EBMF4103 (Chapter 2) Fluid Mechanics for Mechanical Engineering

Copyright © ODL Jan 2005 Open University Malaysia

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Subject Matter Expert/Author: Assoc Prof Dr Shahrir Abdullah

Faculty of Engineering andTechnical Studies

Fluid Mechanics for Mechanical Engineering TUTORIAL 1 – UNIT 1:Chapter 2: Fluid Static

Assoc Prof Dr Shahrir AbdullahEBMF4103 Fluid Mechanics for Mechanical EngineeringJan 2005

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SEQUENCE OF CHAPTER 2

Introduction

Objectives

2.1 Force Equilibrium of a fluid element

2.2 Hydrostatic Pressure Distribution

2.3 Standard Atmosphere

2.4 Pressure Measurement

2.5 Hydrostatic Forces

2.6 Buoyancy and Stability of Floating Bodies

Summary

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Introduction

• This chapter will begin with basic concepts in fluid static, which is the hydrostatic pressure distribution.

• It brings the student into the use of this principle in the measurement of pressure via manometer and in the determination of hydrostatic forces.

• This chapter also include buoyancy and stability which is crucial in designing submerged and floating bodies.

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Objectives

At the end of this chapter, you should be able to :

understand the concept of hydrostatic pressure distribution,

use the principle in pressure measurement of using manometers,

determine quantitatively hydrostatic forces and centres of pressure,

determine quantitatively buoyant forces and centres of buoyancy,

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2.1 Force Equilibrium of a Fluid Element

• Fluid static is a term that is referred to the state of a fluid where its velocity is zero and this condition is also called hydrostatic.

• So, in fluid static, which is the state of fluid in which the shear stress is zero throughout the fluid volume.

• In a stationary fluid, the most important variable is pressure.

• For any fluid, the pressure is the same regardless its direction. As long as there is no shear stress, the pressure is independent of direction. This statement is known as Pascal’s law

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Fluid surfaces

Figure 2.1 Pressure acting uniformly in all directions

Figure 2.2: Direction of fluid pressures on boundaries

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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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• The formulation for pressure gradient can be written as:

p = (g – a) (2.11)

1. If a = 0, the fluid is stationary.

2. if a ≠ 0, the fluid is in the rigid body motion.

• The formulation for pressure gradient in Eq. (2.11) is derived under the assumption that there is no shear stress present, or in other word, there is no viscous effect.

• The rest of this chapter will only concentrate on the first case, i.e. stationary fluids. The second case is applicable the case for a fluid in a container on a moving platform, such as in a vehicle, and is out of scope of this course.

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For a liquid, usually the position is measured as distance from the free surface, or depth h, which is positive downward as illustrated in Fig. 2.3. Hence,

p2 – p1 = g (h2 – h1) = (h2 – h1)

p = g h = h

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If the atmospheric pressure p0 is taken as reference and is calibrated as zero, then p is known as gauge pressure. Taking the pressure at the surface as atmospheric pressure p0, i.e., p1=p, p2=p0 when h1=h, h2=0, respectively:

p = p0 + gh = p0 + h

This equation produces a linear, or uniform, pressure distribution with depth and is known hydrostatic pressure distribution.

The hydrostatic pressure distribution also implies that pressure is the same for all positions of the same depth.

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• This statement can be explained by using a diagram in Fig. 2.4, where all points, a, b, c and d, have the same value of pressure, that is

pa = pb = pc = pd

• However, the pressure at point D is not identical from those at points, A, B, and C since the fluid is different, i.e.

pA = pB = pB ≠ pD

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Example 2.1An underground gasoline tank is accidentally opened during raining causing the water to seep in and occupying the bottom part of the tank as shown in Fig. E2.1. If the specific gravity for gasoline 0.68, calculate the gauge pressure at the interface of the gasoline and water and at the bottom of the tank. Express the pressure in Pascal and as a pressure head in metres of water. Use water = 998 kg/m3 and g = 9.81 m/s2.

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For gasoline:

g = 0.68(998) = 678.64kg/m3

At the free surface, take the atmospheric pressure to be zero, or p0 = 0 (gauge pressure).

p1 = p0 + pgghg = 0 + (678.64)(9.81)(5.5)= 36616.02 N/m2 = 36.6 kPa

The pressure head in metres of water is:h1 = p1 – p0 = 36616.02 - 0

pwg (998)(9.81)= 3.74 m of water

At the bottom of the tank, the pressure:p2 = p1 + pgghg = 36616.02 + (998)(9.81)(1)

= 46406.4 N/m2 = 46.6 kPaAnd, the pressure head in meters of water is:

h2 = p1 – p0 = 46406.4 - 0

pwg (998)(9.81)= 4.74 m of water

Example 2.1

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A pressure is quoted in its gauge value, it usually refers to a standard atmospheric pressure p0. A standard atmosphere is an idealised representation of mean conditions in the earth’s atmosphere.

Pressure can be read in two different ways; the first is to quote the value in form of absolute pressure, and the second to quote relative to the local atmospheric pressure as reference.

The relationship between the absolute pressure and the gauge pressure is illustrated in Figure 2.6.

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The pressure quoted by the latter approach (relative to the local atmospheric pressure) is called gauge pressure, which indicates the ‘sensible’ pressure since this is the amount of pressure experienced by our senses or sensed by many pressure transducers.

If the gauge pressure is negative, it usually represent suction or partially vacuum. The condition of absolute vacuum is reached when only the pressure reduces to absolute zero.

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Based on the principle of hydrostatic pressure distribution, we can develop an apparatus that can measure pressure through a column of fluid (Fig. 2.7)

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We can calculate the pressure at the bottom surface which has to withstand the weight of four fluid columns as well as the atmospheric pressure, or any additional pressure, at the free surface. Thus, to find p5,

Total fluid columns = (p2 – p1) + (p3 – p2) + (p4 – p3) + (p5 – p4)p5 – p1 = og (h2 – h1) + wg (h3 – h2) +

gg (h4 – h3) + mg (h5 – h4)

The p1 can be the atmospheric pressure p0 if the free surface at z1 is exposed to atmosphere. Hence, for this case, if we want the value in gauge pressure (taking p1=p0=0), the formula for p5 becomes

p5 = og (h2 – h1) + wg (h3 – h2) + gg (h4 – h3) + mg (h5 – h4)

The apparatus which can measure the atmospheric pressure is called barometer (Fig 2.8).

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For mercury (or Hg — the chemical symbol for mercury), the height formed is 760 mm and for water 10.3 m.

patm = 760 mm Hg (abs) = 10.3 m water (abs)

By comparing point A and point B, the atmospheric pressure in the SI unit, Pascal,

pB = pA + gh

pacm = pv + gh

= 0.1586 + 13550 (9.807)(0.760) 101 kPa

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This concept can be extended to general pressure measurement using an apparatus known as manometer. Several common manometers are given in Fig. 2.9. The simplest type of manometer is the piezometer tube, which is also known as ‘open’ manometer as shown in Fig. 2.9(a). For this apparatus, the pressure in bulb A can be calculated as:

pA = p1 + p0

= 1gh1 + p0

Here, p0 is the atmospheric pressure. If a known local atmospheric pressure value is used for p0, the reading for pA is in absolute pressure. If only the gauge pressure is required, then p0 can be taken as zero.

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Although this apparatus (Piezometer) is simple, it has limitations, i.e.

a) It cannot measure suction pressure which is lower than the atmospheric pressure,

b) The pressure measured is limited by available column height,

c) It can only deal with liquids, not gases.

The restriction possessed by the piezometer tube can be overcome by the U-tube manometer, as shown in Fig. 2.9(b). The U-tube manometer is also an open manometer and the pressure pA can be calculated as followed:

p2 = p3

pA + 1gh1 = 2gh2 + p0

pA = 2gh2 - 1gh1 + p0

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If fluid 1 is gas, further simplification can be made since it can be assumed that 1 2, thus the term 1gh1 is relatively very small compared to 2gh2 and can be omitted with negligible error. Hence, the gas pressure is:

pA p2 = 2gh2 - p0

There is also a ‘closed’ type of manometer as shown in Fig. 2.9(c), which can measure pressure difference between two points, A and B. This apparatus is known as the differential U-tube manometer. For this case, the formula for pressure difference can be derived as followed:

p2 = p3

pA + 1gh1 = pB + 3gh3 + 2gh2

pA - pB = 3gh3 + 2gh2 - 1gh1

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If a solid plate is immersed into the fluid, the pressure is also acted upon the surface of the solid.

This pressure acts on the submerged area thus generating a kind of resultant force known as hydrostatic force.

Hence, the hydrostatic force is an integration of fluid pressure on an area.

Similar to pressure, the direction in which the force is acting is always perpendicular to the surface.

To derive the hydrostatic force for a planar surface, consider the solid plate shown in Fig. 2.10. For a small elementary area dA, the force magnitude is:

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If a solid plate is immersed into the fluid, the pressure is also acted upon the surface of the solid.

dF = pdA Thus for the entire area A, the total magnitude of resultant force

is

FA = A ( p0 + gh) dA Here, the specific gravity the

fluid g can be taken as constant, which is valid for liquid, then the resultant force becomes

FR = p0 A + g Ah dA

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From Fig.2.9, a trigonometry relation can be used to represent h, i.e., h = y sin . Knowing that the angle is constant for a planar surface, then, the above expression can be written as

FR = p0 A + g sin A y dA If C is the centroid for the area A, by using the centroidal relationship, i.e.

A y dA = yC A and hc = yc sin , then

FR = p0 A + gyC A sin = p0 A + ghC A For isotropic materials where mass is uniformly distributed, the centroid C

is identical to the centre of gravity CG. Hence, the resultant hydrostatic force FR is a product of pressure p at C and the surface area A, i.e.

FR = ( p0 A + ghC )A = pC A In many applications, hydrostatic forces acting on any surfaces such as

walls of a tank are balanced by opposite forces generated by the atmospheric pressure p0 time the same surface area A.

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Therefore, based on this reason, we can omit the term p0A in Eq. (2.20), and thus it can be reduced to

FR = ghC A

The horizontal depth of the center of pressure, yR, (along y axis) is given by:

yR = 1xx + yC

yC A

The x coordinate for this point can be derived using moment equilibrium of the force FR about the y-axis at centroid C, and the formula can be written as followed,

xR = 1xy + xC

yC A

where Ixy is known as the product of inertia. For your convenience, Table 2.2 shows the formula for centroids and the moments of inertia, Ixx, Iyy and Ixy, for typical shapes.


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The hydrostatic force is the resultant of a linear distributed force formed by the liquid pressure acting perpendicular to the surface.

In the case where the surface is a wall of a liquid tank, the pressure distribution is as illustrated in Fig. 2.11.

Here, the application of Eq. (2.21) leads to the volume of the prism known as hydrostatic prism, which is generated by the linear distributed pressure, i.e.

This prism shape can represent the hydrostatic force for a partially immersed surface. As shown in Fig. 2.11, the centre of pressure CP is actually the centroid of the prism.

FR = volume of prism = ½(gh)(bh) = g • ( ½ h) A

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For a completely immersed surface, the hydrostatic prism becomes as shown in Fig. 2.12(a) for vertical surface and Fig. 2.12(b) for inclined surface, where the cross section of the prism is a trapezium.

For both cases, the hydrostatic force calculated using Eq. (2.21) gives the volume of the trapezoidal prism, i.e.

All the formula in this section are derived for the planar surface with a fixed angle. However, hydrostatic forces for curved surfaces, is out of the scope of this course.

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Example 2.3 A circular door having a diameter of 4 m is positioned at the

inclined wall as shown in Fig. E2.3(a), which forms part of a large water tank. The door is mounted on a shaft which acts to close the door by rotating it and the door is restrained by a stopper. If the depth of the water is 10 m at the level of the shaft, Calculate:

(a) Magnitude of the hydrostatic force acting on the door and its centre of pressure,(b) The moment required by the shaft to open the door.

Use water = 998 kg/m3 and g = 9.81 m/s2.

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(a) The magnitude of the hydrostatic force FR is

FR = ghC A

= (998)(9.81)(10) [ ¼ x(4)2]

= 1.230 x 106 N

= 1.23 MN

For the coordinate system shown in Figure E2.3(b), since circle is a symmetrical shape, Ixy = 0, then xR = 0. For y coordinate,

yR = 1xx + yC = ¼ R4 + yC

yC A yCR2

= ¼ (2)4 + 10 (10/sin 60°)(2)2 sin 60°= 11.6 m

or,

Example 2.3

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Example 2.3yR = 1xx + yC = ¼ (2)4 = 0.0866 m

yC A (10/sin 60°)(2)2

(b) Use moment equilibrium M - 0 about the shaft axis. With reference to Figure E2.3(b), the moment M required to open the door is:

M = FR ( yR - yC )

= (1.230 x 105) (0.0866)

= 1.065 x 105 N • m

= 107 kM • m

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This section will cover the interaction of fluid with the whole mass in a gravitational field which produce another form of force known as buoyant force, and the field in Fluid Mechanics which studies the behaviour of floating bodies is called buoyancy.

Buoyant force can be defined as the resultant fluid force which acts on a fully submerged or floating body.

To derive the formula for the buoyant force, consider Fig. 2.13:

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From Fig. 2.13, if a rectangular block ABCD is drawn to cover the body, by using the force equilibrium in the x and y directions, we should obtain:

Fx = 0 : F3 = F4

Fy = 0 : FB = F3 – F1 – Wwhere W is the weight of the fluid volume in ABCD, and FB is the force exerted by the body to the fluid.

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If the fluid is incompressible, which is a valid assumption for liquid, by taking A as the area for the top AB plane and bottom CD plane and V is the volume of the body, the force equilibrium in the y direction becomes

FR = g (h2 - h1) A - g [(h2 - h1) A – V ]

FB = g V (2.24)

From Fig. 2.13, by using the principle of force equilibrium, the fluid force acting onto the body must be of the same magnitude with the force exerted by the solid to the fluid FB, but in the opposite direction. Hence Eq. (2.24) give the formula for the buoyant force which is always in upward direction, i.e., opposite to the weight of the body.


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The relation also shows that the buoyant force is equal to the weight of the volume of fluid which is displaced by the body.

The point of action for this is the centroid of the displaced fluid, which is also called the centre of buoyancy CB.

The difference between the centre of buoyancy and the centre of gravity of the floating body CG may lead to stability issue.

In general, a body in equilibrium may be in two possible positions, as shown in Fig. 2.14 for a completely submerged body:

(a)Stable equilibrium — a small displacement causing it to return to its original position.

(b)Unstable equilibrium — a small displacement causing it to shift to another position.

Hence, it can be concluded that for a completely submerged body where W FB, the restoring moment produced causes the body to return to its stable condition, and as a result, the centre of buoyancy CB should be always lower than the centre of gravity CG.

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Summary

This chapter has summarized on the aspect below:

Should be able to understand the principle of hydrostatics, The conditions of standard atmosphere, and consequently be

able to distinguish between absolute and gauge pressure readings.

Should also be able to apply this knowledge in calculation of pressure measured using manometers, in evaluating the hydrostatic force acted onto a planar surface and locating the corresponding centre of pressure, as well as in evaluating the buoyant force exerted to a submerged or floating body and locating the corresponding centre of buoyancy.

In addition, you should also be able to analyse the stability of a body immersed in fluid.

This concludes the scope of this course for fluid static.

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Thank You

EBMF4103 (Chapter 2) Fluid Mechanics for Mechanical Engineering

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