REAL & REACTIVE POWER CONCEPTS and

GENERATOR CAPABILITY CURVE

- K. Arputharaju,Assistant Executive Engineer /

Operation,Basin Bridge Gas Turbine Power Station,

TANGEDCO, TNEB Ltd., Chennai – 600 012.

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2

REQUIREMENTS OF POWER SYSTEM

It must supply energy practically everywhere the customer demands. The load demands vary with time. The system must able to supply this ever changing demand. The delivered energy must meet certain minimum requirements in regard to quality. The following factors determine the quality:

a) The system frequency must be kept around 50Hz with a variation of ±0.05Hz.

b) The magnitude of the bus voltages are maintained within prescribed limit around the normal value. Generally the voltage variation should be limited to ±5%.

The energy must be available with high reliability. The energy must be delivered without overloading any element in the power system. The energy must be delivered at minimum cost.

3

REAL POWER (P): The real power, P is defined as the average value of P and therefore, physically, means the useful power being transmitted. Its magnitude depends very strongly on the power factor cosΦ.REACTIVE POWER (Q): The reactive power, Q is by definition equal to the peak value of that power component that travels back & forth on the line, resulting in zero average, and therefore capable of no useful work.

4

TYPE OF LOADS:TYPE OF LOAD PHASOR PHASE

ANGLE

POWER ABSORBED BY THE LOAD

P Q

V

I

R VI Ф = 0° P > 0 Q = 0

V

I

L

V

I

Ф Ф = +90° P = 0 Q > 0

V

IC

V

I

Ф Ф = - 90° P = 0 Q < 0

R

L

R L

I

V

V

ΦV

I 0°<Φ<+90° P > 0 Q > 0

5

TYPE OF LOADS:TYPE OF LOAD PHASOR PHASE

ANGLE

POWER ABSORBED BY THE LOAD

P Q

-90°<Φ<0° P > 0 Q < 0

LI

V -90°<=Φ<=+90° P = 0 Q = 0

R

C

RC

V

V

I

IVΦ

C

Ic IL

Tuned to

Resonance

IL = Ic

PL = Pc

Energy travels

Back & forth

Between C&L

6

TYPE OF LOADS• Inductive load absorbs positive Q. i.e., an inductor consumes reactive power.• Capacitive load absorbs negative Q. i.e., a capacitor generates reactive power.• Sign change in Q simply means a 180° phase shift.• Resistive load consumes real power.• Inductive load consumes positive reactive power• Capacitive load consumes negative reactive power.• Combination of R & L load consumes real & positive reactive power.• Combination of R & C load consumes real & negative reactive power.• Reactive power is bi-directional power. It travels from

source to load as well as load to source.

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FREQUENCY IS RELATED TO REAL POWER ( P – f )

SMALL DROP IN SYSTEM LOAD.

VALVE SETTINGS ARE IGNORANT OF THE LOAD CHANGE.

INPUT TORQUE TO EACH MACHINE REMAINS UNALTERED.

DECREASE IN CURRENT SUPPLIED BY EACH ALTERNATOR.

DECREASE IN ELECTRO-MAGNETIC TORQUE BY EACH ALTERNATOR.

EACH ALTERNATOR EXPERIENCES SURPLES ACCELERATING TORQUE.

SLIGHT INCREASE IN SPEED AND FREQUENCY.

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EFFECT ON OTHER LOADS:

AT HIGHER FREQUENCY, THE REMAINING LOAD ROTATES AT HIGHER SPEED AND TAKES MORE CURRENT.

HENCE THE LOAD DEMAND INCREASES. POWER GENERATION AT HIGHER FREQUENCY EQUALS THE LOAD

DEMAND POWER. TO DECREASE THE FREQUENCY, THE VALVE MUST BE CLOSED

SLIGHTLY.

EXAMPLE: PUMP SET (INDUCTION MOTOR)

At high frequency, the speed of IM increases.

Ns = 120f / P Nr = Ns ( 1 - s )

The current taken by the IM will be more. Hence the demand on the system increases.

ADJUSTINGINPUT VALVES

CONTROLSFREQUENCY

CONTROLSREAL POWER

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VOLTAGE IS RELATED TO REACTIVE POWER ( Q – V )

V1 V21 2

G1

jX

P jQ

I

1. Bus Voltage V1 is kept at constant magnitude.2. Transmission line has reactance only i.e. jX.3. Power flow is P jQ.

Take V1 as reference.V2=V1-jXI -----------------------------------(1)V1 * I = P jQI = (P-jQ) / V1 ------------------------------(2)Substitute (2) in (1)V2 = V1 – jX [(P/V1) – j(Q/V1)]V2 = [V1 – (X/V1)Q] – j(X/V1)P]

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VECTOR DIAGRAMS:

V2 = V1- X Q

V1

- j X P V1

BOTH DROPS EQUAL

V1

V2

DOUBLE P DOUBLE Q

X

V1

P

XV1

QV1 V1

V2

V2

X

V1Q

2 X

V1P

2 X

V1Q

X

V1P

DOUBLE “P ”: VOLTAGE ANGLE WILL CHANGE. NO CHANGE IN MAGNITUDE.

DOUBLE “Q ”: VOLTAGE MAGNITUDE IS VERY MUCH RELATED TO REACTIVE POWER.

MORE “Q ” FLOW WILL AFFECT THE VOLTAGE

EXCITATION MORE

EXCITATION LESS

LAGGING MVAR

LAGGING MVAR

GEN. VOLTAGE

GEN. VOLTAGE

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REACTIVE POWER INJECTION AT LOAD SIDE BY USING SHUNT CAPACITORS, IMPROVES THE VOLTAGE.

UNDER LIGHT LOAD CONDITIONS, RECEIVING END VOLTAGE > SENDING END VOLTAGE (FERRANTI EFFECT) DUE TO CAPACITIVE LOAD. CONNECT SHUNT REACTORS TO CONTROL VOLTAGE.

SYNCHRONOUS CONDENSER IS USED TO ABSORB or TO DELIVER THE REACTIVE POWER.

SYNCHRONOUS MOTOR UNDER NO-LOAD CONDITION IS SYNCHRONOUS CONDENSER.

PEAK LOADCONDITION

LIGHT LOADCONDITION

CONNECTCAPACITORS

CONNECTREACTORS

TO CONTROLVOLTAGE

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SHORT CIRCUIT RATIO ( SCR ):SCR = FIELD CURRENT REQUIRED TO PRODUCE RATED VOLTAGE ON O.C.

FIELD CURRENT REQUIRED TO CIRCULATE RATED CURRENT ON S.C.

SCR = o Fo c Fo c Fo 1 1 1

o Fs b Fs a Fo a Fo / c Fo Per unit voltage on open circuit Xd

Corresponding per unit current on short circuit

= RECIPROCAL OF SYNCHRONOUS REACTANCE

FIELD CURRENT

O.C.C.

S.C.C.

1.0

PE

R U

NIT

VO

LT

AG

E

PE

R U

NIT

CU

RR

EN

T

a b

c

o Fo Fs

A B

C

D

E

AD AE DE

AB AC BC

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TYPICAL S.C.R. VALUES: For 500 MW T.G., SCR= 0.48 For 210 MW T.G., SCR= 0.49 For 110 MW T.G., SCR= 0.50 For 60 MW T.G., SCR= 0.59 The SCR value may have to be raised to 1.0 to 1.5, if the loading is likely to be

capacitive i.e. leading MVAR supply. For modern Turbo-alternator, the SCR is normally between 0.48 to 0.7

EFFECT OF S.C.R. ON MACHINE PERFORMANCE: Higher value of SCR has higher stability limit. Better voltage regulation for high SCR. High value of SCR has a long air gap which means that the mmf required by

field is large. Hence machine with higher SCR is costlier to build.

AIR GAP SCR WEIGHT SIZETRANSPORTATION

PROBLEM

Present trend is to build low value of SCR since fast acting excitation system available.

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POWER DIAGRAM (CAPABILITY DIAGRAM):

• ASSUMPTION: I.R. drop is negligible. CASE-I: In Δ ABC, BC=E Sinδ In Δ BCD, BC=IXd

CosФ E Sinδ = IXd CosФ

Multiply both sides by V Xd EV Sinδ = VI CosФ = REAL Xd POWER

At δ=90°, We get the maximum power i.e. the theoritical stability line.• CASE-I I: In Δ ABC, CD=AC – AD; In Δ BCD, CD=IXd SinФ

In Δ ABC, AC=E Cosδ & AD = VIXd SinФ = E Cosδ - V ; Multiply both sides by V , We get

Xd EV Cos δ – V2 = VI Sin Ф = REACTIVE POWER Xd Xd

E

V

IXd MW

MVARIΦ

δ ΦA

B

CD

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CAPABILITY DIAGRAM OF A 110 MW ALTERNATOR

• I) COLLECT THE INFORMATIONS FROM T.G. NAME PLATE / MANUAL:1. Terminal Voltage : 11,000 V2. Rated MVA : 137.53. Rated p.f. (cos Ф) : 0.8 Lagging4. Rated Armature Current : 7220 A5. Rated Field Current : 1500 A6. Short Circuit Ratio : 0.5

• II) CALCULATED VALUES:1. MW = MVA X p.f. = 137.5 X 0.8 = 110 MW2. MVAR = MVA X SCR = 137.5 X 0.5 = 68.75 MVAR (Max. permissible zero

p.f. leading MVAR)3. Ф = cos-1(0.8) = 36.87°4. To ensure operational safety, there should be a margin of at least 12.5 % (given by the manufacturer) of the power rating of the generator between the working point & the theoretical stability (load angle ‘δ’) limit line. The operational limit of a generator rated at 0.8 p.f. lagging can be tabulated below:p.u. MW 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

p.u. MW + Margin 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

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p.u. MW

Lagging p.f.Leading p.f.

Unity p.f.

0.10.2 0.3 0.4 0.50.60.70.80.91.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

A

B

C

D

O

TOTAL ROTOR (O

R) FIELD C

URRENT

TOTA

L S

TATO

R (

OR

) AR

MAT

URE

CU

RR

ENT

TH

EO

RIT

ICA

L S

TA

BIL

ITY

LI

MIT

LIN

E

0.1

0.2

0.3

0.4

0.5

0.6

0.7

TURBINE LIMIT LINE

SCR MVA X SCRMAXIMUM PERMISSIBLE

MVAR IN ZERO p.f. LEADING.

E F

G

STA

TOR

CU

RR

EN

T LIMIT

PR

AC

TIC

AL

ST

AB

ILIT

Y L

IMIT

WIT

H 1

2.5%

MA

RG

IN (δ=63°)

δ=90°

0.8

0.9

H

RO

TO

R C

UR

RE

NT

LIMIT

Ф=

36.8

7°

P.F.= 0.8 LAGGING

REAL POWER

VAR EXPORTVAR IMPORT

REACTIVE POWERp.u. MVAR (lagging)

REACTIVE POWERp.u. MVAR (leading)

CAPABILITY DIAGRAM OF A 110 MW ALTERNATOR

OE : No-load Field Current

OD : Field Current required for Armature Reaction

FGDHF : Capability Diagram of the 110 MW Alternator

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p.u. MW

Lagging p.f.Leading p.f.

Unity p.f.

0.10.2 0.3 0.4 0.50.60.70.80.91.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

A

B

C

D

O

TOTAL ROTOR (O

R) FIELD C

URRENT

TOTA

L S

TATO

R (

OR

) AR

MAT

UR

E C

UR

REN

T

TH

EO

RIT

ICA

L S

TA

BIL

ITY

LI

MIT

LIN

E

0.1

0.2

0.3

0.4

0.5

0.6

0.7

TURBINE LIMIT LINE

SCR MVA X SCRMAXIMUM PERMISSIBLE

MVAR IN ZERO p.f. LEADING.

E F

G

STA

TOR

CU

RR

EN

T LIMIT

PR

AC

TIC

AL

ST

AB

ILIT

Y L

IMIT

WIT

H 1

2.5%

MA

RG

IN (δ=63°)

δ=90°

0.8

0.9

H

RO

TO

R C

UR

RE

NT

LIMIT

Ф=

36.8

7°

P.F.= 0.8 LAGGING

REAL POWER

VAR EXPORTVAR IMPORT

REACTIVE POWERp.u. MVAR (lagging)

REACTIVE POWERp.u. MVAR (leading)

CAPABILITY DIAGRAM OF A 110 MW ALTERNATOR

OE : No-load Field Current

OD : Field Current required for Armature Reaction

FGDHF : Capability Diagram of the 110 MW Alternator

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

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III) III) COMPARISON

Actual MW=50 (i.e. 50/137.5=0.364p.u.)

Actual MVAR=6 (i.e. 6/137.5=0.044p.u.)

Arm. Current = 0.36p.u. X 7220A=2599A

Field Current=0.475p.u. X 1500A=712.5A

p.f.=cos(6.5°)=0.994 lag

Load Angle ‘δ’=33.4°

V=(MVA X 106)/(√3 X Iarm.)

=(√(MW2+MVAR2) X 106)/(√3 X Iarm.)

=(√(502+62) X 106)/(√3 X 2599) = 11.12KV

E.T.P.S. *** UNIT-5 Actual Readings taken from

Meters:DATE: 09.08.2004TIME: 11:00 Hrs.

MW = 50

MVAR = 6

Armature Current = 2600A

Field Current = 710A

p.f.= 0.98 lag

δ = -- (No measurement)

V = 11.2 KV

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CAPABILITY CURVEA. Rotor current limit Class of insulation (to take care of rotor insulation)B. Stator current limit Class of insulation for stator.C. MW load limit Turbine limit (steam power generation capability) Turbine is designed for MW load only .D. Minimum load angle limit Leading p.f. operation Stability limit of generation E. Stator end heating limit Stressing stator winding & heating of stator 10 to 20 MVAR (leading p.f.) is safe Rotor is relieved from stress Stator end winding heated due to capacitive effect Remove capacitor banks in load centres In NCTPS 210 MW unit, running the generator at -64 MVAR

load for ½ an hour. Not able to reduce the load.

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USEFULNESS OF CAPABILITY DIAGRAM FOR EXCITATION CONTROL

The information given by the capability diagram regarding full load rotor current (excitation), maximum rotor angle during steady state leading p.f. zone operation (<75°) etc., are essential for proper setting of the various limiters in the excitation control system.

Capability diagram give the basic information regarding the limiting zones of the operation so that limiters can be set / commissioned suitably for safe operation of the units.