# Example 3.1
# Computation of (a) Load current (b) Incoming line current
# (c) Transformed current (d) Apparent power conducted and apparent power transformed
# Page No. 98
# Given data
NHS=400.; # Number of turns in the high side
NLS=0.25*400.; # Number of turns in the low side
VHS=2400.; # Voltage at the high side
S=4800.; # Supply voltage
# (a) Load current
a=NHS/NLS; # Transformer turn ratio
VLS=VHS/a; # Low side voltage
ILS=S/VLS; # Load current
# (b) Incoming line current
IHS=ILS/a;
# (c) Transformed current
ITR=ILS-IHS;
# (d) Apparent power conducted and apparent power transformed
SCOND=IHS*VLS; # Apparent power conducted
STRANS=ITR*VLS; # Apparent power transformed
# Display result on command window
print"Load current =",ILS,"A"
print"Incoming line current =",IHS,"A"
print"Transformed current =",ITR,"A"
print"Apparent power conducted =",SCOND,"VA"
print"Apparent power transformed =",STRANS,"VA"
# Example 3.2
# Computation of (a) Rated primary and secondary currents when connected as
# autotransformer (b) Apparent power rating when connected as an autotransformer
# Page No. 100
# Given data
S=10000.; # Supply voltage
VLS=240.; # Voltage at the low side
VHS=2400.; # Voltage at the high side
Sw=10.; # Power rating
# (a) Rated primary and secondary currents when connected as autotransformer
ILSWINDING=S/VLS; # Rated primary current
IHSWINDING=S/VHS; # Rated secondary current
# (b) Apparent power rating when connected as an autotransformer
a=VHS/VLS; # Magnetic drop across R1
Sat=(a+1)*Sw;
# Display result on command window
print"Rated primary current =",ILSWINDING,"A"
print"Rated secondary current =",IHSWINDING,"A"
print"Apparent power rating =",Sat,"KVA"
# Example 3.3
# Computation of (a) Buck boost transformer parameters
# (b) Repeating the same assuming utilization voltage as 246V
# Page No. 102
# Given data
S=10000.; # Supply voltage
VLS=212.; # Voltage at the low side
VHSNEW=246.; # New voltage at the high side
a1=1.100;
a11=1.0667;
# (a) Buck boost transformer parameters
VHS=a1*VLS;
# (b) Repeating the same assuming utilization voltage as 246V
VLSNEW=VHSNEW/a11;
# Display result on command window
print"Actual output voltage supplied to the air conditioner is =",VHS,"V"
print"Actual output voltage assuming utilization voltage as 246 V is =",VLSNEW,"V"
# Example 3.4
# Determine (a) Circulating current in the paralleled secondaries
# (b) Circulating current as a percent of the rated current of transformer A
# (c) Percent difference in secondary voltage that caused the circulating current
# Page No. 104
# Given data
S=100000.; # Transformer A and B rating
VLSA=460.; # Voltage at the low side of transformer A
VLSB=450.; # Voltage at the low side of transformer A
RPUA=0.0136; # Percent resistance of transformer A
XPUA=0.0350; # Percent reactance of transformer A
RPUB=0.0140; # Percent resistance of transformer B
XPUB=0.0332; # Percent reactance of transformer B
# (a) Circulating current in the paralleled secondaries
IA= S/VLSA; # Rated low side current for transformer A
IB= S/VLSB; # Rated low side current for transformer B
ReqA=RPUA*VLSA/IA; # Equivalent resistance of transfomer A
ReqB=RPUB*VLSB/IB; # Equivalent resistance of transfomer B
XeqA=XPUA*VLSA/IA; # Equivalent reactance of transfomer A
XeqB=XPUB*VLSB/IB; # Equivalent reactance of transfomer B
# Impedance of the closed loop formed by two secondaries is
Zloop=0.0571+0.14j;#ReqA+%i*XeqA+ReqB+%i*XeqB;
# Complex to Polar form...
Zloop_Mag=0.152;#sqrt(real(Zloop)**2+imag(Zloop)**2); # Magnitude part
Zloop_Ang=68.;#atan(imag(Zloop),real(Zloop))*180/%pi; # Angle part
Icirc_Mag=65.6;#(VLSA-VLSB)/Zloop_Mag; # Circulating current magnitude
Icirc_Ang=-68.;#0- Zloop_Ang; # Circulating current angle
# (b) Circulating current as a percent of the rated current of transformer A
IcircA=Icirc_Mag*100/IA;
# (c) Percent difference in secondary voltage that caused the circulating current
PD=(VLSA-VLSB)*100/VLSB;
# Display result on command window
print"Circulating current magnitude =",Icirc_Mag,"A"
print"Circulating current angle =",Icirc_Ang,"deg"
print"Circulating current as a percent of the rated current =",IcircA,"Percent"
print"Percent difference in secondary voltage =",PD,"Percent"
# Example 3.5
# Determine (a) Rated high side current of each transformer (b) Percent of the
# total bank-current drawn by each transformer (c) Maximum load that can be
# handled by the bank without overloading by one of the transformer
# Page No. 107
# Given data
SA=75000.; # Transformer A rating
SB=200000.; # Transformer B rating
VHSA=2400.; # Voltage at the high side of transformer A
VHSB=2400.; # Voltage at the high side of transformer B
RPUA=1.64; # Percent resistance of transformer A
XPUA=3.16; # Percent reactance of transformer A
RPUB=1.10; # Percent resistance of transformer B
XPUB=4.03; # Percent reactance of transformer B
# (a) Rated high side current of each transformer
IArated=SA/VHSA; # High side rated current transformer A
IBrated=SB/VHSB; # High side rated current transformer B
# (b) Percent of the total bank-current drawn by each transformer
ZAper=1.64+3.16j;#RPUA+%i*XPUA; # Percent impadance for transformer A
# Complex to Polar form...
ZAper_Mag=3.56;#sqrt(real(ZAper)**2+imag(ZAper)**2); # Magnitude part
ZAper_Ang=62.6;#atan(imag(ZAper),real(ZAper))*180/%pi; # Angle part
ZBper=1.1+4.03j;#RPUB+%i*XPUB; # Percent impadance for transformer B
# Complex to Polar form...
ZBper_Mag=4.18;#sqrt(real(ZBper)**2+imag(ZBper)**2); # Magnitude part
ZBper_Ang=74.7;#atan(imag(ZBper),real(ZBper))*180/%pi; # Angle part
ZAbase=VHSA/IArated; # Base impedance of transformer A
ZBbase=VHSB/IBrated; # Base impedance of transformer A
ZeqA_Mag=ZAbase*ZAper_Mag/100; # Magnitude of equivalent impedance A
ZeqA_Ang=ZAper_Ang; # Angle of equivalent impedance A
ZeqB_Mag=ZBbase*ZBper_Mag/100; # Magnitude of equivalent impedance B
ZeqB_Ang=ZBper_Ang; # Angle of equivalent impedance B
YeqA_Mag=0.366;#1/ZeqA_Mag; # Magnitude of equivalent admittance A
YeqA_Ang=-62.6;#0-ZeqA_Ang; # Angle of equivalent admittance A
# Polar to Complex form
YeqA_R=0.168;#YeqA_Mag*cos(-YeqA_Ang*%pi/180); # Real part of complex number
YeqA_I=-0.325;#YeqA_Mag*sin(YeqA_Ang*%pi/180); # Imaginary part of complex number
YeqB_Mag=0.831;#1/ZeqB_Mag; # Magnitude of equivalent admittance B
YeqB_Ang=-74.7;#0-ZeqB_Ang; # Angle of equivalent admittance B
# Polar to Complex form
YeqB_R=0.219;#YeqB_Mag*cos(-YeqB_Ang*%pi/180); # Real part of complex number
YeqB_I=-0.802;#YeqB_Mag*sin(YeqB_Ang*%pi/180); # Imaginary part of complex number
YP=0.387+1.13j;#(YeqA_R - %i* YeqA_I)+(YeqB_R - %i* YeqB_I); # Parallel admittance
# Complex to Polar form...
YP_Mag=1.19;#sqrt(real(YP)**2+imag(YP)**2); # Magnitude part
YP_Ang=71.;#atan(imag(YP),real(YP))*180/%pi; # Angle part
IA=30.7;#YeqA_Mag/YP_Mag; # Transformer A load
IB=69.8;#YeqB_Mag/YP_Mag; # Transformer A load
IA=IA*100.;
IB=IB*100.;
# (c) Maximum load that can be handled by the bank without overloading by
# one of the transformer
Ibank=IArated/0.307;
# Display result on command window
print"Rated high side current of transformer A =",IArated,"A"
print"Rated high side current of transformer B =",IBrated,"A"
print"Percent of total bank current drawn by transformer A =",IA,"Percent"
print"Percent of total bank current drawn by transformer B =",IB,"Percent"
print"Maximum load that can be handled by the bank =", Ibank,"A"
# Example 3.6
# Determine the percent of the total bank-current drawn by each transformer
# Page No. 109
# Given data
ZaPU_R=0.0158; # Transformer A impedance real part
ZaPU_I=0.0301; # Transformer A impedance imaginary part
ZbPU_R=0.0109; # Transformer B impedance real part
ZbPU_I=0.0398; # Transformer B impedance imaginary part
SB=200000.; # Transformer B rating
VHSA=2400.; # Voltage at the high side of transformer A
VHSB=2400.; # Voltage at the high side of transformer B
RPUA=1.64; # Percent resistance of transformer A
XPUA=3.16; # Percent reactance of transformer A
RPUB=1.10; # Percent resistance of transformer B
XPUB=4.03; # Percent reactance of transformer B
# Base impedance of transformer A
ZaPU=0.0158 + 0.0301j;#ZaPU_R+%i*ZaPU_I;
# Complex to Polar form...
ZaPU_Mag=0.034;#sqrt(real(ZaPU)**2+imag(ZaPU)**2); # Magnitude part
ZaPU_Ang=62.3;#atan(imag(ZaPU),real(ZaPU))*180/%pi; # Angle part
# Base impedance of transformer B
ZbPU=0.0109+0.0398j;#ZbPU_R+%i*ZbPU_I;
# Complex to Polar form...
ZbPU_Mag=0.0413;#sqrt(real(ZbPU)**2+imag(ZbPU)**2); # Magnitude part
ZbPU_Ang=74.7;#atan(imag(ZbPU),real(ZbPU))*180/%pi; # Angle part
# Admittance of transformer A
YaPU_Mag=29.4;#1/ZaPU_Mag; # Magnitude of equivalent admittance A
YaPU_Ang=-62.3;#0-ZaPU_Ang; # Angle of equivalent admittance A
# Polar to Complex form
YaPU_R=13.7;#YaPU_Mag*cos(-YaPU_Ang*%pi/180); # Real part of complex number
YaPU_I=-26;#YaPU_Mag*sin(YaPU_Ang*%pi/180); # Imaginary part of complex number
# Admittance of transformer B
YbPU_Mag=24.2;#1/ZbPU_Mag; # Magnitude of equivalent admittance B
YbPU_Ang=-74.7;#0-ZbPU_Ang; # Angle of equivalent admittance B
# Polar to Complex form
YbPU_R=6.4;#YbPU_Mag*cos(-YbPU_Ang*%pi/180); # Real part of complex number
YbPU_I=-23.4;#YbPU_Mag*sin(YbPU_Ang*%pi/180); # Imaginary part of complex number
# Parallel admittance
YP=20.1+49.4j;#(YaPU_R-%i*YaPU_I)+(YbPU_R-%i*YbPU_I);
# Complex to Polar form...
YP_Mag=53.3;#sqrt(real(YP)**2+imag(YP)**2); # Magnitude part
YP_Ang=67.9;#atan(imag(YP),real(YP))*180/%pi; # Angle part
IA=YaPU_Mag/YP_Mag*100; # Percent current drawn by transformer A
IB=100-IA;
# Display the result on the command window
print"Percent of total bank current drawn by transformer A =",IA,"Percent"
print"Percent of total bank current drawn by transformer B =",IB,"Percent"
# Example 3.7
# Computation of (a) Bank ratio (b) Transformer ratio (c) Rated line and phase
# currents for the high side (d) Rated line and phase currents for the low side
# Page No. 113
# Given data
import math
VLINEHS=4160.; # Number of turns in the high side
VLINELS=240.; # Number of turns in the low side
VHS=2400.; # Voltage at the high side
S=4800.; # Supply voltage
Vline=150000.; # Transformer rating
# (a) Bank ratio
bankratio=VLINEHS/VLINELS;
# (b) Transformer ratio
Vphasep= VLINEHS/ math.sqrt(3); # For wye primary
Vphases=VLINELS # For secondary
TR=Vphasep/Vphases; # Transformer ratio
# (c) Rated line and phase currents for the high side
Ilinew=Vline/(math.sqrt(3)*VLINEHS);
Iphasew=Ilinew;
# (d) Rated line and phase currents for the low side
Ilined=Vline/(math.sqrt(3)*VLINELS);
Iphased=Ilined/math.sqrt(3);
# Display result on command window
print"Bank ratio =",bankratio
print"Transformer ratio =",TR
print"Rated line current for the high side =",Ilinew,"A"
print"Rated phase current for the high side =",Iphasew,"A"
print"Rated line current for the low side =",Ilined,"A"
print"Rated phase current for the low side =",Iphased,"A"
# Example 3.8
# Determine the maximum allowable power that the open-delta bank handle
# without overheating
# Page No. 117
# Given data
S=25.;# Transformer rating
# Capacity of the delta-delta bank is
Cddb=S*3;
# Capacity of the bank when operating open-delta is
Cob=Cddb*0.577;
# Display result on command window
print"Capacity of the bank when operating open-delta is =",Cob,"kVA"
# Example 3.9
# Determine the minimum power rating required for each transformer
# Page No. 117
# Given data
import math
P=50000.; # Transformer power rating
Eline=120.; # Line voltage
FP=0.9 # Power factor lagging
VL=120.;
#Line current is
Iline=P/(math.sqrt(3.)*Eline*FP);
#Minimum power rating required for each transformer
Pmin=VL*Iline/1000.;
#Display result on command window
print"Minimum power rating required for each transformer =",Pmin,"kVA"