powersystem 02 power

Spring 2014

2.

( )

PowerSystemNotation (one-line diagram) .

Smartgrid Lab at Kookmin University

59.7 kV

17.6 MW28.8 MVR

40.0 kV

16.0 MW16.0 MVR

17.6 MW 16.0 MW-16.0 MVR 28.8 MVR

1

59.7 kV

17.6 MW28.8 MVR

40.0 kV

16.0 MW16.0 MVR

17.6 MW 16.0 MW-16.0 MVR 28.8 MVR

: , / : : : :

: , , : ,


: , / : : : :

: , , : ,

2

.

. (Stator) (Rotor) .


.

. (Stator) (Rotor) .

3

4 1800MVA

3

(Stator) (Rotor) (Field winding) ,

(Magnetic Flux) (Armature winding)

3 .


3 (Stator) (Rotor) (Field winding) ,

(Magnetic Flux) (Armature winding)

3 .

4

Faradays Law:

dtdv -=

!


ar SjX

aI

5

gV

( )vgg tcosV(t)v += (DC) . .

(Phasor) (phasor) .

v(t) = Vmax cos(wt + qv)i(t) = Imax cos(wt + qI)


Root Mean Square (RMS) :

(phasor) .v(t) = Vmax cos(wt + qv)i(t) = Imax cos(wt + qI)

6

Root Mean Square (RMS) :

2 max

0

1 ( )2

T Vv t dtT

=

Note) , RMS ) 380V, 22.9kV, 154kV, 345kV, 765kV

PhasorRepresentationSmartgrid Lab at Kookmin University

j

( )

Euler's Identity: e cos sin

Phasor notation is developed by rewriting using Euler's identity

( ) 2 cos( )

( ) 2 Re

(Note: is the RMS voltage)

V

Vj t

j

v t V t

v t V e

V

q

w q

q q

w q+

= +

= + =

7

j

( )

Euler's Identity: e cos sin

Phasor notation is developed by rewriting using Euler's identity

( ) 2 cos( )

( ) 2 Re

(Note: is the RMS voltage)

V

Vj t

j

v t V t

v t V e

V

q

w q

q q

w q+

= +

= + =

PhasorRepresentationSmartgrid Lab at Kookmin University

The RMS, cosine-referenced voltage phasor is:

( ) Re 2cos sincos sin

V

V

jV

jj t

V V

I I

V V e V

v t Ve eV V j VI I j I

q

qw

q

q qq q

= =

== += +

8

(Note: Bar .)

The RMS, cosine-referenced voltage phasor is:

( ) Re 2cos sincos sin

V

V

jV

jj t

V V

I I

V V e V

v t Ve eV V j VI I j I

q

qw

q

q qq q

= =

== += +

PhasorAnalysis

0

2 2

Resistor ( ) ( )( )Inductor ( )

1 1Capacitor ( ) (0)C

Z = ImpedanceR = ResistanceX = Reactance

XZ = =arctan( )

t

v t Ri t V RIdi tv t L V j LIdt

i t dt v V Ij C

R jX Z

R XR

w

wf

f

= =

= =

+ =

= + =

+

Device Time Analysis Phasor

Note) Z () !!

0

2 2

Resistor ( ) ( )( )Inductor ( )

1 1Capacitor ( ) (0)C

Z = ImpedanceR = ResistanceX = Reactance

XZ = =arctan( )

t

v t Ri t V RIdi tv t L V j LIdt

i t dt v V Ij C

R jX Z

R XR

w

wf

f

= =

= =

+ =

= + =

+

Device Time Analysis Phasor

)RLCircuitExample

V(t)V(t)

Instantaneous power =

:


Instantaneous power =

:

11


12

RLC :


13

Average power ()

= (Real power, Active power)


max max

0

max max

1( ) [cos( ) cos(2 )]21 ( )

1 cos( )2

cos( )

= =

V I V I

T

avg

V I

V I

V I

p t V I t

P p t dtT

V I

V I

q q w q q

q q

q q

f q q

= - + + +

=

= -

= -

-

Power Factor

Average

P

Angle

ower

14

max max

0

max max

1( ) [cos( ) cos(2 )]21 ( )

1 cos( )2

cos( )

= =

V I V I

T

avg

V I

V I

V I

p t V I t

P p t dtT

V I

V I

q q w q q

q q

q q

f q q

= - + + +

=

= -

= -

-

Power Factor

Average

P

Angle

ower

Complex power ()


[ ]

*

cos( ) sin( )

P = Real Power (W, kW, MW)Q = Reactive Power (var, kvar, Mvar)S = Complex power (VA, kVA, MVA)Power Factor (pf) = cosIf current leads voltage then pf is leadingIf current

V I V I

V I

S V I jP jQ

q q q q

f

- + -+

=

==

lags voltage then pf is lagging

15

[ ]

*

cos( ) sin( )

P = Real Power (W, kW, MW)Q = Reactive Power (var, kvar, Mvar)S = Complex power (VA, kVA, MVA)Power Factor (pf) = cosIf current leads voltage then pf is leadingIf current

V I V I

V I

S V I jP jQ

q q q q

f

- + -+

=

==

lags voltage then pf is lagging

Note) S !


2

1

Relationships between real, reactive and complex powercos

sin 1

Example: A load draws 100 kW with a leading pf of 0.85.What are (power factor angle), Q and ?

-cos 0.85 31.8100

0.

P S

Q S S pf

S

kWS

f

f

ff -

=

= = -

= = -

= 117.6 kVA85

117.6sin( 31.8 ) 62.0 kVarQ

=

= - = -

16

2

1

Relationships between real, reactive and complex powercos

sin 1

Example: A load draws 100 kW with a leading pf of 0.85.What are (power factor angle), Q and ?

-cos 0.85 31.8100

0.

P S

Q S S pf

S

kWS

f

f

ff -

=

= = -

= = -

= 117.6 kVA85

117.6sin( 31.8 ) 62.0 kVarQ

=

= - = -


17

.

.


.

.

18

. (I2 R )

. : Shunt Capacitor, SVC,

STATCOM, (Synchronous Condenser)

. (I2 R )

. : Shunt Capacitor, SVC,

STATCOM, (Synchronous Condenser)

. ()

. (Short Circuit Capacity)

.

ShuntCapacitor

StaticVar Compensator SVC: Static Var Compensator


22

STATCOM STATCOM: Static Synchronous Compensator

(Inverter) ,


23

? (2013 3 )

. . . /

.

? (2012 4 )


? (2013 3 ) . . . /

.

? (2012 4 )

24

2.1Smartgrid Lab at Kookmin University

25

SignConvention (Sign Convention)

Load convention = Passive sign convention Generator convention


26

() , (),

0. (or = ) 0. (or = ) 0. (or = )


, (), 0. (or = ) 0. (or = ) 0. (or = )

27

/Smartgrid Lab at Kookmin University

2Resistor Resistor

2Inductor Inductor L

2Capacitor Capacitor C

CapaCapacitor

Resistors only consume real power

PInductors only consume reactive power

QCapacitors only generate reactive power

1Q

Q

C

I R

I X

I X XC

V

w

=

=

= - =

= -2

citorC

C(Note-some define X negative)

X

28

2Resistor Resistor

2Inductor Inductor L

2Capacitor Capacitor C

CapaCapacitor

Resistors only consume real power

PInductors only consume reactive power

QCapacitors only generate reactive power

1Q

Q

C

I R

I X

I X XC

V

w

=

=

= - =

= -2

citorC

C(Note-some define X negative)

X

Note) Capacitor Inductor .


29

DistributionSystemCapacitors

3 l 3

3 120 .

l 3 l l 3 2 .

Wye (Y) Delta (D)

l 3 3 120 .

l 3 l l 3 2 .

Wye (Y) Delta (D)

Y(Wye) 3 Y Y

3 4(3-phase 4-wire system): Y 3 3(3-phase 3-wire system): 3


32

Y (phase-to-neutral voltages)

(line-to-line voltages)

3 Y _________ . 3 Y __________ . 3 Y ____

_____ .


(phase-to-neutral voltages)

(line-to-line voltages)

3 Y _________ . 3 Y __________ . 3 Y ____

_____ .

33

Note) ) 380V, 22.9kV, 154kV, 345kV, 765kV

Y

3 Y


3 Y

34

D

(phase current) (line current)


35

3 Assume a D-connected load is supplied from a 3f13.8 kV (L-L) source with Z = 10020W. (1) Find the phase currents.(2) Find the line currents.(3) Find the complex power absorbed by the loads.(4) Find the power factor.

Assume a D-connected load is supplied from a 3f13.8 kV (L-L) source with Z = 10020W. (1) Find the phase currents.(2) Find the line currents.(3) Find the complex power absorbed by the loads.(4) Find the power factor.

Delta-WyeSmartgrid Lab at Kookmin University

37


38

Per-phase Analysis

3 . Y . . 2 120 .


Per-phase Analysis 3 . Y . . 2 120 .

39

PerPhaseAnalysisProcedure To do per phase analysis

1. Convert all D load/sources to equivalent Ys2. Solve phase a independent of the other phases3. Total system power S = 3 Va Ia*4. If desired, phase b and c values can be determined by

inspection (i.e., 120 degree phase shifts)5. If necessary, go back to original circuit to determine line-line

values or internal D values.


To do per phase analysis1. Convert all D load/sources to equivalent Ys2. Solve phase a independent of the other phases3. Total system power S = 3 Va Ia*4. If desired, phase b and c values can be determined by

inspection (i.e., 120 degree phase shifts)5. If necessary, go back to original circuit to determine line-line

values or internal D values.

40

3 3


3

41

3 Y


bd == LaLNan IIVV ,

42

Y Y

:


bd == DIIVV abLLab ,

43

2.5 3


44

3

3 . ( ) 3 3: 3 4: .

.

3 (Constant) 1 2

3 2 .

3 .


3 . ( ) 3 3: 3 4: .

.

3 (Constant) 1 2

3 2 .

3 .

45

3 : 2 .

powersystem 02 power

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