3 Web viewJabatan Sains & Matematik PSA Created Date: 01/12/2013 20:26:00 Title: 3 Last modified...

FLUID STATIC

FLUID STATIC 2.1 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:

p=ρ ghWhere, 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.

17

CHAPTER 2

Ptotal = Patmosphere + Pfluid

Ptotal = Patmosphere + ( ρgh )

h

FLUID STATIC

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 :

p= FA

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 ;

18

FLUID STATIC

ω = Specific weight of the liquid( where is water = 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,

p=weight of liquid in the cylinder

Area of the cylinder base

p= ωhA

A= ωh

which can also be shown as ,

p = ρ gh

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 :

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)

19

p

Cross-sectional area A

Liquid of specific weight w

h

Figure 2.2F

FLUID STATIC

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 :

Force exerted on base = Weight of column of liquid

Intensity of pressure × area base = Weight per unit volume × volume of column

pA = ω . Ahp = ωh =ρ gh since ω =ρg

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)

20

FLUID STATIC

Puttingω = 10000 N/m3 and h = 2.0 m

Therefore,p = 1000 x 20

= 200000 N/m2

= 200 kN /m2

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 = hPutting p = 2 N/m2 and ωwater = 1000 kg/m2 x 9.81 m/s2

Therefore,

h =

29810

h = 2 .04 m

21

w

hSpecific weight of the liquid

Height of liquid in the cylinder A

Area of the cylinder base p

Intensity of pressure

FLUID STATIC

ACTIVITY 2A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

2.1 Match the symbols below with their respective meanings.

22

FLUID STATIC

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.

2. ______ (Pascal) is the measurement unit for the intensity of pressure at any point with reference to the formula p = F / A.

3. p = ρ gh or p = __ h

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 _________.

23

w

hSpecific weight of the liquid

Height of liquid in the cylinder A

Area of the cylinder basep

Intensity of pressure

FLUID STATIC

FEEDBACK ON ACTIVITY 2A

2.1

24

FLUID STATIC

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.

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. p = ρ gh or p = ωh

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.

25

FLUID STATIC

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

26

W1 = 1 kg

W2 = 10 kg

A1 = 1 square metre

A2 = 10 square metre

D1= 10 meter

D2 = 1 meter

FLUID STATIC

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:

p1=p2 (since the pressures are equal throughout)

Since pressure equals force per unit area, then it follows that

F1

A1=

F2

A2

It can be shown by substitution that the values shown above are correct,1 kg

1 square metre=10 kg

10 square metre

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.

V 1=V 2

27

Pressure at any point is the same in all directions.

This is known as Pascal Law and applies to fluids at rest

FLUID STATIC

by substitution,

A1×D1=A2×D2

A = cross sectional area

D = the distance moved

Or

A1

A2=

D2

D1

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)

28

px

Pz

Py

FLUID STATIC

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 = h = 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

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 , h = p

ρg

= 340000

103 x9 . 81 = 34.7 m

29

W

F

Area , a

p2p1Area, A

FLUID STATIC

2.2.2Hydraulic 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.

p1 = p2

Example 2.4

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?

30

Figure 2.6

W

F = 800 N

p2p1

Area, A = 200 cm2

FLUID STATIC


Putting F = 800 N, a = 20/1000 m2 , A = 200 / 1000 m2

p1 = p2

Fa =

WA

So thatW = F A

a

= 800× 2.0

0 .2 = 8000 N

Mass lifted =

Wg

= 80000

9. 81

= 815 .49 kg

31

Area , a = 20 cm2

FLUID STATIC

ACTIVITY 2B


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

32

F

Area , a

W

p2p1Area, A

FLUID STATIC

FEEDBACK ON ACTIVITY 2B

2.3(a) if the pistons are at the same level

Now p1 =

Fa and p2 =

WA

p1 = p2 ,

Fa =

WA

Or

F = W a

AThus, 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

Fa =

WA

So thatW = F × A

a

= 650× 1. 5

0 .15

= 6500 N

33

Area , a

Area, A

W

F

p2

p1h

FLUID STATIC

(b) if the large piston is 0.65 m below the smaller piston ?

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

p2 = p1 + ρ gh

p1 = Fa

= 65015×10−4

= 43 .3×104 N /m2

Putting ρ = 103 kg /m3, h = 0.65 m and g = 9.81 m/s2

p2 = p1 + ρ gh

p2 =43.3×104 + (103×9. 81 )×0 .65

= 43.3 ¿ 104+ 6376 . 5 = 439.38 kN

and

W = p2 A

= 439 . 38×103 × 150 ×10−4

= 6 .59 kN

34

Area , a

Area, A

W

F

p2

p1

h

FLUID STATIC

(c) the small piston is 0.40 m below the larger piston ?

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

p1 = p2+ ρ gh

p2 = WA

Putting ρ = 103 kg /m3, g = 9.81 m/s2 and h = 0.40 m

p1 = p2 + ρ gh

sop1 = W

A+ ρ gh

but p1 =

Fa

Fa =

WA

+ ρ gh

so W = ( F

a− ρ gh) A

W = (8500 .15

− 1000×9. 81×0. 4) 1. 5

W = 2.614 kN

35

FLUID STATIC

SELF-ASSESSMENT

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

36

FLUID STATIC

FEEDBACK ON SELF-ASSESSMENT

Answers :

2.1 68.2 N

2.2 276 N

2.3 a) 97000 kg , b) 1.47 m

a) 118000 kg , b) 118 m3

37

FLUID STATIC

2.3 CONCEPT OF MANOMETER, PIEZOMETER AND BAROMETER

2.3.1 MANOMETERS

The relationship between pressure and head is utilized for pressure measurement in the manometer or liquid gauge. We can measure comparatively high pressures and negative pressures with the manometer. The following are a few types of manometers:a) Simple manometer,b) Differential manometer andc) Inverted differential manometer.

a) SIMPLE MANOMETER

A simple manometer is a tube bent in U-shape. One end of which is attached to the gauge point and the other is open to the atmosphere as shown in (Figure 3.5)

The liquid used in the bent tube or simple manometer is generally mercury which is 13.6 times heavier than water. Hence, it is also suitable for measuring high pressure.

Now consider a simple manometer connected to a pipe containing a light liquid under high pressure. The high pressure in the pipe will force the heavy liquid, in the left-hand limb of the U-tube, to move downward. This downward movement of the heavy liquid in the left-hand limb will cause a corresponding rise of the heavy liquid in the right-hand limb. The horizontal surface, at which the heavy and light liquid meet in the left-hand limb is known as a common surface or datum line. Let B-C be the datum line, as shown in Figure 3.5.

38

FLUID STATIC

Let h1 = Height of the light liquid in the left-hand limb above the common surface in cm.

h2 = Height of the heavy liquid in the right-hand limb above the common surface in cm.

pA = Pressure in the pipe, expressed in terms of head of water in cm.

ωP = Specific weight of the light liquid

sQ = Specific gravity of the heavy liquid.

The pressure in the left-hand limb and the right-hand limb above the datum line is equal.

39

Pressure pB at B = Pressure pC at C

P

sQ

Figure 2.7

FLUID STATIC

Pressure in the left-hand limb above the datum line

pB = Pressure, pA at A + Pressure due to depth, h1 of fluid P

= pA + ωPh1

= pA + ρP gh1

40

Specific weight, = g

P

Imagine that the right limb is hidden.

Imagine that the left limb is hidden.

FLUID STATIC

Thus pressure in the right-hand limb above the datum line;

pC = Pressure pD at D + Pressure due to depth h2 of liquid Q

But pD = Atmospheric pressure = Zero gauge pressure

And so, pC = 0 + ωQ h2

= 0 + ρQ gh2

Since pB = pC ,

pA+ρP gh1=ρQ gh2

so,

pA = ρQ gh2 − ρP gh1

41

sQ

FLUID STATIC

Example 2.4

A U-tube manometer similar to that shown in Figure 3.6 is used to measure the gauge pressure of water (mass density ρ = 1000 kg /m3). If the density of mercury is 13.6 × 103

kg /m3, what will be the gauge pressure at A if h1 = 0.45 m and D is 0.7 m above BC.


Considering

ρQ = 13 .6 × 103 kg /m3

ρP= 1 . 0 × 103 kg /m3

h1 = 0.45 mh2 = 0.7 m

the pressure at left-hand limb;pB = Pressure, pA at A + Pressure due to depth, h1 of fluid P

= pA + ωPh1

= pA + ρP gh1

42

ωwater

ωmercury

Figure 2.8

FLUID STATIC

the pressure at right-hand limb;pC = Pressure pD at D + Pressure due to depth h2 of liquid QpC = 0 + ωQ h2

= 0 + ρQ gh2

Since pB = pC

pA+ρP gh1=ρQ gh2

pA=ρQ gh2−ρP gh1

=13 . 6×103×9 .81×0 .7−1 .0×103×9 .81×0 .45

= 88976 .7 N /m2

=88. 97 × 103 N /m2

If negative pressure is to be measured by a simple manometer, this can be measured easily as discussed below:In this case, the negative pressure in the pipe will suck the light liquid which will pull up the heavy liquid in the left-hand limb of the U-tube. This upward movement of the heavy liquid, in the left-hand limb will cause a corresponding fall of the liquid in the right-hand limb as shown in Figure 3.7.

43

Figure 2.9

FLUID STATIC

In this case, the datum line B-C may be considered to correspond with the top level of the heavy liquid in the right column as shown in the Figure 3.7.

Now to calculate the pressure in the left- hand limb above the datum line.

Let h1 = Height of the light liquid in the left-hand limb above the common surface in cm.

h2 = Height of the heavy liquid in the left-hand limb above the common surface in cm

pA = Pressure in the pipe, expressed in terms of head of water in cm. sP = Specific gravity of the light liquid sQ = Specific gravity of the heavy liquid.

Pressure in the left-hand limb above the datum line;

pB = Pressure pA at A + Pressure due to depth h1 of fluid P + Pressure due to

depth h2 of liquid Q

= pA + ωPh1+ ωQ h2

= pA + ρP gh1+ ρQ gh2

44

Pressure pB at B = Pressure pC at C

FLUID STATIC

Pressure in the right-hand limb above the datum line;

pC = Pressure pD at D

But pD = Atmospheric pressure

And so, pC = patm

Since pB = pC

pA + ρP gh1+ ρQ gh2=pD

pA= pB−(ρP gh1+ ρQ gh2 )

45

FLUID STATIC

Example 2.5

A U-tube manometer similar to that shown in Figure 3.8 is used to measure the gauge pressure of a fluid P of density ρ = 1000 kg/m3. If the density of the liquid Q is 13.6 × 103 kg/m3, what will be the gauge pressure at A if h1 = 0.15 m and h2 = 0.25 m above BC. Take into consideration patm = 101.3 kN/m2.


Putting ,

ρQ = 13.6 ¿ 103

ρP = 1000 kg/m3

h1 = 0.15 mh2 = 0.25 m

pressure at left-hand limb;pB = Pressure pA at A + Pressure due to depth h1 of fluid P + Pressure due to

depth h2 of liquid Q

= pA + ωPh1+ ωQ h2

= pA + ρP gh1+ ρQ gh2

pressure at right-hand limb;pC = Pressure pD at D

pD = Atmospheric pressure

46

Figure 2.10

FLUID STATIC

pC = patm

Since pB = pC ,pA + ρP gh1+ ρQ gh2=pD

pA= pB−(ρP gh1+ ρQ gh2 )=101 . 3−(13 . 6 × 103 × 9 .81 × 0 . 15 + 1000 × 9 . 81 × 0. 25 )=70835 .1 N /m2

=70 . 84 kN /m2

47

FLUID STATIC

ACTIVITY 2B

2.2 A U-tube manometer is used to measure the pressure which is more than the atmospheric pressure in a pipe, the water being in contact with the mercury in the left-hand limb. The mercury is 20 cm below A in the left-hand limb and 25 cm above A in the right-hand limb, sketch the manometer.

48

FLUID STATIC

2.3 The U-tube manometer measures the pressure of water at A which is below the atmospheric pressure. If the specific weight of mercury is 13.6 times that of water and the atmospheric pressure is 101.3 kN/m2, find what is the absolute pressure at A when h1 = 10 cm, h2 = 25 cm and the specific weight of water is 9.81×103 N/m3.

49

FLUID STATIC

FEEDBACK ON ACTIVITY 2B

2.2

50

FLUID STATIC

2.3

pB=p A+ωair h1+ωmercury h2

pC=patm=101. 3 kN /m2

pB=pC

pA+ωair h1+ωmercury h2= patm

pA=patm−ωair h1−ωmercury h2

pA=101. 3×103−9810 (0.1 )−9810 (13 . 6 ) (0 .25 )

=66965 N /m2

=66. 965 kN /m2

51

INPUT

FLUID STATIC

b) DIFFERENTIAL MANOMETER

It is a device used for measuring the difference of pressures, between two points in a pipe, or in two different pipes.

A differential manometer consists of a U-tube, containing a heavy liquid with two ends connected to two different points. We are required to find the difference of pressure at these two points, as shown in Figure 3.9.

A differential manometer is connected to two different points A and B. A little consideration will show that the greater pressure at A will force the heavy liquid in the U-tube to move downwards. This downward movement of the heavy liquid, in the left-hand limb, will cause a corresponding rise of the heavy liquid in the right-hand limb as shown in Figure 3.9.

The horizontal surface C-D, at which the heavy liquid meet in the left-hand limb, is the datum line.

Let h = Height of the light liquid in the left-hand limb above the datum line. h1 = Height of the heavy liquid in the right-hand limb above the datum line h2 = Height of the light liquid in the right-hand limb above the datum line pA = Pressure in the pipe A, expressed in term of head of the liquid in cm pB = Pressure in the pipe B, expressed in term of head of the liquid in cm

52

Figure 2.11

FLUID STATIC

ωP = Specific weight of the light liquid ωQ = Specific weight of the heavy liquid

We know that the pressures in the left-hand limb and right-hand limb , above the datum line are equal.

Pressure in the left-hand limb above the datum line

pC = Pressure pA at A + Pressure due to depth h of fluid PpC= pA+ωPhpC= pA+ ρP gh

Pressure in the right-hand limb above the datum line

53

Pressure pC at C = Pressure pD at D

FLUID STATIC

pD = Pressure pA at A + Pressure due to depth h1 of fluid P + Pressure due to depth h2 of liquid Q

pD=pB+ωQ h1+ωPh2

=pB+ρQ gh1+ρP gh2

Since, pC= p D

pA+ρP gh=pB+ρQ gh1+ ρP gh2

pA−pB=ρQ gh1+ ρP gh2− ρP gh

Example 2.6

54

FLUID STATIC

A U tube manometer measures the pressure difference between two points A and B in a liquid. The U tube contains mercury. Calculate the difference in pressure if h =1.5 m, h2

= 0.75 m and h1 = 0.5 m. The liquid at A and B is water ( ω = 9.81 × 103 N/m2) and the specific gravity of mercury is 13.6.


Since C and D are at the same level in the same liquid at rest

Pressure pP at C = Pressure pQ at DFor the left hand limb

pC= pA+ωhFor the right hand limb

pD=pB+ω (h2−h1)+sωh1

=pB+ωh2−ωh1+sωh1

since pC=pD pA+ωh=pB+ωh2−ωh1+sωh1

Pressure difference pA−pB

=ωh2−ωh1+sωh1−ωh=ωh2−ωh+sωh1−ωh1

=ω (h2−h )+ωh1 ( s−1 )

=9. 81×103 (0 .75−1 .5 )+9. 81×103 (0 .5 ) (13 . 6−1 )

=54445 . 5 N /m2 =54 . 44 kN /m2

55

Figure 2.12

FLUID STATIC

ACTIVITY 2C

2.4 A U tube manometer measures the pressure difference between two points A and B in a liquid. The U tube contains mercury. Calculate the difference in pressure if h = 2.0 m, h2 = 0.35 m and h1 = 0.5 m. The liquid at A and B is oil ( s = 0.85 ) and the specific gravity of mercury is 13.6.

56

FLUID STATIC

FEEDBACK ON ACTIVITY 3C

2.4

Since C and D are at the same level in the same liquid at rest


For the left hand limbpC= pA+ωh

For the right hand limbpD=pB+ωh2+sωh1

since pP=pQ pA+ωh=pB+ωh2+sωh1

Pressure difference pA−pB

=ωoil h2+sωh1−ωoil h

=0 . 85 (9810 ) (0 .35 )+13. 6 (9810 ) ( 0.5 )+0 . 85 (9810 ) (2 .0 )=69626.475 N /m2

=69. 626 kN /m2

57

mercury

oil

FLUID STATIC

c) INVERTED DIFFERENTIAL MANOMETER

It is a particular type of differential manometer, in which an inverted U-tube is used. An inverted differential manometer is used for measuring the difference of low pressure, where accuracy is the prime consideration. It consists of an inverted U-tube, containing a light liquid. The two ends of the U-tube are connected to the points where the difference of pressure is to be found out as shown in Figure 3.10.

Now consider an inverted differential manometer whose two ends are connected to two different points A and B. Let us assume that the pressure at point A is more than that at point B, a greater pressure at A will force the light liquid in the inverted U-tube to move upwards. This upward movement of liquid in the left limb will cause a corresponding fall of the light liquid in the right limb as shown in Figure 3.10. Let us take C-D as the datum line in this case.

Let h = Height of the heavy liquid in the left-hand limb below the datum line,h1= Height of the light liquid in the left-hand limb below the datum line ,

58

INPUT

ωP

ωQ

Figure 2.13

FLUID STATIC

h2= Height of the light liquid in the right-hand limb below the datum line,ωP= Specific weight of the light liquidωQ= Specific weight of the heavy liquid

We know that pressures in the left limb and right limb below the datum line are equal.

Example 2.7

The top of an inverted U tube manometer is filled with oil of specific gravity, soil=0.98 and the remainder of the tube with water whose specific weight of water, ω= 9.81×103 N/m2. Find the pressure difference in N/m2 between two points A and B at the same level at the base of the legs where the difference in water level h is 75 mm.


For the left hand limbpD=pA−ωh2−so ωh1

for the right hand limbpC=pB−ω (h1−h2) =pB−ωh1−ωh2

since, pC=pD

pB−ωh1−ωh2=p A−ωh2−sωh1

59


Figure 2.14

FLUID STATIC

pB−p A=−ωh2−sωh1+ωh1+ωh2

=ωh1−sωh1

=ωh1 (1−s ) =9. 81×103 (0 .075 ) [1−0. 98 ]

=14 .715 N /m2

60

FLUID STATIC

ACTIVITY 2D

2.5 An inverted U tube as shown in the figure below is used to measure the pressure difference between two points A and B which has water flowing. The difference in level h = 0.3 m, a = 0.25 m and b = 0.15 m. Calculate the pressure difference pB – pA if the top of the manometer is filled with:

(a) air (b) oil of relative density 0.8.

61

FLUID STATIC

FEEDBACK ON ACTIVITY 2D

2.5In either case, the pressure at X-X will be the same in both limbs, so that

pXX=pA−ρ ga−ρmano gh=pB−ρg (b+h )pB−p A=ρg (b−a )+gh ( ρ−ρmano )

(a) if the top is filled with air ρmano is negligible compared with ρ. Therefore,

pB−p A= ρg (b−a )+ ρ gh

= ρg (b−a+h )

putting ρ=ρH 2O=103 kg/m3 , b=0. 15 m , a=0. 25 m , h=0 .3 m :

pB−p A=103×9. 81 (0 .15−0 . 25+0 .3 )

=1 .962×103 N /m2

(b) if the top is filled with oil of relative density 0.8, ρmano = 0.8 ρH 2O ,

pB−p A=ρg (b−a )+gh ( ρ−ρmano ) =103×9 .81 (0 .15−0 . 25 )+9 . 81×0 .3×103 (1−0 .8 )

=103×9 .81 (−0 .1+0. 06 )

=−392. 4 N /m2

62

FLUID STATIC

SELF-ASSESSMENT

2.1 What is the gauge pressure of the water at A if h1 = 0.6 m and the mercury in the right hand limb, h2 = 0.9 m as shown in the figure below?

2.2 In the figure below, fluid at A is water and fluid B is mercury (s =13.6). What will be the difference in level h if the pressure at X is 140 kN/m2 and a =1.5 m?

2.3 Assuming that the atmospheric pressure is 101.3 kN/m2 find the absolute pressure at A in the figure below whena) fluid P is water, fluid Q is mercury ω = 13.6, a = 1 m and h = 0.4 m.b) fluid P is oil ω = 0.82, fluid Q is brine ω = 1.10, a = 20 cm and h = 55 cm.

63

X

FLUID STATIC

2.4 In the figure below, fluid P is water and fluid Q is mercury (specific gravity=13.6). If the pressure difference between A and B is 35 kN/m2, a = 1 m and b = 30 cm, what is the difference in level h?

2.5 According to the figure in question 3.4, fluid P is oil (specific gravity = 0.85) and fluid Q is water. If a = 120 cm, b = 60 cm and h = 45 cm, what is the difference in pressure in kN/m2 between A and B?

2.6 In the figure below, fluid Q is water and fluid P is oil (specific gravity = 0.9). If h = 69 cm and z = 23 cm, what is the difference in pressure in kN/m2 between A and B?

64

FLUID STATIC

2.7 In question 6, fluid Q is water and fluid P is air. Assuming that the specific weight of air is negligible, what is the pressure difference in kN/m2 between A and B?

65

FLUID STATIC

FEEDBACK ON SELF-ASSESSMENT

Answers:

1. 114.188 kN/m2

2. 1.164 m

3. a) 38.2 kN/m2

b) 93.8 kN/m2

4. 30.7 cm

5. -5.23 kN/m2

6. -1.57 kN/m2

7. 4.51 kN/m2

66

FLUID STATIC

2.3.2 PIEZOMETER ( Pressure Tube )

Figure 2.15 : Piezometer inside a pipe

A Piezometer is used for measuring pressure inside a vessel or pipe in which liquid is there. A tube may be attached to the walls of the container (or pipe) in which the liquid resides so that liquid can rise in the tube. By determining the height to which liquid rises and using the relation p1 = ρgh, gauge pressure of the liquid can be determined. It is important that the opening of the device is to be tangential to any fluid motion, otherwise an erroneous reading will result.

Although the Piezometer tube is a very simple and accurate pressure measuring device, it has several disadvantages. It is only suitable if the pressure in the container (pipe or vessel) is greater than the atmospheric pressure (otherwise air would be sucked into system), and the pressure to be measured must be relatively small so that the required height of column is reasonable. Also, the fluid in the container in which the pressure is to be measured must be a liquid rather than a gas.

Example 2.1

A pressure tube is used to measure the pressure of oil ( mass density,640 kg /m3) in a

pipeline. If the oil rises to a height of 1.2 above the centre of the pipe, what is the gauge

pressure in N /m2at that point? (gravity = 9.81 m/s2)


Putting ρ =640 kg /m3

and h=1.2 m

67

FLUID STATIC

We know that, p = ρ gh

So, p = 640×9. 81×1. 2

p = 7 .55 kN /m2

68

FLUID STATIC

2.3.3 BAROMETERS

Figure 2.16 : Mercury Barometer

A Barometer is a device used for measuring atmospheric pressure. A simple Barometer consists of a tube of more than 30 inch (760 mm) long inserted into 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. Strictly, the space above the liquid cannot be a true vacuum. It contains mercury vapour at its saturated vapour pressure, but this is extremely small at room temperatures (e.g. 0.173 Pa at 20oC). The atmospheric pressure is calculated from the relation patm = ρgh where ρ is the density of fluid in the barometer. There are two types of Barometer; Mercury Barometer and Aneroid Barometer.

69

h

FLUID STATIC

Example 2.2

Describe with a sketch, one method of measuring atmospheric pressure.


Vacuum (zero pressure)

Mercury sp.wg, ωm

p p

A

pA

Figure 2.17

A Mercury Barometer in its simplest form consists of a glass tube, about 1 m long and closed at one end, which is completely filled with mercury and inverted in a bowl of mercury (Figure 3.3). A vacuum forms at the top of the tube and the atmospheric pressure acting on the surface of the mercury in the bowl supports a column of mercury in the tube height, h.

Example 2.3What is the atmospheric pressure in N/m2 if the level of mercury in a Barometer (Figure 3.3) tube is 760 mm above the level of the mercury in the bowl? Given the specific

gravity of mercury is 13.6 and specific weight of water is 9 .81×103 N /m3

.


If A is a point in the tube at the same level as the free surface outside, the pressure pA at A is equal to the atmospheric pressure p at the surface because, if the fluid is at rest, pressure is the same at all points at the same level.

The column of mercury in the tube is in equilibrium under the action of the force due to pA acting upwards and its weight acting downwards; there is no pressure on the top of the column as there is a vacuum at the top of the tube.

So,

70

FLUID STATIC

pA × area of column A = specific weight of mercury × specific weight of water pA× A= ωm×ah

or

pA= ωm×h

Putting h = 760 mm = 0.76 mm

While ωm= specific gravity of mercury × specific weight of water

ωm= 13 . 6×9.81×103 N /m2

From pA= ωm×h

So

pA= 13 .6×9 . 81×103×0 . 76 N /m2

=101 .3 kN /m2

71

FLUID STATIC

2.4 BOURDON GAUGE

Figure 2.18

Bourdon Gauge is used to measure pressure differences that are more than 1.2 bar. The pressure to be measured is applied to a curved tube, oval in cross section. Pressure applied to the tube tends to cause the tube to straighten out, and the deflection of the end of the tube is communicated through a system of levers to a recording needle. This gauge is widely used for steam and compressed gases. The pressure indicated is the difference between that communicated by the system to the external (ambient) pressure, and is usually referred to as the gauge pressure.

72

The pressure sensing element is a tube of oval cross-section bent to a circular shape.

One end of the tube is fixed to the gauge case and is connected to the fluid whose pressure is to be measured.

The other end is closed and is free to move as it is connected via mechanical linkage and gear sector to a pointer.

As measured fluid pressure increases above the surroundings, the tube cross-section tends to become circular and causes the tube to deflect at the second end.

This motion is transmitted via linkage to the pointer, which would directly indicate on the calibrated scale or dial on the gauge pressure.

FLUID STATIC

ACTIVITY 2E


1. What is the maximum gauge pressure of water of height 1.5 m that can be measured by a Piezometer? If the liquid had a relative density of 0.85 what would the maximum measurable gauge pressure?

FEEDBACK ON ACTIVITY 2E

1.

Gauge pressure, p = gh

For water,p = waterghp = 1000 x 9.81x 1.5p = 14715 N/m2 (or Pa)p = 14.715 kN/m 2 (or kPa)

For Liquid,p = liquidghliquid = water x relative densityp = 1000 x 0.85 x 9.81 x 1.5p = 12507.75 N/m2 (or Pa)p = 12.5 kN/m 2 (or kPa)

2.5 BUOYANCY

73

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.

FLUID STATIC

Principle of ArchimedesUpthrust on body = weight of fluid displaced by the body

Figure 2.19 Buoyancy

If the body is immersed so that part of its volume, v1 is immersed in a fluid of

density, ρ1 and the rest of its volume, v2 in another immiscible fluid of mass densityρ2 .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.

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.

74

Upthrust on upper part, R1 = ρ1 gv1

acting through G1, the centroid of v1,

Upthrust on lower part,R2 = ρ2 gv2

acting through G2, the centroid of v2,

Total upthrust = ρ1 gv1 + ρ2 gv2

FLUID STATIC


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, W = ρ ×g ×B× l ×D

a) In fresh water, ρ = 1000 kg/m3 and D = 1.5 m ;

Therefore,

Weight of pontoon, W = 1000 × 9 .81 × 6 ×12× 1.5 N

W = 1059 .5 kN

b) In the sea water, ρ = 1025 kg/m3 ; therefore,

Draught in sea water, D = W

ρ×g×B×l

=

1059 .5 × 103

1025 × 9 .81 × 6 × 12

= 1.46 m

c) For maximum draught of 2 m in fresh water,

Total uptrust = weight of water displaced = g×B×l×D

= 1000×9 .81×6×12×2 N

= 1412 .6 kN

Load which can be supported = Upthrust – weight of pontoon = 1412.6−1059 .5

= 353 .1 kN

75

FLUID STATIC

ACTIVITY 2F


1. Define the Archimedes Principle.

2. 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

76

FLUID STATIC

FEEDBACK ON ACTIVITY 2F

1.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.

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 = gwater Vbarge

Substituting volume Vbarge with 1 w d, we have:

Wbarge = gwater l w d = (9.8 kN/m3)(17 m)(7 m) d

Or

d=W barg e

( 9. 8 kN /m3 ) (17 m ) (7 m)

Substituting the weight of the barge gives:

d= 2. 0 MN( 9. 8 kN /m3 ) (17 m ) (7 m)

=1 . 71 m

77

FLUID STATIC

78

3 Web viewJabatan Sains & Matematik PSA Created Date: 01/12/2013 20:26:00 Title: 3 Last modified...

Documents

Transcript of 3 Web viewJabatan Sains & Matematik PSA Created Date: 01/12/2013 20:26:00 Title: 3 Last modified...