fundamental power system

REVISION ON FUNDAMENTALREVISION ON FUNDAMENTAL THEORY RELATED TO POWER

SYSTEM

Complex Power• If voltage and current are known, complex power can be

calculated. Suppose voltage and current through a load or acircuit are given bycircuit are given by

and

• Complex power is given by:

β∠= IIα∠= VVp p g y

S i l f d t

βα −∠×== IVVIS * (2.1)

• S is also referred as apparent power.

θθ sincos IVjIVS += (2.2)

)sin()cos( βαβα −+−= IVjIVS

QjPS +=

(2.3)

Where is the angle between voltage and current.2Fundamental Theory Related to Power System |

βαθ −=

Complex Power• The real part of equation i.e P is real power and the imaginary part

is reactive power, Q. The Unit for P is watt or MWatt, and for Q Varor MVar. Cosine θ is known as power factor. Equation 2.3 can beor MVar. Cosine θ is known as power factor. Equation 2.3 can berepresented in a vector form:

θ

Figure 2.1 Phasor diagram

IVIVIVQPS =+=+= 2222 )sin()cos( θθ (2.4)

• From the diagram, , power factor :PQ /tan =θ

)/(tancoscos 1 PQ−=θ 22cos

QP

P

+=θ(2.5) (2.6)

3Fundamental Theory Related to Power System |

QP +

Complex Power• For a simple circuit that consists of an element Z = R + jX,

we can substitute into Equation (2.3), to yield:ZIV =

θcos2 ZIP = θsin2 ZIQ =(2.7) (2.8)

• It is known that and , therefore :θcosZR = θsinZX =

RIP 2= XIQ 2=(2.9) (2.10)


ExampleTwo ideal voltage source machines 1 and 2 are connected as shown in the Fig. 2.2. If V, V, and . Determine

(a) Which machine generate or absorb power and the amount

°∠= 01001E °∠= 301002E Ω+= 50 jZ( ) g p

(b) Whether each machine receive or supply reactive power and the what is the amount

(c) Active and reactive power absorb by the impedance(c) Active and reactive power absorb by the impedance.

Figure 2.2


Example (cont.)Solution:

)506.86(010021 jjEEI +−+

=−

=521 jZ

I ==−

°∠=−−=−

= 19535.1068.210504.13 jj5

jj

2681000)68.210(100*121 jjIES +−=+−==−

68.2105

100)506.86(1212 j

jj

ZEE

I +=−+

=−

=−

2681000)68.210)(506.86(*212 jjjIES +=−+==−

Total Power absorb by impedance: VarXI 53653510 22 =×=


Total Power absorb by impedance: VarXI 536535.10 =×=

Example (cont.)

Solution (cont.):According to solution of power flow from Machine 1 to Machine( ) d h h b b2 ( S1‐2), P is negative and Q is positive. Thus, machine 1 absorbs

energy with the rate of 1000 W and supply reactive power of 268Var. Machine 1 is a motor.

According to solution of power flow from Machine 2 to Machine1 ( S2‐1), P is positive and Q is positive. Thus, Machine 2 is agenerator that generates 1000 W and supply reactive power of268 Var. All the active power P generated by Machine 2 isp g ytransferred to Machine 1.


Three Phase System

Three‐phase system is widely being used in industry ascompared to single phase system. There are few reasons such as:compared to single phase system. There are few reasons such as:

1. Power and torque in a 3 phase motor and generator isconstant as compared to single phase machine, which the

i ib i d bl h ftorque is vibrating double the system frequency.

2. Three phase machine produces huge power.

3 Transmission system based on three phase system able to3. Transmission system based on three phase system able todeliver more power ac compared single phase system.


Three Phase e.m.f/currentThere are two ways of generating three phase e.m.f:

a) A rotating armature coil in magnetic field.

b) A rotating magnetic field in an armature coil.

To understand Three Phase system, consider a single phasegeneration consists of 3 winding: a, b, and c that displaced 1200

between each other as shown in Figure 2.3 below:Y

ωa

°120°120XX’

bc

120

°120


Y’Figure 2.3

Three Phase e.m.f/currentIn this position, winding b needs to rotate 120o to be in position a, winding cneeds to rotate 240 o to be in position a. The generated e.m.f can berepresented as follows:

θθ

Figure 2.4

All the generated e.m.f can be written as follows:

tEe mA ωsin= (3.1)(reference)

)120(sin °−= tEe mB ω)240(sin °−= tEe mC ω )120(sin °+= tEe mC ω

(3.2)

(3.3)

AfA 2h

or


AfNBANBE mmm πω 2==where

Three Phase e.m.f/currentThe generated e.m.f is a balance AC voltage because has the same magnitude and different phase of 120o. e.m.f also can be represented in a vector form as shown in Fig. 2.5. E is the magnitude of the e.m.f:

Figure 2.5

For a balance three phase e.m.f, it can be shown that the sum of these three e.m.f is equal to zero:

M th d 1Method 1:

0]60cossin2[sin]60cos)180(sin2[sin

=°=°°−+=++

ttEttEeee mcba

ωωωω


0]60cossin2[sin =−= ttEm ωω

Three Phase e.m.f/currentMethod 2:

00 jEEEa +=°∠=

120 (0.5 0.866)bE E E j= ∠− ° = −( )b j

)866.005.0(120240 jEEEEc +−=°∠=°−∠=

0)866.005.0()866.05.0()0( =+−+−−++=++∴ jEjEjEEEE cba


Phase Sequence

• Phase sequence refers to the order in which the threephases attain their peak or maximum valuesphases attain their peak or maximum values.

• In the generation of e.m.f of Fig. 2.5, the rotation isassumed to be counter‐clockwise. Thus, phase a attain, pthe first maximum, followed by b and lastly c.

• On the other hand, if the rotation is assumed to bel k i th h b > > b thiclock‐wise, the phase sequence becomes a‐> c‐> b, thisis called negative phase sequence.

• In three phase system, only these two sequences areIn three phase system, only these two sequences arepossible.


Interconnection of Three PhasesTo supply voltage/current, armature winding must be connected to a load. Ifthe three armature coils of the 3‐phase alternator are not connected but arekept separate as shown in Figure 2.6 (a), each phase would need two

d h l b f d ill b i Th d hconductors, the total number of conductors will be six. Thus, to reduce thenumber of conductors, the general methods of interconnection are:


Figure 2.6

(a) (b)

Star/Wye ConnectionIn this interconnection, the similar ends say, ‘start’ ends of three coils (it couldbe finishing ends also) are joined together at point N as shown in Fig. 2.6 (b).The point N is known as star point or neutral point. The three conductors

i i N l d b i l d k lmeeting at point N are replaced by a single conductor known as neutralconductor. Such an interconnected is known as four‐wire 3‐phase systemdiagrammatically shown in Fig. 2.7 (a).

Voltage at each coil is called phase voltage and current at the coil is calledphase current Whereas voltage at the terminal is called Line voltage (V ) V

Figure 2.7(a) Star connection for 3 Phase alternator

phase current. Whereas, voltage at the terminal is called Line voltage (VL)‐Vab,Vac, Vbc, and current at the terminal is called line current (IL).


Star/Wye Connection(a) Relationship between Phase and Line

Figure 2 7(b) Vector diagram for 3 Phase alternator

Consider a 3‐phase balance system i.e Ea = Eb = Ec = Vp (r.m.s) as shown in Fig.2.7 (b). Line voltage and line current is represented as VL and IL respectively.

Figure 2.7(b) Vector diagram for 3 Phase alternator

bb EEV −=


baab EEV

Star/Wye Connection(a) Relationship between Phase and Line (cont.)

Using sin rule,°

=° 30i120i

pL VVg ,

°° 30sin120sin

2/12/3pL VV

=2/12/3

pL VV 3= (r.m.s)

For the line, it can be seen that IL = Ip

***Note***

• In a practical 3‐phase system, the parameter data for alternator and motor is referred to the line voltage, unless otherwise stated.

• The above formula is true for a balanced system only.The above formula is true for a balanced system only.


Star/Wye Connection(b) PowerThe total active or true power in the circuit is the sum of the three phase powers, hence;

It is known that and

powerphase×=3poweractive Total φcos3 pp IVP ×=or

pL VV 3= Lp II =Hence, in term if line values,

φcos3

3 ×××= LL IVP φcos3 LL IVP =or

Similarly, the total reactive power is given by,

3φLL

φsin3 LL IVQ =

Note: is angle between phase voltage and phase current and not between the line voltage and line current.

IVS 3

φ

The total apparent power of the three phases is:


LL IVS 3=

Delta ConnectionFor delta connection, the circuit is shown in Fig. 2.8:

The relationship between phase current and line is:Figure 2.8 Delta interconnection for 3 Phase alternator

Consider the phasor diagram in Fig. 2.7 (b), which can also be used to represent current. Hence; ab a bI I I= −

= pL IIUsing sin rule, °

=° 30sin120sin

2/12/3pL II

=


2/12/3

pL II 3= (r.m.s)

Delta Connection(a) PowerTotal active power for star connection is the sum of power on each phase.

φIVhP φ3 IVPT t l

For Delta connection, and

φcospp IVperphaseP = φcos3 pp IVPowerTotal ×=

pL II 3= φcos3 LL IVP =

;

Hence,

φcos3

3 ×××= LL VIP φcos3 LL IVP =or

Similary, the total reactive power is given by . The total apparent power of the three phases is;

φsin3 LL IVQ =

**It can be seen that the equations for power are the same for star

LL IVS 3=

interconnection


Example (Try this now)A three‐phase line has an impedance of 2 + j4Ω as shown in Figure 1 below.

Ω

Ω

Ω

ΩΩ

The line feeds two balanced three‐phase loads that are connected in parallel.

Figure 1

The line feeds two balanced three phase loads that are connected in parallel.The first load is Y‐connected and has an impedance of 30 + j40Ω per phase.The second load is ∆‐connected and has an impedance of 60 ‐ j45Ω. The lineis energized at the sending end from a three‐phase balanced supply of linevoltage 207.85 V. Taking the phase voltage Van as reference, determine:


Example (cont.)a) The current, real power and reactive power drawn from the supply.

b) The line voltage at the combined loads.

c) The current per phase in each load.c) The current per phase in each load.

d) The total real and reactive powers in each load and the line.

323121

RRRRRRR

RA++

= BA

RRRRR

R++

=13R

2

323121

RRRRRRR

RB++

=

323121 RRRRRRR

++

CBA RRR ++

CBA

CA

RRRRR

R++

=2

CB RRR =3


1

323121

RRC =

CBA RRRR

++3

Example (cont.)Solution:(a) The current, real power and reactive power drawn from the supply.The ∆‐connected load is transformed into an equivalent Y. The impedance perh f h i l Y iphase of the equivalent Y is

Ω−=−

= 15203

45602 jjZ

The phase voltage isVV 120

385.207

1 ==

The single‐phase equivalent circuit is shown in Figure 2Ω

Ω

Ω

Ω

Ω

V0120∠


Figure 2


(a)(cont.):

The total impedance isp

Ω=−++=−++−+

++= 2442242)1520()4030(

)1520)(4030(42 jjjj

jjjZ

With the phase voltage as reference, the current in phase a is

AVI 501201 =∠

==

The three‐phase power supplied is

AZ

I 524

p p pp

( )( ) WIVS 180005012033 *1 =∠∠==



(b) The line voltage at the combined loads;

The phase voltage at the load terminal isp g

The line voltage at the load terminal is

( )( ) VjjV 3.108.11120110054201202 −∠=−=∠+−∠=

The line voltage at the load terminal is

VVV ab 7.1964.1937.19)8.111(33030 22 ∠=∠=∠=



(c) The current per phase in each load;

The current per phase in the Y‐connected load and in the equivalent Y of the p p q∆ load is

Ajjj

ZVI 4.63236.221

403020110

1

21 −∠=−=

+−

==

Th h t i th i i l ∆ t d l d i i i b

Ajjj

ZVI 56.26472.424

152020110

2

22 ∠=+=

−−

==

The phase current in the original ∆‐connected load, i.e., is given by

AIIab 56.56582.2303

56.26472.4303

2 ∠=∠∠

=∠

=303303 −∠−∠



(d) The total real and reactive powers in each load and the line;

The three‐phase power absorbed by each load isp p y

( )( ) var6004504.63236.23.108.11133 *121 jWIVS +=∠−∠==

( )( ) 900120056264724310811133 * jWIVS ∠∠

The three‐phase power absorbed by the line is

( )( ) var900120056.26472.43.108.11133 222 jWIVS +=−∠−∠==

It is clear that the sum of load powers and line losses is equal to the power

( ) ( )( ) var30015054233 22 jWjIjXRS LLL +=+=+=

p q pdelivered from the supply, i.e.,

( ) ( ) ( ) var01800300150900120060045021 jWjjjSSS L +=++−++=++


PER UNITSYSTEM

Per Unit Definition

Quantity in per‐unit = Actual quantityBase value of quantity

• Quantity in per‐unit must follow Kirchhoff law and Ohm law. If

q y

this can be done, all circuit analysis technique is applicable.

• The basic quantity in power system is voltage and current. The relationship between voltage and current is:relationship between voltage and current is:

S=VI*, (3.1) and impedance, Z =VI (3.2)

29Per Unit System |

Per Unit Definition

• This means if we fix S and V, we can find Z and I using basic circuitlaw. The chosen base voltage is according to the rating of thet ftransformer.

T ra n s m is s io n lin e

Figure 3.1

3 3 /1 3 2k V 13 2 /1 1 k V

A B

• In the above figure, base voltage for bus A is 33kV, bus B is11kV and for transmission line is 132kV.

30Per Unit System |

Per Unit DefinitionIn power system, per‐unit is used because of the following advantages: (1) It gives us a clear idea of relative magnitudes of various quantities, such as

voltage, current, power and impedance.

(2) The per‐unit impedance of equipment of the same general type based ontheir own ratings fall in a narrow range regardless of the rating of theequipment. Whereas their impedance in ohms vary greatly with the rating.

(3) The per‐unit values of impedance, voltage and current of a transformer arethe same regardless of whether they are referred to the primary orsecondary side. This is a great advantage since the different voltage levelsdisappeared and the entire system reduces to a system of simple impedance.

(4) The per‐unit systems are ideal for the computerized and analysis andsimulation of complex power problems.simulation of complex power problems.

(5) The circuit laws are valid in per‐unit systems, and the power and voltageequations are simplified since factor of and 3 are eliminated in the per‐unitsystemsystem

31Per Unit System |

Per Unit Definition

General equation to determine the bases:

(i) Single Phase System

LN

1

kV voltage,basekVA base

A current, Base φ=

V voltage,baseimpedanceBase LN=Ω

(3.3)

(3.4)A current, base

impedance, Base =Ω

kVA base kW power, Base 11 φφ =

MVAbaseMWpowerBase

(3.4)

(3.5)

MVAbaseMWpower, Base 11 φφ =

φ1

2LN

MVA )kV voltage,(base

impedance, Base =Ω

(3.6)

(3.7)

φ1

2LN

kVA base1000 )kV voltage,(base impedance, Base ×

=Ω

Ωimpedance, actualelementanofimpedanceunitPer =

(3.8)

(3 9)

32Per Unit System |

Ωimpedance, baseelementan ofimpedanceunit Per = (3.9)

Per Unit Definition

General equation to determine the bases:

(ii) Three Phase System

LL

3φ

kV voltage,base3kVAbase

A current, Base×

=

1000)3/kVvoltage,(basei d2

LL ×

(3.10)

( ) /3kVA base

1000)3/kVvoltage,(baseimpedance Base3

LL

φ

=

kVAbase1000)kV voltage,(baseimpedance Base

2LL ×

=

(3.11)

(3.12)kVA base 3φ

MVA base)kV voltage,(base impedance Base

3φ

2LL= (3.13)

Or in general Eq. (3.13) can be represented as ( )

B

BB S

VZ2

=

33Per Unit System |

Per Unit Definition

(ii) Three Phase System (cont.)

A minimum of four base quantities are required to completelydefine a per‐unit system: volt‐ampere, voltage, current andimpedance Usually the three phase base volt‐ampere (SB inimpedance. Usually, the three phase base volt ampere (SB inMVA) and the line to line base voltage (VB in kV) are selected.

34Per Unit System |

Example 1Calculation of pu for 1 phase and 3 phase system

Three Phase One Phase30 000

If the power is 18MW Hence power for single phase is 6MW

kVA 30,000kVA Base 3 =φ kVA 000,013

30,000 kVA Base 1 ==φ

6.030,00018,000 power unit -Per == 6.0

000,10000,6power unit Per =−

120

If Line to Line Voltage is 108kV Voltage per phase is 62.3kV obtain from

kV120kV Base =LL kV2.693

120kV Base LN ==

⎟⎠

⎞⎜⎝

⎛3

108⎠⎝ 3

up.90.0120108 eP.u voltag == pu90.0

2.693.62 eP.u voltag ==

From above, per‐unit quantity is the same for single phase and 3‐phasesystem.

35Per Unit System |

Change of Base

• The impedance of individual generators and transformers, assupplied by the manufacturer are generally in term of

t it titi b d th i tipercentage or per‐unit quantities based on their own ratings.

• The impedance of transmission lines are usually by theirohmic values.

• For power system analysis, all impedance must be expressedon per unit on a common base.

• In order to do this, an arbitrary base for apparent power mustbe selected (usually 100MVA). Then, voltage bases must bedetermined for the system.determined for the system.

• Once, a voltage base has been selected for a point in thesystem, the remaining voltage bases no longer independent;h d b h fthey are determine by the various transformer turns ratios.

36Per Unit System |

Change of BaseldTo convert to a new bases value, suppose is the per‐unit

impedance on the power base and the voltage base . Thus,

oldpuZ

oldBS old

BVoldSZ

or( )2old

B

Bactualold

B

actualoldpu

VSZ

ZZZ ==

( )oldV 2

(3.14a)

Expressing Zactual to a new power base and new voltage base, result in

( )oldB

Boldpuactual S

VZZ = (3.14b)

the new per unit impedance;

( )2new

newB

actualnewB

actualnewpu

VSZ

ZZZ == (3.15)

Substitute Zactual in Eq. (3.14b) into (3.15), yields;

( )BB V

2

⎟⎟⎞

⎜⎜⎛ old

BnewBoldnew VSZZ (3 16)

37Per Unit System |

⎟⎟⎠

⎜⎜⎝

= newB

BoldB

Boldpu

newpu VS

ZZ (3.16)

Example 2

The reactance of a generator, X” is given as 0.25 pu based on its nameplate 18 kV, 500 MVA. Find the new per unit X” using new bases of 20 kV and 100MVA.

U i (3 16) pu0405010018250''2

=⎟⎞

⎜⎛

⎟⎞

⎜⎛=XUsing (3.16), pu0405.0

5002025.0 =⎟

⎠⎜⎝

⎟⎠

⎜⎝

=X

Or by changing the given per‐unit into its actual ohms quantity and divided using the new base,

pu 0405.0100/20

)500/18(25.0'' 2

2

==X

38Per Unit System |

Per Unit Reactance for Single Phase Transformer

• The leakage resistance and reactance (in ohm) for atransformer depends whether it is referred to Primary ortransformer depends whether it is referred to Primary orSecondary. If these values are stated in per‐unit, the basepower is the rating of the transformer.

• Whereas, the referred base voltage could be from thePrimary or secondary. However, the per‐unit values fortransformer is the same regardless the base voltage istaken from the Primary or Secondarytaken from the Primary or Secondary.

(Attention: If not stated what the referred side, by default, the( , y ,impedance is always referred to the lower voltage)

39Per Unit System |

Example 3A single phase transformer has a rating of 110/440 V, 2.5 kVA. Theleakage reactance measured from the low voltage side is 0.06Ω .Determine the leakage reactance in per‐unit.Determine the leakage reactance in per unit.

Z base referred to low voltage side Ω=×

= 84.45.2100011.0 2

In pu, puX 0124.084.406.0

==

If the leakage reactance measured from the high voltage side, thereactance X is: Ω=⎟

⎠⎞

⎜⎝⎛= 96.0

11044006.0

2

X

Z base referred to high voltage side =In pu

⎠⎝ 110

puX 0124096.0==

Ω=× 5.775.2

1000440.0 2

In pu,

40Per Unit System |

puX 0124.05.77==

Example 3 (cont.)Note on the relationship between primary and secondary side:

11

NN

VV

= 21

NN

II

=(3.17) (3.18)

If the impedance is connected to the secondary,

b d d ( d l ) ( )

22 NV 12 NI

2

22 I

VZ =

( ) ( )

(3.19)

Substituted V2 and I2 (in Eq. 3.17 and 3.18 respectively) into (3.19),yields

( )( )

1122 INN

VNNZ = (3.20)

Hence, if impedance is measured from primary side,

( ) 121 INN

2

2

2

1

1

1'2 Z

NN

IVZ ⎟⎟

⎠

⎞⎜⎜⎝

⎛==

Therefore, if impedance connected at the secondary side is referredto the primary side, it must be multiply to the ratio of primary side

21 ⎠⎝

over secondary side.

41Per Unit System |

Example 4Consider the following simple power system network:

2

⎟⎟⎞

⎜⎜⎛

=old

BnewBoldnew VSZZ

The three‐phase power and line‐line ratings of the electrical powersystem are as follows

⎟⎟⎠

⎜⎜⎝

= newB

oldB

pupu VSZZ

system are as follows:

G1: 60 MVA 20kV X = 9%

T 50 MVA 20/200 kV X 10%T1: 50 MVA 20/200 kV X = 10%

T2: 50 MVA 200/20 kV X = 10%

M : 43.2 MVA 18kV X = 8%M : 43.2 MVA 18kV X 8%

Line: 200kV Z = 120 + j200 Ω42Per Unit System |

Example 4 (cont.)Question:

(a) Draw an impedance diagram showing all impedances in per it 100 MVA b Ch 20 kV th lt bunit on a 100‐MVA base. Choose 20 kV as the voltage base

for generator.

(b) The motor is drawing 45 MVA, 0.80 power factor lagging at a line‐to‐line terminal voltage of 18 kV. Determine the terminal l d h i l f f h i i dvoltage and the internal emf of the generator in per unit and

in kV.

43Per Unit System |

Example 4 (cont.)Solution (a):

The base voltage VBG1 on the LV side of T1 is 20 kV. Hence, the base on its HV side is: 200 ⎞⎛its HV side is:

This fixes the base on the HV side of T2 at VB2 = 200 kV, and on its LV

kVVB 20020

200201 =⎟⎠⎞

⎜⎝⎛=

side at

The generator and transformer reactances in per unit on 100 MVA

kVVB 20200202001 =⎟

⎠⎞

⎜⎝⎛=

The generator and transformer reactances in per unit on 100 MVA base is:

100 ⎞⎛ 100 ⎞⎛G: T2:

T : M:

puX 15.060

10009.0 =⎟⎠⎞

⎜⎝⎛=

puX 2.010010.0 =⎟⎠⎞

⎜⎝⎛=

puX 2.050

10010.0 =⎟⎠⎞

⎜⎝⎛=

puX 150181000802

=⎟⎞

⎜⎛⎟⎞

⎜⎛=T1: M:

44Per Unit System |

pu.050

0.0 ⎟⎠

⎜⎝

puX 15.0202.43

08.0 ⎟⎠

⎜⎝⎟⎠

⎜⎝

Example 4 (cont.)Solution (a)(cont.):

The base impedance for the transmission line is:)200( 2

The per unit impedance is

Ω== 400100

)200( 2

BLZ

p p

Line: pujjZ line 5.03.0400

200120+=⎟

⎠⎞

⎜⎝⎛ +

=

The per unit equivalent circuit is shown below:

45Per Unit System |

Example 4 (cont.)Solution (b):

The motor complex power in per unit is puSm °∠=°∠

= 87.3645.0100

87.3645

and the motor terminal voltage is:

puVm °∠=°∠

= 090.020

01820

puI °−∠=°∠

°−∠= 87.365.0

09.087.3645.0

Thus, the generator line‐to‐line terminal voltage is

pujVg °∠=°−∠++°∠= 82.1131795.1)87.365.0)(9.03.0(09.0

kVVg 359.26)20(31795.1 ==

pujEg °∠=°−∠++°∠= 88.13375.1)87.365.0)(05.13.0(09.0

Thus, the generator line‐to‐line internal emf is:

46Per Unit System |

kVEg 5.27)20(375.1 ==

Example 5A simple power system consisting of one synchronous generator andone synchronous motor connected by two transformers and atransmission line is shown in Fig. 3. Create per phase, per unit circuittransmission line is shown in Fig. 3. Create per phase, per unit circuitfor this power system using a base apparent power of 100 MVA and abase line voltage at generator G1 of 13.8 kV.

Figure 3

47Per Unit System |

g

Example 5 (cont.)Solution:

To create a per‐phase, per unit equivalent circuit, we must firstcalculate the impedances of each component in the power system incalculate the impedances of each component in the power system inper‐unit to the system base. The system base apparent power isSB=100 MVA everywhere in the power system. The base voltage in thethree regions will vary as the voltage ratios of the transformers thatthree regions will vary as the voltage ratios of the transformers thatdelineate the regions. The base voltages are:

Region 1: Vbase,1 = 13.8 kVRegion 2: Vbase,2 = Vbase,1 = 110 kV⎟⎟

⎠

⎞⎜⎜⎝

⎛kVkV

8.13110

Region 3: Vbase,3 = Vbase,2 = 13.2 kV⎟⎟⎠

⎞⎜⎜⎝

⎛kVkV

1204.14

48Per Unit System |


The corresponding base impedances in each region are:

R i 1 Z ( ) ( )2 213 8V kVRegion 1: Zbase,1 =

Region 2: Zbase 2 =

( ) ( ),

3 ,

13.81.904

100LL base

base

V kVS MVAϕ

= = Ω

( ) ( )2 2

, 110121LL baseV kV

= = Ωg base,2

Region 3: Zbase,2 =

3 ,

121100baseS MVAϕ

= = Ω

( ) ( )2 2

,

3

13.21.743

100LL base

b

V kVS MVAϕ

= = Ω

The per‐unit impedance for each component:

G1: Unchanged R = 0.1 pu, X = 0.9 pu

3 , 100baseS MVAϕ

G1: Unchanged R 0.1 pu, Xs 0.9 pu

T1: Unchanged R = 0.01 pu, Xs = 0.05 pu

L : R = X =pu124015=

Ω pu62075=

ΩL1: RL = , XL =

49Per Unit System |

pu124.0121Ω

pu62.0121Ω


T2: The impedance of T2 is specified in per‐unit on a base of 14.4kVand 50 MVA in region 3 Therefore the per‐unit resistances andand 50 MVA in region 3. Therefore, the per‐unit resistances andreactance of this component on the new system base is:

( ) puMVAkVR 238.01004.1401.02

=⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

= ( ) pMVAkV 502.13 ⎟

⎠⎜⎝

⎟⎠

⎜⎝

( ) puMVAMVA

kVkVX 119.0

50100

2.134.1405.0

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

M2: The per‐unit initially is base on 13.8 kV and 50 MVA. Thus, thenew per‐unit values:

2⎞⎛⎞⎛( ) pu

MVAMVA

kVkVR 219.0

50100

2.138.131.0

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

( ) MVAkVX 40521008.13112

⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

50Per Unit System |

( ) puMVAkV

X 405.2502.13

1.1 =⎟⎟⎠

⎜⎜⎝

⎟⎟⎠

⎜⎜⎝

=


Per‐phase, per‐unit equivalent circuit is:

51Per Unit System |

Example 6 – Try thisA one‐line diagram of a simple power system is shown in Figure 3. The generators are running on no‐load at their rated voltage and rated frequency with their emfs in phase. Create per phase, per unit circuit for this power system using a base apparent power f 100 MVA d b li lt t t G f 15 kVof 100 MVA and a base line voltage at generator G1 of 15 kV.

52Per Unit System |

Example 6 – Try this

Equipment Rating:

Synchronous Generators: G1 : 1000 MVA 15 kV 18.021 == XX pu, 07.00 =X puG2 : 1000 MVA 15 kV 20.021 == XX pu, 10.00 =X pu G3 : 500 MVA 13.2 kV 15.021 == XX pu, 05.00 =X pu, Xn =0.05 pu G4 : 750 MVA 13 8 kV 0 30X X= = pu 100=X puG4 : 750 MVA 13.8 kV 1 2 0.30X X= = pu, 10.00 =X puTransformers: T1 : 1000 MVA 15kV Δ /765 kV Y X1 = X2 = 0.1 pu dan X0 = 0.05 pu T2 : 1000 MVA 15kV Δ /765 kV Y X1 = X2 = 0.1 pu dan X0 = 0.02 pu T3 : 500 MVA 15kV Y/765 kV Y X1 = X2 = 0 1 pu dan X0 = 0 06 puT3 : 500 MVA 15kV Y/765 kV Y X1 = X2 = 0.1 pu dan X0 = 0.06 puT4 : 750 MVA 15kV Y/765 kV Y X1 = X2 = 0.1 pu dan X0 = 0.08 pu Transmission lines: 1 – 2 765 kV X1 = X2 = 50 Ω , X0 = 150 Ω 1 3 765 kV X X 40 Ω X 100 Ω1 – 3 765 kV X1 = X2 = 40 Ω , X0 = 100 Ω 2 – 3 765 kV X1 = X2 = 40 Ω , X0 = 100 Ω

53Per Unit System |

fundamental power system

Engineering