- 1. KV HYDROPOWER Keith Vaugh BEng (AERO) MEng
- 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. 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. 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. 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. 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. 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. 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. 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. 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. KV Applying Bernoulli’s equation two reference points ① and ②, up and downstream of the hydroelectric power plant;
- 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. KV
- 14. KV
- 15. KV
- 16. KV
- 17. KV Headwater
- 18. KV Headwater Dam
- 26. KV Headwater Turbine Dam Screen Penstock Generator Power house Stop logs Stop valve Draft tube
- 27. KV Tailwater Headwater Turbine Dam Screen Penstock Generator Power house Stop logs Stop valve Draft tube
- 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. KV PENSTOCK
- 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. KV TURBINE
- 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. KV TAILRACE
- 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. KV
- 40. KV Tailwater Headwater Energy loss Energy line Energy line Datum
- 41. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Datum
- 42. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Geodetic elevation of the stream Datum
- 43. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressure head Geodetic elevation of the stream Datum
- 44. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressurehead Pressure head Geodetic elevation of the stream Datum
- 45. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Pressurehead Pressure head Velocity head Geodetic elevation of the stream Datum
- 46. KV Tailwater Headwater Energy loss Velocity head Energy line Energy line Useablehead Pressurehead Pressure head Velocity head Geodetic elevation of the stream Datum
- 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. 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. 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. 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. KV
- 54. KV Low-head power stations Hydroelectric power stations
- 55. KV Low-head power stations Hydroelectric power stations Run-of-river power stations
- 56. KV Low-head power stations Hydroelectric power stations Run-of-river power stations Detached power stations Joined power stations Submerged power stations
- 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. 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. 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. 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. 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. 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. 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. KV HYDROPOWER {turbines} Keith Vaugh BEng (AERO) MEng
- 65. KV
- 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. KV
- 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. KV
- 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. 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. 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. KV
- 74. KV But the static pressure is given by; p1 − p2( ) ρg = E − u1 2 − u2 2 ( ) 2g
- 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. 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. KV
- 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. 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. 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. KV
- 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. 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. 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. 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. KV Inlet triangle Outlet triangle u1 u1 v1 ur1 v2 uw2 ur2 u2 θ Outlet Nozzle uw1 = u1 v
- 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. 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. KV
- 90. KV E = v g uw1 − uw2( ) v1 = v2 = v, therefore Euler’s equation becomes
- 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. 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. 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. 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. KV
- 96. KV E = v g ⎛ ⎝⎜ ⎞ ⎠⎟ u1 − v( )1− kcosϑ( ) = 1− kcosϑ( ) g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ vu1 − v2 ( )
- 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. 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. KV
- 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. 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. 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. KV
- 104. KV FRANCIS TURBINE
- 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. KV }
- 107. KV
- 113. KV Adjustable guide vanesScroll Stationary vanes Draft tube Flow from penstock Impeller
- 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. KV
- 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. 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. 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. KV
- 120. KV Eliminating uw1 from the previous two equations; tanβ1 = uf1 v1 − uf1 tanϑ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or cot β1 = v1 uf1 − cotϑ
- 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. 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. KV
- 124. KV Euler’s equation yields the energy transferred E, for the maximum energy transfer uw2 = 0, therefore E = uw1 v1 g
- 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. 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. KV
- 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. 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. 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. KV
- 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. KV
- 134. KV Casing Casing
- 138. KV v = ωr ur2 uw1 ϑ u2 =uf v v ur1 u2 uf
- 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. KV
- 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. KV SOCIAL & ENVIRONMENTAL ASPECTS Hydroelectric power is a mature technology used in many countries, producing about 20% of the world’s electric power.
- 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. 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. 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. KV
- 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. 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. 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. 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. 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. 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. KV
- 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. 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. 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. 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. 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. 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. 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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- As fluid passes through the turbine, it spins producing power. The flow of this fluid is normally somewhat less than the flow in the moving body of water. However, due to seasonal variations the flow of the fluid body will vary. Therefore the norm is to match the turbine to the lower flow and ensure the turbine produces useful work year round.\n\nIt is imperative that an accurate measurement of flow rate is completed not only for the technical and economic requirements of the scheme, but also for a environmental impact appraisal. \n\nWhile the turbine is matched to the low flow season, it is also important to examine the high season flow to identify the maximum flow and the possibly of flooding which could cause damage to the installation and the surrounding areas. The measuring the flow rate of the moving body is difficult that measuring H \n\n
- As fluid passes through the turbine, it spins producing power. The flow of this fluid is normally somewhat less than the flow in the moving body of water. However, due to seasonal variations the flow of the fluid body will vary. Therefore the norm is to match the turbine to the lower flow and ensure the turbine produces useful work year round.\n\nIt is imperative that an accurate measurement of flow rate is completed not only for the technical and economic requirements of the scheme, but also for a environmental impact appraisal. \n\nWhile the turbine is matched to the low flow season, it is also important to examine the high season flow to identify the maximum flow and the possibly of flooding which could cause damage to the installation and the surrounding areas. The measuring the flow rate of the moving body is difficult that measuring H \n\n
- As fluid passes through the turbine, it spins producing power. The flow of this fluid is normally somewhat less than the flow in the moving body of water. However, due to seasonal variations the flow of the fluid body will vary. Therefore the norm is to match the turbine to the lower flow and ensure the turbine produces useful work year round.\n\nIt is imperative that an accurate measurement of flow rate is completed not only for the technical and economic requirements of the scheme, but also for a environmental impact appraisal. \n\nWhile the turbine is matched to the low flow season, it is also important to examine the high season flow to identify the maximum flow and the possibly of flooding which could cause damage to the installation and the surrounding areas. The measuring the flow rate of the moving body is difficult that measuring H \n\n
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- This method is similar to that applied when using a pitot tube to measure the pressures in a section of pipe other than a pipe having a circular cross section.\n
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- Transfer losses will arises within a hydroelectric power plant and as a consequence only a portion of the theoretical power will be utilised for the generation of electricity. Bernoulli&#x2019;s equation can be applied to illustrate this.\n
- &#x3B1; is the loss coefficient. The lost energy cannot be utilised and arises as a result of friction, i.e. friction converts it into heat. \n\nRecall the final example in the previous set of slides.\n
- &#x3B1; is the loss coefficient. The lost energy cannot be utilised and arises as a result of friction, i.e. friction converts it into heat. \n\nRecall the final example in the previous set of slides.\n
- &#x3B1; is the loss coefficient. The lost energy cannot be utilised and arises as a result of friction, i.e. friction converts it into heat. \n\nRecall the final example in the previous set of slides.\n
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- A typical hydroelectric power station can be divided into three main sections, the intake works, the penstock and the powerhouse/tailrace. The body of fluid is channeled through the intake works down the penstock into the turbine causing it to rotate. The rotating turbine in turn causes the generator to which it is coupled to rotate and thereby electricity is generated. The fluid flows out of the turbine along the draft tube and into the tail race. \n
- A typical hydroelectric power station can be divided into three main sections, the intake works, the penstock and the powerhouse/tailrace. The body of fluid is channeled through the intake works down the penstock into the turbine causing it to rotate. The rotating turbine in turn causes the generator to which it is coupled to rotate and thereby electricity is generated. The fluid flows out of the turbine along the draft tube and into the tail race. \n
- A typical hydroelectric power station can be divided into three main sections, the intake works, the penstock and the powerhouse/tailrace. The body of fluid is channeled through the intake works down the penstock into the turbine causing it to rotate. The rotating turbine in turn causes the generator to which it is coupled to rotate and thereby electricity is generated. The fluid flows out of the turbine along the draft tube and into the tail race. \n
- The intake structure channels the flow of fluid into the penstock. At the entrance to the converging section, a screen stops floating debris from entering the power station. Stop logs allow the power station to be drained for maintenance, while the stop valve (normally quick action) is included to stop flow in the event of an emergency. \n\nThe converging section of the intake structure (IS) results in a partial conversion of potential energy into kinetic energy. Flow losses arise in the intake structure and the these can be accounted for with the inclusion of the &#x3B1;IS correction factor. The velocity of the headwater entering the intake structure at station &#x2460; can be neglected, hence the term is dropped from the left side of the equation and the density of the fluid will be same upstream as downstream. The losses arising in the intake structure are therefore represented as decrease in the pressure level, \n\n
- The intake structure channels the flow of fluid into the penstock. At the entrance to the converging section, a screen stops floating debris from entering the power station. Stop logs allow the power station to be drained for maintenance, while the stop valve (normally quick action) is included to stop flow in the event of an emergency. \n\nThe converging section of the intake structure (IS) results in a partial conversion of potential energy into kinetic energy. Flow losses arise in the intake structure and the these can be accounted for with the inclusion of the &#x3B1;IS correction factor. The velocity of the headwater entering the intake structure at station &#x2460; can be neglected, hence the term is dropped from the left side of the equation and the density of the fluid will be same upstream as downstream. The losses arising in the intake structure are therefore represented as decrease in the pressure level, \n\n
- The intake structure channels the flow of fluid into the penstock. At the entrance to the converging section, a screen stops floating debris from entering the power station. Stop logs allow the power station to be drained for maintenance, while the stop valve (normally quick action) is included to stop flow in the event of an emergency. \n\nThe converging section of the intake structure (IS) results in a partial conversion of potential energy into kinetic energy. Flow losses arise in the intake structure and the these can be accounted for with the inclusion of the &#x3B1;IS correction factor. The velocity of the headwater entering the intake structure at station &#x2460; can be neglected, hence the term is dropped from the left side of the equation and the density of the fluid will be same upstream as downstream. The losses arising in the intake structure are therefore represented as decrease in the pressure level, \n\n
- The penstock links the intake structure with the powerhouse where the turbine is located. Essentially it is pipe and its length is dependent on the site and plant specifications. The diameter can varied i.e. a larger diameter results in a reduction in friction losses and therefore the turbine power increases. Selecting a larger diameter penstock increases the cost, hence a balance most be achieved technical requirements and economics of the project. \n \nAs fluid falls through the penstock, potential energy is converted into pressure energy. Frictional losses resulting from the friction factor and diameter arise in the penstock and these are accounted for in bernoulli&#x2019;s equation as &#x3B1;PS (the loss coefficient of the penstock). The loss coefficient increases proportionally to the length.\n\n
- The penstock links the intake structure with the powerhouse where the turbine is located. Essentially it is pipe and its length is dependent on the site and plant specifications. The diameter can varied i.e. a larger diameter results in a reduction in friction losses and therefore the turbine power increases. Selecting a larger diameter penstock increases the cost, hence a balance most be achieved technical requirements and economics of the project. \n \nAs fluid falls through the penstock, potential energy is converted into pressure energy. Frictional losses resulting from the friction factor and diameter arise in the penstock and these are accounted for in bernoulli&#x2019;s equation as &#x3B1;PS (the loss coefficient of the penstock). The loss coefficient increases proportionally to the length.\n\n
- The penstock links the intake structure with the powerhouse where the turbine is located. Essentially it is pipe and its length is dependent on the site and plant specifications. The diameter can varied i.e. a larger diameter results in a reduction in friction losses and therefore the turbine power increases. Selecting a larger diameter penstock increases the cost, hence a balance most be achieved technical requirements and economics of the project. \n \nAs fluid falls through the penstock, potential energy is converted into pressure energy. Frictional losses resulting from the friction factor and diameter arise in the penstock and these are accounted for in bernoulli&#x2019;s equation as &#x3B1;PS (the loss coefficient of the penstock). The loss coefficient increases proportionally to the length.\n\n
- The turbine the fluid flows over blades. The force exerted on the blades derivers from the rate of change of momentum of the fluid and this in turn produces a torque on the rotor shaft. \n\nThe equation represents the part of the usable water power that can be converted into mechanical energy at the turbine shaft, PTurbine. In this equation, hutil is the usable head at the turbine, &#x3B7;Turbine the efficiency of the turbine and the product of the remaining terms represents the Pwa,act. The losses arising in the turbine are differentiated as volumetric losses, losses due to turbulance and friction. Therefore the power at the turbine shaft PTurbine is less than the usable water power Pwa,act.\n\n
- The turbine the fluid flows over blades. The force exerted on the blades derivers from the rate of change of momentum of the fluid and this in turn produces a torque on the rotor shaft. \n\nThe equation represents the part of the usable water power that can be converted into mechanical energy at the turbine shaft, PTurbine. In this equation, hutil is the usable head at the turbine, &#x3B7;Turbine the efficiency of the turbine and the product of the remaining terms represents the Pwa,act. The losses arising in the turbine are differentiated as volumetric losses, losses due to turbulance and friction. Therefore the power at the turbine shaft PTurbine is less than the usable water power Pwa,act.\n\n
- The turbine the fluid flows over blades. The force exerted on the blades derivers from the rate of change of momentum of the fluid and this in turn produces a torque on the rotor shaft. \n\nThe equation represents the part of the usable water power that can be converted into mechanical energy at the turbine shaft, PTurbine. In this equation, hutil is the usable head at the turbine, &#x3B7;Turbine the efficiency of the turbine and the product of the remaining terms represents the Pwa,act. The losses arising in the turbine are differentiated as volumetric losses, losses due to turbulance and friction. Therefore the power at the turbine shaft PTurbine is less than the usable water power Pwa,act.\n\n
- The energy line for the tailwater is determined by its geodetic level and the ambient pressure. As the fluid enters the tailwater it losses the remaining kinetic energy through turbulence. When the energy line is constructed a sudden drop in energy is evident at station &#x2464;.\n\nThe draft tube diverges and therefore the velocity at station &#x2464; must be less than the velocity at station &#x2463;. As a consequence the pressure p4 at the turbine outlet will be lower than the pressure at the end of the draft tube, p5, i.e. at station &#x2464;. It can be deduced therefore that the turbulences reduces the losses and the head is better utilised.\n\n
- The energy line for the tailwater is determined by its geodetic level and the ambient pressure. As the fluid enters the tailwater it losses the remaining kinetic energy through turbulence. When the energy line is constructed a sudden drop in energy is evident at station &#x2464;.\n\nThe draft tube diverges and therefore the velocity at station &#x2464; must be less than the velocity at station &#x2463;. As a consequence the pressure p4 at the turbine outlet will be lower than the pressure at the end of the draft tube, p5, i.e. at station &#x2464;. It can be deduced therefore that the turbulences reduces the losses and the head is better utilised.\n\n
- The energy line for the tailwater is determined by its geodetic level and the ambient pressure. As the fluid enters the tailwater it losses the remaining kinetic energy through turbulence. When the energy line is constructed a sudden drop in energy is evident at station &#x2464;.\n\nThe draft tube diverges and therefore the velocity at station &#x2464; must be less than the velocity at station &#x2463;. As a consequence the pressure p4 at the turbine outlet will be lower than the pressure at the end of the draft tube, p5, i.e. at station &#x2464;. It can be deduced therefore that the turbulences reduces the losses and the head is better utilised.\n\n
- Bernoulli&#x2019;s equation allows a graphic representation of the energy and losses to be constructed. The dotted line represents the geodetic level of the fluid flowing through the power station. The energy line which originates at the headwater surface level and terminates at the tailwater surface level shows the locations and respective energy losses. The distance to the dotted lined below the energy line illustrates the kinetic energy of the fluid i.e. the velocity head. The converging section at the intake causes the velocity to increase. The difference between the geodetic level and the dotted line represents the pressure head. \n
- Bernoulli&#x2019;s equation allows a graphic representation of the energy and losses to be constructed. The dotted line represents the geodetic level of the fluid flowing through the power station. The energy line which originates at the headwater surface level and terminates at the tailwater surface level shows the locations and respective energy losses. The distance to the dotted lined below the energy line illustrates the kinetic energy of the fluid i.e. the velocity head. The converging section at the intake causes the velocity to increase. The difference between the geodetic level and the dotted line represents the pressure head. \n
- Bernoulli&#x2019;s equation allows a graphic representation of the energy and losses to be constructed. The dotted line represents the geodetic level of the fluid flowing through the power station. The energy line which originates at the headwater surface level and terminates at the tailwater surface level shows the locations and respective energy losses. The distance to the dotted lined below the energy line illustrates the kinetic energy of the fluid i.e. the velocity head. The converging section at the intake causes the velocity to increase. The difference between the geodetic level and the dotted line represents the pressure head. \n
- In a hydroelectric power plant, the losses mainly occur at the intake structure, the penstock and possibly at the outflow. The actual Pwa,act is calculated by deducting the various losses from the theoretical water power. \n\nLosses are dependent on the flow velocity and therefore these can be minimised with an optimised plant design and layout \n\n
- In a hydroelectric power plant, the losses mainly occur at the intake structure, the penstock and possibly at the outflow. The actual Pwa,act is calculated by deducting the various losses from the theoretical water power. \n\nLosses are dependent on the flow velocity and therefore these can be minimised with an optimised plant design and layout \n\n
- Hydroelectric power plants can be categorised as low, medium or high head power stations. Additionally, these power plants can be categorised as run-of-river or hydroelectric power stations with reservoirs. The definition between small and large power plants is somewhat blurred with different geographical region, e.g. in Germany anything greater than 1MW is categorised as large whereas in Russia anything greater that 10MW gets classification. \n
- Low-head hydroelectric power plants can be further divided into two distinct configurations. \nDiversion type - The power &#x201C;station&#x201D; (as distinct from power house) is located outside the riverbed, typically along the course of a man made canal into which the water flow is diverted. The flow is diverted at the a dam into a head race or pipeline, channeled to the power &#x201C;station&#x201D; where power is extracted by turbines, and the transferred back into the river at the tailrace. \n\nIt can be argued that the configuration of the Bonneville Dam is either a run of river or diversion type. \n
- Run-of-River - The power station is built directly into the riverbed. This configuration services multiple purposes, electrical generation, flood management, navigation and groundwater stabilisation. Run-of-River configurations can have alternative arrangements:\n\nConventional block design - The powerhouse and the dam are perpendicular to the flow of the river. This design is only suitable if there is no risk of upstream flooding.\nIndented power station - In this case the powerhouse is positioned in an artificial bay outside the riverbed and is preferred arrangement for narrow rivers, i.e. the dam can use the entire width of the river.\nTwin block power station - This configuration utilises two power houses, one on either side of the dam. This is attractive arrangement for rivers which form a border between two countries, i.e. both can have an independent powerhouse. \nPower station in pier - As the name suggests, the mechanical systems and powerhouse are build into the piers. This saves space, however it&#x2019;s selection is dependent on favourable flow conveyance characteristics.\nSubmersible - Power station and dam are built in one block.\n
- Auxiliary plants have recently gained popularity. These can be found in drinking water supply systems. The water is transported from a high level reservoir to the consumer via high pressure piping networks. Turbines or pumps operating in reverse are installed into such piping networks and therefore, surplus energy can be extract. These plants are attractive given that the turbine or reversible pump is the only additional costs incurred. The economic and environmental benefits out weight the initial cost. \n
- Auxiliary plants have recently gained popularity. These can be found in drinking water supply systems. The water is transported from a high level reservoir to the consumer via high pressure piping networks. Turbines or pumps operating in reverse are installed into such piping networks and therefore, surplus energy can be extract. These plants are attractive given that the turbine or reversible pump is the only additional costs incurred. The economic and environmental benefits out weight the initial cost. \n
- Auxiliary plants have recently gained popularity. These can be found in drinking water supply systems. The water is transported from a high level reservoir to the consumer via high pressure piping networks. Turbines or pumps operating in reverse are installed into such piping networks and therefore, surplus energy can be extract. These plants are attractive given that the turbine or reversible pump is the only additional costs incurred. The economic and environmental benefits out weight the initial cost. \n
- The Dam is the interface between the reservoir and the penstock. In essence these structures allow a large volume of water to build up. This water can then be released in a controlled manner. It is essential the dam and associated spillway are also capable of handling seasonal variations, maintaining an adequate reservoir level at all times and conveying floods if and when such arise. \n\nDams can be constructed in the form of fixed (and in some cases movable) weirs, barrages, embankments of rock and/or earth, or mass concrete. \n\nIf the head water needs to be kept at a constant in small hydroelectric power plants (typically run of river configurations), weirs or barrages with movable gates are selected. If the flow exceeds the design specification of the turbines then the excess water can be released by opening the gates. \n\nWhere the headwater does not need to be maintained (typically diversion configurations) dams without moveable gates are appropriate. \n\nReservoirs can occur naturally (lakes) or can be man made. They help create a balance between the fluctuating water supply and electrical demand. Pumped storage stations can store surplus supply for peak load power requirements. \n
- The next major component found in a hydroelectric power plant is the rotodynamic element, i.e. the turbine. Before discussing the various configurations it is prudent to review the theory associated with rotodynamic machines, i.e. pumps, centrifugal blowers/fans, turbines, etc....\n
- The next major component found in a hydroelectric power plant is the rotodynamic element, i.e. the turbine. Before discussing the various configurations it is prudent to review the theory associated with rotodynamic machines, i.e. pumps, centrifugal blowers/fans, turbines, etc....\n
- The next major component found in a hydroelectric power plant is the rotodynamic element, i.e. the turbine. Before discussing the various configurations it is prudent to review the theory associated with rotodynamic machines, i.e. pumps, centrifugal blowers/fans, turbines, etc....\n
- Flow through a radial device maybe analysed as shown in the diagram. \n \nFluid moving with absolute velocity u1 enters an impeller at an inlet through a cylindrical surface. This surface has a radius r1 and the fluid entering makes an angle &#x3B1;1 with the tangent to this surface at the entry point. A similar trajectory can be expressed for the fluid leaving the impeller at the exit, i.e. fluid leaves through a cylindrical surface of radius r2 with an absolute velocity u2 inclined to the tangent of this surface at the exit point by an angle &#x3B1;2. &#x201C;Velocity triangles&#x201D; for the inlet and outlet can constructed. \n\n\n
- The velocity triangle for the inlet is obtained by;\n1>Construct the vector representing the absolute velocity u1 at an angle &#x3B1;1 with the tangent to the cylindrical surface at the entry point\n2>Construct the vector representing the impeller velocity, v1 \n3>Subtract vectorially the vector representing the impeller velocity, v1 from the vector representing the absolute velocity u1 which will yield the relative velocity vector, ur1 at radius r1\n4>The velocity vector u1, is resolved into two components, i.e. the velocity of flow uf1 (radial) and perpendicular to this the velocity of whirl uw1 (tangential)\n\n
- The velocity triangle at the outlet is obtained by;\n1>Construct the vector representing the absolute velocity u2 at an angle &#x3B1;2 with the tangent to the cylindrical surface at the exit point i.e. on arc r2\n2>Construct the vector representing the blade velocity, v2 i.e. the tangential velocity vector\n3>Subtract vectorially the vector representing the blade velocity, v2 from the vector representing the absolute velocity u2 which will yield the relative velocity vector, ur2 at radius r2\n4>The absolute fluid velocity is resolved into two components, i.e. the velocity of flow vector uf2 (radial) and perpendicular to this the velocity of whirl uw2 (tangential)\n\n
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- &#x3C9; = v/r, so that &#x3C9;r2 = v2 and &#x3C9;r1 = v1, i.e. the respective angular velocities multiplied by the radius equals the tangential velocity at inlet or outlet.\n\nEuler&#x2019;s equation has units of (j&#x2219;kg-1)/m2. This equation will simplify to m and as all the terms are in the same unit as Bernoulli&#x2019;s equation can be used in conjunction with it. Euler&#x2019;s equation can be applied to pumps and turbines, however in the the case of turbines since uw1v1 > uw2v2, E would be negative indicating the reversed direction of energy. In order to obtain a positive result the order of terms in brackets is often reversed. \n\nGiven that E reduces to meters of fluid handled, it is often referred to as Euler&#x2019;s head. When the equation is being applied to pumps, blowers, fans, etc..., it represents the ideal theoretical head developed Hth\n
- &#x3C9; = v/r, so that &#x3C9;r2 = v2 and &#x3C9;r1 = v1, i.e. the respective angular velocities multiplied by the radius equals the tangential velocity at inlet or outlet.\n\nEuler&#x2019;s equation has units of (j&#x2219;kg-1)/m2. This equation will simplify to m and as all the terms are in the same unit as Bernoulli&#x2019;s equation can be used in conjunction with it. Euler&#x2019;s equation can be applied to pumps and turbines, however in the the case of turbines since uw1v1 > uw2v2, E would be negative indicating the reversed direction of energy. In order to obtain a positive result the order of terms in brackets is often reversed. \n\nGiven that E reduces to meters of fluid handled, it is often referred to as Euler&#x2019;s head. When the equation is being applied to pumps, blowers, fans, etc..., it represents the ideal theoretical head developed Hth\n
- &#x3C9; = v/r, so that &#x3C9;r2 = v2 and &#x3C9;r1 = v1, i.e. the respective angular velocities multiplied by the radius equals the tangential velocity at inlet or outlet.\n\nEuler&#x2019;s equation has units of (j&#x2219;kg-1)/m2. This equation will simplify to m and as all the terms are in the same unit as Bernoulli&#x2019;s equation can be used in conjunction with it. Euler&#x2019;s equation can be applied to pumps and turbines, however in the the case of turbines since uw1v1 > uw2v2, E would be negative indicating the reversed direction of energy. In order to obtain a positive result the order of terms in brackets is often reversed. \n\nGiven that E reduces to meters of fluid handled, it is often referred to as Euler&#x2019;s head. When the equation is being applied to pumps, blowers, fans, etc..., it represents the ideal theoretical head developed Hth\n
- It is more advantageous to express Euler&#x2019;s equation in terms of absolute velocities rather than their components. Referring back to the velocity triangles \n
- It is more advantageous to express Euler&#x2019;s equation in terms of absolute velocities rather than their components. Referring back to the velocity triangles \n
- It is more advantageous to express Euler&#x2019;s equation in terms of absolute velocities rather than their components. Referring back to the velocity triangles \n
- The rotodynamic machine that converts the upstream energy into mechanical energy are turbines. The turbine is directly coupled to the electrical generator. The mechanical energy developed by the rotation of the turbine rotates the stator in the generator which in turn generates electricity. Two distinct categories have a arisen as a result of the different heads and flow rates achievable; Reaction and Impulse turbines. \n
- Impulse turbines use one or more nozzles to convert the total head available into kinetic energy. The issuing jets strike vanes located around the periphery of rotatable wheel. The wheel rotates as the jets strike the vanes resulting in a rate of change of the angular momentum and therefore work is done creating energy. The kinetic energy of fluid reduces as it passes through the impeller (runner), i.e. the velocity at the outlet is less than the velocity at the inlet. There is no increase in fluidic pressure, i.e. the pressure remains at atmospheric pressure through out the process. Friction reduces the relative velocity marginally.\n
- The fluid entering a reaction turbine has both kinetic and pressure energy. These turbines are comprised of two sets of vanes located around the periphery, one stationary whilst the other set rotates. The stationary vanes can be located on either an outer or inner ring and guide the fluid into the impeller runner (again located on an inner or outer ring). Only part of the available total head is converted into kinetic energy here. The fluid discharged from the guide vanes has both pressure and kinetic energy. The pressure energy is converted into kinetic energy in the runner and as a consequence, the relative velocity increases as the fluid passes through. A pressure difference results across the the runner. \n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms 1> the drop in static pressure (the first term), 2> drop in velocity (the second term). If either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms 1> the drop in static pressure (the first term), 2> drop in velocity (the second term). If either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms 1> the drop in static pressure (the first term), 2> drop in velocity (the second term). If either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms \n1> the drop in static pressure (the first term), \n2> drop in velocity (the second term). \nIf either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms \n1> the drop in static pressure (the first term), \n2> drop in velocity (the second term). \nIf either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- The degree of reaction is a parameter which describes reaction turbines. It is derived from bernoulli&#x2019;s equation applied at the inlet and outlet to the turbine and assumes that no losses (ideal fluid) arise. \n\nWhen the equation is rearranged and like terms are gathered it is possible to solve for the energy. The resulting equation has two distinct terms \n1> the drop in static pressure (the first term), \n2> drop in velocity (the second term). \nIf either of these two terms is set to zero, then the extreme solution will result, e.g. if the pressure is constant p1 = p2, E = (u12 - u22)/2g, i.e. an impulse turbine, alternatively if v1 = v2 then E = (p1-p2)/&#x3C1;g which is a true reaction turbine. All intermediate results are described by the degree of reaction (R) defined as static pressure drop divided by Total energy transfer\n
- While a wide variety of turbines are available, they generally be categorised into three distinct grouping, Pelton wheel (impulse), Francis (reaction) and Kaplan (reaction). Pelton wheels can operate up to 2000 m. Kaplan turbines can be divided into either vertical axis or horizontal axis with operating heads of \n
- Reference - Giesecke, J.; Mosony, E.: Wasserkraftanlagen &#x2013; Planung, Bau und Betrieb; Sprin- ger, Berlin, Heidelberg, Germany, 2005, 4. Auflage\n
- Reference - Giesecke, J.; Mosony, E.: Wasserkraftanlagen &#x2013; Planung, Bau und Betrieb; Sprin- ger, Berlin, Heidelberg, Germany, 2005, 4. Auflage\n
- The Pelton wheel - an impulse turbine - having its vanes attached to the periphery of a rotating wheels. The vanes typically have an elliptical form and are often referred to as buckets. The turbine is regulated by one or a number of nozzles with spear like valves. These valves project a jet of fluid tangentially to the wheel into the wanes/buckets. The geometry of the vanes/buckets is such that the jet of fluid is split and leaves symmetrically on both sides of the vane. \n\nUseful resource: https://www.zyba.com/reference/engineering/fluid_mechanics/turbines/impulse_and_reaction_turbines.php\n
- The Pelton wheel - an impulse turbine - having its vanes attached to the periphery of a rotating wheels. The vanes typically have an elliptical form and are often referred to as buckets. The turbine is regulated by one or a number of nozzles with spear like valves. These valves project a jet of fluid tangentially to the wheel into the wanes/buckets. The geometry of the vanes/buckets is such that the jet of fluid is split and leaves symmetrically on both sides of the vane. \n\nUseful resource: https://www.zyba.com/reference/engineering/fluid_mechanics/turbines/impulse_and_reaction_turbines.php\n
- The Pelton wheel - an impulse turbine - having its vanes attached to the periphery of a rotating wheels. The vanes typically have an elliptical form and are often referred to as buckets. The turbine is regulated by one or a number of nozzles with spear like valves. These valves project a jet of fluid tangentially to the wheel into the wanes/buckets. The geometry of the vanes/buckets is such that the jet of fluid is split and leaves symmetrically on both sides of the vane. \n\nUseful resource: https://www.zyba.com/reference/engineering/fluid_mechanics/turbines/impulse_and_reaction_turbines.php\n
- The total head discharging at the nozzle can be determined by taking the gross head available and subtracting losses that are encountered in the pipe network leading to the nozzle.\n\nNote the change in signage within the brackets of Euler&#x2019;s equation - we&#x2019;re dealing with turbine, therefore as discussed previously in order to obtain a positive result in the case of turbines, the signage is changed. \n
- The total head discharging at the nozzle can be determined by taking the gross head available and subtracting losses that are encountered in the pipe network leading to the nozzle.\n\nNote the change in signage within the brackets of Euler&#x2019;s equation - we&#x2019;re dealing with turbine, therefore as discussed previously in order to obtain a positive result in the case of turbines, the signage is changed. \n
- The total head discharging at the nozzle can be determined by taking the gross head available and subtracting losses that are encountered in the pipe network leading to the nozzle.\n\nNote the change in signage within the brackets of Euler&#x2019;s equation - we&#x2019;re dealing with turbine, therefore as discussed previously in order to obtain a positive result in the case of turbines, the signage is changed. \n
- The velocity triangles illustrate that the peripheral vane velocity at the outlets is the same as at the inlet, therefore, v1 = v2 = v. Therefore Euler&#x2019;s equation becomes. \n\nRecall the basics for right angled triangles - soh, cah and toa \n\nThe final equation illustrates that there is no transfer of energy for the instances when the vane velocity zero or equal to the jet velocity. It can therefore be deduced that the maximum energy transfer will occur at some intermediated value of the vane velocity, which can be obtained by differentiation.\n \n\n\n\n
- The velocity triangles illustrate that the peripheral vane velocity at the outlets is the same as at the inlet, therefore, v1 = v2 = v. Therefore Euler&#x2019;s equation becomes. \n\nRecall the basics for right angled triangles - soh, cah and toa \n\nThe final equation illustrates that there is no transfer of energy for the instances when the vane velocity zero or equal to the jet velocity. It can therefore be deduced that the maximum energy transfer will occur at some intermediated value of the vane velocity, which can be obtained by differentiation.\n \n\n\n\n
- The velocity triangles illustrate that the peripheral vane velocity at the outlets is the same as at the inlet, therefore, v1 = v2 = v. Therefore Euler&#x2019;s equation becomes. \n\nRecall the basics for right angled triangles - soh, cah and toa \n\nThe final equation illustrates that there is no transfer of energy for the instances when the vane velocity zero or equal to the jet velocity. It can therefore be deduced that the maximum energy transfer will occur at some intermediated value of the vane velocity, which can be obtained by differentiation.\n \n\n\n\n
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- If no friction existed, there would be no reduction in the relative velocity over the vane and therefore k = 0. If &#x3D1; = 180&#x2DA; the maximum efficiency becomes 100%. \n\nFriction is unavoidable, hence k is normally found in to be in the region of 0.8 to 0.85. Also, in order to avoid interference between the entering and exiting jets, the vane angle is usually set at approximately 165&#x2DA;\n\nThe ratio of the wheel velocity to the jet velocity is therefore found to be smaller than the theoretical in the real world i.e. typically the ratio is 0.46\n\nGiven that pelton wheels drive electrical generators, it is essential that the speed of rotation be maintained regardless of load, i.e. v must be constant. However for maximum efficiency it is imperative that the speed ration be also maintained constant. As the jet velocity depends only and total head H, the velocity ratio may be kept constant provided there is no reduction if head at the nozzle. \n
- If no friction existed, there would be no reduction in the relative velocity over the vane and therefore k = 0. If &#x3D1; = 180&#x2DA; the maximum efficiency becomes 100%. \n\nFriction is unavoidable, hence k is normally found in to be in the region of 0.8 to 0.85. Also, in order to avoid interference between the entering and exiting jets, the vane angle is usually set at approximately 165&#x2DA;\n\nThe ratio of the wheel velocity to the jet velocity is therefore found to be smaller than the theoretical in the real world i.e. typically the ratio is 0.46\n\nGiven that pelton wheels drive electrical generators, it is essential that the speed of rotation be maintained regardless of load, i.e. v must be constant. However for maximum efficiency it is imperative that the speed ration be also maintained constant. As the jet velocity depends only and total head H, the velocity ratio may be kept constant provided there is no reduction if head at the nozzle. \n
- If no friction existed, there would be no reduction in the relative velocity over the vane and therefore k = 0. If &#x3D1; = 180&#x2DA; the maximum efficiency becomes 100%. \n\nFriction is unavoidable, hence k is normally found in to be in the region of 0.8 to 0.85. Also, in order to avoid interference between the entering and exiting jets, the vane angle is usually set at approximately 165&#x2DA;\n\nThe ratio of the wheel velocity to the jet velocity is therefore found to be smaller than the theoretical in the real world i.e. typically the ratio is 0.46\n\nGiven that pelton wheels drive electrical generators, it is essential that the speed of rotation be maintained regardless of load, i.e. v must be constant. However for maximum efficiency it is imperative that the speed ration be also maintained constant. As the jet velocity depends only and total head H, the velocity ratio may be kept constant provided there is no reduction if head at the nozzle. \n
- The Francis turbine is a reaction turbine and operates when the chamber is full of fluid. It can be classified as either an inward or outward flow machine type. In the case of the inward flow, the fluid enters the impeller on its whole periphery i.e. the flow of fluid is directed by the guide vanes located on the outer periphery, flows towards the turbine centre and exits axially. For the outward flow type, the fluid flows over the guide vanes located at the centre along the radius (radially), flows outwards into the scroll case over the impeller blades (in this configuration the impeller is located on an outer runner ring). The fluid then flows out of the turbine in in a circumferential manner. In either configuration a change in direction of flow. Francis turbines can be classified as either low-speed or high speed. High rotational impeller speed is desirable resulting in low torques at the turbine axis thereby impacting on the geometrical elements of the turbine i.e. the size can be minimised and therefore an economic advantage can be gained. \n\nDuring energy transfer in the impeller (runner) a drop in static pressure and velocity head will result. Only a portion of the total head entering the turbine is converted to a velocity head before entering the impeller. The conversion is completed as it passes through the impeller. \n
- Power house \nThese fossil like structures are in face turbines that generate hydroelectric power at the Three Gorges Dam in Yichang, China - currently the world&#x2019;s largest electricity-generating plant.\n\nThe turbines are know as Francis Inlet Scrolls. Each spiral-shaped turbine is up to 10.5 m wide and generates electricity by using the high pressure water flowing through them to turn a wheel attached to a dynamo.\n\nBuilding work for the Three Gorges Dam began in December 1994 and is not expected to be completed until next year, even though it&#x2019;s already generating power. When it&#x2019;s fully operational, the total electric generating capacity will be up to 22.5 GW. It was hoped the dam would provide 10 per cent of China&#x2019;s power, but increased demand means that figure will probably only be three per cent. \n\nDespite being hailed by the Chinese state as a success, the dam is a controversal issue. Important archaeological and cultural sites had to be flooded, and over 1.3 million people were moved from their homes to make way for it. The dam has also been identified as a contributing factor to the extinction of the Yangtze River dolphin.\n\nSource: Focus Magazine November 2010 pages 8-9\n
- image source http://www.sas.usace.army.mil/p\n
- image source http://www.sas.usace.army.mil/p\n
- image source http://www.sas.usace.army.mil/p\n
- image source http://www.sas.usace.army.mil/p\n
- image source http://www.sas.usace.army.mil/p\n
- image source http://www.sas.usace.army.mil/p\n
- Consider an inward flow Francis turbine. The 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; \n\nu0A0 = uf1A1\n\nHowever, uf1= u1sin&#x3D1;, therefore\n\nu0A0 = u1 A1 sin&#x3D1; \n
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- The total energy at inlet to the impeller consists of the velocity head u12/2g and the pressure head H. In the impeller the fluid energy is decreased by E, which is transferred to the runner. Water leaves the impeller with kinetic energy u22/2g\n\nh&#x2019;1 is the loss of head in the guide vane ring and h1 is the loss in the whole turbine. \n\n\n
- The total energy at inlet to the impeller consists of the velocity head u12/2g and the pressure head H. In the impeller the fluid energy is decreased by E, which is transferred to the runner. Water leaves the impeller with kinetic energy u22/2g\n\nh&#x2019;1 is the loss of head in the guide vane ring and h1 is the loss in the whole turbine. \n\n\n
- The total energy at inlet to the impeller consists of the velocity head u12/2g and the pressure head H. In the impeller the fluid energy is decreased by E, which is transferred to the runner. Water leaves the impeller with kinetic energy u22/2g\n\nh&#x2019;1 is the loss of head in the guide vane ring and h1 is the loss in the whole turbine. \n\n\n
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- P is the power, &#x1E41; the mass flow rate (i.e. the product of volumetric flow rate and density) and H is the total head available at the turbine inlet.\n\nThe relationship between the impeller speed and the spouting velocity &#x221A;(2gH), for the Francis turbine is not so rigidly defined as for the Pelton wheel. In practice the speed ratio u2/&#x221A;(2gH) is contained within the limits 0.6 to 0.9\n\nIn a hydroelectric power plant the Francis turbine will be coupled to the generator as a consequence the turbine speed must be constant. \n
- P is the power, &#x1E41; the mass flow rate (i.e. the product of volumetric flow rate and density) and H is the total head available at the turbine inlet.\n\nThe relationship between the impeller speed and the spouting velocity &#x221A;(2gH), for the Francis turbine is not so rigidly defined as for the Pelton wheel. In practice the speed ratio u2/&#x221A;(2gH) is contained within the limits 0.6 to 0.9\n\nIn a hydroelectric power plant the Francis turbine will be coupled to the generator as a consequence the turbine speed must be constant. \n
- P is the power, &#x1E41; the mass flow rate (i.e. the product of volumetric flow rate and density) and H is the total head available at the turbine inlet.\n\nThe relationship between the impeller speed and the spouting velocity &#x221A;(2gH), for the Francis turbine is not so rigidly defined as for the Pelton wheel. In practice the speed ratio u2/&#x221A;(2gH) is contained within the limits 0.6 to 0.9\n\nIn a hydroelectric power plant the Francis turbine will be coupled to the generator as a consequence the turbine speed must be constant. \n
- The power developed by a turbine is proportional to the product of the total head and flow rate. Pelton wheels are selected when the total head is large and flow rate is usually small. If the head available is small, then a Francis turbine is selected. The geometry of the Francis turbine, depends on the flow rate that must pass through it, i.e. for greater flow rates, the runner eye must be increased, thereby affecting the economics. Axial flow turbines are selected when the maximum flow rate maybe passed through when the flow is parallel to the axis, \n\nImage source: Wasserwirtschaftsverband Baden-W&#xFC;rttemberg (Hrsg.): Leitfaden f&#xFC;r den Bau von Kleinwasserkraftwerken; Frankh Kosmos, Stuttgart, Germany, 1994, 2. Auf- lage\n
- The illustration shows the guide vane ring in a plane perpendicular to the shaft so that the flow through the turbine is radial. However the impeller is positioned further downstream so that flow turns through a right angle and effectively enters the impeller parallel to its axis. The guide vanes impart whirl to the flow so as that as it enters the impeller it is of a free vortex type. \n\nThe impeller blades must be long in order to accommodate the large flow rate. High torques arise, therefore strength of significant importance. As a consequence the number of blades rarely exceeds 4 or 5 as the impeller blades have large chords, typically a pitch/cord ratio of 1 to 1.5.\n\nFixed blade configurations are available i.e. the blade angle can not be altered. Such configurations result in a substantial fall in efficiency under partial load as the reduction of flow rate though the turbine results in a miss-match between the direction of flow velocity relative to the impeller and blade angle. \n\nVariable pitch blades are also available, i.e. Kaplan turbines. As the blades can be turned about there axes, the stagger angle can be altered to meet the fluid tangentially. A wide band of high efficiencies can be achieved, with Kaplan turbines having efficiencies of the order of 90% to 94% \n\nImage source: http://www.voithhydro.com/vh_e_prfmc_pwrful_prdcts_turbines_kaplan.htm\n
- The illustration shows the guide vane ring in a plane perpendicular to the shaft so that the flow through the turbine is radial. However the impeller is positioned further downstream so that flow turns through a right angle and effectively enters the impeller parallel to its axis. The guide vanes impart whirl to the flow so as that as it enters the impeller it is of a free vortex type. \n\nThe impeller blades must be long in order to accommodate the large flow rate. High torques arise, therefore strength of significant importance. As a consequence the number of blades rarely exceeds 4 or 5 as the impeller blades have large chords, typically a pitch/cord ratio of 1 to 1.5.\n\nFixed blade configurations are available i.e. the blade angle can not be altered. Such configurations result in a substantial fall in efficiency under partial load as the reduction of flow rate though the turbine results in a miss-match between the direction of flow velocity relative to the impeller and blade angle. \n\nVariable pitch blades are also available, i.e. Kaplan turbines. As the blades can be turned about there axes, the stagger angle can be altered to meet the fluid tangentially. A wide band of high efficiencies can be achieved, with Kaplan turbines having efficiencies of the order of 90% to 94% \n\nImage source: http://www.voithhydro.com/vh_e_prfmc_pwrful_prdcts_turbines_kaplan.htm\n
- The illustration shows the guide vane ring in a plane perpendicular to the shaft so that the flow through the turbine is radial. However the impeller is positioned further downstream so that flow turns through a right angle and effectively enters the impeller parallel to its axis. The guide vanes impart whirl to the flow so as that as it enters the impeller it is of a free vortex type. \n\nThe impeller blades must be long in order to accommodate the large flow rate. High torques arise, therefore strength of significant importance. As a consequence the number of blades rarely exceeds 4 or 5 as the impeller blades have large chords, typically a pitch/cord ratio of 1 to 1.5.\n\nFixed blade configurations are available i.e. the blade angle can not be altered. Such configurations result in a substantial fall in efficiency under partial load as the reduction of flow rate though the turbine results in a miss-match between the direction of flow velocity relative to the impeller and blade angle. \n\nVariable pitch blades are also available, i.e. Kaplan turbines. As the blades can be turned about there axes, the stagger angle can be altered to meet the fluid tangentially. A wide band of high efficiencies can be achieved, with Kaplan turbines having efficiencies of the order of 90% to 94% \n\nImage source: http://www.voithhydro.com/vh_e_prfmc_pwrful_prdcts_turbines_kaplan.htm\n
- The velocity of flow is axial at inlet and outlet. The whirl velocity of is tangential. The blade velocity at inlet and outlet is the same but varies along the blades with radius from hub to tip.\n\nIf the angular velocity of the impeller is &#x3C9; the blade velocity at radius r is given by v = &#x3C9;r; since at maximum efficiency uw2 = 0 and u2 = uf it follows that E = vuw1g, where uw1 = ufcot&#x3D1;. Since E should be the same at the blade tip and at the hub, but v is greater at the tip, it follows that uw1 must be reduced. Similarly, the velocity of flow uf should remain constant along the blade. Therefore cot&#x3D1; must be reduced towards the tip of the blade. This &#x3D1; has to be reduced and consequently the blade must be twisted so that it makes a greater angle with the axis at the tip than it does at the hub.\n
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