A brief description about 1.Unit system. 2.Example

1. CHAPTER 2
The Per-Unit System

2. Per Unit System
 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.

3. Per Unit System
 The voltage, current and impedance values are
divided by base values and expressed in per unit or
percentage values.
 The percent impedance,
 e.g. in a synchronous generator with 13.8 kV as its
nominal voltage, instead of saying the voltage is
12.42 kV, we say the voltage is 0.9 p.u.
Value
Base
Value
Actual
Value
System
Per Unit 
100%
x
in
in
Z
base
actual
%



Z
Z

4. base
base
base
base
base
base
base
base
base
base
S
V
I
V
Z
V
S
I
V
S
2
,



How Are the Base Values
Defined
 For an electric element, we
have : Power, Voltage, Current
and Impedance.
 Usually, the nominal apparent
power (S) and nominal voltage
(V) are taken as the base
values for power and voltage.
 The base values for the current
and impedance can be
calculated.

5. Per Unit values using the base values are:
base
pu
S
S
S

base
pu
I
I
I

base
pu
V
V
V

base
pu
Z
Z
Z

Z
Z
2
base
base
base
pu
V
S
Z
Z 
 pu
base
2
base
pu
base Z
S
V
Z
Z 

Z
Conversion from per unit value to ohm & vise
versa
Per Unit System

6. Transformers Voltage Base
V1/V2
Vb1 Vb2
1
1
2
2 b
b V
V
V
V 










7. Per Unit in 3- Circuits
 Simplified:
 Concerns about using phase or line voltages
are removed in the per-unit system
 Actual values of R, XC and XL for lines, cables,
and other electrical equipment typically phase
values.
 It is convenient to work in terms of base VA
(base volt-amperes)

8. Per Unit System (3 Phase)
,
3
3
B
B
B
B
B
B
V
S
I
I
V
S


• Usually, the 3-phase SB or MVAB and line-to-line VB or kVB
are selected
• IB and ZB dependent on SB and VB
 
B
B
B
B
B
B
B
B
S
V
I
V
Z
Z
I
V
2
3
/
3




9. Change of Base
 The impedance of individual generators &
transformer, are generally in terms of
percent/per unit based on their own ratings.
 Impedance of transmission line – ohmic value
 When pieces of equipment with various
different ratings are connected to a system, it
is necessary to convert their impedances to a
per unit value expressed on the same base.

10.   











 Z
V
S
Z
Z
Z
old
B
old
B
old
B
old
pu 2
  











 Z
V
S
Z
Z
Z
new
B
new
B
new
B
new
pu 2
old
B
old
B
V
base
voltage
&
S
base
power
on the
impedance
unit
per
the
be
old
pu
Z
new
B
new
B
V
base
voltage
new
&
S
base
power
new
on the
impedance
unit
per
new
the
be
new
pu
Z
1
2
Change of Base

11. 2
















 new
B
old
B
old
B
new
B
old
pu
new
pu
V
V
S
S
Z
Z








 old
B
new
B
old
pu
new
pu
S
S
Z
Z
From (1) and (2), the relationship between the
old and the new per unit value
If the voltage base are the same,
Change of Base

12. Example
The one-line diagram of three-phase power
system is shown below. Select a common base
of 100 MVA and 22 kV on the generator side.
Draw an impedance diagram with all impedance
including the load impedance marked in per-
unit.

13. Example
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
65.43 Ω

14.  Voltage base for all sections of the network.
 SB = 100 MVA, VB = 22 kV on Generator side
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
Example
65.43 Ω

15. G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
VB1 on the LV of T1 = 22 kV
50 MVA, 22/220 kV, 10%
40 MVA, 22/110 kV, 6.4%
Example

16. G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
40 MVA, 22 kV, 6.4%
50 MVA, 22/220 kV, 10%
VB2 on the HV of T1 = kV
kV
V
V
V
V B
B 220
22
220
22
1
2
1
2 
















VB3 on the HV of T2 = VB2 = 220 kV
Example

17. G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
40 MVA, 22/110 kV, 6.4%
VB5 on the LV of T3 = kV
kV
V
V
V
V B
B 110
22
110
22
1
2
1
5 
















VB6 on the HV of T4 = VB5 = 110 kV
Example
65.43 Ω

18. G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1
220 kV
Line 2
110 kV Load
90 MVA
22 kV
X = 18%
50 MVA
22/220 kV
X = 10%
40 MVA
22/110 kV
X = 6.4%
40 MVA
110/11 kV
X = 8.0%
40 MVA
220/11 kV
X = 6.0%
66.5 MVA
10.45 kV
X = 18.5%
57 MVA
0.6 pf lag
10.45 kV
X = 48.4 Ω
X = 64.43 Ω
VB4 on the LV of T2 =
kV
kV
V
V
V
V B
B 11
220
11
220
1
2
3
4 
















VB4 on the LV of T4 =
40 MVA, 220/11kV, 6.0%
40 MVA, 110/11kV, 8.0%
or
kV
kV
V
V
V
V B
B 11
110
11
110
1
2
6
4 
















Example

19. Since generator & transformer voltage base are the same as their rated
values, their p.u reactance on a 100 MVA








 old
B
new
B
old
pu
new
pu
S
S
Z
Z
u
p
X
u
p
X
u
p
X
u
p
X
u
p
X
T
T
T
T
G
.
2
.
0
40
100
08
.
0
.
16
.
0
40
100
064
.
0
.
15
.
0
40
100
06
.
0
.
2
.
0
50
100
10
.
0
.
2
.
0
90
100
18
.
0
4
3
2
1








































Generator & Transformer

20. u
p
V
V
S
S
X
X new
B
old
B
old
B
new
B
old
pu
new
pu .
25
.
0
11
45
.
10
5
.
66
100
185
.
0
2
2































MVA
S
MVA
S
X
old
B
new
B
old
pu
5
.
66
100
185
.
0
%
5
.
18




kV
V
kV
V
old
B
new
B
45
.
10
11


Motor

21. Line 1 & 2
MVA
S
kV
V
B
B
100
220


MVA
S
kV
V
B
B
100
110


 
  







121
100
110
484
100
220
2
2
2
2
2
1
MVA
kV
S
V
X
MVA
kV
S
V
X
B
B
l
B
B
B
l
B
Line 1 Line 2
u
p
X
X
X
u
p
X
X
X
l
B
actual
l
u
p
l
B
actual
l
u
p
.
54
.
0
121
43
.
65
.
10
.
0
484
4
.
48
2
2
1
1
.
.






Base Impedance, XB P.U Impedance, Xpu

22. Load
lagging
f
p
kV
V
MVA
S 6
.
0
.
,
45
.
10
,
57 


 
 
 
 
 
u
p
j
j
Z
Z
Z
MV
kV
S
V
Z
j
S
V
Z
MVA
S
Therefore
Base
L
actual
L
u
p
L
B
B
Base
L
o
L
L
L
actual
L
o
L
o
.
2667
.
1
95
.
0
21
.
1
53267
.
1
1495
.
1
21
.
1
100
11
53267
.
1
1495
.
1
13
.
53
57
45
.
10
13
.
53
57
,
13
.
53
6
.
0
cos
)
(
)
(
)
.
(
2
2
)
(
2
*
3
2
)
(
3
1


























23. Per Unit Equivalent Circuit
G M
1 2 3 4
5 6
XG= 0.2p.u
XT1= 0.2p.u XT2= 0.15p.u
XT3= 0.16p.u XT4= 0.20p.u
XL1= 0.10p.u
ZL1= 0.54p.u
XM= 0.25p.u
ZLoad= 0.95+j1.2667

24. Advantages
 Give a clear idea of relative magnitudes of various
quantities, such as V, I, P & Z.
 The per unit values of Z, V & I of transformer are the
same whether they are referred to the primary or
secondary side.
 Ideal for the computerized analysis and simulation of
complex power system problems.
 The circuit laws are valid in per unit systems, and the
power and voltage equation are simplified since the
factor √3 and 3 are eliminates in the p.u systems.