Electric Power System Appendix

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APPENDIX COMPLEX, REAL AND REACTIVE POWER 3 = 3 + 3 for any balanced or unbalanced load. 3 = 3 + 3 =3 ¡ 1 + 1 ¢ for a balanced load and source. 3 = 3 | || | ( ) valid only for a balanced three-phase source or load. One denes as the voltage angle and as the current angle. 3 =3 | || | ( ) valid only for a balanced three-phase source or load. One denes as the voltage angle and as the current angle. || = p 2 + 2 This is true for only sinusoidal steady-state systems. = | = 2 + 2 = cos ( ) = | | 2 = | = p 1 2 = | | 2 POWER FACTOR CORRECTION = r 1 2 2 q 1 2 2 ¸ BALANCED SET OF THREE-PHASE VOLTAGES AND THE A-B-C PHASE SEQUENCE 0 120 cn V V an V V 0 120 bn V 0 0 0 3 30 3 90 3 150 ab an bc bn ca cn V V V V V V phase sequence and a balanced set of voltages 1

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A brief Electric Power System Appendix, contains a lot of useful formulas

Transcript of Electric Power System Appendix

APPENDIXCOMPLEX, REAL AND REACTIVE POWER

3 = 3 + 3 for any balanced or unbalanced load.

3 = 3 + 3 = 3¡ 1 + 1

¢for a balanced load and source.

3 =√3 |−| ||∠ ( − ) valid only for a balanced three-phase source or load. One

defines as the voltage angle and as the current angle.

3 = 3 |− | ||∠ ( − ) valid only for a balanced three-phase source or load. One

defines as the voltage angle and as the current angle.

|| =p 2 +2 This is true for only sinusoidal steady-state systems.

= || × = √ 2+2

= cos ( − )

= ||2 = || × =

p1− 2

= ||2

POWER FACTOR CORRECTION

=

∙r1−2

2

−q

1−22

¸BALANCED SET OF THREE-PHASE VOLTAGES AND THE A-B-C PHASE

SEQUENCE

0120cnV V anV V

0120bnV

0

0

0

3 30

3 90

3 150

ab an

bc bn

ca cn

V V

V V

V V

− − phase sequence and a balanced set of voltages

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BASE, PER-UNIT AND PERCENT FORMULAS FOR A BALANCED

THREE-PHASE SYSTEM

=()

2

where is line-to-line voltage and is total three-phase

complex power.

=

% = 100× where is based upon an equivalent wye system.

=×1000√

3where is the three-phase base and

is the line-to-line base .

= ×

CHANGE OF BASE PER-UNIT OR PERCENT FORMULA FOR A

BALANCED THREE-PHASE SYSTEM

=

׳

´×³

´2where is line-to-line voltage and is total

three-phase complex power.

% = % ׳

´×³

´2ACTUAL IMPEDANCE OF THE LOAD

|| = ()2

where is line-to-line voltage across the load and

is the total three-phase complex power. This formula is valid only

for a balanced three-phase system.

|| =

2

where −

is line-to-neutral voltage across each phase of the

load and

is load for any particular phase. One

uses this for an unbalanced load.

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MILLMAN’S THEOREM

1 1V 2 2V 3 3V 4 4V

1Z 2Z 3Z 4Z

V

Z

N NV

NZ

e∠ =1∠1

³11

´+ 2∠2

³12

´+ 3∠3

³13

´+ 4∠4

³14

´+ ∠

³1

´11+ 1

2+ 1

3+ 1

4+ 1

e =1

11+ 1

2+ 1

3+ 1

4+ 1

DELTA-WYE TRANSFORMATION

1 =

+ + 2 =

+ + 3 =

+ +

=12 + 23 + 31

2 =

12 + 23 + 31

3 =

12 + 23 + 31

1

AZ

BZ

CZ

1Z 2Z

3Z

a b

c

c

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THE GENERAL TRANSMISSION LINE EQUATIONS

The exact transmission line equations are given by:

() = cosh () + sinh () (1)

() = cosh () +

0sinh () (2)

= cosh () + sinh () (3)

= cosh () +

0sinh () (4)

= cosh ()− sinh () (5)

= cosh ()−

0sinh () (6)

where = − . Distance is defined as the distance measured from the receiving end of

the transmission line to the sending end of the transmission line. So when = 0, (0) =

and (0) = . Likewise, when = , () = and () = . One can also rewrite the

characteristic impedance as

=

s(+ )

(+ ) (7)

=

r

(8)

where

= = (+ ) ≡ Ω

and ≡

(9)

= = (+ ) ≡ Ω−1

and ≡

(10)

One can also write the constant as

=p(+ ) (+ ) =

√ (11)

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Substitute (8) and (11) into (3) through (6) to get

= cosh³√

´+

r

sinh

³√

´(12)

= cosh³√

´+

sinh³√

´

(13)

= cosh³√

´+

q

sinh³√

´

(14)

= cosh³√

´+

sinh³√

´

(15)

= cosh³√

´−

r

sinh

³√

´(16)

= cosh³√

´−

sinh³√

´

(17)

= cosh³√

´− q

sinh³√

´

(18)

= cosh³√

´−

sinh³√

´

(19)

The ABCD parameters of a long transmission line are given by:

= = cosh³√

´

= sinh

³√

´√

= sinh

³√

´√

For finding the voltage and current anywhere on the line one can rewrite (1) as

() = cosh () + sinh () (20)

= cosh

µ√

¶+

µ

¶ sinh³√

´√

(21)

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Likewise, (2) can be written as

() = cosh () +

0sinh () (22)

= cosh

µ√

¶+

µ

¶ sinh³√

´√

(23)

Therefore, (21) and (23) can be used to compute the voltage and current anywhere on the line

in the range 0 ≤ ≤ .

The short-line transmission line equations are given by

⎡⎣

⎤⎦ =⎡⎣ 1

0 1

⎤⎦⎡⎣

⎤⎦⎡⎣

⎤⎦ =⎡⎣ 1 −0 1

⎤⎦⎡⎣

⎤⎦and the equivalent circuit is given by

LOADsV

rIsI

rV

R j L

and = + .

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The ABCD parameters of a short transmission line are given by:

= = 1

=

= 0

The medium-line transmission line equations for the nominal T-Network are:

⎡⎣

⎤⎦ =⎡⎣ 1 +

2¡1 +

4

¢ 1 +

2

⎤⎦⎡⎣

⎤⎦⎡⎣

⎤⎦ =⎡⎣ 1 +

2− ¡1 +

4

¢− 1 +

2

⎤⎦⎡⎣

⎤⎦and the equivalent circuit is given by

LOADsV

rIsI

rV

2

R

2

j L

Y

2

j L2

R

2

Z2

Z

where = + and = .

The ABCD parameters of a medium transmission line for the nominal Network are given

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by:

= = 1 +

2

=

µ1 +

4

¶ =

The medium-line transmission line equations for the nominal Π-Network are:

⎡⎣

⎤⎦ =⎡⎣ 1 +

2

¡1 +

4

¢1 +

2

⎤⎦⎡⎣

⎤⎦⎡⎣

⎤⎦ =⎡⎣ 1 +

2−

− ¡1 + 4

¢1 +

2

⎤⎦⎡⎣

⎤⎦

LOADsV

rIsI

rV

R j L

2

Y

2

Y

Z

where = + and = .

The ABCD parameters of a medium transmission line for the nominal Π Network are given

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by:

= = 1 +

2

=

=

µ1 +

4

The relationship which must exist among the ABCD parameters regardless of whether the

line is long, medium or short is the following:

− = 1

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NONSINUSOIDAL CURRENTS AND VOLTAGES

Displacement Factor

= cos ( − )

True Power Factor

=

00 +

X=1

cos ( − )vuut X=0

2

vuut X=0

2

Apparent Power

=

vuut X=0

2

vuut X=0

2

Real power

= 00 +

X=1

cos ( − )

Budeanu Reactive Power

=

X=1

sin ( − )

Budeanu Distortion Power

=

q2 − ( 2 +2)

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SYMMETRICAL COMPONENTS

Let , , and be a set of unbalanced three-phase voltages as shown in Figure A.

Positive-Sequence Voltages

Negative-Sequence Voltages Zero-Sequence Voltages

anVbnV

cnV

1anV

1bnV

1cnV

2anV

2bnV

2cnV

0anV

0bnV

0cnVn

Figure A

= 0 + 1 + 2

= 0 + 21 + 2

= 0 + 1 + 22

0 = + +

3

1 = + + 2

3

2 = + 2 +

3

0 = 0 = 0

|1| = |1| = |1|

|2| = |2| = |2|

where = 1∠1200.

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