Explore the losses in hydro power plants, specifically focusing on the reaction turbine and straight draft tube designs. Learn about energy inefficiencies due to height differences and water speeds at outlet levels. Visual illustrations included.
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Losses of the reaction turbine (1) H Exit surface is located at above the outlet water level: s = H h h 2 s B Energy of water at the runner (exit from the turbine): 2 2 p c = + + 0 P gh 2 2 2 Energy of water on the outlet water level p = + 0 P gh B B Energy loss: 2 2 2 2 c c = = + = + P P P g (h h ) g H 2 2 B B s 2 2 Unused potential energy because of the height different between turbine runner and water outlet levels Unused kinetic energy losses because of the speed of water at the outlet
Losses of the reaction turbine (2) Straight draft tube: 2 3 p c = + + 3 P gh 3 3 2 Losses in this case are equal to: 2 3 p p c = = + 3 0 P P P g (h h ) 3 3 B B 2 With: p p + = + 3 0 gh gh 3 B Energy losses: 2 3 c = = P P P 3 B 2 The height different between turbine runner and water outlet levels is utilized with aspirator Unused kinetic energy losses because of the speed of water at the outlet
Losses of the reaction turbine (3) Draft tube (straight) 2 3 p c = + + 3 P gh 3 3 2 Losses in this case are equal to: 2 3 p p c = = + 3 0 P P P g (h h ) 3 3 B B 2 With: p p + = + 3 0 gh gh 3 B Energy losses: 2 3 c = = P P P 3 B 2 The height different between turbine runner and water outlet levels is utilized with aspirator Unused kinetic energy losses because of the speed of water at the outlet
Straight draft tube (1) Straight draft tube utilizes the potential energy of the water due to height difference between turbine runner and water outlet The lenght is limited with cavitation to 8 meters 8 H m s Power is cross-section 2 2 2 p c = + + ok P gh 2 2 2 Power at water outlet level: p P gh (c = + 0 ok ) B B water outlet 2 P P Energy losses: B 2 2 2 2 c c = = + = P P P g(h h ) g H 2 2 B B s 2 2
Straight draft tube (2) = c c The energy of water on the outlet surface with the assumption that (straight tube with constant cross-section area) 3 2 2 2 p c = + + g h 3 P 3 3 2 For cross-section B p = + g h ok P B B Energy losses: 2 2 p p c = = + g (h 3 ok P P P h ) 3 3 B B 2
Straight draft tube (3) 0 canal c Assuming the speed of water in the outlet canal is the Bernoulli equations for the exit cross-section of the draft tube and for the water outlet height of the lower water level: p g h p + = + g h 3 ok 3 B 2 2 p p c = + g (h 3 ok P P h ) Then combining the losses equation 3 3 B B 2 and Bernoulli equation we obtain: 2 2 c P = 2 If we install the straight draft tube the losses are reduced for the potential energy difference
Draft tube (1) Draft tube reduces the potential energy losses and kinetic energy losses It has larger exit cross-section thus reducing the water speed due to constant mass of water flowing The power of water n the cross- section 3: 2 3 p c W = + + g h + g h 3 P 3 3 Rd 2 The power on the outlet level: p = + g h ok P B B
Draft tube (2) Considering the relation: p p + g h = + g h 3 ok 3 B The losses on the outlet are: 2 3 c = + g h P Rd 2
