Unit2 fluid static

FLUID STATIC J3008/2/1
FLUID STATIC
OBJECTIVES
General Objective : To know, understand and apply the concept of pressure and buoyancy
Specific Objectives : At the end of the unit you should be able to :
 explain the relationship between pressure and depth.
 define Pascal Law and apply the concept into the Hydraulic Jack.
 solve problems related to the Hydraulic Jack.
 explain the concept of Buoyancy.
UNIT 2

2.0 INTRODUCTION
Fluid Pressure and Depth
A fluid is a substance that flows easily. Gases and liquids are fluids, although
sometimes the dividing line between liquids and solids is not always clear. Because of
their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic
systems, allow aircraft to fly and ships to float.
The topic that this unit will explore will be pressure and depth. If a fluid is within a
container then the depth of an object placed in that fluid can be measured. The deeper
the object is placed in the fluid, the more pressure it experiences. This is because of the
weight of the fluid above it. The more dense the fluid above it, the more pressure is
exerted on the object that is submerged, due to the weight of the fluid.
The formula that gives the pressure, p on an object submerged in a fluid is:
ghp ρ=
Where,
 ρ (rho) is the density of the fluid,
 g is the acceleration of gravity
 h is the height of the fluid above the object
If the container is open to the atmosphere above, the added pressure must be included if
one is to find the total pressure on an object. The total pressure is the same as absolute
pressure on pressure gauge readings, while the gauge pressure is the same as the fluid
pressure alone, not including atmospheric pressure.
INPUTINPUT
Ptotal = Patmosphere + Pfluid
Ptotal = Patmosphere + ( ρgh )

Pascal is the unit of pressure in the metric system. It represents 1 Newton/m2
.
2.1 PRESSURE AND DEPTH
In Unit 1, we have defined the meaning of ‘pressure’. In this unit we will learn about
the relationship between pressure and depth.
When a liquid (such as water, oil etc) is contained in a vessel, it exerts force at all points
on the sides and bottom of the container. This force per unit area is called pressure. If F
is the force acting on an area a, then intensity of pressure is :
A
F
p =
The direction of this pressure is always at right angle to the surface, with which the
fluid at rest, comes into contact.
The intensity of pressure at any point is the force exerted on an unit area at that point
and is measured in Newtons per square metre, N/m2
(Pascals). An alternative metric
unit is bar, which is in N/m2
.

In this section, we are going to look into the relationship between depth and pressure.
Figure 2.1
Consider a vessel containing some liquid as shown in Figure 2.1. We know that the
liquid will exert pressure on all sides and the bottom of the vessel. Let a cylinder be
made to stand in the liquid as shown in the figure. The weight of liquid contained in
the cylinder is ωhA where ;
ω = Specific weight of the liquid
( where is ωwater = 1000 kg/m3
x 9.81 m/s2
)
h = Height of liquid in the cylinder
A = Area of the cylinder base
The pressure, at the bottom of the cylinder, will be due to the weight of the liquid
contained in the cylinder. Let this pressure be p.
Then,
basecylindertheofArea
cylindertheinliquidofweight
p =
h
A
Ah
p ω
ω
== which can also be shown as ,
ghp ρ=
This equation shows that the intensity of pressure at any point, in a liquid, is
proportional to its depth as measured from the surface (as ω is constant for the given
liquid).
It is thus obvious, that the pressure can be expressed in either one of the following two
ways :
h

a) As force per unit area ( N/m2
)
b) As height of equivalent liquid column
Example 2.1
a) Find the density of pressure p at a depth below the surface of a liquid of specific
weightω = ρg if the pressure at the free surface is zero.
b) A diver is working at a depth of 20 m below the surface of the sea. How much greater
is the pressure intensity at this depth than at the surface? Take into consideration
specific weight of water is 10000 N/m3
.
Solution to Example 2.1 (a)
The column of liquid (Figure 2.2) of cross-sectional area A extending vertically
from the free surface to the depth h is in equilibrium to the surrounding liquid under the
action of its weight acting downwards. The pressure force on the bottom of the column
acting upwards, and the forces on the sides due to the surrounding liquid must act
horizontally since there can be no tangential (shearing) forces in the liquid at rest.
For vertical equilibrium :
p
Cross-sectional
area A
Liquid of specific
weight w
h
Figure 2.2
F

Force exerted on base = Weight of column of liquid
columnofvolumevolumeunitperWeightbaseareapressureofIntensity ×=×
AhpA .ω=
ghhp ρω == since gρω =
Since the same relation applies wherever the column is taken, it follows that :
The intensity of pressure is the same at all points in the same horizontal plane in
a liquid at rest.
Solution to Example 2.1 (b)
Puttingω = 10000 N/m3
and h = 2.0 m
Therefore,
p = 1000 x 20
= 200000 N/m2
=
2
/200 mkN
Example 2.2
Find the height of a water column which is equivalent to the pressure of 2 N/m2
.
( Take into consideration specific weight of water, ω water = 1000 kg/m2
x 9.81 m/s2
)
Solution to Example 2.2
Taking the formula p = ωh
Putting p = 2 N/m2
and ωwater = 1000 kg/m2
x 9.81 m/s2
Therefore,
h =
9810
2
h = m04.2
ACTIVITY 2A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
2.1 Match the symbols below with their respective meanings.
2.2 Fill in the blanks with suitable words :
1. _______ exerts at all points on the sides and bottom of the container when a liquid
(such as water, oil etc) is contained in a vessel.
w
h
Specific weight of the liquid
Height of liquid in the cylinder
A
Area of the cylinder
base
p
Intensity of pressure

2. ______ (Pascal) is the measurement unit for the intensity of pressure at any point with
reference to the formula p = F / A.
3. ghp ρ= or hp __=
4. Pressure can be expressed in two ways if force per unit area ( N/m2
) and _______ of
equivalent liquid column are given.
5. We can summarize that, intensity of pressure at a point in a fluid at rest is the same in
all _________.
FEEDBACK ON ACTIVITY 2A

2.1
2.2
1. Force exerts at all points on the sides and bottom of the container when a liquid (such
as water, oil etc) is contained in a vessel.
w
hSpecific weight of the liquid
Height of liquid in the cylinder A
Area of the cylinder
base
p
Intensity of pressure

2. N/m2
(Pascal) is the measurement unit for the intensity of pressure at any point with
reference to the formula p = F / A.
3. ghp ρ= or hp ω=
4. Pressure can be expressed in two ways if force per unit area ( N/m2
) and height of
equivalent liquid column are given.
5. We can summarize that, intensity of pressure at a point in a fluid at rest is the same in
all directions.
INPUTINPUT

2.2 PASCAL LAW AND HYDRAULIC JACK
Pascal's Principle and Hydraulics
Hydraulic system uses an incompressible fluid, such as oil or water, to transmit forces
from one location to another within the fluid. Most aircraft use hydraulics in the braking
systems and landing gear. Pneumatic systems use compressible fluid, such as air, in
their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear
and movement of flaps.
Pascal's law states that when there is an increase in pressure at any point in a
confined fluid, there is an equal increase at every other point in the container.
A container, as shown below, contains a fluid. There is an increase in pressure as the
length of the column of liquid increases, due to the increased mass of the fluid above.
For example, in the Figure 2.3, P3 would be the highest value of the three pressure
readings, because it has the highest level of fluid above it.
Figure 2.3
If the container had an increase in overall pressure, that same added pressure would
affect each of the gauges (and the liquid throughout) in the same way. For example p1,
p2, p3 were originally 1, 3, 5 units of pressure, and if 5 units of pressure were added to
the system, the new readings would be 6, 8, and 10.
Applied to a more complex system below (Figure2.4), such as a hydraulic car lift,
Pascal's law allows forces to be multiplied. The cylinder on the left shows a cross-
section area of 1 square meter, while the cylinder on the right shows a cross-section area
of 10 square metre. The cylinder on the left has a weight (force) of 1 kg acting

downward on the piston, which lowers the fluid 10 metres. As a result of this force, the
piston on the right lifts a 10 kg weight a distance of 1 metre.
The 1 kg load on the 1 square metre area causes an increase in pressure on the fluid in
the system. This pressure is distributed equally throughout and acts on every square
metre of the 10 square metre area of the large piston. As a result, the larger piston lifts
up a 10 kg weight. The larger the cross-section area of the second piston, the larger the
mechanical advantage, and the more weight it lifts.
Figure 2.4
The formulas that relate to this are shown below:
21 pp = (since the pressures are equal throughout)
Since pressure equals force per unit area, then it follows that
2
2
1
1
A
F
A
F
=
It can be shown by substitution that the values shown above are correct,
metresquare
kg
metresquare
kg
10
10
1
1
=
Because the volume of fluid pushed down on the left side equals the volume of fluid
that is lifted up on the right side, the following formula is also true.
21 VV =
W1
= 1 kg
W2
= 10 kg
A1
= 1 square metre
A2
= 10 square metre
D1
= 10 meter
D2
= 1 meter

by substitution,
2211 DADA ×=×
A = cross sectional area
D = the distance moved
Or
1
2
2
1
D
D
A
A
=
2.2.1 Pascal Law for Pressure At A Point

By considering a small element of fluid in the form of a triangular prism
which contains a point p, we can establish a relationship between the three
pressures px in the x direction, py in the y direction and pz in the z direction
normal to the sloping face. (Figure 2.5)
Figure 2.5
The pressure p at a point in a fluid can be expressed in terms of the
height h of the column of the fluid which causes the pressure, or which would
cause an equal pressure if the actual pressure is applied by other means.
We know that p = ωh = ρg and the height, h is called the pressure head
at that point. It is measured as a length (e.g in meters) of fluid. The name of fluid
must be given because the mass density is different for each liquid.
Example 2.3
Pressure at any point is the same in all
directions.
This is known as Pascal Law and applies
to fluids at rest
px
Pz
Py

Find the head, h of water corresponding to an intensity of pressure, p of 340 000 N/m2
.
Take into consideration that the mass density, ρ of water is 103 kg/m3
.
Since p = ρgh
Head of water , g
p
h
ρ
=
81.910
340000
3
x
=
= 34.7 m
2.2.2 Hydraulic Jack
A Hydraulic Jack is used to lift a heavy load with the help of a light force.
Below is a diagram of a hydraulic jack (Figure 2.6). A force, F is applied
to the piston of the small cylinder and forces oil or water out into the large
cylinder thus, raising the piston supporting the load, W.
The force, F acting on area, a produced a pressure p1 which is
transmitted equally in all direction through the liquid. If the two pistons are at
the same level, the pressure, p2 acting on the larger piston must equal p1.
21 pp =
Example 2.4
W
F
Area , a
p2
p1
Area, A
Figure 2.6

A force, P of 500 N is applied to the smaller cylinder of a hydraulic jack. The area, a
of a small piston is 20 cm2
while the area, A of a larger piston is 200 cm2
. What mass
can be lifted on the larger piston?
Putting F = 800 N, a = 20/1000 m2
, A = 200 / 1000 m2
21 pp =
a
F
=
A
W
So that
a
A
FW =
= 800
2.0
0.2
× = N8000
Mass lifted = g
W
81.9
80000
=
= kg49.815
ACTIVITY 2B
W
F = 800 N
p2
p1
Area,
A = 200 cm2
Area ,
a = 20 cm2

INPUT…!
2.3 A force, P of 650 N is applied to the smaller cylinder of an hydraulic jack. The area, a
of a small piston is 15 cm2
and the area A of a larger piston is 150 cm2
.
What load, W can be lifted on the larger piston if :
(a) the pistons are at the same level ?
(b) the large piston is 0.65 m below the smaller piston ?
(c) the small piston is 0.40 m below the larger piston ?
Consider the mass density ρ of the liquid in the jack is 103
kg/m3
FEEDBACK ON ACTIVITY 2B

2.3
(a) if the pistons are at the same level
Now p1 =
a
F
and p2 =
A
W
p1 = p2 ,
a
F
=
A
W
Or
F =
A
a
W
Thus, the small force F can raise the larger load W because the jack has a
mechanical advantage of A/a.
Putting F = 650 N, a = 15/1000 m2
, A = 150 / 1000 m2
a
F
=
A
W
So that
a
A
FW ×=
= 650
15.0
5.1
×
= N6500
(b) if the large piston is 0.65 m below the smaller piston ?
F
Area , a
W
p2
p1
Area,
A

If the larger piston is a distance h below the smaller piston, the pressure p2 will be
greater than p1, due to the head, h by an amount ρg, where ρ is the mass density of the
liquid and g is the gravity. Take into consideration g = 9.81 m/s2
ghpp ρ+= 12
24
41 /103.43
1015
650
mN
a
F
p ×=
×
== −
Putting ρ 33
/10 mkg= , h = 0.65 m and g = 9.81 m/s2
ghpp ρ+= 12
( ) 65.081.910103.43 34
2 ××+×=p
= 43.3 4
10× 5.6376+
= 439.38 kN
and
W Ap2=
= 43
101501038.439 −
×××
kN59.6=
(c) the small piston is 0.40 m below the larger piston ?
Area , a
Area, A
W
F
p2
p1 h

If the smaller piston is a distance h below the larger piston, the pressure p1 will be
greater than p2, due to the head, h by an amount ρg, where ρ is the mass density of the
liquid and g is the gravity. Take into consideration g = 9.81 m/s2
ghpp ρ+= 21
A
W
p =2
Putting ρ 33
/10 mkg= , g = 9.81 m/s2
and h = 0.40 m
ghpp ρ+= 21
so gh
A
W
p ρ+=1
but p1 =
a
F
a
F
= gh
A
W
ρ+
so W = Agh
a
F






− ρ
W = 5.14.081.91000
15.0
850






××−
W = 2.614 kN
Area , a
Area, A
W
F
p2
p1
h

2.3 BUOYANCY
Principle of Archimedes
Upthrust on body = weight of fluid displaced by the body
Figure 2.7 Buoyancy
If the body is immersed so that part of its volume, v1 is immersed in a fluid of
density, 1ρ and the rest of its volume, v2 in another immiscible fluid of mass density
2ρ .
The upthrust will act through the centre of gravity of the displaced fluid, which
is called the centre of buoyancy.
The positions of G1 and G2 are not necessarily on the same vertical line, and the
centre of buoyancy of the whole body is, therefore, not bound to pass through the
centroid of the whole body.
INPUTINPUT
Archimedes
Principle states
that the buoyant
force on a
submerged object
is equal to the
weight of the fluid
that is displaced
by the object.
Archimedes
Principle states
that the buoyant
force on a
submerged object
is equal to the
weight of the fluid
that is displaced
by the object.
Upthrust on upper part, R1 = 11 gvρ
acting through G1, the centroid of v1,
Upthrust on lower part,R2 = 22 gvρ
acting through G2, the centroid of v2,
Total upthrust = 11 gvρ + 22 gvρ

Example 2.5
A rectangular pontoon has a width B of 6 m, a length l of 12 m, and a draught D of 1.5
m in fresh water (density 1000 kg/m3
). Calculate :
a) the weight of the pontoon
b) its draught in sea water (density 1025 kg/m3
)
c) the load (in kiloNewtons) that can be supported by the pontoon in fresh water if
the maximum draught permissible is 2 m.
When the pontoon is floating in an unloaded condition,
Uptrust on immersed volume = weight of pontoon
Since the uptrust is equal to weight of the fluid displaced,
Weight of pontoon = weight of fluid displaced,
So, DlBgW ××××=ρ
a) In fresh water, 3
/1000 mkg=ρ and D = 1.5 m ;
Therefore,
Weight of pontoon,
W = N5.112681.91000 ××××
kNW 5.1059=
b) In the sea water, 3
/1025 mkg=ρ ; therefore,
Draught in sea water, lBg
W
D
×××
=
ρ
=
12681.91025
105.1059 3
×××
×
m46.1=
c) For maximum draught of 2 m in fresh water,

Total uptrust = weight of water displaced
= DlBg ×××
= N212681.91000 ××××
= kN6.1412
Load which can be supported = Upthrust – weight of pontoon
= 5.10596.1412 −
= kN1.353
ACTIVITY 2C

INPUT…!
2.4 Define the Archimedes Principle.
2.5 Consider a barge filled with rock. The barge is 7 m wide, 17 m long, and 2.5 m
deep. If the barge and rock weigh 2.0 MN, determine the depth of submergence
of the barge in water as shown in the figure below.
2.5 m
17 m
Rock
Barge
d
Submergence Depth
FEEDBACK ON ACTIVITY 2B

2.4
Archimedes Principle states that the buoyant force on a submerged object is
equal to the weight of the fluid that is displaced by the object.
2.5
The specific weight of water is 9,800 N/m3
(or 9.8 kN/m3
). From the buoyancy
concept, the volume of displaced fluid (water) must balance the weight of the
barge. Thus,
Wbarge = FB = γwater Vbarge
Substituting volume Vbarge with 1 w d, we have:
Wbarge = γwater l w d = (9.8 kN/m3
)(17 m)(7 m) d
Or
( )( )( )mmmkN
W
d
eb
717/8.9 3
arg
=
Substituting the weight of the barge gives:
( )( )( )mmmkN
MN
d
717/8.9
0.2
3
=
m71.1=
SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment section and check
your answers with those given in the Feedback on Self-Assessment. If you face any problems,
discuss it with your lecturer. Good luck.
2.1 In a hydraulic jack a force F, is applied to a small piston that lifts the load on the
large piston. If the diameter of the small piston is 15 mm and that of the large
piston is 180 mm, calculate the value of F required to lift 1000 kg.
2.2 Two cylinders with pistons are connected by a pipe containing water. Their
diameters are 75 mm and 600 mm respectively and the face of the smaller piston
is 6 m above the larger. What force on the smaller piston is required to maintain
a load of 3500 kg on the larger piston?
2.3 A rectangular pontoon 5.4 m wide by 12 m long, has a draught of 1.5 m in fresh
water (density 1000 kg/m3
). Calculate:
(a) the mass of the pontoon,
(b) its draught in the sea water (density 1025 kg/m3
).
2.4 A ship floating in sea water displaces 115 m3
. Find
(a) the weight of the ship if sea water has a density of 1025 kg/m3
,
(b) the volume of fresh water (density 1000 kg/m3
) which the ship would
displace.
FEEDBACK ON SELF-ASSESSMENT

Answers :
2.1 68.2 N
2.2 276 N
2.3 a) 97000 kg ,
b) 1.47 m
1.4 a) 118000 kg ,
b) 118 m3

Unit2 fluid static

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