Relationships between Load, Speed Regulation and Frequency. Slope= -R

Transcription

1 elationships between Load, Speed egulation and Frequency Governor Speed characteristic Slope= - Slope=1/D D=2 Frequency-sensitive Load Characteristic Frequency-insensitive Load Characteristic If D (more frequency-dependent load), then f If (stronger LFC feedback), then f P ( D 1/ ) L ss 9

2 Governor Model Classic Watt Centrifugal Governing System Speed changer Linkage mechanism Speed governor Hydraulic Amplifier 10

3 Governor Model P ref P v Without a governor, the generator speed drops significantly when load increases r / r The speed governor closes the loop of negative feedback control For stable operation, The governor reduces (does not eliminate) the speed drop due to load increase. Usually, speed regulation is 5 6% from zero to full load Governor output r / is compared to the change in the reference power P ref P g = P ref r / The difference P g is then transformed through the hydraulic amplifier to the steam valve/gate position command P v with time constant g r / (s) Governor steady-state speed characteristics (i.e. how the speed drops as load increases) 11

4 Turbine Model P v P m The prime mover, i.e. the source of mechanical power, may be a hydraulic turbine at water falls, a steam turbine burning coal and nuclear fuel, or a gas turbine The model for the turbine relates changes in mechanical power output P m to changes in gate or valve position P V T is in seconds 12

5 Load Frequency Control block Diagram (s) / For a step load change, i.e. = / lim (s) ss PL D 1/ The smaller the better? For n generators supporting the load: ss PL D 1/ eq 1 eq 1 // // n n 13

6 Saadat s Example

7 (s) The open-loop transfer function is A necessary and sufficient condition for a linear system to be stable: All roots of the characteristic equation (i.e. poles of the system transfer function) are in the left-hand portion of the s-plane (i.e. all having negative real parts) 15

8 outh Hurwitz Stability Criterion Characteristic equation a n s n +a n-1 s n-1 + +a 1 s+a 0 =0 (a n >0) outh table: 3 2 s s s K = 0 For i>2, x ij =(x i-2,j+1 x i-1,1 x i-2,1 x i-1,j+1 )/x i-1,1 where x ij is the element in the i-th row and j-th column s s s s K K K 0 outh-hurwitz criterion: No. of roots of the equation with positive real parts = No. of changes in sign of the 1 st column of the outh table Necessary and sufficient condition for a linear system to be stable: The 1 st column only has positive numbers s 1 row>0 if K< s 0 row>0 since K>0 So =1/K>1/73.965=

9 oot Locus Method -z i is the i -th zero and -p j is j -th pole Conclusions (see Saadat s B2.22 for details): The loci of roots of 1+KG(s)H(s) begins at KG(s)H(s) s poles and ends at its zeros as K=0 No. of separate loci = No. of poles; root loci must be symmetrical with respect to the real axis The root locus on the real axis always lies in a section of the real axis to the left of an odd number of poles and zeros Linear asymptotes of loci are centered at a point (x, 0) on the real axis with angle with respect to the real axis When s= j3.25, min =1/K= (>0.0135) x=[ j=1 n (-p j ) - i=1 m (-z i ) ]/(n-m) = (2k+1)/(n-m) k=0, 1,, (n-m-1) 17

10 Closed-loop transfer function with =0.05pu (>0.0135): D wr () s ( s)( s) = -D P ( s) (10s+ 0.8)( s)( s) + 1/ 0.05 L 2 0.1s + 0.7s s s s = Steady-state frequency deviation due to a step input: 1 1 D wss = lim sd wr ( s) =-D PL =- 0.2 = p.u. s 0 D + 1/ 20.8 D f = = Hz Note: The frequency is not restored to 60Hz (there is an offset) 18

11 Compare different values of Without LFC (Open-loop) =0.135 Frequency deviation step response =0.05 = t, s 19

12 Coal fired steam turbine power plant 20

13 IEEE Type 1 Speed Governor Model: IEEEG1/IEEEG1_GE Governor Turbine High pressure Low pressure =1/ 21

14 Composite Governor and Load Characteristic Under steady-state conditions (s=0): f (pu) (pu) ss PL (pu) D 1/ 22

15 Saadat s Example 12.2 (pu) ss P D L (pu) 1 i i Note: two generators use different MVA bases. Select 1000MVA as the common MVA base Dw Dw s S Dw S = -DP S B1 ss s B1 ss B1 = = = B1 B2 B2 -DP S m B2 S m B2 -DP S m B = ( 0.06) = 0.1 pu = ( ) = 0.08 pu 1 2 B1 B2 S B1 = S B2 B1 B2 90 D P L = = 0.09 pu

16 (a) D=0 -DP L D wss = = = pu D f = =-0.24 Hz f = f0 +D f = = Hz Dwss D P m 1 =- =- = 0.04 pu = 40 MW Dw ss D P m 2 =- =- = 0.05 pu = 50 MW (b) D=1.5 (ignoring its change due to load increase) -DPL D wss = = = pu D D f = = Hz f = f0 +D f = = Hz Dwss D P m 1 =- =- = pu=37.5mw Dwss D P m 2 =- =- = pu=46.9mw Unit 1 supplies 540MW and unit 2 supplies 450MW at the new operating frequency of 59.76Hz. Unit supplies 537.5MW and unit 2 supplies 446.9MW at the new operating frequency of Hz. The total change in generation is 84.4MW, i.e. 5.6MW less than 90MW load change, because of the change in load due to frequency drop. Dw ss D = = pu = MW 24