전력시스템 해석 및 설계

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전력시스템 해석 및 설계

제 3 장

– Power Transformer -

성균관대학교 김 철 환

CENTER FOR POWER IT

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Center for Power IT CENTER FOR POWER IT

전력IT인력양성센터

CONTENTS

3.1 이상 변압기 (THE IDEAL TRANSFORMERS)

3.2 실제 변압기의 등가회로(EQUIVALENT CIRCUIT FOR PRACTICAL TRANSFORMERS) 3.3 단위 법(THE PER-UNIT SYSTEM)

3.4 3상 변압기 결선 및 위상 변위(THREE PHASE TRANSFORMER CONNECTION AND PHASE SHIFT)

3.5 평형 3상 2권선 변압기의 단위 법 등가회로(PER-UNIT EQUIVALENT CIRCUITS OF BALANCED THREE-PHASE TWO-WINDING TRANSFORMERS)

3.6 3권선 변압기(THREE-WINDING TRANSFORMERS) 3.7 단권 변압기(AUTOTRANSFORMERS)

3.8 비 공칭 권선비를 갖는 변압기(TRANSFORMERS WITH OFF-NORMINAL TURNS RATIOS)

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3.6 3권선 변압기

▣ 그림 3.20(a) : basic 단상 3권선 변압 기

▣ 2권선 변압기의 ideal transformer relations : (3.1.8) , (3.1.14)

 이상적인 3권선 변압기의 관계식

▣ In actual units, ,

▣ In per-units,

3 3 2 2 1

1I N I N I

N = +

3 3 2

2 1

1

N E N

E N

E = =

. . 3 . . 2 .

.

1pu I pu I pu

I = +

. . 3 .

. 2 .

.

1p u E pu E pu

E = =

3/106

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3.6 THREE-WINDING TRANSFORMERS

▣ 누설 임피던스(Per-unit leakage impedances)

= 권선 1에서 측정한 per-unit 누설 임피던스(권선 2 단락, 권선 3 개방)

* 그림 3-20(c) : 권선 2 단락, 권선 3 개방할 때 권선 1에서 측정한 누설 임피던스(병렬 어드미턴스 분기는 무시)

식 (3.6.5) ~ (3.6.7)을 풀면,

▣ 직렬 임피던스(Per-unit series impedance)

) 7 . 6 . 3 ( ) 5 . 6 . 3 ( ,

, ₁₃ ₁ ₃ ₂₃ ₂ ₃

2 1

12 = Z +Z Z = Z +Z Z = Z +Z −

Z Z12

Z13

Z23

) 8 . 6 . 3 ( ) 2(

1

23 13

12

1 Z Z Z

Z = + −

) 9 . 6 . 3 ( ) 2(

1

13 23 12

2 Z Z Z

Z = + −

) 10 . 6 . 3 ( ) 2(

1

12 23 13

3 Z Z Z

Z = + −

4/106

(5)

▣ EX 3.9)

The ratings of a single-phase three-winding transformer are - winding 1: 300 MVA , 13.8 kV

- winding 2: 300 MVA , 199.2 kV - winding 3: 50 MVA , 199.2 kV

The leakage reactance, from short-circuit tests, are - 0.10 per unit on a 300-MVA , 13.8-kV base - 0.16 per unit on a 50-MVA , 13.8-kV base - 0.14 per unit on a 50-MVA , 199.2-kV base

Winding resistances and exciting current are neglected. Calculate the impedances of the per-unit equivalent circuit using a base of 300MVA and 13.8kV for terminal 1.

12 = X

13 = X

23 = X

5/106

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Sol)

Then, From (3.6.8)-(3.6.10),

unit per 96 . 50 0

) 300 16 . 0

13 (  =



 



=  X

unit per 84 . 50 0

) 300 14 . 0

23 ( =



 



=  X

(⁰^.¹⁰ ⁰^.⁹⁶ ⁰^.⁸⁴) ⁰^.¹¹^per^unit

2 1

1 = + − =

X

(⁰^.¹⁰ ⁰^.⁸⁴ ⁰^.⁹⁶) ⁰^.⁰¹^per^unit

2 1

2 = + − =−

X

(⁰^.⁸⁴ ⁰^.⁹⁶ ⁰^.¹⁰) ⁰^.⁸⁵^per^unit

2 1

3 = + − =

X

6/106

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- The per-unit equivalent circuit of this three-winding transformer is show in Figure 3.21.

- Note that is negative. This illustrates the fact that , and are not leakage reactance, but instead are equivalent reactances derived from the leakage reactances.

Leakage reactances are always positive

- Note also that the node where the three equivalent circuit reactances are connected does not correspond to any physical location within the transformer

X2 X₁ X₂ X₃

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▣ EX 3.10)

Three transformers, each identical to that described in Example 3.9, are connected as a three-phase bank in order to feed power from a 900-MVA , 13.8-kV generator to a 345-kV transmission line and to a 34.5-kv distribution line. The transformer windings are connected as follows:

- 13.8-kV windings(X) : , go generator

- 199.2-kV windings(H) : solidly grounded Y, to 345-kV line - 19.92-kV windings(M) : grounded Y through , to 34.5-kV line

The positive-sequence voltages and currents of the high and medium-voltage Y windings lead the corresponding quantities of the low-voltage winding by

Draw the per-unit network, using a three-phase base of 900 MVA and 13.8 kV for terminal X . Assumed balanced positive-sequence operation.

∆

Ω

= j0.10 Z_n

∆ 30°.

8/106

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Sol)

The per-unit network is shown in Figure 3.22

Since the M and H windings are Y- connected,

- which are the rated line-to-line voltages of the M and H windings.

v k V_baseX =13.8

v k V_baseM = 3(19.92)=34.5

v k V_baseH = 3(199.2)=345

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3.7 단권 변압기

▣ 그림. 3.23(b) : 단권 변압기

: the windings are both electrically and magnetically coupled

▣ smaller per-unit leakage impedance

 smaller series voltage drops (장점)  higher short-circuit currents (단점)

▣ 장점 : lower per-unit losses (고효율), lower exciting current, and lower cost if the turns ratio is not too large.

▣ 단점 : 권선이 전기적으로 연결  transient overvoltages

10/106

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3.7 AUTOTRANSFORMERS

▣ EX 3.11)

The single-phase two-winding 20-kVA , 480/120-volt transformer of Example 3.3 is connected as an autotransformer, as in Figure 3.23(b) , where winding 1 is the 120- volt winding. For this transformer, determine

a. The voltage ratings and of the low and high-voltage terminals

b. The kVA rating

c. The per-unit leakage impedance

EX E_H

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3.7 AUTOTRANSFORMERS

Sol)

a. The voltage ratings and of the low and high-voltage terminals

b. The kVA rating

also,

c. The per-unit leakage impedance

From example 3.3, the leakage impedance is per unit as a normal, two-winding transformer.

EX E_H ,

120 volts

E_X = E_X =E₁ =120volts, E₂ =480volts, E_H = E₁+E₂ =120=480 =600volts

°

∠78.13 0729

. 0

Ω

=

= Ω

=

= 11.52

000 , 20

) 480 , (

4 . 000 14

, 25

) 600

( ² ²

baseHold

baseHnew Z

Z

unit per

Z_p_u_new = ∠ °



 



° 

∠

= 0.05832 78.13

4 . 14

52 . ) 11 13 . 78 0729 . 0

. (

.

, 667 . 41 480 / 000 ,

2 I 20 A

I = _H = = S_H =E_HI_H =(600)(41.667)=25 kVA. A I

I I A I

A I

I_H (41.677) 166.7 , _X 208.3

120 , 480

667 .

41 ₁ ₁ ₂

2 = = = = + =

=

. 25

) 3 . 208 ( ) 120

( kVA

I E

S_X = _X _X = =

12/106

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▣ 변압기의 모델링 : 단위값 사용하는 것이 실제값 사용하는 것보다 간단

▣ 회로상에서 이상 변압기 권선의 제거 : 선택된 기준전압의 비 = 권선의 정격전압 의 비

▣ 일부의 경우, 상기 방법으로 기준전압을 선택하는 것이 불가능  그림 3.24 : 변압기를 병렬로 연결하여 사용할 경우

▣ 변압기의 정격전압이 설정된 기준전압에 비례하지 않는 변압기의 단위법 모델  비 공칭 권선비를 갖는 변압기

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3.8 비 공칭 권선비(OFF-NOMINAL TURNS RATIOS)를 갖는 변압기

, /

/

₂ ₁ ₂ ₁ ₂

1

V V V N N

V

_base _base

=

_rated _rated

=

2 1 2

1

rated rated base

base

V V V

V =

(14)

3.8 TRANSFORMERS WITH OFF-NOMINAL TURNS RATIOS

▣ 그림 3.25(a) : 정격전압 , 를 갖는 변압기

▣ 설정된 기준전압이 다음을 만족하는

것으로 가정

▣ 로 정의하면, 식 (3.8.1)은 다음 식 (3.8.3) 직렬로 연결된 2개의 변압기 (그림 3.25(b) )

2번째 변압기 : 설정된 기준전압 비 b 와 같은 정격의 변압비를 가짐 이 변압기는 그림 3.9 ,3.17과 같은 표준형 단위 모델

rated

V₁ V₂_rated

rated t

rated

a V

V

₁

=

₂

2

1 base

base bV

V =

b a c= _t /

rated rated

t

rated V bcV

b b a

V₁  ₂ = ₂



 



= 

14/106

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▣ The per-unit model : 그림 3.25(c)

▣ An alternative representation : 그림 3.25(c) 에 마디 방 정식 적용

where both and are referenced into their nodes in accordance with the nodal equation method



 



 



 



= 



 





− ₂

1 22

21

12 11

2 1

V V Y

Y

Y Y

I I

I1 −I₂

15/106

(16)

▣ 식 (3.8.4)의 어드미턴스 값은,

eq V eq

Z Y V

Y = I = =

=

1

1 0 1 11

2

eq V eq

Y c c

V Z

Y I ₂ ²

2 0 2

22 /

1

=

− =

=

eq eq

V

V cY Z cV V

Y I − = −

=

= 2

2

2 0 1 12

/

1

eq V

Y V c

I c V

Y I ^*

1 1

*

1 0 2 21

2

−

− =

=

16/106



 



 



 



= 



 





− ₂

1 22

21

12 11

2 1

V V Y

Y

Y Y

I I

(17)

▣ EX 3.12)

A three-phase generator step-up transformer is rated 1000 MVA , 13.8 kV / 345 kV Y with The transformer high-voltage winding has taps. The system base quantities are

Determine the per-unit equivalent circuit for the following tap settings:

a. Rated tap

b. -10% tap (providing a 10% voltage decrease for the high-voltage winding)

Assume balanced positive-sequence operation. Neglect transformer winding resistance, exciting current, and phase shift

∆ .

10 .

0 per unit j

Z_eq = ±10%

MVA S_base₃_φ =500

kV V_baseXLL =13.8

kV V_baseHLL =345

17/106

(18)

Sol)

a. Rated tap

Using (3.8.1) and (3.8.2) with the low- voltage winding denoted winding 1,

From (3.3.11)

The per-unit equivalent circuit, not including winding resistance, exciting current, and shift is shown in Figure

345 1 8 . 04 13

. 345 0

8 .

13 = = = = =

= a c

V b V

a _t

baseHLL baseXLL t

unit per j

j

Z_p_u_new 0.05

1000 ) 500 10 . 0

. (

.  =



 



= 

18/106

(19)

b. -10% tap

Using (3.8.1) and (3.8.2)

From Figure 3.23(d)

The per-unit positive-sequence network is shown in Figure

( )⁰^.⁹ ⁰^.⁰⁴⁴⁴⁴ ¹³³⁴⁵^.⁸ ⁰^.⁰⁴

345 8 .

13 = = =

= b

a_t

unit per j j

cY_eq 22.22

05 . 0 1111 1 .

1  =−



 



= 

1111 . 04 1

. 0 04444 .

0 =

=

= b c a^t

( ^j ) ^j ^per ^unit

Y

c) _eq ( 0.11111) 20 2.222 1

( − = − − =

( j ) j per unit Y

c

c ) _eq (1.2346 1.1) 20 2.469

( ² − = − − =−

19/106

(20)

▣ EX 3.13)

Two buses and are connected by two parallel lines L1 and L2 with positive- sequence series reactance and per unit. A regulating transformer is placed in series with line L1 at bus . Determine the bus admittance matrix when the regulating transformer;

(a) provide a 0.05 per-unit increase in voltage magnitude toward bus (b) advances the phase toward bus .

Assume that the regulating transformer is ideal. Also, the series resistance and shunt admittance of the lines are neglected

c b a ′′ abc

25 .

1 =0

XL X_L₂ =0.20 c

b

a ′′ 2×2

c b a ′′

° 3

20/106

(21)

Sol) This circuit is shown in Figure 3.28.

a. For the voltage-magnitude-regulating transformer,

per unit.

From (3.7.5)-(3.7.8) , the admittance parameters of the regulating transformer in series with line L1 are

0 . 25 4

. 0

1

11 j

Y _L = j = −

628 . 3 )

0 . 4 ( ) 9524 . 0

( ²

1

22 j j

Y _L = − =−

810 . 3 ) 0 . 4 ( ) 9524 . 0

2 (

21 1

12 Y j j

Y _L = _L = − − =

9524 . 0 )

05 . 1 ( ) 1

( +∆ ¹ = ¹ =

= V ⁻ ⁻

c

21/106

(22)

For lone L2 alone,

Combining the above admittance in parallel,

per unit

0 . 20 5

. 0

1

2 22 2

11 j

Y j

Y _L = _L = =−

0 . 5 ) 0 . 5

2 (

21 2

12 Y j j

Y _L = _L = − − =

0 . 9 0

. 5 0 .

2 4

11 1 11

11 Y Y j j j

Y = _L + _L = − − = −

628 . 8 0

. 5 628 .

2 3

22 1 22

22 Y Y j j j

Y = _L + _L = − − =−

2 12 1 12 21

12 Y Y L Y L

Y = = +

810 . 8 0 . 5 810 .

3 j j

j + =

=

22/106

(23)

3.8 TRANSFORMERS WITH OFF-NOMINAL TURNS RATI

OS

23/106

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b. For the phase-angle-regulating transformer, Then, for this regulating transformer in series with line L1,

0 . 25 4

. 0

1

11 j

Y _L = j = −

0 . 4 )

0 . 4 ( 3 0 .

1 ²

1

22 j j

Y _L = ∠− ° − = − .

3 1

1∠− = ∠− °

= α

c

) 0 . 4 ( ) 3 0 . 1

1 (

12 j

Y _L = − ∠− ° −

9945 . 3 2093 . 0 87 0 .

4 ∠ °= + j

=

) 0 . 4 ( ) 3 0 . 1

( ^*

1

12 j

Y _L = − ∠− ° −

9945 . 3 2093 . 0 93

0 .

4 ∠ °= − + j

=

24/106

(25)

The admittance parameters for line L2 alone are given in part (a) above. Combining the admittance in parallel,

per unit

0 . 9 0

. 5 0 .

22 4

11 Y j j j

Y = = − − =−

0 . 5 9945 . 3 2093 .

12 0 j j

Y = + +

9945 . 3 2093 .

0 + j

=

0 . 5 9945 . 3 2093 .

21 0 j j

Y =− + +

9945 . 3 2093 .

0 + j

−

=

25/106

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▣ 3상 전압조정 변압기

A. 3상 전압 크기 조정 변압기

26/106

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27/106

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B. 3상 전압 위상각 조정 변압기

28/106

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Questions

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