Hydro power plants Hydro power plants Inlet gate Air inlet Surge shaft Penstock Tunnel Sand trap Trash rack Self closing valve Tail water Main valve Turbine Draft tube Draft tube gate The principle the water conduits of a traditional high head power plant Ulla- Førre Original figur ved Statkraft Vestlandsverkene gate hoist Intake gate Intake trashrack Tunnel Inlet Surge tank Tunnel Inlet trashrack Penstock inlet Valve Desilting basin Headrace tunnel Anchor block Anchor block Shaddle Power house Exp. joint DT end gate IV G Tailrace SC IV -inlet valve R -turbine runner R SC -spiral case G -generator DT Typical Power House with Francis Turbine Arrangement of a small hydropower plant Ligga Power Plant, Norrbotten, Sweden H = 39 m Q = 513 m3/s P = 182 MW Drunner=7,5 m Borcha Power Plant, Turkey H = 87,5 m P = 150 MW Drunner=5,5 m Water intake • • • • Dam Coarse trash rack Intake gate Sediment settling basement Dams • Rockfill dams • Pilar og platedammer • Hvelvdammer Rock-fill dams 1. 2. 3. 4. Core Filter zone Transition zone Supporting shell Moraine, crushed soft rock, concrete, asphalt Sandy gravel Fine blasted rock Blasted rock Slab concrete dam Arc dam Gates in Hydro Power Plants Types of Gates • Radial Gates • Wheel Gates • Slide Gates • Flap Gates • Rubber Gates Radial Gates at Älvkarleby, Sweden Radial Gate The forces acting on the arc will be transferred to the bearing Slide Gate Jhimruk Power Plant, Nepal Flap Gate Rubber gate Flow disturbance Reinforced rubber Open position Reinforced rubber Closed position Bracket Air inlet Circular gate End cover Hinge Ribs Manhole Pipe Ladder Bolt Fastening element Seal Frame Circular gate Jhimruk Power Plant, Nepal Trash Racks Panauti Power Plant, Nepal Theun Hinboun Power Plant Laos Gravfoss Power Plant Norway Trash Rack size: Width: 12 meter Height: 13 meter Stainless Steel CompRack Trash Rack delivered by VA-Tech Cleaning the trash rack 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 Materials • Steel • Polyethylene, PE • Glass-fibre reinforced Unsaturated Polyesterplastic , GUP • Wood • Concrete Materials Material Steel, St.37 Max. Stresses [MPa] 150 Steel, St.42 190 Steel, St.52 206 PE Max. Diameter Max. Pressure [m] [m] ~ 1,0 160 2,4 320 Max. p = 160 m. Max. D: 1,4 m. Wood ~5 80 Concrete ~5 ~ 400 GUP 5 Steel pipes in penstock Nore Power Plant, Norway GUP-Pipe Raubergfossen Power Plant, Norway Wood Pipes Breivikbotn Power Plant, Norway Øvre Porsa Power Plant, Norway Calculation of the change of length due to the change of the temperature ΔL = α ⋅ ΔT ⋅ L Where: ΔL = L = α = ΔT = Change of length [m] Length [m] Coefficient of thermal expansion [m/oC m] Change of temperature [oC] Calculation of the head loss 2 L c hf = f ⋅ ⋅ D 2⋅g Where: hf = f = L = D = c = g = Head loss Friction factor Length of pipe Diameter of the pipe Water velocity Gravity [m] [-] [m] [m] [m/s] [m/s2] Example Calculation of the head loss Power Plant data: H = 100 m Head Q = 10 m3/s Flow Rate L = 1000 m Length of pipe D = 2,0 m Diameter of the pipe The pipe material is steel c⋅D Re = υ Where: c = 3,2 m/s Water velocity ν = 1,308·10-6 m2/s Kinetic viscosity Reynolds number Re = 4,9 ·106 2 L c hf = f ⋅ ⋅ D 2⋅g Where: Re = 4,9 ·106 Reynolds number ε = 0,045 mm Roughness D = 2,0 m Diameter of the pipe ε/D = 2,25 ·10-5 Relative roughness f = 0,013 Friction factor The pipe material is steel 0,013 Example Calculation of the head loss Power Plant data: H = 100 m Head Q = 10 m3/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 1000 3,2 hf = f ⋅ ⋅ = 0,013 ⋅ ⋅ = 3,4 m D 2⋅g 2 2 ⋅ 9,82 Where: f = 0,013 c = 3,2 m/s g = 9,82 m/s2 Friction factor Water velocity Gravity Calculation of maximum pressure Static head, Hgr (Gross Head) Water hammer, Δhwh Deflection between pipe supports Friction in the axial direction Hgr • • • • Maximum pressure rise due to the Water Hammer Δh wh Δhwh a cmax g = = = = a ⋅ c max = g 2⋅L IF TC << a Pressure rise due to water hammer Speed of sound in the penstock maximum velocity gravity c Jowkowsky [mWC] [m/s] [m/s] [m/s2] Example Jowkowsky a = 1000 [m/s] cmax = 10 [m/s] g = 9,81 [m/s2] TC << 2⋅L a Δh wh c=10 m/s a ⋅ c max = = 1020 m g Maximum pressure rise due to the Water Hammer Δh wh 2⋅L c max ⋅ 2 ⋅ L a ⋅ c max a = ⋅ = TC g ⋅ TC g Where: Δhwh a cmax g L TC = = = = = = Pressure rise due to water hammer Speed of sound in the penstock maximum velocity gravity Length Time to close the main valve or guide vanes C L IF TC ≥ [mWC] [m/s] [m/s] [m/s2] [m] [s] 2⋅L a Example L TC cmax g = = = = 300 10 10 9,81 [m] [s] [m/s] [m/s2] Δh wh C =1 0 L m/s c max ⋅ 2 ⋅ L = = 61 m g ⋅ TC Calculation of the pipe thickness L ⋅ Di ⋅ p ⋅ Cs = 2 ⋅ σ t ⋅ L ⋅ t • Based on: ⇓ σt = – Material properties – Pressure from: p ⋅ ri ⋅ Cs t • Water hammer • Static head p = ρ ⋅ g ⋅ (H gr + h wh ) Where: L = Length of the pipe Di p σt t Cs ρ Hgr Δhwh = Inner diameter of the pipe = Pressure inside the pipe = Stresses in the pipe material = Thickness of the pipe = Coefficient of safety = Density of the water = Gross Head = Pressure rise due to water hammer [m] [m] [Pa] [Pa] [m] [-] [kg/m3] [m] [m] σt σt p ri t Example Calculation of the pipe thickness L ⋅ Di ⋅ p ⋅ Cs = 2 ⋅ σ t ⋅ L ⋅ t ⇓ p ⋅ ri ⋅ Cs t= = 0,009 m σt p = ρ ⋅ g ⋅ (H gr + h wh ) = 1,57 MPa • Based on: – Material properties – Pressure from: • Water hammer • Static head σt Where: L = 0,001 m Length of the pipe Di σt ρ Cs Hgr Δhwh = = = = = = Inner diameter of the pipe Stresses in the pipe material Density of the water t Coefficient of safety Gross Head Pressure rise due to water hammer 2,0 m 206 MPa 1000 kg/m3 1,2 100 m 61 m σt p ri Calculation of the economical correct diameter of the pipe Cost [$] dK tot d(K f + K t ) = =0 dD dD T K tot , s t os c l ota ost Installation c s, Kt Costs for h ydraulic lo sses, K f Diameter [m] Example Calculation of the economical correct diameter of the pipe Hydraulic Losses Power Plant data: H = 100 m Q = 10 m3/s ηplant = 85 % L = 1000 m PLoss Head Flow Rate Plant efficiency Length of pipe L Q2 C2 = ρ ⋅ g ⋅ Q ⋅ hf = ρ ⋅ g ⋅ Q ⋅ f ⋅ = 5 2 4 2⋅r 2⋅g⋅π ⋅r r Where: PLoss = ρ = g = Q = = hf f = L = r = C2 = Loss of power due to the head loss Density of the water gravity Flow rate Head loss Friction factor Length of pipe Radius of the pipe Calculation coefficient [W] [kg/m3] [m/s2] [m3/s] [m] [-] [m] [m] Example Calculation of the economical correct diameter of the pipe Cost of the Hydraulic Losses per year K f = PLoss ⋅ T ⋅ kWh price Where: = Kf PLoss = T = kWhprice = r = C2 = C2 = 5 ⋅ T ⋅ kWh price r Cost for the hydraulic losses Loss of power due to the head loss Energy production time Energy price Radius of the pipe Calculation coefficient [€] [W] [h/year] [€/kWh] [m] Example Calculation of the economical correct diameter of the pipe Present value of the Hydraulic Losses per year Where: = Kf T = kWhprice = r = C2 = C2 K f = 5 ⋅ T ⋅ kWh price r Cost for the hydraulic losses Energy production time Energy price Radius of the pipe Calculation coefficient [€] [h/year] [€/kWh] [m] Present value for 20 year of operation: n K f pv Where: Kf pv n I Kf =∑ i ( ) 1 + I i =1 = Present value of the hydraulic losses = Lifetime, (Number of year ) = Interest rate [€] [-] [-] Example Calculation of the economical correct diameter of the pipe Cost for the Pipe Material p⋅r m = ρm ⋅ V = ρm ⋅ 2 ⋅ π ⋅ r ⋅ t ⋅ L = ρm ⋅ 2 ⋅ π ⋅ r ⋅ ⋅ L = C1 ⋅ r 2 σ K t = M ⋅ m = M ⋅ C1 ⋅ r Where: m = ρm = V = r = L = p = σ = C1 = Kt = M = Mass of the pipe Density of the material Volume of material Radius of pipe Length of pipe Pressure in the pipe Maximum stress Calculation coefficient Installation costs Cost for the material 2 [kg] [kg/m3] [m3] [m] [m] [MPa] [MPa] [€] [€/kg] NB: This is a simplification because no other component then the pipe is calculated Example Calculation of the economical correct diameter of the pipe • Installation Costs: – Pipes – Maintenance – Interests – Etc. Example Calculation of the economical correct diameter of the pipe K f pv C2 ⋅ T ⋅ kWh price n 5 =∑ r i ( ) 1 + I i =1 K t = M ⋅ C1 ⋅ r 2 d (K t + K f ) 5 n C 2 ⋅ T ⋅ kWh price = 2⋅M⋅C⋅r − 6 ⋅∑ =0 i r i =1 dr (1 + I ) Where: Kf = = Kt T = kWhprice = r = C1 = = C2 M = n = I = Cost for the hydraulic losses Installation costs Energy production time Energy price Radius of the pipe Calculation coefficient Calculation coefficient Cost for the material Lifetime, (Number of year ) Interest rate [€] [€] [h/year] [€/kWh] [m] [€/kg] [-] [-] Example Calculation of the economical correct diameter of the pipe d(K t + K f ) = 2⋅M ⋅C⋅r − dr C 2 ⋅ T ⋅ kWh price 5 =0 ⋅ i 6 ∑ r i =1 (1 + I ) ⇓ 5 n C 2 ⋅ T ⋅ kWh price r = 7 ⋅∑ 2 i =1 M ⋅ C ⋅ (1 + I )i n Calculation of the forces acting on the anchors Calculation of the forces acting on the anchors F5 F1 F F2 F3 F4 F1 = Force due to the water pressure [N] F2 = Force due to the water pressure [N] F3 = Friction force due to the pillars upstream the anchor F4 = Friction force due to the expansion joint upstream the anchor F5 = Friction force due to the expansion joint downstream the anchor [N] [N] [N] Calculation of the forces acting on the anchors F R G Valves Principle drawings of valves Open position Closed position Spherical valve Hollow-jet valve Gate valve Butterfly valve Spherical valve Bypass system Butterfly valve Butterfly valve Butterfly valve disk types Hollow-jet Valve Pelton turbines • Large heads (from 100 meter to 1800 meter) • Relatively small flow rate • Maximum of 6 nozzles • Good efficiency over a vide range Jostedal, Norway *Q = 28,5 m3/s *H = 1130 m *P = 288 MW Kværner Francis turbines • Heads between 15 and 700 meter • Medium Flow Rates • Good efficiency η=0.96 for modern machines SVARTISEN P = 350 MW H = 543 m Q* = 71,5 m3/S D0 = 4,86 m D1 = 4,31m D2 = 2,35 m B0 = 0,28 m n = 333 rpm 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