δy θ Pressure is used to indicate the normal force per unit area at a given point acting on a given plane.

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1 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). 2.1 Pressure at a Point Pressure is used to indicate the normal force per unit area at a given point acting on a given plane. By considering the equilibrium of a small fluid element in the form of a triangular prism in the fluid subject to a pressure p x in x-direction, p y in y- direction and p s in normal to any plane inclined at an angle θ to the horizontal. δz ps δs px δy θ δx py Fig. 1 Equality of pressure in all directions at a point For simplicity, the forces in the z direction are not shown. P.2-1

2 The equation of motion in the x and y directions are respectively, ΣF x = p x δy δz - p s δz δs sinθ ΣF y = p y δx δz - p s δz δs cosθ - γ δxδyδz/2 By geometry, δx = δs cosθ; δy = δs sinθ Since the fluid element is in equilibrium, i.e. ΣF x = 0 & ΣF y = 0 p x δy δz - p s δy δz = 0 p x = p s and p y δx δz - p s δx δz -γ δxδyδz/2 = 0 p y - p s = γ δy/2 s δy approaches to zero, p z = p s Therefore, the pressure at a point in a fluid is the same in all direction - Pascal s Law. Fig. 2 General Case of Direction of Force of Pressure P.2-2

3 2.2 Pressure Variation in a Fluid with Depth Considering an element of vertical column of constant cross-sectional area and totally surrounded by the same fluid of mass density ρ. rea, p+dp Fluid density ρ h+dh h p Fig. 3 Vertical variation of pressure Suppose Pressure at h = p Pressure at h + δh = p + δp (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 acting up Force due to p + δp on area acting down Force due to the weight of element = p = (p + δp) = ρg(δh) p - (p + δp) - ρg(δh) = 0 δp δh = -ρg or dp dh = -ρg Thus, in any fluid under gravitational attraction, pressure decreases with increase of height h. P.2-3

4 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 will be in equilibrium. The forces acting on this element horizontally are P 1 at P and P 2 at Q. rea Fluid density ρ p1 p2 mg Fig. 4 Equality of pressure at the same level 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. p 1 = p 2 p 1 = p 2 Thus the pressure at any two points at the same level in a body of fluid at rest will be the same. 2.4 Pressure and Head In a fluid of constant density, dp = -ρg can be integrated immediately to dh give p = -ρgh + 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 P.2-4

5 Patm h Liquid Density ρ p Fig. 5 Pressure and head p = ρgh + constant and since the pressure at the free surface will normally be atmospheric pressure p atm (i.e. at h = 0, p = p atm ), p = ρgh + p atm It is often convenient to take atmospheric pressure as a datum. Pressure measured above atmospheric pressure are known as gauge pressure. Pressure measured above perfect vacuum are called absolute pressure. bsolute pressure = Gauge pressure + tmospheric pressure Gauge pressure tmospheric Pressure B Gauge pressure B bsolute pressure B Barometer reading bsolute pressure Vaccum Fig. 6 Relationship between Pressures 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. P.2-5

6 If the pressure is between atmospheric pressure and absolute zero, it is called partial vacuum. By considering the gauge pressure only, p = ρgh which indicates that, if g is assumed constant, the pressure increases linearly with depth. The gauge pressure at a point can be defined by stating the vertical height h, called the head, of a column of a given fluid of mass density ρ. i.e. h = p ρg m Note that when pressure is expressed as head, it is essential that the mass density ρ is specified. P.2-6

7 Worked examples: 1. Calculate the pressure at a point on the sea bed 1 km deep. The density of sea water is 1025 kg/m 3. nswer ρ = 1025 kg/m 3 g = 9.81 m/s 2 h = 1000 m (pressure head) Since p = ρgh = 1025 * 9.81 * 1000 = 10,055,000 N/m 2 = bar (1 bar = 10 5 N/m 2 ) 2 The pressure at a point on the sea bed is 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? nswer (a) ρ water = 1000 kg/m 3 Since p = ρgh x 10 5 = 1000 * 9.81 * h h = 1025 m, i.e m head of water. (b) ρ Hg = 13.6 * 1000 kg/m x 10 5 = 13.6 * 1000 * 9.81 * h h = m of mercury. P.2-7

8 2.5 Measurement of Pressure Many instruments for pressure measurement use the fact that a pressure is equivalent to a head of liquid Piezometer 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 open open Pressure h Vaccum h p = γh vacuum = γh or pressure = -γh Fig. 7 Piezometers 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 U-Tube Manometer 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. P.2-8

9 open open 13.6h water h Level of separation Heavier liquid Hg (S.G. = 13.6) Fig. 8 Manometers 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. γ1 open 1 h1 2 3 γ2 h2 By starting at point and work around to the open end. p = p 1 (pressure at equal elevations in a continuous mass of fluid at rest must be the same) P.2-9

10 s we move from point (1) to point (2) the pressure will increase by γ 1 h 1. lso p 2 = p 3 t open end, pressure is zero. Therefore the pressure decreases by γ 2 h 2. p + γ 1 h 1 - γ 2 h 2 = 0 or p = γ 2 h 2 - γ 1 h 1 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 Bourdon Pressure Gauge High air pressures are more conveniently measured using a bourdon pressure gauge. Pointer Flattened tube Fig. 9 Bourdon Pressure Gauge 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. P.2-10

11 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. Worked examples: 1 In the following figure, determine the pressure of the water flowing in the pipeline at point based on the manometer reading shown. (S.G. Hg = 13.6) nswer 60mm B D C 40mm Hg Let pressure at be p p B = p + (60+40)/1000*γ w = p + 0.1*γ w p C = p B = p + 0.1*γ w (+ means going down) (level of separation) p D = p C 0.04*γ Hg = p + 0.1*γ w 0.04*γ Hg (-ve means going upward) Considering gauge pressure, p D = p atm = 0 p + (9.81 * 0.1) + (13.6*9.81* -0.04) = 0 p = 4.36 kn/m 2 or 4.36 kpa P.2-11

12 2 Determine the difference in pressure between pipeline and pipeline B in the following figure. 10mm 30mm Ethylene glycol SG=1.1 Gasoline SG=0.72 B 50mm C E D Hg nswer Let pressure at be p p C = p + ( )/1000*γ eg (+ means going down) = p *γ eg p D = p C = p *γ eg (level of separation) p E = p D 0.05*γ Hg = p *γ eg 0.05*γ Hg (-ve means going upward) p B = p E 0.04*γ gas = p *γ eg 0.05*γ Hg 0.04*γ gas p + (1.1*9.81*0.08) - (13.6*9.81* 0.05) - (0.72*9.81*0.04) = p B or p - p B = 6.09 kpa P.2-12

13 3 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 water 0.12m 0.6m mercury nswer Let the pressure at point 1 be p 1. p = p 1 + ( )*γ w *γ Hg = p 1 + ( )* *13.6*9.81 = p kpa p = p *γ w = p *9.81 = p kpa Hence p = p p 2 p 1 = kpa P.2-13

14 Class Exercise 2.1: mercury manometer connects the entrance 1 and throat 2 of a Venturi meter. Find the head difference (h 1 h 2 ) between the entrance and the thoat if a liquid of specific gravity, s is flowing through the meter. (h 1 and h 2 are measured as heads of the liquid in the meter) z Hg 1 2 S.G. = s 7 venturi meter h [( s 1)*h] P.2-14

15 Class Exercise 2.2: For a gauge pressure at of kpa, find the specific gravity of the gauge liquid L in the figure below. 3.2m air 3.429m E D F 2.743m B C S.G.= m G Liquid L (0.994) P.2-15

16 Tutorial: Fluid Statics - Pressure 1 For the inclined-tube manometer shown below, the pressure in pipe is 5kPa. The fluid in both pipes 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? water 75mm water 200mm 30 S.G.=2.6 B 75mm 2 Vessels 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? 4.877m h B 3.048m 3 For a gauge reading at 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. E F G 12.5m 9.5m 6m 4m ir S.G. 0.7 Water S.G. 1.6 h L N M Q R 2m C D h1 S.G P.2-16