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Learning Outcomes
Students should be able to
Identify the symbols and draw the one-line
diagram Construct the impedance diagram and find
the per unit values.
Solve the per unit problems of a single-phaseand three-phase circuit.
Identify the advantages of per unit system.
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Power System Representation :
One Line Diagram Represent or portray the interconnection of the power system
components used extensively in power flow studies
Also referred as Single-line Diagram
In a one line diagram, lines on paper (or on a computer screen)represent wires.
Advantage: Simplicity Rather than drawing all three wires in a three-phase
system, it is normal to simplify things by representing allthree phases with one line thus the name one linediagram
Equivalent circuit of the components are replaced by their
standard symbols The completion of the circuit through the neutral isomitted.
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One Line Diagram - Symbols
two-winding
transformercurrent transformer
two-winding
transformer
generator
bus
voltage transformer
capacitor
circuit breaker
transmission line
delta connection
wye connection
circuit breaker
fuse
surge arrestor
static load disconnect
Symbols used in one line diagram (from ANSI and IEEE)
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Exercise
Draw a one line diagram for a simple
power system which composes of the
following equipments :
Generator (1 unit), transformer,
transmission line, transformer and load
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L7
L6
L2
L1
L3
1 2
3
54
L5L4
generator
transformer
bus coding
circuit breaker
bus
load
line coding
tie line connection
with neighboring
system
transmission line
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One-line diagram - symbols
The main component of a one-line (or
single line) diagram are : Buses,
Branches, Loads, Machines, 2 windingTransformers, Switched Shunts,
Reactor and Capacitor Banks.
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One-line diagram - explanation
Buses are represented as a dot, circle or a thick line.
The bus name (EAST500) and number (202) are given, aswell as the voltage measured on the line (510.5kV and1.021V in per unit).
Final characteristic given is the angle(-26.1 degree). The voltage is indicated by the color of the bus. Red
indicates 500kV
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One-line diagram - explanation
Branches are represented as a thin line.
The real power P, as shown on the branch below, flowsfrom 478MW to -473MW and the reactive power Q, flowsfrom 89.9MVA flow to -229.4MVA.
The power flows from positive number to the negativenumber, and the number on top is the real power whilethe number on the bottom is the reactive power.
The voltage is indicated by the color of the bus. Redindicates 500kV
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One-line diagram - explanation
Loads are represented as a triangle with the IDnumber located inside the triangle.
The real power, PLOAD, is donated by the numberon top(250MW), and the reactive power QLoad, is
denoted by the number on bottom (100Mvar) The voltage is indicated by the color of the load. In
this example, black indicates 230kV
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One-line diagram - explanation
Machines are represented as a circle with the ID numberlocated inside the circle.
The real power PGEN, is denoted by the number on top(321.0MW), and the reactive power QGEN, is denoted by
the number on bottom (142.3RMVAR). The R indicatesthis machine is in voltage regulation mode, and it iscontrolling a specific bus to a voltage set point whichrequires it to generate 142.3MVAR.
The voltage is indicated by the color of the machine. In
this example, red indicates 500kV.
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One-line diagram - explanation Reactor bank are represented as an inductor at the end
of a line. The number on the top (or on the left if thereactor is shown vertical) indicates that the real chargeadmittance, GSHUNT of the line (3.6MW).
If you see a SW instead of a number, you are looking ata switched shunt compensator.
The number on bottom indicates the initial reactivecharge admittance, BSHUNT, of the line ( 490.0MVAR )
The voltage is indicated by the color of the reactor. In
this example, red indicates 500kV
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One-line diagram - explanation
Capacitor bank are represented as a capacitor at the endof a line. The number on the top (or on the left if thecapacitor is shown vertical) indicates that the realcharge admittance, GSHUNT of the line (0.0MW).
If you see a SW instead of a number, you are looking ata switched shunt compensator.
The number on bottom indicates the initial reactivecharge admittance, BSHUNT, of the line (-1080.0MVAR ).
The voltage is indicated by the color of the capacitor.
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One-line diagram - explanation
Switched shunts are represented as either a capacitor orinductor at the end of a line. The SW shown on top (oron the left if the shunt is shown vertical) indicates thatthis unit is a switched shunt compensator. Permanently
installed reactor or capacitor bank. The number on bottom indicates the initial reactive
charge admittance, BINIT(253.MVAR or 599.6MVAR)
The voltage is indicated by the color of the shunt. In thisexample, red indicates 21.6kV
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T i l di i d i i i d b i i
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Typical medium sized city transmission and sub-transmission system.
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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 andAC machines
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One line Diagram and Per unit system
A one-line diagram is a simplified
graphical representation of a three
phase system, used extensively inpower flow studies
Per unit system used extensively in
one-line diagram to further simplifythe process
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Per-Unit System
An interconnected power system typically consistsof many different voltage levels (sizes and nominalvalues) given a system containing severaltransformers and/or rotating machines.
A simple power system
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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
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Base values
Generally the following two base values are chosen :
The base power = nominal power of the equipment
The base voltage = nominal voltage of theequipment
The base current and The base impedance are
determined by the natural laws of electrical circuits
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Base values
Usually, the nominal apparent power
(S) and nominal voltage (V) are taken
as the base values for power (Sbase) andvoltage (Vbase).
The base values for the current (Ibase)
and impedance (Zbase) can becalculated based on the first two base
values.
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Base value
For single phase system
A,currentbase
V,voltagebase,impedanceBase
kV,voltagebase
kVAbaseA,currentBase
LN
LN
1
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Base value
For single phase system
11
11
1
2
LN
MVAbaseMW,powerBase
kVAbasekW,powerBase
MVA)KV,voltagebase(,impedanceBase
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Base value - example
Base kVA3 = 30,000 kVA
and Base kVLL = 120 kVA
therefore Base kVA 1 = 30,000 / 3 = 10,000 kVA
and Base kV LN= 120 / 3 = 69.2 kVA
valuebased
quantitytheofvalueactualquantityanyofvalueunitper
For actual line-to-line voltage 108 kV, the line-to-neutral
voltage, VLNis 108/ 3 = 62.3
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Per unit value - example
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
valuebased
quantitytheofvalueactualquantityanyofvalueunitper
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Per unit value
e.g. in a synchronous generator with
13.8 kV as its nominal voltage, instead
of saying the voltage is 12.42 kV, wesay the voltage is 0.9 p.u.
p. u voltage = 12.42/13.8 = 0.9 p.u
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Per Unit System
base
base
base
basebase
base
basebase
basebase
SV
IVZ
V
SI
SV
2
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exercise
A generator has an impedance of 2.65
ohms. What is its impedance in per-
unit, using bases 500MVA and 22kV
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Base value in 3 Circuit
,3
3
B
BB
BBB
V
SI
IVS
Usually, the 3-phase SB or MVAB and line-to-line VB or kVB
are selected
IB and ZB dependent on SB and VB
B
B
B
BB
BBB
S
V
I
VZ
ZIV
23/
3
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base
pu SS
S
basepu II
I
basepu
VV
V
base
puZZ
Z
ZZ
2
base
base
base
puV
S
ZZ
pu
base
2
basepubase Z
S
VZZ Z
Per Unit System
Voltage, current, kilovoltamperes and impedance
are quantities often expressed in per unit value
Conversion of Per Unit Values
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Transformer Voltage Base
V1/V2
Vb1 Vb2
1
1
22 bb V
V
VV
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Per Unit System
The percent impedance
e.g. in a synchronous generator with 13.8 kV
as its nominal voltage, instead of saying thevoltage is 12.42 kV, we say the voltage is 0.9
p.u.
100%Z
base
actual%
Z
Z
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Example : Three-Phase Transformer
Consider, for example, a three-phase two-windingtransformer. The following typical parameters couldbe provided by the manufacturer:
Nominal power = 300 kVA total for three phases
Nominal frequency = 60 HzWinding 1: connected in wye, nominal
voltage = 25 kV RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Winding 2: connected in delta, nominal
voltage = 600 V RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Magnetizing losses at nominal voltage in % ofnominal current:
Resistive 1%, Inductive 1%
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Example : Three-Phase Transformer
Base power 300 kVA/3 = 100e3 VA/phase
Base voltage 25 kV/sqrt(3) = 14434 V RMS
Base current 100e3/14434 = 6.928 A RMS
Base impedance 14434/6.928 = 2083
Base resistance 14434/6.928 = 2083
Base inductance 2083/(2*60)= 5.525 H
The base values for each single-phase transformer are first
calculated: For winding 1:
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p.u. to SI
The value of the winding resistance and leakage
inductances expresses in SI units are
For winding 1: R1=0.01 * 2083 = 20.83 L1 = 0.02 * 5.525 = 0.1105 H
For winding 2: R2 = 0.01 * 3.60 = 0.0360
L1 = 0.02 * 0.009549 = 0.191 mH
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p.u. to SI
For the magnetizing branch, magnetizing losses of 1%
resistive and 1% inductive mean a magnetizing resistance
Rm of 100 pu and a magnetizing inductance Lm of 100 pu.
Therefore, the values expressed in SI units referred towinding 1 are
Rm = 100*2083 = 208.3 k
Lm = 100*5.525 = 552.5 H
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excercise
Region 1
Vbase1 = 13.8 kV
Region 2
Vbase2 = Vbase1(110kV/13.8kV) = 110 kV
Region 3
Vbase3 = Vbase2(14.4kV/120kV) = 13.2 kV
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Excercise
2. Calculate corresponding base impedances for each region
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exercise
Region 1
Region 2
Region 3
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Per Unit in 3 Circuit
Simplified:
Concerns about using phase or line
voltages are removed in the per-unitsystem
Actual values of R, XC and XL for lines,
cables, and other electrical equipment
typically phase values.
It is convenient to work in terms of base
VA (base volt-amperes)
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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 ofequipment 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.
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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.
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ZV
S
Z
ZZ
old
B
oldB
old
B
old
pu 2
Z
V
S
Z
ZZ
new
B
new
B
new
B
new
pu 2
then,valueactualisZifand,Vbasevoltage&
Sbasepoweron theimpedanceunitperthebeZ
oldB
oldB
oldpu
new
B
new
B
Vbasevoltagenew&
Sbasepowernewon theimpedanceunitpernewthebenewpuZ
1
2
Change of Base
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Change of base
oldB
2oldBold
puS
VZZ
From 1 and 2
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2
new
B
old
Bold
B
new
Boldpu
newpu
VV
SSZZ
old
B
new
Bold
pu
new
puS
SZZ
Then, the relationship between the old and the
new per unit value is obtained as
If the voltage base are the same,
Change of Base
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Example
The reactance of a generator designated X is given as
0.25 per unit based on the generators nameplate
rating of 18 kV, 500 MVA. The base for calculations is
20kV, 100 MVA. Find X on the new base2
new
B
old
B
old
B
new
Bold
pu
new
puV
V
S
SZZ
25.0Zoldpu kV18VoldB MVA500S
oldB
kV20VnewB MVA100SnewB
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Example
unitper0405.020
18
500
100
25.0"XZ
2newpu
25.0Zoldpu kV18VoldB MVA500SoldB
kV20VnewB MVA100SnewB
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Change of base
4. Calculate the p.u. resistances and
inductances at G1 and T1
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Exercise
RG1,pu = 0.1 per unit
XG1,pu = 0.9 per unit
RT1,pu = 0.01 per unit
XT1,pu = 0.05 per unit
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Excercise
5. Calculate per unit value for resistance and inductance of
transmission line
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Exercise
6. Calculate per unit value for resistance and inductance of
T2 and M2
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Exercise
unitper119.0MA50
MA100
kV2.13
kV4.14)05.0(X
unitper238.0
MA50
MA100
kV2.13
kV4.14)01.0(R
2
pu,2T
2
pu,2T
unitper405.2MA50
MA100
kV2.13
kV8.38)05.0(X
unitper219.0MA50
MA100
kV2.13
kV8.38)01.0(R
2
pu,2M
2
pu,2M
At T2
At M2
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Per-Unit System the per-unit-system which choosing a common set of
base parameter in term of which quantities are defined.The different voltage levels disappear and the overallsystem reduces to a set of impedances.
Per-phase, per unit equivalent circuit of the simple power system
G1s
impedance
T1s
impedance
Transmission
lines impedance
T2s
impedance
Motors
impedance
Writing Node Equations for
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Writing Node Equations for
Equivalent Circuit
Once the per-phase, per unit
equivalent circuit of a power system is
created, it may be used to find thevoltages, currents, and powers present
at various points in a power system.
Most common technique used to solvesuch circuit is nodal analysis
Writing Node Equations for
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Writing Node Equations for
Equivalent Circuit
The simple three-
phase power
system
containing three
busses connectedby three
transmission
lines, generator
to bus 1, a load
to bus 2 and a
motor to bus 3.
Draw the
impedance
diagram and
respective nodes
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Writing Node Equations for
Equivalent CircuitThe diagram showing current sources at nodes 1, 2and 3; all other equipments are admittance
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Writing Node Equations for
Equivalent Circuit
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Advantages
Transformer equivalent circuit can besimplified by properly specifying basequantities.
Give a clear idea of relative magnitudes ofvarious quantities such as voltage,current, power and impedance.
Avoid possibility of making seriouscalculation error when referring quantitiesfrom one side of transformer to the other.
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Advantages
Per-unit impedances of electricalequipment of similar type usually liewithin a narrow numerical range when
the equipment ratings are used as basevalues. Manufacturers usually specify the
impedances of machines and transformers
in per-unit or percent in nameplaterating.
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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 isstraightforward.
For example, a transient voltage reaching a maximum of 1.42 pu indicates immediately that this voltageexceeds the nominal value by 42%.
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 variesapproximately between 0.01 pu and 0.03 pu, whereas the winding resistances vary between 0.01 pu and0.005 pu, whatever the nominal voltage. For transformers in the 300 kVA to 300 MVA range, the leakagereactance varies approximately between 0.03 pu and 0.12 pu, whereas the winding resistances varybetween 0.005 pu and 0.002 pu.
Similarly, for salient pole synchronous machines, the synchronous reactanceXd is generally between 0.60and 1.50 pu, whereas the subtransient reactanceX'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 majorerror by assuming an average value of 0.02 pu for leakage reactances and 0.0075 pu for windingresistances.
The calculations using the per unit system are simplified. When all impedances in a multivoltage powersystem are expressed on a common power base and on the nominal voltages of the differentsubnetworks, the total impedance in pu seen at one bus is obtained by simply adding all impedances inpu, without taking into consideration the transformer ratios.
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Review
Study the examples.
Try to solve the problems in your
tutorials.
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Impedance Diagram
G1
Load A
1
G2
Load B
G3
2
G4
T1 T2
G2G1 G3 G4