Successfully reported this slideshow.
Upcoming SlideShare
×

# Hydropower

23,542 views

Published on

Hydro Power Lecture slides

• Full Name
Comment goes here.

Are you sure you want to Yes No
• send the file please at (mambo652@gmail.com) Thank You!

Are you sure you want to  Yes  No

Are you sure you want to  Yes  No
• lovely.

Are you sure you want to  Yes  No
• Hi Can you please mail us your presentation. It is required to impart training to Hydro power engineers in India.
My email ID: vkabbey@gmail.com

Are you sure you want to  Yes  No
• Thanks for sharing

Are you sure you want to  Yes  No

### 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
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
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
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