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Fluid discharge
1) The document discusses fluid flow through orifices and mouthpieces. It describes the theory of small orifices discharging fluid using Bernoulli's equation and defines relevant terms like coefficient of velocity and coefficient of discharge. 2) Torricelli's theorem states the velocity of a discharging jet is proportional to the square root of the head producing flow. The theoretical discharge can be calculated using the orifice area and velocity. 3) Examples are provided to demonstrate calculating coefficients of velocity, discharge, and contraction for given orifice dimensions and fluid flow values.
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Fluid discharge
- 1. KV FLUID DISCHARGE Keith VaughBEng (AERO) MEng
- 2. KV OBJECTIVES Identify the uniquevocabulary used in the description and analysis of fluid flow with an emphasis on fluid discharge Describe and discuss how fluid flow discharge devices affects fluid flow. Derive and apply the governing equations associated with fluid discharge Determine the power of flow in a channel or a stream and how this can be affected by height, pressure, and/or geometrical properties.
- 3. KV FLOW THROUGH ORIFICES and MOUTHPIECES Anorifice Is an opening having a closed perimeter A mouthpiece Is a short tube of length not more than two to three times its diameter
- 4. KV THEORY OF SMALL ORIFICES DISCHARGING u1 z1 z2② H u = u2 ① Orifice area A Applying Bernoulli’s equation to stations ① and ②
- 5. KV THEORY OF SMALL ORIFICES DISCHARGING u1 z1 z2② H u = u2 ① Orifice area A z1 + p1 ρg + u1 2 2g = z2 + p2 ρg + u2 2 2g Applying Bernoulli’s equation to stations ① and ②
- 6. KV THEORY OF SMALL ORIFICES DISCHARGING u1 z1 z2② H u = u2 ① Orifice area A z1 + p1 ρg + u1 2 2g = z2 + p2 ρg + u2 2 2g Applying Bernoulli’s equation to stations ① and ② putting z1 - z2 = H, u1 = 0, u2 = u and p1 = p2 Velocity of jet, u = 2gH( )
- 7. KV TORRICELLI’s THEOREM Torricelli’s theorem statesthat the velocity of the discharging jet is proportional to the square root of the head producing flow.This is support by the preceding derivation;
- 8. KV TORRICELLI’s THEOREM Torricelli’s theorem statesthat the velocity of the discharging jet is proportional to the square root of the head producing flow.This is support by the preceding derivation; Velocity of jet, u = 2gH( )
- 9. KV TORRICELLI’s THEOREM Torricelli’s theorem statesthat the velocity of the discharging jet is proportional to the square root of the head producing flow.This is support by the preceding derivation; Velocity of jet, u = 2gH( ) The discharging flow rate can be determined theoretically if A is the cross-sectional area of the orifice !V = Area × Velocity = A 2gH( )
- 10. KV TORRICELLI’s THEOREM Torricelli’s theorem statesthat the velocity of the discharging jet is proportional to the square root of the head producing flow.This is support by the preceding derivation; Velocity of jet, u = 2gH( ) The discharging flow rate can be determined theoretically if A is the cross-sectional area of the orifice !V = Area × Velocity = A 2gH( ) Actual discharge !Vactual = Cd A 2gH( )
- 11. KV The velocity ofthe jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity, u = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98
- 12. KV The velocity ofthe jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity, u = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98 The paths of the particles of the fluid converge on the orifice and the area of the discharging jet at B is less than the area of the orifice A at C actual area of jet at B = Cc A where Cc is the coefficient of contraction - determined experimentally - typically 0.64 Vena contracta C B pB u uC pC
- 13. KV Actual discharge =Actual area at B × Actual velocity at B
- 14. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( )
- 15. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( ) we see that the relationship between the coefficients is Cd = Cc × Cu
- 16. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( ) we see that the relationship between the coefficients is Cd = Cc × Cu To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge.
- 17. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( ) we see that the relationship between the coefficients is Cd = Cc × Cu To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge. Coefficient of discharge, Cd = Actual measured discharge Theoretical discharge
- 18. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( ) we see that the relationship between the coefficients is Cd = Cc × Cu To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge. Coefficient of discharge, Cd = Actual measured discharge Theoretical discharge Coefficient of contraction, Cc = Area of jet at vena contracta Area of orifice
- 19. KV Actual discharge =Actual area at B × Actual velocity at B = Cc × Cu A 2gH( ) we see that the relationship between the coefficients is Cd = Cc × Cu To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge. Coefficient of discharge, Cd = Actual measured discharge Theoretical discharge Coefficient of contraction, Cc = Area of jet at vena contracta Area of orifice Coefficient of velocity, Cu = Velocity at vena contracta Theoretical velocity
- 20. KV EXAMPLE 1 (a) Ajet of water discharges horizontally into the atmosphere from an orifice in the side of large open topped tank. Derive an expression for the actual velocity, u of a jet at the vena contracta if the jet falls a distance y vertically for a horizontal distance x, measured from the vena contracta. (b) If the head of water above the orifice is H, calculate the coefficient of velocity. (c) If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a horizontal distance x of 1.5 m from the vena contracta, calculate the values of the coefficients of velocity, discharge and contraction, given that the volumetric flow is 0.117 m3 and the head H above the orifice is 1.2 m
- 21. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B.
- 22. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t
- 23. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟
- 24. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟
- 25. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟ Theoretical velocity = 2gH( )
- 26. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟ Theoretical velocity = 2gH( )
- 27. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟ Theoretical velocity = 2gH( ) Coefficient of velocity, Cu = Actual velocity Theoretical velocity
- 28. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟ Theoretical velocity = 2gH( ) Coefficient of velocity, Cu = Actual velocity Theoretical velocity = u 2gH( )
- 29. KV Let t bethe time taken for a particle of fluid to travel from the vena contracta A to the point B. x = ut → u = x t Velocity at the vena contracta, u = gx2 2y ⎛ ⎝⎜ ⎞ ⎠⎟ y = 1 2 gt2 → t = 2y g ⎛ ⎝⎜ ⎞ ⎠⎟ Theoretical velocity = 2gH( ) Coefficient of velocity, Cu = Actual velocity Theoretical velocity = u 2gH( ) = x2 4yH
- 30. KV putting x =1.5m, H = 1.2m and Area,A = 650×10-6m2
- 31. KV Coefficient of velocity,Cu = x2 4yH putting x = 1.5m, H = 1.2m and Area,A = 650×10-6m2
- 32. KV Coefficient of velocity,Cu = x2 4yH putting x = 1.5m, H = 1.2m and Area,A = 650×10-6m2 = 1.52 4 × 0.5 ×1.2( ) = 0.968
- 33. KV Coefficient of velocity,Cu = x2 4yH putting x = 1.5m, H = 1.2m and Area,A = 650×10-6m2 = 1.52 4 × 0.5 ×1.2( ) = 0.968 Coefficient of discharge, Cd = !V A 2gH( )
- 34. KV Coefficient of velocity,Cu = x2 4yH putting x = 1.5m, H = 1.2m and Area,A = 650×10-6m2 = 1.52 4 × 0.5 ×1.2( ) = 0.968 Coefficient of discharge, Cd = !V A 2gH( ) = 0.117 60 ⎛ ⎝⎜ ⎞ ⎠⎟ 650 ×10−6 2 × 9.81×1.2( ) = 0.618
- 35. KV Coefficient of velocity,Cu = x2 4yH putting x = 1.5m, H = 1.2m and Area,A = 650×10-6m2 = 1.52 4 × 0.5 ×1.2( ) = 0.968 Coefficient of discharge, Cd = !V A 2gH( ) = 0.117 60 ⎛ ⎝⎜ ⎞ ⎠⎟ 650 ×10−6 2 × 9.81×1.2( ) = 0.618 Coefficient of contraction, Cc = Cd Cu = 0.618 0.968 = 0.639
- 36.
- 37. KV (a) A reservoirdischarges through a sluice gate of width B and height D.The top and bottom openings are a depths of H1 and H2 respectively below the free surface. Derive a formula for the theoretical discharge through the opening (b) If the top of the opening is 0.4 m below the water level and the opening is 0.7 m wide and 1.5 m in height, calculate the theoretical discharge (in meters per second) assuming that the bottom of the opening is above the downstream water level. (c) What would be the percentage error if the opening were to be treated as a small orifice? EXAMPLE 2
- 38. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface
- 39. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface Area of strip = Bδh
- 40. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface Area of strip = Bδh Velocity of flow through strip = 2gH
- 41. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface Area of strip = Bδh Velocity of flow through strip = 2gH Discharge through strip, δ !V = Area × Velocity = B 2g( )h 1 2 δh
- 42. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface For the whole opening, integrating from h = H1 to h = H2 Area of strip = Bδh Velocity of flow through strip = 2gH Discharge through strip, δ !V = Area × Velocity = B 2g( )h 1 2 δh
- 43. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface For the whole opening, integrating from h = H1 to h = H2 Discharge !V = B 2g( ) h 1 2 dh H1 H2 ∫ Area of strip = Bδh Velocity of flow through strip = 2gH Discharge through strip, δ !V = Area × Velocity = B 2g( )h 1 2 δh
- 44. KV Given that thevelocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface For the whole opening, integrating from h = H1 to h = H2 Discharge !V = B 2g( ) h 1 2 dh H1 H2 ∫ Area of strip = Bδh Velocity of flow through strip = 2gH Discharge through strip, δ !V = Area × Velocity = B 2g( )h 1 2 δh = 2 3 B 2g( ) H2 3 2 − H1 3 2 ( )
- 45. KV putting B =0.7 m, H1 = 0.4 m and H2 = 1.9 m
- 46. KV putting B =0.7 m, H1 = 0.4 m and H2 = 1.9 m Theoretical discharge !V = 2 3 × 0.7 × 2 × 9.81( ) 1.9 3 2 − 0.4 3 2 ( ) = 2.067 2.619 − 0.253( ) = 4.891m3 s
- 47. KV putting B =0.7 m, H1 = 0.4 m and H2 = 1.9 m Theoretical discharge !V = 2 3 × 0.7 × 2 × 9.81( ) 1.9 3 2 − 0.4 3 2 ( ) = 2.067 2.619 − 0.253( ) = 4.891m3 s For a small orifice, !V = A 2gh A = BD = 0.7 ×1.5 h = 1 2 H1 + H2( ) = 1 2 0.4 +1.9( )=1.15m !V = 0.7 ×1.5 2 × 9.81×1.15 = 4.988m3 s
- 48. KV putting B =0.7 m, H1 = 0.4 m and H2 = 1.9 m Theoretical discharge !V = 2 3 × 0.7 × 2 × 9.81( ) 1.9 3 2 − 0.4 3 2 ( ) = 2.067 2.619 − 0.253( ) = 4.891m3 s For a small orifice, !V = A 2gh A = BD = 0.7 ×1.5 h = 1 2 H1 + H2( ) = 1 2 0.4 +1.9( )=1.15m !V = 0.7 ×1.5 2 × 9.81×1.15 = 4.988m3 s error = 4.988 − 4.891( ) 4.891 = 0.0198 =1.98%
- 49.
- 50. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface.
- 51. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface. Area of strip = bδh Velocity of flow through strip = 2gh Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )
- 52. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface. Integrating from h = 0 at the free surface to h = H at the bottom of the notch Area of strip = bδh Velocity of flow through strip = 2gh Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )
- 53. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface. Integrating from h = 0 at the free surface to h = H at the bottom of the notch Total theoretical discharge !V = 2g( ) bh 1 2 dh 0 H ∫ Area of strip = bδh Velocity of flow through strip = 2gh Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )
- 54. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface. Integrating from h = 0 at the free surface to h = H at the bottom of the notch Total theoretical discharge !V = 2g( ) bh 1 2 dh 0 H ∫ Area of strip = bδh Velocity of flow through strip = 2gh Discharge through strip, δ !V = Area × Velocity = bδh 2gh( ) b must be expressed in terms of h before integrating
- 55. KV Consider a horizontalstrip of width b and height δh at a depth h below the free surface. Integrating from h = 0 at the free surface to h = H at the bottom of the notch Total theoretical discharge !V = 2g( ) bh 1 2 dh 0 H ∫ Area of strip = bδh Velocity of flow through strip = 2gh Discharge through strip, δ !V = Area × Velocity = bδh 2gh( ) b must be expressed in terms of h before integrating !V = B 2( )g h 1 2 0 H ∫ = 2 3 B 2g( )H 3 2
- 56. KV For a rectangularnotch, put b = constant = B b = constant B H
- 57. KV For a rectangularnotch, put b = constant = B !V = B 2g( ) h 1 2 dh 0 H ∫ = 2 3 B 2gH 3 2 b = constant B H
- 58. KV For a rectangularnotch, put b = constant = B !V = B 2g( ) h 1 2 dh 0 H ∫ = 2 3 B 2gH 3 2 b = constant B H For a vee notch with an included angle θ, put b = 2(H - h)tan(θ⁄2)b = 2(H - h)tan(θ⁄2) H h θ
- 59. KV For a rectangularnotch, put b = constant = B !V = B 2g( ) h 1 2 dh 0 H ∫ = 2 3 B 2gH 3 2 b = constant B H For a vee notch with an included angle θ, put b = 2(H - h)tan(θ⁄2) !V = 2 2g( )tan θ 2 ⎛ ⎝⎜ ⎞ ⎠⎟ H − h( )h 1 2 dh 0 H ∫ = 2 2g( )tan θ 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 3 Hh 3 2 − 2 5 h 5 2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 0 h = 8 15 2g( )tan θ 2 ⎛ ⎝⎜ ⎞ ⎠⎟ H 5 2 b = 2(H - h)tan(θ⁄2) H h θ
- 60. KV In the foregoinganalysis it has been assumed that • the velocity of the liquid approaching the notch is very small so that its kinetic energy can be neglected • the velocity through any horizontal element across the notch will depend only on the depth below the free surface These assumptions are appropriate for flow over a notch or a weir in the side of a large reservoir If the notch or weir is located at the end of a narrow channel, the velocity of approach will be substantial and the head h producing flow will be increased by the kinetic energy;
- 61. KV In the foregoinganalysis it has been assumed that • the velocity of the liquid approaching the notch is very small so that its kinetic energy can be neglected • the velocity through any horizontal element across the notch will depend only on the depth below the free surface These assumptions are appropriate for flow over a notch or a weir in the side of a large reservoir If the notch or weir is located at the end of a narrow channel, the velocity of approach will be substantial and the head h producing flow will be increased by the kinetic energy; x = h + αu2 2g where ū is the mean velocity and α is the kinetic energy correction factor to allow for the non-uniform velocity over the cross section of the channel
- 62. KV Therefore δ !V =bδh 2gx( ) = b 2g( )x 1 2 dx
- 63. KV Therefore δ !V =bδh 2gx( ) = b 2g( )x 1 2 dx at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and x = H + αū2/2g. Integrating between these two limits !V = 2g( ) bx 1 2 αu2 2g H+ αu2 2g ∫ dx
- 64. KV Therefore δ !V =bδh 2gx( ) = b 2g( )x 1 2 dx at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and x = H + αū2/2g. Integrating between these two limits !V = 2g( ) bx 1 2 αu2 2g H+ αu2 2g ∫ dx For a rectangular notch, putting b = B = constant !V = 2 3 B 2g( )H 3 2 1+ αu2 2gH ⎛ ⎝⎜ ⎞ ⎠⎟ 3 2 − αu2 2gH ⎛ ⎝⎜ ⎞ ⎠⎟ 3 2⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥
- 65. KV Pressure, p, velocity,u, and elevation, z, can cause a stream of fluid to do work.The total energy per unit weight H of a fluid is given by H = p ρg + u2 2g + z THE POWER OF A STREAM OF FLUID
- 66. KV Pressure, p, velocity,u, and elevation, z, can cause a stream of fluid to do work.The total energy per unit weight H of a fluid is given by H = p ρg + u2 2g + z If the weight per unit time of fluid is known, the power of the stream can be calculated; THE POWER OF A STREAM OF FLUID Power = Energy per unit time = Weight Unit time × Energy Unit weight
- 67. KV Weight per unittime = ρg!V Power = ρg!VH =ρg!V p ρg + u2 2g + z ⎛ ⎝⎜ ⎞ ⎠⎟ = p!V+ 1 2 ρu2 !V+ρg!Vz
- 68.
- 69.
- 70. KV In a hydroelectricpower plant, 100 m3 /s of water flows from an elevation of 12 m to a turbine, where electric power is generated.The total irreversible heat loss is in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine-generator is 80%, estimate the electric power output. EXAMPLE 2
- 71. KV Assumptions 1. The flowis steady and incompressible 2. Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 (Çengel, et al 2008)
- 72. KV Assumptions 1. The flowis steady and incompressible 2. Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 The mass flow rate of water through the turbine is (Çengel, et al 2008)
- 73. KV Assumptions 1. The flowis steady and incompressible 2. Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 !m = ρ !V = 1000kg m3 ⎛ ⎝⎜ ⎞ ⎠⎟ 100m3 s( )=105 kg s The mass flow rate of water through the turbine is (Çengel, et al 2008)
- 74. KV Assumptions 1. The flowis steady and incompressible 2. Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 !m = ρ !V = 1000kg m3 ⎛ ⎝⎜ ⎞ ⎠⎟ 100m3 s( )=105 kg s The mass flow rate of water through the turbine is We take point ➁ as the reference level, and thus z2 = 0.Therefore the energy equation is (Çengel, et al 2008)
- 75. KV Assumptions 1. The flowis steady and incompressible 2. Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 !m = ρ !V = 1000kg m3 ⎛ ⎝⎜ ⎞ ⎠⎟ 100m3 s( )=105 kg s The mass flow rate of water through the turbine is We take point ➁ as the reference level, and thus z2 = 0.Therefore the energy equation is P1 ρg +α1 V1 2 2g + z1 + hpump,u = P2 ρg +α2 V2 2 2g + z2 + hturbine,e + hL (Çengel, et al 2008)
- 76. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to
- 77. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL
- 78. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are
- 79. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine,e = z1 − hL =120 − 35 = 85m
- 80. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine,e = z1 − hL =120 − 35 = 85m !Wturbine,e = !mghturbine,e = 105 kg s ⎛ ⎝⎜ ⎞ ⎠⎟ 9.81m s2( ) 85m( ) 1kJ kg 1000m2 s2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = 83,400kW
- 81. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine,e = z1 − hL =120 − 35 = 85m !Wturbine,e = !mghturbine,e = 105 kg s ⎛ ⎝⎜ ⎞ ⎠⎟ 9.81m s2( ) 85m( ) 1kJ kg 1000m2 s2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = 83,400kW Therefore, a perfect turbine-generator would generate 83,400 kW of electricity from this resource.The electric power generated by the actual unit is
- 82. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine,e = z1 − hL =120 − 35 = 85m !Wturbine,e = !mghturbine,e = 105 kg s ⎛ ⎝⎜ ⎞ ⎠⎟ 9.81m s2( ) 85m( ) 1kJ kg 1000m2 s2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = 83,400kW Therefore, a perfect turbine-generator would generate 83,400 kW of electricity from this resource.The electric power generated by the actual unit is !Welectric = ηturbine−gen !Wturbine,e = 0.80( ) 83.4MW( )= 66.7MW
- 83. KV Also, both points➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 =V2 = 0).Then the energy equation for steady, incompressible flow reduces to hturbine,e = z1 − hL Substituting, the extracted turbine head and the corresponding turbine power are hturbine,e = z1 − hL =120 − 35 = 85m !Wturbine,e = !mghturbine,e = 105 kg s ⎛ ⎝⎜ ⎞ ⎠⎟ 9.81m s2( ) 85m( ) 1kJ kg 1000m2 s2 ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = 83,400kW Therefore, a perfect turbine-generator would generate 83,400 kW of electricity from this resource.The electric power generated by the actual unit is !Welectric = ηturbine−gen !Wturbine,e = 0.80( ) 83.4MW( )= 66.7MW Note that the power generation would increase by almost 1 MW for each percentage point improvement in the efficiency of the turbine-generator unit.
- 84. KV Flow through orificesand mouthpieces Theory of small orifice discharge ! Torricelli’s theorem Theory of large orifices Notches and weirs The power of a stream of fluid
- 85. 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.F., Gasoriek, J.M., Swaffield, J., Jack, L. (2011), Fluid Mechanics, sisth edition, Prentice Hall Turns, S. (2006) Thermal fluid sciences:An integrated approach, Cambridge University Press Young, D., Munson, B., Okiishi,T., Huebsch,W., 2011Introduction to Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc. Some illustrations taken from Fundamentals of thermal fluid sciences
Editor's Notes
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- #4 Orifices and mouthpieces can be used to measure flow rate.\n\nAn orifice is an opening which has a closed perimeter. They are generally made in the walls or the bottom of a tank containing fluid and as a consequence the contents can flow through opening thereby discharging. Orifice’s may be classified by size, shape, shape of the upstream edges and the conditions of discharge.\n\nA mouthpiece is a tube not more than two to three times its diameter. Such would be fitted to a circular opening or orifice of the same diameter of a tank thereby creating an extension of the orifice. The contents of the tank could be discharged through this. \n
- #5 An orifice is an opening found in the base or side of a tank. The pressure acting on the fluid surface forces a discharge through this opening, therefore the volumetric flow rate discharged will depend on the the head of fluid above the level of the opening. The term “small orifice” is used when the opening is small compared to the head producing flow, i.e. it can be assume that this head does not vary appreciably from point to point across the orifice.\n\nA small orifice in the side of a large tank is with a free surface open to the atmosphere is illustrated. At point on the free surface ① the pressure is atmospheric pressure p1. As the large, the velocity u1 can be considered to be negligible. The conditions in the region of the Orifice are uncertain, but at a point ② in the jet, the pressure p2 will again be atmospheric pressure with a velocity u2 which equals the jet velocity u. Establishing the datum for potential energy at the centre of the orifice and applying Bernoulli’s equ (assuming no loss in energy)\n
- #6 An orifice is an opening found in the base or side of a tank. The pressure acting on the fluid surface forces a discharge through this opening, therefore the volumetric flow rate discharged will depend on the the head of fluid above the level of the opening. The term “small orifice” is used when the opening is small compared to the head producing flow, i.e. it can be assume that this head does not vary appreciably from point to point across the orifice.\n\nA small orifice in the side of a large tank is with a free surface open to the atmosphere is illustrated. At point on the free surface ① the pressure is atmospheric pressure p1. As the large, the velocity u1 can be considered to be negligible. The conditions in the region of the Orifice are uncertain, but at a point ② in the jet, the pressure p2 will again be atmospheric pressure with a velocity u2 which equals the jet velocity u. Establishing the datum for potential energy at the centre of the orifice and applying Bernoulli’s equ (assuming no loss in energy)\n
- #7 An orifice is an opening found in the base or side of a tank. The pressure acting on the fluid surface forces a discharge through this opening, therefore the volumetric flow rate discharged will depend on the the head of fluid above the level of the opening. The term “small orifice” is used when the opening is small compared to the head producing flow, i.e. it can be assume that this head does not vary appreciably from point to point across the orifice.\n\nA small orifice in the side of a large tank is with a free surface open to the atmosphere is illustrated. At point on the free surface ① the pressure is atmospheric pressure p1. As the large, the velocity u1 can be considered to be negligible. The conditions in the region of the Orifice are uncertain, but at a point ② in the jet, the pressure p2 will again be atmospheric pressure with a velocity u2 which equals the jet velocity u. Establishing the datum for potential energy at the centre of the orifice and applying Bernoulli’s equ (assuming no loss in energy)\n
- #8 An orifice is an opening found in the base or side of a tank. The pressure acting on the fluid surface forces a discharge through this opening, therefore the volumetric flow rate discharged will depend on the the head of fluid above the level of the opening. The term “small orifice” is used when the opening is small compared to the head producing flow, i.e. it can be assume that this head does not vary appreciably from point to point across the orifice.\n\nA small orifice in the side of a large tank is with a free surface open to the atmosphere is illustrated. At point on the free surface ① the pressure is atmospheric pressure p1. As the large, the velocity u1 can be considered to be negligible. The conditions in the region of the Orifice are uncertain, but at a point ② in the jet, the pressure p2 will again be atmospheric pressure with a velocity u2 which equals the jet velocity u. Establishing the datum for potential energy at the centre of the orifice and applying Bernoulli’s equ (assuming no loss in energy)\n
- #9 u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg)\n\nThe actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd\n\nThere are two reasons for the difference between the theoretical and actual. What are these?\n\nThe velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98\n
- #10 u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg)\n\nThe actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd\n\nThere are two reasons for the difference between the theoretical and actual. What are these?\n\nThe velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98\n
- #11 u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg)\n\nThe actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd\n\nThere are two reasons for the difference between the theoretical and actual. What are these?\n\nThe velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98\n
- #12 u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg)\n\nThe actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd\n\nThere are two reasons for the difference between the theoretical and actual. What are these?\n\nThe velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98\n
- #13 u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg)\n\nThe actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd\n\nThere are two reasons for the difference between the theoretical and actual. What are these?\n\nThe velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98\n
- #14 In the plane of the of the orifice the streamlines have a component of velocity towards the centre and the pressure at C is greater than atmospheric pressure. At B, a small distance outside the orifice, the streamlines have become parallel. This section through B is referred to as the vena contracta\n
- #15 In the plane of the of the orifice the streamlines have a component of velocity towards the centre and the pressure at C is greater than atmospheric pressure. At B, a small distance outside the orifice, the streamlines have become parallel. This section through B is referred to as the vena contracta\n
- #16 In the plane of the of the orifice the streamlines have a component of velocity towards the centre and the pressure at C is greater than atmospheric pressure. At B, a small distance outside the orifice, the streamlines have become parallel. This section through B is referred to as the vena contracta\n
- #17 The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. \n\nIf the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.\n
- #18 The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. \n\nIf the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.\n
- #19 The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. \n\nIf the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.\n
- #20 The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. \n\nIf the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.\n
- #21 The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. \n\nIf the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.\n
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- #35 If orifice has a large opening compared to the head producing flow, then the discharge calculated using the formula for a small orifice and the head measured from the centre line of the orifice will yield an inaccurate result, i.e. the velocity will vary substantially from top to bottom. \n\nThere for a large orifice the preferred method to is calculate the flow through a thin horizontal strip across the orifice opening an integrate from top to bottom in order to determine the theoretical discharge. From this the actual discharge can be determined providing the coefficient of discharge is known.\n
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- #48 An opening in the side of a tank or a reservoir which extends above the free surface is referred to as a Notch. It is essentially a large orifice which does not have an upper edge and therefore has a variable area depending upon the level of the free surface. \n\nA weir is a notch on much larger scale i.e. it may have a large width in the direction of flow. \n\nThe method developed for determining the theoretical flow through a large orifice is also used to determine the theoretical flow for a notch.\n\n
- #49 This theory applies to a notch of any shape.\n
- #50 This theory applies to a notch of any shape.\n
- #51 This theory applies to a notch of any shape.\n
- #52 This theory applies to a notch of any shape.\n
- #53 This theory applies to a notch of any shape.\n
- #54 This theory applies to a notch of any shape.\n
- #55 This theory applies to a notch of any shape.\n
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- #60 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
- #61 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
- #62 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
- #63 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
- #64 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
- #65 NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch). \n
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