This document discusses fluid statics and pressure measurement. It defines concepts like absolute pressure, gauge pressure, atmospheric pressure, and Pascal's law. It describes devices used to measure pressure like manometers, piezometers, and Bourdon gauges. Specifically, it provides details on how liquid manometers and differential manometers work, including the principles, setup, and equations to calculate pressure. It also lists the advantages and limitations of using manometers for pressure measurement applications.

1. Fluid Statics: Pressure intensity and
pressure head: pressure and specific weight
relationship, absolute and gauge pressure
Dr. Mohsin Siddique
Assistant Professor
1
Fluid Mechanics

2. Fluid Statics
2
Fluid Statics: It is the branch of fluid mechanics that deals
with the behavior/response of fluid when they are at rest.
Pressure, (average pressure intensity): It is the normal force
exerted per unit area. It is denoted by P and is given by;
Units
SI: N/m2 (called Pascal)
BG: lb/ft2 or lb/in2 (called psi)
CGS: dyne/cm2
1 bar=105N/m2=105Pascal
A
F
area
force
P ==

3. Pressure vs Water depth/height
3
Consider a strip or column of a cylindrical fluid,
h= height or depth of strip of fluid
γ= specific weight of fluid
dA=cross-sectional area of strip
dV=volume of strip
dW=weight of strip
Pressure at base of strip=dF/dA=dW/dA
P= γdV/dA
P= γdA.h/dA
P=γh
h

4. Pressure vs Water depth/height
4
P=γh
P α h
For h=0, P=0
For h=h, P=γh
h
Pressure distribution
diagram/pressure profile
As you know atmospheric pressure reduces, as we move to
higher elevations. Is it because of h, as h reduces, P also reduces.

5. PASCAL’S LAW
5
“Pressure at any point in fluid is same in all directions when
the fluid is at rest”
Consider a wedge shape element of fluid
having dimension dx, dy and dz along x, y
and z axis.
dl= dimension of inclined plane making
an angle α with the vertical
Px, Py, Pz and P are pressure acting in x, y,
z and perpendicular to inclined surface
dW=weight of element
zy
x
Py(dxdz)
Px(dydz)
P(dldz)
dx
dy
dz
α
α α P(dldz)
cosα
P(dldz)
sinα
dW
zy
x

6. PASCAL’S LAW
6
Py(dxdz)
Px(dydz)
P(dldz)
α α P(dldz)
cosα
P(dldz)
sinα
PP
odydzPdydzP
dldyodldydldzPdydzP
odldzPdydzP
F
x
x
x
x
x
=
=−
==−
=−
=∑
)(
/cos/)(
cos)(
0
α
α
Q
dW
PP
odxdzPdxdzP
dwdldxodldxdldzPdxdzP
odldzPdWdxdzP
F
y
y
y
y
y
=
=−
≅==−
=−−
=∑
)(
0&/sin/)(
sin)(
0
α
α
Q

7. PASCAL’S LAW
7
Similarly by applying the conditions in z direction, it can be proved
that
Hence,
The above states that the pressure acting on fluid particle is same in
all directions when the fluid is at rest.
PPz =
PPPP zyx ===

8. Absolute and Gauge Pressure
8
Atmospheric pressure
Gauge pressure
Vacuum/negative pressure
Absolute pressure
Atmospheric pressure: Pressure exerted by atmosphere
Gauge pressure: Pressure more than atmospheric pressure
Vacuum/negative pressure: Pressure less than atmospheric pressure
Absolute pressure: Pressure measure relative to absolute zero
vacatmabs
gatmabs
PPP
PPP
−=
+=

9. Atmospheric Pressure
9
It is defined as weight of air per unit
surface area of earth.
It decreases with increase in
elevation w.r.t. surface of earth.
Standard atmospheric pressure at
mean-sea-level is
=101.3KN/m2
=1.013bar
=14.7psi
=760mm of Hg
=33.9ft of water
=10.3m of water

10. Measurement of Atmospheric Pressure
10
Barometer: It is device used
to measure the atmospheric
pressure at any point on the
earth.
There are two types of
barometer
(i) Liquid barometer
It measures the pressure with
help of column of liquid
(ii) Aneroid barometer
It measures atmospheric
pressure by its action on an
elastic lid of evacuated box.

11. Liquid Barometer
11
It consists of a transparent tube which
is open from one end only. The tube is
filled with liquid and is inserted in a jar
also containing same liquid.The liquid
initially drops in tube due to gravity but
stabilizes at certain level under the
action of atmospheric forces.The
atmospheric pressure is then measured
as height of liquid at which it stabilizes.
Three forces acting on fluid are
Patm(A)=Force of atmospheric Pressure
W=Weight of liquid
Pvap(A)= Force of vapour pressure
A=Cross-sectional area of tube
Weight
of liquid
Force of
Patm
Vacuum
Pressure
Liquid/

12. Liquid Barometer
12
Three forces acting on fluid are
Patm(A)=Force of atmospheric Pressure
W=Weight of liquid
Pvap(A)= Force of vapour pressure
A=Cross-sectional area of tube
For Equilibrium
W
PatmA
PvapA
Liquid/
h
0; =−−=∑ APWAPoFy vapatm
0=−− APAhAP vapatm γ AhW γ=Q
vapatm PhP += γ

13. Liquid Barometer
13
Generally, mercury is preferred liquid because its vapour pressure is
minimum. Moreover, its specific gravity is very high so that size
(height) of barometer required is small.
However, for other liquid vapour pressure must be considered in
estimation.
The barometer using mercury is called mercury barometer and
while using water is called water barometer.
Size of barometer tube should be more than ½ inches (or 13mm)
to avoid capillarity.

14. Absolute Pressure
14
Gauge Pressure (Pg): It is the pressure measured relative to
atmospheric pressure (Patm) and is always above the atmospheric
pressure
It may be defined as normal compressive force per unit area
Vacuum Pressure (Pvac): It is the pressure measured relative to
atmospheric pressure and is less than the atmospheric pressure
It may be defined as normal tensile force per unit area
Absolute Pressure(Pabs): It is the pressure measured from absolute
zero
vacatmabs
gatmabs
PPP
PPP
−=
+=

15. Problem
15
Q.3.2.2
4600m
Surface
γ=10.05kN/m3
( )
2
/46700
460005.10
mkN
P
hP
=
= γ

16. Problem
16
Q.3.2.4
( )
psi
P
hSPhPP woil
9.57
)12/(684.6288.032 2
2
112
+=
+=+= γγ
P2=?
P1=32psi Stream
68ft
S=0.88
γ=Sγw

17. Problem
17

18. Problem
18
Q.3.3.1
5m
Surface
γ=10 kN/m3
( ) 2
/1628 mkNhPi === γ
γ=8 kN/m3
2m
interface
Pi
Pb
( ) 2
2211
/1.655)81.9(28 mkNP
hhP
b
b
=+=
+= γγ

19. NUMERICALS
19
Q.3.3.3
h=4600γ=10.05
kN/m3
2
/______4600)05.10( mkNP
hP
==
= γ
P=?

20. Problem
20
Q.3.3.3
h=?
Surface
γ=12N/m3
( ) ( )
mh
h
hP
8442
1210003.101
=
=×
= γ
P=101.3kPa
Determine the depth of gaseous atmosphere to cause 101.3kN/m2 over surface of
earth?

21. Problem
21
3.5.1
psiaP
psiaOHftP
vap
atm
09.2
47.144.33 2
=
==
Weight
of liquid
Force of
Patm
Vacuum
Pressure
Alcohol
w
vapatm
vapatmvapatm
vapatm
vapatm
S
PP
h
PP
A
APAP
h
APAhAP
APWAP
oFy
γ
γγ
γ
−
=
−
=
−
=
=−−
=−−
=∑
0
0
;
h=?

22. 22

23. Measurement of Pressure
23
The following devices are used for pressure measurement
1. Piezometer
2. Manometer
a) Simple manometer
B) Differential manometer
3. Mechanical PressureTransducer (Bourden gauge)
4. Electrical PressureTransducer

24. 1. Piezometer
24
It is used to measure pressure in
pipes or vessels.
In it simplest form, it consists of a
transparent tube open from other
ends
The diameter of tube should > ½” to
avoid capillarity action
Piezometers may be connected to
sides or bottom of pipe to avoid
eddies that are produced in the top
region of pipe
Limitations:
It must only be used for liquids
It should not be used for high pressure
It cannot measure vacuum (-ve) pressure
When connected to pipes, the
water level rises in it which
gives a measure of pressure.

25. 2. Manometer
25
a). Simple Manometer
Figure shows a set up of simple
manometer.
It consists of a U shaped tube, part
of which is filled with manometric
fluid.
One end of tube is connected
with the pipe whose pressure is
required to be determined.
Due to pressure, level of
manometric fluid rises on one side
while it falls on other side.
The difference in levels is
measured to estimate the
pressure.
A
Fluid, γf
Manometric
fluid, γm
Y
z
Y=Manometric reading
γf =Specific weight of fluid in pipe
γm =Specific weight of
manometric fluid

26. 2. Manometer
26
Manometric Fluids
1. Mercury
2. Oils
3. Salt solution etc
Properties of manometric
Fluid
1. Manometric fluid should not be
soluble/intermixale with fluid
flowing in pipe whose pressure is
required to be determined.
2. Lighter fluid should be used if
more precision is required.
A
Fluid,
Manometric
fluid,
y
z

27. 2. Manometer
27
Pressure measurement
A
Fluid, γf
Manometric
fluid, γm
Y
z
Sign Convention
-ve: upward direction
+ve: downward direction
AatmAabs
atmmfAabs
PPP
PYZP
+=
=−+
Q
γγ
Patm
ZYP
YZP
fmA
mfA
γγ
γγ
−=
=−+ 0
PA
Above is a gauge pressure
equation

28. 2. Manometer
28
b). Differential manometer
It is used to measure difference of pressure.
Case 1: when two vessels/pipes are at same level
ZA
ZB
Y
A B
PA
PB
Fluid A, γA
Fluid B, γB
Manometric
Fluid , γm
AAmBBBA
AAmBBBA
ZYZPP
ZYZPP
γγγ
γγγ
−++=−
−++=

29. 2. Manometer
29
( )
( )
( )( )YPP
YYPP
ZZYPP
ZYZPP
if
fmBA
fmBA
BAfmBA
AfmBfBA
fAB
γγ
γγ
γγ
γγγ
γγγ
−=−
−=−
−−=−
−++=−
==

30. 2. Manometer
30
b). Differential Manometer
Case 1I: when two vessels/pipes are at different level
ZA ZB
Y
A
B
PA
PB
Fluid A, γA
Fluid B, γB
Manometric
Fluid , γm
AAmBBBA
AAmBBBA
ZYZPP
ZYZPP
γγγ
γγγ
−++=−
−++=

31. 2. Manometer
31
( )BAfmBA
AfmBfBA
fAB
ZZYPP
ZYZPP
if
−−=−
−++=−
==
γγ
γγγ
γγγ

32. 2. Manometer
32
b). Differential manometer
Case 1II: when manometer is inverted
ZA
ZB
Y
A
B
PA
PB
Fluid A, γA
Fluid B, γB
Manometric
Fluid , γm
AAmBBBA
AAmBBBA
ZYZPP
ZYZPP
γγγ
γγγ
+−−=−
+−−=

33. 2. Manometer
33
( )BAfmBA
AfmBfBA
fAB
ZZYPP
ZYZPP
if
−+−=−
+−−=−
==
γγ
γγγ
γγγ

34. Advantages and Limitation of Manometers
34
Advantages
Easy to fabricate
Less expansive
Good accuracy
High sensitivity
Require little maintenance
Not affected by vibration
Specially suitable for low
pressure and low differential
pressure
Easy to change sensitivity by
changing manometric fluid
Limitations
Usually bulky and large in size
Being fragile, get broken easily
Reading of manometer is get
affected by temperature,
altitude and gravity
Capillary action is created due
to surface action
Meniscus has to be measured
accurately for better accuracy.

35. 3. Mechanical Pressure Transducer
35
Transducer is a device which is used to transfer energy from one
system to other
Mechanical pressure transducer converts pressure system to
displacement in mechanical measuring system
Bourden Gauge is used to measure high pressure either positive
or negative. It gives pressure directly in psi of Pascal units

36. Bourden Gauge
36
The essential mechanical
element in this gage is the
hollow, elastic curved tube
which is connected to the
pressure source as shown
in Fig.
As the pressure within the
tube increases the tube tends
to straighten, and although
the deformation is small, it
can be translated into the
motion of a pointer on a
dial as illustrated.
Fig. Bourden gauge

37. Bourden Gauge
37
Since it is the difference in pressure between the outside
of the tube and the inside of the tube that causes the
movement of the tube, the indicated pressure is gage
pressure.
The Bourdon gage must be calibrated so that the dial
reading can directly indicate the pressure in suitable units
such as psi, psf, or pascals.
A zero reading on the gage indicates that the measured
pressure is equal to the local atmospheric pressure.
This type of gage can be used to measure a negative gage
pressure (vacuum) as well as positive pressures.

38. 3. Mechanical Pressure Transducer
38
Elevation Correction
Bourden gauge gives pressure at the
center of dial. So to calculate pressure
at point A,
Where
γz=elevation correction
zPP gA γ+=
z
A
Pg

39. 4. Electrical Pressure Transducer
39
It converts displacement of mechanical measuring system to an
electrical signal.
Its gives continuous record of pressure when converted to a strip
chart recorder.
Data can be displayed using computer data acquisition system.

40. Problem
40
Q 3.5.4
x= ??in
New Manometric reading=4+2x=34.7in
OHftpsiaPatm 292.337.14 ==
( )
psiaP
PP
A
atmmwA
??
4125
=
=−×+ γγ
( ) ( ) atmmwA PxxP =+−+×+ γγ 241252

41. 41

42. Problem
42
Q. 3.5.7
0=+=
=−−
atmwmmA
atmmmwA
PhRP
PRhP
Qγγ
γγ
56.13,/31.6268@
45.13,/20.61150@
3
3
==
==
m
o
w
m
o
w
SftlbF
SftlbF
γ
γ

43. 43

44. Problem
44
3.5.8

45. 45

46. Problem
46
3.5.10
Two vessels are connect to a differential
manometer using mercury (S=13.56), the
connecting tubes being filled with water. The
higher pressure vessel is 5ft lower in elevation
than the other. Room temperature prevails. If the
mercury reading is 4.0in, what is the pressure
difference in feet of water and in psi ? (b) if
carbon tetrachloride (S=1.59) were used instead
of mercury what would be manometric reading
for the same pressure difference.

47. Problem
47

48. Problem
48
3.12

49. Problem
49

50. Forces on Immersed Bodies: Forces on
submerged planes & curved surfaces and
their applications, Drag and Lift forces
Dr. Mohsin Siddique
Assistant Professor
UOS Sharjah, UAE
50
Date:
Fluid Mechanics
University of Sharjah
Dept. of Civil and Env. Engg.

51. Forces on Immersed Bodies
51
Hydrostatic Force: It is the resultant force of pressure exerted by
liquid at rest on any side of submerged body.
It is the summation of product of uniform pressures and elementary
areas of submerged body
It is equal to the product of submerged area and pressure at the
centroid of the submerged area
pAdAppdAF === ∫∫

52. Forces on Plane Area
52
Center of pressure
The point of application of resultant
force of pressure on a submerged area
is called center of pressure.

53. Forces on Plane Area
53
A=total submerged area
F=hydrostatic force
θ=angle of submerged plane with
free surface
hc=depth of center of area
hp=depth of center of pressure
yc=inclined depth to center of area
yp=inclined depth to center of
pressure
dA=elemtry area
dF=force of pressure (hydrostatic
force) on elementry area
( ) ( )dAhdApdF γ==

54. Forces on Plane Area
54
Lets’ choose an elementary area so that pressure over it is uniform. Such
an element is horizontal strip, of width, x so . The pressure force,
dF on the horizontal strip is
Integrating
Where, yc is the distance OX along the sloping plane to the centroid C of
the area A. If hc is the vertical depth to the centriod, then we have
( ) ( )dAhdApdF γ==
xdydA =
∫∫∫∫∫ ==== ydAdAyhdApdAdF θγθγγ sinsin
θ
γ
sinyh
hp
=
=
A
ydA
yc
∫=
( )AyF cθγ sin=
AhF cγ=

55. Forces on Plane Area
55
Thus, we find the total force on any plane area submerged in a liquid
by multiplying the specific weight of the liquid by the product of the
area and the depth of its centriod.
The value of F is independent of the angle of inclination of the plane
so long as the depth of its centroid is unchanged.
Since is pressure at the centroid, we can also say that total
pressure force on any plane area submerged is a liquid is the
product of the area and the pressure at the centroid
AhF cγ=
chγ

56. Center of Pressure
56
In order to determine location of center of pressure, yp, from OX,
let’s take the moment of elementary area around OX
Integrating
Where, I is the 2nd moment of submerged area about OX
Where ycA is called static moment of area
( ) ( ) ( )dAyyhdAypdAyydFdM θγγ sin====
( ) ( )
( )IFy
dAydAyyydF
p θγ
θγθγ
sin
sinsin 2
=
== ∫∫∫
( ) ( )
Ay
I
Ay
I
F
I
y
cc
p ===
θγ
θγθγ
sin
sinsin ( )AyF cθγ sin=Q

57. Center of Pressure
57
Now, according to parallel axis theorem,
Where, Ic is 2nd moment of area about centroidal axis.
From this equation we again see that the location of center of pressure, P,
is independent of the angle θ.
When the plane is truly vertical, i.e., θ=90o
Ay
AyI
Ay
I
y
c
cc
c
p
2
+
== cc AyII 2
+=Q
Ay
I
yy
c
c
cp +=
Ah
I
hh
c
c
cp +=

58. Lateral Position of Center of Pressure
58
To find the lateral position of center of pressure P, consider the area
is made up of series of elemental strips.The center of pressure for
each strip is at the mid point of the strip. Since the moment of the
resultant force F must be equal to the moment of distributed force
system about any axis, say, the y-axis
Where, Xp is the lateral distance from the selected y-axis to the
center of pressure P of the resultant force F, and xp is the lateral
distance to the center of any elemental horizontal strip of area dA
on which the pressure p is acting
∫= pdAxFX pp
∫= pdAxFX pp

59. Forces on Curved Surface
59
Horizontal force on curved surface
Vertical force on curved surface
areahorizontalequivalentonforcechydrostati'' === FFxFx
surfaceaboveliquidofvolumeofWeight' === WFzFz

60. Forces on Curved Surface
60
Resultant Force
22
zx FFF += 





= −
x
z
F
F1
tanθ

61. Hydrostatic force formulas
61
Ay
I
yy
c
c
cp +=
AhF cγ= ( )AyF cθγ sin=

62. 62

63. Problem
63
Q 3.7.6:A plane surface is circular with a diameter of 2m. If it is vertical and
the top edge is 0.5m below the water surface, find the magnitude of the
force on one side and the depth of center of pressure.
Solution:
Ah
I
hh
c
c
cp +=
( )AyF cθγ sin=
AhF cγ=
Ay
I
yy
c
c
cp +=
Free surface
0.5m
D=2m
m
D
hc 5.1
2
5.0 =+=
( )
kNF
F
AhF c
2.46
2
4
5.1810.9 2
=






=
=
π
γ
( ) ( )
mh
DDh
Ah
I
hh
p
p
c
c
cp
667.1
4/5.1/64/5.1 24
=
×+=
+=
ππ

64. Problem
64
Q 3.7.8: A rectangular plate 5ft by 4ft is at an angle of 30o with the
horizontal, and the 5 ft side is horizontal. Find the magnitude of force on
one side of the plate and the depth of its center of pressure when the top
edge is (a) at the water surface (b) 1 ft below water surface
(a)
4ft
5ft
30o
hp
yp
hc
yc
4ft
θsinyh =
ftyh o
cc 130sin2sin === θ ( )( ) lbAhF c 12484514.62 =×== γ
( ) ( ) ftbdbd
Ay
I
yy
c
c
cp 67.22/12/2 3
=×+=+= fthp 33.1=

65. Problem
65
(b)
4ft
5ft
hp
yp
hc
yc=4ft
6ft
θsinyh =
fth o
c 230sin21sin21 =+=+= θ ( )( ) lbAhF c 25004524.62 =×== γ
( ) ( ) ftbdbd
Ay
I
y
c
c
p 33.44/12/44 3
=×+=+=
ftyh o
pp 167.230sin33.4sin === θ
1ft
ft
hy cc
4
sin/
=
= θ

66. Problem
66
Q 3.7.9
(a) For critical stability resultant of
all hydrostatic for must pass
through B.
Thus B will be at a/3=5.4/3=1.8m
above N
Schematic pressure
distribution
diagram
a/3

67. Problem
67
(a)
Since B point is hinge, therefore, moment
of all force at there must be zero
Schematic pressure
distribution diagram
1m
4.4/3
RB
RN
( )
mkNRN
RNMB
/59.17
08.13/4.43/4.595;0
=
=−−=∑
For reaction at B
mkNRN
FRBRNFx
/4.7759.1795
;0
=−=
=+=∑
mkNAhF c /954.4)
2
4.4
(81.9 === γ
For a water depth of 4.4m

68. Problem
68
3.7.12

69. Problem
69
Q 3.8.8. Cross section of tank is shown in figure., where r=2m and tank is
open and contain water to a depth h=3.5m. Determine the magnitude and
location of horizontal and vertical force components acting on unit width of
tank wall ABC.
( ) mh
mkNF
AhF
p
x
cx
33.25.3
3
2
/1.605.3)2/5.3(81.9
==
==
= γ
hp
xp

70. `
70
hp
xp
( )ALVOLWFz γγ ===
86.3)2(5.3
4/)4(
4
)2(5.3 2
=−=






−=
π
π
A
A
2m
d=4m( )
mkN
AFz
/9.37
)86.3(81.9
=
== γ
Lets take moment about AB
( ) ( )
mx
x
p
p
123.1
849.0)1(25.386.3
=
−×= π
m
r
xc 849.0
3
4
==
π

71. Drag and Lift Force
71
Lift is the component of aerodynamic
force perpendicular to the relative wind.
Drag is the component of aerodynamic
force parallel to the relative wind.
Weight is the force directed downward
from the center of mass of the airplane
towards the center of the earth.
Thrust is the force produced by the
engine. It is directed forward along the
axis of the engine.

72. Drag and Lift Force
72
The drag force acts in a direction that is
opposite of the relative flow velocity.
Affected by cross-section area (form
drag)
Affected by surface smoothness
(surface drag)
The lift force acts in a direction that is
perpendicular to the relative flow.
CD= Coefficient of drag
CL= Coefficient of lift
A=projected area of body
normal to flow
V= relative wind velocity

73. Buoyancy and Floatation
73
When a body is immersed wholly or partially in a fluid, it is subjected to an
upward force which tends to lift (buoy)it up.
The tendency of immersed body to be lifted up in the fluid due to an
upward force opposite to action of gravity is known as buoyancy.
The force tending to lift up the body under such conditions is known as
buoyant force or force of buoyancy or up-thrust.
The magnitude of the buoyant force can be determined by Archimedes’
principle which states
“ When a body is immersed in a fluid either wholly or partially, it is
buoyed or lifted up by a force which is equal to the weight of fluid
displaced by the body”

74. Buoyancy and Floatation
74
Lets consider a body
submerged in water as shown
in figure.
The force of buoyancy
“resultant upward force or
thrust exerted by fluid on
submerged body” is given
Water surface
11 hP γ=
( )212 hhP += γ
2h
1h 1F
2F
dA=Area of cross-section of
element
γ= Specific weight of liquid
( ) ( )
( )[ ]
[ ]volumeF
dAhF
dAhdAhhF
FFF
B
B
B
B
γ
γ
γγ
=
=
−+=
−=
2
121
12

75. Buoyancy and Floatation
75
=Weight of volume of liquid displaced by
the body (Archimedes's Principle)
Force of buoyancy can also be determined as difference of
weight of a body in air and in liquid.
Let
Wa= weight of body in air
Wl=weight of body in liquid
FB=Wa-Wl
[ ]volumeFB γ=

76. Buoyancy and Floatation
76
Center of Buoyancy (B): The point of application
of the force of buoyancy on the body is known as the
center of buoyancy.
It is always the center of gravity of the volume of fluid
displaced.
Water surface
CG or G
C or B
CG or G= Center of gravity
of body
C or B= Centroid of
volume of liquid displaced
by body

77. Thank you
Questions….
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77