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 to -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 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.

1

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 cos . REACTIVE POWER (Q):The reactiveΦ 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.2

TYPE OF LOADS:TYPE OF LOAD I V I V I V L I C I V R I V PHASOR PHASE ANGLE POWER ABSORBED BYФ Ф THE LOAD P Q P>0 Q=0

= 0°Ф

V

= +90°Ф

P=0

Q>0

= - 90°Ф

P=0

Q<0

I V

R V L I 0°< <+90° P>0 Q>0Φ Φ

V

R

L3

TYPE OF LOADS:TYPE OF LOAD I V R C R C PHASOR PHASE ANGLE POWER ABSORBED BY THE LOAD P QΦ

I V

-90°< <0°Φ

P>0

Q<0

V

Tuned to Resonance I V C Ic IL L IL = Ic PL = Pc Energy travels Back & forth Between C&L4-90°<= <=+90°Φ

P=0

Q=0

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.5

CAPABILITY DIAGRAM OF A 110 MW ALTERNATOR• I) COLLECT THE INFORMATIONS FROM T.G. NAME PLATE / MANUAL: 1. Terminal Voltage : 11,000 V 2. Rated MVA : 137.5 3. Rated p.f. (cos ) : 0.8 Lagging 4. RatedФ Armature Current : 7220 A 5. Rated Field Current : 1500 A 6. Short Circuit Ratio : 0.5 II) CALCULATED VALUES: 1. MW = MVA X p.f. = 137.5 X 0.8 = 110 MW 2. 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 ‘ ’) limit line.δ 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 6

•

CAPABILITY DIAGRAM OF A 110 MW ALTERNATORREAL POWER p.u. MW Unity p.f. Leading p.f. VAR IMPORT VAR EXPORT Lagging p.f.

OE : No-load Field Current OD : Field Current required for Armature Reaction FGDHF : Capability Diagram of the 110 MW Alternator

=90°δ1.0 0.9 0.9

BP.F.= 0.8 LAGGING

GTHEORITICAL STABILITY LIMIT LINE 0.7 0.6 0.5MAR GIN ( =6 3 °)δ

TURBINE LIMIT LINE 0.8 0.7

0.8

D

OR AT ST

0.6 0.5 0.4 R (O TO ROR) FIE EN RR CU

ITY LIM IT W ITH 12.5 %

T

CU RR E

NT

R CU

AR MA TU RE

NT RE

UR R C ROTO

LD

IT LIM

0.4 0.3 0.2

ST AT OR

(O R)

IT T LIM REN

0.2 0.1

0.1

A1.0 0.9 0.8 0.7 0.6 SCR

E0.5

PR A CT IC

F

=36 .87°Ф

BIL

L TA 0.3 TO

AL STA

TO TA L

H0.3 0.2 0.1

C1.0

0.4

REACTIVE POWER p.u. MVAR (leading)

O

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

MVA X SCR MAXIMUM PERMISSIBLE MVAR IN ZERO p.f. LEADING.

REACTIVE POWER p.u. MVAR (lagging)

7

CAPABILITY DIAGRAM OF A 110 MW ALTERNATORREAL POWER p.u. MW Unity p.f. Leading p.f. VAR IMPORT VAR EXPORT Lagging p.f.

OE : No-load Field Current OD : Field Current required for Armature Reaction FGDHF : Capability Diagram of the 110 MW Alternator

=90°δ1.0 0.9 0.9

BP.F.= 0.8 LAGGING

GTHEORITICAL STABILITY LIMIT LINE 0.7 0.6 0.5( =6 3°)δ

TURBINE LIMIT LINE 0.8 0.7

0.8

D

MA R GIN

OR AT ST

0.6 0.5R) 0.4 R (O TO RO FI

PRA CTIC AL STA BIL IT Y L IMIT WIT H 1 2.5 %

D EL

0.4 0.3 0.2

0.2 0.1 0.3

0.1

= 36 .8Ф

TO TA L

7°

L TA 0.3 TO

ST AT OR

(O R)

A1.0 0.9 0.8 0.7 0.6 SCR

E0.5

AR MA TU RE

T EN RR CU

CU RR E

NT

IT L IM T EN LIMIT RR CU RENT CUR OR ROT

0.5

0.4

0.6

0.2

0.7

0.8

0.9

1.0

F

H0.9

C1.0

0.1

0.4

0.3

0.2

0.1

REACTIVE POWER p.u. MVAR (leading)

O

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

MVA X SCR MAXIMUM PERMISSIBLE MVAR IN ZERO p.f. LEADING.

REACTIVE POWER p.u. MVAR (lagging)

8

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 ‘ ’=33.4° V=(MVA X 106)/(√3 Xδ2 2

E.T.P.S. *** UNIT-5DATE: TIME: MW = 50 MVAR = 6 Armature Current = 2600A Field Current = 710A p.f.= 0.98 lag = -- (No measurement) V = 11.2 KV 09.08.2004 11:00 Hrs.δ

Iarm.)6

=(√(MW +MVAR ) X 10 )/(√3 X2 2 6

Iarm.)

=(√(50 +6 ) X 10 )/(√3 X 2599) = 11.12KV

9

• • • • K.

Rotor current limit Class of insulation (to take care of rotor insulation) Stator current limit Class of insulation for stator. MW load limit Turbine limit (steam power generation capability) Turbine is designed for MW load only . Minimum load angle limit Leading p.f. operation Stability limit of generation 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.

CAPABILITY CURVE

10

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.11

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.

12

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.ADJUSTING INPUT VALVES CONTROLS FREQUENCY CONTROLS REAL POWER

13

VOLTAGE IS RELATED TO REACTIVE POWER ( Q – V )G1

1

V1

I

V2 jX P jQ

2

1. 2. 3.

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

Take V1 as reference. V2=V1-jXI -----------------------------------(1) V1 * I = P jQ I = (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]

14

VECTOR DIAGRAMS:V2 = V1- X Q - j X P V1 V1BOTH DROPS EQUAL DOUBLE P DOUBLE Q

V1X Q V1 X P V1

V1XV1

V1QXP V1 2X Q V1

V2

V2 V22X P V1

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. VOLTAGE15

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. PEAK LOAD CONDITION CONNECT CAPACITORS TO CONTROL VOLTAGE

LIGHT LOAD CONDITION

CONNECT REACTORS

SYNCHRONOUS CONDENSER IS USED TO ABSORB or TO DELIVER THE REACTIVE POWER. SYNCHRONOUS MOTOR UNDER NO-LOAD CONDITION IS SYNCHRONOUS CONDENSER.16

POWER DIAGRAM (CAPABILITY DIAGRAM):

•

ASSUMPTION: I.R. drop is negligible. Sin E Cos Aδ Фdδ

CASE-I: In ABC, BC=EΔ

B ΦIX

MW C

In BCD, BC=IXdΔ

E Sin = IXd Cos I MVAR Multiply both sides by V Xd EV Sin = VI Cos = REAL Xdδ Ф δ Ф POWER At =90°, We get the maximum power i.e. the theoritical stability line. V Dδ• In ABC, CD=AC – AD; In BCD, CD=IXd Sin In ABC, AC=E Cos & AD = V IXdΔ Δ Ф Δ δ Sin = E Cos - V ; Multiply both sides by V , We get Xd EV Cos – V2 = VI Sin Ф δ δ Ф = REACTIVE POWER Xd Xd CASE-I I:

Φ

17

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.

S.C.C.

O.C.C. PER UNIT CURRENT PER UNIT VOLTAGE 1.0 a b C EAD AE AC DE BC

c

A

D

B

AB

oSCR = o Fo 1 o Fs Xd c Fo b Fs c Fo a Fo

Fo1

Fc

FIELD CURRENT1 Per unit voltage on open circuit Corresponding per unit current on short circuit 18

a Fo / c Fo

= RECIPROCAL OF SYNCHRONOUS REACTANCE

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. TRANSPORTATION PROBLEM

SCR

AIR GAP

WEIGHT

SIZE

Present trend is to build low value of SCR since fast acting excitation system available.19

• T.G. CAPACITY IN M.W.: 50 60 62.5 100 110 120 200 210 – Weight: 250 tonnes 235 250 500 800 future 1000 future

GENERATOR – IMPORTANT TIPS

20

GENERATOR – IMPORTANT TIPS• T.G. TERMINAL VOLTAGE IN KV : 10.5 11 – ETPS 60 MW, 110 MW 13.8 15 – Neyveli-Stage I 15.75 – BHEL 210 MW 16 – Nuclear 235 MW 18.4 – NTPC 210 MW 21 – 500 MW 22 - 500 MW 33 (or) 34 – Future (800 MW/1000 MW) requires 800 KV line (year 2010)21

GENERATOR – IMPORTANT TIPS Higher capacity Hydro machine in India : 250 MW, KOINA (Maharastra), Air cooled.

Higher capacity T.G. in India Higher capacity T.G. : 500 MW. : Advantage : Reduction of cost of Generation. Limitations : (i) Transportation problem (bigger size) (ii) Do not have adequate transmission lines. : 315 MVA, 3 phase, single unit, 400 KV. : 400 KV AC. : 800 or 765 KV line – year 2010. : 400 KV line. : FRANCE, 1500 MW T.G., Nuclear , 1600 MVA, 1200 KV. 22

Higher capacity G.T. in India

Maximum voltage National Grid Regional Grid World highest

GENERATOR – IMPORTANT TIPSSPECIFICATION FOR ROTATING MACHINES:IEC 34 Part – I, II, III (International Electro-Technical commission) IS 5422 2*105 hours guaranteed operating time (23 years) 8760 hrs/year. 104 start/stop times. Total life time : 25 years. Capital O/H : Once in 4 to 5 years (25 days). Annual O/H : < 10 days.23