Chapter Eight - University of · PDF fileThe concept of self and Mutual inductance can be extended to a group of ... GMD is the Geometric mean Distance ... GMR is the Geometric Mean

ANALS POWER SYSTEM ANALYSIS MISSAN UNIVERSITY\ COLLEGE OF ENGINEERING

Chapter Eight

Transmission Line parametersResistance, Inductance , Capacitance

-Resistance :The resistance of a solid round conductor at specified temperature isgiven by :-

(36)

Where is the conductor resistivity

Is the conductor length Is the cross-section area

When ac ( alter native current ) flow in an conductor the currentdistribution is not uniformly over the conductor cross-sectional area andthe current density is greatest at the surface of the conductor. This causes the ac resistance to be somewhat high. Than the d.cresistance. This behavior is known as skin effect. At 60 HZ the a.cresistance is about 2 percent higher than dc resistance.

*The conductor resistance increase as temper due increase.


Where are conductor resistance at respectively. T isthe temperature constant that depend on the conductor material . forexample T for the aluminum is T=228.

Inductance (L)Any conductor carrying a current will be surrounded by a magnetic fieldchange with the change of the current.

Where is the magnetic linkage flux and this will be change as thechanging of the current ( and this lead to make the permeabilityconstant for all flux level as in the air.

Henery (H) (1)

Where is the flux linkage of the conductor ( web.turn). I is the current flow in the conductor

*In any conductor carrying a current there is :-

1- Internal fluxThis will be as a concentric circles inside the conductor and caused bythe current that flow in the same conductor.2- External fluxThis will be as a concentric circles outside or around the conductor andcaused by the same current .


Inductance of conductor to external flux :-

After the derivative of external linkage flux between

The inductance between theTwo point that location outsideOr external the conductor.

H/m

Flux outside the conductor Due to the current I that Flow in the same conductor

*Inductance between two single- phase conductor :-

=

rX

D


H/m where

(3)And in the same manner

H/m where (4)

And the inductance of the Total system is

And if

Also we can write eq. (3) and (4) by this form

(5)


*Flux linkage in term of self and Mutual inductanceAs in eq. (1)

(6)

Whereis the Total linkage flux in the conductor 1 is the Total linkage flux in the conductor 2 is the self inductance due to is the Mutual inductance due to is the mutual inductance due to is the self inductance due to

Since

(7)

By comparing eq.(7) with eq. (5)

(8)

} Self inductance

} Mutual inductance


The concept of self and Mutual inductance can be extended to a group ofn conductors. Consider n conductors carrying phase currentsSuch that

(9)

Generalizing eq. (6), the linkage of conductor are

(10)

And from eq. (8)

(11)


*Inductance of composite conductor :-The current in eachSu b conductor ofGroup A is

And the current in eachSub conductor ofGroup B is

Due to eq. (11) andIts application

(12) (12)

The inductance of sub conductor a is

+1n

+1n

a

b

c

d

nm

ConductorA

Has acurrent +I

ConductorA

Has acurrent -I


As so on

The average conductor in any sub conductor in group A

Since all the sub conductor of conductor A are electrically parallel, so thatthe inductance o A will be

(13)(13)

(14) (14)

(15) (15)

…(

…(


Where

GMD is the Geometric mean Distance And it is equal for A and BGMR is the Geometric Mean radius And it is not equal for A and B in any time


Ex1) A single phase has two group of conductor A and B where Aconsists of 3- wires (sub conductors) a ,b, c each of its have a radius of0.25 cm ,and the group B consists of two wires d ,e each of its have aradius of 0.5cm. calculate the inductance of the Total system where thedistance between the sub conductor is as follow :.

So1)

Where

=0.4 m

b

c

d

e6 m

6 m

GroupA

GroupB



*Inductance of Transmission Line :-a- Symmetrical spacing :-

consider one meter length of line with three conductors, each with radius r , symmetrically spacing in a triangle (as example )configuration as shown in the Fig. belowAssume we have a balancedCurrent

(16)

From eq. (11). The Total flux linkage of phase a is

(17) (17)

Since

Since we have a symmetrical spacing

H/m (18) (18)

D D

D


b) A symmetrical spacing :-with a symmetrical spacing even with balance currents, the voltage dropdue to line inductance with be unbalanced, because the inductance of eachphase will be not equal. Consider one meter length of of line withthree conductor each of radius r as shown in the fig. below .The application of eq. (11)

1

2

3


since we have a balance system

If we have the parameter aWhich is

The equations of inductance a symmetrical spacing as in theFig above are :-

120 120

120


(18)

*Transpose System :-One way to solve the problem of a symmetrical spacing and the differentinductance for each phase is :-Transpose system which it is :-Interchanging the phase configuration every one – third the length so thateach conductor is moved to occupy the next physical in regular sequence.And by this method the inductance for each phase will be equal.Such a transposition arrangement is shown in the Fig. below :-

a

a

a

b

b

b

c

c

c

-1- -2- -3-First

TransposeSecond

Transposethird

Transpose

3 2

a


Now let take this example

Since in a transposed line each phase take all three position. Theinductance per phase can be obtained by finding the average value ofeq. (18)

H/m(19) (19)

3

1

2

3

1

2


1 represent first transpose2 represent second transpose3 represent third transpose

This is another way to solve the problem of transpose system. From eq.(18)

( First transpose )

( second transpose )

(third transpose )

1

2

3

3

1

2


(20) (20)


*Bundle System :-Bundle system or bundle conductor consists of a two, three, four subconductors symmetrically arranged in configuration as in the fig. below :-

The geometric Mean Radius of each fig. is as below

Two sub(conductor) =

And so on

(three-sub conductor)

The advantage of the bundle system is1- Reduce the inductance of the line, which is reduce the line losses andimproves the line capacity and system stability over the use of singleconductor.2- Reduce the corona losses .3-Reuce the Radio in deference.

d d d

d

d

d

d

-1- -2--3-


Ex2) A double circuit of conductors as shown in the Fig. below.Calculate the inductance per phase per km if the system is transpose atequal distance. The diameter of the conductor is 5 cm.So1)In order to solve the problemWe must first know theThere are two way to solve theProblem.

From the fig. above we see that

So that in order to get a per phase inductance (Balance system) and wehave a transpose systemWe must take an equivalent GMR

Where

a

b

c

1.5m

1.5m

3 m


In all cases

Where

(first transpose) (second transpose) (third transpose) (general from)

OR in other meaning


Ex3) line, each phase has two parallel conductor and the radius ofthe conductor is 0.205 cm. If the distance between the different conductoris as in the fig. below. Calculate the reactance in ( ohm ) per one km. if thefrequency is 50 HZ and the phases are transpose the location between itsat equal distance.

So1)Transpose states

Not

1) The Distance of sub conductor for each phase is equally

2) The Distance of sub conductor for each phases for the three stages ofcomplete transpose are equally

0.4 m

6.050.4 m

6.050.4 m

-1-

-2-

-3-


Also

3) for all cases of tanspose stages ( general form)


Ex4) what is the bandle system, what is the advantage of it, show thatfrom the two fig. below by calculating the inductance per phase per km ofthe line in each fig. below. Note that the diameter of the conductor is 5 cm

So1)

From the Fig. (1)

From the fig. (2)

a b c4 m 4 mFig-1-

0.6m 4 m 0.6m 0.6m4 m

Fig-2-


The bundle system cause a reduction in the inductanceBy


-Capacitance :

Where C is the capacitance is the charge

V is the voltage is the radius of the

Conductor.For one meter length of the conductor.The electric flux density (D) at a radius X is given

Where is the charge ( coulomb / one meter ) and A is the area of the surface of an axial length of One meter of the conductor.Along uniformly charged conductor has lines of electric flux extendingradials out from the conductor surface

The electric field intensity E may be found from the relation

r


Where E is the electric field intensity V/m

Is the relative permittivity of the vacuum

is the relative permittivity of free space = 8.85 is the relative permittivity of the air = 1

V/M(21)

The potential difference between cylinder from position isdefine as the work done in moving a unit charge of coulomb from

Through the electric field produced by the charge on the conductor

(22)


f/m(23)

-phase lines :–* Capacitance of singleAssuming conductor q alone toHave a charge ofThe voltage between conductor1 and 2 is

Now assume only conductor 2 having a charge of . The voltagebetween conductor 2 and 1 is

Since

From the principle of superposition, the potential difference due topresence of both charge is

r r

D


For single – phase line

v/m(23)

f/m(24)

Equation (24) give the line – to line capacitance between the conductor

(between conductor ) ( between each

Conductor and the neutral )

f/m(25)

MF/KM(26)

n


-lines :*Capacitance ofAfter take the complete transposeOf this fig. and make a dative forThe capacitance

F/M(27)

Mf/km(28)

* if we have bundle system. The same process as in the inductance

MF/KM(29)

Note : In the inductance the radius is . But the capacitance contain radius r .


*The derive ofConsider one meter length of line with 3 conductors each of its hasa radius of r as shown in the fig. below

Since we have a balanced system

(30)

For complete transpose as shown in the fig. above


Also , similarly we find the average voltage

Where

For balance 3 voltage

(31)

Mf/km(32)

130

130a

b

c


transmossionThe effect of earth on the capacitance of

F/M(33)

is the capacitance of each phase ( conductor ) To the neutral .( earth)Note :-

earth


(34)

phase-The effect of earth on the capacitance of single*

(35)

Also


Ex5) transmission line of 10km , as shown in the fig. below. The radiusof each conductor is 0.885 cm and the phase are transpose at equal

distance also the resistivity of the line is (Find the resistance, inductance, capacitance and the charging current ofthe line .

Note that :The voltageThe line is 33kvAnd the frequencyIs 50 Hz

So1)

Note that

So that we must takeBut in the first let calculate

5.4

3 m

3 m

5.4 m

6.3m


In order to get GMR, as we say before there are two way to the solution :-

1)

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Chapter Eight - University of · PDF fileThe concept of self and Mutual inductance can be extended to a group of ... GMD is the Geometric mean Distance ... GMR is the Geometric Mean