Hydro power plants. Inlet gate Surge shaft. Air inlet. Penstock. Tunnel Sand trap Trash rack Self closing valve. Tail water. Main valve Turbine

Transcription

1 Hydro power plants

2 Hydro power plants Air inlet Inlet gate Surge shaft Penstock Tail water Tunnel Sand trap Trash rack Self closing valve Draft tube Draft tube gate Main valve Turbine

3 The principle the water conduits of a traditional high head power plant

4 Ulla- Førre Original figur ved Statkraft Vestlandsverkene

5 Intake gate gate hoist Intake trashrack Tunnel Inlet Tunnel Inlet trashrack Surge tank Penstock inlet Valve Desilting basin Anchor block Headrace tunnel Anchor block Shaddle Exp. joint Power house IV G SC DT end gate Tailrace IV -inlet valve R -turbine runner SC -spiral case G -generator R DT Typical Power House with Francis Turbine

6 Arrangement of a small hydropower plant

7 Ligga Power Plant, Norrbotten, Sweden H = 39 m Q = 513 m 3 /s P = 182 MW D runner =7,5 m

8 H = 87,5 m P = 150 MW D runner =5,5 m Borcha Power Plant, Turkey

9 Water intake Dam Coarse trash rack Intake gate Sediment settling basement

10 Dams Rockfill dams Pilar og platedammer Hvelvdammer

11 Rock-fill dams 1. Core Moraine, crushed soft rock, concrete, asphalt 2. Filter zone Sandy gravel 3. Transition zone Fine blasted rock 4. Supporting shell Blasted rock

12 Slab concrete dam

13 Arc dam

14 Gates in Hydro Power Plants

15 Types of Gates Radial Gates Wheel Gates Slide Gates Flap Gates Rubber Gates

16 Radial Gates at Älvkarleby, Sweden

17 Radial Gate The forces acting on the arc will be transferred to the bearing

18 Jhimruk Power Plant, Nepal Slide Gate

19 Flap Gate

20 Rubber gate Reinforced rubber Open position Flow disturbance Bracket Reinforced rubber Closed position Air inlet

21 Circular gate End cover Ribs Pipe Hinge Manhole Bolt Ladder Fastening element Seal Frame

22 Jhimruk Power Plant, Nepal Circular gate

23 Trash Racks Panauti Power Plant, Nepal

24 Theun Hinboun Power Plant Laos

25 Gravfoss Power Plant Norway Trash Rack size: Width: 12 meter Height: 13 meter Stainless Steel

26 CompRack Trash Rack delivered by VA-Tech

27 Cleaning the trash rack

28 Pipes Materials Calculation of the change of length due to the change of the temperature Calculation of the head loss Calculation of maximum pressure Static pressure Water hammer Calculation of the pipe thickness Calculation of the economical correct diameter Calculation of the forces acting on the anchors

29 Materials Steel Polyethylene, PE Glass-fibre reinforced Unsaturated Polyesterplastic, GUP Wood Concrete

30 Materials Material Max. Diameter Max. Pressure Max. Stresses [m] [m] [MPa] Steel, St Steel, St Steel, St PE ~ 1, GUP 2,4 320 Max. p = 160 m. Max. D: 1,4 m. Wood ~5 80 Concrete ~5 ~ 400

31 Steel pipes in penstock Nore Power Plant, Norway

32 GUP-Pipe Raubergfossen Power Plant, Norway

33 Wood Pipes Breivikbotn Power Plant, Norway Øvre Porsa Power Plant, Norway

34 Calculation of the change of length due to the change of the temperature ΔL = α ΔT L Where: ΔL = Change of length [m] L = Length [m] α = Coefficient of thermal expansion [m/ o Cm] ΔT = Change of temperature [ o C]

35 Calculation of the head loss L c 2 h f = f D 2 g Where: h f = Head loss [m] f = Friction factor [ - ] L = Length of pipe [m] D = Diameter of the pipe [m] c = Water velocity [m/s] g = Gravity [m/s 2 ]

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37 Example Calculation of the head loss Power Plant data: H = 100 m Head Q = 10 m 3 /s Flow Rate L = 1000 m Length of pipe D = 2,0 m Diameter of the pipe The pipe material is steel h f = f L D c 2 2 g Re = c D υ Where: c = 3,2 m/s Water velocity ν = 1, m 2 /s Kinetic viscosity Re = 4, Reynolds number

38 0,013 Where: Re = 4, Reynolds number ε = 0,045 mm Roughness D = 2,0 m Diameter of the pipe ε/d = 2, Relative roughness f = 0,013 Friction factor The pipe material is steel

39 Example Calculation of the head loss Power Plant data: H = 100 m Head Q = 10 m 3 /s Flow Rate L = 1000 m Length of pipe D = 2,0 m Diameter of the pipe The pipe material is steel 2 2 L c ,2 h f = f = 0,013 = D 2 g 2 2 9,82 3,4 m Where: f = 0,013 Friction factor c = 3,2 m/s Water velocity g = 9,82 m/s 2 Gravity

40 Calculation of maximum pressure Static head, H gr (Gross Head) Water hammer, Δh wh Deflection between pipe supports Friction in the axial direction H gr

41 Maximum pressure rise due to the Water Hammer Δh wh = a c g max IF T C << 2 L a Jowkowsky Δh wh = Pressure rise due to water hammer [mwc] a = Speed of sound in the penstock [m/s] c max = maximum velocity [m/s] g = gravity [m/s 2 ] c

42 a = 1000 [m/s] c max = 10 [m/s] g = 9,81 [m/s 2 ] T C 2 L << a Example Jowkowsky Δh a c max wh = = g 1020 m c=10 m/s

43 Maximum pressure rise due to the Water Hammer Δh wh = a c g max 2 L a T C = c 2 L g T max Where: Δh wh = Pressure rise due to water hammer [mwc] a = Speed of sound in the penstock [m/s] c max = maximum velocity [m/s] g = gravity [m/s 2 ] L = Length [m] T C = Time to close the main valve or guide vanes [s] C IF T C 2 L a C L

44 L = 300 [m] T C = 10 [s] c max = 10 [m/s] g = 9,81 [m/s 2 ] Example Δh c 2 L g T max wh = = C 61 m C=10 m/s L

45 Calculation of the pipe thickness L D σ t = i p C s p ri C t = 2 σ s t L t ( ) p = ρ g H gr + h wh Based on: Material properties Pressure from: Water hammer Static head Where: L = Length of the pipe [m] D i = Inner diameter of the pipe [m] p = Pressure inside the pipe [Pa] σ t = Stresses in the pipe material [Pa] t = Thickness of the pipe [m] C s = Coefficient of safety [ - ] ρ = Density of the water [kg/m 3 ] H gr = Gross Head [m] Δh wh = Pressure rise due to water hammer [m] t σ t p ri σ t

46 Example Calculation of the pipe thickness L D t = i p C p ri C σ p = ρ g t s s = 2 σ L t = 0,009 m t ( H + h ) 1,57 MPa gr wh = Based on: Material properties Pressure from: Water hammer Static head σ t Where: L = 0,001 m Length of the pipe D i = 2,0 m Inner diameter of the pipe σ t = 206 MPa Stresses in the pipe material ρ = 1000 kg/m 3 Density of the water C s = 1,2 Coefficient of safety H gr = 100 m Gross Head Δh wh = 61 m Pressure rise due to water hammer t p ri σ t

47 Calculation of the economical correct diameter of the pipe ( K + K ) dk d f t = dd dd tot = 0 Total costs, Ktot Installation costs, K t Costs for hydraulic losses, K f Diameter [m] Cost [$]

48 Example Calculation of the economical correct diameter of the pipe Hydraulic Losses Power Plant data: H = 100 m Head Q = 10 m 3 /s Flow Rate η plant = 85 % Plant efficiency L = 1000 m Length of pipe 2 L Q P Loss = ρ g Q h f = ρ g Q f = r 2 g π r Where: P Loss = Loss of power due to the head loss [W] ρ = Density of the water [kg/m 3 ] g = gravity [m/s 2 ] Q = Flow rate [m 3 /s] h f = Head loss [m] f = Friction factor [ - ] L = Length of pipe [m] r = Radius of the pipe [m] C 2 = Calculation coefficient C r 2 5

49 Example Calculation of the economical correct diameter of the pipe Cost of the Hydraulic Losses per year K f = P Loss T kwh price = C r 2 5 T kwh price Where: K f = Cost for the hydraulic losses [ ] P Loss = Loss of power due to the head loss [W] T = Energy production time [h/year] kwh price = Energy price [ /kwh] r = Radius of the pipe [m] C 2 = Calculation coefficient

50 Example Calculation of the economical correct diameter of the pipe Present value of the Hydraulic Losses per year K = C T kwh 2 f 5 price Where: r K f = Cost for the hydraulic losses [ ] T = Energy production time [h/year] kwh price = Energy price [ /kwh] r = Radius of the pipe [m] C 2 = Calculation coefficient Present value for 20 year of operation: K f pv = n f i= 1 + K ( 1 I) Where: K fpv = Present value of the hydraulic losses [ ] n = Lifetime, (Number of year ) [-] I = Interest rate [-] i

51 Example Calculation of the economical correct diameter of the pipe Cost for the Pipe Material m K = ρ t m V = ρ m = M m = 2 π r t L = ρ M C 1 r 2 m 2 π r p r L = σ C 1 r 2 Where: m = Mass of the pipe [kg] ρ m = Density of the material [kg/m 3 ] V = Volume of material [m 3 ] r = Radius of pipe [m] L = Length of pipe [m] p = Pressure in the pipe [MPa] σ = Maximum stress [MPa] C 1 = Calculation coefficient K t = Installation costs [ ] M = Cost for the material [ /kg] NB: This is a simplification because no other component then the pipe is calculated

52 Example Calculation of the economical correct diameter of the pipe Installation Costs: Pipes Maintenance Interests Etc.

53 Example Calculation of the economical correct diameter of the pipe K d f K t = M C Where: K f = Cost for the hydraulic losses [ ] K t = Installation costs [ ] T = Energy production time [h/year] kwh price = Energy price [ /kwh] r = Radius of the pipe [m] = Calculation coefficient C 1 C 2 pv = n C r ( K + K ) 2 5 T kwh ( 1 I) i= 1 + i price T kwh ( 1+ I) = Calculation coefficient M = Cost for the material [ /kg] n = Lifetime, (Number of year ) [-] I = Interest rate [-] 5 6 r n t f 2 price = 2 M C r = i dr i= 1 C 1 r 0 2

54 Example Calculation of the economical correct diameter of the pipe ( ) ( ) ( ) 7 n 1 i i price 2 n 1 i i price 2 6 f t I 1 C M kwh T C 2 5 r 0 I 1 kwh T C r 5 r C M 2 dr K K d = = + = = + = +

55 Calculation of the forces acting on the anchors

56 Calculation of the forces acting on the anchors F 5 F F 1 F 2 F 3 F 4 F 1 = Force due to the water pressure [N] F 2 = Force due to the water pressure [N] F 3 = Friction force due to the pillars upstream the anchor F 4 = Friction force due to the expansion joint upstream the anchor F 5 = Friction force due to the expansion joint downstream the anchor [N] [N] [N]

57 Calculation of the forces acting on the anchors F G R

58 Valves

59 Principle drawings of valves Open position Closed position Spherical valve Gate valve Hollow-jet valve Butterfly valve

60 Spherical valve

61 Bypass system

62 Butterfly valve

63 Butterfly valve

64 Butterfly valve disk types

65 Hollow-jet Valve

66 Pelton turbines Large heads (from 100 meter to 1800 meter) Relatively small flow rate Maximum of 6 nozzles Good efficiency over a vide range

67 Jostedal, Norway Kværner *Q = 28,5 m 3 /s *H = 1130 m *P = 288 MW

68 Francis turbines Heads between 15 and 700 meter Medium Flow Rates Good efficiency η=0.96 for modern machines

69 P = 350 MW H = 543 m Q* = 71,5 m 3 /S D0 = 4,86 m D1 = 4,31m D2 = 2,35 m B0 = 0,28 m n = 333 rpm SVARTISEN

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72 Kaplan turbines Low head (from 70 meter and down to 5 meter) Large flow rates The runner vanes can be governed Good efficiency over a vide range

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