This document discusses fluid mechanics and hydraulics concepts including: 1. Definitions of density, specific gravity, atmospheric pressure, absolute and gauge pressure. 2. Descriptions of viscosity, laminar flow, turbulent flow, continuity equation, and steady vs unsteady flow. 3. Explanations of surface tension, capillarity, hydrostatic pressure, buoyancy, and center of pressure. 4. Discussions of manometers, energy equations, forces on submerged surfaces, and fluid static forces.

1. 9/21/2022
Revision of “ Hydraulics I - CVE 215”
Winter Semester 2015 - 2016
By: Dr. Ezzat El -Sayed G. SALEH

3. 3
Hydraulic I

4. 4

5. FLUID MECHANICS
3
m
kg
V
m
=
ρ 
kg
m
m
V
= 3

υ
3
m
KN
1000
g
V
g
m
V
W
= 




γ

6. Its specific gravity (relative density) is equal to the
ratio of its density to that of water at standard temperature and
pressure.
W
γ
γL

W
L
L
ρ
ρ
=
S
Its specific gravity (relative density) is equal to the ratio
of its density to that of either air or hydrogen at some specified
temperature and pressure.
ah
G
γ
γ
=
ah
G
G
ρ
ρ
=
S
where: At standard condition:
W = 1000 kg/m3
W = 9.81 KN/m3

7. Atmospheric pressure:
The pressure exerted by the
atmosphere.
At sea level condition:
Pa = 101.325 KPa
= .101325 Mpa
= 1.01325Bar
= 760 mm Hg
= 10.33 m of water
= 1.133 kg/cm2
= 14.7 psi
= 33.878 ft of water
Absolute and Gage Pressure
Absolute Pressure: is the pressure measured referred to
absolute zero and using absolute zero as the base.
Gage Pressure: is the pressure measured referred to
atmospheric pressure, and using atmospheric pressure as
the base

8. Atmospheric Pressure
Atmospheric pressure is
normally about 100,000 Pa
Differences in atmospheric
pressure cause winds to blow
Low atmospheric pressure
inside a hurricane’s eye
contributes to the severe
winds and the development of
the storm surge.
‫الجوي‬ ‫الضغط‬
‫اإلعصار‬ ‫عين‬ ‫داخل‬ ‫الجوي‬ ‫المنخفض‬
‫العواصف‬ ‫تطوير‬ ‫و‬ ‫شديدة‬ ‫رياح‬ ‫في‬ ‫يساهم‬

9. Moving Plate
Fixed Plate
y
dy
V + dv
V
V
  dv/dx
 =   (dv/dy)
 =   (V/y)
  =  /(V/y)
where:
 - absolute or dynamic viscosity
in “Pa.sec”
 - shearing stress in “Pascal”
V - velocity in “m/sec”
y - distance in “meters”.
Force F Area =
A

11. Surface tension is related to the cohesive properties of
water. Capillarity action however, is related to the
adhesive properties of water.
You can see capillarity action “in action” by placing a
straw into a glass of water. The water “climbs” up the
straw. Capillarity action is limited by gravity and the size
of the straw.
Plants take advantage of capillarity action to pull water
from the soil into themselves. From the roots water is
drawn through the plant by another force, transpiration.

12. The surface tension acts to hold the
surface intact, so instead of just the
edges moving upward, the whole liquid
surface is dragged upward
Surface Tension
Cohesive forces between molecules cause the
surface of a liquid to contract to the smallest
possible surface area. This general effect is
called surface tension

13. Cohesive Forces
Attractive forces between molecules of the same type are
called cohesive forces.
Adhesive Forces
Attractive forces between molecules of different types
are called adhesive forces.

14. 
r h
 

Where:
 - surface tension, N/m
 - density of liquid, kg/m3
g - acceleration of gravity
r - radius, m
H - capillary rise, m
C 
0 0.0756
10 0.0742
20 0.0728
30 0.0712
40 0.0696
60 0.0662
80 0.0626
100 0.0589
r
g
cos
2
h




Surface Tension of Water for
different temperature degrees

15. Surface Tension

16. The soap bubbles in this photograph are caused
by cohesive forces among molecules in liquids.
The soap bubbles that
the child blows into the
air maintain their shape
because of the cohesive
forces among
molecules in liquids..

17. 17

18. MANOMETERS
Manometer is an instrument used in measuring gage pressure in
length of some liquid column.
 Open Type Manometer : It has an atmospheric surface and is
capable in measuring gage pressure.
 Differential Type Manometer : It has no atmospheric surface and
is capable in measuring differences of pressure.
h
g
P



19. Open Type Manometer
Open
Indicating Liquid
Fluid A
Differential Type Manometer
Fluid B
Indicating Liquid
Fluid A

20. In steady flow the velocity of the fluid particles at any point
is constant as time passes.
 Unsteady flow exists whenever the velocity of the fluid
particles at a point changes as time passes.
 Turbulent flow is an extreme kind of unsteady flow in
which the velocity of the fluid particles at a point change
erratically in both magnitude and direction.

21. Steady Flow Through Pipes:

22. When the flow is steady, streamlines are often used to
represent the trajectories of the fluid particles.

23.  For steady flow, streamlines, path-lines, and streak-lines are
identical.
 For unsteady flow, they can be very different
 Streamlines are an instantaneous picture of flow field
 Path-lines and streak-lines are flow patterns that have a
time history associated with them.
 Streak-lines: instanteous snapshot of a time-integrated
flow pattern.
 Path-lines: time-exposed flow path of an individual
particle.
Comparisons

25.  Fluid flow can be compressible or incompressible.
 Most liquids are nearly incompressible.
 Fluid flow can be viscous or non-viscous.
 An incompressible, nonviscous fluid is called an ideal fluid.
More types of fluid flow

26. 2
2
2
2
V
A
t
m



1
1
1
1
V
A
t
m



V
m 
 
distance
t
V
A 


The Continuity Equation

27. Continuity equation gives:

28. 2
2
2
1
1
1 V
A
V
A 


The mass flow rate has the same value at every position
along a tube that has a single entry and a single exit for fluid
flow.
SI Unit of Mass Flow Rate: kg/s
The Continuity Equation

29. x
y
Open
Open
Fluid A
Fluid B
SAx = SB.y
Determination of S using a U - Tube

30. Reference Datum (Datum Line)
1
2
z1
Z2
HL = 1  2
Energy and Head
Bernoulli’s Energy equation:

31. 31
Comparison between Absolute and Gage Pressures

32. 32
A closed tank is filled with water
and has a 1.2 m-diameter
hemispherical dome as shown. A
U-tube manometer is connected to
the tank. If the air pressure at the
upper end of the manometer is 87
KPa, determine:
3)
(SG
Liquid
indicating

m
5
.
1
m
6
.
0
m
6
.
0
Water
diameter
m
1.2
f
o
dome
cal
Hemispheri Air
A
P
????
PB 
a
A kP
87
P 
"
B
"
The gage reading at “B” ,
The vertical force of the water
on the dome

33. 33
3)
(SG
Liquid
indicating

m
5
.
1
m
6
.
0
m
6
.
0
Water
diameter
m
1.2
f
o
dome
cal
Hemispheri Air
A
P
????
PB 
a
A kP
87
P 
"
B
"
1 2
i
i
air
2
1 h
g
P
P
P 



If we start at “A” and move vertically downward to level “1”, the pressure will increase
by PA +i g hi and will be equal to the pressure at “2”. Thus, in equation form
……………….. (1)

34. 34
B
w
w
i
i
air P
h
g
h
g
P 





B
3
3
3
3
P
kPa
137
10
)
6
.
0
6
.
0
(
81
.
9
10
)
6
.
0
5
.
1
(
81
.
9
10
3
10
87













If we move from “1” vertically upward to level “B”, the pressure will decrease by:
w g hw . Eq. (1) gives
For the data given the pressure at point “B” is
)
kP
137
P
(
at
that
as
same
the
is
level
l
horizonata
same
the
at
pressure
The
a
B 
"
B
"
The vertical force of the water on the dome is
kN
55
.
1
)
4
/
2
.
1
(
37
.
1
)
area
projected
(
P
F 2
B
Dome 







35. 35
A
P
A
h
g
A
)
sin
y
(
g
F 








 sin
y
h
35
Hydraulic I – CVE 215 (6) Fluid Static Forces
Substituting into Eq. (1) gives:
y
h 
Hydrostatic Force on an Inclined Plane Surface (magnitude)

36. 36
 The location of the center of pressure is
independent of the angle ,
 The center of pressure is always below the
centroid,
 As the depth of immersion increase, the depth
of the center of pressure approaches the
centroid..
y
A
I
y
y
g
.
c
cp 

36
Hydraulic I – CVE 215 (12) Fluid Static Forces
Line of action of F

37. 37

C.G
C.P
Free Surface
A Completely Submerged Tilted Rectangular Plate
)
b
a
(
sin
)
2
/
a
S
(
g
A
h
g
F 








)
b
a
(
)
2
/
a
S
(
)
12
/
a
b
(
)
2
/
a
S
(
y
3
cp










 sin
)
2
/
a
S
(
h
h
y
cp
y
S
F
a




38. 38

C.G
C.P
Free Surface


 sin
)
2
/
a
(
h y
a
cp
y
F
)
b
a
(
sin
)
2
/
a
(
g
A
h
g
F 








h
 
 
a
3
2
)
b
a
(
2
/
a
)
12
/
a
b
(
2
/
a
y
3
cp 





When the Upper Edge of the Submerged Tilted Rectangular Plate is
Free Surface and thus (S = 0)



h

39. 39
 =90o
C.G
C.P
Free Surface
A Completely Submerged Vertical Rectangular and thus ( = 0)
a
S
1
90
sin
:
where
)
2
/
a
S
(
h
o



cp
cp h
y 
h
y 
F
  )
b
a
(
2
/
a
S
g
A
h
g
F 







 
)
b
a
(
)
2
/
a
S
(
)
12
/
a
b
(
2
/
a
S
y
3
cp











40. 40
 =90o
C.G
C.P
Free Surface
h
y 
cp
cp h
y 
0
S
,
1
90
sin
:
where
2
/
a
h
o



a
F
When the Upper Edge of the Submerged Vertical Rectangular
Plate is at the Free Surface and thus (S = 0 &  = 90o)
)
b
a
(
)
2
/
a
(
g
A
h
g
F 






a
3
2
)
b
a
(
)
2
/
a
(
)
12
/
a
b
(
)
2
/
a
(
y
3
cp 







41. 41
A
B
C
FH, U
D
FH, U
FH, D
FV, U FV, D
O
E


)
W
"
AD
("
"
AD
"
g
F U
,
H 



 )
W
"
OA
("
"
OA
"
g
F D
,
H 




  W
BCD
area
ABCD
area
g
F U
,
V 




  W
AOE
area
g
F D
,
V 




3
/
r
4
“Static Forces on Curved Surfaces”

42. 42
Sketch Area
Location
of
Centroid
Ic.g or I
Rectangle
b
h
c
y
g
.
c
I
h g
.
c
I
c
y
b
h
b
2
h
yc 
12
h
b
I
3
g
.
c 
Triangle
b
h
c
y
g
.
c
I
h g
.
c
I
c
y
b
2
h
b
3
h
yc 
36
h
b
I
3
g
.
c 
Ellipse
D
c
y
g
.
c
I
h g
.
c
I
c
y
b
4
h
b 

2
h
yc  64
h
b
I
3
g
.
c


Semi ellipse
h
c
y
I
h
I
c
y
b
b
c
x
b
4
h
b



3
h
4
yc
16
h
b
I
3


Circle D
c
y
g
.
c
I
h g
.
c
I
c
y
b
4
/
D2

 2
/
D
yc 
64
D
I
2
g
.
c


Parabola
h
c
y
I
c
x
b
h
b
3
2
8
b
3
x
5
h
3
y
c
c


7
h
b
2
I
3


43. Inclined Surface Findings
 The horizontal center of pressure and the horizontal
centroid coincide when the surface has either a
horizontal or vertical axis of symmetry.
 The center of pressure is always below the centroid
 The vertical distance between the centroid and the
center of pressure decreases as the surface is lowered
deeper into the liquid
 What do you do if there isn’t a free surface?




y
A
I
y
y
G
.
C
cp

44. Forces on Curved Surfaces:
Horizontal Component
 The horizontal component of pressure force on a curved
surface is equal to the pressure force exerted on a
horizontal projection of the curved surface
 The horizontal component of pressure force on a closed
body is always zero.
 The center of pressure is located on the projected area
using the moment of inertia.

45. Forces on Curved Surfaces: Vertical
Component
The vertical component of pressure force on a
curved surface is equal to the weight of liquid
vertically above the curved surface and
extending up to the (virtual or real) free surface.
Streeter, et. al

46. C
0.948 m
1.083 m
89.7kN
78.5kN
Cylindrical Surface Force Check
•All pressure forces pass
through point.
•The pressure force C
applies no moment about
point C.
•The resultant must pass
through point C.

47. 47
o
5
4
A
B
m
4
m
2
water

Gate AB shown in the figure
is a three-eights (3/8) circle,
3 m wide, hinged at “B”, and
resting against a smooth wall
at “A”. Compute the reaction
force at points A and B.

48. 48
o
5
4
A
B
m
2
m
2
water

C
g
w
)
AC
curve
the
above
water
of
weight
he
T
(
Fy 




0
Fx 

x
F
y
F y
F
x
F
m
2
M
For the curve CMA

49. 49
To compute the summation of the vertical
forces on the curve CMA
o
5
4
A
m
2
m
2
water

C
x
F
y
F y
F
x
F
2
/
2
2
/
2 2
/
2
D E
M
y
F

B

50. 50
B"
"
hinge
to
right
m
2.0
acts
It
kN
98
.
132
1000
)
14
.
1
66
.
5
(
81
.
9
1000
3
81
.
9
1000
3
2
2
2
4
2
1
4
2
)
2
/
4
(2
g
w
)
AMCA
Area
ACDE
area
The
(
g
w
)
AMCDE
area
The
(
F
2
)
CMA
(
y




































 














o
5
4
A
m
2
m
2
water

C
W
= 3 m
x
F
y
F y
F
x
F
2
/
2
2
/
2 2
/
2
D E
M
y
F

B

51. 51
o
5
4
A
B
m
4
m
2
water

C
x
F
y
F
For the curve CB
x
F G
C
P
C
2
/
2
m
29
.
3
2
2
2
4
h




















 kN
137
)
2
/
2
3
(
2
2
2
4
81
.
9
1000
A
h
g
Fx
= 3.29 m
It acts a vertical distance = hCP down the free surface, where

52. 52
o
5
4
A
B
m
4
m
2
water

C
)
2
/
2
(
2
m
59
.
0
)
2
/
2
(
2 

F
G F
G
B
m
41
.
1
2
/
2 
C
2
/
2
m
4
N
2
,
y
F
1
,
y
F
2
,
y
1
,
y
y F
F
F 


  

















kN
88
.
57
81
.
9
1000
3
)
59
.
0
41
.
1
3
2
(
)
59
.
0
4
(
g
W
)
BCN
area
BGFN
area
(
F
F
F 2
,
y
1
,
y
y

53. 53

54. 54
The kinetic and potential energies of the fluid can be
converted to flow energy (and vice versa) during flow,
causing the pressure to change. Multiplying the
Bernoulli's equation by the density “” , each term in the
equation has pressure.
.
const
z
g
V
2
1
P 2





…… . (along a stream
line)

55. 55
It is often convenient to represent the level of mechanical
energy graphically using heights to facilitate visualization
of the various terms of the Bernoulli equation. This is
done by dividing each term of the Bernoulli's equation by
g to give:
.
const
H
z
g
2
V
g
P 2





Total Head
Potential Head
Velocity Head
Pressure Head

56. 56
)
ft
&
m
(
L
)
T
L
/
M
(
)
L
T
/
M
(
)
T
/
L
(
)
L
/
M
(
)
L
T
/
L
M
(
)
T
/
L
(
)
L
/
M
(
L
/
F
g
P
2
2
2
2
3
2
2
2
3
2








)
ft
&
m
(
L
)
T
/
L
(
)
T
/
L
(
)
T
/
L
(
)
T
/
L
(
g
2
V
2
2
2
2
2
2



)
ft
&
m
(
L
Z 

Pressur
e Head

Velocity
Head 
Elevatio
n Head

To check the dimensions of each separate component of the
Bernoulli's equation,
Thus, all he constituent parts of the equation have units of length.
For this reason, each term may be regard as “ a head “

57. 57
Z: is the Elevation head;
 It represents the potential head of the fluid.
 V 2/2g : is the velocity head;
 It represents the elevation needed for a fluid to reach
the velocity V during frictionless free fall.
 P/ g : is the pressure head
It represents the height of a fluid column the produces
the static pressure

58. 58
The sum of pressure, velocity, and elevation heads along a
stream line is constant during steady flow when the
compressibility and frictional effects are negligible.
Bernoulli's equation can be expressed in terms of head
as:
H: is the total head

59. 59
 Is a line that represents the total head of the fluid,
Energy grad line (EGL)
Z
g
2
V
g
P 2




60. 60
The difference between the heights (EGL) and (HGL) is
equal the dynamic head,
g
2
V2

61. 61
For stationary bodies such as reservoirs or lacks,
the EGL and HGL coincide with the free surface of
the liquid. (The elevation of the free surface z in
such cases represents both the EGL and HGL since
the velocity is zero and the state pressure “gage” is
zero).

62. 62
g
2
V
g
p
Z
EGL
2





Pressure head
(w.r.t. reference pressure)
EGL (or TEL) and HGL
velocity
head
Elevation head
(w.r.t. datum)
Piezometric head
Energy
Grade Line
Hydraulic
Grade Line
What is the difference between EGL defined by Bernoulli
and EGL defined here?
g
p
Z
HGL




63. Orifice: An orifice is an any opening with a closed perimeter (without
considering Head Loss)
1
2
a
a
Vena Contracta
h
By applying Bernoulli's Energy theorem:
2
2
2
2
1
1
g
2
v
g
P
Z
g
2
v
g
P
Z 2
2
1 






But P1 = P2 = Pa and v1is negligible, then
2
1
2
2
Z
Z
g
2
v


and from figure: Z1 - Z2 = h,
therefore
h
g
2
v
2
2

h
g
2
v2

Let v2 = vt
h
g
2
vt

where:
vt - theoretical velocity, m/sec
h - head producing the flow, meters
g - gravitational acceleration, m/sec2

64. velocity
l
theoretica
velocity
actual
v
C 
th
act
v
v
v
C 
orifice
the
of
area
contracta
vena
@
jet
of
area
Cc 
A
a

Cc
discharge
l
theoretica
discharge
actual
C 
d
th.
act
Q
Q
C 
d
v
c
d C
C 

C
where:
Vact : actual velocity
Vth : theoretical velocity
a : area of jet at vena contracta
A : area of orifice
Qact : actual flow
Qth: theoretical flow
Cv: coefficient of velocity
Cc : coefficient of contraction
Cd : coefficient of discharge

66.   meters"
"
H
Z
Z
g
2
V
V
P
P
H L
1
2
2
1
2
2
1
2
t 







g
/s"
m
"
Q 3
d
d
S
S V
A
V
A 



KW"
"
W T
H
Q
g
p 


KW"
"
W
000
.
60
N
T
2
p




67. HYDRO ELECTRIC POWER PLANT
Headrace
Tailrace
H – Gross Head
Penstock turbine
1
2



68. B. Reaction Type turbine (Francis Type
Headrace
Tailrace
H – Gross Head
Penstock
ZB
1
2
Draft Tube
B
Generator
B – turbine inlet

69. 69
69
Purpose
 The energy grade line may never slope upward (in
direction of flow) unless energy is added pump
 The decrease in total energy represents the head loss
or energy dissipation per unit weight
 EGL and HGL are coincident and lie at the free
surface for water at rest (reservoir)
 Whenever the HGL falls below the point in the
system for which it is plotted, the local pressures are
lower than the reference pressure

70. 70
Example: Energy Equation
(Hydraulic Grade Line - HGL)
₪ We would like to know if there are any places in
the pipeline where the pressure is too high (Pipe
burst) or too low (water might boil - cavitations).
₪ Plot the pressure as piezometric head (height
water would rise to in a piezometers
₪ How?

71. EGL (or TEL) and HGL
• The energy grade line may never be horizontal or slope
upward (in direction of flow) unless energy is added
“Pump”.
• The decrease in total energy represents the head loss or
energy dissipation per unit weight
• EGL and HGL are coincident and lie at the free surface
for water at rest “reservoir”
• Whenever the HGL falls below the point in the system for
which it is plotted, the local pressures are lower than the
reference pressure.
L
T
2
2
2
2
2
2
p
2
1
1
1
1
1 h
H
g
2
V
g
p
Z
H
g
2
V
g
p
Z 












75. Lower Reservoir
Upper
Reservoir
Suction Gauge Discharge Gauge
Gate Valve
Gate
Valve



76. Pump-Storage Hydroelectric power plant:
During power generation the turbine-pump
acts as a turbine and during off-peak period
it acts as a pump, pumping water from the
lower pool (tailrace) back to the upper pool
(headrace).
Turbine-Pump


(headrace).
(Tailrace).

77. • Venturi meter
• Orifice meter
• Nozzle meter
Flow
Measurement
Devices

78. • Venturi meter
• Orifice meter
• Nozzle meter
Flow
Measurement
Devices

79. B) Orifice-meter

80. Ventur-imeter is an instrument
used to measure fluid flow
velocity in the pipeline

81. Ventury Meter
 A. Without considering Head loss
flow
l
theoretica
Q
v
A
v
A
Q
Z
g
2
v
g
P
Z
g
2
v
g
P
th
2
2
1
1
2
2
2
2
1
2
1
1










Manometer
1
2
B. Considering Head loss
flow
actual
Q
v
A
v
A
Q
H
Z
g
2
v
g
P
Z
g
2
v
g
P
act
2
2
1
1
act
L
2
2
2
2
1
2
1
1











Meter Coefficient
th
act
d
Q
Q
C 

82. A venture- meter is used to
measure the flow speed of
a fluid in a pipe. The meter
is connected between two
sections of the pipe (see the
figure).

83. 83
P
V
f
,
L
h
g
2
/
V2
P
.
ent
,
L
h

H EGL
HGL
The pipe has a sharp entry
The energy “H” of the water in the reservoir is dissipated in:
(a) Overcoming head loss at. entrance to the pipe hL,entr,
(b) Overcoming friction loss hLf. In the pipe, and
(c) Imparting velocity energy VP
2 / 2g to the water at exit (outlet).
How to Draw or at least Sketch the HGL & EGL

84. 84
P
V
f
,
L
h
g
2
/
V 2
n

H
EGL
HGL
n
V
g
2
/
V 2
P
 The energy “H” of the water in the reservoir is divided between a friction loss
hL,f and velocity energy (head) at exit Vn
2 / 2g .
The pipe has a rounded entry and nozzle at outlet
How to Draw or at least Sketch the HGL & EGL

85. A Pitot static Tube

86. Pitot-
Probe
D =……..
 
f = … & L =….
 Pitot Probes “also called
Pitot”: are widely used for
flow velocity measurements.
It is a small tube with its
open end aligned into the
flow so as to sense the full
impact pressure of the
flowing fluid.
 It measure the stagnation
pressure.
A Pitot Tube
Stagnation
Point

87. A Pitot probe: is just a tube with a pressure tap at the stagnation point
that measures stagnation pressure, while a Pitot-static probe has both a
stagnation pressure tap and several circumferential static pressure taps
and it measures both stagnation and static pressures..
(a) A Pitot probe measures stagnation pressure at the nose of the probe,
while
(b) a Pitot-static probe measures both stagnation pressure and static
pressure, from which the flow speed is calculated.

88. Measuring flow velocity with a
Pitot-static probe. (A manometer
may be used in place of the
differential pressure transducer).
Close-up of a Pitot-static probe, showing
the stagnation pressure hole and two of the
five static circumferential pressure holes.

89. A Pitot tube is mounted in the
airplane body to register the
airspeed.

90. 90
A siphon is a defined upward sweep of a pipe above its hydraulic gradient and
below again as shown in the above figure. The purpose of the siphon is to
carry the pipe over an intervening bank.


S
b
a
c
Flow through Siphon
The limiting factor in the flow through a siphon is the negative head at “SC”, so
long as this is not too close to the maximum possible negative head, there will
be a hydraulic gradient “abc” just as though the pipe head been driven straight
through the bank.
Intervening
bank

91. 91
Since the pipe has a constant diameter in all reaches of 6.0 m, the velocity head
terms are constant and canceled. One can easily conclude that:
"1"
section
at
EGL
of
Elevation
"2"
section
at
EGL
of
Elevation 
i.e the liquid flows from “2” to “1”
Now to calculate the head loss over the 6.0m pipe length
m
50
.
1
50
.
1
0
.
3
)
elevation
EGL
(
)
elevation
EGL
(
h 1
2
2
1
L 





The velocity head
Terms canceled
Energy Equation and its Applications

92. 92
h 100 mm
300 mm
0.6 m
Oil (SG = 0.9) flows downward through a
vertical pipe contraction as shown in the
figure. If the mercury manometer reading,
h, is 100 mm,
Determine the flow rate for frictionless
flow.
Is the actual flow rate more or less than
the frictionless value? Explain.
Energy Equation and its Applications

93. 93
With the assumptions of steady, incompressible
flow, the energy equation can be written as
2
1
10 cm
A B
100 mm
300 mm
2
1
L
2
2
oil
2
2
2
1
oil
1
1 h
g
2
V
g
P
Z
g
2
V
g
P
Z 








A
A
V
V
or
A
V
A
V
2
1
1
2
2
2
1
1 




2
2
2
1 V
9
1
300
100
V
V 









)
9
1
1
(
g
2
V
)
Z
Z
(
g
P
P 2
2
1
2
oil
2
1






 0
x
0.6 m
Energy Equation and its Applications
Assuming a uniform flow, the continuity equation
provides us with the following equation,
By combining these two equations we obtain

94. 94
g
2
V
89
.
0
60
.
0
g
P
P 2
2
oil
2
1





Energy Equation and its Applications
Substituting Z1= 0.60 m and Z2 = 0 gives
………. (1)
The pressure difference is measured by the given mercury manometer, thus
B
A P
P 
h
g
x
g
P
g
)
h
x
60
.
0
(
P m
oil
2
oil
1 











or
Substituting Z1- Z2= 0.60 m, h = 0.10m, oil = 0.90 and m = 13.60, Eq. (2)
becomes
………. (2)
060
41
.
1
60
.
0
10
.
0
)
1
90
.
0
60
.
13
(
60
.
0
h
)
1
(
g
P
P
oil
m
oil
2
1














… (3)

95. 95
………. (4)
60
.
0
41
.
1
81
.
9
2
V
89
.
0
60
.
0
2
2






s
/
m
58
.
5
89
.
0
81
.
9
2
41
.
1
V2 




s
/
L
8
.
43
s
/
m
0438
.
0
m
58
.
5
10
.
0
4
V
A
Q 3
2
2
2 














Energy Equation and its Applications
Using Eq. (1) and Eq. (3) to have the desired result, we have
Or the average velocity at section “2”
or
The above theoretical result would undoubtedly larger than the actual flow rate
because of various difference between the real and the assumptions used in the
derivation of the equations. This difference depends on the pipe contraction
geometry used.

96. 96
An incompressible liquid flows steadily along the pipe shown in the
figure.
Determine the direction of flow and the head loss
over the 6 m length of pipe.
Energy Equation and its Applications
1.5 m
1.0 m 3.0 m
0.75 m

97. 97
Energy Equation and its Applications
1.5 m
1.0 m 3.0 m
0.75 m
 
Application of the energy equation to the contents of the shown control volume, we
can find the total head at section “1” as
g
2
V
50
.
2
g
2
V
0
.
1
50
.
1
g
2
V
g
P
Z
"1"
section
at
EGL
of
Elevation
2
2
1
2

















g
2
V
00
.
3
g
2
V
0
.
3
0
.
0
g
2
V
g
P
Z
"2"
section
at
EGL
of
Elevation
2
2
2
2

















Similarly,

98. pump
power
input
p
h
w
power
input
power
output
ffeciency
Pump p



 A
 B
HP
EGL
P
H
)
A
(
)
B
( 

  p
P /
H
Q
g
power
Required 


Neglecting all Losses (minor & main)

99. pump
power
input
P
H
W
power
input
power
output
ffeciency
Pump



 A
 B
HP
P
L H
h
)
A
(
)
B
( 

 
P
H
Q
g
Power 

hL (2) 
hL (1)


Neglecting Minor Losses Only (friction losses only)

100. 100
Turbine
p
h
w
power
output
power
input
power
output
ffeciency
Turbine T



 A
 B
HT
EGL
T
H
)
A
(
)
B
( 

 
T
H
Q
g
power
Output T


 
Neglecting all Losses Only (minor & main)

101. 101
El. 50
El. 40
El. 150 ft
1000 ft &
D1=8 in
P


A
B
HP
hL(2)
hL(1)
p
)
2
(
L
)
1
(
L H
)
B
A
(
h
h 



2
5
2
2
L Q
D
g
L
f
8
g
2
V
D
L
f
h 



c
/
H
Q
g
power p


Neglecting all losses except
main losses (friction losses)
EGL
Rise in EGL and HGL due to Pump
HGL

102. 102
El. 50
El. 40
El. 150 ft
1000 ft &
D1=8 in
T


A
B
HT
hL(1)
hL(2)
)
B
A
(
H
h
h T
)
2
(
L
)
1
(
L 



2
5
2
2
L Q
D
g
L
f
8
g
2
V
D
L
f
h 



c
/
)
H
Q
g
(
power T


Neglecting all losses except
main losses (friction losses)
EGL
Drop in EGL and HGL due to Pump
HGL

103. 103
Cross Section in a Power Station

104. 104
If there is no pump, 0.14 m3/s of water
would flow through this pipe system.
Calculate the pump power required to
maintain the same flow rate in the
opposite direction.
Draw the EGL and HGL
30.00) 
(42.00) 
300 m- 30 cm dia.
600 m- 30 cm dia.
Pum
p
Faculty of Engineering (Hydraulic I)- CVE215- Feb. 2012

105. 105
(30)  1
1  (42)
300 m- 30 cm dia.
600 m- 30 cm dia.
Pum
p
 There is no pump
EGL
HGL
Reference datum
Z1
Z2
hL 2 1 Vp
2 / 2g
Faculty of Engineering (Hydraulic I)- CVE215- Feb. 2012

106. 10
6

107. 107
• Newton’s second law relates net force and
acceleration. A net force on an object will accelerate
it - that is, change its velocity. The acceleration will
be proportional to the magnitude of the force and in
the same direction as the force. The proportionality
constant is the mass, m, of the object.
dt
)
V
m
(
d
dt
V
d
m
a
m
F









Newton’s Second Law

108. 108
dt
)
V
m
(
d
dt
V
d
m
a
m
F









Net force
Rate of change
of momentum
Momentum = Mass in
motion
Momentum

109. 109

110. 110
• Momentum is a vector quantity.
The direction of the
momentum vector is the
same as the direction of
the velocity vector
Momentum

111. 111
Momentum
 All objects have mass; so if an object is moving,
then it has momentum - it has its mass in
motion.
 The amount of momentum which an object has
is dependent upon two variables:
How much matter is moving?
How fast the matter is moving?

112. 112
The momentum equation may be used directly to evaluate
the force causing a change of momentum in a fluid:
Applications :determining forces on:
• Pipe bends,
•Junctions,
•Nozzles, and
•Hydraulic machines.
Linear Momentum Equation

113. 113
In addition, the momentum equation is used directly to
evaluate to solve problems in which energy losses occur that
cannot be evaluated directly, or when the flow is unsteady
Examples of such problems include:
•Local head losses in pipes,
•The hydraulic jump, and
•Unsteady flow in pipes and channels.
Linear Momentum Equation

114. 114
Expansion Joints
Concrete Anchors
Bend
30o

P1
P2
V1
V2
y
x
The general momentum equation
for steady one-dimensional flow is:
30o
The continuity equation gives:
A pipe has a 30o horizontal bend in it
anchor
F

115. 115
Control Surface




The continuity equation gives:
The general momentum equation
for steady one-dimensional flow is:
x
y
P1
V1 V2
P2
)
V
V
(
Q
F
p
p
1
2
bold
2
1





A Fitting between Two Pipes of Different Size (TRANSITION)
bold
F

116. 116
450




30 cm-
dia. pipe
V2 cos
V2 sin
FX
Fy
V1

X
y
P1
P2
A 30 cm diameter pipe carries water under a head of 20 m
with a velocity of 3.5 m/s. If the axis of the pipe turns through
45o, find the magnitude and direction of the resultant force
on the bend. Neglect friction forces.
Worked Example

117. 117
7.5 cm-dia
Nozzle




V1
V2
30 cm-dia
A 30 cm diameter horizontal pipe terminates in a nozzle with
the exit diameter of 7.5 cm. If the water flows through the
pipe at a rate of 0.15 m3/s, what force will be exerted by the
fluid on the nozzle?
Worked Example (cont.)

118. 118
Consider a water jet is deflected by a stationary
vane as shown. Determine the force acting on
the vane by the jet if the jet speed is 100 ft/s
and the diameter of the jet is 2 in. and there is
no significant divergence of the jet flow during
impact.
75°
Vin=100 ft/s
Fin
Fout
Fx
Fy
 Assume steady state, shear action on the
fluid does not slow down the jet
significantly and the jet velocity is uniform.
 Fin & Fout will be static pressure acting on
the jet.
 W is the weight of the water jet inside the
CV.
 Assume all three contributions are also
small relative to the momentum of the jet.
W
Worked Example

119. 119
A closed tank on wheels 1 m x 1.25 m in plan, 4.5 m high and weighing
1175 N is filled with water to a depth of 3 m. A hole in one of the side wall
has an effective area of 7.5 cm2 and is located at 20 cm above the tank
bottom. If the coefficient of friction between the ground and the wheels is
0.012, determine the air pressure in the tank that is required to set it into
motion.
Air, P
3 m
4.5 m
V
20 cm
Water
Worked Example

120. 120
o
45
o
45
s
/
m
30
cm
5
.
2
o
45
o
45
s
/
m
30
cm
5
.
2
s
/
m
3
A Pelton wheel vane directs a horizontal, circular cross sectional jet of water
symmetrically as indicated in the figure. The jet leaves the nozzle with a velocity
of 30 m/s. Determine the x-direction component of anchoring force required to:
hold
 Hold the vane stationary
 Confines the speed of the vane to a value of 3 m/s to the right (the fluid speed
remains constant along the vane surface).
Impulse Momentum Equation Applications

121. 121
o
45
o
45
cm
5
.
2
s
/
m
30
V1 
s
/
m
30
0
30
V
V
V 2
1
r





0
V2 
V
.
C


Impulse Momentum Equation
Applications
The fluid speed remains
constant along the vane
surface
x
y
.
anch
F
With x positive to the right, equal to
the net horizontal force to the x-
direction momentum change
)
V
V
(
Q
F
F in
out
.
anch
x 



 ………. (1)
With Fx represents the net force acting on the control volume due to the given
causes; it is needed to balance the change in momentum of the fluid as it turns and
decelerates while passing through the control volume.
s
/
m
30
Vr 

122. 122
Impulse Momentum Equation Applications
 The relative velocity Vr = 30 – 0 = 30 m/s
 The discharge Q= A x Vr = (/4x0.025)2 x30 =0.0147 m3/s
There are one inlet and two outlets and the pressure force is zero in the uniform
atmosphere. Therefore the simple form for the desired result is
)
45
cos
1
(
V
Q
)
Vr
45
cos
V
(
Q
F
o
r
o
r
.
anch











………. (2)
Substituting the above values into Eq. (2) gives
N
753
)
45
cos
1
(
30
0147
.
0
10
F o
3
.
anch 





Holding the vane stationary

123. 123
o
45
o
45
s
/
m
30
V1 
cm
5
.
2
s
/
m
3
V2 
s
/
m
27
3
30
V
V
V 2
1
r





V
.
C


s
/
m
27
Vr 
s
/
m
27
3
30
V
V
V 2
1
r





Impulse Momentum Equation Applications
 The relative velocity Vr = 30 – 3 = 27 m/s
 The discharge Q= A x Vr = (/4x0.025)2 x27 =0.0132 m3/s, and
N
611
)
45
cos
1
(
27
0132
.
0
10
F o
3
.
anch 





Confines the speed of the vane
to a value of 3 m/s to the right

124. 124


d
ft
90
.
0
h 
ft
0
.
1
H 
②
①
A
B
A jet of water is discharging at a constant rate of 1.2 ft3/s from
the upper tank. If the jet diameter at section ① is 4 in.
Assume the empty tank weights 300 Ibf, the cross-sectional
area of the tank is 4 ft2, h = 0.90 ft, and H =1.0 ft.
 What forces will be measured by scales A and B?????

125. 125


d
ft
90
.
0
h 
ft
0
.
1
H 
②
①
A
B
 
s
/
ft
2
.
1
Q
&
ft
)
12
/
4
(
d 3


x
,
1
1 V
V 
s
/
ft
75
.
13
)
12
/
4
(
2
.
1
A
/
Q
V
section
at
velocity
mean
The
2
1





V
V
V
,
resistance
air
the
Negligting x
,
2
x
,
1
1 


126. 126


d
ft
90
.
0
h 
ft
0
.
1
H 
②
①
A
B s
/
ft
75
.
13
V
V x
,
1
x
,
2


y
,
2
V
s
/
ft
93
.
15
V2 

x
,
1
1 V
V 
o
1
2
2
,
x
1
33
.
30
)
93
.
15
/
75
.
13
(
cos
)
V
/
V
(
cos






s
/
ft
31
.
9
33
.
30
tan
93
.
15
tan
V
V
o
2
,
x
2
,
y






:
gives
"2"
and
"1"
between
equation
s
Bernoulli'
Applying
g
2
V
g
P
Z
g
2
V
g
P
Z
2
2
1
2























Reference datum
x
y

127. 127
:
gives
eq.
energy
the
???,
V
and
ft/s,
13.75
V
(gage),
0
P
P
P
0,
Z
ft,
1.0
H
Z
Since
2
1
atm.
2
1
2
1








g
2
V
0
0
2
.
32
2
75
.
3
1
0
1
2
2






















s
/
ft
93
.
15
2
.
32
2
94
.
3
V
Then, 2 



2.94
ft/s
31
.
9
V
ft/s,
13.75
V
V
above,
the
in
shown
geometry
the
From
y
2,
x
,
1
x
2,















Ibf
64
.
524
)
90
.
0
4
(
4
.
62
300
tank)
the
inside
water
of
volume
(
g
300
water
of
weight
the
tank
the
of
weight
The
w

128. 128
Ibf
32
)
75
.
13
0
(
2
.
1
94
.
1
)
V
V
(
Q
F
obtain
we
data
given
the
With
direction
x
in
in
out
x 







 








 








 

"
B
"
scale
Ibf
32
B"
"
scale
by
measured
orce
f
The 
  Ibf
54
.
21
)
25
.
9
(
0
2
.
1
94
.
1
)
V
V
(
Q
F
Similarly,
direction
y
in
in
out
y 







 












 












 

"
A
"
scale
Ibf
1
.
503
54
.
21
64
.
24
5
A"
"
scale
by
measured
orce
f
The 



129. 129

130. 130
 Obsorn Reynolds was the first to demonstrate that laminar or
turbulent flow can be predicted if the magnitude of a dimensionless
number, now called the Reynolds number, Re is known. The basic
definition of the Reynolds number is,
How to Determine the Type of Flow???








D
V
D
V
Re
Critical Reynolds Number
 For practical applications in pipe flow, it was found that if:

131. 131
Critical Reynolds Number
 Re is less than 2000  The flow will be laminar
 Re is between 2000 and 4000  It is impossible to predict which type of
flow exists
Re is greater than 4000  The flow is turbulent
 It is impossible to predict which type of flow that of Reynolds
number in the range between 2000 and 4000

132. 132
Darcy’s Equation for Energy Loss
 For the case of flow in pipes and tubes, Darcy’s equation for energy
loss is expressed mathematically as,
g
2
V
D
L
f
h
2
L 


where
hL : energy loss due to friction (m or ft),
L : length of pipe flow (m or ft),
D : diameter of pipe flow (m or ft), and
V : the average velocity of flow (m/s or ft/s), and
F : friction factor (dimensionless).
……………………..…( )

133. Minor Losses
some valves + flow meters +
pipe fittings (elbows, bends,
..etc).

134. 134
Valves + Pipe fittings (elbows, bends, ..etc).

135. 135

136. 13
6
Flow Through a Valve

137. 13
7
where:
VV = flow velocity in valve throat
Vp = flow velocity in pipeline.
K = resistance coefficient which is found
experimentally by the valve manufacturer for
different valve openings.
Note: K depends greatly on the %age of valve
opening (as shown in the next Table)
g
2
V
h
2
V
L 

138. Lower Reservoir
Upper
Reservoir
Suction Gauge Discharge Gauge
Gate Valve
Gate
Valve



139. 139


flow
Reservoir
S
t
r
e
a
m
Pump
Steel pipeline
8
-in
dia.
and
2500
ft long
210 ft
Water is being pumped from a
stream to a reservoir whose
surface is 210 ft above the
pump. The pipeline from the
pump to the reservoir is 8-in
diameter steel pipe, “ f = 0.016
“ and 2500 ft long. If 4.0 ft3/s is
being pumped,
- compute the pressure at the
outlet of the pump. Consider
only the friction loss in the
pipeline, neglecting other
losses.
- For the same pump, if the
pressure at the pump inlet is
– 2.36 psi, compute the power
delivered by the pump to the
water.

140. 140


flow
Reservoir
S
t
r
e
a
m
Pump
Steel pipeline
8
-in
dia.
and
2500
ft long
210 ft
②
①
 Applying Bernoulli's Equation
between points ① and ② at the
surfaces of both the stream
and the reservoir gives
2
2
2
1
L
1
2
g
2
V
g
P
Z
h
g
2
V
g
P
Z
























 
 Taking the reference datum passes through outlet of the pump,
Z1 = 210 m P1 = Patm  0, V1 = 0
Z2 = 0 V2 = Q/A2=4/{(8/12)2} =2.86 m/s
2
5
2
f
L
2
1
L Q
D
g
L
f
8
h
h 



 
…(1)
…(1)

141. 141
Substituting into Eq. (1) the given data for this condition,
  






















2
.
32
2
86
.
2
g
P
0
4
)
12
/
8
(
2
.
32
2500
016
.
0
8
0
0
210
2
2
2
5
2
psi
2
.
144
12
2
.
32
94
.
1
4
.
332
P
m
4
.
332
)
127
.
0
34
.
122
210
(
g
P
or
2
2
2










122.34
….. (2)
⍟If the pressure at the pump inlet is – 2.36 p.s.i
 Neglecting the difference elevation between the inlet and outlet of the pump and
applying Bernoulli’s equation

142. 142
Pump
③
 Taking the reference datum
passes through outlet of the pump,
Z2 = 0 P2 = 144.2 p.s.i V2 = V3 the inlet and outlet
pipes have the same diameters.
Z3 = 0 P3 = -2.36 p.s.i
(3)
② p
3
2
2
2
h
g
2
V
g
P
Z
g
2
V
g
P
Z 
























 Substituting into Eq. (3) gives
p
2
3
2
2
h
g
2
V
g
36
.
2
0
g
2
V
g
2
.
144
0 

























143. 143
m
328
2
.
32
94
.
1
12
56
.
146
g
)
36
.
2
2
.
144
(
h
2
p 







p
p h
150
550
328
4
2
.
32
94
.
1
h
Q
g
pump
the
by
delivered
power
The









144. 144
 650 m
 653 m
Length L, D = 7 cm
and f = 0.02
Q = 11 L /s
Water is pumped from a lake as shown. If the
flow rate is 0.011 m3/s, what is the maximum
length inlet pipe, L, that can used without
cavitations occurring?
(Vapor pressure = 1.228x 103 N/m2 (abs).

145. 145
 650 m
 653 m
Length L, D = 7 cm
and f = 0.02
Q = 11 L /s
1
2
Reference datum
Applying the energy equation (Bernoulli’s equation) between points  and ,
2
1
L
2
2
1
2
h
g
2
V
g
P
Z
g
2
V
g
P
Z 























 ……….…. (1)
Taking a reference datum passing through the free water surface, we have
Z1 = 650m, Z2 = 653 m, P1 = Patm = 0, V1 = 0, P2 =1.228x103 (abs.) and
s
/
m
86
.
2
)
4
/
07
.
0
(
10
11
A
Q
V
V 2
3
2 








146. 146
water
of
m
33
.
10 water
of
09
.
10
81
.
9
10
10
338
.
2
m
33
.
10
g
P
3
3
2
















Absolute zero reading
Gage zero reading
Substituting into Eq. (1) gives
  L
119
.
0
81
.
9
2
86
.
2
09
.
10
653
0
0
650
2















L
119
.
0
81
.
9
2
86
.
2
07
.
0
L
02
.
0
g
2
V
D
L
f
h
"2"
till
"1"
point
from
pipeline
the
in
loss
head
The
2
2
2
1
L 







The maximum length inlet pipe, L, that can used without cavitations occurring
m
65
L 

147. 147
 (30)
 (00)
Power house (turbine)
of 35 kW
Pipeline of L = 90 m,
D = 0.30 m and f = 0.02
1
2
Free
jet
The turbine shown in the figure extracts 35 KW from the
water flowing through it. The 0.30 m-diameter, 90 m-long
pipe is assumed to have a friction factor of 0.02. Minor losses
are negligible.
• Determine the flow rate through the pipe and the turbine,
• Draw the EGL & HGL.

148. 148
 (30)
 (00)
Power house (turbine)
of 35 kW
Pipeline of L = 90 m,
D = 0.30 m and f = 0.02
1
2
Free
jet
Reference datum
Applying the energy equation (Bernoulli’s equation) between points  and ,
Z1 = 30m, Z2 = 0, P1 = P2 = Patm = 0,
V1 = 0, V2 = V = ??
2
2
T
L
1
2
g
2
V
g
P
Z
h
h
g
2
V
g
P
Z 
























Taking a reference datum passing through
the free jet, we have
……….…. (1)
Substituting into Eq. (1) gives
  














g
2
V
0
0
h
h
0
0
30
2
T
L ……….…. (2)

149. 149
The friction losses through the pipeline can be computed in terms of V as:
2
2
2
L V
31
.
0
81
.
9
2
V
3
.
0
90
02
.
0
g
2
V
D
L
f
h 





The head extracted from water by the turbine,
V
47
.
50
)
4
/
3
.
0
V
(
81
.
9
10
10
35
)
A
V
(
g
Power
Q
g
Power
h 2
3
3
T 












……….…. (3)
….…. (4)
Substituting Eqs. (3) and (4) into Eq. (21), we have
0
47
.
50
V
30
V
36
.
0 3



A trial and error solution gives,
negligted.
be
can
which
value
negative
a
and
s
/
m
13
.
8
V
&
s
/
m
75
.
1
V 


150. 150
V = 1.75 m /s V = 8.13 m /s
Q = 0.124
m3/s
Q = 0.575 m3/s
hL = 0.95 m hL = 2.52m
hT = 28.84 m hT = 6.21 m
The obtained data can be tabulate in the form,
For a relatively high velocity in the pipeline, large head loss due to friction leaving
a small head for the turbine.

151. 151
 (30)
 (00)
Power house (turbine)
of 35 kW
Pipeline of L = 90 m,
D = 0.30 m and f = 0.02
1
2
Free
jet
EGL
Free jet
HGL
T
h
g
2
/
V 2
g
2
/
V 2
HGL & EGL ( Neglecting minor losses)
L
h

152. On to Examination problems...

153. 153
‫يذل‬ ‫فلن‬ ‫بك‬ ‫اعزت‬ ‫من‬ ‫للهم‬‫ا‬
‫يضل‬ ‫فلن‬ ‫بك‬ ‫تدي‬‫ه‬‫ا‬ ‫ومن‬
‫يقل‬ ‫فلن‬ ‫منك‬ ‫تكرث‬‫س‬‫ا‬ ‫ومن‬
‫يضعف‬ ‫فلن‬ ‫بك‬ ‫تقوي‬‫س‬‫ا‬ ‫ومن‬
‫يفتقر‬ ‫فلن‬ ‫بك‬ ‫تغين‬‫س‬‫ا‬ ‫ومن‬
‫خيذل‬ ‫فلن‬ ‫بك‬ ‫تنرص‬‫س‬‫ا‬ ‫ومن‬
‫يغلب‬ ‫فلن‬ ‫بك‬ ‫تعان‬‫س‬‫ا‬ ‫ومن‬
‫فلن‬ ‫يك‬‫عل‬ ‫تولك‬ ‫ومن‬
‫يب‬‫خي‬
‫هدي‬ ‫فقد‬ ‫بك‬ ‫تصم‬‫ع‬‫ا‬ ‫ومن‬ ‫يع‬‫يض‬ ‫فلن‬ ‫مالذه‬ ‫جعكل‬ ‫ومن‬
.

154. NUST Institute of Civil Engineering/Engr
Sajjad Ahmad
154