EE2022 Electrical Energy Systems
Lecture 13: Transformer and Per Unit Analysis
02-03-2012
Panida Jirutitijaroen
Department of Electrical and Computer Engineering
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
1
Detailed Syllabus
20/01/2012
20/01/2012
27/01/2012
27/01/2012
30/01/2012
03/02/2012
03/02/2012
06/02/2012
10/02/2012
10/02/2012
13/02/2012
17/02/2012
17/02/2012
27/02/2012
02/03/2012
02/03/2012
2/11/2012
Three-phase circuit analysis: Introduction to three-phase circuit. Balanced three-phase systems.
Three-phase circuit analysis: Delta-Wye connection, Relationship between phase and line quantities
Three-phase circuit analysis: Per-phase analysis, Three-phase power calculation
Guest Lecture “Energy & Environment, Smart Grid & Challenges Ahead” by Prof. J Nanda (IIT Delhi,
IEEE Fellow)
Generator modeling: Simple generator concept
Generator modeling: Equivalent circuit of synchronous generators
Generator modeling: Operating consideration of synchronous generators, i.e. Excitation voltage
control, real power control, and loading capability
Generator modeling: Principle of asynchronous generators
Transmission line modeling: Overhead VS Underground cable
Transmission line modeling: Four basic parameters of transmission line
Transmission line modeling: Long transmission line model, Medium-length transmission line model,
Short transmission line model
Transmission line modeling: Operating consideration of transmission lines i.e. voltage regulation,
line loadability, efficiency
Transformer and per unit analysis: Principle of transformer, Single-phase transformer
Transformer and per unit analysis: Single-phase per unit analysis
Transformer and per unit analysis: Three-phase transformer, Three-phase per unit analysis
Review : if time permits.
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
2
Learning Outcomes
• Formulate equivalent circuits of various
components in electrical energy systems
– Equivalent circuit of transformer
• Analyze three-phase balanced circuits
– Per-unit analysis for three-phase circuit
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
3
Outline
• Three Phase Transformer
– Delta/Wye Connections
• Three-phase per unit analysis
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
4
References
•
Glover, Sarma, and Overbye, “Power System Analysis and
Design”.
–
2/11/2012
Chapter 3
EE2022: Three-phase circuit by P. Jirutitijaroen
5
Three-phase transformer connections
Phase shift
THREE PHASE TRANSFORMER
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
6
Three Single-Phase Transformers
Wye-connection
Delta-connection
n
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
7
3Ф Transformer Connection
• The voltage rating of a three
phase transformer is the ratio
between line-to-line voltage
at the primary side and lineto-line voltage at the
secondary side.
• The single-phase equivalent
shows line-to-neutral voltage.
• For Y-Y and ∆-∆ transformers,
voltage and current in both
primary and secondary are in
phase. The ratio of the
voltage and current follows
the turn ratio of the
transformer.
• The same does not apply to Y∆ and ∆-Y connections.
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
8
3Ф Transformer Connection
i₁ (p.u.) Zeq (p.u.)
Per phase
per unit
circuit
i₂(p.u.)
+
+
V₁(p.u.) Y (p.u.)
V₂(p.u.)
-
-
Wye-Delta and Delta-Wye
transformer connections
introduce some phase shift
to the voltage.
i₁ (p.u.) Zeq (p.u.)
Per phase
per unit
circuit
2/11/2012
+
V₁(p.u.) Y (p.u.)
-
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
i₂(p.u.)
+
V₂(p.u.)
9
Wye-Delta 3Ф Transformer
Wye-connected
A1
Delta-connected
A2
VC1n
Positive
sequence
VB 2C 2
VA1n
VB1n
B1
B2
C1
C2
VC 2 A2
VA2 B 2
n
Line-to-neutral voltage
2/11/2012
VA1n
 N1 
VA2 B 2
 
 N2 
Line-to-line voltage!!
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
10
Wye-Delta 3Ф Transformer
VC 2 A2
VA1n
VC 2n
 N1 
VA2 B 2
 
 N2 
VA2B 2  VA2n  VB 2n
VA2 B 2
VB 2n
For a positive sequence,
VA2n
VB 2C 2
VA1n
 VA2n  VA2n 120
 3VA2 n 30
 N1 
 N1 
VA2 B 2  
 3VA2n 30
 
 N2 
 N2 
This value becomes the given turn ratio of the transformer.
The line-to-neutral voltage at the primary side leads the line-to neutral
voltage at the secondary side by 30 degrees. The same is true for the
current.
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
11
Y-Δ 3Ф Transformer: Per Phase Model
For a positive sequence voltage source,
VA1n
 N1 
 3VA2n 30
 
 N2 
i₁ (p.u.) Zeq (p.u.)
+
V₁(p.u.) Y (p.u.)
-
2/11/2012
i₂’(p.u.)
+
V₂’(p.u.)
-
VA1n : VA2n  130 : 1
i₂(p.u.)
Voltage
and current
phase shift of
30 degrees
130 : 1
+
V₂(p.u.)
i2  i2   30
V2  V2  30
-
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
12
3Ф Transformer Per Unit Model
i₁ (p.u.)
Zeq (p.u.)
i₂(p.u.)
+
+
V₁(p.u.) Y p.u.)
V₂(p.u.)
-
-
i₁ (p.u.)Zeq (p.u.)
+
i₂(p.u.)
V, I phase
shift
V₁(p.u.)
Y (p.u.)
i₁ (p.u.)Zeq (p.u.)
V₁(p.u.)
Y (p.u.)
-
i₁ (p.u.)
-
i₂(p.u.)
V, I phase
shift
+
V₂(p.u.)
1  30 : 1
Zeq (p.u.)
+
V₁(p.u.) Y p.u.)
2/11/2012
V₂(p.u.)
130 : 1
-
+
+
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
-
i₂(p.u.)
+
V₂(p.u.)
13
Three-Phase Transformer
Source: www.mathworks.com
Source: http://www.meidensg.com.sg
2/11/2012
Source:
http://www.youtube.com/watch?v=9Y958Vc5ohI&feature=BFa&list=PLC7
AEFDAFD9223780&lf=rellist
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
14
5.5 MVA
ABB and 10
MVA GE
Substation
Transformer
Nameplate
Source:
http://www.tu
csontransform
er.com
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
15
Per Unit Quantity: 3Ф
Base Value: 3Ф
Steps of Calculation: 3Ф
THREE PHASE PER UNIT ANALYSIS
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
16
Recall: Per Unit Quantity
• The per unit quantity of voltage, current, power and
impedance is found from dividing the actual quantity by a
base value of that quantity.
actual quantity
per  unit quantity 
base value of quantity
• Per unit value is denoted by ‘p.u.’.
• The base value quantity typically follows transformer
rating.
• The per unit values are the same irrespective of the sides of
the transformer.
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
17
Per Unit Analysis in 3Ф Circuit
• Recall that in three phase circuit:
– Voltage is given as line-to-line voltage.
– Current is given as line current.
– Apparent power is given as three-phase power.
Phase a
3Φ
(Y/∆)
Ia
Phase b
+
-
Vab
S3  3 VLine To  Line I Line
Phase c
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
18
Base Value Selection: 3Ф Case
• Base values are real numbers, denote by subscript ‘B’.
• Voltage base values follow transformer voltage ratings.
– The voltage ratios are given as line-to-line voltage at both ends.
This means that it already incorporates effects of the Y or ∆
configurations.
line to line
line to  neutral
B
V

VB
3
• Only single base complex power SB3 in the system.
• The base value of power is used to normalize the quantity.
Thus, the base values of real power, reactive power, and
complex power are all the same real number.
PB3  QB3  SB3
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
19
Base Value Selection: 3Ф Case
• Current base values are calculated from the base power
and base voltage.
IB 
S B3
line to line
B
3V

S B1
VBline to neutral
• Impedance base values (same value for impedance,
resistance, or reactance) are calculated from voltage and
current.
ZB 
2/11/2012
line to neutral
B
V
IB

V

  V
line to neutral 2
B
1
B
S

line to line 2
B
3
B
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
S
20
Steps of Calculation: 3Ф Case
1. Choose SB3 for the system.
2. Select VBlinetoneutral or VBlinetoline for different
zones.
3. Calculate Z B for different zones.
4. Express all quantities in p.u.
5. Draw impedance diagram and solve for p.u.
quantities.
6. Convert back to actual quantities if needed.
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
21
Example 1: Δ-Y 3Ф Transformer
• Three-phase generator rated 300 MVA,23 kV supplying a
system load of 240 MVA, 0.9 power factor lagging at 230 kV
through a 330 MVA 23Δ/ 230Y kV step-up transformer of
leakage reactance 11%.
• Use base values at the load of 100 MVA and 230 kV, find the
current supplied by the generator. Use load voltage for an
angle reference.
line  to line
Vload
 230 kV
3
Sload
 240 MVA,0.9 pf lagging
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
22
Example 1: Base Values
330 MVA
23Δ/ 230Y kV
SB3  100 MVA
X T 1  0.11 p.u.
toline
VBline
 230 kV
2
230
 to neutral
VBline

kV
2
3
toline
VBline
 23 kV
1
23
 to neutral
VBline

kV
1
3
Z B1 
I B1 
V
  23 10 
line  to line 2
B1
3
B
S
S B3
line to line
B1
3V
2/11/2012
3 2
100  10
6
 5.29  Z B 2 
100  106

 2510 .2 A
3
3  23  10
I B2 
V
  230 10 
line  to line 2
B2
3
B
S
S B3
to line
3VBline
2
3 2
100  10
6
 529 
100  106

 251.02 A
3
3  230  10
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
23
Example 1: Transformer Reactance
330 MVA
23Δ/ 230Y kV
X T 1  0.11 p.u.
SB3  100 MVA
toline
VBline
 23 kV
1
Z B1  5.29 
I B1  2510.2 A
toline
VBline
 230 kV
2
Z B 2  529 
I B2  251.02 A
 230  103 2 

0.11
6 
old
old

330  10 
X p.u. X B
new

X T, p.u. 

 0.03333 p.u.
X Bnew
529
 23  103 2 

0.11
6 
old
old

330  10 
X p.u. X B
new

X T, p.u. 

 0.03333 p.u.
X Bnew
5.29
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
24
Example 1: Load Voltage and Current
SB3  100 MVA
toline
VBline
 230 kV
2
Z B 2  529 
I B2  251.02 A
At the load, use load voltage
as reference.
line  to line
Vload
 2300 kV
Load current magnitude is found from the three-phase apparent power and lineto-line voltage.
I load 
3
Sload
line- to-line
3 Vload
240 106

 602.45 A
3
3  230 10
Load current angle can be found from power factor. Note that the reference
angle of load voltage is zero degree.
I load    cos1 0.9  25.842
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
25
Example 1: Per Unit Values
toline
VBline
 230 kV
2
Z B 2  529  I B 2  251.02 A
Vload, p.u.
line  to  line
Vload
230 0 kV
 line  to line 
 10 p.u.
VB 2
230 kV
I load, p.u. 
2/11/2012
I load
602.45  25.842 A

 2.4  25.842 p.u.
I B2
251.02 A
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
26
Example 1: Per Unit Circuit
I B1  2510.2 A
I B2  251.02 A
I gen,p.u.  ?? Zeq = j0.03333 p.u.
~
Vgen, p.u.
-
Voltage
and current
1  30 p.u. phase shift of
-30 degrees
- 1  30 : 1
+
+
I load, p.u.  2.4  25.842 p.u.
+
Vload, p.u.  10 p.u.
-
1  30

I load,p.u.
1
 I load, p.u.   30  2.4  55.842
I gen,p.u.
I gen,p.u.
I gen  I gen,p.u.  I B1  2.4  55.842  2510.2  6024.5  55.842 A
Can you find the line-to-line voltage supplied by the generator?
line  to line
 23.86  26.02 kV
Ans: Vgen
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
27
Summary
• Per unit system helps to eliminate transformer in the circuit
analysis.
• Per unit value is found by normalize the actual value by
base value.
• We can divide circuits into zones according to transformer
voltage ratings.
• Choose only single base power. Voltage, current, and
impedance base value is calculated for each zone.
• Actual value is found from multiplying the per unit value to
its corresponding base value in its zone.
• Three-phase transformer Y-∆ and ∆-Y connections
introduces phase shift of 30 and -30 degrees
correspondingly.
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
28
Subsequent Lectures
2/11/2012
EE2022: Transformer and Per Unit Analysis by P. Jirutitijaroen
29
1st Mid-term Test
• March 5th 5:15pm to 6:00pm @LT 4
• Materials (Lecture 1-5, Tutorials 1-3):
– AC circuit analysis
– Apparent power, power factor correction
– Three-phase voltage and current
– Three-phase complex power
• Format:
– 4 questions, closed book.
– 45 minutes.
2/11/2012
EE2022: Transmission Line Modeling by P. Jirutitijaroen
30