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by

Akshoy Ranjan PaulAssistant Professor

Email: arpaul2k@gmail.com

DEPARTMENT OF APPLIED MECHANICSDEPARTMENT OF APPLIED MECHANICSMOTI LAL NEHRU NATIONAL INSTITUTE OF TECHNOLOGY MOTI LAL NEHRU NATIONAL INSTITUTE OF TECHNOLOGY

ALLAHABADALLAHABAD

Fluid Fluid Mechanics (Mechanics (AM1401)AM1401)B B Tech. Tech. Mech.& Prod. (4th semesterMech.& Prod. (4th semester))

Chapter: Fluid Statics Chapter: Fluid Statics

Chapter 2: Fluid Statics

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Pressure concepts and problems Pressure concepts and problems

4Copyright © ODL Jan 2005 Open University Malaysia

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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Force Equilibrium of a Fluid Element

Fluid surfaces

Figure 2.1 Pressure acting uniformly in all directions

Figure 2.2: Direction of fluid pressures on boundaries

Force Equilibrium of a Fluid ElementPressure 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

AFP

appliedisforcethewhichofAreaForceessure

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Force Equilibrium of a Fluid Element• 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.

Hydrostatic Pressure Distribution•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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Hydrostatic Pressure Distribution 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.

Hydrostatic Pressure Distribution• 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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Standard Atmosphere 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.

Standard Atmosphere 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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Pressure Measurement 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)

Pressure Measurement 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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Pressure Measurement 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 + ghpacm = pv + gh

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

Pressure Measurement 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 pAis in absolute pressure. If only the gauge pressure is required, then p0 can be taken as zero.

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Pressure Measurement 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

Pressure Measurement 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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Example 2.5•An 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.

•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 kPa•And, the pressure head in meters of water is:• h2 = p1 – p0 = 46406.4 - 0• pwg (998)(9.81)• = 4.74 m of water

Example 2.5

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• Example 2.6 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)

Example 2.6

• 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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23Copyright © ODL Jan 2005 Open University Malaysia

•Example 2.7•Determine 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)

•Thus• Hggh2 = 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=PatmP2 = P3P4 = PA

A

MicromanometerThe micro manometers are used for measuring small pressure difference.The micro manometers utilizes two manometric liquids, which are immiscible with each other and also with the fluid whose pressure difference is to be measured.When PA > PB , the liquid levels will be as shown in fig.The volume of the liquid displaced in each tank is equal to the volume of liquid displaced in the U-tube.If a= cross-sectional area of the U-tube and A= cross-sectional area of tank then,

ahzA 2/

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• Example 2.8 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

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 Pa

PA = 15.7 kPa (gauge)Pressure at the bottom of the tank;

PB = PA + waterghwater

PB = 15.7x1000 + 1000 x 9.81 x 4 = 54940 Pa

PB = 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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Hydrostatic forces

Figure 3.1 (p. 31)

Pressure is a scalar quantity

Force balance in the x-direction:

PASCAL’S LAWPASCAL’S LAW

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Force balance in the z-direction:

Vertical force on A Vertical force on lower boundary

Total weight of wedge element

= specific weight

From last slide:

Divide through by to get

Now shrink the element to a point:

This can be done for any orientation a, so

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Pressure Variation with Elevation

Figure 3.4 (p. 35)

Static fluid:All forces must balance as there are no accelerations.

Look at force balance in

direction of l

From figure, note that

Shrink cylinder to zero length:

(from previous slide)

or

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Hydrostatic Forces

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

To derive the hydrostatic force for a planar surface, consider the solid plate shown in following slide..

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CONTINUE…….

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CONTINUE…….

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Example 2.1 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 center of pressure,

(b) The moment required by the shaft to open the door.

Use water = 1000 kg/m3 and g = 9.81 m/s2.

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(a) The magnitude of the hydrostatic force FR isFR = 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 + yCyC A yCR2

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

= 11.6 m

or,

Example 2.1

Example 2.1

yR = 1xx + yC = ¼ (2)4 = 0.0866 myC 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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Example 2.2

Find the normal force required to open the elliptical gate if it is hinged at the top.

First find Ftotal, the total hydrostatic force acting on the plate:

With (Appendix p. A-5) we get

Now calculate the slant distance between and

The slant distance to the hinge is 8m x 5m/4m = 10m, and the slant distance from the hinge to the centroid is 2.5m. Hence,

The two moments about the hinge must add to zero:

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Example 2.3

Find magnitude and line of action of equivalent force F.

Force balance in x and y:

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The line of action of the horizontal force is

Where we just read directly off the figure.

The line of action for the vertical force can be found by summing the moments about C (or any other point…)

(notice that we could add a constant to every x-coordinate since )

Distance from C to centroid is:

So that xcp is found to be

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The complete result is summarized below: