CHAPTER 2

FLUID PRESSURE CONCEPT

&
MEASUREMENT
 INTRODUCTION

 Forces exerted by fluids is important to

be calculated in designing the water
structures and/or structures that deal/in-
contact with fluid.
 False design and error in design can
cause damage/failure. For example:
Collapse of wall, dam, water tank,
leakage or damage in joints/pipe fittings.
 INTRODUCTION

 Therefore, it is important to understand

the concept of pressure and avoid any
mistakes in measuring the pressure.
 INTRODUCTION
What is Pressure?
Force (F) exerted by a liquid on a plane
area (A) formulated as:
P = F/A (N/m2) ……….(i) or,
The product of the unit weight ρg of the
liquid, the depth hcg of the centre of the
gravity of the area, and the area (A). The
equation is:
P = ρghcg (N/m2)………...(ii)
TYPES OF PRESSURE

 Atmospheric Pressure
 Surface of the earth depends upon head of air above the surface.
 At sea level (101.3 kN/m2) = 10.35 m of water or 760 mmHg.

 Vacuum
 Pressure is zero (perfectly empty in space).

Vacuum Pressure = Atmospheric Pressure – Absolute Pressure

 Gauge Pressure.
 Measured above or below atmospheric pressure.

 Absolute Pressure.

Absolute Pressure = Gauge Pressure + Atmospheric Pressure

Hydrostatic Pressure

Pressure Gauge

•The pressure at a given depth

is independent of direction. It is
the same in all directions.
* The pressure on a submerged
object is always perpendicular
to the surface at each point on
the surface.

10
 Positive Pressure Gauge

Pressure

Atmospheric Pressure
P = 0 k Pa Gauge or
P = 101 kPa Absolute

Absolute Pressure

Vacuum Pressure or
Negative Pressure Gauge
Absolute Zero Pressure

Relationship between types of pressure

Unit conversion for pressure as follows:
1 bar = 100,000 N/m2 = 0.1 MPa
1 kgf/m2 = 9.806614 N/m2 = 9.806614 Pa
1 atm (standard) = 101325 Pa = 1.01325 bar

12
EXAMPLE 2.1

Find the absolute pressure in kPa unit if

Barometer shows the reading of 60 kPa.
Given that Barometer reading at sea level
is 740 mmHg and specific gravity of
mercury is 13.6.

SOLUTION:

Given:-
Gauge pressure, Pgage = 60 kPa
Atmospheric pressure, Patm = 740 mmHg
Mercury specific gravity, s.gmercury = 13.6
SOLUTION .... Cont ‘

Known that,
Patm = mgh (m = mercury density)

Patm = (13.6)(1000)(9.81)(740/1000)

Patm = 98.728 kPa

x
s.g x 
w

Hence,
Pabs = Pgauge + Patm

Pabs = 60 + 98.73

Pabs = 158.728 kPa

EXAMPLE 2.1A
 PRESSURE VARIATION SUBJECT TO
FLUID DEPTH
FORCE AND PRESSURE AT A POINT

 External forces acting on the fluid element are due to the

pressure and weight.

 Force  distributed to all mass or volume of the element.

 Cause by external factors ie: gravity, electromagnet etc.

 Pressure can be represented as:

Where:
F
P
A …………….. ( 2.1 )

F = Force (N)
A = Area (m2)
PRESSURE VARIATION SUBJECT TO
FLUID DEPTH
PRESSURE VARIATION SUBJECT TO
FLUID DEPTH
PRESSURE VARIATION SUBJECT TO
FLUID DEPTH
PRESSURE VARIATION SUBJECT TO
FLUID DEPTH
PASCAL LAW

 Was inroduced by Blaise Ascal – French (1623 – 1662).

 States - the pressure at a point in a fluid at rest, or in motion, is

independent of direction as long as there are no shearing stress
present.

P3 z
B
q
P1

x
A C
W
P2

Figure 2.2: Pressure at a point in fluid at rest

PASCAL LAW... continued
 Consider the free-body diagram obtained by removing a small triangle wedge of
fluid from some arbitrary location within a fluid mass;
(a) Summation of horizontal forces,
(P1  AB  1) – (P3  BC  1) cos q = 0 (i)
where,
cos q = AB / BC
 P1 = P3 (ii)

(b) Summation of vertical forces,

(P2  AB  1) – (P3  BC  1) sin q – W = 0 (iii)
where,
sin q = AC / BC
W=0  (one point is considered)

 P2 = P3 (iv)
From equation (ii) and (iv), it was found that;

P1 = P2 = P3

Hence, pressures at a point in fluid at rest is same at any direction

PRESSURE HEAD

 Pressure increase with depth.

 Consider a container consist of fluid and cylinder was put
inside as shown in figure below:
Cylinder

Container

h

Pressure head
 Sum of pressure at sylinder base = Fluid weight inside cylinder
PA=  A h

P  h
where,
h = Fluid height in cylinder
 = Specific weight

Hence, pressure in fluid is proportioned with depth

PRESSURE HEAD …. continued

 Pressure is not influenced by the size or shape.

Liquid surface
( P = Po )

h
(specific weight = )
A B

Fluid equilibrium in a container or arbitrary shape

 The actual value of pressure along AB depends only on the depth,

h, the surface pressure, Po and the specific weight,  of the
liquid in the container, thus;
P  h
 Pressure also can be expressed in height of free surface, where;
P P
h 
 g
EXAMPLE 2.2

Find the pressure at 25 m depth below

water free surface inside a reservoir.

SOLUTION:

Given:-
Depth of water, h = 25 m
Specific weight of water, w = 9.81 kN/m3

Hence,
P = h

P = (9.81)(25)

P = 245.25 kN/m2
EXAMPLE 2.2A
EXAMPLE 2.3
Figure below shows 4 different fluid inside a glass
container at 20C. If atmospheric and absolute
pressure at the surface and base are 101.03 kPa
and 231.3 kPa, find the specific graviy of olive
oil.

SAE 30 Oil 1.5 m

Water 2.5 m

Olive oil 2.9 m

Mercury 0.4 m
SOLUTION:

Given:-
Patm = 101.03 kPa
Pabs = 231.3 kPa

Refer Chapter 1 to obtain specific gravity value for different fluids

PSAE 30 = (sgSAE 30)(ρw)(g)(h)

PSAE30 = (0.92)(1000)(9.81)(1.5)
PSAE30= 13537.8 Pa

Pw = (sgw) (ρw) (g)(h)

Pw = (1)(1000)(9.81)(2.5)
Pw = 24525 Pa
SOLUTION .... Cont ‘

Polive = (sgolive)(ρw) (g)(h)

Polive = (sgolive)(1000)(9.81)(2.9)
Polive = 28449 sgolive

Pm = (sgm)(ρw) (g)(h)
Pm= (13.56)(1000)(9.81)(0.4) = 53209.4 Pa

Known that,

Patm + PSAE30 + Pw + Polive + Pm = Pabs

Polive = Pabs – (Patm + PSAE30 + Pw + Pm)
28449 sgolive = (231.3)(1000) – [(101.03)(1000) +
13537.8 + 24525 + 53209.3

sgolive = 1.37
EXAMPLE 2.3A
 PRESSURE MEASUREMENT

Device :-

i. Piezometer
ii. Manometer
iii. Bourdon Gauge
iv. Tansducer
 PIEZOMETER
 Example type of manometer.
 To measure pressure in pipe.
 Suitable for measuring moderate pressure of fluids.
 Fluid in pipe is same in piezometer tube.
 Tube diameter should at least 0.5 inch or 12 mm  avoid
capillary error.

h h

P  h

Figure 2.5: Piezometer

EXAMPLE 2.4
Water flow from point A to B in a pipe as shown
in figure below. Find pressure at point A and B.

Hydraulic Grade Line

Pitot tube
2.1 m

B
4.8 m
ater
w

8m
A
6m
Datum
SOLUTION:

At point A pressure head, hA :-

PA
hA 
g
PA
6  4.8 
(1000)( 9.81)
 PA  105948 Pa  105.95 kPa

At point B pressure head, hB :-

PB
hB 
ρg
PB
8  2.1 
(1000)(9.8 1)
 PB  99081 Pa  99.08 kPa
 MANOMETER
Types of manometers :-

 Simple U-Tube Manometers

 Differential Manometers

 Inverted Manometers
SIMPLE U-TUBE MANOMETERS

PB
PA B

A
h1

h2
x x

Figure 2.6: Simple U-Tube Manometers

 SIMPLE U-TUBE MANOMETERS
 Usually mercury is used to measure high gauge pressure.
 Not suitable for gas pressure.
 Consider left and right side;

Px-left = Px-right
PB
where, PA B

Px-left = PA + A gh2 A
h1

Px-right = Patm + B g(h1+h2) h2

x x

Patm = 0 (as datum) therefore,

PA + A gh2 = B g(h1+h2)

PA = B g(h1+h2) - A gh2
EXAMPLE 2.5
Calculate pressure at A (kN/m2) if h1 = 50 mm,
h2 = 120 mm, w = 1000 kg/m3 and
m = 13560 kg/m3
w
(water) B

h1
A

h2

x x
m
(mercury)
SOLUTION:
Consider left and right side of manometers,

Px-left = Px-right
where,

Px  left  PA   w gh 2
Px  left  PA  (1000)( 9.81)( 0.12)
Px  left  PA  1177.2

Px  right  Patm  m g(h1  h 2 )

Px  right  Patm  (13560)( 9.81)( 0.12  0.05)
Px  right  Patm  22614.012
SOLUTION .... Cont ‘

(Patm = 0  expose to air) therefore;

PA  1177.2  0  22614.012
PA  21436.8 N/m 2  21.44 kN/m 2
DIFFERENTIAL MANOMETERS

A

B

Figure 2.7: Differential Manometers

 DIFFERENTIAL MANOMETERS

 Used to measure pressure difference between 2 points.

 Consider left and right side;

Px-left = Px-right
where,
Px-left = PA + A gh2

Px-right = PB + B gh1 + A gh1 (h3 – h1)

hence,

PA + A gh2 = PB + B gh1 + A g (h3 –h1)

PA - PB = B gh1 + Ag (h3 - h1) - A gh2

PA - PB = B gh1 + Ag (h3 - h1 - h2)

EXAMPLE 2.6
Figure below show diffential manometers used to
measure pressure inside a pipe. If PA – PB = 120
kPa, find h.

Water
(s.g = 1.0) B

1.6 m
A

Mercury
h (s.g = 13.6)

x x
SOLUTION:
Consider left and right side of manometers,

Remenber:
Px-left = Px-right x
where, s.g x 
w

Px  left  PA   w g(ha  h )
Px  left  PA  (1000)( 9.81)(ha  h )
Px  left  PA  9810ha  9810h

Px  right  PB  m gh   w g(h a  h b )
Px  right  PB  s.g m . w gh   w g(h a  h b )
Px  right  PB  (13.6)(1000)( 9.81)(h )  (1000)( 9.81)(h a  1.6)
Px  right  PB  133416h  9810h a  15696
SOLUTION .... Cont ‘

Equal the pressure at x – x line left and right side therefore;

PA  9810h a  9810h  PB  133416h  15696  9810h a

PA  PB  133416h  9810h  15696
120 x 10 3  123606h  15696
h  0.844 m
 INVERTED MANOMETERS
air

x x
h1

h2
PA
A
h3

PB
B

a

Figure 2.8: Inverted Manometers

 INVERTED MANOMETERS
 Mercury is used to measure pressure if fluid is low density, ie oil.
 Consider left and right side;

Px-left = Px-right
where,
Px-left = PA - a gh2 – air gh1  (neglect air)

Px-right = PB - agh3 - ρagh1 = PB - ρag (h3 + h1)

air
hence,

PA - agh2 = PB - ag (h3 + h1) x

h1
x

h2
PA
PA - PB = agh2 - ag (h3 + h1) A
h3

PB
B

a
EXAMPLE 2.7
Figure below show inverted manometers used to
measure pressure inside a pipe. Find the pressure
difference of PA – PB.
Fluid X
(s.g = 0.9)

x x

0.25 m
1.625 m

0.5 m

A Water
(s.g = 1.0)
SOLUTION:
Consider left and right side of manometers,

Px-left = Px-right
where,

Px  left  PA   w g(ha )
Px  left  PA  (1000)( 9.81)(1.625)
Px  left  PA  15941.25

Px  right  PB   x ghb   w ghc

Px  right  PB  s.g x . w ghb   w ghc
Px  right  PB  (0.9)(1000)( 9.81)( 0.25)  (1000)( 9.81)(1.625  0.25  0.5)
Px  right  PB  2207.25  8583.75
Px  right  PB  10791
SOLUTION .... Cont ‘

Equal the pressure at x – x line left and right side therefore;

PA  15941.25  PB  10791
PA  PB  5150.25 N/m 2
EXAMPLE 2.8
Calculate force is required to balance the weight
of the cylinder if the weight of the plunger is
negligible?
 BOURDON GAUGE
 Commonly measure gauge pressures or vacuum.
 A curved tube of elliptical cross section tend to straighten if
subjected to higher pressure.

Figure 2.9: Bourdon gage

 TRANSDUCER
 Converts pressure into electric signal in digital form.
 Commonly used  “Strain-gage base transducer”.
 As pressure changes  the deflection of the diaphragm
changes  changes the electrical output  provide
pressure.

Figure 2.10: Transducer