Power Systems Elements Lines

7/27/2019 Power Systems Elements Lines

1/80

ECE4334

Dr. C.Y. Evrenosoglu

ECE4334MODELS OF POWER SYSTEM ELEMENTS

PART B: TRANSMISSION LINES

Dr. E


2/80

ECE4334Outline

Power Transformers Transmission Lines

Generators

Loads



3/80

ECE4334Outline

Power Transformers Transmission Lines

Generators

Loads



4/80

ECE4334Development of Line Models

Goals of this section are

1) develop simple models for transmission

lines

2) gain an intuitive feel for how the geometry

of the transmission line affects the modelparameters

Thanks to Dr. Tom Overbye, University of Illinois for the content


5/80

ECE4334Transmission Lines

ACSR Aluminum conductor steel-reinforced AAC all-aluminum conductor

AAAC all-aluminum-alloy conductor

ACAR aluminum-clad steel conductor



6/80

ECE4334ACSR



7/80

ECE4334Line Parameters

Resistance Inductance

Capacitance



8/80

ECE4334


Resistance

Conductor resistance depends on Spiraling

Temperature

Frequency (skin effect)

Current magnitude (magnetic conductors)


9/80

ECE4334Resistance



10/80

ECE4334Resistance



11/80

ECE4334Inductance

The inductance of a magnetic circuit that has aconstant permeability, can be obtained bydetermining

Magnetic field density,Hfrom Amperes law

Magnetic flux density,B (B = H)

Flux linkages

Inductance from flux linkages per ampere ( = L i L= /i )



12/80

ECE4334Inductance

Dr. C.Y. Evrenosoglu Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content

= r0r= relative permeability 1

0 = permeability of free space=410-7[H/m]


13/80

ECE4334Inductance



14/80

ECE4334Inductance of a single conductor

Infinite straight wire is an approximation of areasonable long wire (in order to use

superposition)

infinite assumption is similar to a one-turn coilwith the return path at infinity

non-magnetic with radius r Uniform current density in the wire (skin effect is

ignored)

Flux lines form concentric (having a commoncenter) circles

Angular symmetryDr. C.Y. Evrenosoglu


15/80


16/80

ECE4334Inductance of a single conductor


Case 2: Flux linkages outside the conductor (x > r):

Thanks to Dr. Thomas Baldwin, FAMU/FSU for slide content

[H/m]The flux linkage caused by the conductor

at an external point at distances D1 and

D2from the conductor


17/80

ECE4334Inductance of a single-phase circuit


Conductors of same radius, rand separated by a distanceD;D1= randD2=D

Thanks to Dr. Thomas Baldwin, FAMU/FSU or slide content

[H/m] per conductor

Ltotal (Loop inductance) = 2L = 4 [H/m] per circuit


18/80

ECE4334Self & mutual inductance in single-phase circuit

If the conductors are identicalL11 =L22



19/80

ECE4334Inductance of three-phase lines

Asymmetrical spacing



20/80

ECE4334Inductance of three-phase lines

Symmetrical spacing


La=Lb =Lc= [H/m] per phase


21/80

ECE4334Important points

Although there is magnetic coupling between phases, for

balanced system with equilateral spacing , we can model the

magnetic effect using only self-inductances and the self

inductances are equal. We can then use per-phase analysis.

To reduce the inductance per meter we can try to reduce thespacing between the conductors and increase their radii.

Reducing the spacing,D, can only go so far because of

considerations of voltage flashover There are cost and weight problems associated with

increasing the radii, r, of solid conductors

Hollow conductors have problems with flexibility and ease ofhandling

What is the practical approach?



22/80

ECE4334Conductor bundling

Suppose that instead of one conductor per phase there are b

conductors in close proximity as compared with the spacing

between the phases. Such a conductor is said to be made up of

bundled conductors. (b=4 in the following example)

These conductors are effectively in parallel. All the conductorshave the same radius r.


D

D

D

Phase a

Phase b

Phase c

1 2

4 3

9 10

12 11

5 6

8 7

Conducting frame

supporting the conductors


23/80


Consider the flux linkages of conductor 1 inphase a bundle.

Assumption: The current in each phase splits equally among the

four parallel branches.

dij = the distance between conductors i andj.


D

D

D

Phase a

Phase b

Phase c1 2

4 3

9 10

12 11

5 6

8 7

18

12 13 14

01

15 16 17

19 1,10 1,11 1,12

1 1 1 1ln ln ln ln

4 '

1 1 1 1

ln ln ln ln2 4

1 1 1 1ln ln ln ln

4

a

b

c

i

r d d d

i

d d d d

i

d d d d

+ + + +

= + + + +

+ + +



24/80



D

D

D

Phase a

Phase b

Phase c

1 2

4 3

9 10

12 11

5 6

8 7



25/80



D

D

D

Phase a

Phase b

Phase c

1 2

4 3

9 10

12 11

5 6

8 7


geometric mean radius (GMR) of bundle geometric mean distance (GMD) from

conductor 1 to phase b

, , , GMD from conductor 1 to phase c


26/80


27/80


Dr. C.Y. Evrenosoglu Thanks to Dr. Tom Overbye, University of Illinois for the content

geometric mean radius (GMR) of bundle

geometric mean distance (GMD) from conductor 1 tophase b

, , , GMD from conductor 1 to phase c


28/80



Remember that each bundle has b conductors and

in our example b = 4

L1 = L2 = L3 = L4

2 10

[H/m]


29/80

E l


30/80

ECE4334Example


a

0a

0

7

0 3

6

Substituting

i

Hence

1 1ln ln

2 '

ln2 '

4 10 5ln ln2 ' 2 9.67 10

1.25 10 H/m

b c

a a

a

a

i i

i ir D

Dir

D

Lr

=

=

=

= =

=


31/80


32/80

ECE4334M b t b dli


33/80

ECE4334More about bundling

If we view the bundle as an approximation of a hollow conductor, the

reason for the increased radius is clear.

The larger radius helps in another respect. At high voltages, above

approximately 230 kV, the electric field strength near conductors is

sufficiently high to ionize the air nearby. This phenomenon is called

Corona and has an undesirable effect since it is associated with

Line losses

Radio interference

Audible noise

All other things being equal, the lager the conductor radius, the less

electric field strength at the surface of the conductor. Bundling is

beneficial since it effectively increases the conductor radius

Compared with a single conductor of the same cross-sectional area,

bundled conductors, having a larger surface area exposed to the air,

are better cooled. Higher currents may be carried without exceeding

the thermal limits.Dr. C.Y. Evrenosoglu Bergen & Vittal, Power System Analysis 2

ndedition, 2000

ECE4334Transposition


34/80


The practice of equilateral arrangement is not convenient

Horizontal or vertical configurations are most popular

Symmetry is lostDab Dac Dbc unbalance

Symmetry is regained by the method of transposition Average inductance of each phase will be the same

We can think of this as a top view of three conductors in the

same horizontal plane. It could also be a side view of three

conductors in the same vertical plane.



35/80


36/80



37/80



Section 1 Section 2 Section 3

2 10 1 1 1 2 10 1

1 1

2 10 1 1

1

The average of the above flux linkages for phase a is:

/3 /3 /3 3


38/80



39/80

Transposition


2 1 0

33 1 1 1

2 1 0

3

3 1

1 2 10

2 10

(GMD between the phases)For solid conductors use

For stranded conductors use GMR of the phase


40/80


41/80

ECE4334Impedance of three-phase lines including


42/80

Impedance of three phase lines including

the ground return

In some cases we cannot assume balanced operation

due to lack of transposition or lack of load balance.

In faulted conditions there might be a neutral currentalso.

The effect of earth and neutral return on the

impedance of a transmission line have to be modeled.



43/80


44/80



45/80

p p g

the ground return

J. R. Carson determined in 1923 that the earthresistance rd is a function of the frequency and

derived the given empirical formula.

In the formula forDearth, if the actual earth

resistance is unknown, it is common to assume

to be 100 -m. Dearth, GMRi and dij are only involved as ratios in

the formulas forZii andZij. As long as the units are

consistent, either feet or meters can be used.



46/80



47/80

the ground return

IfVn = 0, then


0

0


48/80


49/80


50/80



51/80

the ground return

phasor voltage drop of phase i, i = a, b, c, n

phasor current flowing in phase i self-impedance of conductor i including the

effect of ground return length of transmission line in meters

2 1 0 /m resistance of phase i in /m 9.869 10 is the earth resistance in /m



h d


52/80

the ground return

operating frequency in hertz

658.368 m = resistivity of the earth in -m

geometric mean radius of conductor i in meters is the mutual impedance between conductor

i and conductorj including the effect of the ground return

2 1 0

/m

distance between conductors i andj in meters


ECE4334Capacitance


53/80

The capacitance of between conductors in a

medium with constant permittivity, can beobtained by determining

Electric flux density,D (Cuolomb/m2

), from Gaussslaw

Electric field strengthE(V/m) fromD = E

Voltage between conductors Capacitance from charge per unit volt (C = q/V)

permittivity in farads/m (F/m) o ro permittivity of free space (8.85410

-12 F/m)

r relative permittivity or the dielectric constant

(1 for dry air, 2 to 6 for most dielectrics)Dr. C.Y. Evrenosoglu


54/80

ECE4334Electric field and voltage

lid i fi it t i ht i


55/80

a solid infinite straight wire

D = 0 in the wire (x < r) D = q / (2x) (x r)

The voltage drop V12 between two points

with radial distances asD1 andD2 fromthe wire due to a charge (q):

Voltage is defined as the energy (in Joules)

required to move a 1-Coulomb charge against anelectric field.

Voltage drop between two points can be + or

due to the sign of the charge or whetherD2 >D1

or not.Dr. C.Y. Evrenosoglu

D1

D2

P1P2

+q

ECE4334Capacitance of a two-wire line


56/80

For two-wire line qa = - qb


D

ra rb

qa qb

2

2

due to qa due to qb

2

2

2


57/80


58/80

ECE4334Capacitance of three-phase lines

id il l i d


59/80

Consider equilateral spacing and qa + qb + qc = 0


a

b

c

2 2

2

2

2 +

3

3

1

3

2

2

2

2

1

3

22

2

2

ECE4334Capacitance of three-phase lines

C id il l i d 0


60/80

Consider equilateral spacing and qa + qb + qc = 0


a

b

c

1

3

22

2

2

1

3

2

2

2

2

2

[F/m] to neutral

ECE4334

C id l i d 0

Capacitance of three-phase lines


61/80

Consider unequal spacing and qa + qb + qc = 0

If the conductors are bundled; replace rwith the bundle GMRb,Rb.

Note that we use r (outside radius) instead of r which is in the

case of inductors. We are neglecting the effect of the earth!


a

b

c

2

[F/m] to neutral

ECE4334Example

C l l t th h li t t l it d iti


62/80

Calculate the phase line-to-neutral capacitance and capacitive

reactance (or shunt admittance) of a balanced three-phase, 60 Hz,220 kV transmission line with horizontal spacing between

conductors in bundles of 4 of 0.5m bundle spacing. Assume that the

line is uniformly transposed and the solid conductors have a 2cm

radius. If the line length is 175 mi find the charging current per

mile, charging current for the entire length and total 3 chargingreactive power.


10 m 10 m

0.5m


63/80


64/80

ECE4334Using the ACSR tables


65/80



Inductive reactance


66/80

Inductive reactance


2 2 2 1 0 410

410 1609

, /


Inductive reactance


67/80

Inductive reactance


410 1609

2.02 10

2.02 10

Xa from table inductive

reactance at 1ft spacing

Xd inductive reactance

spacing factor to be

calculated


68/80


Capacitive reactance


69/80

Capacitive reactance


.

.

.

M-mile to neutral

Xa from table

capacitive reactance at 1ftspacing; rbeing the outside

diameter

Xd capacitive reactance

spacing factor to becalculated


70/80

ECE4334Example

Find the inductive reactance per phase in ohms per mile and the


71/80

p p p

capacitive reactance to neutral in ohm-miles for a three phase linethat has three equilaterally spaced conductors of ACSR Dove. The

conductors are 10 ft apart and the operating frequency is 60 Hz.


Dove GMR = 0.0313, D=10ft, r=(0.927/2)1/12=0.0386 ft

XC=.

.

.

0.1648 M-mile to neutral

From the table

Xa= 0.0965 M-mile,Xd =.

.

10= 0.0683 M-mile

XC=Xa+Xd= 0.0965+0.0683 = 0.1648 M-mile


72/80


73/80

ECE4334Additional Transmission topics

Shunt conductance: Usually ignored. A small


74/80

Shunt conductance: Usually ignored. A small

current may flow through contaminants on

insulators.

DC Transmission: Because of the large fixedcost necessary to convert AC to DC and then back

to AC; DC transmission is only practical for

several specialized applications such as long distance overhead power transfer (> 400 miles)

long cable power transfer such as underwater

providing an asynchronous means of joining different

power systems (such as the Eastern and Western grids).



75/80

ECE4334Homework 4.1 (due next week)

Calculate the per phase inductance (per meter) of 765-kV line below.


76/80

Note the flat horizontal spacing of 45 ft between the phases. Assumethat4 (four) conductors per bundle are placed at the corners of a square

(with 18 in on a aside). Use the conductor GMR value in place ofr.


ECE4334Homework 4.2 (due next week

A. Calculate the phase-neutral capacitance per meter of the 765-kV


77/80

line in HW 4.1.B. Calculate the product of inductance and capacitance values found

or the 765-kV line in 4.1 and 4.2A and compare with the value of

00.



78/80


79/80


80/80

Documents

Power Systems Elements Lines