Sistem Per Unit

Electric Power System

Figure 1.8 Map of National Grid, Malaysia and South-East Asia

1.5.2 Tenaga Nasional Berhad

Tenaga Nasional Berhad is the largest electricity utility company in Malaysiaand also the largest power company in Southeast. TNB's core activities are in thegeneration, transmission and distribution of electricity.

TNB was formed in 1990 by the Electricity Supply Successor Company Act 1990,as the successor of the National Electricity Board of the States of Malaya(Lembaga Letrik Negara Tanah Melayu).

To date, TNB remains a major player in electricity generation which form asignificant part of the Group's diversified range of business activities.

Currently, the TNB Group has a complete power supply system, including theNational Grid which is energised at 132, 275 and 500 kV. The National Grid islinked via 132 kV HVAC and 300 kV HYDe interconnection to Thailand and 230kV cables to Singapore.

The TNB Group has a generation capacity of 11,296 MW. TNB generateselectricity mainly from two major types of plant; hydroelectric plants and thermalplants.

1.6 Representation of Electric Power System

1.6.1 One-line Diagram

In solving a power system network, it is often involve a single-phase and three-phase equivalent circuit. For the three-phase system, that may consist of three

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T

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Introduction To Power System

lines and a neutral return when drawing a diagram of the circuit. Thesecomplicated diagrams of the circuits can be simplified and called as a single-lineor one-line diagram. One-line diagram is a simplified single-phase circuit diagramof a balanced three-phase electric power system as given in Figure 1.9. It isindicated by a single line and standard apparatus symbols as shown in Table 1.3.

RYB

3-phase system One-line system

Figure 1.9 Three-phase system to one-line diagram

Table 1.3 Apparatus symbols of one-line diagram

~ 0 Machine or rotating armatureor

-1~ --(Q)- Two-winding power transformeror

~~~-&- Three-winding power transformer

or

1 Loador

-D- XPower circuit breaker, oil/ liquid

Air circuit breaker

Busbar

Transmission line

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:~.'


--rr==n- Fuse

OF'\ Current transformerf r+t

-3 or -3t- Potential transformer

~Three-phase, three-wire delta connection

y .Three-phase wye, neutral ungrounded

n Three-phase wye, neutral grounded

-@- -0- Ammeter and Voltmeter

The information on a one-line diagram is vary according to the problem at handand the practice of the particular company preparing the diagram. The one-linediagram has its largest application in power flow and transient stability studies.

The advantages of the one-line diagram are:

(i) Simplicity.(ii) One phase represents all three phases of the balanced system.(iii) The equivalent circuits of the components are replaced by their standard

symbols.(iv) The completion of the circuit through the neutral is omitted.

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!

J

I,

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Bus 1 Bus 2

Figure 1.10 A one-line diagram of a simple power system.

1.6.2 Impedance and Reactance Diagrams

The impedance (Z = R + jX) diagram is converted from one-line diagram showingthe equivalent circuit of each component of the system. It is needed in order tocalculate the performance of a system under load conditions (Load flow studies)or upon the occurrence of a short circuit (fault analysis studies).

Reactance (jX) diagram is further simplified from impedance diagram by omittingall static loads, all resistances, the magnetizing current of each transformer, andthe capacitance of the transmission line. It is apply to fault calculations only, andnot to load flow studies.

The impedance and reactance diagrams sometimes called the Positive-sequencediagram.

Example 1.1

Simplify the following one-line diagram into impedance diagram and reactancediagram.

Load .•.

f

Figure 1.11 One-line diagram of an electric power system SB

1.J

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1Generators Load

1 and 2 ATransformer

T1Transformer

T2

~Load Gen.

B 3

Figure 1.12(a) Impedance diagram corresponding to the one-line diagramof Example 1.2

--~ ~-~/ '--y---J

Generators TransformerTransmission Transformer Gen.1 and 2 T1 Line T2 3

Figure 1-12(b) Reactance diagram corresponding to the one-line diagram

of Example 1.2

1.6.3 Per-Unit Representation

The Common quantities used in power system analysis are voltage (kV), current(kA), voltamperes (kVA or MVA), and impedance (n). It is very cumbersome toconvert currents to a different voltage level in a power system having two or morevoltage levels. Per-unit representation is introduced such that the various physicalquantities are expressed as a decimal fraction or multiples of base quantities anddefined as:

actual quantityQuantity in per-unit = ----~---.:.--base value of quantity

(1.1)

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In orther word, the per-unit value of any quantity is the ratio of the quantity to itsbase expressed as a decimal add denoted by p.u.

The Advantages of Per-Unit Quantities are:

(i) The apparatus of the sJme general type of p.u. volt drops and losses are inthe same order, regardl~ss of size.

(ii) The use of .J3 in three~Phase calculations is reduced.(iii) By the choice of appropriate voltage bases, the solution of networks

containing several tran~forrners is facilitated.(iv) Per-unit quantities more readily to digital computation.

The formulas relate the various quantities for single-phase system:I

base kVM1Base current, A = : ¢base voltagj' kV LN

B . d. base voltage, VLNase zmpe ance = Ibase current, A

. d. (base voltale, kVLNlBase zmpe ance = i

base .NfVA1",

Base power, kW;", = base kVAl~

Base power, Mw;", = base MVIA1",

(1.2)

The formulas relate the various quantities for three-phase system:

base lVA3",Base current, A = --:::::-----Ir-----'----.J3 X base lo/tage, kVLL

. d (base voltage, kVLLlBase impe ance = i

base MVA3",

Base power, kW3c/> = base kVAL .I

Base power, MW3c/> = base MVL 3¢

(1.3)

The base impedance and base voltage for a given power system are 10nand 400, respectively. Calcula Iethe base kVA and the base current.

Example 1.2:

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Solution:

From Ohm's law,

400Base current = - = 40A

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BasekVA= 40X400 = 16kVA1000

Example 1.3:

The base current and the base voltage of a 345 kV system are chosen to be3000A abd 300 kV, respectively. Determine the per-unit voltage and the, baseimpedance for the system.

Solution:

. 300Xl03

Base Impedance = = lOOn3000

Per-unit voltage = 345 = 1.ISpu300

Example 1.4:

A three-phase, wye-connected system is rated at 100 MVA and 132 kY.Express 80 MVA of three-phase apparent power as a per-unit value referred to (a)the three-phase system MVA as base and (b) the single-phase system MVA as base.

Solution:

(a) For the three-phase base,

Base MVA= 100 MVA=lpu

and Base kV = 132 kV (LL) = 1 pu

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so Per-unit MVA = 80 = 0.8pu100

(b) For the single-phase base,

1Base MVA= -X100MVA = 33.333MVA =lpu

3

132Base kV = .J3 = 76.21 kV = 1 puand

!'. I

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so Per-unit MVA =.!.X 80 = 0.8pu3 33.333

Changing the Base of Per-Unit Quantities

The impedance of individual generators and transformers are generally in terms of% or pu quantities based on their own ratings (By manufacturer) .. For powersystem analysis, all impedances must be expressed in pu on a common systembase. Thus, it is necessary to convert the pu impedances from one base to another(common base, for example: 100 MVA). Per-unit impedance of a circuit elementis given by:

P .. d (actual impedance, 0) X (base MVA)er-unit tmpe ance = 2

(base voltage, kV). (1.4)

The equation shows that pu impedance is directly proportional to base MVA andinversely proportional to the square of the base voltage. Therefore, to change fromold base pu impedance to new base pu impedance, the following equation applies:

( )2( )basekV baseMVAPer-unit Znew = per-unit Zold old new

base kVnew baseMVAold

(1.5)

Example 1.5:

The reactance x" of a generator is given as 0.20 pu based on thegenerator's nameplate rating of 13.2 kV, 30 MVA. The base for calculations is13.8 kV, 50 MVA. Find x" on this new base.

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Solution:

x"= 0.20(13.2)2 50 = 0.306 pu13.8 30

Example 1.6:

A 30 MVA 13.8 kV three-phase generator has a subtransient reactance of15%. The generator supplies two motors over a transmission line havingtransformers at both ends, as shown in the one-line diagram in Figure 1-9. Themotors have rated inputs of 20 MVA and 10 MVA, both 12.5 kV with x" = 20%.The three-phase transformer T1 is rated 35 MVA, 13.2Ll- 115Y kV with leakagereactance of 10%. Transformer T2 is composed of three single-phase transformerseach rated at 10 MVA, 12.5 - 67 kV with leakage reactance of 10%. Seriesreactance of the transmission line is 80 Q. Draw the reactance diagram with allreactances marked in per unit. Select the generator rating as base in the generatorcircuit.

(12.9 kV)

m(120 kV)

Figure 1.13 One-line diagram ofthe system

Solution:

The three-phase rating of transformer T2 is 3 X 10 MVA= 30 MVA.

Line-to-line voltage ratio is 12.5 -.J3 X 67 = 12.5 - 116 kV

A base of 30 MVA, 13.8 kV in the generator circuit requires a 30 MVA base in allparts of the system and the following voltage bases:

In transmission line: 13.8 (115) = 120 kV13.2

In motor circuit: 120( 115) = 12.93 kV13.2

The reactances of the transformers converted to the proper base are:

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lntroduction To Power System

Transformer Tl: x = 0.1 (13.2)2(30) = 0.0784 pu.13.8 35

(125)2X = 0.1 -' = 0.0940 pu.12.9

Transformer T2:

The base impedance of the transmission line is

(120 kVY = 480 n30MVA

and the reactance of the line is (~) = 0.167 pu.480

(12.5)2(30)Reactance of motor 1 = 0.2 -- -- = 0.282 pu.12.9 20

(12.5)2(30)Reactance of motor 2 = 0.2 -- -- = 0.563 pu.12.9 10

k jO.0784 I jO.167 m jO.0940 n

Figure 1.14 Reactance diagram for Example 1.7

1.7 Basic Computer Analysis of Electric Power System

For a power system to be practical, it must be safe, reliable, and economical." Hence, many analyses must be performed to design and operate a power system.

However, the pre-requisite before the analyses can be performed is the modellingtasks to model all components of electric power systems. The modelling andanalysis of a power system require the aid of personal computer. Some of thebasic analyses of an electric power system are:

(i) Load flow/ power flow analysis.(ii) Fault analysis.(iii) Stability analysis (steady state, dynamic, and transient stabilities).

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Sistem Per Unit

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