EBVF4103 (Chapter 2) Fluid Mechanics for Civil Engineering

Copyright © ODL Jan 2005 Open University Malaysia

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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)

Faculty of Engineering andTechnical Studies

FLUID MECHANICS FOR CIVIL ENGINEERING TUTORIAL 1 – UNIT 1: Elements of Fluid MechanicsChapter 2: Fluid Statics

Assoc. Prof. Dr Othman A. KarimEBVF4103 Fluid Mechanics for Civil EngineeringJan 2005

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SEQUENCE OF CHAPTER 2Introduction

Objectives

2.1 Pressure2.2 Absolute and Gauge Pressure2.3 Variation of Pressure with Elevation

2.3.1 Pressure and Head2.3.2 Equality of pressure at the same level in static fluid2.3.3 Pascal’s Paradox

2.4 Pressure Measurement2.5 Hydrostatic Force on Plane Surface2.6 Hydrostatic Buoyant ForceSummary

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• Fluid static is the study of pressures throughout a fluid at rest and the pressure forces on finite surface.

• The general rule applies to fluid at rest:1. No shears stress /force acting on it.

2. Any force between the fluid and the boundary must be acting at right angles to the boundary (normal to the surface).

Figure 2.1 Pressure forces normal to the boundary

F F

Introduction

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Objectives

1. Define the relationship between absolute pressure, gauge pressure and atmospheric pressure.

2. Define the relationship between change in elevation and change in pressure.

3. Describe pressure measurements.

4. Compute the hydrostatic pressures and forces on submerged surfaces on static fluid.

5. Use the principle of static equilibrium to solve for the forces involves in buoyancy problems.

6. Define the condition for stability of submerged and floating bodies.


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2.1 Pressure

The basic property of a static fluid is pressure. Pressure is defined as the amount of surface force

exerted by a fluid on any boundary it is in contact with. It can be written as:

Unit: N / m2 or Pascal (Pa). (Also frequently used is bar, where 1 bar = 105 Pa).

)1.2(

Pr

A

FP

appliedisforcethewhichofArea

Forceessure

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2 (TWO) important principles about pressure;• Pressure acts uniformly in all directions on a small

volume of fluid.

• In a fluid confined by solid boundaries, pressure acts perpendicular to the boundary.

• These principles, called Pascal’s Law,Fluid surfaces

Figure 2.2 Pressure acting uniformly in all directions

Figure 2.3: Direction of fluid pressures on boundaries

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2.2 Absolute and Gauge Pressure

• Pressure measurements are generally indicated as being either absolute or gauge pressure.

Gauge pressure

• is the pressure measured above or below the atmospheric pressure (i.e. taking the atmospheric as datum).

• can be positive or negative. • A negative gauge pressure is also known as vacuum

pressure.

Absolute pressure

• uses absolute zero, which is the lowest possible pressure.• Therefore, an absolute pressure will always be positive. • A simple equation relating the two pressure measuring

system can be written as:

Pabs = Pgauge + Patm (2.2)

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Atmospheric pressure• refers to the prevailing pressure in the air around us.

• It varies somewhat with changing weather conditions, and it decreases with increasing altitude.

• At sea level, average atmospheric pressure is 101.3 kPa (abs), 14.7 psi (abs), or 1 atmosphere (1 bar = 1x105 Pa).

• This is commonly referred to as ‘standard atmospheric pressure’.

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Example 2.1 Express a pressure of 155 kPa (gauge) as an absolute

pressure. Express a pressure of –31 kPa (gauge) as an absolute

pressure. The local atmospheric pressure is 101 kPa (abs).

Solution:

Pabs = Pgauge + Patm

Pabs = 155 + 101 = 256 kPa

Pabs = -31 + 101 = 70 kPa

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2.3 Variations of Pressure with Elevation

To find the variations of pressure with elevation, let’s consider a small cylindrical element of fluid of cross-sectional area A, and height (h = Z2 –Z1), surrounded by the same fluid of mass density, .

Fluid Density

Area, A

P2, A

P1, A

Z1

Z2

h

Reference/datum

Figure 2.4: Small cylindrical element of fluid

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The pressure at the bottom of the cylinder is P1 at level Z1, and at the top is P2 at level Z2. The fluid is at rest and in equilibrium so all the forces in the vertical direction sum to zero.

Force due to P1 (upward) = P1A Force due to P2 (downward) = P2A Force due to weight of element = mg = gA(Z2-Z1) Taking the summation of forces (upward as positive);

()F=0P1A – P2A - gA(Z2-Z1) = 0P1 – P2 = g (Z2-Z1) = gh (2.3)

or P2 – P1 = - g(Z2-Z1) = - gh (2.3a)

Thus, in any fluid under gravity, an increase in elevation causes a decrease in pressure. a decrease in elevation causes an increase in pressure.

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2.3.1 Pressure and head

In a liquid with a free surface the pressure at any depth h measured from the free surface can be found by applying equation (2.3) to the figure.

From equation (2.3): P1 – P2= g (ya-y)

But ya-y = h , and P2 = Patm (atmospheric pressure since it is at free surface).

Thus,P1 – Patm= gh

or P1 = Patm + gh (abs) (2.4) or in terms of gauge pressure (Patm= 0),:

P1 = gh = h (2.5)

h

P1

P2 = Patm

y

ya

Free surface

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From the above equations, it can be concluded that the change in pressure is directly proportional to the specific weight of the liquid, and pressure varies linearly with the change of elevation or depth.

The linear variation with depth below the free surface is known as hydrostatic pressure distribution.

Hydrostatic pressure increases with the depth of fluid. Notice that in Figure 2.5 below, the reading on the pressure gauge of tank A is lower than the reading of tank B. The gauges show the pressure created by the depth and specific weight of the liquid.

Tank A Tank B

liquidliquid

Figure 2.6: Different pressuredue to different depthPA < PB

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As g is assumed constant, the gauge pressure can be given by stating the vertical height, h, of any fluid density, , which would be necessary to produce this pressure. This vertical height, h, is known as pressure head or just head of fluid, and can be written as;

h = P/g (2.6)

Note that when pressures are expressed as head, the density of fluid must be given or the fluid is named.

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2.3.2 Equality of Pressure at the Same Level in a Static Fluid

Consider the horizontal cylindrical element of fluid with cross sectional area, A, in a fluid of density , pressure PL at the left end and PR at the right end.

Fluid is at equilibrium, so the sum of forces acting on the x-direction is zero.

() ΣF =0.PLA – PRA = 0

PL = PR (2.7) This proof that pressure in the horizontal direction is constant.

W = mg

PRPL

Fluid density, ρ

AAFigure 2.7 Horizontal element cylinder of fluid

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Applying equation (2.4) PL = PP + ρgh ---------------- (1)

andPR = PQ + ρgh ---------------- (2)

Earlier, we have shown that PL = PR (refer equation 2.6), therefore equating (1) and (2) will give

Pp + ρgh = PQ + ρgh

PP = PQ (2.8) This shows that the pressures at two equal levels P & Q are the

same. This is an important concept when dealing with manometers (see Section 2.4).

Figure 2.8: Two tanks of different cross-section connected by a pipe

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2.3.3 Pascal’s Paradox

Earlier we have shown that the change in pressure depends only on the change of elevation and the type of fluid, not on the weight of the fluid present.

Therefore, all the containers shown in Figure 2.8 would have the same pressure at the bottom – no matter what the size or shape of container and how much fluid they contained.

This observation is called Pascal’s Paradox.

hh

Pressure is the same at the bottom of container: P=gh

Figure 2.9: Illustration of Pascal’s Paradox

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Example 2.2 What will be the gauge pressure and absolute

pressure of water at a depth 12m below the surface? Take ρwater = 1000 kg/m3 and Patm = 101 kN/m2

Solution:

Pgauge = ρgh = 1000 x 9.81 x 12= 117.7 kN/m2 (kPa)

Pabs = Pgauge + Patm = (117.7 + 101) kN/m2

= 218.7 kN/m2

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Example 2.3 A cylinder contains a fluid at a gauge pressure of 200

kN/m2. Express this pressure in terms of head of water ( =1000 kg/m3) head of mercury (SG=13.6) What would be the absolute pressure if the

atmospheric pressure is, Patm = 101.3 kN/m2. Solution:

h= P/ga) for water: h = 200x103/(1000x9.81)

= 20.39 m of water. b) for mercury h = 200x103/(13.6x1000x9.81)

= 1.5 m of mercuryAbsolute pressure = Patm + Pgauge

= 101.3 + 200 = 301.3 kN/m2.

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Example 2.4 Figure below shows a tank with one side open to the

atmosphere and the other side sealed with air above the oil (SG=0.90). Calculate the gauge pressure at points A,B,C,D,E.

2 m

3 m

1 m

E

A

C

B D

Oil (SG = 0.90)

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Solution:At point A, the oil is exposed to the atmosphere

thus PA=Patm = 0 (gauge)Point B is 3 m below point A,

Thus PB = PA + oilgh= 0 + 0.9x1000x9.81x3= 26.5 kPa (gauge)

Point C is 5 m below point A, Thus PC = PA + oilgh

= 0 + 0.9x1000x9.81x5= 44.15 kPa (gauge)

Point D is at the same level of point B,thus PD = PB

= 26.5 kPa (gauge)Point E is higher by 1 m from point A,

Thus PE = PA - oilgh= 0 - 0.9x1000x9.81x1= -8.83 kPa (gauge).

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2.4 Pressure Measurement

Atmospheric pressure is usually measured by a mercury barometer.

A simple barometer consists of a tube more than 760 mm (30 inch) long inserted in an open container of mercury with a closed and evacuated end at the top and open end at the bottom with mercury extending from the container up into the tube.

A void is produced at the top of the tube which is very nearly a perfect vacuum. Figure 2.10 below shows an example of a barometer.

Mercury rises in the tube to a height of approximately 760 mm (30 in.) at sea level.

The level of mercury will rise and fall as atmospheric pressure changes; direct reading of the mercury level gives prevailing atmospheric pressure as a pressure head (of mercury), which can be converted to pressure using the relation:

Patm = ρgh.

Figure 2.10:Barometer

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Piezometer tube

A simple vertical tube open at the top, which is attached to the system containing the liquid where the pressure (higher than atmospheric pressure) to be measured.

As the tube is open to the atmosphere, the pressure measured is the gauge pressure.

A

B

h1

h2

Liquiddensity,

Pressure at A = pressure due to column of liquid above APA = gh1

Pressure at B = pressure due to column of liquid above B

PB = gh2

Figure 2.11: Piezometer tube

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U-tube Manometer

One end of the U-tube is connected to the pressure that is to be measured, while the other end is left open to atmosphere.

The tube contains a liquid, which is called the manometric fluid, which does not mix with the fluid whose pressure is to be measured.

The fluid whose pressure is being measured should have a lesser density than the manometric fluid. (ρ < ρman )

B

PA

C

D

h1

h2

Pa

Manometric fluid density, man

Fluid density,

Figure 2.12: U-tube manometer

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Pressure in a continuous static fluid is the same at any horizontal level so,

Pressure at B = Pressure at CPB = Pc (1)

For the left hand arm:Pressure at B = Pressure at A + pressure due to h1

PB = PA + ρgh1 (2)For right hand arm:Pressure at C = Pressure at D + pressure due to height h2

PC = PD + ρmangh2 but PD = Patm (atmospheric pressure due to open end)As we are measuring gauge pressure, Patm = 0

PC= ρmangh2 (3)But PC = PB, substitute in (3); will give;

PB = ρmangh2 (4)Equating (2) and (4);

PA + ρgh1= ρmangh2

PA = ρmangh2 - ρgh1 (2.9)

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Differential Manometer

In some cases, the different between the pressures at two different points is desired rather than the actual value of the pressure at each point.

A manometer to determine this pressure difference is called the differential manometer (see figure below).

The liquids in manometer will rise or fall as the pressure at either end (or both ends) of the tube changes.

Figure 2.13: Differential manometer

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In the above figure:

P1 = PA + 1ga

P2 = PB + 1g(b-h) + mangh

But P1 = P2 (same horizontal level)

Thus PA + 1ga = PB + 1g(b-h) + mangh

or PA - PB = 1g(b-h) + mangh - 1ga

PA- PB = 1g(b-a) + gh(man - 1)(2.10)

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Pressure Gauges

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.

Figure 2.16: Bourdon tube pressure gauges

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Example 2.5 What height would a barometer need to be to

measure atmospheric pressure?

Solution:

Patm = 1 bar = 1 x 105 Pa

Patm = gh

Water barometer:

Mercury barometer:

waterofmx

x

g

Ph

water

atm 2.1081.91000

101 5

mercuryofmxx

x

g

Ph

Hg

atm 75.081.910006.13

101 5

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Example 2.6Determine the pressure at point A in the figure below if h1 = 0.2 m and h2 = 0.3 m. Use water = 1000 kg/m3.

Solution:P2 = P1 + Hggh2

But P1 = Patm (open to atmosphere) ==>P1 = 0 (gauge)

P2 = Hggh2

P3 = PA + waterg(h1+h2)

We know that P2 = P3 (same horizontal level)

ThusHggh2 = PA + waterg(h1+h2)

PA = Hggh2 - waterg(h1+h2)

PA = 13.54x1000x9.81x0.3 – 1000x9.81x(0.2+0.3)

PA = 39, 848 - 4905

PA = 34.9 kPa (gauge)

Points to be selected:

1 – at the open end of the manometer2 – at the right leg of the manometer3 – same level with point 2 but at left

leg of the manometer4 – same level as point A

Pressure at the points:P1=Patm

P2 = P3

P4 = PA

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2.5 Hydrostatic Force on Plane Surface

Pressure has been defined as force divided by the area on which it acts. This principle can be restated as when a fluid is adjacent to a fixed surface, it exerts a force on the surface because of the pressure in the liquid. For fluid at rest, the force always act at right angles to the surface.

For horizontal plane submerged in a liquid, the pressure, P, will be equal at all points of the surface. This leads to the conclusion that the resultant force on horizontal surface due to that pressure can be computed from the simple product of pressure times the area of interest, i.e.

Force = Pressure x Area of plane

F = PA

This force will act vertically downward and through the center of pressure.

hF=PA=gh

Figure 2.17: Resultant force on horizontal plane

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2.5.1 Resultant Force and Center of Pressure on a Submerged Plane Surface in a Liquid

Figure 2.18 below shows a plane surface PQ of an area A submerged in a liquid of density, , and inclined at an angle to the free surface.

Considering one side only, there will be a force due to fluid pressure, acting on each element of area A, the magnitude of the pressure will depend on the vertical depth y of the element below the free surface. Taking the pressure at the free surface as zero, and from equation (2.5), the pressure at a distance y below the free surface can be written as:

p = gy. (2.13)

Figure 2.18: Resultant force on a plane surface immersed in a fluid

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Force on elemental area A: dF = PA = gyA The resultant force acting on the plane can be found by summing all the

forces on the small element:

F = ΣPA = Σgy A Assuming that and g are constant,

F = g Σy A (2.14) The quantity Σy A is the first moment of area under the surface PQ about the

free surface of the liquid and is equal to Aŷ, where A = the area of the whole immersed surface and ŷ = vertical distance from the free surface to the centroid of the area, G, of the immersed surface.

Centroid of the area is defined as the point at which the area would be balanced if suspended from that point. It is equivalent to the center or gravity of a solid body.

Substituting into equation (2.13) will giveF = gŷA (2.15)

It may be noted that the resultant force, F, is independent of the angle of inclination so long as the depth of the centroid ŷ is unchanged.

The point of application of the resultant force on the submerged area is called the center of pressure. This resultant force will act perpendicular to the immersed surface at the center of pressure, C.

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The vertical depth of the center of pressure, y0, below the free surface can be found using the following:

(2.16)

where Ig = second moment of plane area about its center of gravity

A = the area of the whole immersed surface

ŷ = vertical distance from the free surface to the centroid of the area A

The above equation implies that the center of pressure is always below the centroid.

Table 2.1 below gives the second moment of area about a line through the centroid of some common shapes.

yA

Iyy

go ˆ

ˆ

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Table 2.1 Second Moments of Area

G G

h

b

G

G

hh/3

G

d

G

Rectangle

Triangle

d4/64d2/4Circle

bh3/36bh/2

bh3/12bh

IgAreaShape

G G

h

h/3G h

G

b

GG

d

b

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The area of this triangle (RST) represents the resultant force per unit width on the vertical wall. So;

Area of pressure diagram = Therefore, the resultant force per unit width,(2.17)

This force acts through the centroid of the pressure diagram. For a triangle, the centroid is located at 2/3 its height, thus the resultant force acts at a depth of 2/3 H from R.

The total resultant force can be obtained by multiplying the above equation with the width of the surface, B.

F = ½ pgH2B (2.17a) The same pressure diagram technique can be used when combinations of

liquid are held in tanks (e.g. oil floating on water).

2.5.2 Pressure Diagram In Figure, the triangle on the right hand side

(RST) is a graphical representation of the (gauge) pressure change with depth on one side of the vertical wall of the tank containing a liquid with density . At the free surface the gauge pressure is zero. It increases linearly from zero at the surface by P = gy, to a maximum of at the base of P = gH. S

Fw

T

y

H

Liquid density,

2/3 H

P = gy

P = gH

R

)/(2

1 2 mNgHFw

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Example 2.7 A 6-m deep tank contains 4 m of water and 2-m of oil

as shown in the diagram below. Determine the pressure at point A and at the bottom of the tank. Draw the pressure diagram.

Aoil

water

2 m

4 m

water = 1000 kg/m3

SG of oil = 0.98

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Solution:Pressure at oil water interface (PA)

PA = Patm + Poil (due to 2 m of oil) = 0 + oilghoil = 0 + 0.98 x 1000 x 9.81 x 2 = 15696 PaPA = 15.7 kPa (gauge)

Pressure at the bottom of the tank;PB = PA + waterghwater

PB = 15.7x1000 + 1000 x 9.81 x 4 = 54940 PaPB = 54.9 kPa (gauge)

Patm = 0

4 m

2 m

PA

PA=15.7 kPa

B

Aoil

water

PB = 54.9 kPA

Pressure Diagram

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2.6 Hydrostatic Buoyant Force

When a body is submerged or floating in a static fluid, the resultant force exerted on it by the fluid is called the buoyancy force. This buoyancy force is always acting vertically upward, and has the following characteristics;

The buoyancy force is equal to the weight of the fluid displaced by the solid body.

The buoyancy force acts through the centroid of the displaced volume of fluid, called the center of buoyancy.

A floating body displaces a volume of fluid whose weight is equal to the weight of the body

The above principle is known Archimedes’ principle and can be defined mathematically as demonstrated below (see Figure 2.21);

W = mg

Fb= W Fb = W

GB

GB

W = mg

Volume of displaced fluid

Figure 2.21: Buoyancy force

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For equilibrium: + ΣFy = 0

Fb – W = 0 or Fb = W

Therefore we can write ;

Fb = weight of fluid displaced by the body

or Fb = W = mg = g (2.21)

where Fb = buoyant force

= displaced volume of fluid

W = weight of fluid The buoyant force will act through the center of

buoyancy (B), while the weight will act through the center of gravity (G) of the body.

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Summary

The chapter has summarized the important of:1. Pressure as basic property of fluid and the equation2. Principal of Pascal Law3. Indicating the gauge pressure and absolute pressure 4. The variations of pressure with elevation and the

calculation involved5. Pressure measurement of Piezometer Tube, U-tube

Manometer, Differential Manometer, Advances of U-Tube Manometer and Pressure Gauges

6. The Resultant force and the application to the pressure diagram

7. Demonstrating the technique of determining the forces acting on submerged or partially submerged surfaces

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Thank You

EBVF4103 (Chapter 2) Fluid Mechanics for Civil Engineering

Documents

subject matter

small cylindrical

pressure due

mangh

5 hydrostatic

force due

immersed surface

resultant