Power System Calculations in D sharp

8/12/2019 Power System Calculations in D sharp

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1

Formulary Power System Calculations

version: 0.1

Dirk Van Hertem

[email protected]

I. INTRODUCTION

A. Three phase quantities

In a balanced three phase power system:

|U a 0| = |U b 0| = |U c 0| and ∠(U a 0,U b 0) = ∠(U b 0,U c 0) =∠(U c 0,U a 0).

The line voltages are the sum of the phase voltages:U ab =U an +U nb =U an −U bn .

a = e ·2·π

3 (1)

U bn = a 2 ·U an (2)

U cn = a ·U an (3)

U ab =U an − a 2 ·U an = (1− a 2) ·U an (4)

P = 3 · |U L | · |I L | · cosφp (5)

Q =

3 · |U L | · |I L | · sinφp (6)

S =

3 · |U L | · |I L | =

P 2 +Q 2 (7)

Star-delta transformations:

Delta to star (8)

Z a =Z ab · Z ac

Z ab + Z ac + Z bc (9)

Star to Delta (10)

Z ab =Z a · Z b + Z a · Z c + Z b · Z c

Z c (11)

in balanced systems: (12)

Z ∆ = 3 · Z Y (13)

B. Per Unit

per unit value=

real value

base or reference value(14)

C. Symmetrical components

Transformation from symmetrical components to

phasor quantities:

U a

U b

U c

=

1 1 1

1 a 2 a

1 a a 2

·

U 0U 1U 2

(15)

U p = A ·U s (16)

Inverse:

U s = A −1 ·U p (17)U 0

U 1U 2

= 1

3

1 1 1

1 a a 2

1 a 2 a

·U a

U b

U c

(18)

II. LINE PARAMETERS

A. Voltage profile in a line

U G

I G

=

cosh(γ · (L )) Z 0 ·sinh(γ · (L ))sinh(γ·(L ))

Z 0cosh(γ · (L ))

·

U C

I C

(19)

u (x , t ) = ℜ

U (x ) ·

2 ·e ·ω·t

(20)

= |U 1| ·

2 ·e α·(L −x ) ·cos

ω · t +β · (L − x )+arg |U 1|(21)

+|U 2| ·

2 ·e −α·(L −x ) ·cosω · t −β · (L − x )+arg |U 2|

(22)

|U (x )|2 = |U 1|2 +|U 2|2 ·cosh (2 ·α · (L − x )) (23)

+|U 1|2 −|U 2|2

· sinh(2 ·α · (L − x )) (24)

+ (2 · |U 1| · |U 2|) ·cos2 ·β · (L − x )−δ

(25)

|Z 0 · I (x )|2 =|U 1|2 +|U 2|2

·cosh (2 ·α · (L − x )) (26)

+|U 1|2 −|U 2|2

· sinh(2 ·α · (L − x )) (27)

− (2 · |U 1| · |U 2|) ·cos2 ·β · (L − x )−δ

(28)



2

For a lossless line:

Z 0 =

c (29)

γ= ·ω · ·c (30)

v wave =

1

· c (31)

B. Resistance of transmission lines

r = ρ

S (32)

R ac

R dc

= 1+ 1

194·α4 (33)

α=D

2 · ω ·µ

ρ (34)

C. Capacitance value of two parallel conductors

Capacitance value of two conductors at distance D

and of radius R 1 and R 2:

U 12 = V 1 −V 2 = q

2 ·π · ·

ln

D

R 1

+ ln

D

R 2

(35)

=q

2 ·π · · ln D 2

R 1 · R 2

(36)

C = q

U = 2 ·π ·

ln

D 2

R 1·R 2

(37)

0 = 8.8537 ·10−12 F /m (38)

In general, the potential at the surface of a conductor

i =

V i =m

j =1

q j

2 ·π · · ln 1

d j i

(39)

D. Inductance of a line

λi nt =r

0

µ · I · x 3

2 ·π · r 4 ·d x = µ · I

8 ·π (40)

Φu =D

R 1

d Φu =D

R 1

µu ·I

2 ·π · x ·d x (41)

=µu

2 ·π ·I

·ln

D

R 1(42)

Inductance of two conductors in parallel:

(1)u = µu

2 ·π · ln D

R 1(43)

(2)u = µu

2 ·π · ln D

R 2(44)

and the internal resistance against changing flux

(45)

i =µi

8 ·π (46)

= 2 ·i +(1)u +(2)

u (47)

= µi

4 ·π + µu

2 ·π · ln

D 2

R 1 ·R 2

[H /m ] (48)

= 1

2 ·π ·µu · ln D 2

R eq 1 · R

eq 2

(49)

R eq = R ·e − µi

4·µu (50)

= 0.7788 ·R (in air) (51)

Three phase lines:

Φ= L · I (52)

L k j =1

2 ·π ·µ · ln 1

d j k (53)

U = K ·q (54)

K k j =1

2

·π

·

· ln 1

d j k (55)

E. Cables with common screen

c 0 = 2 ·π ·ln

27·R 6−a 6

27·R 3·a 2·r

(56)

c 1 = c 2 = 2 ·π ·

ln

a ·(3·R 2−a 2)

32

r ·

27·R 6−a 6

(57)

With a the distance between two phases, R the radius

of the screen and r the radius of the conductors.

F. Power and voltage drop in a line

Power flow in a purely inductive line:

P S =| U S | · | U R |

X · sin(δ) (58)

Q S

=| U S | · | U R |

X ·cos(δ)

−| U S |2

X (59)



3

Voltage drop in a line with series impedance Z :

∆U =Z ·I

= R + · X

· I · cosφR − · I · cosφR −R · I ·sinφR

(60)

∆|U | ≈ R · I ·cosφR + X · I · sinφR (61)

III. SHORT CIRCUIT CALCULATIONS

Largest fault current:

I RM S (τ) =κ(τ) · I ac (62)

κ(τ) = 1+2 ·e −

4πτ X /R (63)

κ≈ 1.02+0.98 ·e −3· R X (64)

AC and DC short circuit current:

I ac (t ) =

2 ·

I k − I k

· e

− t τ +

I k − I k

·e

− t τ + I k

·sin

ω · t +α−φ

(65)

I dc (t ) =

2 ·E

Z ·sin(α−φ) · e

− t τdc (66)

Switching current:

I b (t ) =

I 2ac (t )+ i 2dc

(t ) (67)

I ac =µ · I k (68)

i dc (t ) =

2 · I k ·e −ω·R X

·t (69)

Force between two conductors:

F = µ · i a · i b

2 ·π · d [N /m ] (70)

Thermal effect:

C ·S ·∆T = R · I 2eq ·∆ (71)

I eq

= 1

∆t ∆t

0

i (t )d t (72)

A. From the exercise session

Z source =U 2

nominal

S 2short

−circuit

−power

(73)

Z transf ormer =u k ·U 2

rated

100% · S r a t e d (74)

R transf ormer =u r ·U 2

rated

100% · S r a t e d (75)

X generator ,1 =

x d

·U 2rated

100% · S r a t e d (76)

X generator ,1 =

x d

·U 2rated

100% · S r a t e d (77)

X generator ,1=

x d

·U 2rated

100% · S r a t e d

(78)

X generator ,2 ≈ X

generator ,2 ≈ X generator ,2 ≈ X

generator ,1

(79)

Z motor =U 2

rated I s t a r t

I nominal ·S r a t e d

=I nominal ·U 2

rated

I s t a r t ·

P rated η·cos(φ)

(80)

IV. POWER FLOW

Admittance matrix:

I m

I n

=

y mn − y mn

− y mn y mn

·

U m

U n

(81)

Y bu s ,i j = − y i j (82)

Y bu s ,i i =n

j =1

y i j (83)

Y bu s = A T ·C · A (84)

Power flow equations:

I i =n

j =1

Y i j ·U j ∀i ∈N≤ n (85)

S ∗i = U ∗i · I i (86)

= P i − Q i (87)

=U ∗i ·n

j =1

Y i j ·U j (88)

Power injections in a node in rectangular and polar



4

coordinates:

P i =U i

n j =1

U j ·G i j ·cos(θi −θ j )+B i j ·sin(θi −θ j )

(89)

Q i =U i

n

j =

1

U j · G i j ·sin(θi −θ j )− B i j · cos(θi −θ j )(90)

P i =U i

n j =1

U j ·Y i j · cos(θi −θ j −φi j ) (91)

Q i =U i

n j =1

U j ·Y i j · sin(θi −θ j −φi j ) (92)

V. OPTIMIZATION IN POWER SYSTEMS

Lagrangian function:

L ( x ,λ,µ) = f ( x )+µT

· (h ( x ))+λT

· g ( x )

(93)∂L ( x ,λ,µ)

∂x i = 0 ∀i (94)

∂L ( x ,λ,µ)

∂λi = 0 ∀i (95)

∂L ( x ,λ,µ)

∂µi = 0 ∀i (96)

VI . POWER SYSTEM STABILITY

dδi

dt =ω (97)

J · dωi

dt = T mech −T elec (98)

J · dωi

dt = P mech −P elec

ωi (99)

H = J ·ω2s

2 ·S rated (100)

2 · H

ωs · dωi

dt ≈ P mech −P elec (101)

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Power System Calculations in D sharp