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Fluid statics

1. Fluid Statics: Pressure intensity and pressure head: pressure and specific weight relationship, absolute and gauge pressure Dr. Mohsin Siddique Assistant Professor 1 Fluid Mechanics
2. Fluid Statics 2 Fluid Statics: It is the branch of fluid mechanics that deals with the behavior/response of fluid when they are at rest. Pressure, (average pressure intensity): It is the normal force exerted per unit area. It is denoted by P and is given by; Units SI: N/m2 (called Pascal) BG: lb/ft2 or lb/in2 (called psi) CGS: dyne/cm2 1 bar=105N/m2=105Pascal A F area force P ==
3. Pressure vs Water depth/height 3 Consider a strip or column of a cylindrical fluid, h= height or depth of strip of fluid γ= specific weight of fluid dA=cross-sectional area of strip dV=volume of strip dW=weight of strip Pressure at base of strip=dF/dA=dW/dA P= γdV/dA P= γdA.h/dA P=γh h
4. Pressure vs Water depth/height 4 P=γh P α h For h=0, P=0 For h=h, P=γh h Pressure distribution diagram/pressure profile As you know atmospheric pressure reduces, as we move to higher elevations. Is it because of h, as h reduces, P also reduces.
5. PASCAL’S LAW 5 “Pressure at any point in fluid is same in all directions when the fluid is at rest” Consider a wedge shape element of fluid having dimension dx, dy and dz along x, y and z axis. dl= dimension of inclined plane making an angle α with the vertical Px, Py, Pz and P are pressure acting in x, y, z and perpendicular to inclined surface dW=weight of element zy x Py(dxdz) Px(dydz) P(dldz) dx dy dz α α α P(dldz) cosα P(dldz) sinα dW zy x
6. PASCAL’S LAW 6 Py(dxdz) Px(dydz) P(dldz) α α P(dldz) cosα P(dldz) sinα PP odydzPdydzP dldyodldydldzPdydzP odldzPdydzP F x x x x x = =− ==− =− =∑ )( /cos/)( cos)( 0 α α Q dW PP odxdzPdxdzP dwdldxodldxdldzPdxdzP odldzPdWdxdzP F y y y y y = =− ≅==− =−− =∑ )( 0&/sin/)( sin)( 0 α α Q
7. PASCAL’S LAW 7 Similarly by applying the conditions in z direction, it can be proved that Hence, The above states that the pressure acting on fluid particle is same in all directions when the fluid is at rest. PPz = PPPP zyx ===
8. Absolute and Gauge Pressure 8 Atmospheric pressure Gauge pressure Vacuum/negative pressure Absolute pressure Atmospheric pressure: Pressure exerted by atmosphere Gauge pressure: Pressure more than atmospheric pressure Vacuum/negative pressure: Pressure less than atmospheric pressure Absolute pressure: Pressure measure relative to absolute zero vacatmabs gatmabs PPP PPP −= +=
9. Atmospheric Pressure 9 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 is =101.3KN/m2 =1.013bar =14.7psi =760mm of Hg =33.9ft of water =10.3m of water
10. Measurement of Atmospheric Pressure 10 Barometer: 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) Aneroid barometer It measures atmospheric pressure by its action on an elastic lid of evacuated box.
11. Liquid Barometer 11 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 Patm(A)=Force of atmospheric Pressure W=Weight of liquid Pvap(A)= Force of vapour pressure A=Cross-sectional area of tube Weight of liquid Force of Patm Vacuum Pressure Liquid/
12. Liquid Barometer 12 Three forces acting on fluid are Patm(A)=Force of atmospheric Pressure W=Weight of liquid Pvap(A)= Force of vapour pressure A=Cross-sectional area of tube For Equilibrium W PatmA PvapA Liquid/ h 0; =−−=∑ APWAPoFy vapatm 0=−− APAhAP vapatm γ AhW γ=Q vapatm PhP += γ
13. Liquid Barometer 13 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.
14. Absolute Pressure 14 Gauge Pressure (Pg): It is the pressure measured relative to atmospheric pressure (Patm) and is always above the atmospheric pressure It may be defined as normal compressive force per unit area Vacuum Pressure (Pvac): 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 Absolute Pressure(Pabs): It is the pressure measured from absolute zero vacatmabs gatmabs PPP PPP −= +=
15. Problem 15 Q.3.2.2 4600m Surface γ=10.05kN/m3 ( ) 2 /46700 460005.10 mkN P hP = = γ
16. Problem 16 Q.3.2.4 ( ) psi P hSPhPP woil 9.57 )12/(684.6288.032 2 2 112 += +=+= γγ P2=? P1=32psi Stream 68ft S=0.88 γ=Sγw
17. Problem 17
18. Problem 18 Q.3.3.1 5m Surface γ=10 kN/m3 ( ) 2 /1628 mkNhPi === γ γ=8 kN/m3 2m interface Pi Pb ( ) 2 2211 /1.655)81.9(28 mkNP hhP b b =+= += γγ
19. NUMERICALS 19 Q.3.3.3 h=4600γ=10.05 kN/m3 2 /______4600)05.10( mkNP hP == = γ P=?
20. Problem 20 Q.3.3.3 h=? Surface γ=12N/m3 ( ) ( ) mh h hP 8442 1210003.101 = =× = γ P=101.3kPa Determine the depth of gaseous atmosphere to cause 101.3kN/m2 over surface of earth?
21. Problem 21 3.5.1 psiaP psiaOHftP vap atm 09.2 47.144.33 2 = == Weight of liquid Force of Patm Vacuum Pressure Alcohol w vapatm vapatmvapatm vapatm vapatm S PP h PP A APAP h APAhAP APWAP oFy γ γγ γ − = − = − = =−− =−− =∑ 0 0 ; h=?
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23. Measurement of Pressure 23 The following devices are used for pressure measurement 1. Piezometer 2. Manometer a) Simple manometer B) Differential manometer 3. Mechanical PressureTransducer (Bourden gauge) 4. Electrical PressureTransducer
24. 1. Piezometer 24 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 Piezometers 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.
25. 2. Manometer 25 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. A Fluid, γf Manometric fluid, γm Y z Y=Manometric reading γf =Specific weight of fluid in pipe γm =Specific weight of manometric fluid
26. 2. Manometer 26 Manometric Fluids 1. Mercury 2. Oils 3. Salt solution etc Properties of manometric Fluid 1. Manometric fluid should not be soluble/intermixale with fluid flowing in pipe whose pressure is required to be determined. 2. Lighter fluid should be used if more precision is required. A Fluid, Manometric fluid, y z
27. 2. Manometer 27 Pressure measurement A Fluid, γf Manometric fluid, γm Y z Sign Convention -ve: upward direction +ve: downward direction AatmAabs atmmfAabs PPP PYZP += =−+ Q γγ Patm ZYP YZP fmA mfA γγ γγ −= =−+ 0 PA Above is a gauge pressure equation
28. 2. Manometer 28 b). Differential manometer It is used to measure difference of pressure. Case 1: when two vessels/pipes are at same level ZA ZB Y A B PA PB Fluid A, γA Fluid B, γB Manometric Fluid , γm AAmBBBA AAmBBBA ZYZPP ZYZPP γγγ γγγ −++=− −++=
29. 2. Manometer 29 ( ) ( ) ( )( )YPP YYPP ZZYPP ZYZPP if fmBA fmBA BAfmBA AfmBfBA fAB γγ γγ γγ γγγ γγγ −=− −=− −−=− −++=− ==
30. 2. Manometer 30 b). Differential Manometer Case 1I: when two vessels/pipes are at different level ZA ZB Y A B PA PB Fluid A, γA Fluid B, γB Manometric Fluid , γm AAmBBBA AAmBBBA ZYZPP ZYZPP γγγ γγγ −++=− −++=
31. 2. Manometer 31 ( )BAfmBA AfmBfBA fAB ZZYPP ZYZPP if −−=− −++=− == γγ γγγ γγγ
32. 2. Manometer 32 b). Differential manometer Case 1II: when manometer is inverted ZA ZB Y A B PA PB Fluid A, γA Fluid B, γB Manometric Fluid , γm AAmBBBA AAmBBBA ZYZPP ZYZPP γγγ γγγ +−−=− +−−=
33. 2. Manometer 33 ( )BAfmBA AfmBfBA fAB ZZYPP ZYZPP if −+−=− +−−=− == γγ γγγ γγγ
34. Advantages and Limitation of Manometers 34 Advantages Easy to fabricate Less expansive Good accuracy High sensitivity Require little maintenance Not affected by vibration Specially suitable for low pressure and low differential pressure Easy to change sensitivity by changing manometric fluid Limitations Usually bulky and large in size Being fragile, get broken easily 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.
35. 3. Mechanical Pressure Transducer 35 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 Bourden Gauge is used to measure high pressure either positive or negative. It gives pressure directly in psi of Pascal units
36. Bourden Gauge 36 The essential mechanical element in this gage is the hollow, elastic curved tube which is connected to the pressure source as shown in Fig. As 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. Fig. Bourden gauge
37. Bourden Gauge 37 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 Bourdon gage must be calibrated so that the dial reading can directly indicate the pressure in suitable units such as psi, psf, or pascals. A 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.
38. 3. Mechanical Pressure Transducer 38 Elevation Correction Bourden gauge gives pressure at the center of dial. So to calculate pressure at point A, Where γz=elevation correction zPP gA γ+= z A Pg
39. 4. Electrical Pressure Transducer 39 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.
40. Problem 40 Q 3.5.4 x= ??in New Manometric reading=4+2x=34.7in OHftpsiaPatm 292.337.14 == ( ) psiaP PP A atmmwA ?? 4125 = =−×+ γγ ( ) ( ) atmmwA PxxP =+−+×+ γγ 241252
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42. Problem 42 Q. 3.5.7 0=+= =−− atmwmmA atmmmwA PhRP PRhP Qγγ γγ 56.13,/31.6268@ 45.13,/20.61150@ 3 3 == == m o w m o w SftlbF SftlbF γ γ
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44. Problem 44 3.5.8
45. 45
46. Problem 46 3.5.10 Two vessels are connect to a differential manometer using mercury (S=13.56), the connecting tubes being filled with water. The higher pressure vessel is 5ft lower in elevation than the other. Room temperature prevails. If the mercury reading is 4.0in, what is the pressure difference in feet of water and in psi ? (b) if carbon tetrachloride (S=1.59) were used instead of mercury what would be manometric reading for the same pressure difference.
47. Problem 47
48. Problem 48 3.12
49. Problem 49
50. Forces on Immersed Bodies: Forces on submerged planes & curved surfaces and their applications, Drag and Lift forces Dr. Mohsin Siddique Assistant Professor UOS Sharjah, UAE 50 Date: Fluid Mechanics University of Sharjah Dept. of Civil and Env. Engg.
51. Forces on Immersed Bodies 51 Hydrostatic Force: It is the resultant force of pressure exerted by liquid at rest on any side of submerged body. 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 pAdAppdAF === ∫∫
52. Forces on Plane Area 52 Center of pressure The point of application of resultant force of pressure on a submerged area is called center of pressure.
53. Forces on Plane Area 53 A=total submerged area F=hydrostatic force θ=angle of submerged plane with free surface hc=depth of center of area hp=depth of center of pressure yc=inclined depth to center of area yp=inclined depth to center of pressure dA=elemtry area dF=force of pressure (hydrostatic force) on elementry area ( ) ( )dAhdApdF γ==
54. Forces on Plane Area 54 Lets’ choose an elementary area so that pressure over it is uniform. Such an element is horizontal strip, of width, x so . The pressure force, dF on the horizontal strip is Integrating Where, yc is the distance OX along the sloping plane to the centroid C of the area A. If hc is the vertical depth to the centriod, then we have ( ) ( )dAhdApdF γ== xdydA = ∫∫∫∫∫ ==== ydAdAyhdApdAdF θγθγγ sinsin θ γ sinyh hp = = A ydA yc ∫= ( )AyF cθγ sin= AhF cγ=
55. Forces on Plane Area 55 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 centriod. The value of F is independent of the angle of inclination of the plane so long as the depth of its centroid is unchanged. Since is pressure at the centroid, we can also say that total pressure force on any plane area submerged is a liquid is the product of the area and the pressure at the centroid AhF cγ= chγ
56. Center of Pressure 56 In order to determine location of center of pressure, yp, from OX, let’s take the moment of elementary area around OX Integrating Where, I is the 2nd moment of submerged area about OX Where ycA is called static moment of area ( ) ( ) ( )dAyyhdAypdAyydFdM θγγ sin==== ( ) ( ) ( )IFy dAydAyyydF p θγ θγθγ sin sinsin 2 = == ∫∫∫ ( ) ( ) Ay I Ay I F I y cc p === θγ θγθγ sin sinsin ( )AyF cθγ sin=Q
57. Center of Pressure 57 Now, according to parallel axis theorem, Where, Ic is 2nd moment of area about centroidal axis. From this equation we again see that the location of center of pressure, P, is independent of the angle θ. When the plane is truly vertical, i.e., θ=90o Ay AyI Ay I y c cc c p 2 + == cc AyII 2 +=Q Ay I yy c c cp += Ah I hh c c cp +=
58. Lateral Position of Center of Pressure 58 To find the lateral position of center of pressure P, consider the area is made up of series of elemental strips.The center of pressure for each strip is at the mid point of the strip. Since the moment of the resultant force F must be equal to the moment of distributed force system about any axis, say, the y-axis Where, Xp is the lateral distance from the selected y-axis to the center of pressure P of the resultant force F, and xp is the lateral distance to the center of any elemental horizontal strip of area dA on which the pressure p is acting ∫= pdAxFX pp ∫= pdAxFX pp
59. Forces on Curved Surface 59 Horizontal force on curved surface Vertical force on curved surface areahorizontalequivalentonforcechydrostati'' === FFxFx surfaceaboveliquidofvolumeofWeight' === WFzFz
60. Forces on Curved Surface 60 Resultant Force 22 zx FFF +=       = − x z F F1 tanθ
61. Hydrostatic force formulas 61 Ay I yy c c cp += AhF cγ= ( )AyF cθγ sin=
62. 62
63. Problem 63 Q 3.7.6:A 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: Ah I hh c c cp += ( )AyF cθγ sin= AhF cγ= Ay I yy c c cp += Free surface 0.5m D=2m m D hc 5.1 2 5.0 =+= ( ) kNF F AhF c 2.46 2 4 5.1810.9 2 =       = = π γ ( ) ( ) mh DDh Ah I hh p p c c cp 667.1 4/5.1/64/5.1 24 = ×+= += ππ
64. Problem 64 Q 3.7.8: A rectangular plate 5ft by 4ft is at an angle of 30o with the horizontal, and the 5 ft 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) 1 ft below water surface (a) 4ft 5ft 30o hp yp hc yc 4ft θsinyh = ftyh o cc 130sin2sin === θ ( )( ) lbAhF c 12484514.62 =×== γ ( ) ( ) ftbdbd Ay I yy c c cp 67.22/12/2 3 =×+=+= fthp 33.1=
65. Problem 65 (b) 4ft 5ft hp yp hc yc=4ft 6ft θsinyh = fth o c 230sin21sin21 =+=+= θ ( )( ) lbAhF c 25004524.62 =×== γ ( ) ( ) ftbdbd Ay I y c c p 33.44/12/44 3 =×+=+= ftyh o pp 167.230sin33.4sin === θ 1ft ft hy cc 4 sin/ = = θ
66. Problem 66 Q 3.7.9 (a) For critical stability resultant of all hydrostatic for must pass through B. Thus B will be at a/3=5.4/3=1.8m above N Schematic pressure distribution diagram a/3
67. Problem 67 (a) Since B point is hinge, therefore, moment of all force at there must be zero Schematic pressure distribution diagram 1m 4.4/3 RB RN ( ) mkNRN RNMB /59.17 08.13/4.43/4.595;0 = =−−=∑ For reaction at B mkNRN FRBRNFx /4.7759.1795 ;0 =−= =+=∑ mkNAhF c /954.4) 2 4.4 (81.9 === γ For a water depth of 4.4m
68. Problem 68 3.7.12
69. Problem 69 Q 3.8.8. Cross section of tank is shown in figure., where r=2m and tank is open and contain water to a depth h=3.5m. Determine the magnitude and location of horizontal and vertical force components acting on unit width of tank wall ABC. ( ) mh mkNF AhF p x cx 33.25.3 3 2 /1.605.3)2/5.3(81.9 == == = γ hp xp
70. ` 70 hp xp ( )ALVOLWFz γγ === 86.3)2(5.3 4/)4( 4 )2(5.3 2 =−=       −= π π A A 2m d=4m( ) mkN AFz /9.37 )86.3(81.9 = == γ Lets take moment about AB ( ) ( ) mx x p p 123.1 849.0)1(25.386.3 = −×= π m r xc 849.0 3 4 == π
71. Drag and Lift Force 71 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.
72. Drag and Lift Force 72 The drag force acts in a direction that is opposite of the relative flow velocity. Affected by cross-section area (form drag) Affected by surface smoothness (surface drag) The lift force acts in a direction that is perpendicular to the relative flow. CD= Coefficient of drag CL= Coefficient of lift A=projected area of body normal to flow V= relative wind velocity
73. Buoyancy and Floatation 73 When a body is immersed wholly 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 Archimedes’ principle which states “ When a body is immersed in a fluid either wholly or partially, it is buoyed or lifted up by a force which is equal to the weight of fluid displaced by the body”
74. Buoyancy and Floatation 74 Lets consider a body submerged in water as shown in figure. The force of buoyancy “resultant upward force or thrust exerted by fluid on submerged body” is given Water surface 11 hP γ= ( )212 hhP += γ 2h 1h 1F 2F dA=Area of cross-section of element γ= Specific weight of liquid ( ) ( ) ( )[ ] [ ]volumeF dAhF dAhdAhhF FFF B B B B γ γ γγ = = −+= −= 2 121 12
75. Buoyancy and Floatation 75 =Weight of volume of liquid displaced by the body (Archimedes's Principle) Force of buoyancy can also be determined as difference of weight of a body in air and in liquid. Let Wa= weight of body in air Wl=weight of body in liquid FB=Wa-Wl [ ]volumeFB γ=
76. Buoyancy and Floatation 76 Center of Buoyancy (B): 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 B CG or G= Center of gravity of body C or B= Centroid of volume of liquid displaced by body
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