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
z2 ②
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
z2 ②
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
z2 ②
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 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 = A 2gH( )

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 = A 2gH( )
Actual discharge
!Vactual
= Cd
A 2gH( )

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
× 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) 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 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 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 =
x
t

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 =
x
t
y =
1
2
gt2
→ t =
2y
g
⎛
⎝⎜
⎞
⎠⎟

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 =
x
t
Velocity at the vena contracta, u =
gx2
2y
⎛
⎝⎜
⎞
⎠⎟
y =
1
2
gt2
→ t =
2y
g
⎛
⎝⎜
⎞
⎠⎟

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 =
x
t
Velocity at the vena contracta, u =
gx2
2y
⎛
⎝⎜
⎞
⎠⎟
y =
1
2
gt2
→ t =
2y
g
⎛
⎝⎜
⎞
⎠⎟
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 =
x
t
Velocity at the vena contracta, u =
gx2
2y
⎛
⎝⎜
⎞
⎠⎟
y =
1
2
gt2
→ t =
2y
g
⎛
⎝⎜
⎞
⎠⎟
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 =
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 be the 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 be the 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. 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 surface
Area 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 surface
Area of strip = Bδh
Velocity 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 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 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 = Bδh
Velocity of flow through strip = 2gH
Discharge through strip, δ !V = Area × Velocity = B 2g( )h
1
2
δh

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 = 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 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
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. 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 = bδh
Velocity of flow through strip = 2gh
Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )

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 = bδh
Velocity of flow through strip = 2gh
Discharge through strip, δ !V = Area × Velocity = bδh 2gh( )

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
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 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
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 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
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 rectangular notch, put b = constant = B
b = constant
B
H

57. KV
For a rectangular notch, 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 rectangular notch, 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 rectangular notch, 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 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 +
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 unit time = ρg!V
Power = ρg!VH
=ρg!V
p
ρg
+
u2
2g
+ z
⎛
⎝⎜
⎞
⎠⎟
= p!V+
1
2
ρu2 !V+ρg!Vz

68. KV

69. KV

70. KV
In a hydroelectric power 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 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/m3
(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/m3
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/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 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/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 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/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 points ➀ 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 points ➀ 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 points ➀ 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 points ➀ 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 points ➀ 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 points ➀ 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 points ➀ 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 points ➀ 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 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) 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