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Per-Unit System
1.0 Introduction
The per-unit system is a widely system of
normalization. Being familiar with it is
essential to functioning in the world of
electric power engineering.
2.0 What is per-unit? (Section 5.5, 5.6)
The per-unit system is a way to transform
the numerical quantities (voltages, currents,
powers, and impedances) to gain certain
advantages while maintaining the basic
relations between them (Ohm’s laws).
The main advantage is that voltages tend to
be very close to 1.0 and consequently
numerical algorithms are more reliable. In
addition, the range of values for voltage,
currents, powers, and impedances tends to
be narrower, offering opportunities to detect
bad data.
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The basic idea in the per-unit system is that
all voltage, currents, powers, and
impedances are normalized according to
Value pu
Valueunits

BaseValue
(1)
In order to remain consistent with Ohm’s
laws, we are allowed to choose the base
values for any two quantities we want
(voltage, current, power, or impedance).
Then the other two values must be
computed.
For example, it is common to choose the
voltage and power bases.
 Voltage base is chosen as nominal voltage
 Power base is chosen as multiple of 10: 1,
10, or 100 kVA or MVA
Then, we compute:
I base 
Z base 
S1base
(2)
VLNbase
VLNbase 2
S1base
VLNbase

I base
2
(3)
The above works for single phase systems or
3-phase systems. In addition, for 3-phase
systems, we also have:
VLLbase  3VLNbase
S 3base  3 S1base
I base  I base,Y 
Z base  Z base,Y 
(4)
(5)
S 3base
3V LLbase
(6)
S 3base
(7)
VLLbase 2
In addition:
Z base,  3 Z base,Y
I base,  
1
3
(8)
I base,Y
(9)
3.0 Changing base
A calculation that is done very often is to
convert a per-unit impedance on one base to
a per-unit impedance on another base.
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A common situation when this calculation is
required is when the purchasing a new piece
of equipment (e.g., transformer, generator)
from a manufacturer (e.g., ABB, GE,
Toshiba, etc.), that has to be modeled in a
system (e.g. power flow) study.
The manufacturer will typically supply data
in per unit, but on bases that are chosen
consistent with the component ratings.
These bases are unlikely to be the same as
the bases used in the system study.
These formula are easy to derive, by
equating the impedance in ohms expressed
as a function of the per-unit impedance, e.g.,
Z   Z pu1 Z base1  Z pu2 Z base2
where
Z base1 
VLLbase1 2
S 3base1
,
Z base2
(10)
2

VLLbase 2 

S 3base2 (11)
Substituting eqs. (11) into (10) and solving
for Zpu1 yields
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Z pu1  Z pu2
2

Z base2
VLLbase 2  S 3base1
 Z pu2
Z base1
S 3base2 VLLbase 1 2
2

VLLbase 2 
 Z pu2
VLLbase 1 2
S 3base1
(12)
S 3base2
It may be easier for you to remember it
when written in terms of the “new” base
(base you are converting to) and the “old”
base (base you are converting from):
2

VLLbase ,old  S 3base,new
Z pu,new  Z pu,old
VLLbase ,new 2 S3base,old (13)
It may be the case that only voltage base is
changed, or that only power base is changed.
The part of eq. (13) that does not get
changed will simply be 1.0, and so can be
ignored in the expression.
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4.0 Perunitization of transmission systems
We will assume our system is “normal” as
defined in Section 5.4. This means that the
voltage gain for any paralleled transformers
is the same.
We will need to per-unitize the data for an
entire transmission system. The difficulty
here is that, because of transformers, there
are different nominal voltages.
The solution to this is to choose the voltage
base for one “section” of the system. A
section is a set of interconnected
components not separated by a transformer.
Then compute the voltage bases for all other
sections of the system so that voltage bases
of different sections are in the same ratio as
the line-to-line voltages.
Let’s work a problem.
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Problem 5.19:
Draw an impedance diagram for the system
whose one-line diagram is shown in Fig. 1.
T1
T2
G1
Line 1
G2
Section D
Section A
Line 2
Section B
Line 3
T3
Section C
M
Fig. 1
Data for the system is:
G1: 50 MVA, 13.8 kV, X=0.15 pu
G2: 20 MVA, 14.4 kV, X=0.15 pu
M : 20 MVA, 14.4 kV, X=0.15 pu
T1 : 60 MVA, 13.2kV/161kV, X=0.10 pu
T2 : 25 MVA, 13.1kV/161kV, X=0.10 pu
T3 : 25 MVA, 13.2kV/160kV, X=0.10 pu
(I changed voltage rating on LV side of T2 and
HV side of T3).
Line 1: 20+80 ohms
Line 2: 10+j40 ohms
Line 3: 10+j40 ohms
Load: 20+j15 MVA at 12.63 kV
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We begin by choosing the system power
base as S3φ,base=100 MVA.
We must also choose the voltage base in one
section of the system. We will select 161kV
in Section D.
Now we compute the voltage bases in the
other three sections of the system.
13.2 VLLbaseA
Section A: 161  161 , VLLbaseA  13.2kV
13.1 VLLbaseB
Section B: 161  161 , VLLbaseB  13.1kV
13.2 VLLbaseC
Section C: 160  161 , VLLbaseC  13.2825kV
Now we can use eq. (13) to convert the
given per-unit impedances for G1, G2, M, T1,
T2, and T3 into per-unit impedances on our
new bases.
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G1:
2

13.8  100
Z pu,new  0.15
 0.3279
2
13.2 50
G2:
2

14.4 100
Z pu,new  0.15
 0.9062
2
13.1 20
M:
2

14.4 100
Z pu,new  0.15
 0.8815
2
13.285 20
T1:
Z pu,new  0.10
100
 0.1667
60
T2:
Z pu,new
T3:
100
 0.10
 0.4
25
13.22
100
Z pu,new  0.10
 0.3950
2
13.2825 25
Note that the last calculation (for T3) could
have been done as follows:
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2


160 100
Z pu,new  0.10
 0.3950
2
161 25
Now let’s per-unitize the lines. The line
impedances are all in ohms. So we need to
find the impedance base for Section D. From
eq. (7), we get:
Z baseD
2

VLLbaseD 
(161E 3) 2


 259.21
S 3base
100 E 6
Then we compute
Z Line 1, pu 
Z Line 2, pu 
Z Line 3, pu 
Z Line 1,
Z baseD
Z Line 2,
Z baseD
Z Line 3,
Z baseD
20  j 80

 0.07716 + j0.3086
259.21
10  j 40

 0.03858 + j 0.1543
259.21
10  j 40

 0.03858 + j 0.1543
259.21
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Load:
The load requires a bit of thinking. We are
told that it is 20+j15 MVA at 12.63 kV. Now
we could convert it to per-unit power by
dividing by 100. However, the problem
requires that we develop the impedance
diagram. So we need to convert this powerspecification to an impedance specification.
We can do this because we know the voltage
at which the given power is consumed. But
one question remains. Since the power is
complex, there needs to be an R and an X. But
is the R and X in series or in parallel?
We are not given this information and so it is
up to us to assume one or the other. We will
assume a series combination. In this case,
2
Z Load
V
(12.63  103 )2
 * 
 5.1045 + j3.8284
6
S
(20  j15)  10
Now we need to per-unitize it. To do this,
we need the impedance base of Section C.
This is computed as:
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Z baseC
2

VLLbaseC 
(13.2825 E 3) 2


 1.7642
S 3base
Now we may
impedance as
Z Load , pu 
100 E 6
compute
the
per-unit
Z Load 5.1045 + j3.8284

 2.8934 + j2.1700
Z baseC
1.7642
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