Department of Electrical & Communication Engineering
CHAPTER 1
1.
POWER SYSTEMS
Per Unit Calculations
Power System Representation
Power Component
M
Symbol
Power Component
Symbol
= Generator
= Circuit breaker
= Transformer
=
= Motor
= Feeder + load
Transmission
line
= Busbar (substation)
Power components and symbols
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POWER SYSTEMS
Interconnections among these components in the power system may be shown is a so-called one-line diagram
or single-line diagram. Single-line diagram represents all 3- of balanced system. For the purpose of analysis,
the single-line diagram of a particular power system network is represented to its equivalent reactance or
impedance diagram. A sample of a interconnected of individual power component is shown in Figure 1.1. This
represent a circuit diagram of a power network which is referred to as a single-line diagram.
M
Figure 1.1 – Single-line diagram
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POWER SYSTEMS
Impedance diagram
In power system fault calculations it is often that a single-line diagram representing a typical power network in
3- be converted into its per phase impedance diagram. Some assumptions for converting from single-line
diagram into its equivalent impedance diagram needed to be considered.
(i)
A generator can be represented by a voltage source in series with an inductive reactance. The internal
resistance of the generator is assumed to be negligible compared to the reactance.
(ii) The loads are usually inductive represented by resistance and inductance.
(iii) The transformer core is assumed to be ideal, and the transformer may be represented by a reactance
only.
(iv) The transmission line is represented by its resistance and inductance, the line-to-ground capacitance is
assumed to be negligible.
Let us consider the following on how the single-line diagram of Figure 1.2 converted into its impedance diagram
counterpart.
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Generator
G1
Station A
Transformer
T1
Transmission
Line
TL
POWER SYSTEMS
Transformer
T2
Station B
G3
G2
G4
Load
L1
Load
L2
Figure 1.2 – Single-line diagram of a power network
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Transformer
T1
Transmission
Line
TL
POWER SYSTEMS
Transformer
T2
Station B
Station A
j XT1
j X1
j X2
G1
G2
RL1
j XL1
RTL
j XTL
j XT2
RL2
j XL2
j X3
G3
j X4
G4
Figure 1.3 – Impedance diagram of Figure 1.2
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POWER SYSTEMS
Per-Unit Quantities
Per unit quantities are quantities that have been normalized to a base quantity. In general,
Z pu 
Z actual
Z base
per-unit (p.u)
Choice of the base value Zbase is normally a rated value which is often one of the normal full-load operations of
power component in a power network.
Let us look at two of the most common per unit formula which are widely used when per unit calculations are
involved.
(i)
Base impedance (Zbase)
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. Let begin with 3- base
quantity of
Sbase  3Vbase I base
----- (i)
where Vbase = line voltage, Ibase= line or phase current
Per phase base impedance,
V base
Z base 
I base
3
-----(ii) This is line-to-neutral impedance
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POWER SYSTEMS
Combining (i) and (ii) yields,
V base
Z base 
S base
3
Z base
3V base

kV base 2

MVAbase
where kVbase and MVAbase are 3- qualtities
(ii)
Changing base impedance (Znew]
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,
2
Z NEW
 kVbase OLD 
MVAbase NEW
pu

Z


  OLD
2
MVAbase OLD
 kVbase NEW 
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POWER SYSTEMS
Example 1
Determine the per-unit values of the following single-line diagram and draw the impedance diagram.
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
Transmission line
j 3.48 
50 MVA
132 kV/66 kV
XT2 = 0.04 p.u
XT1 = 0.1 p.u
Load
40 MW, 0.8 p.f. lagging
Solution:
Chosen base: Always choose the largest rating, therefore Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV
Per-unit calculations:
Transformer T1:
Generator G1:
2
Z NEW
 kVbase OLD 
MVAbase NEW
pu

Z


  OLD
2
MVAbase OLD
 kVbase NEW 
X g ( pu )  0.16 
100
 0.32
50
p.u.
X T 1 ( pu )  0.1
p.u.
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Transmission line TL:
Z base
Transformer T2:

kV base 2

MVAbase
3.4  100
X TL ( pu ) 
POWER SYSTEMS
1322
Z pu 
Z actual
Z base
X T 2 ( pu )  0.04 
100
 0.08
50
p.u.
 0.0195 p.u.
Inductive load:
66  103
Z actual 
Z L ( pu ) 
40  10
3
 87.1236.87o
6

3  66  103  0.8
87.1236.87o  100
2
66
 236.87o or (1.6  j1.2)
p.u.
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POWER SYSTEMS
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.
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
Transmission line
j 3.48 
50 MVA
132 kV/66 kV
XT2 = 0.04 p.u
XT1 = 0.1 p.u
Load
40 MW, 0.8 p.f. lagging
Transformer
T1
j 0.1 p.u.
j 0.32 p.u.
Transmission Line
TL
j 0.0195 p.u.
Transformer
T2
j 0.08 p.u.
1.6 p.u..
Generator
G
Load
j 1.2 p.u.
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POWER SYSTEMS
Load
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POWER SYSTEMS
Summarise:
Load
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POWER SYSTEMS
Load
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