Fluid Engineering Mechanics

Transcription

1 Fluid Engineering Mechanics Chapter 3 Fluid Statics: ressure intensity and pressure head: pressure and specific weight relationship, absolute and gauge pressure, Forces on submerged planes & curved surfaces and their applications, drag and lift forces, uoyancy and Floatation Dr. Muhammad shraf Javid ssistant rofessor Department of Civil and Environmental Engineering 1

2 Fluid Statics Fluid Statics: It is the branch of fluid mechanics that deals with the behavior/response of fluid when they are at rest. ressure, (average pressure intensity): It is the normal force exerted per unit area. It is denoted by and is given by; Units SI: N/m 2 (called ascal) force area G: lb/ft 2 or lb/in 2 (called psi) F CGS: dyne/cm bar=10 5 N/m 2 =10 5 ascal

3 ressure vs Water depth/height Consider a strip or column of a cylindrical fluid, h= height or depth of strip of fluid γ= specific weight of fluid d=cross-sectional area of strip dv=volume of strip dw=weight of strip h ressure at base of strip=df/d=dw/d = γdv/d = γd.h/d =γh 3

4 ressure vs Water depth/height =γh α h h For h=0, =0 For h=h, =γh ressure distribution diagram/pressure profile s you know atmospheric pressure reduces, as we move to higher elevations. Is it because of h, as h reduces, also reduces. 4

5 SCL S LW ressure at any point in fluid is same in all directions when the fluid is at rest dz y 5 z x Consider a wedge shape element of fluid having dimension dx, dy and dz along x, y and z axis. di= dimension of inclined plane making an angle α with the vertical x, y, z and are pressure acting in x, y, z and perpendicular to inclined surface dw=weight of the element dy y x(dydz) α z x dx (dldz) sinα y(dxdz) (dldz) α α (dldz) cosα dw

6 SCL S LW 6 y(dxdz) x(dydz) (dldz) α α (dldz) cosα (dldz) sinα o dy dz dydz dl dy o dl dy dldz dydz o dldz dydz F x x x x x ) ( / cos / ) ( )cos ( 0 dw o dx dz dxdz dw dl dx o dl dx dldz dxdz o dldz dw dxdz F y y y y y ) ( 0 & / sin / ) ( )sin ( 0

7 SCL S LW Similarly by applying the conditions in z direction, it can be proved that Hence, z x y z The above states that the pressure acting on fluid particle is same in all directions when the fluid is at rest. 7

8 bsolute and Gauge ressure tmospheric pressure Gauge pressure Vacuum/negative pressure bsolute pressure tmospheric pressure: ressure exerted by atmosphere Gauge pressure: ressure more than atmospheric pressure Vacuum/negative pressure: ressure less than atmospheric pressure bsolute pressure: ressure measure relative to absolute zero 8 abs abs atm atm g vac

9 tmospheric ressure It is defined as weight of air per unit surface area of earth. It decreases with increase in elevation w.r.t. surface of earth. Standard atmospheric pressure at mean-sea-level (MSL) is =101.3KN/m 2 =1.013bar =14.7psi =760mm of Hg =33.9ft of water =10.3m of water 9

10 Measurement of tmospheric ressure arometer: It is device used to measure the atmospheric pressure at any point on the earth. There are two types of barometer (i) Liquid barometer It measures the pressure with help of column of liquid (ii) neroid barometer It measures atmospheric pressure by its action on an elastic lid of evacuated box. 10

11 Liquid arometer It consists of a transparent tube which is open from one end only. The tube is filled with liquid and is inserted in a jar also containing same liquid. The liquid initially drops in tube due to gravity but stabilizes at certain level under the action of atmospheric forces. The atmospheric pressure is then measured as height of liquid at which it stabilizes. Three forces acting on fluid are atm()=force of atmospheric ressure W=Weight of liquid vap()= Force of vapour pressure =Cross-sectional area of tube Liquid/ Vapor pressure Weight of liquid Force of atm 11

12 Liquid arometer Three forces acting on fluid are atm()=force of atmospheric ressure vap. W=Weight of liquid vap()= Force of vapour pressure =Cross-sectional area of tube For Equilibrium Liquid/ h W Fy o; W 0 atm vap atm. atm h vap 0 W h h atm vap 12

13 Liquid arometer Generally, mercury is preferred liquid because its vapour pressure is minimum. Moreover, its specific gravity is very high so that size (height) of barometer required is small. However, for other liquid vapour pressure must be considered in estimation. The barometer using mercury is called mercury barometer and while using water is called water barometer. Size of barometer tube should be more than ½ inches (or 13mm) to avoid capillarity. 13

14 bsolute ressure Gauge ressure (g): It is the pressure measured relative to atmospheric pressure ( atm ) and is always above the atmospheric pressure It may be defined as normal compressive force per unit area Vacuum ressure ( vac ): It is the pressure measured relative to atmospheric pressure and is less than the atmospheric pressure It may be defined as normal tensile force per unit area bsolute ressure( abs ): It is the pressure measured from absolute zero abs abs atm atm g vac 14

15 roblem Q h kN / m 2 γ=10.05kn/m 3 Surface 4600m 15

16 roblem Q

17 roblem Q

18 roblem Q oil 57.9 psi h S h w 68/(12 2 ) 68ft 1=32psi Stream 2=? S=0.88 γ=sγ w 18

19 roblem Q i b b h h kN / m h (9.81)5 65.1kN / m Surface γ=8 kn/m 3 2m i interface 5m γ=10 kn/m 3 b 19

20 roblem Q Determine the depth of gaseous atmosphere to cause 101.3kN/m 2 over surface of earth? h h 8442m h γ=12n/m 3 h=? =101.3ka Surface 20

21 roblem 21 psia psia ft H O vap atm Weight of liquid Force of atm Vapor ressure lcohol w vap atm vap atm vap atm vap atm vap atm S h h h W o Fy 0 0 ; h=?

22 Measurement of ressure The following devices are used for pressure measurement 1. iezo-meter 2. Manometer a) Simple manometer ) Differential manometer 3. Mechanical ressure Transducer (ourden gauge) 4. Electrical ressure Transducer 22

23 1. iezometer 23 It is used to measure pressure in pipes or vessels. In it simplest form, it consists of a transparent tube open from other ends The diameter of tube should > ½ to avoid capillarity action iezometers may be connected to sides or bottom of pipe to avoid eddies that are produced in the top region of pipe Limitations: It must only be used for liquids It should not be used for high pressure It cannot measure vacuum (-ve) pressure When connected to pipes, the water level rises in it which gives a measure of pressure.

24 2. Manometer a). Simple Manometer Figure shows a set up of simple manometer. It consists of a U-shaped tube, part of which is filled with manometric fluid. One end of tube is connected with the pipe whose pressure is required to be determined. Due to pressure, level of manometric fluid rises on one side while it falls on other side. The difference in levels is measured to estimate the pressure. z Fluid, γ f Manometric fluid, γ m Y=Manometric reading γ f =Specific weight of fluid in pipe γ m =Specific weight of manometric fluid Y 24

25 2. Manometer Manometric Fluids 1. Mercury 2. Oils 3. Salt solution etc roperties of manometric Fluid z Fluid, Y 1. Manometric fluid should not be soluble/intermixalbe with fluid flowing in pipe whose pressure is required to be determined. Manometric fluid, 2. Lighter fluid should be used if more precision is required. 25

26 2. Manometer ressure measurement abs abs f Z Y atm m atm z Fluid, γ f Y atm Z Y 0 f Y m Z It is also equation of gauge pressure m f Manometric fluid, γ m Sign Convention -ve: upward direction 26 +ve: downward direction

27 2. Manometer b). Differential manometer It is used to measure difference of pressure. Case 1: when two vessels/pipes are at same level Fluid, γ Z Z Fluid, γ Y? Z Y m Z Y m Z Z Manometric Fluid, γ m 27

28 2. Manometer 28 Y Y Y Z Z Y Z Y Z if f m f m f m f m f f

29 2. Manometer b). Differential Manometer Case 1I: when two vessels/pipes are at different level Z Fluid, γ Y Fluid, γ Z Z Z Y m Y m Z Z Manometric Fluid, γ m 29

30 2. Manometer 30 f m f m f f Z Z Y Z Y Z if

31 2. Manometer b). Differential manometer Case 1II: when manometer is inverted In this case, lighter fluid should be used as manometeric fluid. Y Manometric Fluid, γ m Fluid, γ Z Fluid, γ Z 31 Z Z Y m Y m Z Z

32 2. Manometer 32 b). Differential manometer Case 1II: when manometer is inverted m m Z Y Z Z Y Z f m f m f f f m f m Z Z Y Z Y Z if Z Y Z Z Y Z In this case, lighter fluid should be used as manometeric fluid.

33 dvantages and Limitation of Manometers dvantages Limitations Easy to fabricate Usually bulky and large in size Less expansive eing fragile, get broken easily Good accuracy High sensitivity Require little maintenance Not affected by vibration Specially suitable for low pressure and low differential pressure Reading of manometer is get affected by temperature, altitude and gravity Capillary action is created due to surface action Meniscus has to be measured accurately for better accuracy. Easy to change sensitivity by changing manometric fluid 33

34 3. Mechanical ressure Transducer Transducer is a device which is used to transfer energy from one system to other Mechanical pressure transducer converts pressure system to displacement in mechanical measuring system ourden Gauge is used to measure high pressure either positive or negative. It gives pressure directly in psi or ascal units 34

35 ourden Gauge The essential mechanical element in this gage is the hollow, elastic curved tube which is connected to the pressure source as shown in Fig. s the pressure within the tube increases the tube tends to straighten, and although the deformation is small, it can be translated into the motion of a pointer on a dial as illustrated. 35 Fig. ourden gauge

36 ourden Gauge Since it is the difference in pressure between the outside of the tube and the inside of the tube that causes the movement of the tube, the indicated pressure is gage pressure. The ourdon gage must be calibrated so that the dial reading can directly indicate the pressure in suitable units such as psi, psf, or pascals. zero reading on the gage indicates that the measured pressure is equal to the local atmospheric pressure. This type of gage can be used to measure a negative gage pressure (vacuum) as well as positive pressures. 36

37 3. Mechanical ressure Transducer Elevation Correction ourden gauge gives pressure at the center of dial. So to calculate pressure at point, g z g Where (γ)z=elevation correction z 37

38 4. Electrical ressure Transducer It converts displacement of mechanical measuring system to an electrical signal. Its gives continuous record of pressure when converted to a strip chart recorder. Data can be displayed using computer data acquisition system. 38

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45 45 Forces on Immersed odies

46 Forces on Immersed odies Hydrostatic Force: It is the resultant force of pressure exerted by liquid at rest on any side of submerged body. F pd p d p It is the summation of product of uniform pressures and elementary areas of submerged body It is equal to the product of submerged area and pressure at the centroid of the submerged area (to be discussed later) 46

47 Forces on lane rea Center of pressure The point of application of resultant force of pressure on a submerged area is called center of pressure. 47

48 Forces on lane rea = total submerged area F = hydrostatic force θ =angle of submerged plane with free surface h c =depth of center of area h p =depth of center of pressure rojection of area of vertical plane y c =inclined depth to center of area y p =inclined depth to center of pressure d=elementary area df=force of pressure (hydrostatic force) on elementary area 48 df p d hd

49 Forces on lane rea Lets choose an elementary area so that pressure over it is uniform. Such an element is horizontal strip, of width, x so d xdy. The pressure force, df on the horizontal strip is Integrating df p d hd p h h y sin df pd hd F y sin d sin sin y c yd y c yd Where, y c is the distance along the sloping plane to the centroid, C, of the area. If h c is the vertical depth to the centroid, then we have; 49 F h c

50 Forces on lane rea F h c Thus, we find the total force on any plane area submerged in a liquid by multiplying the specific weight of the liquid by the product of the area and the depth of its centroid. The value of F is independent of the angle of inclination of the plane as long as the depth of its centroid is unchanged. Since h c is pressure at the centroid, we can also say that total pressure force on any plane area submerged in a liquid is the product of the area and the pressure at the centroid. 50

51 Center of ressure In order to determine location of center of pressure, y p, from OX, let s take the moment of elementary area around OX dm ydf y pd yhd yy sind Integrating ydf F * y p y sin 2 y sind sin y d I F * y p ydf Where, I is the 2 nd moment of submerged area about OX y p sin F I c I sin y sin I y c F sin y c 51 Where y c is called static moment of area

52 Center of ressure 52 Now, according to parallel axis theorem, Where, I c is 2 nd moment of area about centroidal axis. From this equation we again see that the location of center of pressure,, is independent of the angle θ. When the plane is truly vertical, i.e., θ=90 o It is concluded that center of pressure is always below the center of area except when the plane is horizontal. When the plane is horizontal center of pressure and center of area lies at the same time. y y I y I y c c c c p 2 c I c y I 2 y I y y c c c p h I h h c c c p

53 Lateral osition of Center of ressure x p I xy y c When at least one centroidal axis is the axis of symmetry then I xy becomes equal to zero. i.e. x p = 0 It means center of pressure is lying on the symmetrical axis, just below the center of area. 53

54 Hydrostatic force formulas F h F sin y c c h p h c Ic h c y p y c I y c c 54

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56 roblem 56 plane surface is circular with a diameter of 2m. If it is vertical and the top edge is 0.5m below the water surface, find the magnitude of the force on one side and the depth of center of pressure. Solution: D h c m 2 F F F h h h p p p h c kN h c 1.5 Ic h c 1.667m D / 64 / 1.5D / 4 0.5m D=2m Free surface Ic hp hc hc Ic y p yc yc F hc F sin y c

57 roblem rectangular plate 1.5 m by 1.8 m is at an angle of 30 o with the horizontal, and the 1.5 m side is horizontal. Find the magnitude of force on one side of the plate and the depth of its center of pressure when the top edge is (a) at the water surface (b) 0.3m below water surface (a) h ysin hp hc 30 o yc yp 1.8 ft 4ft 1.5 m h o y sin 2sin 30 ft F hc lb c c bd /12/ 2bd 2. ft Ic y p yc 2 67 y c h p ft

58 roblem (b) 0.3 m h yc h c ysin / sin hp hc y c yp 1.8 m 1.5 m h o 1 2sin 1 2sin 30 ft F hc lb c bd /12/ 4bd 4. ft Ic y p y c o hp yp sin 4.33sin ft

59 Forces on Curved Surface Horizontal force on curved surface F x Fx' F' hydrostati c force on equivalent Vertical force on curved surface F z horizontalarea Fz' W Weight of volume of liquid above surface 59

60 Forces on Curved Surface Resultant Force F 2 F x F z 2 tan 1 F F z x 60

61 Drag and Lift Force Lift is the component of aerodynamic force perpendicular to the relative wind. Drag is the component of aerodynamic force parallel to the relative wind. Weight is the force directed downward from the center of mass of the airplane towards the center of the earth. Thrust is the force produced by the engine. It is directed forward along the axis of the engine. 61

62 Drag and Lift Force The drag force acts in a direction that is opposite of the relative flow velocity. ffected by cross-section area (form drag) ffected by surface smoothness (surface drag) The lift force acts in a direction that is perpendicular to the relative flow. C D = Coefficient of drag C L = Coefficient of lift =projected area of body normal to flow V= relative wind velocity 62

63 roblem =1.2m =1.8m =.45m a = 0.45 m, d = 1.8 m, and b = 1.2 m 63

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65 65 uoyancy and Floatation

66 uoyancy and Floatation When a body is immersed fully or partially in a fluid, it is subjected to an upward force which tends to lift (buoy) it up. The tendency of immersed body to be lifted up in the fluid due to an upward force opposite to action of gravity is known as buoyancy. The force tending to lift up the body under such conditions is known as buoyant force or force of buoyancy or up-thrust. The magnitude of the buoyant force can be determined by rchimedes principle which states 66 When a body is immersed in a fluid either fully or partially, it is buoyed or lifted up by a force which is equal to the weight of fluid displaced by the body

67 uoyancy and Floatation Lets consider a body submerged in water as shown in figure. F h 1 1 Water surface 1 h 1 67 The force of buoyancy resultant upward force or thrust exerted by fluid on submerged body is given F F F F F 2 F 1 h1 h2 d h1 h d 2 volume d h 2 F 2 d=rea of cross-section of element γ= Specific weight of liquid 2 h1 h2

68 uoyancy and Floatation F volume =Weight of volume of liquid displaced by the body (rchimedes's rinciple) Force of buoyancy can also be determined as difference of weight of a body in air and in liquid. Let W a = weight of body in air W l =weight of body in liquid F =W a -W l 68

69 uoyancy and Floatation Center of uoyancy (): The point of application of the force of buoyancy on the body is known as the center of buoyancy. It is always the center of gravity of the volume of fluid displaced. Water surface CG or G C or 69 CG or G= Center of gravity of body C or = Centroid of volume of liquid displaced by body

70 Types of equilibrium of Floating odies Stable Equilibrium: If a body returns back to its original position due to internal forces from small angular displacement, by some external force, then it is said to be in stable equilibrium. Note: Center of gravity of the volume (centroid) of fluid displaced is also the center of buoyancy. 70

71 Types of equilibrium of Floating odies Unstable Equilibrium: If the body does not return back to its original position from the slightly displaced angular displacement and heels farther away, then it is said to be in unstable equilibrium Note: Center of gravity of the volume (centroid) of fluid displaced is also center of buoyancy. 71

72 Types of equilibrium of Floating odies Neutral Equilibrium: If a body, when given a small angular displacement, occupies new position and remains at rest in this new position, it is said to be in neutral equilibrium. W CG C F Note: Center of gravity of the volume (centroid) of fluid displaced is also center of buoyancy. 72

73 Metacenter and Metacentric Height Center of uoyancy () The point of application of the force of buoyancy on the body is known as the center of buoyancy. Metacenter (M): The point about which a body in stable equilibrium start to oscillate when given a small angular displacement is called metacenter. It may also be defined as point of intersection of the axis of body passing through center of gravity (CG or G) and original center of buoyancy () and a vertical line passing through the center of buoyancy ( ) of tilted position of body. 73 F F

74 Metacenter and Metacentric Height Metacentric height (GM): The distance between the center of gravity (G) of floating body and the metacenter (M) is called metacentric height. (i.e., distance GM shown in fig) GM=M-G F 74

75 Condition of Stability For Stable Equilibrium osition of metacenter (M) is above of center of gravity (G) For Unstable Equilibrium osition of metacenter (M) is below of center of gravity (G) For Neutral Equilibrium osition of metacenter (M) coincides center of gravity (G) Restoring moment Overturning moment 75

76 Determination of Metacentric height The metacentric height may be determined by the following two methods 1. nalytical method 2. Experimental method 76

77 Determination of Metacentric height In Figure shown C is the original waterline plane and the center of buoyancy in the equilibrium position. When the vessel is tilted through small angle θ, the center of buoyancy will move to as a result of the alteration in the shape of displaced fluid. F C is the waterline plane in the displaced position. 77

78 Determination of Metacentric height 78 To find the metacentric height GM, consider a small area in plan d at a distance x from O. The height of elementary area is given by xθ. Therefore, volume of the elementary area becomes dv x d The upward force of buoyancy on this elementary area is then df x d Moment of df generated by element about an axis passing through O is given by; x. df x x d x x. df I 2 d rism OC C has gone inside the liquid and O has come out from liquid. There is an increase in force of buoyancy on right side and decrease in left side. df = gain or loss in F = Couple formed by triangular prism d x F

79 F F Determination of Metacentric height for Moment developed by F due to shift from to is ' V ' V M Sin M small angle and stable equilibrium,sin For equilibrium, Moment developed by F due to shift from to = Couple formed by triangular prism I M GM V I V M M G F 79

80 NUMERICLS Determine the volume of an object that weights 22 N in water and 30 N in oil (s=0.82). What is the specific weight of the object? Solution: For water F = W in air W water γ w (Vol.) displaced = W in air W water γ w (Vol.) object = W in air 22 (1) For Oil F = W in air W oil γ oil (Vol.) displaced = W in air W oil γ oil (Vol.) object = W in air 30 (II) y subtracting (II) from 1 (Vol.) object = 4.53 x 10-3 m 3 W in air = N γ object = N/m3 80

81 NUMERICLS Q. 1 wooden block of specific gravity 0.75 floats in water. If the size of block is 1mx0.5mx0.4m, find its meta centric height Solution: Given Data: Size of wooden block= 1mx0.5mx0.4m, Specific gravity of wood=0.75 Specific weight of wood=0.75(9.81)=7.36kn/m 3 Weight of wooden block=(specific weight)x(volume) Weight of wooden block=7.36(1x0.5x0.4)=1.47kn Let h is depth of immersion=? For equilibrium Weight of water displaced = weight of wooden block 9.81(1x0.5xh)= >> h=0.3m 0.4m 0.5m 1m h

82 NUMERICLS Distance of center of buoyancy=o=0.3/2=0.15m Distance of center of gravity=og=0.4/2=0.2m Now; G=OG-OM= =0.05m lso; M=I/V I=moment of inertia of rectangular section I=(1)x(0.5) 3 /12= m 4 V=volume of water displaced by wooden block V=(1)x(0.5)x(0.3)=0.15m3 M=I/V=0.0104/0.15=0.069m Therefore, meta centric height=gm=m-g GM= =0.019m 0.4m 0.5m G O 1m h m

83 NUMERICLS Q 2. solid cylinder 2m in diameter and 2m high is floating in water with its axis vertical. If the specific gravity of the material of cylinder is 0.65, find its meta-centric height. State also whether the equilibrium is stable or unstable. Solution: Given Data: 2m Size of solid cylinder= 2m dia, & 2m height Specific gravity solid cylinder=0.65 Let h is depth of immersion=? For equilibrium Weight of water displaced = weight of wooden block 9.81(π/4(2) 2 (h))=9.81(0.65).(π/4(2) 2 (2)) h=0.65(2)=1.3m 2m G 1.3m 83 O

84 NUMERICLS Center of buoyancy from O=O=1.3/2=0.65m Center of gravity from O=OG=2/2=1m G=1-0.65=0.35m lso; M=I/V Moment of inertia=i=(π/64)(2) 4 =0.785m 4 Volume displaced=v=(π/4)(2) 2 (1.3)=4.084m 3 M=I/V=0.192m GM=M-G= =-0.158m -ve sign indicate that the metacenter (M) is below the center of gravity (G), therefore, the cylinder is in unstable equilibrium 2m 2m G 1.3m 84 O