POWER SYSTEM ANALYSIS

Power System Representation and Per Unit System

Introduction
 It is clear that a detailed representation of each of
the three phases in the system is cumbersome and
can also obscure information about the system.

 A balanced three-phase system is solved as a

single-phase circuit made of one line and the
neutral return; thus a simpler representation.

 Further simplification by omitting the neutral.

Introduction
 The simplified diagram is called the single-line diagram
or one-line diagram.

 The one-line diagram summarizes the relevant

information about the system for the particular problem
studied.

 For example, relays and circuit breakers are not

important when dealing with a normal state problem.
However, when fault conditions are considered, the
location of relays and circuit breakers is important and
is thus included in the single-line diagram.
Introduction
 The International Electrotechnical Commission (IEC),
the American National Standards Institute (ANSI),
and the Institute of Electrical and Electronics
Engineers (IEEE) have published a set of standard
symbols for electrical diagrams.

 The main component of a one-line (or single line)

diagram are : Buses, Branches, Loads, Machines, 2
winding Transformers, Switched Shunts, Reactor and
Capacitor Banks.
One Line Diagram - Symbols

two-winding
transformer current transformer

two-winding
transformer voltage transformer

generator capacitor

bus circuit breaker

transmission line circuit breaker

delta connection fuse

wye connection surge arrestor

static load disconnect

Symbols used in one line diagram (from ANSI and IEEE)

Single-line diagram
 It is important to know the location of points where
a system is connected to ground in order to
calculate the amount of current flow when an
unsymmetrical fault involving ground occurs. The
standard symbol to designate a three-phase Y with
the neutral solidly grounded is shown in Figure
below.
G1 T1 T2
G3

G2 Load B

Load A
One-line diagram
 IEEE 30bus system
Impedance and Reactance Diagrams

 The impedance (Z = R + jX) diagram is converted

from one-line diagram showing the equivalent circuit
of each component of the system.

 It is needed in order to calculate the performance

of a system under load conditions (Load flow
studies) or upon the occurrence of a short circuit
(fault analysis studies).
Impedance and Reactance Diagrams

1 T1 T2 3

2 Load B

Load A

One-Line Diagram of an Electric Power System

E1 E2 E3

Generators Load Transformer Transmission Transformer Load Gen.

1 and 2 A T1 Line T2 B 3

Impedance Diagram Corresponding to the One-Line Diagram above

Impedance and Reactance Diagrams
 Reactance (jX) diagram is further simplified from
impedance diagram by omitting
🞑 all static loads,
🞑 all resistances,
🞑 the magnetizing current of each transformer, and
🞑 the capacitance of the transmission line.

E1 E2 E3

Generators Transformer Transmission Transformer Gen.

1 and 2 T1 Line T2 3

Reactance Diagram Corresponding to the One-Line Diagram of Example

Per Unit System
 In power systems large amounts of power being
transmitted in the range of kilowatts to megawatts,
at different voltage levels.

 As a result, in analysis, it is useful to scale, or

normalize quantities with large physical values and
this is commonly called per unit system in power
system analysis.
Per Unit System
 The per unit system is widely used in the power
system industry to express values of voltages,
currents, powers, and impedances of various power
equipment.
 It is mainly used for transformers and AC machines
 Per unit system used extensively along with one-line
diagram to further simplify the process.
Per Unit Calculations
 All base values are only magnitude. They are not associated with any
angle.
 The per unit values, however, are phasors.
 The phase angles of the currents and voltages and the power factor of the
circuit are not affected by the conversion to per unit values.
 In general, the per unit value is the ratio of the actual value and the base
value of the same quantity.

actual value
per unit value =
base value

 The per unit system values can also be expressed as per cent values.
Base value
 Specify the base values of current and voltage,
base impedance, kilovoltamperes can be
determined
 Quantities and base value selected
voltage, base value in kilovolts, kV
current, base value in ampere, A
Base values
Generally the following two base values are chosen :
The base power = nominal power of the equipment
The base voltage = nominal voltage of the
equipment
The base current and The base impedance are determined by
the natural laws of electrical circuits
Base values
 Usually, the nominal apparent power (S) and
nominal voltage (V) are taken as the base values
for power (Sbase) and voltage (Vbase).

 The base values for the current (Ibase) and

impedance (Zbase) can be calculated based on the
first two base values.
Per-unit System for 1-  Circuits

PB,1 = QB,1 = SB,1

SB,1
IB =
VB,LN
2
VB,LN VB,LN
Z B = RB = X B = =
IB S B,1
Base value

 For single phase system

(base voltage, KVLN )2

Baseimpedance,  =
MVA1
Base power, kW1 = base kVA1
Base power, MW1 = base MVA1
Per-Unit System
Per-unit system:
Vactual Iactual
V p.u. = I p.u. =
VB IB
Sactual Zactual
S p.u. = Z p.u. =
SB ZB

Z % = Z p.u. 100% Percent of base Z

Per-unit System for 1-  Circuits

Z
For actual quantity that has complex number,
Z pu  =
Per unit can be expressed as:
ZB
Per unit can be expressed in rectangular Z pu = Rpu + jX pu
form:
S pu = Rpu + jQpu
Power: S
S pu = = V pu I *pu
Sb
P
Ppu = = V pu I pu cos
Sb
Q
Q pu = = V pu I pu sin
The picture can't
be displayed.

Sb
exercise
 A generator has an impedance of 2.65 ohms. What
is its impedance in per-unit, using bases 500MVA
and 22kV
Per Unit system for 3-  Circuits
 Have the same per unit values for line to line and line
to neutral quantities.
 Make everything look like a single phase circuit.
 Balanced three phase circuits can be solved in per
unit on a per phase basis after converting delta load
impedance to equivalent Y impedance.
 Base value can be selected on a per phase basis or
on a three phase basis.
Per Unit system for 3-  Circuits (Voltage)

 In a three phase system, we have:

VLLpu = VLN
pu

 Consider Y connected: VLL = 3VLN

VLL VLN
 So: VLLpu = VLNpu  =
VB,LL VB,LN
  VB,LL = 3VB,LN
Per Unit system for 3-  Circuits (Power)

 In a three phase system, we have:

S3pu = S1pu


S3 = 3S1
 We know:

S3 S1
S =S
pu
3
pu
1  =
SB,3 SB,1
 So:
  SB,3 = 3SB,1
Per Unit system for 3-  Circuits (Current)

 In a three phase system, we have:

S3 = 3VLN LN = 3VLLI LL

I *

 We know:

S pu = V pu I pu
 So:

S VLL ILL
S =V I
pu pu pu
 =
SB,3 VB,LL IB
SB,3
  IB =
3VB,LL
Per Unit system for 3-  Circuits (Impedance)

VLN ZLN ILN

V pu
=Z I
pu pu
 =
VB,LN ZB IB
VB,LL
(V )
2
VB,LN 3
  ZB = = = B,LL

IB SB,3 SB,3
3VB,LL
Example 1
 Given base kVA for 3 phase systems is 30 000kVA and
voltage base line to line 120kV. Find:

SB,1
Vpu
VB,LN
VLN

 Given actual line to line voltage is 108kV.

 Example 1

Base kVA3Φ = 30,000 kVA

and Base kVLL = 120 kVA
therefore Base kVA 1Φ = 30,000 / 3 = 10,000 kVA

and Base kV LN = 120 / √3 = 69.2 kVA

actual valueof thequantity

per unit valueof anyquantity =
based value

For actual line-to-line voltage 108 kV, the line-to-neutral voltage, VLN is 108/ √3 = 62.3
 Per unit value - example

actual valueof thequantity

per unit valueof anyquantity =
based value
and
Per-unit voltage = 108/120 (3) OR
= 62.3/69.2 (1)
= 0.9
For three-phase power of 18,000 kW,
Per-unit power = 18,000/30,000 (3 ) OR
= 6,000/10,000(1)
= 0.6
Change of Base

 The impedance of individual generators &

transformer, are generally in terms of percent/per
unit based on their own ratings.
 Impedance of transmission line in ohmic value
 When pieces of equipment with various different
ratings are connected to a system, it is necessary to
convert their impedances to a per unit value
expressed on the same base.
Change of Base
 In other word, since all impedances in any one part of the a
system must be expressed on the same impedance base
when making computations, it is necessary to have a means
of converting per-unit impedances from one based to
another.
Z = Z pu
Z = Z pu
Z
actual new B,new old B,old

ZB,old
 Z npeuw = Z opl ud
Z B,new
V 2
But we always have : Z B = B

SB
2
S B,new V 
Z npeuw = Z o lpdu  B,old 
S B,old V 
 B,new 
Change of Base
 If VB,new = VB,old
 So
SB,new
Z npeuw = Z opl ud
S B,old
Example 2
The reactance of a generator designated X” is given as 0.25
per unit based on the generator’s nameplate rating of 18 kV,
500 MVA. The base for calculations is 20kV, 100 MVA. Find
X” on the new base

2
 S new
 V old 
new
Z pu = Z pu
old
 Bold  Bnew 
 S  V 
 B  B 

pu = 0.25
Zold VBold = 18 kV B = 500 MVA
Sold

VBnew = 20 kV Snew
B = 100 MVA
Example 2

pu = 0.25
Zold VBold = 18 kV B = 500 MVA
Sold

VBnew = 20 kV Snew
B = 100 MVA

2
 100  18 
Znew
pu = X"= 0.25   = 0.0405 per unit
 500  20 
Exercise 2
 Generator rated at 10MVA, 20kV
 XS = 0.9pu on the basis of the generator rating
 Given SB,new =100MVA and VB,new=20kV
 Find: X new
S

Solution:
Exercise 3
 Transformer rated at 10MVA, 33/11kV
 Z = 10% and R = 1%
 Given SB,new=200MVA and VB,new=22kV (HV side transformer)
 Find: i. Zbase(HV and LV sides)
ii. actual Z and R referred to primary and secondary
iii. Transformer losses in kW, if 0.033 p.u (selected base) of current
flow through R
Solution:
Procedure for Per Unit Analysis
1. Pick SBase for the system.
2. PickVBase according to line-to-line voltage.
3. Calculate Z for different zones.
Base

4. Express all quantities in p.u.

5. Draw impedance diagram and solve for p.u.
quantities.
6. Convert back to actual quantities if needed.
Source

How to solve problems containing multiple

transformer?
Transformer Voltage Base

Vb1 Vb2

V1/V2

 V2 
Vb 2 =  •V b1
 V1 
How to Choose Base Values ?
 Divide circuit into zones by transformers.
 Specify two base values out of IB ,VB , ZB , SB ; for
example, SBase and VBase
 Specify voltage base in the ratio of zone line to line
voltage.
V1 :V2 V2 :V3 V3 :V4
Source

Zone 1 Zone 2 Zone 3 Zone 4

VBase1 VBase2 VBase3 VBase4
S Base VBase1
I Base1 = Z Base1 =
VBase1 I Base1
 Example 5
 Given a one line diagram,

Vg = 13.2kV
Ig Zline = 10 + j100
~ Zload = 300

5 MVA 10 MVA
13.2 Δ – 132 Y kV 138 Y - 69 Δ kV
X1 = 0.1p.u. X 2 = 0.08p.u.

Choosing a base apparent power of 10MVA and a base line

voltage L1 of 69 kV ; find
Ig I t-line Iload Vload Pload
 Step 1, 2, and 3: Base Values
Vg = 13.2kV T1 T2
Ig Zline = 10 + j100
~ Zload = 300

5 MVA 10 MVA
13.2 Δ – 132 Y kV 138 Y - 69 Δ kV
X1 = 0.1p.u. X 2 = 0.08p.u.

SB,new = 10MVA VB,new = 69kV

Zone 1 Zone 2 Zone 3
VB1 = 13.8kV VB2 = 138kV VB3 = 69kV

13.2kV 138kV
VB = x138kV = 13.8kV VB2 = x69kV = 138kV
1
132kV 69kV

 Calculate the base voltage in each zone

 Step 1, 2, and 3: Base Values
Vg = 13.2kV
Ig Zline = 10 + j100
~ Zload = 300

5 MVA 10 MVA
13.2 Δ – 132 Y kV 138 Y - 69 Δ kV
X1 = 0.1p.u. X 2 = 0.08p.u.

SB,new = 10MVA VB,new = 69kV

Zone 1 Zone 2 Zone 3
VB1 = 13.8kV VB2 = 138kV VB3 = 69kV
2 2 2

Z B1 =
VB1
=
(13.8k ) 2
= 19.04 Z B2 =
VB2
=
(138k ) 2
= 1904 Z B3 =
VB3
=
(69k )2
= 476
SB 10M SB 10M SB 10M
SB31 10M S B32 10M SB33 10M
I B1 = = = 418.4 I B2 = = = 41.84 I B3 = = = 83.67
3 VB 1 3 13.8k 3 VB 2 3 138k 3 VB 3 3  69k
Step 4: All in Per Unit Quantities
Vg = 13.2kV T1 T2
Ig Zline = 10 + j100
~ Zload = 300

5 MVA 10 MVA
13.2 Δ – 132 Y kV 138 Y - 69 Δ kV
X1 = 0.1p.u. X 2 = 0.08p.u.

+
-
 Calculate Vg in per unit
Calculate the new reactance for X1 and X2 in per unit
Calculate the impedance in per unit
 Draw the impedance diagram
 Step 4: All in Per Unit Quantities
X 2, pu = 0.08p.u.

2
+ S B,new  V B,old 
- Z npeuw = Z o lpdu   Zload 300

S B,old  V B , n e w  Z load,p.u. = = = 0.63
 Z B3 476
2
 10MVA  13.2k 
X 1,p.u. = 0.1   = 0.183
 5MVA  13.8k 

Vg 13.2kV
Vg,p.u. = = = 0.960
V B1 13.8kV Zline 10 + j100
Z line,p.u. = = = 5.2510−3 (1+ j10)
Z B2 1904
Step 5: One Phase Diagram & Solve
Zline,p.u. = 5.2510 −3 (1+ j10)
X1,p.u. = j0.183 X 2, pu = j0.08

+ Vg,p.u. = 0.960
- Zload,p.u. = 0.63

Vg,p.u. 0.960
I load,p.u. = = = 1.35 − 26.4
Z total,p.u. 0.70926.4
Vload,p.u. = Iload,p.u.Zload,p.u. = 0.8505 − 26.4
S load,p.u. = Vload,p.u. I load,p.u.
*
= 1.148
Ig,p.u. = I t-line,p.u. = Iload,p.u. = 1.35 − 26.4
 Step 6: Convert back to actual quantities

Vg = 13.2kV
Ig Zline = 10 + j100
~ Zload = 300

5 MVA 10 MVA Vload,p.u. = 0.8505 − 26.4

13.2 Δ – 132 Y kV 138 Y - 69 Δ kV
X1 = 0.1p.u. X 2 = 0.08p.u. Sload,p.u. = 1.148

Ig,p.u. = I t-line,p.u. = Iload,p.u. = 1.35 − 26.4

Zone 1 Zone 2 Zone 3

I g = I g,p.u. I B1 I t-line = I t-line,p.u. I B2 Iload = Iload,p.u. IB 3

Vload = Vload,p.u.VB3
Sload = Sload,p.u.SB

HFAW'09
Advantages of P.U. system
 Transformer equivalent circuit can be simplified by
properly specifying base quantities.
🞑 Give a clear idea of relative magnitudes of
various quantities such as voltage, current, power
and impedance.
🞑 Avoid possibility of making serious calculation error
when referring quantities from one side of transformer
to the other.
Advantages of P.U. system
 Per-unit impedances of electrical equipment of
similar type usually lie within a narrow numerical
range when the equipment ratings are used as base
values.
🞑 Manufacturers usually specify the impedances of
machines and transformers in per-unit or percent in
nameplate rating.
Advantages of P.U. system
 The circuit laws are valid in per unit systems, and
the power and voltage equation are simplified since
the factor √3 and 3 are eliminates in the per-unit
systems.
 Ideal for the computerized analysis and simulation
of complex power system problems.
Advantages
 Why Use the Per Unit System Instead of the
Standard SI Units?
 Here are the main reasons for using the per unit
system:
 When values are expressed in pu, the comparison of
electrical quantities with their "normal" values is
straightforward.
 For example, a transient voltage reaching a
maximum of 1.42 pu indicates immediately that this
voltage exceeds the nominal value by 42%.
 Advantages
 The values of impedances expressed in pu stay fairly
constant whatever the power and voltage ratings.
 For example, for all transformers in the 3 kVA to 300
kVA power range, the leakage reactance varies
approximately between 0.01 pu and 0.03 pu, whereas
the winding resistances vary between 0.01 pu and
0.005 pu, whatever the nominal voltage. For
transformers in the 300 kVA to 300 MVA range, the
leakage reactance varies approximately between 0.03
pu and 0.12 pu, whereas the winding resistances vary
between 0.005 pu and 0.002 pu.
Advantages
 Similarly, for salient pole synchronous machines, the
synchronous reactance Xd is generally between
0.60 and 1.50 pu, whereas the subtransient
reactance X'd is generally between 0.20 and 0.50
pu.
 It means that if you do not know the parameters for
a 10 kVA transformer, you are not making a major
error by assuming an average value of 0.02 pu for
leakage reactances and 0.0075 pu for winding
resistances.
Advantages
 The calculations using the per unit system are
simplified. When all impedances in a multivoltage
power system are expressed on a common power
base and on the nominal voltages of the different
subnetworks, the total impedance in pu seen at one
bus is obtained by simply adding all impedances in
pu, without taking into consideration the transformer
ratios.
In a balanced 3-Φ system, the magnitude of voltages and
currents in every phase, is the same and is displaced by
120o.

Therefore in a balanced condition, any 3-Φ system can be

represented by one of the phases and the neutral
connection.

To simplify further we can remove the neutral connection,

thus leaving only the live connection as a single line.

This kind of diagram is called, the single line diagram.

Symbol for common components in power system
The parameters supplied with each component in the one
line diagram are:

i. rated voltage,
ii. rated power,
iii. phase impedance,
iv. type of connection (either Wye or Delta connection).

For 3-Φ system, – Vline , P3Φ, Z1Φ (in pu or percentage)

Types of connection should be stated for the three phase

generators, motors, transformers, and loads. Typically,
the connection is assumed in wye.
Three phase system
Single line diagram
The one line diagram for a power system can be converted
into per phase equivalent circuit which is used to analyse
the performance of the system.

The equivalent circuit is known as ‘Impedance Diagram’.

The impedance diagram is an equivalent single phase

circuit with neutral line taken as the return path.
The analysis on the impedance diagram is similar to that on
the alternating current circuit.
Example, of one line diagram has to be converted into
the impedance diagram.

One line diagram

Impedance diagram
For most components of power system, the value of
resistance is very small compared to that of inductive
reactance.

Thus, by neglecting resistance in the impedance diagram

we can constructreactancediagram representation for
simpler analysis of the system.
Reactance diagram
Q1. Obtain impedance and reactance diagram for power
system in Figure shown below.
Y Y
Y
Y Y
Per unit (p.u.) value for a quantity is defined as a ratio of
the actual value over the base value.

The p.u. value is commonly given either in fractional or

percentage forms.
Four main parameters in power system analysis:

i. Voltage : Line voltage (VL) or phase voltage (VP)

ii. Power : 3Φ apparent power (VA3Φ) or 1 Φ (VA1Φ)
iii. Current : Line current (IL) or phase current (IP).
iv. Impedance : Phase impedance (ZP).
By taking two parameters as the base parameters, the
other base parameters can be obtained.

VP-base and VA1Φ-base,

Relationship between base current and base impedance.

When all base parameters are obtained, any

parameter in the system can be converted into p.u. using
the above equations.
The following items are to be noted when there is
transformer connected in the system:

i. The base parameter for power is applicable for both

sides of the transformer since the transformer ratio
does not affect the power through it.

i. The base parameter for voltage changes according to

the transformer ratio.
eg : for a 10/20 kV transformer, if the base voltage
at 10 kV is taken as 10 kV, then the base voltage at
the 20 kV side should be 20 kV.
Transformer impedance in Ohm, referred to the primary
winding is not equal to that referred to the secondary
winding.

As a result, the analysis involving transformer is

complicated since it has to be done by referring all
parameters to either side of the transformer.

However, in per unit representation, the equivalent

impedance values are the same at both sides of the
transformer.
To simplify the analysis, assume that the resistance of the
transformer can be neglected.
The reactance referred to the primary side is
The final equivalent is circuit is as shown below
If the impedance is referred to the secondary :
The final equivalent is circuit is as shown below
with base voltage of 10 V and 20 V respectively at the
primary and secondary windings and base power of 100
VA, thus the base reactance at the primary side is

and base reactance at the secondary side is

The reactance referred to the primary side in per unit is
And the reactance referred to the secondary side is
Thus the equivalent circuit in pu is:
Example : Find the per unit value for each component and
draw the impedance diagram.
The base parameters taken for the system, both at the
generator, are:

VL-base = 10 kV.
VA3Φ-base = 100 MVA.
The impedance diagram :
Example :
Find the load current, actual current in every component
and voltage drop in every component if the load voltage
is 1pu.

The load current value in per unit is the same throughout

the system.
The base current value at each component is
The actual current value at each component is
The voltage drop at every componenr-tUinsit:
The generated voltage is :

The actual value of the generated voltage is

it can be observed that the generated voltage is far

beyond the nominal voltage. This occurs due to the excessive
voltage drop in the components of the system.
Question 2.

Repeat the previous by replacing the load with a new value

of 50 + j 100 Ω.
List of advantages of per unit compared to normal unit :

i. Manufacturers of power system components

conveniently provide impedance data for the
components in per unit based on the voltage and power
ratings of the components.

i. The calculation for system parameters such as current

and voltage is relatively simple and straightforward
using per unit system in the impedance

ii. The impedance of a transformer in per unit is the same

for both side.
iv. per unit are scaling mechanisms, useful to power
system engineers because it can easily help indicate
overvoltage, over current and overloading.

v. Simplify the relationship involves the multiplication of

current and voltage.
In certain situation, conversion of impedance in per unit
from a given base to another need to be done.

For instance, the impedance for an individual component is

initially based on the ratings of the component.

However, if the base values used for the whole system

differ from the component’s ratings, then the per unit value
of the impedance has to be corrected to the new system
base.
A three phase generator has the following voltage and
power ratings:

Vbase = 10 kV
VAbase = 50 MVA

Given the impedance of the generator

In Ohm, the impedance is

if the following new bases are used,

Vbase-new = 20 kV
VAbase-new = 10 MVA,
Then

Thus, the impedance value in the new bases

Given the old base values for voltage and power and the
impedance based on these values,

The new base values

Therefore, the impedance value in the new base is :
Question 3

A generator has reactance of 0.25 p.u. based on 18 kV

500 MVA. The generator is connected to a power system
that uses base values of 20 kV 100 MVA. Calculate

i. reactance of the generator in Ohm.

ii. the reactance in per unit based on 20 kV 100 MVA
base values.
Example :
Data :

Find the new per-unit impedance for each component by

taking 11 kV at the motor side and 100MVA as the new
base.
The new per unit impedance
A 3 phase Wye connected load consists of three
impedances, each with a value of 20∠30o Ω. The line
voltage at the load terminal is 4.4 kV. The line impedance
is 1.4 ∠ 75o Ω. Obtain the line voltage at the source end.
Example :
This figure shows a sample of power system network. Find
the current supplied by the generator, the transmission
line current, the load current, the load voltage and the
power consumed by the load. Choose base 100MVA and
138kV at the line.
This figure shows a single line diagram of a network.
Select a common base of 100MVA and 13.8kV on the
generator side. Draw the per unit impedance diagram.
The figure shows a single line diagram of a single phase
circuit.
Using the base value of 3 kVA and 230V.
Draw the per unit circuit diagram.
Also calculate the load current both in per unit and amperes.
For generator

For Line
For load
Fault calculation

Per unit calculation is used extensively in fault

calculation. The idea is to convert all quantities to per
unit.

The normal circuit analysis is used to calculate current

and voltage under short-circuit condition.

Xpu= Base MVA/ SC MVA

As an example, consider the following network
Changing impedances into pu with 100MVA base,