Per Unit Analysis of Three Phase System

1. Lecture 14
Per Unit Analysis of Three
Phase System

2. 2
After completing this unit you will be able to:
• Apply per unit models of power system components such as
generators, transformers, transmission lines, etc. to analyse a
balanced three-phase power system.
Learning Outcomes

3. 3
Per-Unit System for Three Phase System
For a given single-line (one-line) diagram of a power network, all
component parameters are expressed in 3- quantity whether it is
the rating (capacity) expressed as MVA or voltage as kV. Lets begin
with 3- base quantity of
base
base
base I
V
S 3

where Vbase = line voltage, Ibase = line or phase current
Per phase base impedance,
base
base
base
I
V
Z
3

This is line-to-neutral impedance.
(i)
(ii)

4. 4
Combining (i) and (ii) yields,
 
base
base
base
base
base
base
MVA
kV
V
S
V
Z
2
3
3 

where kVbase and MVAbase are 3- qualtities

5. 5
Sometimes the parameters for two elements in the same circuit
(network) are quoted in per-unit on a different base. The changing
base impedance is given as,
Changing base impedance (Znew]
 
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
 
 
  
 
 

6. 6
Determine the per-unit values of the following single-line diagram
and draw the impedance diagram. From the impedance diagram
obtained find the transmission line current.
Example
XT1 = 0.1 p.u
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
50 MVA
132 kV/66 kV
Transmission line
j 3.48 
XT2 = 0.04 p.u Load
40 MW, 0.8 p.f. lagging

7. 7
Chosen base: Always choose the largest rating, therefore
Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV
Solution
Per-unit calculations
Generator G1:
 
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
 
 
  
 
 
32
.
0
50
100
275
275
16
.
0
)
( 2
2



pu
X g p.u.

8. 8
Transformer T1:
1
.
0
)
(
1 
pu
XT
p.u.
Transmission line TL:
 
base
base
base
MVA
kV
Z
2

actual
pu
base
Z
Z
Z
 where
Therefore,
 
0195
.
0
132
100
4
.
3
2
2
.)
.
( j
j
kV
MVA
Z
Z
base
base
actual
u
p
TL 



 p.u.

9. 9
Transformer T2:
 
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
 
 
  
 
 
08
.
0
50
100
66
66
04
.
0 2
2
.)
.
(
2 


u
p
T
X
Now, we have all the impedance values in per-unit with a common
base and we can now combine all the impedances and determine
the overall impedance.

10. 10
Inductive load:
A
39
.
437
8
.
0
10
66
3
10
40
cos
3 6
6










LL
L
V
MVA
I
Magnitude of load current is
Since the load is inductive, so we can write
Given:
8
.
0
cos 

Therefore,
  

 
87
.
36
8
.
0
cos 1

A
87
.
36
39
.
437 



L
I

11. 11












 87
.
36
12
.
87
87
.
36
39
.
437
0
3
10
66 3
)
(
L
Phase
actual
L
I
V
Z
Therefore,
 






 2
.
1
6
.
1
56
.
43
87
.
36
12
.
87
)
(
.)
.
( j
Z
Z
Z
base
actual
L
u
p
L
and
  


 56
.
43
100
662
2
base
base
base
MVA
kV
Z
Therefore,

12. 12
Load
G
j 0.32 p.u.
j 0.1 p.u. j 0.0195 p.u.
Transformer
T1
Transformer
T2
Transmission Line
TL
j 0.08 p.u.
1.6 p.u..
j 1.2 p.u.
Generator
XT1 = 0.1 p.u
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
50 MVA
132 kV/66 kV
Transmission line
j 3.48 
XT2 = 0.04 p.u Load
40 MW, 0.8 p.f. lagging

13. 13
Solution
Transmission line current
p.u.
06
.
47
426
.
0
j1.72
1.6
0
1
.)
.
(
.)
.
(
.)
.
( 








u
p
total
u
p
S
u
p
TL
Z
V
I
 
 
A
39
.
437
10
132
3
10
100
3 3
6
)
( 






base
base
base
TL
V
MVA
I
Therefore,
A
06
.
47
3
.
186
39
.
437
06
.
47
426
.
0
.)
.
(
)
(










 base
u
p
TL
actual
TL I
I
I

14. 14
END