Turbines and speed governors

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1 ELEC Power system dynamics, control and stability Turbines and speed governors Thierry Van Cutsem t.vancutsem@ulg.ac.be November / 36

2 Steam turbines Turbines and speed governors Steam turbines SG: speed governor measures speed and adjusts steam valves accordingly CV: control (or high pressure) valves maneuvered by speed governor in normal operating conditions IV: intercept valves fully opened in normal operating conditions; closed in case of overspeed MSV, RSV: main stop valve and reheater stop valve used as back-up in case of emergency 2 / 36

3 Steam turbines 3 / 36

4 Steam turbines Assumptions: power developed in one turbine stage steam flow at exit of that stage steam flow at entry of HP vessel valve opening z steam pressure p c steam flow at exit of a vessel follows steam flow at entry with a time constant Per unit system: each variable is divided by the value it takes when the turbine operates at its nominal power P N. Time constants are kept in seconds. T HP s T R 4 11 s T LP s f HP 0.3 f MP 0.4 f HP + f MP + f LP = 1 4 / 36

5 Steam turbines Interactions between turbine and boiler for large disturbances, the change in steam flow d HP results in an opposite change in steam pressure p c taking this into account requires to model the boiler and its controllers we only mention the boiler and turbine control modes Boiler-following regulation 5 / 36

6 Turbine-following regulation Turbines and speed governors Steam turbines Coordinated or integrated regulation 6 / 36

7 Steam turbines Example Responses to a demand of large production increase: comparison of the three regulations 7 / 36

8 Speed governors of steam turbines Speed governors of steam turbines Servomotor modelling z: opening of control valves (0 < z < 1 in per unit) z o : valve opening setpoint (changed when power output of unit is changed) σ: permanent speed droop (or statism) 8 / 36

9 Speed governors of steam turbines The non-windup integrator ẋ = 0 if x = x max and u > 0 = 0 if x = x min and u < 0 = u otherwise 9 / 36

10 Equivalent block-diagram Turbines and speed governors Speed governors of steam turbines T sm = 1/(Kσ) servomotor time constant ( a few 10 1 s) A little more detailed model T r : time constant of speed relay (additional amplifier) ( 0.1 s) a transfer function (1 + st 1 )/(1 + st 2 ) may be used to improve dynamics block 2 accounts for nonlinear variation of steam flow with valve opening block 1 compensates block 2 10 / 36

11 Speed governors of steam turbines Steady-state characteristics turbine: p c = 1 pu P m = z speed governor: assuming z is not limited: z = z o ω 1 σ and referring to the system frequency f (in Hz) with nominal value f N (in Hz): combining both: z = z o f f N σf N P m = z o f f N σf N z o seen as a power setpoint, in pu on the turbine power. 11 / 36

12 Turbines and speed governors Hydraulic turbines Hydraulic turbines Action (or impulse-type) turbines The potential energy of water is converted into pressure and then into kinetic energy by passing through nozzles. The runner is at atmospheric pressure. The high-velocity jets of water hit spoon-shaped buckets on the runner. Pelton turbine used for large water heights (300 m or more) 12 / 36

13 Hydraulic turbines Reaction turbines The potential energy of water is partly converted into pressure. The water supplies energy to the runner in both kinetic and pressure forms. Pressure within the turbine is above atmospheric. Require large water flows to produce significant powers. Rotation speeds are lower than with impulse turbines. 13 / 36

14 Turbines and speed governors Hydraulic turbines Francis turbine for water heights up to ' 360 m 14 / 36

15 Turbines and speed governors Hydraulic turbines Kaplan turbine For water heights up to ' 45 m Variable-pitch blades can be used (angle adjusted to water flow to maximize efficiency) Mainly used in run-of-river hydro plants 15 / 36

16 Hydraulic turbines Bulb turbine For small water heights Mainly used in run-of-river hydro plants 16 / 36

17 Hydraulic turbines Simple model of a hydro turbine Assumptions: water assumed incompressible pressure travelling waves (hammer effect) neglected H s : water height ρ specific mass of water (kg/m 3 ) g gravity acceleration (m/s 2 ) Q water flow (m 3 /s) E energy provided by 1 m 3 of water (J/m 3 ) 17 / 36

18 Hydraulic turbines Potential energy contained in 1 m 3 of water in upper reservoir: E pot = ρgh s Total power provided by water (a part of which goes in losses): P = ρgh s Q Let s define the head: H = E ρg (m) where E is the energy delivered by 1 m 3 of water. Total power provided by water (a part of which goes in losses): P = EQ = ρghq in steady state : H = H s during transients : H H s 18 / 36

19 Hydraulic turbines Basic relationships: 1 mechanical power provided by turbine, taking into account losses in conduites, etc.: P m = ρgh(q Q v ) < P 2 water flow: z : section of gate (0 z A) Q = k Q z H 3 acceleration of water column in conduite: ρla dv dt = ρga(h s H) Q = Av dq dt = ga L (H s H) 19 / 36

20 Passing to per unit values Turbines and speed governors Hydraulic turbines base of a variable = value taken by variable at nominal operating point of turbine: mechanical power P m = nominal power P N of turbine head H = height H s gate opening z = A water flow Q = nominal value Q N water speed v = Q N /A At nominal operating point: P N = ρgh s (Q N Q v ) Q N = k Q A H s Normalizing the power equation: P m pu = H H s with K P = 1 Q Q v Q N Q v = H H s Q N Q N Q v Q Q v Q N = K P H pu (Q pu Q v pu ) 1 Q v pu 20 / 36

21 Normalizing the flow equation: Turbines and speed governors Hydraulic turbines Q pu = z pu Hpu Normalizing the water acceleration equation: dq pu dt where T w = L Q N g AH s = L v N g H s = g AH s L Q N H s H H s = 1 T w (1 H pu ) is the water starting time at nominal operating point. T w = time taken by water, starting from standstill, to reach nominal speed under the effet of head H s (0.5-4 s) 21 / 36

22 Hydraulic turbines Response of a hydro turbine to small disturbances Small disturbances around operating point (z o, H o = 1, Q o ). Transfer function between z and P m? Q = H o z + zo 2 H o H st w Q = H P m = K P H o Q + K P (Q o Q v ) H Eliminating Q and H yields: where T w = T w H o If Q v is neglected: z o o Q v ) (Q 1 P m = K P (H o 3/2 z ) T o H o w s z 1 + s T w 2 is the water starting time at the considered operating point. P m = K P (H o ) 3/2 1 st w 1 + s T w 2 z 22 / 36

23 Hydraulic turbines non-minimum phase system: zero in right half complex plane initial reaction opposite to final reaction Example: response P m to step change in gate opening of magnitude Z: lim P m(t) = lim t 0 s sk P(H o ) 3/2 1 (Q o Q v ) z st o H o w 1 + s T w 2 Z s = 2K P H o (Qo Q v ) z o Z initial behaviour: inertia of water speed v and flow Q do not change head H decreases mechanical power P m decreases after some time: Q increases and H comes back to 1 P m increases non-minimum phase systems may bring instability when embedded in feedback system (one branch of the root locus ends up on the zero) 23 / 36

24 Hydraulic turbines Speed governors of hydro turbines Presence of a pilot servomotor: T p 0.05 s K 3 5 pu/pu with σ , the turbine and speed governor would be unstable when the hydro plant is in isolated mode or in a system with a high proportion of hydro plants first solution: increase σ the power plant will participate less to frequency control : not desirable other solution: add a compensator that temporarily increases the value of σ 24 / 36

25 Hydraulic turbines In the very first moment after a disturbance: lim σ + sδt r = σ + δ s 1 + st r σ = 0.04, δ , temporary statism = 6-26 permanent statism In steady state: lim σ + sδt r = σ s st r T r : reset time : s characterizes the time to come back to steady-state statism. In some speed governors, the transfer function 1 + st r K 1 + s(δ/σ)t r is used in the feed-forward branch of the speed governor 25 / 36

26 Case study. Frequency regulation in an isolated system Case study. Frequency regulation in an isolated system Hydro plant: generator: 300 MVA, 3 rotor winding model turbine: 285 MW, T w = 1.5 s Q v = 0.1 automatic voltage regulator: static gain G = 150 exciter: time constant T e = 0.5 s speed governor: σ = 0.04 mechanical-hydraulic : K = 4 ż min = 0.02 ż max = 0.02 pu/s T p = 0 PI controller: see slide 29 Load: behaves as constant impedance, insensitive to frequency 5 % step increase of admittance at t = 1 s 26 / 36

27 Case study. Frequency regulation in an isolated system Mechanical-hydraulic speed governor with compensation: δ = 0.5 T r = 5 s 27 / 36

28 Case study. Frequency regulation in an isolated system Mechanical-hydraulic speed governor without compensation (δ = 0.) 28 / 36

29 Case study. Frequency regulation in an isolated system Speed governor with PI control servomotor: K = 4 ż min = 0.02 pu/s ż max = 0.02 pu/s T p = 0 PI controller: T m = 1.9 s K p = 2 K i = 0.4 σ = / 36

30 Case study. Frequency regulation in an isolated system 30 / 36

31 Case study. Primary and secondary frequency regulation Case study. Primary and secondary frequency regulation Primary frequency control: left area: generator 2 (P N = 850 MW, σ = 0.05) β left = MW.s right area: generator 4 (P N = 850 MW, σ = 0.05) β right = MW.s Secondary frequency control: left area: generator 1 (P N = 850 MW) right area: generator 3 (P N = 850 MW) regulates P 7 8, the active power flow in tie-lines 7-8, to P o 7 8 = 400 MW 31 / 36

32 Case study. Primary and secondary frequency regulation 5 % step increase of load at bus 9. Primary frequency control only 32 / 36

33 Case study. Primary and secondary frequency regulation f 1 (resp. f 3) : frequency in left (resp. right) area (Hz) obtained from rotor speed of gen. 1 (resp. 3) (available in pu) ACE left (resp. ACE right ) : Area Control Error of left (resp. right) area (MW) P 1 c (resp. P 3 c ) : power setpoint correction sent to gen. 1 (resp. 3) (MW) 33 / 36

34 Case study. Primary and secondary frequency regulation Same load increase. Primary and secondary frequency control (λ left = λ right = Ki left = K right i = 0.02) 34 / 36

35 Case study. Primary and secondary frequency regulation Same load increase. Adjustment of power of gener. 1 by secondary frequency controller of left area, for various values of λ left 35 / 36

36 Case study. Primary and secondary frequency regulation At t = 1 s, the areas decide to decrease their power exchange to 300 MW. is set to 300 MW. Load demand is unchanged. P o / 36

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