Fluid Mechanics for Construction Chapter 2 – Fluid Pressures

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2 FLUID PRESSURES By definition, a fluid must deform continuously when a shear stress of any magnitude is applied. Therefore when a fluid is either at rest or moving in such a manner that there is no relative motion between adjacent particles, there will be no shear forces acting and, therefore all forces exerted between a fluid and a solid boundary must be normal (i.e. right angle to the given surface). Pressure is used to indicate the normal force per unit area at a given point acting on a given plane.

i.e. A

Fp normal

Pressure is a scalar quantity. Since it is a ratio of force and area, therefore the unit of pressure in SI is pascal (Pa). Which is defined as N/m2.

Hence Pa = 2m

N;

kPa = 2m

kN;

MPa = 2mm

N

2.1 Pressure at a Point By considering the equilibrium of a small fluid element in the form of a triangular prism in the fluid subject to a pressure px in x-direction, py in y-direction and ps in normal to any plane inclined at an angle to the horizontal.

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px

x

z

s

y

ps

py

For simplicity, the forces in the z direction are not shown. The equation of motion in the x and y directions are respectively, Fx = px y z - ps z s sin Fy = py x z - ps z s cos - xyz/2 By geometry, x = s cos; y = s sin Since the fluid element is in equilibrium, i.e. Fx = 0 & Fy = 0

px y z - ps y z = 0 px = ps

and py x z - ps x z - xyz/2 = 0

py - ps = y/2 As y approaches to zero, py = ps Hence px = py = ps Therefore,

Pressure at a point in a fluid is the same in all direction - Pascal’s Law

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2.2 Pressure Variation in a Fluid with Depth Considering an element of vertical column of constant cross-sectional area A and totally surrounded by the same fluid of mass density .

p

p+dp

h

h+dh

Area, A

Fluid density

Suppose Pressure at h = p Pressure at h + h = p + p (h increases in upward direction) Since the fluid is at rest, the element must be in equilibrium with no shearing force and the summation of vertical force must be zero. Force due to p on area A acting up = pA Force due to p + p on area A acting down = (p + p)A Force due to the weight of element = gA(h) pA - (p + p)A - gA(h) = 0

h

p

= -g

or dh

dp = -g

If one travels upward in the fluid (positive in h direction), the pressure decreases; if one goes downward (negative h), the pressure increases.

For any fluid under gravitational attraction, pressure increases with increase of water depth.

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2.3 Equality of Pressure at the Same Level in a Static Fluid If P and Q are two points at the same level in a fluid at rest, a horizontal prism of fluid of constant cross-sectional area A will be in equilibrium. The forces acting on this element horizontally are p1A at P and p2A at Q.

p1 p2

Area A

mg

Fluid density

Since the fluid is at rest, there will be no horizontal shear stresses on the sides of the element. For static equilibrium the sum of the horizontal forces must be zero. p1A = p2A p1 = p2

2.4 Pressure and Head

In a fluid of constant density, dh

dp= -g can be integrated immediately to

give p = -gh + constant However in practice, the depth of liquid is usually measured from the top free water surface downward, i.e. h = -h, the pressure will then be

The pressure at any two points at the same level in a body of fluid at rest will be the same

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p

h

Patm

Liquid Density

p = gh + constant and since the pressure at the free surface will normally be atmospheric pressure patm (i.e. at h = 0, p = patm), p = gh + patm It is often convenient to take atmospheric pressure as a datum. Pressures measured above atmospheric pressure are known as gauge pressure. Pressures measured above perfect vacuum are called absolute pressure.

Atmospheric Pressure

A

B Absolutepressure A

Absolutepressure B

Barometerreading

Gauge pressure A

Gauge pressure B

Vaccum

The region of pressure below atmospheric pressure is generally referred to as vacuum. If the pressure is at absolute zero, it is called perfect vacuum. If the pressure is between atmospheric pressure and absolute zero, it is called partial vacuum.

Absolute pressure = Gauge pressure + Atmospheric pressure

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Standard atmospheric pressure, which is the air pressure at sea level, can be considered as 1.0 atm = 1.0 bar = 760 mm Hg = 101.4 kPa By considering the gauge pressure only, then p = gh which indicates that The gauge pressure at a point can be defined by stating the vertical height h, called the head or pressure head, of a column of a given fluid of mass density .

i.e. h = m g

p

Note that when pressure is expressed as head, it is essential that the mass density is specified.

If g is assumed constant, the pressure increases linearly with depth

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Worked examples: 1. Calculate the pressure at a point on the sea bed 1 km deep. The density of sea water is 1025 kg/m3. Answer = 1025 kg/m3 g = 9.81 m/s2 h = 1000 m (pressure head) Since p = gh = 1025 * 9.81 * 1000 = 10,055,000 N/m2 = 100.55 bar (1 bar = 105 N/m2) 2 The pressure at a point on the sea bed is 100.55 bar, (a) express this pressure as a head of fresh water , and (b) what is the pressure as a head of mercury of S.G. = 13.6? Answer (a) water = 1000 kg/m3 Since p = gh 100.55 x 105 = 1000 * 9.81 * h h = 1025 m, i.e. 1025 m head of water. (b) Hg = 13.6 * 1000 kg/m3 100.55 x 105 = 13.6 * 1000 * 9.81 * h h = 75.37 m of mercury.

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2.5 Measurement of Pressure Many instruments for pressure measurement use the fact that a pressure is equivalent to a head of liquid. 22..55..11 PPiieezzoommeetteerr If a transparent tube is inserted into some point of a liquid under pressure, then the liquid will be seen to rise in the tube until its height balances the pressure in the liquid. This is the simplest pressure-measuring instrument, the piezometer.

open

open

Pressure Vaccum

open open

hh

p = h vacuum = hor pressure = -h

This device is only suitable if the pressure in the container is greater than atmospheric pressure, and the pressure to be measured must be relatively small so the required height of column is reasonable. 22..55..22 UU--TTuubbee MMaannoommeetteerr If a heavier liquid is used to balance the pressure, the gauge will become more compact. For example, 2 m of water is equivalent to only 147 mm of mercury. However a different arrangement is necessary in order to prevent the mixing of two liquids of different densities.

open

open

Heavierliquid

h

13.6h

Level of separation

water

Hg (S.G. = 13.6)

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Pressure measuring instruments using the U-tube are called manometers. In the design of a U-tube manometer, it is essential to provide sufficient of the heavier liquid to ensure that it always occupies the bend of the tube. Notice that it is only necessary to consider the equalizing of pressure due to the liquids above the level of separation. When equating pressures it is convenient to work in terms of pressure heads, converting all heads to one specified liquid. For a U-tube manometer shown below

Ah1 h2

1

2

1

2 3

open

By starting at point A and work around to the open end. pA = p1 (pressure at equal elevations in a continuous mass of fluid at rest must be the same) As we move from point (1) to point (2), the pressure will increase by 1h1. Hence p2 = p1 + 1h1 = pA + 1h1 Also p3 = p2 (level of separation) = pA + 1h1 From point (3) to point (4), the pressure will decrease by 2h2.

p4 = p3 - 2h2 = pA + 1h1- 2h2

4

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At open end, pressure is zero when considering as gauge pressure. pA + 1h1 - 2h2 = 0 or pA = 2h2 - 1h1 A major advantage of the U-tube manometer lies in the fact that the gauge fluid can be different from the fluid in the container in which the pressure is to be determined. 22..55..33 BBoouurrddoonn PPrreessssuurree GGaauuggee High air pressures are more conveniently measured using a bourdon pressure gauge.

Pointer

Flattenedtube

The instrument consists of a hollow coil closed at one end and the other end being connected to the pressure being measured. When the internal pressure is greater than the outside pressure, the tube tends to straighten, causing the pointer to move. This gauge measures pressure relative to the pressure surrounding the tube, and therefore gives values of gauge pressure. One disadvantage of Bourdon gauge is that it is limited to the measurement of pressure that are static or only changing slowly. Because of the relatively large mass of the Bourdon tube it cannot respond to rapid changes in pressure.

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Worked examples: 3. In the following figure, determine the pressure of the water flowing in

the pipeline at point A based on the manometer reading shown. (S.G.Hg = 13.6)

Answer

A

60mm

40mm

Hg

B C

D

Let pressure at A be pA

pB = pA + (60+40)/1000*w (+ means going down) = pA + 0.1*w pC = pB (level of separation) = pA + 0.1*w pD = pC – 0.04*Hg = pA + 0.1*w – 0.04*Hg (-ve means going upward) Considering gauge pressure, pD = patm = 0 pA + (9.81 * 0.1) + (13.6*9.81* -0.04) = 0 pA = 4.36 kN/m2 or 4.36 kPa

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4 Determine the difference in pressure between pipeline A and pipeline B in the following figure.

A

B

10mm

30mm

50mmHg

GasolineSG=0.72

Ethylene glycolSG=1.1

C D

E

Answer Let pressure at A be pA

pC = pA + (30 + 50)/1000*eg (+ means going down) = pA + 0.08*eg pD = pC (level of separation) = pA + 0.08*eg pE = pD – 0.05*Hg = pA + 0.08*eg – 0.05*Hg (-ve means going upward) pB = pE – 0.04*gas = pA + 0.08*eg – 0.05*Hg – 0.04*gas pA + (1.1*9.81*0.08) - (13.6*9.81* 0.05) - (0.72*9.81*0.04) = pB or pA - pB = 6.09 kPa

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5. Water is flowing through a pipe device as shown below. The pressure between the two pipes is measured by a mercury manometer . Determine the pressure difference between point 1 and 2.

Answer Let the pressure at point 1 be p1. pA = p1 + (0.6-0.12)*w + 0.12*Hg = p1 + (0.6-0.12)*9.81 + 0.12*13.6*9.81 = p1 + 20.72 kPa pA = p2 + 0.6*w = p2 + 0.6*9.81 = p2 + 5.89 kPa Hence p1 + 20.72 = p2 + 5.89 p2 – p1 = 14.83 kPa

water

mercury

0.6m

0.12m

1 2

A A

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6. In the arrangement shown below, the two open-ended limbs are initially subjected to atmospheric pressure with the inside diameter of the tube equal to 7 mm, while the open-ended diameters at A and B are both 44 mm. Find the pressure difference between A and B in mm water head due to an increased pressure applied to side B if the surface of separation moves 100 mm. The oil has a specific gravity of 0.83.

z1z2

h

A

B

1 2 1 2

4 3

Patm

Patm PatmPatm +p

water

oil

initially finally Answer

Let the ratio of the areas of the enlarged ends A and B to the small-bore tube C be K.

Then K = (44/7)2 = 39.5 (A d2) Initially at the common level, p1 = p2, or wgz1 = ogz2, (1) or z1 = 0.83 z2. Finally at the new common level, p3 = p4,

or wg (z1 + h + K

h ) = og (z2 + h - K

h ) + p (2)

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(2) - (1)

wg (h + K

h ) = og (h - K

h ) + p

or h [(1 + 1

39 5. ) - 0.83 (1 - 1

39 5. )] = p

gw

and for h = 100 mm,

g

p

w = 100 * (1.0253 - 0.809) mm

= 21.63 mm water

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Class Exercise 2.1: A mercury manometer connects the entrance 1 and throat 2 of a Venturi meter. Find the head difference (h1 – h2) between the entrance and the thoat if a liquid of specific gravity, s is flowing through the meter. (h1 and h2 are measured as heads of the liquid in the meter)

1 2

A Ah

venturi meter

Hg

S.G. = s

3 45

6 7

z

[(136.

s– 1)*h]

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Class Exercise 2.2: For a gauge pressure at A of –10.89 kPa, find the specific gravity of the gauge liquid L in the figure below.

A3.2m

2.743m

3.048m

3.429m

Liquid LS.G.=1.6

air

B C

DE

F G

(0.994)

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Tutorial: Fluid Statics - Pressure 1 The tank in figure below is at 20C. If the pressure at level A is 200

kPa, determine the pressures at level B and C. 2. The tube in the figure below is filled with oil. Determine the pressure

at A and the pressure head at B in m of water.

3. For the inclined-tube manometer shown below, the pressure in pipe A

is 5kPa. The fluid in both pipes A and B is water and the gauge fluid in the manometer has a S.G. of 2.6. What is the pressure in pipe B corresponding to the differential reading shown?

3075mm

75mm

water

water

S.G.=2.6

200mmA

B

Water

Air 2m P=200kPa Air

6m

C

B

A

2m

Air

4m

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4. Vessels A and B contain water under pressure of 276 kPa and 138 kPa, respectively. What is the deflection of the mercury, h in the differential gauge?

h

A

B

4.877m

3.048m

5. Calculate the pressure difference between A and B for the setup shown

in the figure below.

6. Determine the pressure pA for the setup shown below if the S.G. of oil

is 0.8.

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7. In the system shown in figure below, the pressure gauge at A reads 2 kN/m2. Specific gravity, S.G. of oil is 0.8. Determine

(i) the length of the water column y in the open piezometer B, and (ii) the reading of the mercury, Hg column h in the U-tube

manometer.

A

air

oil

water

y

0.6m

0.5m

0.8m

h

Hg

B

8. For a gauge reading at A of -15 kPa, determine (a) the levels of the

liquids in the open piezometer columns E, F, and G and (b) the deflection of the mercury in the U-tube gauge in the figure below.

AE F G

L

C D

MN

Q

R

2m

4m

6m

9.5m

12.5m Air

S.G. 0.7

Water

S.G. 1.6

S.G. 13.6h1

h

END