There is a simplest explaination of fluid mechanics

1. Bernoulli’s Equation
By: Dr. Ezzat Elsayed G. SALEH
Civil Eng. Dept.
‫يا‬‫ن‬‫مل‬‫ا‬ ‫جامعة‬
‫ندسة‬‫له‬‫ا‬ ‫ية‬‫لك‬
CVE 215

4. Mass, Bernoulli and
Energy Equations
By: Dr. Ezzat El-Sayed G. SALEH
Faculty of Engineering
Civil Eng. Dept.

5. 5
Giving credit where credit is due:
 Most of the lecture notes are based on the
slides from the course notes available to
students,
 Some slides are from Different Website related
to the same subjects,
 I have modified them and added new slides.
Fluid Mechanics (CVE 215)

6. Continuity Equation for “Constant Density and Uniform Flow”
Simple version of the continuity equation
for conditions of constant density. It is
understood that the velocities are either
uniform or spatially averaged
Q
A
V
A
V 2
2
1
1 

Q
A
V
A
V 2
2
2
1
1
1 



Q
A
V
A
V 2
2
1
1 
  L3/T
1 = 2
The continuity equation states that mass cannot be created or destroyed.
To conserve mass, the inflow rate must be equal the outflow rate.
Thus, conservation of mass
If the density remains constant, then 1 = 2, and the above
equation becomes the continuity equation for incompressible flow
or

7. where:
 V : is the fluid flow Speed at a point on a streamline,
 g : is the local acceleration due to gravity,
 Z : is the elevation of the point above a reference plane, with
the positive z-direction pointing upward - so in the direction
opposite to the gravitational acceleration,
 P : is the pressure at the point, and
  : is the density of the fluid at all points in the fluid.
constant
p
Z
g
2
V2




Daniel Bernoulli
 Daniel Bernoulli was a Dutch-Swiss mathematician who is most
known for his work in fluid mechanics and for his boom
Hydrodynamics.
 Bernoulli came up with the equitation

8. What does it mean?.
 In a simplified non-mathematical explanation, Bernoulli’s
equation basically means that if velocity (the speed of
motion) of a fluid increases, the pressure within (the force
per unit area applied to an object) the fluid decreases.
 Bernoulli’s principle is used in aerodynamics and help
explains how airplane get lift, and how race cars stay on
the ground. It also has uses in lots of different machines
such as a Hozon for agricultural use.
The Bernoulli’s Equation

9.  It is often convenient to plot mechanical energy graphically
using heights.
 Hydraulic Grade Line
Energy Grade Line (or total energy)
g
p
Z
HGL



g
2
V
g
p
Z
EGL
2




HGL & EGL

10.  The Bernoulli equation is
an approximate relation
between pressure, velocity,
and elevation and is valid
in regions of:
The Bernoulli’s Equation
 Equation is useful in flow regions outside of boundary
layers and wakes.
»Steady,
»Incompressible where net
frictional forces are negligible

11.  Limitations on the use of the Bernoulli Equation
o Steady flow: dV/dt = 0,
o Frictionless flow,
o No shaft work: Wpump= Wturbine= 0,
o Incompressible flow: r = constant,
o Applied along a streamline (except for irrotational
flow).
The Bernoulli’s Equation
Attention for the limitation of applicability of the Bernoulli
equation!

12. • If we neglect piping losses, and have a system without pumps or
turbines
• This is the Bernoulli equation,
• It can also be derived using Newton's second law of
motion (see Lecture notes),
• 3 terms correspond to: Static, dynamic, and potential
head .
2
2
1
2
g
2
V
g
p
Z
g
2
V
g
p
Z 


















The Bernoulli’s Equation

13. 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




14. Bernoulli Assumptions
Key Assumption  1 “Velocity = 0”
Imagine a swimming pool with a small “1 cm”
hole on the floor of the pool. If you apply the Bernoulli equation at the
surface, and at the hole, we assume that the volume exiting through the
hole is trivial (
‫طفيف‬
) compared to the total volume of the pool.
 Therefore the Velocity of a water particle at the surface can be
assumed to be zero
 There are three main variables in the Bernoulli Equation
Pressure – Velocity – Elevation
 To simplify problems, assumptions are often made to
eliminate one or more variables:

15. Key Assumption  2 “Pressure= 0”
Whenever the only pressure acting on a point is the
standard atmospheric pressure, then the pressure at that
point can be assumed to be zero because every point in
the system is subject to that same pressure.
 Therefore, for any free surface or free jet, pressure at
that point can be assumed to be zero.
Bernoulli Assumptions

16. Key Assumption  3 “ The Continuity Equation”
In cases where one or both of the previous assumptions do
not apply, then we might need to use the continuity equation
to solve the problem
A1V1=A2V2
 Which satisfies that inflow and outflow are equal at any
section
Bernoulli Assumptions

17. 17
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.

18. 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 - cavitation).
 Plot the pressure as Piezometric head (height
water would rise to in a piezometer)
 How?

19. Flow Rates
₪ Volume flow rate: is the volume of fluid flowing past a
section per unit time.
₪ Mass flow rate: is the mass of fluid flowing past a
section per unit time.
₪ Weight flow rate: is the weight of fluid flowing past a
section per unit time.

20. Artificial Pump
Water Pump
Pump Instillation

21. Hp
H
H
g
2
V
g
p
Z
H
g
2
V
g
p
Z L
T
out
2
p
in
2




























g
2
V2
in
in

g
2
V2
out
out

Pump Head

22. pressure head
Example: HGL and EGL
z = 0
pump
energy grade line
hydraulic grade line
velocity head
datum
 V2/ 2g
Elevation head
P / g
z
L
T
2
2
p
1
2
H
H
g
2
V
g
p
Z
H
g
2
V
g
p
Z 




























23. ( 2.4 m)
L
T
p h
h
g
V
g
p
Z
h
g
V
g
p
Z 



















2
2
1
2
2
2




Worked Example: Energy Equation (energy loss)
reference datum
( 2 m)
( 4 m)
An irrigation pump lifts 50 L/s of water from a reservoir and
supplies a total head of 10 m. How much mechanical energy
is lost? What is hL?
cs1
cs2
Why can’t I draw the cs at the end of the pipe?
L
2
p h
Z
h 
 2
p
L Z
h
h 
 m
6
4
10
hL 




24. ( 2.4 m)
L
T
2
2
p
1
2
h
h
g
2
V
g
p
Z
h
g
2
V
g
p
Z 























Example: Energy Equation (pressure at pump out let)
reference datum
( 2 m)
( 4 m)
The total pipe length is 50 m and is 20 cm in diameter. The
pipe length to the pump is 12 m. What is the pressure in the
pipe at the pump outlet? You may assume (for now) that the
only losses are frictional losses in the pipeline.
cs1
cs2
We need velocity in the pipe, , and loss

25. ₪ To get the velocity in the pipe, the continuity equation
gives
    s
/
m
6
.
1
2
.
0
05
.
0
4
4
/
D
Q
A
Q
V 2
2
2
2 







₪ To get the frictional losses, (Expect losses to be
proportional to length of the pipe)
m
44
.
1
50
12
6
hL 


₪ How about “”

26. ₪ “” is a function of the velocity distribution in the pipe.
₪ For a uniform velocity distribution “ = 1.0”
₪ For laminar flow “ = 2.0”
₪ For turbulent flow “1.01    1.10”
₪ Often  is neglected in calculations because it is so close to
“1”.
 









cs
3
3
dA
V
V
A
1
Kinetic Energy Correction Term : 

27. reference datum
(2 m)
4 m
50 L/s
hP = 10 m
(2.4 m)

28. L
T
p h
h
g
V
g
p
Z
h
g
V
g
p
Z 



















2
2
1
2
2
2




Since, p1 = 0 (gag), Z1 = 0, V1 = 0, V2 = 1.6m, Z2= 2.4m,
hL= 1.44m and taking  =1.05












g
V
Z
h
h
g
p
L
p
2
2
2
2
2
2
















81
.
9
2
6
.
1
05
.
1
4
.
2
44
.
1
10
2
2
g
p

water
of
m
g
p
02
.
6
2



  2
2 06
.
59
1000
81
.
9
1000
02
.
6
kNlm
p 





kPa

29. 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 - cavitation).
₪ Plot the pressure as Piezometric head (height
water would rise to in a piezometer)
₪ How?

30. Example: Energy Equation (Energy Grade Line - EGL)
datum
2 m
4 m
50 L/s
2.4 m
hP = 10 m
p = 59 kPa
What is the pressure at the pump intake?
Entrance loss
Exit loss
Loss due to shear
L
T
2
2
p
1
2
h
h
g
2
V
g
p
Z
h
g
2
V
g
p
Z 



























g
V
g
p
2
2




31. Powerhouse
River
Reservoir Example: Hydro-plant
2100
kW
180 rpm
116 kN·m
50 m
Water power = 2.45MW
Overall efficiency = 0.857
efficiency of turbine = 0.893
efficiency of generator = 0.960

32. Time of Empting and Filling

33. The flow out of a reservoir is 2 L/s. The reservoir surface is
5 m x 5 m. How fast is the reservoir surface dropping?
dh
Velocity of the reservoir surface
Example: Conservation of Mass??
dt
/
V
d
Q
Q in
out 


0
Q
For in 
dt
dh
A
dt
V
d
Q
tank
out 



tank
A
Q
dt
dh

 
 


2
1
H
H
tank
t
0
dh
Q
A
dt
H2
H1

34. Atank
H1
H2
dh
H
g
2
A
C
Q orifice
d
out 
dh
A
dt
.
Q tank
out 

dh
.
Q
A
dt
t
0
H
H
tank
2
1
 


dh
.
h
g
2
A
.
C
A
dt
t
0
H
H
5
.
0
orifice
d
tank
2
1
 




0

in
Q
Qout

35. Atank
H1
H2
dh
g
V
D
L
f
h f
L
2
2
.
)
( 
dh
A
dt
.
Q tank


dh
.
Q
A
dt
t
0
H
H
tank
2
1
 


dh
.
h
g
2
A
D
L
f
5
.
1
A
dt
t
0
H
H
5
.
0
pipe
tank 2
1
 




0
Qin 
g
V
2
2
g
V
2
5
0
2
.











D
L
f
5
.
1
g
2
V
g
2
V
g
2
V
D
L
f
g
2
V
5
.
0
h
2
2
2
2
5
.
0
.
5
.
1
2
h
D
L
f
g
V


5
.
0
.
5
.
1
2
.
. h
D
L
f
g
A
V
A
Q pipe
pipe




Q
F, L, D
V : the velocity through the pipe

36. The velocity of a jet of water
is clearly related to the depth
of water above the hole. The
greater the depth, the higher
the velocity. Similar behavior
can be seen as water flows
at a very high velocity from
the reservoir behind a large
dam such as Hoover Dam
Free Jets

37. Water flows from a 10-cm diameter container that contains three holes as
shown in the figure. The diameter of each fluid stream is 0.4cm, and the
distance between holes is 5cm. If viscous effects are negligible and quasi-
steady conditions are assumed.
Determine the time at which the water stops draining from the top hole.
Compare it with the time required to stop draining from the container.
Quasi-steady flow: means that at
any moment of time, we can solve
the problem as if the flow were
steady.
5 cm
5 cm
5 cm 0.4 cm dia.

38. Siphon
Liquid

40. Siphon

41. 4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
wate
r
For the siphon shown in the
figure, calculate:
 The volume flow-rate of water
from the nozzle.
 The pressure at points “A” and
“B”.

42. 4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
wate
r
» Because p1 = p2 = patm. = 0,
the remaining terms are
2
1
2
2
1
2
2
2

















g
V
g
p
Z
g
V
g
p
Z


» Applying Bernoulli’s Equation
between points “1” and “2”
 
 
  s
m
Z
Z
g
V
/
5
.
9
6
.
4
0
81
.
9
2
2 2
1
2








Reference
Datum
» The volume flow rate of water, Q
s
L
V
D
V
A
Q
/
66
.
4
5
.
9
4
025
.
0
4
2
2
2
2
2
2









 















43. The continuity equation gives:
water
of
m
Z
g
p
g
p A
B
19
.
1
9
.
0
29
.
0 








B
A
g
V
g
p
Z
g
V
g
p
Z 
















2
2
2
2


s
m
D
D
V
V
siph
Siph /
39
.
2
25
50
5
.
9
2
2
.
2
2
. 



















» To determine the pressure at point “A”, apply Bernoulli’s Equation
between points “1” and “A” gives,
water
of
m
g
V
g
p siph
A
29
.
0
81
.
9
2
39
.
2
2
0
2
2









» Similarly, since the siphon cross section is constant (VA =VB), the
pressure at point “B” can be determined as,
2
3
3
/
85
.
2
10
81
.
9
10
29
.
0
m
kN





2
3
3
/
67
.
11
10
81
.
9
10
19
.
1
m
kN






44. 1.0 m
25mm dia.
h
wate
r »
A siphon is used to draw water
from a large container as indicated
in the figure.
 Does changing the elevation, h,
of the siphon centerline above
the water level in the tank vary
the volume flow-rate through the
siphon? Explain.
 What is the maximum allowable
value of h ?

45. A
7.5cm dia.
wate
r
A water siphon having a constant
inside diameter of 7.5cm is arranged
as shown in the figure. If the friction
losses between A and B is 0.8 V2/
2g, where “V” is the velocity of flow
in the siphon.
 Determine the flow rate involved?
1.0 m
1.0 m
3.0 m
B

47. Pitot Tube

48. Q
Measures the
static pressure
Pitot measures
the total head

Z
P/g
V2/2g
EGL
HGL

: Static Pressure Tap
Measures the sum of the
elevation head and the
pressure Head.
: Pitot Tube
Measures the Total Head
EGL : Energy Line
Total Head along a system
HGL : Hydraulic Grade
line
Sum of the elevation and the
pressure heads along a
system
Pitot Tube
Reference Datum

49. Q
Z
P/ g
V2/2g
EGL
HGL
Understanding the graphical approach of Energy
Line and the Hydraulic Grade line is key to
understanding what forces are supplying the
energy that water holds.
V2/2g
P/g
Z


Point 1:
Majority of energy stored
in the water is in the
Pressure Head
Point 2:
Majority of energy stored
in the water is in the
elevation head
If the tube was
symmetrical, then the
velocity would be
constant, and the HGL
would be level
EGL & HGL
Reference Datum

51. Nozzle-meter
Typical Nozzle-meter construction
(Typically placed between
flanges of pipe sections)
The nozzle-meter is more
efficient than the orifice-meter

52. Venturi-meter

53. Example: Venturi-meter

54. Venturi-meter

55. Example: Venturi-meter
Find the flow (Q) given the pressure drop between section “1”
and “2” and the diameters of the two sections. Draw an
appropriate control volume. You may assume the head loss is
negligible. Draw the EGL and the HGL.

h


56. Venturi-meter
The Venturi-meter consists of a short converging conical tube
leading to a cylindrical portion called “throat” which is followed by a
diverging section
The entrance and exit diameter
is the same as that of the
pipeline into which it is inserted
Pressure tapings

57. 2
2
1
2
2
2

















g
V
g
p
Z
g
V
g
p
Z


57
Applying Bernoulli’s Eq. between points “1” and “2”
For Z1 = Z2 = 0
or
The continuity equation gives,
or
….. (1)
…(2)
…(3)
Substituting Eq. (2) into Eq. (1) gives,
Worked Example: Venturi-meter
2
2
1
2
2
2















g
V
g
p
g
V
g
p

 g
V
V
g
p
p
2
2
1
2
2
2
1 



2
2
1
1 V
A
V
A
Q .
. 

4
1
2
2
2
2
1 








D
D
V
V

58. 58
 The theoretical discharge Qth is
 and the actual discharge Qact is
 4
1
2
2
1
2
1
2
D
D
g
p
p
g
V








 


 4
1
2
2
1
2
2
2
2
1
2
4 D
D
g
p
p
g
D
V
A
Qth








 











.
 4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q
C
Q d
th
d
act








 













  









 1
/
2
1


 i
h
g
p
p 
D2 / D1= 

59. Flexible Joints
 
 
Q
h
y
i
L
4
3 p
p 
   h
g
y
g
p
h
y
g
p i
L
L 
 

 



 2
1
h
g
p
p
L
i
L











 1
2
1




If the measured liquid is water
and the indicating liquid is
mercury, the above equation
gives,
 
g
s
h
g
p
p
w
1
.
2
1



 

Thus
,
 13.6

60.  4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act








 












What throat diameter is needed for a Venturi-meter in a 200cm horizontal
pipe carrying water with a discharge of 10m3/s if the differential pressure
between the throat and upstream section is to be limited to 200kPa at this
discharge? (Cd = 0.98)
Solution
The basic discharge equation for a Venturi-meter is:
m
g
p
39
.
20
81
.
9
1000
10
200 3






D1 =2.0m
= 0.98
10 m3/s
………….. (1)

61. Eliminating Eq. (1) gives: 2
2
2
16
8
.
6400
13
D
D


Solving by trial and error:  D2 = 0.80m
………….. (2)

62. Water flows through a Venturi-meter that has a 30cm throat. The
Venturi-meter is installed in a 60cm pipe. What deflection occur in a
water-mercury manometer connected between the upstream and
throat sections if the discharge of 0.57m3/s. (Cd = 0.98)
Solution
 4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act








 












The basic discharge equation for a Venturi-meter is:
D1 =0.60 m
= 0.98
0.57 m3/s
………….. (1)
  









 1
/
2
1


 i
h
g
p
p 
D2 =0.30 m

63. For the given case,
6
.
13
0625
.
0
16
1
60
.
0
30
.
0
0706
.
0
4
3
.
0
4
4
4
1
2
2
2
2
2
2
























w
m
L
i
D
D
m
D
A







Substituting these values into Eq. (1) gives
 
0625
.
0
1
1
6
.
13
81
.
9
2
0706
.
0
98
.
0
57
.
0







h
 ………….. (2)
Solving for h,  h (the deflection in the manometer) = 25.70cm

65. Water Bridge in Germany

67. Hozon – Siphon Mixer
A Hozon siphon mixer is used
agriculturally to fertilize plants.
This device operates somewhat with
Bernoulli’s principle.
Water flows through the device at a
certain velocity. As you can see the
water flows over tube where fertilizer is
projected into the water.

68. Hozon – Siphon Mixer (con’t)
This is where Bernoulli’s principle steps in.
Because the water is flowing over the top of
the fertilizer tube it creates pressure around
the stream of moving water.
This pressure around the stream of water
causes the fertilizer to be projected upwards
towards the stream of water.

69. Hozon – Siphon Mixer (con’t)
Why doesn’t the water just spray into
the fertilizer tube?
The pressure of the stream is very low
since it’s velocity is very high.
But, the velocity around the out side of
the stream of water is very low, thus
causing the pressure outside the
stream of water to be very high.
The high pressure that is projecting
fertilizer into the water doesn’t allow the
water to enter the fertilizer tube.

70. Racecars!
Racecars designs such as the ones the perform in the Indianapolis
500 use Bernoulli’s principle to keep the car on the ground, and allow
the car to make sharper turns.
Notice the shape of the
spoiler.
The spoiler starts out
thin and gradually curves
upward.
This causes the velocity to
decrease and pressure to
increase on the top.

71. Racecars! (cont’d)
The spoiler of the car is shaped so that the velocity of air on the top of
the spoiler is moving slower, thus creating more pressure on the top of
the car pushing the car downward. The velocity of air on the bottom of
the spoiler is much faster, thus having less pressure pushing upward
on the car.

72. Airplanes
Notice the shape of
the airplane’s wings.
They are shaped
opposite of a racecars
spoiler.
The top of an airplane’s wings are shaped so that the velocity of air
moves faster over the top and slower on the bottom. Since the
velocity of air is faster on top that means the pressure on top is low.
The velocity of air is slower on the bottom so the pressure on
bottom is greater helping give lift (the force that pulls an object
outward, or upward.) to the airplane keeping it up in the air.

73. Airplane Wing
The top of an airplane’s wing is shaped
so that the velocity of air moves faster
over the top and slower on the bottom.
Since the velocity of air is faster on top
that means the pressure on top is low.
The velocity of air is slower on the bottom
so the pressure on bottom is greater
helping give lift to the airplane keeping it
up in the air.
* The same concept of low velocity and high pressure on the bottom of an
airplane’s wings also applies to the wings of flying birds.

74. Summary
 What did we learn?
 Daniel Bernoulli was a Dutch-Swiss mathematician who came
up with the principle of fluid mechanics known as Bernoulli’s
Principle.
 Bernoulli’s principle basically states that as a fluid’s velocity
increases the pressure within that fluid decreases.

75. Summary (con’t)
 Devices such as Hozon siphon mixers use Bernoulli’s
principle to cause fertilizer to mix with water by the pressure
of the water it is being mixed with.
 Racecars use a spoiler shaped so that pressure stays on top
of the car holding down.
 Airplane wings are shaped opposite of racecars to have
pressure on the bottom of the wing to give it lift.