Per Unit Analysis
The per unit method of power system analysis eliminates the need for conversion of voltages, currents
and impedances across every transformer in the circuit. In addition, the need to transform from 3-phase
to 1-phase equivalent and vice versa is avoided with the use of per unit quantities.
Single-Phase systems
In the per unit system, any electrical quantity may be expressed in per unit as the ratio of the actual
quantity and the chosen base value for that quantity.
per unit value =
actual value
base value
Four base quantities must be considered, that are: apparent power, voltage, current and impedance. In
single-phase systems, the relationships among the base quantities are:
S base = Vbase I base
Vbase = I base Z base
With only two equations relating the four quantities, it is necessary to specify two base values. The
apparent power and the voltage bases are usually chosen equal to the rated values and the other two
base values are computed accordingly.
I base =
S base
Vbase
(V ) (kVbase )
V
= base = base =
I base
S base
MVAbase
2
Z base
2
The specified power base is applicable to all parts of the power system. However, the voltage base
varies across a transformer and so does the current and impedance bases. Note that the actual voltage,
current and impedances change across the transformer. However, because the bases also change across
the transformer, the per unit values are the same on both sides of the transformer.
31
Example
A generator supplies a load through a feeder whose impedance is Z f = 1 + j 2Ω . The load impedance is
Z L = 8 + j 6Ω . The voltage across the load is 120V. Find the real power and reactive power supplied by
the generator. Take the load voltage as the reference phasor for per unit analysis. Assume that the
generator has zero impedance and that S base = 1500VA & Vbase = 120V .
Solution
Ig
ZF
VL
Eg
The base current is: I base =
IL
ZL
S base 1500
=
= 12.5 A
Vbase 120
The base impedance is: Z base
2
2
Vbase )
(
120 )
(
=
=
S base
1500
= 9 .6 Ω
The per unit values are computed as follows:
VL =
120∠0
= 1∠0 pu
120
ZL =
10∠36.9
= 1.04∠36.9 pu
9 .6
IL =
1.0∠0
= 0.96∠ − 36.9 pu
1.04∠36.9
ZF =
2.24∠63.4
= 0.233∠63.4 pu
9 .6
32
The generator voltage in per unit is:
E g = VL + I L Z F
E g = 1∠0 + (0.96∠ − 36.9 )(0.233∠63.4 ) = 1.204∠4.8 pu
E g = 1.204∠4.8 × 120∠0 = 144.5∠4.8 V
The complex power is:
S g = E g I g*
S g = (1.204∠4.8)(0.96∠36.9 ) = 1.156∠41.7 = 0.863 + j 0.769 pu
Pg = 0.863 × 1500 = 1295 Watt
Qg = 0.769 × 1500 = 1154 VAR
Three-Phase systems
In three-phase systems, the base power is the total three-phase power and the base voltage is the lineto-line voltage. With the Star connection assumed, the base line current is equal to the base phase
current. The base impedance is per phase. The three-phase base quantities are related to the single
phase bases as follows:
Base Power, S base
3−φ
= 3S base 1−φ
Base Voltage, VL
base
= 3V phase
base
=
Base Current, I L
Base Impedance, Z base
S total
=
base
3VL
(V
=
base
base
S phase
base
V phase
base
) = (V
= I phase
S total
base
)
2
2
L base
base
phase base
S phase
base
33
The impedance characteristic of electrical equipment is usually expressed as a percentage based on its
ratings. When such a device is connected in a power system in which the selected base values are
different from the machine ratings, the per-unit quantities have to be expressed in terms of the system
bases. The per unit values of any impedance may be converted to the new base as follows:
Z pu
new
= Z pu
æ S base new öæ Vbase old ö
ç
֍
÷
old ç S
÷
ç
÷
V
è base old øè base new ø
2
Example
A three-phase, 60 Hz, 30 MVA, 13.8kV, wye connected synchronous generator has an armature
resistance Ra = 2Ω per phase and a synchronous reactance X s = 10Ω per phase.
a) Express the machine impedance in per unit based on the machine ratings.
b) Using the results in part (a), find the per unit impedance based on S base = 50MVA and
Vbase = 34.5kV .
Solution
a) Z base
2
13.8)
(
=
30
= 6.35Ω
Z s = Ra + jX s = 2 + j10 = 10.2∠78.7Ω
Zs
pu
=
Zs
10.2∠78.7
=
= 1.606∠78.7 pu
Z base
6.35
2
b) Z s
pu − new
Zs
pu − new
æ 50 öæ 13.8 ö
= 1.606∠78.7ç ÷ç
÷ = 0.428∠78.7 pu
è 35 øè 34.5 ø
= 0.084 + j 0.42 pu
34
NOTE
In large power systems, the power base is usually chosen to be the rating of one of the major
equipment. The power base is the same for all parts of the power system. The voltage bases are
normally chosen as the nominal voltages in the various parts of the system or are selected to be the
voltage ratings of the transformer windings. The current and impedance bases are computed based on
the previously selected values of the power and voltage bases.
Example
A transformer of 30MVA, 60Hz, 15/138kV has equivalent impedance referred to the primary side of
Z = 0.75 Ω . Calculate the Pu impedance of the transformer referred to both primary and secondary
windings, taking the power base as 30MVA.
Solution
From the primary side
Vbase 1 = 15kV and S base = 30MVA
S base 30,000
=
= 2,000 A
Vbase
15
I base 1 =
Z base 1 =
Z pu =
1
(Vbase )2 = (15)2
S base
Z1
Z base 1
30
=
= 7 .5 Ω
0.75
= 0.1 pu
7.5
35
From the secondary side
2
æN ö
æ 138 ö
Z = Z1 çç 2 ÷÷ = 0.75ç
÷ = 0.75 × 84.64 = 63.48 Ω
è 15 ø
è N1 ø
2
\
1
Vbase 2 = 138kV and S base = 30MVA
I base 2 =
Z base
2
2
2
Vbase )
(
(
138)
=
=
Z pu =
2
S base 30,000
=
= 217.39 A
Vbase
138
S base
Z2
Z base
∴ Z pu 1 = Z pu
30
=
2
2
= 634.8 Ω
63.48
= 0.1 pu
634.8
= Z pu = 0.1 pu
Example
The one-line diagram of a three-phase power system is shown in the figure below. Select a common
base of 100MVA and 22kV on the generation side and draw the impedance diagram with all
impedances including the load impedance in per unit. The manufacturer data for each device is given in
the table below. The three-phase load at bus #4 absorbs 57MVA, 0.6 power factor lagging at 10.45kV.
Lines 1 and 2 have reactances of 48.4 and 65.43 ohms, respectively.
Device Power (MVA) Voltage (kV) Reactance (%)
G1
90
22
18.0
T1
50
22/220
10.0
T2
40
220/11
6.0
T3
40
22/110
6.4
T4
40
110/11
8.0
M
66.5
10.45
18.5
36
1
T1
T2
2
Line #1
220 kV
3
4
G
M
T3
5
Line #2
110 kV
6
T4
Load #1
Solution
Vbase 1 (Low voltage of T1 ) = 22kV
Vbase
2
(High voltage of T1 ) = 22æç 220 ö÷ = 220kV
Vbase
3
(High voltage of T2 ) = 220kV
Vbase
4
(Low voltage of T2 ) = 220æç
è 22 ø
11 ö
÷ = 11kV
è 220 ø
æ 110 ö
Vbase 5 = Vbase 6 = 22ç
÷ = 110kV
è 22 ø
The new per unit values for the reactances of G, T1, T2, T3 and T4 are: (Note: there is no change in the
voltage base)
æ 100 ö
G: X = 0.18ç
÷ = 0.2 pu
è 90 ø
æ 100 ö
T1: X = 0.10ç
÷ = 0.2 pu
è 50 ø
æ 100 ö
T2: X = 0.06ç
÷ = 0.15 pu
è 40 ø
æ 100 ö
T3: X = 0.064ç
÷ = 0.16 pu
è 40 ø
æ 100 ö
T4: X = 0.08ç
÷ = 0.2 pu
è 40 ø
37
The new per unit reactance for the motor M is:
2
æ 100 öæ 10.45 ö
M: X = 0.185ç
֍
÷ = 0.25 pu
è 66.5 øè 11 ø
Z base
Z base
2
2
2
(
Vbase )
(
220 )
=
=
= 484 Ω ,
Line #1: X =
48.4
= 0.1 pu
484
5
2
2
Vbase )
(
(
110 )
=
=
= 121Ω ,
Line #2: X =
65.43
= 0.54 pu
121
100
S base
S base
100
The load impedance Z L =
∴ZL =
V phase
I phase
Since S L
∴ZL =
3−φ
=
S L*
∴ZL
pu
3(V phase )
=
S L*
3−φ
= 3V
*
phase phase
I
ÞI
*
phase
=
SL
3−φ
3V phase
Þ I phase =
S L*
3−φ
*
3V phase
2
(
VLine )
=
Ω
3−φ
S L*
3−φ
= 57∠ cos −1 0.6 = 57∠53.13 MVA
= Z base 4 =
=
Ω , but S L
2
*
3V phase
3−φ
57∠ − 53.13
base
I phase
V phase
(10.45)2
∴ZL
V phase
= 1.1495 + j1.53267Ω
(11)2
100
= 1.21Ω
1.1495 + j1.53267
= 0.95 + j1.2667 pu
1.21
The per unit impedance diagram can be drawn as shown in the figure below:
J0.2
J0.1
J0.15
J0.16
J0.54
J0.2
J0.2
J0.25
0.95
G
M
J1.2667
38