Dr. Mohamed Taha Mouwafi
Department of Electrical Engineering
Menoufia University, Egypt
Lecture: 3
The Per-Unit System
 In the analysis of power networks, instead of using actual values of
quantities it is usual to express them as fractions of reference
quantities, such as rated or full-load values. These fractions of
reference are called per unit (denoted by p.u.) and the p.u. value of
any quantity is defined as:
actual value (in any unit )
base or reference value in the same unit
where actual quantity is the value of the quantity in the actual units.
The base value has the same units as the actual quantity, thus making
the per-unit quantity dimensionless. Also, the base value is always a
real number. Therefore, the angle of the per-unit quantity is the same
as the angle of the actual quantity.
 Power-system quantities such as voltage, current, power, and
impedance are often expressed in per-unit or percent of specified
base values. For example, if a base voltage of 20 kV is specified, then
the voltage 18 kV is (18/20)= 0.9 per unit or 90%. Calculations can
then be made with per-unit quantities rather than with the actual
quantities.
 For example,
S pu 
S
S base
,
V
V pu 
,
V base
I pu 
I
I base
Z
and Z pu 
Z base
Also,
Example 1
A d.c. machine rated at 200 V, 100 A has an armature resistance of 0.1 Ω
and field resistance of 0.15 Ω. The friction and windage loss is 1500 W.
Calculate the efficiency when operating as a generator.
Solution
Choose Vb = 200 V and Ib = 100 A,
Then Vpu = Vr/Vb=1 p.u.
and
Ipu = Ir/Ib=1 p.u.
Also, (VA)b = 200 x 100 = 20,000 VA and Rb = Vb / Ib = 200/100 = 2 Ω
and, Rw = Ra + Rf =0.1 + 0.15 = 0.25 Ω
and, Pw =I2 x Rw = (100)2 x 0.25 = 2500 W
Hence,
Total
PLoss
 1500  2500  4000 W

Therefore,
4000
 0.2
20, 000
p .u .
V pu  I pu
Pout
% 
 100 
 100
Total
Total
Pout  PLoss
V pu  I pu  PLoss

11
 100  83%
11  0.2
Three-phase circuits
It is a convenient in a.c. circuit calculations to work in terms of base voltamperes (VAbase). Thus,
V Abase 
3 V base  I base
When Vbase is the line voltage and Ibase is the line current in a threephase system. Hence,
I base 
Therefore,
V Abase
3 V base
2
V base
3
V base
3
V base
Z base 


I base
V Abase
3 V base V Abase
Hence,
Z pu 
Z 
Z base
Z 
V Abase
 2
 Z  
2
V base V Abase
V base
(1)
(2)
(3)
Per-Unit Representation of Transformer
Consider a single-phase transformer in which the total series impedance
of the two windings referred to the primary is Z1 as shown in Figure
below.
V1 1
V2

N
I1 N

I2 1
Then the p.u. impedance =Z1 I1/V1, where I1 and V1 are the rated or base
values.
The total impedance referred to the secondary = Z1N2
and this in p.u. notation is
2  I2 
 Z 1N 

V 2 
I1
Z 1 I1
1
 Z 1N 


N V 1N
V1
Hence the p.u. impedance of a transformer is the same whether
considered from the primary or the secondary side.
2
Change from an old base to a new base
We know that,
Z pu 
Z 
Z base
V Abase
 Z  
2
V base
Then, to change from the old base to a new one:
Z
new
pu
Z
old
pu
V A

V A
new
base
old
base
 V

 V
old
base
new
base



2
(4)
Example 2
In the network of Figure below, two single-phase transformers supply a
10 kVA resistance load at 200 V. Show that the p.u. load is the same for
each part of the circuit and calculate the voltage at point D.
Solution
Choose
VAb  10000
VAb  10000
Vb  100
Vb  400
VAb  10000 VA
Vb  200
V
RB  Rc  N 2
V2
VA  V  I 
RC
RA  RB  N
2
 16  (100 400) 2
 RA  1 
10000
RA  1
 1 p.u.
100 2
 4  (400 200) 2
 RB  16 
RB  16 
10000
 1 p.u.
2
400
V2
200 2
 RC 

4
VA 10000
RC 
4  10000
 1 p.u.
2
200
The equivalent circuit is shown in Figure below.
Equivalent circuit
Hence the supply voltage Vs = VR + I X
= 1+1x(j0.1+j0.15)
Hence,
p.u.
Vs = 1.03 p.u.
Therefore, the voltage at point D = VR +(j0.15) x I
= 1+j0.15 x 1
= 1.012
p.u.
= 1.012 x 400 = 404.8 V
Example 3
The schematic diagram of a radial transmission system is shown in
Figure below. The ratings and reactances of the various components as
shown, along with the nominal transformer line voltages. A load of 50
MW at 0.8 p.f. lagging is taken from the 33 kV substation which is to be
maintained at 30 kV. It is required to calculate the terminal voltage of
the synchronous machine. The line and transformers may be
represented by series reactances. The system is three-phase.
Solution
Choose
MVAb  100
kVb  11
MVAb  100
kVb  132
MVAb  100 MVA
kVb  33
kV
For the load:
50
Pr ( p.u.) 
 0. 5
100
30
Vr 
 0.910.0
33
 Pr ( p.u.)  V r  I r  cos( r )
Pr   cos 1 (0.8)
so, I r ( p.u.) 
V r  cos( r )
0.5  36.87 o

 0.687  36.87 o
(0.910.0)  0.8
Example 4
Three zones of a single-phase circuit are identified in Figure below. The
zones are connected by transformers T1 and T2, whose ratings are also
shown. Using base values of 30 kVA and 240 volts in zone 1, draw the
per-unit circuit and determine the per-unit impedances and the per-unit
source voltage. Then calculate the load current both in per-unit and in
amperes. Transformer winding resistances and shunt admittance
branches are neglected.
Solution
First the base values in each zone are determined. Sbase = 30 kVA is the
same for the entire network. Also, Vbase1 = 240 volts, as specified for
zone 1. When moving across a transformer, the voltage base is changed
in proportion to the transformer voltage ratings. Thus,
and,
The base impedances in zones 2 and 3 are,
and,
and the base current in zone 3 is,
Next, the per-unit circuit impedances are calculated using the system
base values. Since Sbase = 30 kVA is the same as the kVA rating of
transformer T1, and Vbase1 = 240 volts is the same as the voltage rating of
the zone 1 side of transformer T1, the per-unit leakage reactance of T1 is
the same as its nameplate value, XT1p.u. = 0.1 per unit. However, the perunit leakage reactance of transformer T2 must be converted from its
nameplate rating to the system base. Using Equation (4) and Vbase2 = 480
volts,
Alternatively, using Vbase3 = 120 V,
which gives the same result. The line, which is located in zone 2, has a
per-unit reactance,
and the load, which is located in zone 3, has a per-unit impedance,
The per-unit circuit is shown in Figure below, where the base values for
each zone, per-unit impedances, and the per-unit source voltage are
shown. The per-unit load current is then easily calculated from the
Figure as follows:
Per-unit circuit