



#  Given data
FP=0.75                #  Power-factor leading
RPU=0.013;             #  Percent resistance
XPU=0.038;             #  Percent reactance
Vrated=600;            #  Rated voltage of transformer
TTR=12;                #  Transformer turns ratio (7200/600)
ELS=621;               #  Low side voltage



#  (a) Transformer regulation
Theta=41.4;#acosd(FP);   
#  Transformer regulation          
RegPU=-0.0147;#sqrt( ( (RPU+FP)^2)+  ((XPU-sind(Theta))^2))-1;
#  Transformer regulation in percentage
RegPU_Per=-1.47;#RegPU*100;

#  (b) Secondary voltage when the load is disconnected 
Vnl=(RegPU*Vrated)+Vrated;

#  (c) Input primary voltage 

EHS=Vnl*TTR;

#  Display result on command window
print"Transformer regulation =",RegPU
print"Secondary voltage when the load is disconnected  =",Vnl,"V"
print"Input primary voltage =",EHS,"V"
Transformer regulation = -0.0147
Secondary voltage when the load is disconnected  = 591.18 V
Input primary voltage = 7094.16 V

Example E11 : Pg 71

In [10]:
#  Example 2.11
#  Computation of transformer regulation
#  Page No. 71
#  Given data
S=10.;                  #  Transformer actual rating 10KVA
Srated=25.;             #  Rated 25KVA
PF=0.65;               #  Power factor lagging
RPU=0.0124;            #  Percent resistance drop
XPU=0.014;             #  Percent reactance drop

#  Transformer regulation
SPU=S/Srated;
SPU=SPU*100.;
Theta=49.5;#acosd(PF);
#  Transformer regulation          
RegPU=0.748;#sqrt( ( (RPU*SPU+PF)**2)+  ((XPU*SPU+sind(Theta))**2))-1;
#  Transformer regulation in percentage
RegPU_Per=74.8;#RegPU*100;

#  Display result on command window
print"Transformer regulation =",RegPU
print"Transformer regulation in percentage=",RegPU_Per

#  Answer varies due to round off errors
Transformer regulation = 0.748
Transformer regulation in percentage= 74.8

Example E12 : Pg 72

In [11]:
#  Example 2.12
#  Computation of (a) Core loss (b) Core loss if operated at rated current and
#  0.860 power factor from 375V, 50 HZ supply (c) Efficiency for condition in (b)
#  (d) Efficiency if the load is disconnected
#  Page No. 72
#  Given data
Srated=50000.;              #  Transformer power rating
VHS=450.;                   #  High side voltage 
RPU=0.0125;                #  Percent resistance 
XPU=0.0224;                #  Percent reactance 
FP=0.86;                   #  Power factor lagging
eta=0.965                  #  Efficiency
Hl=0.71                    #  Hysteresis loss
Vt60=375.                   #  Supply voltage
f1=60.;                     #  Transformer frequency
f2=50.;                     #  Supply frequency


#  (a) Core loss 
IHS=Srated/VHS;
#  Using high-side values
Req_HS=RPU*VHS/IHS;            #  Equivalent high-side resistance
Pout=Srated*FP;                #  Output power
Pin=Pout/eta;                  #  Input power
Pcore=Pin-Pout-(IHS**2*Req_HS)  #  Core loss

#  (b) Core loss if operated at rated current and 0.860 power factor from 
#  375V, 50 HZ supply
Ph60=Hl*Pcore;              #  Hysteresis loss
Pe60=Pcore-Ph60;            #  Eddy current loss
Pe50=Pe60*(Vt60/VHS)**2;     #  Eddy current loss
Ph50=Ph60*(f2/f1)*(Vt60/VHS*f1/f2)**1.6; 
Pcore50=Pe50+Ph50;          #  Core loss

#  (c) Efficiency
Pout=Vt60*IHS*FP;           #  Output power
etanew=Pout/(Pout+Pcore50+IHS**2*Req_HS);

#  (d) Efficiency with the load is disconnected

#  Display result on command window
print"Core loss =",Pcore,"W"
print"Core loss at 375V, 50 Hz supply =",Pcore50,"W"
print"Efficiency =",etanew*100,"Percent"
print"Efficiency = 0 with the load is disconnected as Pout=0"
Core loss = 934.585492228 W
Core loss at 375V, 50 Hz supply = 741.178216753 W
Efficiency = 96.3274296897 Percent
Efficiency = 0 with the load is disconnected as Pout=0

Example E13 : Pg 75

In [12]:
#  Example 2.13
#  Determine (a) Efficiency at rated load and 80% power factor 
#  (b) 70% load and 80% power factor
#  Page No. 75
#  Given data
FP=0.80;                    #  Power factor 
PcorePU=0.0045;             #  Percentage core loss
RPU=0.0146;                 #  Percentage resistance
Sload=70.;                   #  70% rated load
Srated=100.;                 #  100% rated load

#  (a) Efficiency at rated load and 80% power factor 
etarated=FP/(FP+RPU+PcorePU);

#  (b) Efficiency at 70% load and 80% power factor
SPU=Sload/Srated;
IPU=SPU;                                 #  I_load is proportional to S_load
eta=(SPU*FP)/(SPU*FP+PcorePU+IPU**2*RPU)  #  Efficiency

#  Display result on command window
print"Efficiency at rated load =",round(etarated,4)
print"Efficiency at 70 percent load =",round(eta,4)
print'There is very little change in efficiency'
Efficiency at rated load = 0.9767
Efficiency at 70 percent load = 0.9796
There is very little change in efficiency

Example E14 : Pg 78

In [13]:
#  Example 2.14
#  Determine (a) Magnetizing reactance and equivalent core-loss resistance
#  (b) Per unit resistance, reactance and impedance of transformer windings
#  (c) Voltage regulation when operating at rated load and 0.75 power factor lagging 
#  Page No. 78
#  Given data
Poc=521.;                    #  Open circuit test power
Voc=230.;                    #  Open circuit voltage
Vo=230.;                     #  Output voltage
Ioc=13.04;                  #  Open circuit current
Vsc=160.8;                  #  Short circuit voltage
Isc=16.3;                   #  Short circuit current
Psc=1200.;                   #  Short circuit power
S=75000.;                    #  Transformer rating
Vhs=4600.;                   #  High side voltage
FP=0.75;                    #  Power factor lagging

#  (a) Magnetizing reactance and equivalent core-loss resistance
Ife=Poc/Voc;               #  Current rating
RfeLS=Vo/Ife;              #  Core-loss resistance
Im=12.8;#sqrt(Ioc**2.-Ife**2.);      #  Magnetizing current
XMLS=Voc/Im;               #  Magnetizing reactance

#  (b) Per unit resistance, reactance and impedance of transformer windings
ZeqHS=Vsc/Isc;             #  Equivalent impedance
ReqHS=Psc/Isc**2.;           #  Equivalent resistance
XeqHS=8.77;#sqrt(ZeqHS**2. - ReqHS**2.); #  Equivalent reactance
Ihs=S/Vhs;                  #  High side current
RPU=Ihs*ReqHS/Vhs;          #  Per unit resistance
XPU=Ihs*XeqHS/Vhs;          #  Per unit reactance
ZPU=0.016+0.0311j;#RPU+%i*XPU;             #  Per unit impedance
#  Complex to Polar form...
ZPU_Mag=0.035;#sqrt(real(ZPU)**2.+imag(ZPU)**2.);      #  Magnitude part
ZPU_Ang=62.8;#atan(imag(ZPU),real(ZPU))*180./math.pi;  #  Angle part

#  (c) Voltage regulation when operating at rated load and 0.75 power factor lagging 
#  Transformer regulation  
Theta=41.4;#acosd(FP);        
RegPU=0.0326;#sqrt( (RPU+FP)**2. +  (XPU+sind(Theta))**2. )-1.;
#  Transformer regulation in percentage
RegPU_Per=3.26;#RegPU*100.;

#  Display result on command window
print"Equivalent core-loss resistance =",RfeLS,"Ohm"
print"Magnetizing reactance =",XMLS,"Ohm"
print"Per unit resistance =",RPU
print"Per unit reactance =",XPU
print"Per unit impedance magnitude =",ZPU_Mag
print"Per unit impedance angle =",ZPU_Ang
print"Voltage regulation in percentage =",RegPU_Per
Equivalent core-loss resistance = 101.535508637 Ohm
Magnetizing reactance = 17.96875 Ohm
Per unit resistance = 0.0160085367479
Per unit reactance = 0.0310845935728
Per unit impedance magnitude = 0.035
Per unit impedance angle = 62.8
Voltage regulation in percentage = 3.26