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

Fp =

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 cylinderA = 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

cylindertheinliquidofweightp =

hA

Ahp ωω == 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 2.2F


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

hSpecific 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/m 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 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

ph

ρ=

81.910

3400003 x

=

= 34.7 m

2.2.2 Hydraulic JackA 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 thata

AFW =

= 8002.0

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



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

aW

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 thata

AFW ×=

= 65015.0

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

2441 /103.43

1015

650mN

a

Fp ×=

×== −

Putting ρ 33 /10 mkg= , h = 0.65 m and g = 9.81 m/s2

ghpp ρ+= 12

( ) 65.081.910103.43 342 ××+×=p

= 43.3 410× 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

Wp =2

Putting ρ 33 /10 mkg= , g = 9.81 m/s2 and h = 0.40 m

ghpp ρ+= 21

so ghA

Wp ρ+=1

but p1 = a

F

a

F = gh

A

W ρ+

so W = Agha

F

− ρ

W = 5.14.081.9100015.0

850

××−

W = 2.614 kN

Area , a

Area, A

W

F

p2

p1

h


2.3 BUOYANCY

Principle of ArchimedesUpthrust 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 = 11gvρacting through G1, the centroid of v1,

Upthrust on lower part,R2 = 22 gvρacting through G2, the centroid of v2,

Total upthrust = 11gvρ + 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 pontoonb) 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

WD

×××=

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



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


2.4Archimedes 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

Wd eb

717/8.9 3

arg=

Substituting the weight of the barge gives:

( )( )( )mmmkN

MNd

717/8.9

0.23

=

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

Engineering

Transcript of Unit2 fluid static