Chapter-2

FLUID PRESSURE &

ITS MEASUREMENT
Prof. Suresh Ramaswwamyreddy
Department of Civil Engineering
BMSCE, Bangalore-19.
Contents:
Definition of pressure, types of pressures, Pressure at a point in a static
fluid, Pascal’s law, Hydrostatic pressure law. Measurement of fluid pressure-
Simple & Differential manometers and Mechanical gauge. 6 Hours

2
INTRODUCTION:
Pressure or intensity of pressure may be defined as the force exerted on an unit area. If F
be the force uniformly acting over an area, ‘A’, then the pressure at any point is given by p=
F/A.
If the force is not uniformly distributed then the Force, F
expression gives an Average value only. When Area, A
the force varies from point to point on an area,
the magnitude of pressure at any point can be obtained as p= dF/dA
Where dF is the force acting on an infinitely small area dA.
Definition of Pressure: Pressure is one of the basic properties of all fluids. Pressure (p) is
the force (F) exerted on or by the fluid on a unit of surface area (A).
Mathematically expressed: p = F/A = N/m2
The basic unit of pressure is Pascal (Pa).
When a fluid exerts a force of 1 N over an area of 1m2, the pressure equals one Pascal,
i.e., 1 Pa = 1 N/m2.
 Fluid is a substance capable of moving. If a mass of fluid is held in static equilibrium by
confining within solid boundaries, it exerts a force against the solid boundaries(Fig.1), it
exerts force along direction perpendicular to the boundary in contact. This force so exerted is
NORMAL to the surface in contact.
This is because, the fluid at rest cannot sustain the shear stress
and hence the forces cannot have tangential components. The
Normal force exerted on the solid boundary per unit area is called
‘Fluid Pressure’.

Fluid, 𝛾

4
VARIATION OF PRESSURE IN A STATIC MASS OF FLUID
Consider a fluid element of size 𝛿𝑥, 𝛿𝑦 𝑎𝑛𝑑 𝛿𝑧
at any point in a static mass of fluid. Since the fluid is at rest, it is
𝑍
under equilibrium by the action of various forces acting on it. p+
𝜕𝑝
𝛿𝑧/2
𝜕𝑧
The forces acting on the fluid element are : 1. The pressure P+
𝜕𝑝
𝛿𝑦/2
𝜕𝑦
forces and the self weight of the element.
Now let ‘p’, the intensity of pressure at point ‘o’, the mid point of
the fluid element. 𝛿𝑧
𝜕𝑝
Then the intensity of pressure on LHS of the fluid element in the p- 𝛿𝑥/2
𝜕𝑥 𝑜 𝜕𝑝
X direction is p-(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2, similarly on RHS, it is p+(𝜕𝑝/ 𝑝 P+ 𝜕𝑥
𝛿𝑥/2

𝜕𝑥) 𝛿𝑥/2 𝑊

Then the pressure force on LHS is : [p-(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 𝛿𝑦

𝛿𝑥
On RHS: [p+(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 𝑋
𝜕𝑝
Similarly in the other 2 directions, the pressure forces are: 𝑌
p- 𝜕𝑦
𝛿𝑦/2
𝜕𝑝
Y Direction: [p-(𝜕𝑝/𝜕𝑦) 𝛿𝑦/2]* 𝛿𝑥 𝛿𝑧 and [p+(𝜕𝑝/𝜕𝑦) 𝛿𝑦/2]* 𝛿𝑥 𝛿𝑧 p- 𝜕𝑧
𝛿𝑧/2

Z direction: [p-(𝜕𝑝/𝜕𝑧) 𝛿𝑧/2]* 𝛿𝑥 𝛿𝑦 and [p+(𝜕𝑝/𝜕𝑧) 𝛿𝑧/2]* 𝛿𝑥 𝛿𝑦

For equilibrium, the forces acting in any direction must be Zero.

i.e., 𝐹𝑥 = 0
∴ [p-(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 - [p+(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 = 0

5
∴ [p-(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 - [p+(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2]* 𝛿𝑦 𝛿𝑧 = 0

= (-2(𝜕𝑝/𝜕𝑥) 𝛿𝑥/2)* 𝛿𝑦 𝛿𝑧= 0

= (-(𝜕𝑝/𝜕𝑥) ∗ 𝛿𝑥 𝛿𝑦 𝛿𝑧 )= 0
= - (𝜕𝑝/𝜕𝑧) 𝛿𝑉 = W
= (𝜕𝑝/𝜕𝑥) ∗ 𝛿𝑉= 0 Consider W:
WKT, 𝛾 =W/V
= (𝜕𝑝/𝜕𝑥)= 0 ----------------------(1) W=𝛾V
= -(𝜕𝑝/𝜕𝑧) 𝛿𝑉 = 𝛾𝛿V
Similarly in the Y direction, (𝜕𝑝/𝜕𝑦)= 0 ----------------------(2) = -(𝜕𝑝/𝜕𝑧) = 𝛾
Consider Z Direction, Fz = 0
= [p-(𝜕𝑝/𝜕𝑧) 𝛿𝑧/2]* 𝛿𝑥 𝛿𝑦 - [p+(𝜕𝑝/𝜕𝑧) 𝛿𝑧/2]* 𝛿𝑥 𝛿𝑦 − W = 0 = (𝜕𝑝/𝜕𝑧) = −𝛾 −−−−−−−− − 3
From Equation (1), (2) and (3), it
=-2 (𝜕𝑝/𝜕𝑧) 𝛿𝑧/2* 𝛿𝑥 𝛿𝑦 − W = 0 can be concluded that the pressure
= - (𝜕𝑝/𝜕𝑧) 𝛿𝑧 𝛿𝑥 𝛿𝑦 − W = 0 varies only in the vertical direction.
-ve sign indicates pressure
= - (𝜕𝑝/𝜕𝑧) 𝛿𝑉 − W = 0 decreases as Z moves towards the
= - (𝜕𝑝/𝜕𝑧) 𝛿𝑉 = W free surface.

6
PRESSURE AT A POINT IN A FLUID
Consider the equation, (𝜕𝑝/𝜕𝑧) = −𝛾
If the fluid is incompressible for which the specific weight is constant, the integration of the above
equation becomes p = −𝛾z+c ------------(1)
P= pressure at any point in a static mass of fluid.
c- Constant of Integration. Free surface

At the free surface of the fluid, the pressure is atmospheric. The depth of h
the fluid from the free surface is Z= Zo+H H

If pa is the atmospheric pressure, then pa = −𝛾(Zo+H)+c

C= pa +𝛾(Zo+H)
Substitute the value of c in equation (1), we have Zo

p = −𝛾z+ pa +𝛾(Zo+H)------ (2) Datum

Now consider a point in a fluid at a depth ‘h’ from the free surface, then
Z= Zo+H-h
Substitute the value of z in equation (2), we have
p = −𝛾(Zo+H-h)+ pa +𝛾(Zo+H)
p= 𝛾 h+ pa -------------------------(3)
Since atmospheric pressure at a point is constant, the above equation
reduces to p = 𝛾 h.

7
Pressure Head
The vertical height of the free surface above any point in a liquid at rest is known as pressure
head.
Pressure is same in all directions-Pascal’s Law:
The pressure at any point in a static mass of fluid at rest has the h = pressure head
same magnitude in all the directions. i.e., when a certain force is
applied at any point in a fluid, it is transmitted equally in all
directions.
Consider a small triangular wedge of fluid element ABC having
an unit width in a static mass of fluid(Fig.2). The pressures on the y
three faces of the wedge are equal in magnitude of px = py = pz pz
Let px = the intensity of pressure in the x direction.
A 90-θ
py = the intensity of pressure in the y direction
θ
pz = the intensity of pressure in the z direction
px dy
Θ – angle of the triangular element. dz
x
Now the pressure force on the vertical side AB of the fluid θ
element, Px = px*(AB*1) = px*AB dx θ
B C

py
8
Similarly the pressure force on the horizontal side BC of the fluid element, Py = py*BC*1 = py*BC
On the diagonal side, Pz = pz*AC
Since the fluid is under rest, the sum of the components of the horizontal and vertical forces must
be ZERO.
Resolving the forces horizontally, we have,
Px = Pz*cos(90-θ)
Px = Pz*Sinθ
px*AB = pz*AC Sinθ
But from the figure, AC Sin θ = AB
∴ px=pz ---- (1)
Now in the vertical direction,
Py = Pz*Sin(90-θ)
Py = Pz*Cosθ
py*BC=pz*AC Cosθ
AC Cosθ = BC
∴ py=pz ---- (2)
From (1) and (2), we have px= py=pz.

9
Application of Pascal’s Law:
Pascal’s law is made use in the Hydraulic Press, Hydraulic Jack, Hydraulic Lift, Hydraulic
Crane, etc.,

Q. A large piston supports a car. The total mass of the

piston and car is 3200 𝑘𝑔. What force must be applied to
the smaller piston ?
Pressure at the same height is the same! (Pascal’s Law)

𝐹1 𝐹2 𝐴1 𝜋 × 0.152
= 𝐹1 = 𝑚𝑔 = × 3200 × 9.8 = 490 𝑁
𝐴1 𝐴2 𝐴2 𝜋 × 1.202
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Types of Pressure:
1. Atmospheric Pressure
2. Absolute Pressure
3. Vacuum Pressure
Atmospheric pressure
Air above the surface of liquids exerts pressure on the exposed surface
of the liquid and normal to the surface.
The pressure exerted by the atmosphere is called atmospheric
pressure.
Atmospheric pressure at a place depends on the elevation of the
place and the temperature. Atmospheric pressure is measured using an
instrument called ‘Barometer’ and hence atmospheric pressure is also
called Barometric pressure. At sea level, the atmospheric pressure is
10.3m of water/76cm of Mercury/1.03kg/cm2
However, for engineering purposes, it is more convenient to measure
the pressure above a datum pressure at atmospheric pressure.
By setting patmophere = 0, p = ρgh
Unit: kPa. ‘bar’ is also a unit of atmospheric pressure 1-bar = 100 kPa.

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Fluid pressures may be measured with
respect to 2 arbitrary datums.
1. Absolute Zero pressure Datum
2. Local atmospheric pressure Datum
Absolute pressure at a point is the intensity of
pressure at that point measured with reference to
absolute vacuum or absolute zero pressure.
Absolute pressure at a point can never be negative
since there can be no pressure less than absolute
zero pressure.
Gauge Pressure: If the intensity of pressure at a point is measured with reference to
atmospheric pressure datum , and the pressure is more than local atmospheric pressure,
then it is called gauge pressure at that point.
Vaccum pressure at a point is measured with reference to atmospheric pressure datum , and
the pressure is less than local atmospheric pressure.
Absolute pressure at a point = Atmospheric pressure ± Gauge/vacum pressure
NOTE: If we measure absolute pressure at a Point below the free surface of the liquid, then,
p2 (absolute) = 𝛾. h + patm
p1 = patm
If gauge pressure at a point is required, then atmospheric pressure is
taken as zero, then, p2 (gauge) = 𝛾. h = 𝜌gh

Also, the pressure is the same at all points with the same depth from the
free surface regardless of geometry, provided that the points are
interconnected by the same fluid. However, the thrust due to pressure is
perpendicular to the surface on which the pressure acts, and hence its
direction depends on the geometry.

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Measurement of Pressure:
Measurement Devices:
 Barometer
 Manometers
 Mechanical Gauges
Barometer: A barometer is a device for measuring atmospheric pressure. A simple
barometer consists of a tube more than 760 mm long inserted in an open container
of mercury with a closed and evacuated end at the top and open tube end at the
bottom and with mercury extending from the container up into the tube.
Manometers: These are devices based on the Principle of balancing the
column of liquid whose pressure is to be measured by the same fluid or another
column of fluid. Manometers are classified into:
 Simple Manometers
 Differential Manometers
Simple Manometer: Simple monometers are used to measure intensity of pressure at a point. It consists
of a glass tube having one of its end connected to the gauge point, where pressure is to be measured
and the other end is open to atmosphere.
Types of Simple Manometers
Common types of simple manometers are: a) Piezometers b) U-tube manometers c) Single tube
manometers d) Inclined tube manometers

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Piezometers
Open
 Piezometer consists of a glass tube inserted in the wall of the
vessel or pipe at the level of point at which the intensity of
pressure is to be measured.
 The other end of the piezometer is exposed to air.
 The height of the liquid in the piezometer gives the pressure head
from which the intensity of pressure can be calculated.
Fluid  To minimize capillary rise effects the diameters of the tube is kept
more than 12mm.
 In case of Negative pressure, the fluid will not rise in the
piezometer.
 In such a case the piezometer is bent in the shape of U.
 One end is connected to the point where pressure is to be
h
measure and the other end is dipped in a fluid whose specific
weight is different from the fluid present in the pipe.
Fluid,
A
 In this case the fluid rises in the piezometer at the other end
to a height, ‘h’.
 knowing the value of ‘h’ at the other end, pressure in the pipe
is given by p=-𝛾ℎ
15
Merits
 Simple in construction
 Economical
Demerits
 Not suitable for high pressure intensity.
 Gas Pressure cannot be measured.
U-Tube Manometer
A U-tube manometers consists of a glass tube bent in U-Shape, one end of
which is connected to gauge point and the other end is open to atmosphere.
U-tube consists of a fluid having specific of gravity higher than that of fluid
whose pressure intensity is to be measured and is called manometric liquid.
Manometric Fluids:
 Manometric fluids should neither mix nor have any chemical reaction with the fluid whose
pressure intensity is to be measured.
 It should not undergo any thermal variation.
 Manometric fluid should have very low vapour pressure.

16
To write the gauge equation for manometers
Steps:
1. Convert all given intensity of pressure to pressure head, i.e., meters of water.
2. Starting from one end move towards the other end, keeping the following points in mind.
 Any horizontal movement inside the same liquid will not cause change in pressure.
 Vertically downward movement causes increase in pressure and upward motion cause
decrease in pressure.
 Convert all vertical columns of liquids to meters of water by multiplying them by
corresponding specify gravity.
OPEN
 Take atmospheric pressure as zero (gauge pressure computation).
3. Solve for the unknown quantity and convert it into the required unit. B

Refer to the fig. Let us find the pressure at A:

1. Starting from A: Fluid, 𝛾1

2. Pressure at A, pA A
A’

3. A and A’ are at the same level, so whatever the pressure acting at A, Manometric Fluid,
𝛾2

same pressure is at A’.

4. Pressure at A’, pA’= pA

17
From A’, moving towards X, pressure increases. Hence, pressure at X,
pX= Pressure at A’+𝛾1ℎ1 ; 𝛾1 = Specific weight of fluid present in the
manometer from A’ to X.
pX= pA’+𝛾1ℎ1
pX= pA +𝛾1ℎ1
Pressure at X’ = Pressure at X, because same level
pX ‘ =pX= pA +𝛾1ℎ1
From X’ to B, moving upwards, pressure decreases.
Pressure at B, pB = (pX–(𝛾2ℎ2)) ; 𝛾2 = Specific weight of the
manometric fluid
pB = pA +𝛾1ℎ1 –(𝛾2ℎ2)
But at B, pB pressure is atmospheric and is equal to Zero.
pB =0= pA +𝛾1ℎ1 –(𝛾2ℎ2)
The unknown is pA
Hence pA =(𝛾2ℎ2)-𝛾1ℎ1

18
Start from the other end:
Pressure at B, pB = pressure is atmospheric and is equal to Zero.
pB =0
From B, move towards X’ , pressure increases and hence
pX ‘ = pB + 𝛾2ℎ2
Pressure at X = Pressure at X’, because same level
pX=pX ‘ = pB + 𝛾2ℎ2
From X, moving towards A’, pressure decreases. Hence,
pressure at A’, pA’ = pX-𝛾1ℎ1
pA’ = pB + 𝛾2ℎ2 -𝛾1ℎ1
Pressure at A, pA = pA’ A and A’ are at the same level, so whatever the pressure acting at A,
same pressure is at A’.
pA = pB + 𝛾2ℎ2 −𝛾1ℎ1
But pB = 0
pA = 𝛾2ℎ2 −𝛾1ℎ1

19
Single Column Manometer
 In case of a U tube manometer, one must note down the level
of manometric fluid in both limbs.
 This difficulty may overcome by using a single column
manometer.
 It is a modified form of U tube manometer.
 A shallow reservoir having a large c/s area of about 100 times
the diameter of the tube is introduced in one of the limbs.
 For any variation in the pressure, change in the liquid level in
the reservoir is so small that it may be neglected and the
pressure is indicated approximately by the height of the liquid in
the other limb.
DIFFERENTIAL MANOMETERS
Differential manometers are used to measure pressure difference between any two points.
Common varieties of differential manometers are:
a) Two Piezometers
b) U-tube differential manometer.
c) Inverted U-tube manometer.
d) Micro manometers. 20
Two Piezometers Inverted U-tube manometers:
The arrangement consists of two Inverted U-tube manometer is used to measure
piezometers at the two points between small difference in pressure between any two
which the pressure difference is required. points. It consists of an inverted U-tube
The liquid will rise in both the connecting the two points between which the
piezometers. The difference in elevation pressure difference is required. In between there
of liquid levels can be recorded and the will be a lighter sensitive manometric liquid.
pressure difference can be calculated. Pressure difference between the two points can
be calculated by writing the gauge equations for
the system.

21
Let ‘pA’ and ‘pB’ be the pressure at ‘A’ and
‘B’ in N/m2 of water Let,
pA – (𝛾1h1) + (𝛾mh ) + (𝛾2h2) = pB. 𝛾1 = Specific weight of liquid in pipe A
(pA – pB ) = (𝛾1h1) - (𝛾mh ) - (𝛾2h2) 𝛾2 = Specific weight of liquid in pipe B
(pA – pB) = (𝛾1h1) - (𝛾mh ) - (𝛾2h2) 𝛾m = Specific weight of manometeric liquid
pA be the pressure in the pipe at point A
U-tube Differential manometer- pB be the pressure in the pipe at point B
h is the distance of mercury level in the right limb from
Two points are at different level the datum line XX’
h1 is the height of manometer liquid level in the left limb from
the from the datum line XX’
h2 is the height of liquid level in the right limb from the center of
pipe at point B to the top level of manometric fluid in the right limb.
Pressure at X= Pressure at X’ (same level)
Left limb: pA + 𝛾1h1.
Right limb: pB + 𝛾2h2+ 𝛾mh
pA + 𝛾1h1=pB + 𝛾2h2+ 𝛾mh
(pA – pB) = 𝛾2h2+ 𝛾mh-𝛾1h1

22
U-tube Differential manometer-
Let,
Two points are at the same level 𝛾1 = Specific weight of liquid flowing in the pipeline
𝛾2 = Specific weight of manometric liquid (assume
A B

mercury)
pA be the pressure in the pipe at point A
pB be the pressure in the pipe at point B
h is the distance of mercury level in the right limb
from the datum line XX’
h1 is the height of manometer liquid level in the
right limb from the center of pipe at point B.
Pressure at X= Pressure at X’ (same level)
Left limb: pA + 𝛾1 (h1.+h)
Right limb: pB + 𝛾1h1+ 𝛾mh
pA + 𝛾1 (h1.+h)= pB + 𝛾1h1+ 𝛾mh
(pA – pB) = 𝛾1h1+ 𝛾mh - 𝛾1 (h1.+h)
(pA – pB) = 𝛾mh - 𝛾1 h
(pA – pB) = h(𝛾m- 𝛾1 )

23
MECHANICAL GAUGES:
Pressure gauges are the devices used to
measure pressure at a point. They are used
to measure high intensity pressures where
accuracy requirement is less. Pressure
gauges are separate for positive pressure
measurement and negative pressure
measurement. Negative pressure gauges are
called Vacuum gauges.
Mechanical gauge consists of an elastic
The arrangement consists of a pressure
element which deflects under the action of responsive element made up of phosphor bronze
applied pressure and this deflection will move or special steel having elliptical cross section. The
a pointer on a graduated dial leading to the element is curved into a circular arc, one end of
measurement of pressure. Most popular the tube is closed and free to move and the other
pressure gauge used is Bordon pressure end is connected to gauge point. The changes in
pressure cause change in section leading to the
gauge. movement. The movement is transferred to a
needle using sector pinion mechanism. The
24 needle moves over a graduated dial.
1. Calculate intensity of pressure due to a column of 0.3m of (a) water (b) Mercury
(c) Oil of specific gravity-0.8.
Soln:
(a) Given: h = 0.3m of water
𝛾wáter = 9.81kN/m3
p=?
p = 𝛾wáter *h = 9810*0.3=2943N/m2

(b) Given: h = 0.3m of Mercury

𝛾Mercury = 𝛾wáter * S Mercury = 9810*13.6 N/m3 = 133.416k N/m3
p Mercury =?
p Mercury = 𝛾 Mercury *h = 133.416*0.3=4.025KN/m2
(c) Given: h = 0.3m of oil of specific gravity, 0.9
𝛾oil = 𝛾wáter * S Oil = 9810*0.9 N/m3 = 7.848kN/m3
p oil =?
p Oil = 𝛾 Oil *h = 7.848*0.3=2.354KN/m2

25
Intensity of pressure required at a point is 40kPa. Find corresponding head in
(a) water (b) Mercury (c) oil of specific gravity-0.9.
Solution: Given Intensity of pressure at a point 40 kPa i.e. p = 40 kN/m2
(a) Head of water hwater =?
p = 𝛾wáter * h
h = p/𝛾wáter
h = (40*1000)/9810 = 4.077m of water

(b)Head of Mercury hHg =?

p = 𝛾Hg * h
h = p/𝛾Hg
h = (40*1000)/(9810*13.6) = 0.3m of Mercury
(c) Head of oil hoil =?
p = 𝛾oil * h
h = p/𝛾oil
h = (40*1000)/9810*0.9 = 4.53m of oil
An open container has water to a depth of 2.5m and above this an oil of S = 0.85 for a depth
of 1.2m. Find the intensity of pressure at the interface of two liquids and at the bottom of the
tank.

(i) At the Oil - water interface

pA = 𝛾0il* hoil = (0.85 x 9.81) x 1.2
Oil, 0.85 1.2m
pA = 10 k Pa
*A (ii)At the bottom of container
Water, 1.0 2.5m pB = 𝛾0il* hoil + 𝛾 water * hwater
pB = pA + 𝛾 water * hwater
*B pB = 10 k Pa + 9.81 x 2.5
pB = 34.525 k Pa

27
Convert the following absolute pressure to gauge pressure:(a) 120kPa (b) 3kPa (c) 15m of
H2O (d) 800mm of Hg.
Solution:
(a) pabs = patm + pgauge
Atmosperic pressure = 10.3m of water;
p= 9810*10.3 = 101043N/m2 =101.04kN/m2
Pgauge = pabs - patm = 120 – 101.04 = 18.96 kPa

(b) pgauge = 3-101.3 = -98.3 kPa ; pgauge = 98.3 kPa (vacuum)

(c) habs = hatm + hgauge
(15/100) =10.3 +hgauge

hgauge = 10.15m of water

(d) habs = hatm + hgauge

800 =760 + hgauge
hgauge = 40 mm of mercury
28
Determine the pressure at A for the U- tube manometer shown in fig. Also calculate the
absolute pressure at A in kPa.
Let ‘pA’ be the pressure at ‘A’
pA + (0.75 *1*9810)-(0.5 x13.6*9810)= 0
pA = 59.35 kPa(gauge pressure)
pabs= patm + p gauge
=101.04 + 59.35
p abs =160.39 kPa

X’
c
a
b

29
For the arrangement shown in figure, determine gauge and absolute
pressure at the point M.

c Let ‘pM’ be the pressure at the point ‘M’

b
b’ b’’ pM – (0.75 x 0.8*9810) – (0.25 x 13.6*9810) = 0
(atmosperic pressure)
a
pM= (9810*0.8)*4
pM = 31.39 kPa
pabs =101.04 + 39.24
p abs= 140.28 kPa

30
If the pressure at ‘A’ is 10 kPa (Vacuum) what is the value of ‘x’?

pA = -10*103 N/m2
B
c =[pA + (0.2*1.2*9810) + x* (13.6*9810) = 0
=[-10*103 + (0.2*1.2*9810) + x* (13.6*9810) = 0
x = 0.0572 m
The tank in the accompanying figure consists of oil of S = 0.75. Determine the pressure
gauge reading in kN/m2

Let the pressure gauge reading be ‘h’ m of water

a b
c = p – (3.75* 0.75*9810) + (0.25* 13.6*9810)= 0
p=(9810*0.75)*-0.5875
p p = -4.322 kPa
p = 4.3223 kPa (Vacuum)
A closed tank is 8m high. It is filled with Glycerine up to a depth of 3.5m and linseed oil to
another 2.5m. The remaining space is filled with air under a pressure of 150 kPa. If a
pressure gauge is fixed at the bottom of the tank what will be its reading. Also calculate
absolute pressure. Take relative density of Glycerine and Linseed oil as 1.25 and 0.93
respectively.
Solution:
pH =150 kPa
Let ‘pN’ be the pressure gauge reading.

pN –(3.5*1.25*9810) –(2.5*0.93*9810) =150*103

p = 215.72 kPa (gauge)
pabs = 215.72+101.04=316.76 kPa

33
A vertical pipe line attached with a gauge and a manometer contains oil and
Mercury as shown in figure. The manometer is opened to atmosphere at the
other end. What is the gauge reading at ‘A’? Assume no flow in the pipe.

pA-(3 x 0.9*9810) + (0.375 x 0.9*9810) – (0.375 x 13.6*9810) = 0

pA = 20233.125N/m2
p = 20.33kPa
p abs = 101.04+20.23
p abs = 121.27kPa
b c C’

34
An inverted U-tube manometer is shown in figure. Determine the pressure difference
between A and B in N/m2.

Let pA and pB be the pressure at A and B in meters of water.

c C’
pA – (1.90 x 9810) + (0.3 x 0.9*9810) + (0.4) (0.9*9810 = pB
pA – pB = 120663N/m2 d
pA – pB = 12.06 x 103 N/m2 e

35
In the arrangements shown in figure. Determine the ‘h’.

20*103 + (1.5*9810) -(h*1.5*9810)– (4 + 1.5 – h) (0.8*9810) = – 3.4*9810

h = 3.6 m

36
In figure given, the air pressure in the left tank is 230 mm of Mercury (Vacuum).
Determine the elevation of gauge liquid in the right limb at A. If liquid in the right tank is
water.

pB = 230mm of Hg
E = -230*(13.6/1000)*9810
pB = - 3.128*9810 = 30685.68N/m2
= – (3.128*9810) + (5 x 0.8*9810) + (y x 1.6*9810) – (y + 2)*9810 = 21*103
D y = 5.446 m
Elevation of A = 100 – 5.446
Elevation of A = 94.553m

37
Compute the pressure different between ‘M’ and ‘N’ for the system shown in figure.

a c
y
B’ b Let ‘pM’ and ‘ pN’ be the pressure at M and N.
pM + y (1.15*9810) – 0.2 (0.92*9810) + (0.3 – y + 0.2) 1.15*9810 = pN
pM + y (11281.5) – 1805.04+ (0.5 – y ) 11281.5 = pN
pM + y (11281.5) – 1805.04+ 0.5*11281.5 – y * 11281.5 = pN
pM – pN = 1805.04 – (0.5*11281.5) = -3835.71N/m2
pN – pM = 3835.71N/m2

38
Petrol of specify gravity 0.8 flows up through a vertical pipe. A and B are the two points in the pipe, B
being 0.3 m higher than A. Connection are led from A and B to a U–tube containing Mercury. If the
pressure difference between A and B is 18 kPa, find the reading of manometer.

pA – pB = 18kPa
B
𝛾hA – 𝛾hB = 9810(hA-hB) = 18*103
hA-hB = 1.835m of water of m

pA + y x 0.8*9810 – x* 13.6*9810 – (0.3 + y – x) 0.8*9810 = pB

pA – pB = – (0.8*9810)y + (13.66*9810) x + (0.24*9810) + (0.8*9810) y –
c e (0.8*9810) x
x = 0.1246 m
d

39
A cylindrical tank contains water to a height of 50mm. Inside is a small open cylindrical tank
containing kerosene having a specify gravity 0.8. The following pressures are known from indicated
gauges. pB = 13.8 kPa (gauge); pC = 13.82 kPa (gauge);Determine the gauge pressure pA and
height h. Assume that kerosene is prevented from moving to the top of the tank.

pC = 13.82 kPa
pB = 13.8 kPa
pA = 13.82*103 -9810*(50/1000)
pA = 13.33 kPa

pB – h *(9810* 0.8) – (0.05 – h)*9810 = pA

13.8*103 – (0.8*9810) h – 0.05*9810 + h*9810 = 13.33*103
0.2 h = 0.002
h = 0.02 m

40
Find the pressure different between A and B if d1 = 300mm, d2 = 150mm,d3 = 460mm, d4 =
200mm and Specific gravity of Mercury is 13.6.

Let pA and pB be the pressure at A and B.

pA+ 0.3*9810 – (0.46 + 0.2 Sin 45)* 13.6*9810 = pB
pA – pB = (7.88 ) (9.81)
B pA – pB = 77.29 kPa

41
What is the pressure pA in the fig given below? Take specific gravity of oil as 0.8.

pA + (3* 0.8*9810) + (4.6 - 0.3) (13.6*9810) = 0

pA = 9810 x 2.24
pA = 21.97 kPa

42
Find ‘d’ in the system shown in fig. If pA = 2.7 kPa

pA= 2.7*103 Pa
pA+9810*0.6*0.05+9810*.6*(10/1000)-9810*13.6*(300-10)/1000 -9810*1.4*d=0
d= 0.0494m

43
Determine the absolute pressure at ‘A’ for the system shown in fig.

pA - (0.25* 0.8*9810) + (0.15* 0.7*9810) + (0.3*0.8*9810)-(0.6*9810) = 0

pA = 4.464 kPa
pabs = 101.04 + 4.464
pabs = 105.704 kPa

44
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