Fluid discharge

Lecture slides coverings essential fluid discharge.

1. KV
FLUID DISCHARGE
Keith Vaugh BEng (AERO) MEng

2. KV
OBJECTIVES
Identify the unique vocabulary 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
An orifice
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 Az1+p1ρg+u122g=z2+p2ρg+u222g
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 Az1+p1ρg+u122g=z2+p2ρg+u222g
Applying Bernoulli’s equation
to stations ① and ②
putting
z1 - z2 = H, u1 = 0, u2 = u
and p1 = p2Velocity of jet, u=2gH()

7. KV
TORRICELLI’s
THEOREM
Torricelli’s theorem states that 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 states that 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 states that 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=A2gH()

10. KV
TORRICELLI’s
THEOREM
Torricelli’s theorem states that 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=A2gH()
Actual discharge!Vactual=CdA2gH()

11. KV
The 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, 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 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, 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×CuA2gH()

15. KV
Actual discharge = Actual area at B × Actual velocity at B=Cc×CuA2gH()
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×CuA2gH()
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×CuA2gH()
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 dischargeTheoretical discharge

18. KV
Actual discharge = Actual area at B × Actual velocity at B=Cc×CuA2gH()
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 dischargeTheoretical dischargeCoefficient of contraction, Cc=Area of jet at vena contractaArea of orifice

19. KV
Actual discharge = Actual area at B × Actual velocity at B=Cc×CuA2gH()
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 dischargeTheoretical dischargeCoefficient of contraction, Cc=Area of jet at vena contractaArea of orificeCoefficient of velocity, Cu=Velocity at vena contractaTheoretical velocity

20. KV
EXAMPLE 1
(a)A jet 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 mm
2
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 m
3
and the head H above the orifice is 1.2 m

21. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.

22. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xt

23. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xty=12gt2→t=2yg⎛⎝⎜⎞⎠⎟

24. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟

25. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟Theoretical velocity =2gH()

26. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟Theoretical velocity =2gH()

27. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟Theoretical velocity =2gH()Coefficient of velocity, Cu=Actual velocityTheoretical velocity

28. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟Theoretical velocity =2gH()Coefficient of velocity, Cu=Actual velocityTheoretical velocity=u2gH()

29. KV
Let t be the time taken for a particle of fluid to travel from the vena
contracta A to the point B.x=ut→u=xtVelocity at the vena contracta, u=gx22y⎛⎝⎜⎞⎠⎟y=12gt2→t=2yg⎛⎝⎜⎞⎠⎟Theoretical velocity =2gH()Coefficient of velocity, Cu=Actual velocityTheoretical velocity=u2gH()=x24yH

30. KV
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2

31. KV
Coefficient of velocity, Cu=x24yH
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2

32. KV
Coefficient of velocity, Cu=x24yH
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2= 1.524×0.5×1.2( )=0.968

33. KV
Coefficient of velocity, Cu=x24yH
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2= 1.524×0.5×1.2( )=0.968Coefficient of discharge, Cd= !VA2gH()

34. KV
Coefficient of velocity, Cu=x24yH
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2= 1.524×0.5×1.2( )=0.968Coefficient of discharge, Cd= !VA2gH()= 0.11760⎛⎝⎜ ⎞⎠⎟650×10−62×9.81×1.2( )=0.618

35. KV
Coefficient of velocity, Cu=x24yH
putting x = 1.5m, H = 1.2m and Area, A = 650×10
-6
m
2= 1.524×0.5×1.2( )=0.968Coefficient of discharge, Cd= !VA2gH()= 0.11760⎛⎝⎜ ⎞⎠⎟650×10−62×9.81×1.2( )=0.618Coefficient of contraction, Cc=CdCu=0.6180.968=0.639

36. KV
THEORY OF
LARGE
ORIFICES
H1
H2
h
δh
D
B

37. KV
(a)A reservoir discharges 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 the velocity 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 the velocity 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 surfaceArea of strip=Bδh

40. KV
Given that the velocity 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 surfaceArea of strip=BδhVelocity of flow through strip=2gH

41. KV
Given that the velocity 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 surfaceArea of strip=BδhVelocity of flow through strip=2gHDischarge through strip, δ!V=Area×Velocity=B2g()h12δh

42. KV
Given that the velocity 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=BδhVelocity of flow through strip=2gHDischarge through strip, δ!V=Area×Velocity=B2g()h12δh

43. KV
Given that the velocity 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 = H2Discharge !V=B2g()h12dhH1H2∫
Area of strip=BδhVelocity of flow through strip=2gHDischarge through strip, δ!V=Area×Velocity=B2g()h12δh

44. KV
Given that the velocity 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 = H2Discharge !V=B2g()h12dhH1H2∫
Area of strip=BδhVelocity of flow through strip=2gHDischarge through strip, δ!V=Area×Velocity=B2g()h12δh
=23B2g()H232−H132( )

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 mTheoretical discharge !V=23×0.7×2×9.81( )1.932−0.432( )=2.0672.619−0.253( )=4.891m3s

47. KV
putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 mTheoretical discharge !V=23×0.7×2×9.81( )1.932−0.432( )=2.0672.619−0.253( )=4.891m3sFor a small orifice, !V=A2ghA=BD=0.7×1.5h=12H1+H2( )=120.4+1.9( )=1.15m!V=0.7×1.52×9.81×1.15=4.988m3s

48. KV
putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 mTheoretical discharge !V=23×0.7×2×9.81( )1.932−0.432( )=2.0672.619−0.253( )=4.891m3sFor a small orifice, !V=A2ghA=BD=0.7×1.5h=12H1+H2( )=120.4+1.9( )=1.15m!V=0.7×1.52×9.81×1.15=4.988m3serror=4.988−4.891( )4.891 =0.0198=1.98%

49. KV
NOTCHES
& WEIRS
H
h
δh
H
b

50. KV
Consider a horizontal strip of width b and height δh at a depth h below the
free surface.

51. KV
Consider a horizontal strip of width b and height δh at a depth h below the
free surface. Area of strip=bδhVelocity of flow through strip=2ghDischarge through strip, δ!V=Area×Velocity=bδh2gh()

52. KV
Consider a horizontal strip 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=bδhVelocity of flow through strip=2ghDischarge through strip, δ!V=Area×Velocity=bδh2gh()

53. KV
Consider a horizontal strip 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
notchTotal theoretical discharge !V=2g()bh12dh0H∫
Area of strip=bδhVelocity of flow through strip=2ghDischarge through strip, δ!V=Area×Velocity=bδh2gh()

54. KV
Consider a horizontal strip 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
notchTotal theoretical discharge !V=2g()bh12dh0H∫
Area of strip=bδhVelocity of flow through strip=2ghDischarge through strip, δ!V=Area×Velocity=bδh2gh()
b must be expressed in terms of h before integrating

55. KV
Consider a horizontal strip 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
notchTotal theoretical discharge !V=2g()bh12dh0H∫
Area of strip=bδhVelocity of flow through strip=2ghDischarge through strip, δ!V=Area×Velocity=bδh2gh()
b must be expressed in terms of h before integrating!V=B2()gh120H∫=23B2g()H32

56. KV
For a rectangular notch, put b = constant = B
b = constant
B
H

57. KV
For a rectangular notch, put b = constant = B!V=B2g()h12dh0H∫=23B2gH32
b = constant
B
H

58. KV
For a rectangular notch, put b = constant = B!V=B2g()h12dh0H∫=23B2gH32
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 rectangular notch, put b = constant = B!V=B2g()h12dh0H∫=23B2gH32
b = constant
B
H
For a vee notch with an included angle θ, put
b = 2(H - h)tan(
θ
⁄2)!V=22g()tanθ2⎛⎝⎜⎞⎠⎟H−h( )h12dh0H∫=22g()tanθ2⎛⎝⎜⎞⎠⎟23Hh32−25h52⎡⎣⎢ ⎤⎦⎥0h=8152g()tanθ2⎛⎝⎜⎞⎠⎟H52
b = 2(H - h)tan(
θ
⁄2)
H
h
θ

60. KV
In the foregoing analysis 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 foregoing analysis 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+αu22g
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
Thereforeδ!V=bδh2gx()=b2g()x12dx

63. KV
Thereforeδ!V=bδh2gx()=b2g()x12dx
at the free surface, h = 0 and x = αū
2
/2g, while at the sill , h = H and
x = H + αū
2
/2g. Integrating between these two limits!V=2g() bx12αu22gH+αu22g∫ dx

64. KV
Thereforeδ!V=bδh2gx()=b2g()x12dx
at the free surface, h = 0 and x = αū
2
/2g, while at the sill , h = H and
x = H + αū
2
/2g. Integrating between these two limits!V=2g() bx12αu22gH+αu22g∫ dx
For a rectangular notch, putting b = B = constant!V=23B2g()H321+αu22gH⎛⎝⎜ ⎞⎠⎟32−αu22gH⎛⎝⎜⎞⎠⎟32⎡⎣⎢⎢ ⎤⎦⎥⎥

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=pρg+u22g+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=pρg+u22g+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=WeightUnit time×EnergyUnit weight

67. KV
Weight per unit time=ρg!VPower=ρg!VH=ρg!Vpρg+u22g+z⎛⎝⎜ ⎞⎠⎟=p!V+12ρu2!V+ρg!Vz

68. KV

69. KV

70. KV
In a hydroelectric power plant, 100 m
3
/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 flow is 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/m
3
(Çengel, et al 2008)

72. KV
Assumptions
1.The flow is 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/m
3
The mass flow rate of water through the turbine is
(Çengel, et al 2008)

73. KV
Assumptions
1.The flow is 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/m
3!m=ρ!V=1000kgm3⎛⎝⎜ ⎞⎠⎟100m3s( )=105kgs
The mass flow rate of water through the turbine is
(Çengel, et al 2008)

74. KV
Assumptions
1.The flow is 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/m
3!m=ρ!V=1000kgm3⎛⎝⎜ ⎞⎠⎟100m3s( )=105kgs
The mass flow rate of water through the turbine is
We take point ➁ as the reference level, and thus z
2 = 0. Therefore the energy
equation is
(Çengel, et al 2008)

75. KV
Assumptions
1.The flow is 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/m
3!m=ρ!V=1000kgm3⎛⎝⎜ ⎞⎠⎟100m3s( )=105kgs
The mass flow rate of water through the turbine is
We take point ➁ as the reference level, and thus z
2 = 0. Therefore the energy
equation isP1ρg+α1V122g+z1+hpump,u=P2ρg+α2V222g+z2+hturbine,e+hL
(Çengel, et al 2008)

76. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces to

77. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL

78. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power are

79. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power arehturbine,e=z1−hL=120−35=85m

80. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power arehturbine,e=z1−hL=120−35=85m!Wturbine,e=!mghturbine,e=105kgs⎛⎝⎜ ⎞⎠⎟9.81ms2( )85m()1kJkg1000m2s2⎛⎝⎜⎜⎜ ⎞⎠⎟⎟⎟=83,400kW

81. KV
Also, both points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power arehturbine,e=z1−hL=120−35=85m!Wturbine,e=!mghturbine,e=105kgs⎛⎝⎜ ⎞⎠⎟9.81ms2( )85m()1kJkg1000m2s2⎛⎝⎜⎜⎜ ⎞⎠⎟⎟⎟=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 points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power arehturbine,e=z1−hL=120−35=85m!Wturbine,e=!mghturbine,e=105kgs⎛⎝⎜ ⎞⎠⎟9.81ms2( )85m()1kJkg1000m2s2⎛⎝⎜⎜⎜ ⎞⎠⎟⎟⎟=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 points ➀ and ➁ are open to the atmosphere (P
1 = P
2 = P
atm)
and the flow velocities are negligible at both points (V
1 = V
2 = 0). Then
the energy equation for steady, incompressible flow reduces tohturbine,e=z1−hL
Substituting, the extracted turbine head and the corresponding turbine
power arehturbine,e=z1−hL=120−35=85m!Wturbine,e=!mghturbine,e=105kgs⎛⎝⎜ ⎞⎠⎟9.81ms2( )85m()1kJkg1000m2s2⎛⎝⎜⎜⎜ ⎞⎠⎟⎟⎟=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 orifices and 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) Mechanical Technology, 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