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– Transmission Lines : Steady –State Operation -

성균관대학교

김 철 환

(2)

l Previous lectures have covered

how to calculate the distributed inductance,

capacitance and resistance

of transmission lines.

l In this section we will use

these distributed parameters to develop the transmission line models

used in power system analysis.

© 2012 Cengage Learning Engineering. All Rights Reserved.

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5.1 Medium and Short Line Approximations 5.2 Transmission-Line Differential Equations 5.3 Equivalent π-Circuit

5.4 Lossless Lines

5.5 Maximum Power Flow 5.6 Line Loadability

5.7 Reactive Compensation Techniques

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▣ 단거리 및 중거리 송전선로(short and medium-length transmission-line) - ABCD parameters

▣ the two-port network

▣ 단거리 송전선로 - 80 km

▣ 중거리 송전선로 - 80 ~ 250 km

▣ 장거리 송전선로 - 250 km 이상 :

, s

s I

V sending-end voltage and current :

, R

R I

V receiving-end voltage and current

(5)

▣ The relationship between the sending-end and receiving-end quantities can be written as

Or, in matrix format, where A, B, C, and D are parameters the

depend on the transmission-line constant R, L, C, and G.

▣ Network theory texts [5] show that ABCD parameters apply to linear, passive, bilateral two-port networks, with the following general relation :

R R

S

R R

S

DI CV

I

BI AV

V

+

=

+

=

ú û ù ê ë

é ú û ù ê ë

= é ú û ù ê ë é

R R S

S

I V D

C

B A

I V

1

=

- BC

AD

(6)

▣ The circuit in Figure 5.2 represents a short transmission line, usually applied to overhead 60-Hz lines less than 80km long.

Only the series resistance and reactance are included.(shunt admittance is neglected.)

The circuit applies to either single-phase or completely transposed three-phase lines operating under balanced conditions

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For a completely transposed three-phase line, Z is the series impedance,

are positive-sequence line-to-neutral voltages, and are positive-sequence line currents.

▣ Parameter’s notation is followings.

z = R+jωL Ω/m, series impedance per unit length y = G+jωC S/m, shunt admittance per unit length Z = zl Ω, total series impedance

Y= yl S, total shunt admittance l = line length m

R S and V

V IS and IR

(8)

▣ ABCD parameters for the short line is obtained by KVL, KCL.

▣ By comparing (5.1.7) and (5.1.3), (A,B,C,D) and (1,Z,0,1)

For medium-length lines, typically ranging from 80 to 250 km at 60Hz, it is

common to lump the total shunt capacitance and locate half at each end of the line.

R S

R R

S

I I

ZI V

V

=

+

=

ú û ù ê ë é ú û ù ê ë

= é ú û ù ê ë é

R R S

S

I V Z

I V

1 0 1

0

1

=

=

=

=

\

C

Z B

D

A

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▣ 중거리 송전선로

- lump the total shunt capacitance - a nominal π-circuit

- Figure 5.3.

▣ ABCD parameter of π-circuit (using KVL, KCL)

R R

R R

R

S V Y YZ V ZI

I Z V

V = + + = + ) +

1 2 ( 2 )

(

R R

R R

R R

S R

R S

YZ I YZ V

Y

ZI Y YZ V

Y I V

Y V Y I V

I

2 ) 1 ( 4 )

1 (

] 2 2 )

1 2 [(

2 2

+ + +

=

+ +

+ +

= +

+

=

(Writing a KVL equation)

(Writing a KCL equation)

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Writing these equations in matrix format,

Thus, comparing them

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(12)

▣ ABCD parameters of common networks in Figure 5.4

Including a series impedance

network that approximates a short line and a π circuit that

approximates a medium-length line.

A medium-length line could also be approximated by the T circuit shown in Figure 5.4, lumping half of the series impedance at each end of the line.

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▣ 전압변동(Voltage regulation )

The change in voltage at the receiving end of the line when the load varies from no-load to a specified full load at a specified power factor, while the sending-end voltage is held constant.

▣ Phasor diagram for short transmission line

| 100

|

|

|

|

| - ´

=

RFL RFL RNL

V V VR V

percent VRNL = no-load receiving-end voltage (=Vs/A) VRFL = full-load receiving-end voltage

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▣ 송전능력(line loadability) : 선로가 얼마만큼의 전력을 송전할 수 있는가 하는 능력

▣ 3가지 line-loading limits

열적 한계(Thermal limit)- 도체의 최대 허용온도(단거리 선로)

전압강하 한계(Voltage-drop limit) - (중거리 선로)

정상상태 안정도 한계(Steady-state stability limit) – 동기 발전기의 동기 유지 능력 (장거리 선로)

▣ Example 5.1

Q) A three-phase, 60-Hz, completely transposed 345-kV, 200-km line has two 795,000-cmil 26/2 ACSR conductors per bundle and the following positive- sequence line constants :

z = 0.032+j0.35 Ω/km y = j4.2 [s/km]

Full load at the receiving end of the line is 700MW at 0.99 p.f. leading and at 95% of rated voltage. Assuming a medium-length line, determine the

following :

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a. ABCD parameters of the nominal π circuit

b. Sending-end voltage , current , and real power c. Percent voltage regulation

d. Thermal limit, based on the approximate current-carrying capacity listed in Table A.4

e. Transmission-line efficiency at full load

Sol) a. The total series impedance and shunt admittance values are

Vs

I

s Ps

S C

Z B

pu D

A

S j

yl Y

j j

zl Z

° Ð

´

=

° Ð

+

° Ð

´

=

W

° Ð

=

=

° Ð

=

° Ð

° Ð

´ +

=

=

° Ð

´

=

´

=

=

W

° Ð

= +

= +

=

=

- -

-

- -

08 . 90 10

277 . 8 ) 78 . 174 01476

. 0 1 )(

90 10

4 . 8 (

78 . 84 29 . 70

159 . 0 9706 .

0 ) 5 . 0 )(

78 . 84 29 . 70 )(

90 10

4 . 8 ( 1

90 10

4 . 8 ) 200 )(

10 2 . 4 (

78 . 84 29 . 70 70

4 . 6 ) 200 )(

35 . 0 032 . 0 (

4 4

4

4 6

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b. The receiving-end voltage and current quantities are

° Ð

= +

=

° Ð

´ =

= Ð

° Ð

=

° Ð

=

=

=

-

14 . 26 6

. 199 )

)(

( ) )(

(

11 . 8 246 . ) 1

99 . 0 )(

345 95

. 0 )(

3 (

99 . 0 cos 700

0 2 . 189 3 0

8 . 327

8 . 327 )

345 )(

95 . 0 (

1

R R

S R R R

I Z V

A V

I V

V kVLL

kVLN

kA

kVLN

현재 이 이미지를 표시할 수 없습니다 .

(17)

c. The no-load receiving-end voltage is

MW P

I

pu kV

V

S S

LL S

5 . 730

) 5 . 15 14

. 26 cos(

) 241 . 1 )(

8 . 345 )(

3 (

5 . 15 241 . 1

) 11 . 8 246 . 1 )(

159 . 0 9706 .

0 ( ) 0 2 . 189 )(

08 . 90 10

277 . 8 (

00 . 1 8

. 345 3

6 . 199

4

=

° -

°

=

° Ð

=

° Ð

° Ð

+

° Ð

° Ð

´

=

»

=

=

-

% 7 . 8

% 8 100

. 327

8 . 327 3

. 356

3 . 356

= - ´

=

=

=

VR percent

A kV

VRNL VS LL

(18)

d. From Table A.4, the approximate current-carrying capacity of two 795,000- cmil 26/2 ACSR conductors is

e. The full-load line losses are

and the full-load transmission efficiency is kA 8 . 1 9 . 0 2´ =

MW P

PS - R =730.5-700 =30.5

% 8 . 95 100 =

´

=

S R

P EFF P

percent

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