MEC2404_Week 1 - Fluid Properties and Fluid Statics Part 1 (1).pdf

Fluid properties Week - 1
Mechanical Engineering

1. Week 1 -Fluid Properties and Fluid
Statics Part 1

2. These slides include copyrighted material from the course textbook:
B.R. Munson, T.H. Okiishi, W.W. Huebsch and A.P. Rothmayer. Fundamentals of
Fluid Mechanics 7th Edition SI Version, John Wiley & Sons, Inc., 2013.
These slides are for the use of Monash University students registered in this course
and are not to be further distributed."
The original development of this slide presentation was done by Dr Josie Carberry.
This slide presentation was enhanced and added to by Thomas Simko.
2

3. We will study the behaviour of fluids when subject to applied forces.
Two subcategories:
•Fluid statics–Behaviour of fluids at rest
•Fluid dynamics–Behaviour of fluids in motion
Fluids are everywhere
•Everyday phenomenon
•Environmental flows
•Biological flows
•Medical devices
•Aerodynamics
3

4. A fluid is a substance that continuously deforms (i.e., it is strained) when subjected to a
shear (tangential) stress.
Although solids deform initially, they do not deform continuously.
The term “fluid” generally apples to liquids and gases.
F
What is a fluid?
4

5. Fluid as a Continuum
All fluids (gases & liquids) are made up of atoms or molecules.
But generally we do not analyse fluids at the molecular level, i.e., by considering the
interaction of individual molecules.
Instead, we look at very small volumes containing large numbers of molecules.
We call these small volumes fluid particles or fluid elements.
The number of molecules per mm
3
is of the order of 10
18
for gases, and 10
21
for liquids.
Assume that using averaged values of important properties (pressure, velocity,
temperature etc.) over very small volumes is reasonable.
That is, we treat the fluid as a continuum.
5

6. Different fluids flow differently.
This is because different fluids have
different characteristics (e.g., compare
water, oil, honey, and air).
The quantification of fluids therefore
requires the definition of fluid properties:
•Density, specific volume,
specific gravity
•Viscosity
•Vapour pressure
Fluid Properties
6

7. Density, specific volume, specific gravity
A measure of a fluid’s mass or weight per volume (or the inverse)
Viscosity
A measure of how easily a fluid “flows”
Vapour pressure
A measure of when a liquid evaporates (becomes a gas)
Fluid Properties
7

8. 1.1b Density, Specific Volume and
Specific Gravity

9. Density = Mass (m)per unit volume (V)
SI Units: kg/m
3
Dependent on temperature and pressure
Can be inferred from a hydrometer (which
actually measures specific gravity). The
device sinks to an equilibrium position, at
which point the weight balances the buoyancy
force (upthrust).
Archimedes principle: buoyancy force = Vρg
If we know V (volume of fluid displaced), then
ρto be calculated.
DensityV
m

9

10. Specific volume (units: m
3
/kg)
Volume V per unit mass m
Reciprocal of density:
Specific gravity or relative density(dimensionless)
Density relative to density of water at 4
o
C (1000 kg/m
3
):
Specific weight (units: kg/m
2
s
2
)
Weight per unit volume:
Specific Volume & Specific Gravity
1
=
m
V
v OH
2


SG g
10

11. 1.2 Viscosity

12. Recall the definition of a fluid:
A substance that continuously deforms when subjected to
a shear (tangential) stress
We will now introduce the concept of viscosity to describe
the ‘fluidity’of a fluid, i.e., how easily it flows.
12

13. Normal stress (pressure) σ:
The force Fis applied normally to an area A.
13
F
A
F
AAF/=

14. 14
F
A
σ
ϵ(Δl/l)
Δl
lAF/= E=

15. Shear stress,τ
The force Fis applied tangentially to an area A.
15AF/=
F
A
How much shearing stress (or force) must be applied to cause a fluid to deform/move?
Note: when you deform a fluid you change its velocity, i.e., you set up a velocity gradient.

16. Consider fluid trapped between two plates:
▪Continuity of velocity boundary condition
▪Fluid directly adjacent to moving upper plate
moves at velocity U, i.e.:()Uhyu ==
u(y)
U
y = 0
y = h
F
u = 0
u = U
16

17. Consider fluid trapped between two plates:
▪Continuity of velocity boundary condition
▪Fluid directly adjacent to moving upper plate
moves at velocity U, i.e.:
▪Fluid directly adjacent to stationary lower plate
is stagnant (no slip condition), i.e.:()00==yu ()Uhyu ==
u(y)
U
y = 0
y = h
F
u = 0
u = U
17

18. Consider fluid trapped between two plates:
▪Continuity of velocity boundary condition
▪Fluid directly adjacent to moving upper plate
moves at velocity U, i.e.:
▪Fluid directly adjacent to stationary lower plate
is stagnant (no slip condition), i.e.:
▪Thus, velocity uis a function of y
▪Assume linear dependence:
▪Velocity gradient is then:()00==yu ()Uhyu ==
u(y)
U
y = 0
y = h
F
u = 0
u = U()
h
U
yyu= h
U
dy
du
=
18

19. Consider two situations:
•Water trapped between the plates
•Honey trapped between the plates
If the plate area is the same, and you want to set
up the same velocity gradient, say 1 m/s (i.e.,
move the top plate at 1 m/s) do you need to apply
a different force to the plate for each situation?
Yes. The honey requires more force.
So there is a relationship among the fluid type (or a
property), force and velocity gradient:
u(y)
U
y = 0
y = h
F
u = 0
u = U
19dy
du
A
F
=

20. Define a constant of proportionality, μ, which we
will call the dynamic (or absolute) viscosity.dy
du
A
F
 == dy
du
A
F
=
20
τ
du/dy

21. Define a constant of proportionality, μ, which we
will call the dynamic (or absolute) viscosity.
Units for dynamic viscosity:
kg/m·s, N·s/m
2
or Pa·s
Common to use Poise (P): 1 P = 0.1 Pa·s
Viscosity of water at room temperature is 1 cP = 0.001 Pa·s
Also define kinematic viscosity, ν, (incorporates fluid density):

=
Units: m
2
/s or Stoke (St) where 1 St = 1 cm
2
/s = 0.0001 m
2
/sdy
du
A
F
=
21dy
du
A
F
 ==

22. u(y)
U
y = 0
y = h
F
u = 0
u = U
Note: parabolic velocity profile here due to
absence of upper boundary
Assuming a linear velocity profile is
not always appropriate.
See the velocity profile (below) for flow over a wall.
22

23. 1.3 Viscosity –Temperature Dependence

24. Viscosity is a measure of a fluid’s resistance to deformation and hence flow.
It acts like friction between layers of fluid when they move relative to each other.
The temperature dependence is different for liquids and gases.
24

25. Liquids
•Viscosity is due to cohesive intermolecular forces between liquid molecules.
•At higher temperatures, molecules have higher energy and are able to overcome
these cohesive forces, thus being able to move more freely → T ↑ μ↓.
Gases
•Gas molecules more widely spaced –cohesive forces relatively small.
•Resistance to flow arises from random molecular collisions.
•At higher temperatures, gas molecules experience increased collisions with increased.
transfer of momentum btw molecules, decreasing coherent flow motion → T↑ μ↑.
T
μ Oil
Water
Air
25

26. Source: University of Alberta
http://www.uofainsideout.ca/news/fun-facts-about-edmonton-winter-/
Viscosity is affected by temperature
For example, SAE Code: 5W-30
26

27. 1.4 Viscosity –Types of Fluids

28. The viscosity can be determined from the slopeof a plot of the shear
stress versus strain rate (deformation rate or velocity gradient du/dy).dy
du
A
F
 ==
28

29. The viscosity can be determined from the slopeof a plot of the shear
stress versus strain rate (deformation rate or velocity gradient du/dy).
The slope (i.e., the viscosity) is:
•Linear for most common fluids (Newtonianfluids)
•Non-linear for some fluids (Non-Newtonian fluids)
Shear/Deformation rate (du/dy)
Shear
stress
Newtonian
Shear Thickening
Shear Thinning
Bingham
plastic
μdy
du
A
F
 ==
29

30. With a shear thinning fluid, the
viscosity decreases with shear –i.e.,
the slope decreases with increasing
shear.
Example: Latex paint.
Latex paint clings to a stationary
paintbrush because there is little
shear rate and high viscosity.
However, once the brush is applied
against a wall, there is a large shear
rate (velocity gradient) across the thin
layer of paint. The viscosity drops and
the paint flows easily.
Shear/Deformation rate (du/dy)
Shear
stress
Newtonian
Shear Thickening
Shear Thinning
Bingham
plastic
μ
30

31. Shear/Deformation rate (du/dy)
Shear
stress
Newtonian
Shear Thickening
Shear Thinning
Bingham
plastic
μ
With a shear Thickening fluid, the
viscosity increases with shear –i.e., the
slope decreases with increasing shear.
Example: Quicksand (sand and water) or
a corn starch and water mixture.
The more you fight against quicksand,
quickly moving to try to get escape
(thereby increasing the velocity gradient
or shear rate), the more viscous it
becomes, and the harder it becomes to
extract yourself. The best approach is to
use slow movements.
31

32. A Bingham plastic does not move (in this
respect, it’s like a solid) until the shear
stress exceeds a certain yield stress.
Then it flows like a fluid.
Example: Toothpaste
Toothpaste will stably sit in a pile until
spread by the toothbrush onto a surface like
your teeth, when it will flow like a fluid.
Shear/Deformation rate (du/dy)
Shear
stress
Newtonian
Shear Thickening
Shear Thinning
Bingham
plastic
μ
32

33. 1.5 Viscosity –Example

34. Viscosity can be measured using a viscometer.
The viscometer below consists of two 400 mm long concentric
cylinders.
The inner cylinder has a radius of 60 mm, and it is rotated at
300 rpm by applying a torque of 1.8 N∙m.
The fluid flowing in the 1.5 mm annular gap is oil.
What is the viscosity of the oil?
34

35. Two 400 mm long concentric cylinders. Inner radius 60 mm rotating
at 300 rpm by applying a torque of 1.8 Nm. Fluid flowing in the
1.5 mm annular gap is oil.
Linear velocity on surface of rotating cylinder:m/s 885.1
60
2300
06.0 =

==


iRU
35

36. Assuming a linear velocity profile, the velocity gradient across the gap is:
This assumption is only valid if curvature effects are negligible, which is
true only if l<< R
i1
3
s 1257
105.1
885.1
−
−
=

=


l
u
dr
du
36

37. Shear stress defined as:
Thus, the shear force is:
Since the applied torque is:
Solve for μ(Equation 1 = Equation 2):l
u
dr
du
=  LDAF
i1257== ( )( )( )
skg/m 158.0
106010400101201257
8.1
333
=

=
−−−

 mN 8.1==
iFR
37iRF / mN 8.1=
1
2

38. 1.6 Vapour Pressure

39. Consider a liquid in a sealed container with an air space above the liquid surface. The air is
evacuated to leave a vacuum above the liquid.
Liquid molecules near the liquid surface with sufficient momentum to overcome intermolecular
cohesive forces will escape the liquid and form a gas. They exert a pressure on the liquid surface.
Some molecules in the gas will return to the liquid.
At equilibrium, # of molecules leaving surface (evaporation) equals # of molecules arriving at the
surface (condensation). The corresponding pressure at this condition is known as the vapour
pressure. Since temperature governs the energetics of the molecules, vapour pressure is a
function of temperature.
39

40. Boiling is the formation of vapour bubbles in a liquid when the
absolute pressure in the liquid equals the vapourpressure.
It occurs when the vapourpressure of the liquid reaches the
surrounding pressure above the liquid surface, e.g., water
boiling in a pot on the stove.
In this case, at sea level, P
vap= P
atm= 101 kPa. The
temperature corresponding to this vapourpressure –i.e.,
required to initiate boiling at this pressure –is 100 °C.
But if we go to a higher altitude (e.g., to 9000 m, where
P
atm= 30 kPa) the temperature required for a vapourpressure
of 30 kPais 69 °C. So water will boil at this temperature.
Thus, one can force a liquid to boil not only by raising the
temperature at constant pressure, but also by lowering the
pressure at a given temperature.
40

41. There are certain instances in flowing liquids where regions of low pressure are
encountered, where the absolute pressure in the liquid can drop below the vapor pressure.
For example, at the neck of a converging-diverging nozzle:
•Fluid velocity is higher at the neck to maintain mass flow conservation.
•Thus pressure is low in this region (Bernoulli’s Equation).
•If this pressure decreases below the vapour pressure of the liquid, boiling occurs,
i.e., vapour bubbles form in the neck region.
•These bubbles are swept into other regions by the flow.
•If they enter high pressure regions, they will be forced to collapse.
•This sudden bubble collapse is known as cavitation.
High velocity, low pressure region
Vapour bubble formation
41

42. One should avoid cavitation especially in
pumps since bubble collapse can occur with
such intensities that it can cause structural
damage.
42
Attribution: US Navy
https://en.wikipedia.org/wiki/File:Cavitating-prop.jpg
Attribution: Erik Axdahl
https://commons.wikimedia.org/wiki/
File:Cavitation_Propeller_Damage.JPG

43. 1.7 Pressure

44. Pressure in a fluid at rest is the normal force per
unit area exerted on a plane surface (real or
imaginary) immersed in a fluid.
Units: N/m
2
or Pa
If the pressure on every surface is equal, the net
force generated by the pressure is zero –
irrespective of the pressure magnitude, i.e.,
relative pressure is very important.
44

45. Pressure is measured with respect to a reference level.
If the reference level is a perfect vacuum, the pressure is known as absolute pressure,p
abs.
This is what you are familiar with, and what you must use in the ideal gas equation.
It is often convenient to measure the pressure relative to atmospheric pressure
(p
0 = 1 atm = 1.01 x 10
5
Pa (101 kPa) = 14.7 psi above a perfect vacuum).
This is known as the gauge pressurep
gauge:0ppp
absgauge −=
45

46. The gauge pressure,p
gauge , is the difference from the reference atmospheric pressure:
For example, a gage pressure of 9 kPa means an absolute (“actual”) pressure of:
101 kPa + 9 kPa = 110 kPa.
We can also have a “negative gage pressure,” i.e., pressure below atmospheric pressure.
But this is properly referred to as vacuum pressure.
For example, a vacuum pressure of 11 kPa means an absolute (“actual”) pressure of:
101 kPa -11 kPa = 90 kPa.0ppp
absgauge −=
46

47. p
abs= p
0
p
gauge = 0
When measuring tyre pressures, the reading shows the gauge pressure (amount by which
the pressure exceeds the atmospheric pressure).
At liquid surfaces, the gauge pressure is zero –the absolute pressure is equal to the
atmospheric pressure.
47

48. 1.8 Fluid Statics
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

49. The study of the behavior of fluids at rest or when there is no relative
motion.
Topics:
•Concept of pressure at a point
•How pressure varies with depth in a fluid at rest
•Hydrostatic forces (forces that arise from the pressure)
-On planar surfaces -On curved surfaces
•Buoyancy
•Rigid body motion (fluids in relative motion with no shear
stresses present)
•Linear rigid body motion
•Rotational rigid body motion
Fluid Statics/Hydrostatics

50. Recall the definition of a fluid:
A fluid is a substance that deforms
continuously whenever a shear stress is
applied to it.
Thus, when a fluid is at rest, the shear
(tangential) stress is zero.
The only stress acting on the fluid is then just
the pressure (normal stress).
http://www.scienceimage.csiro.au/tag
/rivers/i/3191/aerial-view-of-river/
F
50

51. Knowledge of the hydrostatic forces acting on a submerged surface is
important in the design of ships, dams, storage tanks and other hydraulic
systems.
Since a fluid at rest does not experience any shear stresses, the only force
to consider is that due to pressure (hydrostatic force), which always acts
normally to the surface
Pressure also varies linearly with depth as ρgh, as shown below
Hydrostatic forces
http://www.scienceimage.csiro.au/image
/1010/discharging-ballast-water/
51

52. 1.9 Pressure is isotropic
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

53. Pressure is a normal stress
Pressure at a point
Perpendicular Force per unit area
We will show pressure is a scalar –the pressure value at any point in space is independent of direction.
Pressure Force
When there is a pressure at a surface, it generates a force that is always normal to that surface
The magnitude of the pressure force on a surface of area A is:
F
pressure= p xA
The direction of the pressure force is defined such that a positive pressure generates a force inwards on
the surface (as in the above figure).
53

54. Consider a small infinitesimal fluid element in the form of a wedge. Assume no shear stresses acting on the
wedge. True if fluid element is not moving. This can be generalised to the case where the element is moving as
a rigid body (no relative motion between fluid elements).
If there are no shear stresses, the only external forces acting on the wedge are:
Force due to pressure (force = pressure x area)
Weight (force = mass x g)
p
s
–Average pressure
on slope face
Pressure at a point
54

55. We’re going to rely on Newton’s 2
nd
law:
ΣF= ma
Which we’ll apply in elemental form:
F
yand F
z
Mass = (density)(volume of cube/2)
= ρ(dxdydz)/2
Weight = mass x g
55

56. Let’s start by breaking F
sinto components:
θ
θ
90-θ
θ
F
z=F
scosθ
F
s
F
y=F
ssinθ
y
z
56

57. sum of forces = mass x acceleration
y
z
F
y= F
ssinθ= p
sA
ssinθy
a
zyx
sx
s
pzx
y
p
y
F
2
sin

 =−=
δz
δx
δs
F
y= p
yA
y
F = pA
δy
57

58. sum of forces = mass x acceleration
y
z
F
z= F
scosθ= p
sA
scosθ
δz
δx
δs
F
z= p
zA
yWeightz
a
zyx
g
zyx
sx
s
pyx
z
p
z
F
22
cos



 =−−=
F = pA
δy
58

59. sum of forces = mass x acceleration
y
z0=
x
F
δz
δx
δs
F = pA
F
x= p
xA
x
F
x= p
xA
x
(they cancel)
δy
59

60. Since:
δy = δs cos θ
δz = δs sin θ2
y
y
a
s
p
y
p

=− 2
z
z
ag
s
p
z
p

 




+=− y
a
zyx
sx
s
pzx
y
p
y
F
2
sin

 =−= z
a
zyx
g
zyx
sx
s
pyx
z
p
z
F
22
cos



 =−−=
60

61. To determine the pressure at a point, we let
δx,δy, δz → 0 (whilst maintaining the angle θ)
and hence:
Since θis arbitrary, the above means that the
pressure is independent of direction (isotropic) –
as long as there are no shear stresses acting on
the fluid (Pascal’s Law).s
p
z
p
y
p ==
61

62. 1.10 Pressure Depth Variation (Fluids at Rest)
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

63. For a stationary fluid, we have shown that the pressure
components at a point do not vary with direction.
What about the magnitude of pressure from point to point?
In particular, how does the pressure vary with depth?
Consider again a small fluid element –this time a rectangular
one.
Consider the forces acting on this element and apply Newton’s
2
nd
Law.
A
z
δz
p
p+(∂p/∂z)δz
63

64. Two types of forces acting on the element:
•Body forces: forces that act on the entire volume of
the fluid (e.g. gravity/weight, electric, magnetic,
centripetal inertia forces)
•Surface forces: forces that act on the surface (e.g.,
shear stress and pressure)
Since there are no shear stresses in fluids at rest, we only
need to consider pressure forces. Also, we only need to
consider the weight of the fluid. There are no other
external forces.
A
z
δz
p
p+(∂p/∂z)δz
64

65. Also, we can neglect any forces in the x-and y-directions
since we are only interested in depth variation.
We’ll show later there are no variations in the pressure in
these directions anyway for fluids at rest. A
z
δz
p
p+(∂p/∂z)δz
65

66. Newton’s 2
nd
Law:
Forces in z-direction:
Equating the above two equations gives:gAzAz
dz
dp
ppA
z
F  −+−=









 z
am
z
F= 0==−+−










z
amzAgAz
dz
dp
ppA  g
dz
dp
−=
A
z
δz
p
p+(∂p/∂z)δz
66

67. The negative sign indicates that pressure decreases with
increasing elevation for a fluid at rest (we defined z=0 at
the bottom).
Note that ρis constant for an incompressible fluid.
The pressure depends only on z(dp/dx = dp/dy= 0)since
gravity only acts in the vertical direction).
A
z
δz
p
p+(∂p/∂z)δzg
dz
dp
−=
67

68. Integrating from point 1 to point 2:
−=
2
1
2
1
z
z
p
p
dzgdp ( )
1221
zzgpp −=− 12
zzhletting −= Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi
Copyright ©2005 by John Wiley & Sons, Inc. All rights reserved.
Figure 2.3 (p. 43)
Notation for pressure variation in a fluid at rest with a free surface. p
1
-p
2
=rgh g
dz
dp
−=
68

69. If point 2 (z= habove the bottom z=
0) is at the surface where p
2= 0(recall
gauge pressure is zero at the liquid
surface), then:
On the other hand, we can obtain the
absolute pressure at point 1 by setting
p
2= p
atm:ghp=
1 p
1=p
atm+rgh ghpp =−
21 Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi
Copyright ©2005 by John Wiley & Sons, Inc. All rights reserved.
Figure 2.3 (p. 43)
Notation for pressure variation in a fluid at rest with a free surface.
69

70. We can also use depth, h, as a measure of the pressure:
h is known as the pressure head or hydrostatic head.
It is the height of the liquid column of specific weight ρg required
to give rise to a pressure difference p
1–p
2g
pp
h

21−
=
70ghpp =−
21

71. For example, what is the gage pressure (in kPa) for a head
of 10.33 m of water?
P = ρgh
= (1000 kg/m
3
) (9.81 m/s
2
) (10.33 m)
=101.3 kPa(1 atm)
So, for every 10.33 m you go down in a lake, the pressure
on you increases by 1 atm.g
pp
h

21−
=
10.33 m
71

72. For example, blood pressure is measured in terms of millimetresof
mercury (Hg) even though there is no mercury in blood.
If your blood pressure is 120 mm Hg, noting that the specific gravity of Hg
is 13.6, determine the equivalent pressure in kPa.
1 atm= 101.3 kPa= 760 mm Hg = 10.33 m of waterkPa 16101208.910006.13
3
===
−
ghp g
pp
h

21−
=
72

73. 1.11 Pressure and Hydraulic Jacks
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

74. Pressure in a homogeneous incompressible fluid at rest depends only on the depth of
the fluid relative to a reference plane. It is independent of the size and shape of the
container.
The pressure is therefore the same at all points along ABdespite the shape
irregularities of the liquid column above this plane.
74

75. Hydraulic jacks take advantage of the pressure equality at the same elevation in a fluid
system.
Consider the fluid surfaces right below the pistons of areas A
1and A
2, as shown below.
75

76. A small force applied at the piston with area A
1
can be used to develop a larger force at the
piston with area A
2.
Since the fluid surfaces at 1 and 2 are at the same level:
P
1= P
2
F
1/A
1= F
2/A
2
F
1= (A
1/A
2) F
2
So a small force at 1 can be
used to oppose a larger force,
e.g., the weight of a car, at 2.
76

77. 1.12 Open End Manometers
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

78. A pressure difference generates a net force.
Pressure can be reported in terms of the
displacement of a particular fluid.
78

79. Manometry is a technique to measure pressure using stationary liquid columns. If we
know the pressure at a point and the density of all of the fluids in the manometer, then we
can work out the pressure at any other point in the system.
Remember that for a fluid at rest:
•Pressure decreases as we move upward
•Pressure increases as we move downward
79

80. •Pressure decreases as we move upward
•Pressure increases as we move downward
In this case, the open manometer tells us that the pressure at point 3 is:
Or with specific weight:223 hpp
atm+= 223 ghpp
atm+=
80

81. Example:
Derive the equation for the pressure at point A
in terms of the atmospheric pressure.
81

82. Looking at points 2 & 3 we see they are at the same height.
Also, all of the fluid below points 2 & 3 is the gage fluid (blue).
Therefore, as you go down from point 3 and back up to point 2 the pressure first
increases by a certain amount and then decreases by exactly that amount.
Thusp
2= p
3.
82

83. From a reference point, if you go down, add the pressure of the column of fluid.
If you go up, subtract the pressure of the column of fluid.
P
A= P
2–p
greygh
1
P
2= P
3= P
atm+ p
bluegh
2
So the absolutepressure at A is:
P
A= P
atm+ p
bluegh
2–p
greygh
1
83

84. Let’s say:
the grey fluid is pressurized air (p
grey=1.23 kg/m
3
);
the blue fluid is water; and
h
1= 0.1 m and h
2= 0.3 m.
What is P
A?
P
A= P
atm+ p
bluegh
2–p
greygh
1
84

85. P
A= P
atm+ p
bluegh
2–p
greygh
1
= 101 kPa+ (1000 kg/m
3
)(9.81 m/s
2
)(0.3 m)
–(1.23 kg/m
3
) (9.81 m/s
2
)(0.1 m)
= 104 kPa
(> 1 atmas expected)
85

86. 1.13 U-Tube Manometers
AERIAL VIEW OF THE CATARACT DAM AND RESERVOIR, NSW. 1999.
CSRIO: HTTP://WWW.SCIENCEIMAGE.CSIRO.AU/TAG/LAKES/I/4531/AERIAL-VIEW-OF-THE-CATARACT-DAM-AND-RESERVOIR-NSW-1999-/

87. U-tube manometers are commonly used to
measure the pressure drop of a fluid flowing
in a pipe due to viscous losses
Since the fluid in the U-tube is stationary,
we can use fluid statics equations.
The pressure (in a given fluid) is the same at
positions with the same elevation. Only
true when considering the same fluid.
Therefore, P
C= P
D
h
A
ρ
l
B
F
C
E
D
H
ρ
m
87

88. For the liquid column in the left arm of the
tube:
Similarly, for the right arm of the tube:()Hhgpp
lAC
++= gHghpp
mlFD
++=
88
h
A
ρ
l
B
F
C
E
D
H
ρ
m

89. Left arm of the tube:
Right arm of the tube:
SinceP
C = P
D()Hhgpp
lAC
++= gHghpp
mlFD
++= gHpgHp
mFlA
 +=+ ( )gHpp
lmFA
−=−
89
h
A
ρ
l
B
F
C
E
D
H
ρ
m