ECE 420
Homework #2
Transformers
Due: Lesson 12
(20 points) Ideal Transformer.
a. (10 points) Beginning with the voltage and current relations of an ideal transformer
(equations 2-4 and 2-5 in the textbook), derive the formula for reflection of impedance across
an ideal transformer.
b. (10 points) Figure 2-37c on page 122 of the text shows a wiring diagram for a Delta-Wye
three phase transformer. The phase shift, according to the text below the figure, is 30 degrees
with the secondary lagging the primary. Draw a wiring diagram for a Delta-Wye three phase
transformer where the secondary leads the primary by 30-degrees.
The voltage and current relations are
VP
VS
IP
=a
IS
=
1
a
Rearrange the current relation
IS
IP
=a
Multiply the voltage relation by the rearranged current relation
VP IS

= a a
VS IP
Rearrange this expression
VP IS
2

=a
IP VS
VP
IP
2
=a 
VS
IS
Define the load impedance as the ratio of VS to IS.
VS
ZS =
IS
Substitute for VS and IS.
VP
IP
2
= a Z
Q. E. D.
b. (10 points) Figure 2-37c on page 122 of the text shows a wiring diagram for a Delta-Wye
three phase transformer. The phase shift, according to the text below the figure, is 30 degrees
with the secondary lagging the primary. Draw a wiring diagram for a Delta-Wye three phase
transformer where the secondary leads the primary by 30-degrees.
The given Fig 2-37c shows Y lagging ∆ by 30 degrees.
c
a
A
b
b
a
ab
B
C
c
A
N
C
AB
N
B
As discussed in class, the oppositely directed phase shift is shown below.
c
a
A
a
c
b
B
C
b
C
N
N
A
B
The dots show that we performed a 180 degree shift. We renamed the lines on the delta side to
account for a -120 degree shift. The net on the delta side is 180-120=+60 degrees. That +60
degrees shift from the given circuit is the change necessary to create a delta leading by 30 degrees.
2. (15 points) Single Phase Steinmetz Model. Do Problem 2.1 on page 144 of the
textbook.
2.1 A 100kVA, 8000/277V distribution tranformer has the following resistances and reactances:
RP  5  Ω
RS  0.005  Ω
XP  6  Ω
XS  0.006  Ω
RC  50kΩ
XM  10 kΩ
The excitation branch impedance are given referred to the high voltage side of the transformer.
a. Find the equivalent circuit for thistransformer referred to the low voltage side.
RP
N:1
jXP
RC
N 
8000 V
277  V
R'p 
RP
2
RC
2
 28.881
3
 5.994  10
Ω
X'P 
XP
 7.193  10
2
3
Ω
N
 59.945 Ω
X'M 
N
XM
 11.989 Ω
2
N
3
RS  5  10
R`P
RS
jXM
N
R'C 
jXS
3
XS  6  10
Ω
jX`P
Ω
jXS
R`C
jX`M
RS
b. Find the per unit equivalent circuit of the transformer.
Set up the bases on the low voltage side: S, V, and Z.
Sbase  100  kV A
pu  1
VbaseL  277  V
2
VbaseL
ZbaseL 
ZbaseL  0.767 Ω
Sbase
Recall
R'p  5.994  10
3
3
X'P  7.193  10
3
Ω
R'C  59.945 Ω
RS  5  10
Ω
X'M  11.989 Ω
XS  6  10
3
Ω
Ω
These are all reflected to the low voltage side. Calculating the per unit values of each:
R'p
R'C
RS
3
3
R'Ppu 
 7.813  10  pu R'Cpu 
 78.125 pu RSpu 
 6.516  10  pu
ZbaseL
ZbaseL
ZbaseL
X'Ppu 
X'P
ZbaseL
 9.375  10
R`Ppu
3
 pu
X'Mpu 
X'M
ZbaseL
 15.625
jX`Ppu
XSpu 
Z
jXSpu
R`Cpu
XS
 7.82  10
3
 pu
baseL
RSpu
jX`Mpu
c. Assume that his transformer is supplying rated load at 277 V and 0.85 pf lagging. What is this
transformer's input voltage? What is its voltage regulation?
lagging  1
j  1
SL  100  kV A pfL  0.85 lagging
VL  277  V
Find the power factor angle and the current.

 
θL  acos pfL  31.788 deg
j  θL 

 SL e

IL  
  ( 306.859  190.174i) A IL  361.011 A
 VL 
Work our way back through the circuit to find the input voltage.


E2  VL  RS  j  XS  IL  ( 279.675  0.89i ) V
E2  279.677 V
 
arg E2   0.182  deg
arg IL  31.788 deg


3
 
E1  N E2  8.077  10  25.712i V
E1  8.077  kV
arg E1  0.182  deg
E1
IL
E1
I1 


 ( 10.789  7.392i) A
RC
N
j  XM
I1  13.078 A
arg I1  34.416 deg



3

V1  E1  RP  j  XP  I1  8.176  10  53.487i V
V'1 
V1
N
 ( 283.079  1.852i) V
 
 
V1  8.176  kV
arg V1  0.375  deg
V'1  283.085 V
arg V'1  0.375  deg
 
Find the no load current and voltage by similar circuit analysis, but with load current at zero.
V1
INL 
 ( 0.169  0.816i) A
INL  0.833 A

 RC j  XM 
RP  j  XP  

arg INL  78.293 deg

 RC  j  XM 
 


1
VNL   V1  INL RP  j  XP   ( 282.88  1.958i) V

N 
Calculate the voltage regulation.
VNL  VL
Vreg 
VL
VNL  282.887 V
Some students may estimate the no load
voltage as the input reflected voltage.
 100  %  2.125  %
V1
N
 VL
VL
 2.197  %
d. What are the copper losses and core losses under the conditions of part (c)?
Pcu 
 I1   RP   IL   RS  1.507  kW
2
2
 E1   1.305  kW

2
Pcore
RC
e. What is the transformer's efficiency under the conditions of part (c)?
PL  SL pfL  85 kW
η 
PL
PL  Pcu  Pcore
 0.968
3. (15 points) Three Phase Transformer Connections and Ratings. Do Problem
2.10 on page 147 of the textbook. For each of the six configurations, the problem
asks for the voltage, turns ratio, and apparent power rating of the individual
transformer windings. Find also the current rating of each winding.
A three phase tranformer bank is to handle 500kVA and have a 34.5kV / 11kV voltage ratio. Find
the rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and
apparent power) if the transformer is connected to...
S3  500  kV A
VP  34.5 kV
VS  11 kV
No other information is given about the voltages, so they are assumed to be line-to-line.
a. Y - Y
V1 
VP
 19.919 kV
3
b. Y - ∆
VP
V1 
 19.919 kV
3
V1
Na 
 3.136
V2
S3
Sa 
 166.667  kW
3
V1
Na 
 1.811
V2
S3
Sa 
 166.667  kW
3
V1
Na 
 5.432
V2
S3
Sa 
 166.667  kW
3
V2  VS  11 kV
V1
Na 
 3.136
V2
S3
Sa 
 166.667  kW
3
V2  VS  11 kV
V1
Na 
 3.136
V2
Sa 
V2  VS  11 kV
V1
Na 
 1.811
V2
Sa 
V2 
VS
 6.351  kV
3
V2  VS  11 kV
c. ∆ - Y
V1  VP  34.5 kV
V2 
VS
 6.351  kV
3
d. ∆ - ∆
V1  VP  34.5 kV
e. open -∆
V1  VP  34.5 kV
f. open - Y - open - ∆
VP
V1 
 19.919 kV
3
S3
 288.675  kW
3
S3
3
 288.675  kW
4. (15 points) Three Phase Transformers. Do Problem 2.11 on page 147 of the
textbook.
A 100 KVA, 230kV / 115 kV, ∆ Y three phase transformer has a per unit resistance of 0.015pu and
a per unit reactance of 0.06pu. The excitation branch elements are RC=100pu and XM=20pu.
Let's define the impedance parameters referred to the 230kV side of the transformer as
Req  7.935  Ω
Xeq  31.74  Ω
R eq
RC  52900  Ω
2:1
jX eq
IL
I1
+
RC
V1
230  kV
N4 
2
115  kV
XM  10580  Ω
jX M
‐
+
+
E1
VL
‐
‐
a. If this transformer supplies a load of 80MVA at 0.8pf lagging, draw the phasor diagram of one
phase of the transformer.
6
lagging  1 MVA  10  V A
VL  115  kV
SL  80 MVA
pfL  0.8 lagging
 
θL  acos pfL  36.87  deg
We assume here that the high side VLL=230kV and the low side VLL=115kV. Other
interpretations are both possible and reasonable. If you made a different interpretation, calculate
VLL on the low voltage side, declare it here, and proceed with the problme
Find the current in the load.

j  θL 

 SL e

IL  
  ( 321.308  240.981i)  A IL  401.635 A
 3 VL 
 
arg IL  36.87  deg
Find the internal voltage on the primary side
E1  VL N4  230  kV
Sum the currents to get the input current.
IL
E1
E1
I1 


 ( 165.002  142.23i ) A
N4
RC
j  XM
 
arg I1  40.761 deg
Find the input voltage.


V1  E1  Req  j  Xeq  I1  ( 235.824  4.109i)  kV
V1  235.859  kV
 
arg V1  0.998  deg
A phasor diagram shows the sum of the voltages.
0 V




Re E1 



Vre  Re E1  I1  Req 


Re V1 




0 V


0 V




Im E1 



Vim  Im E1  I1  Req 


Im V1 




0 V


3
6 10
3
4 10
Vim
V1  ( 235.824  4.109i)  kV
I1  j  Xeq
3
2 10
0
I1  Req
E1  230  kV
3
 2 10
0
5
5
1 10
2 10
Vre
It is also possible to get a phasor diagram of the sum of the currents.
0
0





 IL 


Re



 N4 


E
I


 L
1
Re

IRe  


 N4 RC 


  IL E1
E1  
 Re



  N4 RC j  XM  


0




I
 L


Im



 N4 


E
I


 L
1
Im

IIm  


 N4 RC 


  IL E1
E1  
 Im



  N4 RC j  XM  


0


5
3 10
0
IL
N4
 50
IIm
 ( 160.654  120.49i ) A
I1  ( 165.002  142.23i ) A
 100
E1
RC
 4.348 A
E1
 150
j  XM
0
50
100
 21.739i A
150
200
IRe
b. What is the voltage regulation under these conditions.
V1
VNL 
 ( 117.912  2.054i)  kV
N4
Vreg 
VNL  VL
VL
VNL  117.93 kV
VL  115  kV
 100  %  2.548  %
c. Sketch the equivalent circuit referred to the low voltage side of one phase of this transformer.
Calculate all the transformer impedances referred to the low voltage side.
2:1
R`eq
+
+
V1
V`1
‐
‐
Req  7.935 Ω
Xeq  31.74 Ω
jX`eq
I1
IL
+
R`C
jX`M
VL
‐
RC  52.9 kΩ
XM  10.58  kΩ
Reflection of impedance is by the square of the turns ratio; The corresponding impedances will
be smaller on the low voltage side.
R'eq 
Req
N4
R'C 
2
RC
N4
2
 1.984 Ω
X'eq 
Xeq
N4
 13.225 kΩ
X'M 
XM
N4
+
118.6kV
-
 7.935 Ω
 2.645  kΩ
1.98 Ohm
j7.94 Ohm
+
237.2kV
-
2
2
Load
j2.65 kOhm
+
237kV
-
13.2 kOhm
d. Determine the losses in the transformer and the efficiency of the transformer under the
conditions of part b.
Find the losses in the resistors using the currents and voltages already found.

Pwinding  3  I1
  Req  1.13 MW
2
 E1   3 MW
 3 
2
Pcore
 VL 
 IL  64 MW
 3 
Pout  Re 3 
RC
Efficiency
η 
Pout
Pout  Pwinding  Pcore
 0.939
5. (15 points) Three Phase Steinmetz Model. Do Problem 2.12 on page 147 of the
textbook. Work in Volts, Amps, and Ohms. State which side of the transformer
your diagram is referred.
Three 20kV, 24000 / 277V distribution transformers are connected in ∆ - Y. The open circuit test
was performed on the low voltage side of this transformer bank and the following data were
recorded:
VlineOC  480  V
IlineOC  4.10 A
P3φOC  945  W
The short circuit test was performed on the high voltage side of this transformer bank and the
following data were recorded:
I assume here that the ratings are the nominal line-to-line voltage on each side of the
transformer. Other interpretations exist. If you assumed something else, state your
assumption and proceed with the problem.
VlineSC  1400 V
IlineSC  1.8 A
P3φSC  912  W
a. Find the per unit equivalent circuit of this transformer bank.
YOC 
GC 
IlineOC
VlineOC
2
 4.102  10
3 1
Ω
BM 
1
YOC YOC  GC GC  0.014
Ω
ZSC 
VlineSC  1.347  kΩ
 IlineSC 


 3 
24000
277
1
YOC  0.015
Ω
 VlineOC 


3 

P3φOC
N5 
RC 
1
GC
 243.81 Ω
1
XM 
 70.35 Ω
BM
Reflect to low voltage side.
 86.643
RSC 
P3φSC
IlineSC
XSC 
2
 281.481 Ω
R'SC 
RSC
N5
ZSC ZSC  RSC RSC  1.317  kΩ
X'SC 
2
 0.037 Ω
XSC
N5
2
 0.175 Ω
Reflect to the low voltage side.
0.037 Ohm
j0.175Ohm
j70.4 Ohm
244 Ohm
b. Find the voltage regulation of this transformer bank at the rated load and 0.90 power factor
lagging.
S5  20 kV A
VL5  277  V
pf5  0.90 lagging

j  θ5 

 S5 e 
I5  
  ( 64.982  31.472i ) A
 VL5 
 
θ5  acos pf5  25.842 deg
VL5
VL5
IP5  I5 

 ( 66.118  35.41i) A
RC
j  XM


E25  VL5  IP5 R'SC  j  X'SC  ( 285.693  10.276i ) V
VNL5  E25
Vreg5 
VNL5  VL5
VL5
 100  %  3.205  %
c. What is the transformer bank's efficiency under these conditions.
Find the losses.
 E25   1.006  kW
 3 
2

  R'SC  0.633  kW
2
Pwinding5  3  IP5
Pcore5
RC
Find the power out
Pout5  3  S5  pf5  54 kW
Efficiency
η 
Pout5
Pout5  Pwinding5  Pcore5
 0.971
6. (10 points) Transformers in a Distribution System. Do Problem 2.14 on page
148 of the textbook.
A 13.8kV single phase generator supplies power to a load through a transmission line. The load's
impedance is 500 Ohms at 36.87 degrees. The transmission line's impedance is 60 Ohms at 60
degrees.
a. If the generator is directly connected to the load,what is the ratio of the load voltage to the
generated voltage? What are the transmission losses in the system?
VG  13.8 kV
Zline  60 e
j  60 deg
Ω
Zload  500  e
j  36.87 deg
Ω
Use a voltage division to find the load voltage.
VL6  VG
Z
Zload
line  Zload

4

 1.241  10  526.681i V
VL6  12.417 kV
The ratio of load voltage to generated voltage is
Vratio 
VL6
VG
 0.9
To find losses, first find the current and then calculate the losses as |i|2 r.
VL6
IL6 
 ( 19.218  15.73i) A
Zload
Ploss6 
 IL6   ReZline  18.503 kW
2
b. What percentage of the power supplied by the source reaches the load?


arg VL6  2.431  deg
Calculate power in and load power, then take the ratio:



Pin6  Re VG IL6  265.205  kW
PL6  Pin6  Ploss6  246.702  kW
PL6
Pin6
 93.023 %
c. If a 1:10 step up transformer is placed at the output of the generator and a 10:1 step down
transformer is placed at the load end of the transmission line, what is the new ratio of the load
voltage to the generated voltage? What are the transmission losses in the system? Assume the
transformers to be ideal.
Reflect the line impedance to the load side.
10
N6 
1
Zline  ( 30  51.962i ) Ω
Z'line 
Zline
N6
2
 ( 0.3  0.52i ) Ω
Reflect the generator across two transformers, which gives the same numerical value of the
generator as a voltage source.
V'G  VG N6 
1
N6
 13.8
1
A
 kW
Use a voltage division to find the load voltage.


Zload
4
VL6  V'G
 1.378  10  6.491i V
Z'line  Zload
VL6  13.785 kV
The ratio of load voltage to generated voltage is
Vratio 
VL6
V'G
 0.999
To find losses, first find the current and then calculate the losses as |i|2 r.
VL6
IL6 
 ( 22.048  16.552i ) A
Zload


arg VL6  0.027  deg
Ploss6 
 IL6   ReZ'line  228.024  W
2
d. What percentage of the power supplied by the source reaches the load?
Calculate power in and load power, then take the ratio:



Pin6  Re V'G IL6  304.26 kW
PL6  Pin6  Ploss6  304.032  kW
PL6
Pin6
 99.925 %
e. Compare the efficiencies of the transmission system with and without transformers.
With the transformers in place, the transmission system efficiency is significantly improved. The
losses are more than a factor of 80 lower with the transformer, so the efficiency improves from 93%
to 99.9%.
7. (10 points) Transformer Polarity. A single phase transformer is said to have subtractive
polarity. Go to the library or the Internet and look up a description or definition of this term
“subtractive polarity”.
a. (5 points) Explain what this means and how verify it.
b. (5 points) Show a simple transformer diagram with the polarity dots properly placed
and the four terminals labeled (H1-H2-X1-X2) for this subtractive polarity transformer.
This is a way to specify the polarity of a transformer. It describes a test for polarity. High voltage
terminals are designated as H teminals, e.g., H1 and H2. Low voltage terminals are designated X,
e.g., X1 and X2.
To perform the test, connect adjacent H and X terminals together as shown below. Connect a
voltmeter to the other two H and X terminals.
V oltm eter
N :1
Subtractive polarity
H1
+
+
VH
VL
‐
‐
X1
X2
H2
If the voltage measured is less than the input voltage VH, then the transformer is said to be
subtractive. Its polarity is as shown above.
If the voltage measured is greater than the input voltage VH, then the transformer is said to be
additive. Its polarity is as shown below.
Voltmeter
N:1
+
‐
VH
VL
‐
+