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Hydropower

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Hydro Power Lecture slides

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Hydropower

  1. 1. KV HYDROPOWER Keith Vaugh BEng (AERO) MEng
  2. 2. KV Utilise the vocabulary associated with Hydro Electric power plants and power generation Develop a comprehensive understanding of flow measurement, the workings of a Hydroelectric power plant, the various configurations and the components associated with the plant Derive the governing equations for the power plant and the associated components Determine the forces acting impeller’s and the power that can be achieved OBJECTIVES
  3. 3. KV MEASUREMENT OF FLOW RATE The turbine produces power as fluid flows through it. Flow rate is prone to seasonal variations Turbine’s are normally matched to low season flow Flow measurement is necessary for environmental impact High season maximum flow measurement is essential
  4. 4. KV Basic method The moving body of fluid is either diverted or stopped by a dam The trapped volume is then used to measure the flow rate No assumptions are made about the flow This method is ideal for low flow rates Basic flow measurement,Twidell, J. andWeir, T. (2006)
  5. 5. KV Refined method I The mean speed ū will be marginally less than the surface speed, us due to viscous friction Therefore, ū ≈ 0.8×us us is measured by timing the duration a float takes to pass two defined points. Ideally the stream should be as close to uniform in the region of measure and relatively straight The cross sectional area is estimated by measuring at several points across the moving body of fluid and integrating Refined method I flow measurement, Twidell, J. andWeir,T. (2006)
  6. 6. KV Refined method II On fast flowing bodies of fluid, a float is realised from a specified depth below the surface The time interval it takes to rise to the surface is independent of its horizontal motion The horizontal distance required for the float to rise yields the speed In this case the mean speed ū is measured being averaged over depth rather than cross section Refined method II flow measurement, Twidell, J. andWeir,T. (2006)
  7. 7. KV Sophisticated method The preferred choice of hydrologists, as its the most accurate A two-dimensional grid through a section of the stream. The forward speed u is measured at each grid point using a flow meter The integral is then evaluated Refined method II flow measurement, Twidell, J. and Weir, T. (2006)
  8. 8. KV Using a weir Measuring the volumetric flow rate over an extended period of time, requires the construction of a dam with a specially shaped calibration notch The height of the flow through the notch yields a measure of the flow Calibration of this arrangement is achieved by the utilisation of a laboratory model with the same form of notch The calibrations are tabulated in standard handbooks. Refined method II flow measurement, Twidell, J. andWeir,T. (2006)
  9. 9. KV HYDRO-ELECTRIC POWER GENERATION Hydropower plants harness the potential energy within falling water and utilise rotodynamic machinery to convert that energy to electricity The theoretical water power Pwa,th between two points for a moving body of water can be determined by:
  10. 10. KV HYDRO-ELECTRIC POWER GENERATION Hydropower plants harness the potential energy within falling water and utilise rotodynamic machinery to convert that energy to electricity The theoretical water power Pwa,th between two points for a moving body of water can be determined by: Pwa,th = ρwa g!Vwa hhw − htw( )
  11. 11. KV Applying Bernoulli’s equation two reference points ① and ②, up and downstream of the hydroelectric power plant;
  12. 12. KV Applying Bernoulli’s equation two reference points ① and ②, up and downstream of the hydroelectric power plant; where; p1 ρwa,1 g + z1 + uwa,1 2 2g = p2 ρwa,2 g + z2 + uwa,2 2 2g +α uwa,2 2 2g = const. p ρwa g = pressure head z = potential energy head uwa 2 2g = kinetic energy α uwa 2 2g = lost energy
  13. 13. KV
  14. 14. KV
  15. 15. KV
  16. 16. KV
  17. 17. KV Headwater
  18. 18. KV Headwater Dam
  19. 19. KV Headwater Dam Screen
  20. 20. KV Headwater Dam Screen Stop logs
  21. 21. KV Headwater Dam Screen Stop logs Stop valve
  22. 22. KV Headwater Dam Screen Penstock Stop logs Stop valve
  23. 23. KV Headwater Turbine Dam Screen Penstock Stop logs Stop valve
  24. 24. KV Headwater Turbine Dam Screen Penstock Generator Stop logs Stop valve
  25. 25. KV Headwater Turbine Dam Screen Penstock Generator Stop logs Stop valve Draft tube
  26. 26. KV Headwater Turbine Dam Screen Penstock Generator Power house Stop logs Stop valve Draft tube
  27. 27. KV Tailwater Headwater Turbine Dam Screen Penstock Generator Power house Stop logs Stop valve Draft tube
  28. 28. KV INTAKE STRUCTURE Headwater Dam Screen Stop logs Stop valve
  29. 29. KV INTAKE STRUCTURE Headwater Dam Screen Stop logs Stop valve ① ② p1 ρwa g + z1 = p2 ρwa g + z2 + 1+αIS( ) uwa,2 2 2g Applying Bernoulli’s equation between stations ① and ②
  30. 30. KV PENSTOCK
  31. 31. KV PENSTOCK Penstock
  32. 32. KV PENSTOCK ② p2 ρwa g + z2 + uwa,2 2 2g = p3 ρwa g + z3 + 1+αPS( ) uwa,3 2 2g Penstock Applying Bernoulli’s equation between stations ② and ③ ③
  33. 33. KV TURBINE
  34. 34. KV TURBINE PTurbine = ηTurbine ρwa g!Vwa hutil In the turbine, pressure energy is converted into mechanical energy as the fluid passes from ③ to ④ Conversion losses are described by the efficiency of the turbine ③ ④
  35. 35. KV TAILRACE
  36. 36. KV TAILRACE Applying Bernoulli’s equation between stations ④ and ⑤ p4 ρwa g + uwa,4 2 2g = p5 ρwa g + uwa,5 2 2g⑤ ④
  37. 37. KV
  38. 38. KV Tailwater Headwater
  39. 39. KV Tailwater Headwater Energy line Energy line Datum
  40. 40. KV Tailwater Headwater Energy loss Energy line Energy line Datum
  41. 41. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Datum
  42. 42. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Geodetic elevation of the stream Datum
  43. 43. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressure head Geodetic elevation of the stream Datum
  44. 44. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressurehead Pressure head Geodetic elevation of the stream Datum
  45. 45. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressurehead Pressure head Velocity head Geodetic elevation of the stream Datum
  46. 46. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Useablehead Pressurehead Pressure head Velocity head Geodetic elevation of the stream Datum
  47. 47. KV COMPLETE SYSTEM ① ② ③ ④ ⑤
  48. 48. KV COMPLETE SYSTEM Applying Bernoulli’s equation between stations ① and ⑤ Pwa,act = ρwa g!Vwa hhw − htw( )−αIS uwa,2 2 2g −αPS uwa,3 2 2g − uwa,5 2 2g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ① ② ③ ④ ⑤
  49. 49. KV CATEGORISATION Low-head plants: Are categorised by large flow rates and relatively low heads (less than 20 m). Typically these are run-of-river power plants i.e. harness the flow of the river Medium-head plants: This category of plant uses the head created by a dam (20 - 100 m) and the average discharges used by the turbines result from reservoir management High-head plants: Found in mountainous regions with typical heads of 100 - 2,000 m. Flow rates are typically low and therefore the power results from high heads
  50. 50. KV } name: Bonneville Dam river: Columbia River location: Oregon, USA head: 18 m no. turbine’s: 20 capacity: 1092.9 MW DIVERSION TYPE Source: http://maps.google.com/maps?f=q&source=s_q&hl=en&geocode=&q=45%C2%B038%E2%80%B239%E2%80%B3N+121%C2%B056%E2%80%B226%E2%80%B3W&aq=&sll=37.052985,37.890472&sspn=1.008309,1.767426&ie=UTF8&ll=45.644288,-121.940603&spn=0.027602,0.055232&t=k&z=15
  51. 51. KVSource: http://maps.google.com/maps?f=q&source=s_q&hl=en&geocode=&q=46%C2%B035%E2%80%B215%E2%80%B3N+118%C2%B001%E2%80%B234%E2%80%B3W&aq=&sll=24.943901,105.113523&sspn=0.035799,0.059094&ie=UTF8&t=k&z=15 name: Little Goose Dam river: Lake Bryan location: Washington, USA head: 30 m no. turbine’s: 6 capacity: 932 MW RUN-OF-RIVER
  52. 52. KV
  53. 53. KV Hydroelectric power stations
  54. 54. KV Low-head power stations Hydroelectric power stations
  55. 55. KV Low-head power stations Hydroelectric power stations Run-of-river power stations
  56. 56. KV Low-head power stations Hydroelectric power stations Run-of-river power stations Detached power stations Joined power stations Submerged power stations
  57. 57. KV Low-head power stations Hydroelectric power stations Run-of-river power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations
  58. 58. KV Low-head power stations Hydroelectric power stations Medium-head power stations High-head power stations Run-of-river power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations
  59. 59. KV Low-head power stations Hydroelectric power stations Medium-head power stations High-head power stations Run-of-river power stations Storage power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations
  60. 60. KV Low-head power stations Hydroelectric power stations Medium-head power stations High-head power stations Run-of-river power stations Storage power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations Storage power stations
  61. 61. KV Low-head power stations Hydroelectric power stations Medium-head power stations High-head power stations Run-of-river power stations Storage power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations Storage power stations Series of power stations with head reservoir
  62. 62. KV Low-head power stations Hydroelectric power stations Medium-head power stations High-head power stations Run-of-river power stations Storage power stations Detached power stations Joined power stations Submerged power stations Run-of-river power stations Storage power stations Series of power stations with head reservoir
  63. 63. KV SYSTEM COMPONENTS Dams - are fixed structure and enables a controlled flow of water from the reservoir to the powerhouse. Weirs - can be either fixed or movable Barrages - have moveable gates Reservoirs - A supplementary supply of water Intake, penstock, powerhouse, tailrace (discussed above)
  64. 64. KV HYDROPOWER {turbines} Keith Vaugh BEng (AERO) MEng
  65. 65. KV
  66. 66. KV No increase in pressure, i.e. atmospheric pressure is maintained throughout the process Use nozzles to convert total head into kinetic energy Jets of fluid strikes vanes located on the periphery of a rotatable disk The rate of change of angular momentum results in work being done, thereby creating energy IMPULSE TURBINE’s
  67. 67. KV
  68. 68. KV REACTION TURBINE’s Fluid entering has both kinetic and pressure energy Two sets of vanes located around the periphery of rings, one being fixed, the other rotatable The relative velocity of the fluid increases as it passes through the runner A pressure differential arises across the runner.
  69. 69. KV
  70. 70. KV p1 ρg + u1 2 2g = E + p2 ρg + u2 2 2g Applying Bernoulli’s equation at the inlet ① and outlet ② of a reaction turbine
  71. 71. KV p1 ρg + u1 2 2g = E + p2 ρg + u2 2 2g Applying Bernoulli’s equation at the inlet ① and outlet ② of a reaction turbine where E is the energy transferred by the fluid to the turbine per unit weight, therefore E = p1 − p2( ) ρg + u1 2 − u2 2 ( ) 2g
  72. 72. KV p1 ρg + u1 2 2g = E + p2 ρg + u2 2 2g Applying Bernoulli’s equation at the inlet ① and outlet ② of a reaction turbine where E is the energy transferred by the fluid to the turbine per unit weight, therefore E = p1 − p2( ) ρg + u1 2 − u2 2 ( ) 2g Degree of reaction (R) R = Static pressure drop Total energy transfer
  73. 73. KV
  74. 74. KV But the static pressure is given by; p1 − p2( ) ρg = E − u1 2 − u2 2 ( ) 2g
  75. 75. KV But the static pressure is given by; therefore; p1 − p2( ) ρg = E − u1 2 − u2 2 ( ) 2g R = E − u1 2 − u2 2 ( ) 2g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ E =1− u1 2 − u2 2 ( ) 2gE
  76. 76. KV But the static pressure is given by; therefore; p1 − p2( ) ρg = E − u1 2 − u2 2 ( ) 2g R =1− u1 2 − u2 2 ( ) 2guw1 v1 R = E − u1 2 − u2 2 ( ) 2g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ E =1− u1 2 − u2 2 ( ) 2gE Substituting from Eular’s equation for E = uw1v1/g
  77. 77. KV
  78. 78. KV Table 1: Comparison of water turbines, Douglas et al (2005) PELTON WHEEL FRANCIS KAPLAN Number ωs range (rad) 0.05 - 0.4 0.4 - 2.2 1.8 - 4.6 Operating total head (m) 100 - 1700 80 - 500 Up to 400 Maximum power output (MW) 55 40 30 Best efficiency (%) 93 94 94 Regulation mechanism Spear nozzle and deflector plate Guide vanes, surge tanks Blade stagger
  79. 79. KV V Francis turbines approximately 30 to 700 m V Kaplan turbines, vertical axis approximately 10 to 60 m V Kaplan turbines, horizontal axis approximately 2 to 20 m Cross flow turbine Diagonal turbine Bulb turbine Pelton turbine 50 kW 100 kW 200 kW 500 kW 1 MW 2 M W 5 MW 10 MW 20 MW 50 MW 100 MW 200 MW 500 MW 1,000 MW Power2,000 M W 2Nozzles 4Nozzles 6Nozzles 1Nozzle 200 140 100 50 20 10 5 300 500 700 1,000 1,400 2,000 20105210.5 50 100 500200 1,000 Vertical Kaplan turbine Headinm Flow rate in m³/s Francis turbine Fig. 8.8 Application of different turbine types (see /8-6/) Application of different turbine types, Giesecke, J. et al (2005)
  80. 80. KV Efficiency curve for different turbine types, Giesecke, J. et al (2005) Layout and function of different turbine types are discussed in more detail be ow. 0.0 0.2 0.4 0.6 0.8 1.0 0 20 40 60 80 100 Ratio of flow to design flow Efficiencyin% Pelton turbine Kaplanturbine Francis turbine (Low-speed)Francis turbine (High-speed) Propellerturbine Crossflowturbine 10 30 50 70 90 0.1 0.3 0.5 0.7 0.9 ig. 8.9 Efficiency curve of different turbine types (see /8-6/) aplan, propeller, bulb, bevel gear, S and Straflo-turbines. The Kaplan turbin
  81. 81. KV
  82. 82. KV PELTON WHEEL 8 Hydroelectric Power Generation the buckets (Fig. 8.12). During this process the entire pressure energy of the r is converted into kinetic energy when leaving the nozzle. This energy is erted into mechanical energy by the Pelton wheel; the water then drops more ss without energy into the reservoir underneath the runner. 8.12 Power station with a Pelton turbine (see /8-11/)
  83. 83. KV PELTON WHEEL 8 Hydroelectric Power Generation the buckets (Fig. 8.12). During this process the entire pressure energy of the r is converted into kinetic energy when leaving the nozzle. This energy is erted into mechanical energy by the Pelton wheel; the water then drops more ss without energy into the reservoir underneath the runner. 8.12 Power station with a Pelton turbine (see /8-11/) Generator
  84. 84. KV PELTON WHEEL 8 Hydroelectric Power Generation the buckets (Fig. 8.12). During this process the entire pressure energy of the r is converted into kinetic energy when leaving the nozzle. This energy is erted into mechanical energy by the Pelton wheel; the water then drops more ss without energy into the reservoir underneath the runner. 8.12 Power station with a Pelton turbine (see /8-11/) Generator Nozzle
  85. 85. KV PELTON WHEEL 8 Hydroelectric Power Generation the buckets (Fig. 8.12). During this process the entire pressure energy of the r is converted into kinetic energy when leaving the nozzle. This energy is erted into mechanical energy by the Pelton wheel; the water then drops more ss without energy into the reservoir underneath the runner. 8.12 Power station with a Pelton turbine (see /8-11/) Generator Pelton Wheel Nozzle
  86. 86. KV Inlet triangle Outlet triangle u1 u1 v1 ur1 v2 uw2 ur2 u2 θ Outlet Nozzle uw1 = u1 v
  87. 87. KV u = Cu 2gH( )Recall Inlet triangle Outlet triangle u1 u1 v1 ur1 v2 uw2 ur2 u2 θ Outlet Nozzle uw1 = u1 v
  88. 88. KV u = Cu 2gH( )Recall The total energy transferred to the wheel is given by Euler’s equation E = 1 g v1 uw1 − v2 uw2( ) Inlet triangle Outlet triangle u1 u1 v1 ur1 v2 uw2 ur2 u2 θ Outlet Nozzle uw1 = u1 v
  89. 89. KV
  90. 90. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes
  91. 91. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes however; uw2 = v − ur2 cos 180 −ϑ( )= v + ur2 cosϑ and ur2 = kur1 = k u − v( )
  92. 92. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes however; uw2 = v − ur2 cos 180 −ϑ( )= v + ur2 cosϑ and ur2 = kur1 = k u − v( ) k represents the reduction of the relative velocity due to friction, therefore
  93. 93. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes however; uw2 = v − ur2 cos 180 −ϑ( )= v + ur2 cosϑ and ur2 = kur1 = k u − v( ) k represents the reduction of the relative velocity due to friction, therefore uw2 = v + k u1 − v( )cosϑ and uw1 = u1
  94. 94. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes however; uw2 = v − ur2 cos 180 −ϑ( )= v + ur2 cosϑ and ur2 = kur1 = k u − v( ) k represents the reduction of the relative velocity due to friction, therefore uw2 = v + k u1 − v( )cosϑ and uw1 = u1 E = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v − k u1 − v( )cosϑ⎡ ⎣ ⎤ ⎦ = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 1− kcosϑ( )− v 1− kcosϑ( )⎡ ⎣ ⎤ ⎦ = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v( )1− kcosϑ( )
  95. 95. KV
  96. 96. KV E = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v( )1− kcosϑ( ) = 1− kcosϑ( ) g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ vu1 − v2 ( )
  97. 97. KV Therefore for a maximum, i.e. E = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v( )1− kcosϑ( ) = 1− kcosϑ( ) g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ vu1 − v2 ( ) dE dv = 1− kcosϑ( ) g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ u1 − 2v( )= 0
  98. 98. KV Therefore for a maximum, i.e. E = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v( )1− kcosϑ( ) = 1− kcosϑ( ) g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ vu1 − v2 ( ) dE dv = 1− kcosϑ( ) g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ u1 − 2v( )= 0 u1 − 2v = 0 v = 1 2 u1 Hence,
  99. 99. KV
  100. 100. KV Substituting for v back into Euler’s modified equation an expression for maximum energy transfer can be obtained E = u1 2g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − 1 2 u1 ⎛ ⎝⎜ ⎞ ⎠⎟ 1− kcosϑ( ) = u1 2 4g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− kcosϑ( )
  101. 101. KV Substituting for v back into Euler’s modified equation an expression for maximum energy transfer can be obtained E = u1 2g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − 1 2 u1 ⎛ ⎝⎜ ⎞ ⎠⎟ 1− kcosϑ( ) = u1 2 4g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− kcosϑ( ) The energy from the nozzle is kinetic energy,KEjet = u1 2 2g
  102. 102. KV Substituting for v back into Euler’s modified equation an expression for maximum energy transfer can be obtained E = u1 2g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − 1 2 u1 ⎛ ⎝⎜ ⎞ ⎠⎟ 1− kcosϑ( ) = u1 2 4g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− kcosϑ( ) The energy from the nozzle is kinetic energy,KEjet = u1 2 2g The maximum theoretical efficiency of the Pelton wheel becomes ηmax = Emax KEjet = u1 2 4g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1− kcosϑ( ) u1 2 2g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1− kcosϑ( ) 2
  103. 103. KV
  104. 104. KV FRANCIS TURBINE
  105. 105. KV FRANCIS TURBINE eads as low as 2 m (see Fig. 8.8). If these plants were newly built nowadays, ouble regulation Kaplan tubular turbines or S-turbines would be used. With their ery good efficiency curve over a broad range of discharges, they guarantee an ptimum exploitation of energy. If old plants are reactivated, the entire in and out- low areas would have to be adjusted. The work involved in this is often so ex- ensive that Francis machines are put in again, although they have a slightly less avourable efficiency curve. ig. 8.11 Power station with a vertical Francis turbine (see /8-10/)
  106. 106. KV }
  107. 107. KV
  108. 108. KV Flow from penstock
  109. 109. KV Scroll Flow from penstock
  110. 110. KV Scroll Stationary vanes Flow from penstock
  111. 111. KV Scroll Stationary vanes Flow from penstock Impeller
  112. 112. KV Adjustable guide vanesScroll Stationary vanes Flow from penstock Impeller
  113. 113. KV Adjustable guide vanesScroll Stationary vanes Draft tube Flow from penstock Impeller
  114. 114. KV R2 R1 β2 v1 ω β1 Ro Guide vane ring Runner Runner blade uf1 v2 ur2 uw1 u2=uf2 u1 ur1 u0 u0 ϑ
  115. 115. KV
  116. 116. KV Total head available is H and the fluid velocity entering is u0.The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation;
  117. 117. KV However, Total head available is H and the fluid velocity entering is u0.The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0 A0 = uf1 A1 u0 A0 = u1 A1 sinϑ uf1 = u1 sinϑ
  118. 118. KV However, Total head available is H and the fluid velocity entering is u0.The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0 A0 = uf1 A1 u0 A0 = u1 A1 sinϑ uf1 = u1 sinϑ The direction of u1 is governed by the guide vane angle ϑ.The angle ϑ is selected so that the relative velocity meets the runner tangentially, i.e. it makes an angle β1 with the tangent at blade inlet.Therefore; tanϑ = uf1 uw1 and tanβ1 = uf1 v1 − uw1( )
  119. 119. KV
  120. 120. KV Eliminating uw1 from the previous two equations; tanβ1 = uf1 v1 − uf1 tanϑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or cot β1 = v1 uf1 − cotϑ
  121. 121. KV Therefore, Eliminating uw1 from the previous two equations; tanβ1 = uf1 v1 − uf1 tanϑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or cot β1 = v1 uf1 − cotϑ v1 uf1 = cot β1 + cotϑ or v1 = uf1 cot β1 + cotϑ( )
  122. 122. KV Therefore, Eliminating uw1 from the previous two equations; The total energy at inlet to the impeller consists of the velocity head and the pressure head H1. In the impeller the fluid energy is decreased by E, which is transferred to the runner.Water leaves the impeller with kinetic energy tanβ1 = uf1 v1 − uf1 tanϑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or cot β1 = v1 uf1 − cotϑ v1 uf1 = cot β1 + cotϑ or v1 = uf1 cot β1 + cotϑ( ) H = u1 2 2g + H1 + h1 ' and H = E + u2 2 2g + h1
  123. 123. KV
  124. 124. KV Euler’s equation yields the energy transferred E, for the maximum energy transfer uw2 = 0, therefore E = uw1 v1 g
  125. 125. KV The condition of no whirl component at outlet may be achieved by making the outlet blade angel β2, such that the absolute velocity at outlet u2 is radial. Therefore from the velocity triangle it follows; Euler’s equation yields the energy transferred E, for the maximum energy transfer uw2 = 0, therefore E = uw1 v1 g tanβ2 = u2 v2
  126. 126. KV The condition of no whirl component at outlet may be achieved by making the outlet blade angel β2, such that the absolute velocity at outlet u2 is radial. Therefore from the velocity triangle it follows; Euler’s equation yields the energy transferred E, for the maximum energy transfer uw2 = 0, therefore E = uw1 v1 g uf1 A1 = uf 2 A2 tanβ2 = u2 v2 since uw2 = 0, then u2 = uf2 and by the continuity equation; so that β2 can be determined
  127. 127. KV
  128. 128. KV If the condition of no whirl at outlet is satisfied, then the second energy equation takes the form H = uw1 v1 g + u2 2 2g + h1
  129. 129. KV The hydraulic efficiency is given by, If the condition of no whirl at outlet is satisfied, then the second energy equation takes the form ηh = E H = uw1 v1 gH H = uw1 v1 g + u2 2 2g + h1
  130. 130. KV The hydraulic efficiency is given by, If the condition of no whirl at outlet is satisfied, then the second energy equation takes the form the overall efficiency is given by ηh = E H = uw1 v1 gH H = uw1 v1 g + u2 2 2g + h1 η = P !mgH
  131. 131. KV
  132. 132. KV AXIAL FLOW TURBINE’s turbine. The disadvantage of this design is the costly sealing between the runner and the generator. Because of their design a flat efficiency curve at an overall high level can be realised for Straflo-turbines (Fig. 8.9). Due to their double regulation, all Kaplan turbines and derived designs can be operated over a broad partial-load range (30 to 100 % of the design power) with comparably high efficiency levels. Trash rack Guide vanes Runner Draft tube Generator stator Mounting cran Generator rotor Fig. 8.10 Run-of-river power station with Straflo turbine (see /8-9/)
  133. 133. KV
  134. 134. KV Casing Casing
  135. 135. KV Casing Casing Guide vanes
  136. 136. KV Casing Casing Guide vanes Impeller
  137. 137. KV Casing Casing Guide vanes Impeller Draft tube
  138. 138. KV v = ωr ur2 uw1 ϑ u2 =uf v v ur1 u2 uf
  139. 139. KV v = ωr ur2 uw1 ϑ u2 =uf v v ur1 u2 uf If the angular velocity of the impeller is ω then blade velocity at radius r is given by v = ωr Since at maximum efficiency uw2 = 0 and u2 = uf it follows E = vuw1 g
  140. 140. KV
  141. 141. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power.
  142. 142. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power.
  143. 143. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power. Hydroelectric power accounts for over 90% of the total electricity supply in some countries including Brazil & Norway,
  144. 144. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power. Hydroelectric power accounts for over 90% of the total electricity supply in some countries including Brazil & Norway,
  145. 145. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power. Hydroelectric power accounts for over 90% of the total electricity supply in some countries including Brazil & Norway, Long-lasting with relatively low maintenance requirements: many systems have been in continuous use for over fifty years and some installations still function after 100 years.
  146. 146. KV
  147. 147. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant.
  148. 148. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant.
  149. 149. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant. The complications of hydro-power systems arise mostly from associated dams and reservoirs, particularly on the large-scale projects.
  150. 150. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant. The complications of hydro-power systems arise mostly from associated dams and reservoirs, particularly on the large-scale projects.
  151. 151. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant. The complications of hydro-power systems arise mostly from associated dams and reservoirs, particularly on the large-scale projects. Most rivers, including large rivers with continental-scale catchments, such as the Nile, the Zambesi and theYangtze, have large seasonal flows making floods a major characteristic.
  152. 152. KV The relatively large initial capital cost has long since been written off, the ‘levelised’ cost of power produced is less than non-renewable sources requiring expenditure on fuel and more frequent replacement of plant. The complications of hydro-power systems arise mostly from associated dams and reservoirs, particularly on the large-scale projects. Most rivers, including large rivers with continental-scale catchments, such as the Nile, the Zambesi and theYangtze, have large seasonal flows making floods a major characteristic.
  153. 153. KV
  154. 154. KV Therefore most large dams are (i.e. those >15m high) are built for more than one purpose, apart from the significant aim of electricity generation, e.g. water storage for potable supply and irrigation, controlling river flow and mitigating floods, road crossings, leisure activities and fisheries.
  155. 155. KV Therefore most large dams are (i.e. those >15m high) are built for more than one purpose, apart from the significant aim of electricity generation, e.g. water storage for potable supply and irrigation, controlling river flow and mitigating floods, road crossings, leisure activities and fisheries.
  156. 156. KV Therefore most large dams are (i.e. those >15m high) are built for more than one purpose, apart from the significant aim of electricity generation, e.g. water storage for potable supply and irrigation, controlling river flow and mitigating floods, road crossings, leisure activities and fisheries. Countering the benefits of hydroelectric power are excessive debt burden (dams are often the largest single investment project in a country), cost over-runs, displacement and impoverishment of people, destruction of important eco-systems and fishery resources, and the inequitable sharing of costs and benefits.
  157. 157. KV Therefore most large dams are (i.e. those >15m high) are built for more than one purpose, apart from the significant aim of electricity generation, e.g. water storage for potable supply and irrigation, controlling river flow and mitigating floods, road crossings, leisure activities and fisheries. Countering the benefits of hydroelectric power are excessive debt burden (dams are often the largest single investment project in a country), cost over-runs, displacement and impoverishment of people, destruction of important eco-systems and fishery resources, and the inequitable sharing of costs and benefits.
  158. 158. KV Therefore most large dams are (i.e. those >15m high) are built for more than one purpose, apart from the significant aim of electricity generation, e.g. water storage for potable supply and irrigation, controlling river flow and mitigating floods, road crossings, leisure activities and fisheries. Countering the benefits of hydroelectric power are excessive debt burden (dams are often the largest single investment project in a country), cost over-runs, displacement and impoverishment of people, destruction of important eco-systems and fishery resources, and the inequitable sharing of costs and benefits. For example, over 3 million people were displaced by the construction of the Three Gorges dam in China....
  159. 159. KV Measurement of flow Hydroelectric power generation ! Plant configuration ! Governing equations ! Energy line ! Plant components - hydro elements ! Categorisation Euler’s equation Turbine’s ! Pelton wheel ! Francis turbine ! Axial flow turbine Social and environmental aspects
  160. 160. KV Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University Press Bacon, D., Stephens, R. (1990) MechanicalTechnology, second edition, Butterworth Heinemann Boyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University Press Çengel,Y.,Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw Hill Douglas, J., Gasiorek, J., Swaffield, J., Jack, L. (2005) Fluid mechanics, fifth edition, Pearson Education Turns, S. (2006) Thermal fluid sciences:An integrated approach, Cambridge University Press Twidell, J. and Weir,T. (2006) Renewable energy resources, second edition, Oxon:Taylor and Francis Illustrations taken from Energy science: principles, technologies and impacts & Fundamentals of thermal fluid science

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