Per unit systems in power systems

1
POWER SYSTEMS
SUBJECT NAME : POWER SYSTEMS
CHAPTER NO: 1
CHAPTER NAME : POWER SYSTEM FUNDAMENTALS
LECTURE : 1

India grid
Southern grid
SRLDC

1. Basic Concepts
2. Synchronous Machines and
Transformer Modelling
3. Transmission Line Modelling
4. Power Flow Analysis
5. Fault Analysis
6. Power System Stability
7. Economic Operation of Power
System,

G M
50 MVA
11 kV /132kV
50 MVA
11 kV
50 MVA
132kV /11 kV
50 MVA
11 kV

G M
50 MVA
11 kV
/132kV
50
MVA
11 kV
50 MVA
132kV /11
kV
50
MVA
11 kV

G M
50 MVA
11 kV /132kV
50 MVA
11 kV
50 MVA
132kV /11 kV
50 MVA
11 kV
0.15 PU
0.12 PU
0.08 PU 0.08 PU
0.12 PU

27
Per-Unit System
In the per-unit system, the voltages, currents, powers,
impedances, and other electrical quantities are expressed on a per-
unit basis by the equation:
Quantity per unit =
Actual value
Base value of quantity
It is customary to select two base quantities to define a given per-
unit system. The ones usually selected are voltage and power.

28
Per-Unit System
Assume:
Then compute base values for currents and impedances:
rated
b V
V 
rated
b S
S 
b
b
b
V
S
I 
b
b
b
b
b
S
V
I
V
Z
2



29
Per-Unit System
And the per-unit system is:
b
actual
u
p
V
V
V 
.
.
b
actual
u
p
I
I
I 
.
.
b
actual
u
p
S
S
S 
.
.
b
actual
u
p
Z
Z
Z 
.
.
%
100
% .
. 
 u
p
Z
Z
Percent of base Z
rated
b V
V 
rated
b S
S 
b
b
b
V
S
I 
b
b
b
b
b
S
V
I
V
Z
2



30
Example 1
An electrical lamp is rated 120 volts, 500 watts. Compute the per-
unit and percent impedance of the lamp. Give the p.u. equivalent
circuit.
Solution:
(1) Compute lamp resistance
(2) Select base quantities
(3) Compute base impedance
(4) Find the per-unit impedance
120 volts
500 watts

31
Example 1
An electrical lamp is rated 120 volts, 500 watts. Compute the per-
unit and percent impedance of the lamp. Give the p.u. equivalent
circuit.
Solution:
power factor = 1.0





 8
.
28
500
)
120
( 2
2
2
P
V
R
R
V
P


 0
8
.
28
Z
120 volts
500 watts

32
Example 1
(4) The per-unit impedance is:
VA
Sb 500

V
Vb 120




 8
.
28
500
)
120
( 2
2
b
b
b
S
V
Z
.
.
0
1
8
.
28
0
8
.
28
.
. u
p
Z
Z
Z
b
u
p 





33
Example 1
(5) Percent impedance:
(6) Per-unit equivalent circuit:
%
100
% 
Z
.
.
0
1 u
p
Z 

.
.
0
1 u
p
VS 


34
Example 2
An electrical lamp is rated 120 volts, 500 watts. If the voltage
applied across the lamp is twice the rated value, compute the
current that flows through the lamp. Use the per-unit method.
Solution:
(4) Find the per-unit impedance
120 V
500 W
2 X 120 V
240 V

35
Example 2
An electrical lamp is rated 120 volts, 500 watts. If the voltage
applied across the lamp is twice the rated value, compute the
current that flows through the lamp. Use the per-unit method.
Solution:
V
Vb 120

.
.
0
2
120
240
.
. u
p
V
V
V
b
u
p 



.
.
0
1
.
. u
p
Z u
p 


36
Example 2
The per-unit equivalent circuit is as follows:
.
.
0
1 u
p
Z 

.
.
0
2 u
p
VS 

.
.
0
2
0
1
0
2
.
.
.
.
.
. u
p
Z
V
I
u
p
u
p
u
p 





A
V
S
I
b
b
b 167
.
4
120
500



A
I
I
I b
u
p
actual 0
334
.
8
167
.
4
0
2
.
. 






37
Per-unit System for 1-  Circuits
One-phase circuits
LV
bLV V
V 



 I
V
S
Sb 
 
1
where
neutral
to
line
V
V 



current
line
I
I 


HV
bHV V
V 

bLV
b
bLV
V
S
I 
bHV
b
bHV
V
S
I 
Transformer

38
b
bLV
bLV
bLV
bLV
S
V
I
V
Z
2
)
(


b
bHV
bHV
bHV
bHV
S
V
I
V
Z
2
)
(


*
pu
pu
b
pu I
V
S
S
S 


cos
pu
pu
b
pu I
V
S
P
P 


sin
pu
pu
b
pu I
V
S
Q
Q 


39
Transformation Between Bases
Selection 1
A
b V
V 
1
A
b S
S 
1
Then
1
1
b
L
pu
Z
Z
Z 
1
2
1
1
b
b
b
S
V
Z 
Selection 2
B
b V
V 
2
B
b S
S 
2
Then
2
2
b
L
pu
Z
Z
Z 
2
2
2
2
b
b
b
S
V
Z 

40
2
2
2
1
2
1
2
1
1
2
1
2
b
b
b
b
b
b
L
b
b
L
pu
pu
V
S
S
V
Z
Z
Z
Z
Z
Z
Z
Z























1
2
2
2
1
1
2
b
b
b
b
pu
pu
S
S
V
V
Z
Z
“1” – old
“2” - new


















old
b
new
b
new
b
old
b
old
pu
new
pu
S
S
V
V
Z
Z
,
,
2
,
,
,
,

41
Generally per-unit values given to another base can be converted
to new base by by the equations:
2
1
1
_
_
_
2
_
_
_ )
,
,
(
)
,
,
(
base
base
base
on
pu
base
on
pu
S
S
S
Q
P
S
Q
P 
2
1
1
_
_
_
2
_
_
_
base
base
base
on
pu
base
on
pu
V
V
V
V 
1
2
2
2
2
1
1
_
_
_
2
_
_
_
)
(
)
(
)
,
,
(
)
,
,
(
base
base
base
base
base
on
pu
base
on
pu
S
V
S
V
Z
X
R
Z
X
R 
When performing calculations in a power system, every per-unit value
must be converted to the same base.

42
Per-unit System for 1- Transformer
Consider the equivalent circuit of transformer referred to LV side and
HV side shown below:
LV
V HV
V LV
V HV
V
S
S jX
R 
1
N 2
N
2
2
a
X
j
a
R S
S

(1) Referred to LV side (2) Referred to HV side
Define 1
2
1



N
N
V
V
a
HV
LV
S

43
Choose:
rated
LV
b V
V ,
1 
rated
b S
S 
Compute:
1
1
2
1
b
b
LV
HV
b V
a
V
V
V
V 

b
b
b
S
V
Z
2
1
1 
b
b
b
S
V
Z
2
2
2 
2
2
1
2
1
2
2
2
1
2
1
)
1
(
a
V
a
V
V
V
Z
Z
b
b
b
b
b
b



Normally choose rated
values as base values

44
Per-unit impedances are:
1
1
.
.
b
S
S
u
p
Z
jX
R
Z


1
2
1
2
2
2
2
2
2
.
.
b
S
S
b
S
S
b
S
S
u
p
Z
jX
R
a
Z
a
jX
a
R
Z
a
jX
a
R
Z






So:
2
.
.
1
.
. u
p
u
p Z
Z 
Per-unit equivalent
circuits of transformer
referred to LV side and
HV side are identical !!

45
Per-unit Eq. Circuit for 1- Transformer
LV
V HV
V
S
S jX
R 
1
N 2
N
Fig 1. Eq Ckt referred to LV side
1
2
1



N
N
V
V
a
HV
LV
S
1
b
Z
1
b
V 2
b
V
Fig 2. Per-unit Eq Ckt referred to LV side Fig 3.
pu
S
Z ,
1
:
1
1
b
V 2
b
V
pu
S
Z ,
1
b
V 2
b
V
b
S

46
LV
V HV
V
1
N 2
N
Fig 4. Eq Ckt referred to HV side
1
2
1



N
N
V
V
a
HV
LV
S
2
b
Z
2
b
V
Fig 5. Per-unit Eq Ckt referred to HV side Fig 6.
pu
S
Z ,
1
:
1
1
b
V 2
b
V
pu
S
Z ,
1
b
V 2
b
V
1
b
V
2
2
a
X
j
a
R S
S

b
S

47
Voltage Regulation
%
100






load
full
load
full
load
no
V
V
V
VR
Voltage regulation is defined as:
%
100
,
,
,






load
full
pu
load
full
pu
load
no
pu
V
V
V
VR
In per-unit system:
Vfull-load: Desired load voltage at full load. It may be equal
to, above, or below rated voltage
Vno-load: The no load voltage when the primary voltage is
the desired voltage in order the secondary voltage
be at its desired value at full load

48
Voltage Regulation Example
A single-phase transformer rated 200-kVA, 200/400-V, and 10% short circuit
reactance. Compute the VR when the transformer is fully loaded at unity PF
and rated voltage 400-V.
Solution:
Fig 7. Per-unit equivalent circuit
P
V S
V
1
.
0
j
load
S
V
Vb 400
2 
kVA
Sb 200

pu
S pu
load 0
1
, 

pu
j
X pu
S 1
.
0
, 
S
X

49
Rated voltage:
pu
V pu
S 0
0
.
1
, 

pu
j
j
X
I
V
V
o
pu
S
pu
pu
S
pu
P
7
.
5
001
.
1
1
.
0
1
1
.
0
0
0
.
1
0
0
.
1
,
,
,











pu
V
S
I
pu
S
pu
load
pu
load 0
0
.
1
0
0
.
1
0
0
.
1
*
*
,
,
, 




















50
pu
V
V o
pu
P
load
no
pu 7
.
5
001
.
1
,
, 



pu
V
V pu
S
load
full
pu 0
0
.
1
,
, 



Secondary side:
Voltage regulation:
%
1
.
0
%
100
0
.
1
0
.
1
001
.
1
%
100
,
,
,










load
full
pu
load
full
pu
load
no
pu
V
V
V
VR

51
Problem 1
Select Vbase in generator circuit and Sb=100MVA,
compute p.u. equivalent circuit.
G

100
j
20 kV 22kV/220kV
80MVA
14%
220kV/20kV
50MVA
10%
50MVA
0.8 PF
lagging

52
Per-unit System for 3- Circuits
Three-phase circuits
LV
L
bLV V
V ,




 I
V
S
S
Sb 3
3 1
3 

 

where
3
/
)
(line
L
neutral
to
line V
V
V 
 


L
current
line I
I
I 
 

HV
L
bHV V
V ,

L
L
b I
V
S 3

bHV
bHV
bLV
bLV
b I
V
I
V
S 3
3 


53
b
bLV
b
bLV
bLV
LV
LV
bLV
S
V
S
V
V
I
V
Z
2
)
(
3
3





b
bHV
bHV
S
V
Z
2
)
(

*
*
3
3
3
pu
pu
b
b
L
L
b
pu I
V
I
V
I
V
S
S
S 



bLV
b
bLV
V
S
I
3

bHV
b
bHV
V
S
I
3


54
Three 25-kVA, 34500/277-V transformers connected in -Y. Short-
circuit test on high voltage side:
Determine the per-unit equivalent circuit of the transformer.
V
V SC
Line 2010
, 
A
I SC
Line 26
.
1
, 
W
P SC 912
,
3 


55
(a) Using Y-equivalent
3
34500
277
S
S jX
R 
VA
Sb 25000

3
2010

SC
V
26
.
1

SC
I


 00
.
921
26
.
1
47
.
1160
SC
Z
V
VSC 47
.
1160
3
2010



56
So





 86
.
900
48
.
191
921 2
2
2
2
S
SC
S R
Z
X
W
P 304
3
912


 


 48
.
191
26
.
1
304
2
2
SC
S
I
P
R 


 86
.
900
48
.
191 j
ZSC
VA
Sb 25000
 V
V HV
b 58
.
19918
3
34500
, 



 99
.
15869
25000
58
.
19918 2
,HV
b
Z
pu
j
j
Z Y
pu
SC 0568
.
0
012
.
0
99
.
15869
86
.
900
48
.
191
,
, 




57
(b) Using -equivalent
34500 277

,
SC
Z
VA
Sb 25000

2010

SC
V
3
26
.
1

SC
I



 79
.
2764
727
.
0
2010
,
SC
Z
V
VSC 2010
 A
ISC 727
.
0
3
26
.
1



58
So





 

 30
.
2704
18
.
575
79
.
2764 2
2
2
,
2
,
, S
SC
S R
Z
X
W
P 304
3
912


 



 18
.
575
727
.
0
304
2
2
,
SC
S
I
P
R 


 86
.
900
48
.
191 j
ZSC
VA
Sb 25000
 V
V HV
b 34500
, 


 47610
25000
34500 2
,HV
b
Z
pu
j
j
Z pu
SC 0568
.
0
012
.
0
47610
30
.
1704
18
.
575
,
, 





59
In power systems there are so many different elements such as Motors, Generators and Transformers with very
different sizes and nominal values.
To be able to compare the performances of a big and a small element, per unit system is used
Physical Components in the system are represented by a mathematical model.
Mathematical models of components are connected in exactly the same way as the physical components to
obtain the system representation.
Various physical components have different ratings or basis.
It is convenient to obtain the representation with respect to a common basis.

60
The numerical per unit value of any quantity is its ratio to the chosen base quantity of the same dimensions.
Thus a per unit quantity is a normalized quantity with respect to a chosen base value.
Percent is the per unit quantity multiplied by a 100
In the per-unit system of representation, device parameters tend to fall in a relatively fixed range, making
erroneous values prominent.
Ideal transformers are eliminated as circuit elements. This results in a large saving in component representation
and reduces computational burden.
The voltage magnitude throughout a given power system is relatively close to unity in the per-unit system for a
power system operating normally. This characteristic provides a useful check on the calculations.

61
The numerical per unit value of any quantity is its ratio to the chosen base quantity of the same dimensions.
In power system calculations the nominal voltage of lines and equipment is almost always known, so the voltage
is a convenient base value to choose.
The apparent power (volt-ampere) is usually chosen as a second base. In equipment this quantity is usually
known and makes a convenient base.
The choice of these two base quantities will automatically fix the base of current, impedance, and admittance.
In a system study, the volt-ampere base can be selected to be any convenient value such as 100 MVA, 200 MVA,
etc
The same volt-ampere base is used in all parts of the system. One base voltage in a certain part of the system is
selected arbitrarily. All other base voltages must be related to the arbitrarily selected one by the turns ratio of
the connecting transformers.

62
Power system quantities such as voltage, current and impedance are often expressed in per unit or percent of
specified values.
Per unit quantities are calculated as:
Quantity per unit =
Actual value
Base value of quantity

63
Per-Unit System
Assume:
Then compute base values for currents and impedances:
rated
b V
V 
rated
b S
S 
b
b
b
V
S
I 
b
b
b
b
b
S
V
I
V
Z
2



64
Per-Unit System
And the per-unit system is:
b
actual
u
p
V
V
V 
.
.
b
actual
u
p
I
I
I 
.
.
b
actual
u
p
S
S
S 
.
.
b
actual
u
p
Z
Z
Z 
.
.
%
100
% .
. 
 u
p
Z
Z
Percent of base Z

65
One-phase circuits
LV
bLV V
V 



 I
V
S
Sb 
 
1
where
neutral
to
line
V
V 



current
line
I
I 


HV
bHV V
V 

bLV
b
bLV
V
S
I 
bHV
b
bHV
V
S
I 

66
b
bLV
bLV
bLV
bLV
S
V
I
V
Z
2
)
(


b
bHV
bHV
bHV
bHV
S
V
I
V
Z
2
)
(


*
pu
pu
b
pu I
V
S
S
S 


cos
pu
pu
b
pu I
V
S
P
P 


sin
pu
pu
b
pu I
V
S
Q
Q 


67
Selection 1
A
b V
V 
1
A
b S
S 
1
Then
1
1
b
L
pu
Z
Z
Z 
1
2
1
1
b
b
b
S
V
Z 
Selection 2
B
b V
V 
2
B
b S
S 
2
Then
2
2
b
L
pu
Z
Z
Z 
2
2
2
2
b
b
b
S
V
Z 

68
2
2
2
1
2
1
2
1
1
2
1
2
b
b
b
b
b
b
L
b
b
L
pu
pu
V
S
S
V
Z
Z
Z
Z
Z
Z
Z
Z























1
2
2
2
1
1
2
b
b
b
b
pu
pu
S
S
V
V
Z
Z
“1” – old
“2” - new


















old
b
new
b
new
b
old
b
old
pu
new
pu
S
S
V
V
Z
Z
,
,
2
,
,
,
,

69
Generally per-unit values given to another base can be converted
to new base by by the equations:
2
1
1
_
_
_
2
_
_
_ )
,
,
(
)
,
,
(
base
base
base
on
pu
base
on
pu
S
S
S
Q
P
S
Q
P 
2
1
1
_
_
_
2
_
_
_
base
base
base
on
pu
base
on
pu
V
V
V
V 
1
2
2
2
2
1
1
_
_
_
2
_
_
_
)
(
)
(
)
,
,
(
)
,
,
(
base
base
base
base
base
on
pu
base
on
pu
S
V
S
V
Z
X
R
Z
X
R 
When performing calculations in a power system, every per-unit value
must be converted to the same base.

70
Ideal Transformer


 2
2
1
1 I
N
I
N 
2
2
1
1
N
V
N
V

0

Rc
0

Rw

71
*
2
2
2
*
1
1
1 I
V
S
I
V
S 



72
Real Transformer
000
,
10

 0

Rw 0

Rc
C
M
E I
I
I 
 M
M L
X 

2
2
2
1 ' N
I
I
N  Where LM is the
magnetizing inductance

73
t
V
V m
s 
cos

m
s
JX
V 


0
Im 


 90
m
s
X
V

74
RP + JXP
RC JXM
VP VS
NS
NP
+
+
-
-
RS + JXS
Schematic of a Real Transformer

75
s
s
p
s
N
V
N
V

'
s
s
p
s I
N
N
I 
'
s
p
N
N
a 
2
2
)
(
s
p
N
N
a  s
s
p
s V
N
N
V )
(
' 
S
S aV
V 
'
S
P
s
S I
N
N
I 
'
a
I
I
s
s 
'
Referring to primary side
JXM
RC
RP + JXP a2RS + Ja2XS
V’s
+
-
+
-
VP
VS
+
-
NS
NP
Is’ Is

76
Referring to secondary side
p
p aI
I 
'
p
p
s
p V
N
N
V 
'
s
p
p
p
N
V
N
V '
 p
s
p
p I
N
N
I '

a
V
V p
p 
'
IP
VP
+
-
I’P RP/a2 + JXP/a2 RS + JXS
+
-
VS
JXM/a2
RC/a2
IS
NS
NP
VP’

77
Approximate Equivalent circuit
Place the magnetizing branch at the primary side
VP VS
+ +
- -
JXM
RC
IE
RP+JXP RS+JXS
JXM
RC
VP
-
+
+
VS
-
NP NS
NP NS
REQP=RP+a2RS
XEQP=XP+a2XS
REQ+JXEQ
Original
Circuit
Looking
From
primary

78
a
V
V
p
p 
' s
p
EQS R
a
R
R 
 2
p
p aI
I 
' s
P
EQS X
a
X
X 
 2
Place the magnetizing branch at the primary side
VP VS
+ +
- -
JXM/a2
RC/a2
REQ+JXEQ
V’P
+
-
IE
NP NS
I’P
IP
Looking
From
Secondary

79
LOAD
FULL
e I
I _
)
05
.
0
(

s
P
EQP R
a
R
R 2


s
P
EQP X
a
X
X 2


c
m R
X || Very Large
REQP+JXEQP
VP VS
+ +
- -
IP IS
NP NS
I’S=IS/a

80
s
P
EQS R
a
R
R 
 2
s
P
EQS X
a
X
X 
 2
IP
VP VS
IS
REQS+JXEQS
V’P
+ +
- -

81
240
480
2
240
480


a
10
)
240
480
(
' 2
2

 L
L R
a
R 40
)
10
(
4 

Example
From High Voltage Side
Refine
Ideal Transformer
100VA

10
L
R

82
4
10
10
' 2


a
R L
2
240
480


a
Define
From Low Voltage Side
From Low Voltage Side R’L=?

10
L
R
480
240
480
240
R’L

83
4
.
)
1
(.
4
' 2


 L
L X
a
X
2
240
480


a
From High Voltage Side
J 0.1

10
L
R
480
240
XL

84
Consider the equivalent circuit of transformer referred to LV side and
HV side shown below:
LV
V HV
V LV
V HV
V
S
S jX
R 
1
N 2
N
2
2
a
X
j
a
R S
S

(1) Referred to LV side (2) Referred to HV side
Define 1
2
1



N
N
V
V
a
HV
LV
S

85
Choose:
rated
LV
b V
V ,
1 
rated
b S
S 
Compute:
1
1
2
1
b
b
LV
HV
b V
a
V
V
V
V 

b
b
b
S
V
Z
2
1
1 
b
b
b
S
V
Z
2
2
2 
2
2
1
2
1
2
2
2
1
2
1
)
1
(
a
V
a
V
V
V
Z
Z
b
b
b
b
b
b



Normally choose rated
values as base values

86
Per-unit impedances are:
1
1
.
.
b
S
S
u
p
Z
jX
R
Z


1
2
1
2
2
2
2
2
2
.
.
b
S
S
b
S
S
b
S
S
u
p
Z
jX
R
a
Z
a
jX
a
R
Z
a
jX
a
R
Z






So:
2
.
.
1
.
. u
p
u
p Z
Z 
Per-unit equivalent
circuits of transformer
referred to LV side and
HV side are identical !!

87
LV
V HV
V
S
S jX
R 
1
N 2
N
Fig 1. Eq Ckt referred to LV side
1
2
1



N
N
V
V
a
HV
LV
S
1
b
Z
1
b
V 2
b
V
Fig 2. Per-unit Eq Ckt referred to LV side Fig 3.
pu
S
Z ,
1
:
1
1
b
V 2
b
V
pu
S
Z ,
1
b
V 2
b
V
b
S

88
LV
V HV
V
1
N 2
N
Fig 4. Eq Ckt referred to HV side
1
2
1



N
N
V
V
a
HV
LV
S
2
b
Z
2
b
V
Fig 5. Per-unit Eq Ckt referred to HV side Fig 6.
pu
S
Z ,
1
:
1
1
b
V 2
b
V
pu
S
Z ,
1
b
V 2
b
V
1
b
V
2
2
a
X
j
a
R S
S

b
S

89
Voltage Regulation
%
100






load
full
load
full
load
no
V
V
V
VR
%
100
,
,
,






load
full
pu
load
full
pu
load
no
pu
V
V
V
VR
In per-unit system:

90
Voltage Regulation
%
100






load
full
load
full
load
no
V
V
V
VR
%
100
,
,
,






load
full
pu
load
full
pu
load
no
pu
V
V
V
VR
In per-unit system:

91
A single-phase transformer rated 200-kVA, 200/400-V, and
10% short circuit reactance. Compute the VR when the
transformer is fully loaded at unity PF and rated voltage
400-V.
Solution:
Fig 7. Per-unit equivalent circuit
P
V S
V
1
.
0
j
load
S
V
Vb 400
2 
kVA
Sb 200

pu
S pu
load 0
1
, 

pu
j
X pu
S 1
.
0
, 
S
X

92
Rated voltage:
pu
V pu
S 0
0
.
1
, 

pu
j
j
X
I
V
V
o
pu
S
pu
pu
S
pu
P
7
.
5
001
.
1
1
.
0
1
1
.
0
0
0
.
1
0
0
.
1
,
,
,











pu
V
S
I
pu
S
pu
load
pu
load 0
0
.
1
0
0
.
1
0
0
.
1
*
*
,
,
, 




















93
pu
V
V o
pu
P
load
no
pu 7
.
5
001
.
1
,
, 



pu
V
V pu
S
load
full
pu 0
0
.
1
,
, 



Secondary side:
Voltage regulation:
%
1
.
0
%
100
0
.
1
0
.
1
001
.
1
%
100
,
,
,










load
full
pu
load
full
pu
load
no
pu
V
V
V
VR

94
Select Vbase in generator circuit and Sb=100MVA,
compute p.u. equivalent circuit.
Problem 1
G

100
j
20 kV 22kV/220kV
80MVA
14%
220kV/20kV
50MVA
10%
50MVA
0.8 PF
lagging

95
Three-phase circuits
LV
L
bLV V
V ,




 I
V
S
S
Sb 3
3 1
3 

 

where
3
/
)
(line
L
neutral
to
line V
V
V 
 


L
current
line I
I
I 
 

HV
L
bHV V
V ,

L
L
b I
V
S 3

bHV
bHV
bLV
bLV
b I
V
I
V
S 3
3 


96
b
bLV
b
bLV
bLV
LV
LV
bLV
S
V
S
V
V
I
V
Z
2
)
(
3
3





b
bHV
bHV
S
V
Z
2
)
(

*
*
3
3
3
pu
pu
b
b
L
L
b
pu I
V
I
V
I
V
S
S
S 



bLV
b
bLV
V
S
I
3

bHV
b
bHV
V
S
I
3


97
Three 25-kVA, 34500/277-V transformers
connected in -Y. Short-circuit test on high voltage
side:
Determine the per-unit equivalent circuit of the
transformer.
V
V SC
Line 2010
, 
A
I SC
Line 26
.
1
, 
W
P SC 912
,
3 


98
(a) Using Y-equivalent
3
34500
277
S
S jX
R 
VA
Sb 25000

3
2010

SC
V
26
.
1

SC
I


 00
.
921
26
.
1
47
.
1160
SC
Z
V
VSC 47
.
1160
3
2010



99
So





 86
.
900
48
.
191
921 2
2
2
2
S
SC
S R
Z
X
W
P 304
3
912


 


 48
.
191
26
.
1
304
2
2
SC
S
I
P
R 


 86
.
900
48
.
191 j
ZSC
VA
Sb 25000
 V
V HV
b 58
.
19918
3
34500
, 



 99
.
15869
25000
58
.
19918 2
,HV
b
Z
pu
j
j
Z Y
pu
SC 0568
.
0
012
.
0
99
.
15869
86
.
900
48
.
191
,
, 




100
(b) Using -equivalent
34500 277

,
SC
Z
VA
Sb 25000

2010

SC
V
3
26
.
1

SC
I



 79
.
2764
727
.
0
2010
,
SC
Z
V
VSC 2010
 A
ISC 727
.
0
3
26
.
1



101
So





 

 30
.
2704
18
.
575
79
.
2764 2
2
2
,
2
,
, S
SC
S R
Z
X
W
P 304
3
912


 



 18
.
575
727
.
0
304
2
2
,
SC
S
I
P
R 


 86
.
900
48
.
191 j
ZSC
VA
Sb 25000
 V
V HV
b 34500
, 


 47610
25000
34500 2
,HV
b
Z
pu
j
j
Z pu
SC 0568
.
0
012
.
0
47610
30
.
1704
18
.
575
,
, 





102
Related Materials in Textbook
(1) Section 2.6 and 2.7, page 83~90, Chapman
book
(2) Section 2.10, page 113~116, Chapman book

103
•More meaningful when comparing different voltage levels
•The per unit equivalent impedance of the transformer remains the same when referred to either the primary or the
secondary side
•The per unit impedance of a transformer in a three-phase system is the same, regardless the winding connection
•The per unit method is independent of voltage changes and phase shifts through transformers
•Manufacturers usually specify the impedance of the equipment in per unit or percent on the base of its nameplate
ratings
•The per unit impedance values of various ratings of equipment lie in a narrow range
Transformer equivalent circuit can be simplified by properly specifying base quantities.
Give a clear idea of relative magnitudes of various quantities such as voltage, current, power and impedance.
Avoid possibility of making serious calculation error when referring quantities from one side of transformer to
the other
Per-unit impedances of electrical equipment of similar type usually lie within a narrow numerical range when the
equipment ratings are used as base values.
Manufacturers usually specify the impedances of machines and transformers in per-unit or percent in
nameplate rating
The circuit laws are valid in per unit systems, and the power and voltage equation are simplified since the factor
√3 and 3 are eliminated in the per-unit systems.
Ideal for the computerized analysis and simulation of complex power system problems.

Per unit systems in power systems

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