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Chapter 2 Fluid at Rest – Pressure and it Effects (Chapter 2 Fluid Statics)

Chapter 2 Fluid at Rest – Pressure and it Effects (Chapter 2 Fluid Statics) Fluid is either at rest or moving -- no relative motion between adjacent particles -- no shearing stresses in the fluid → surface force will be due to the pressure

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Chapter 2 Fluid at Rest – Pressure and it Effects (Chapter 2 Fluid Statics)

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  1. Chapter 2 Fluid at Rest – Pressure and it Effects (Chapter 2 Fluid Statics) Fluid is either at rest or moving -- no relative motion between adjacent particles -- no shearing stresses in the fluid → surface force will be due to the pressure In the chapter, the principal concern is to investigate (1) pressure and its variation throughout a fluid, (2) the effect of pressure on submerged surface. 林再興教授編

  2. §2.1 Presure at a point At rest, on any plane Gas P2 Shear stress = 0 A or Normal stress   Liquid P1 P3 =P1=P2=P3=P4=P =Pressure  P4 ( positive for compression ) θ 林再興教授編

  3. normal force   Fluid pressure : (1) = P ( fluid pressure ) (2)  the surface of contact In an area P = At any point P = How the pressure at a point do varies with the orientation of the plane passing through the point ? 林再興教授編

  4. Equation of motion ( or equilibrium equation) In y-direction, ΣFy = 0 Py δxδz – ( Ps δxδs)sin θ = 0 Py δxδz – Psδxδz = 0 Py– Ps = 0 or Py = Ps 林再興教授編

  5. In z-direction ΣFz = 0 Pz δxδy – ( Ps δxδs)cosθ –ρg δxδy δz(1/2) = 0 Since δscos θ = δy Pz δxδy – Ps δxδy –1/2 ρg δxδyδz = 0 Pz = Ps + 1/2 ρgδz For δxδyδz→ 0 Pz = Ps =>Py = Ps = Pz 林再興教授編

  6. Pascal's law • Since the angle θ was arbitrarilychosen , we can obtain • Py = Px = Pz • Py = Px = Pz= PS ~ Pascal's law 林再興教授編

  7. §2-2 Basic Equation for Pressure Field pressure net force pressure gradient net force Let P = the pressure at the center of the element. (x,y,z) pressure force in y direction on a fluid element 林再興教授編

  8. In the same manner In x direction In z direction The total net pressure force on the element 林再興教授編

  9. Equilibrium of a fluid Force on a fluid element ─Surface force ─ acting on the sides of the element i,e, pressure gradient and viscous stress ( not included in this chapter; will be considered in chapter 6) ─Body force ─ acing on the entire mass of the element, i.e, gravitational potential and electromagnetic potential ( neglected in this chapter ) 林再興教授編

  10. Surface forces (1)pressure gradient (2) viscous stress Body force (gravitational potential) 林再興教授編

  11. Force balance -----------------(2.2) Eq(2.2) is the general equation of motion for a fluid in which there is no shearing stress. The following equation is the general equation of motion for a fluid in which there is shearing stress 林再興教授編

  12. §2.3 Pressure Variation in a Fluid at Rest From Eq.(2.2) Such as ---------(2.2) For fluid at rest 林再興教授編

  13. Pressure does not change in a horizontal plane ( or x-y plane) (2.3) z p2─z2 p1─z1 …(2.4) 林再興教授編

  14. §2.3.1 Incompressible Fluid Incompressible fluid, ρ=const. & assuming g = Const. z p2─z2 p1─z1 Hydrostatic pressure distribution ─ the pressure varies linearly with depth  h= where h is called pressure head For H2O h = 23.1 ft for ΔP = 10 psia For Hg h = 518mm for ΔP = 10 psia 林再興教授編

  15.   P = P0 + ρgh where P0: pressure at free surface; h: Depth below the free surface) same h  same P 林再興教授編

  16. Example 2.1 Given: As figure on the right SG(gasoline)=0.68 h1 = 17 ft (Gasoline) h2 = 3 ft (water) Find: pressure at point (1) & (2) in units of lb/ft2, lb/in2, and as a pressure head in feet of water 林再興教授編

  17. Solution: • P1 = P0 + ρgh1 • P1 = P0 + ρH2OghH2O • hH2O = • P2 = P1 + ρH2Ogh2 • = 722.13(lbf/ft2) + 1.94(slug/ft3)  32.2(ft/s2)  3(ft) • = 722.13 + 187.4 = 909.53 林再興教授編

  18. (d) P2 = P0 + ρH2OghH2O hH2O = Transmission of fluid pressure since p1 = p2 林再興教授編

  19. §2.3.2 Compressible Fluid --- Compressible fluid air, oxygen(O2), hydrogen(N2) (ρg)gas = f(p,T) Eq(2.4)  dp = -ρgdz (ρg)air = 0.076 lbf/ft3 at p=14.7psia & T=60℉ (ρg)H2O = 62.4 lbf/ft3 at p=14.7 psia &T=60℉ 林再興教授編

  20. If dz is small, dp=-(ρg)air dz → 0  If dz is large Eq(2.4)  dp =-ρgdz Eq. of state for ideal gas  p = ρRT 林再興教授編

  21. Example2.2 • The Empire State Building in New York city, one of the tallest building in the world • Given: h = 1250 ft • p2(at top) / p1(at bottom) = ? if air is compressible fluid • at T = 59℉ • (2) p2 / p1 = ? if air is assumed to be incompressible fluid • (ρg)air = 0.076 lbf/ft3 at 14.7psia 林再興教授編

  22. Solution: • From Eq (2.10) • (2) If air is incompressible fluid • dp = -ρgdz  p2 –p1 = -ρg(z2 – z1) 林再興教授編

  23. §2.4 Standard Atmosphere From Eq.(2.9) where Ta = temp. at z = 0 (sea level) ;β= lapse rate where the parameters in Eq(2.10) are shown in Table 2.1 (P.50) R=286.9 J/kg.k or 1716 ft-lb/slug.0R --- (2.10) 林再興教授編

  24. Fig 2.6 P.51 (fig. 2.6) 林再興教授編

  25. §2.5 Measurement of Pressure pressure measurement ─absolute pressure (with respective to a zero pressure reference) ─gage pressure (with respective to local atm pressure) Absolute pressure ─ PAbs > 0 psia ; Patm ~ 14.7 psia (atmosphere pressure) Gage pressure ─ Pgage = 0 psig (atmosphere pressure) P =Pgage+14.7 psia [=] psia Pgage> 0  P > Patm Pgage< 0  P < Patm 林再興教授編

  26. 林再興教授編

  27. 20.83 psia 7.29 psig 13.54 psia - 5.21 psig 5.21 psi vacuum 20.83 psia 8.33 psia 0 psia -13.54 psig 13.54 psi vacuum 林再興教授編

  28. Mercury barometer Patm = ρgh + Pvapor ρgh because Pvapor = 0.000023 psia at T=68℉ P = 14.7 psia h=760mmHg(abs.) =29.9in Hg(abs.) =34ft H2O(abs.) 林再興教授編

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  31. §2.6 Manometry ─A standard technique for measuring pressure involves the use of liquid columns in vertical or inclined tubes. Pressure measuring devices based on the technique are called manometers. Piezometer tube ─Manometer U-tube Inclined-tube 林再興教授編

  32. §2.6.1 Piezometer tube P = PA = ρgh + Po [=] psia (absolute pressure) P = PA = ρgh [=] psig (gage pressure) Disadvantages 1. PA > Patm 2. h1 should be reasonable 3. Fluid in container must be a pressure at point(A) liquid is desired 林再興教授編

  33. §2.6.2 U-Tube manometer ─To overcome the difficulties note previously, another type of manometer which is widely used consisted of a tube formed into the shape of a U as is show in Fig.2.10. ─A major advantage of the U-tube manometer Gage fluid can be different from the fluid in the container in which the pressure is to be determined ─"Jump across" Same elevation within the same continuous mass of fluid. 林再興教授編

  34. Method(A) PA-P2 = -ρ1g(z1 - z2) +) P2-P4 = -ρ2g(z3 - z4) PA-P4= -ρ1g(z1-z2)-ρ2g(z3-z4) => PA= -ρ1gh1-ρ2g(-h2) or PA= -ρ1gh1+ρ2gh2 => PA+ρ1gh1 -ρ2gh2 = 0 => PA=ρ2gh2-ρ1gh1 Method(B) 0 PA+ρ1gh1-ρ2gh2=P4 =>PA+ρ1gh1-ρ2gh2=0 =P2=P3 =P4 =>PA=ρ2gh2-ρ1gh1 (4) 林再興教授編

  35. Example 2.4 Give Right figure SG(oil) = 0.9 SG(Hg) = 13.6 h1 = 36 inch h2 = 6 inch h3 = 9 inch Find：PA=？ Solution PA+ρ0g(h1+h2) -ρHggh3 = pgage pA= -ρ0gh(h1+h2) +ρHggh3 = -0.9 * 1.94 slug/ft3 * 32.2 ft/s2*(36+6)in * 1ft / 12inch + 13.6 * 1.94 * 32.2 * 9/12 = 440lbf / ft2*(1ft2 / 144in )= 3.06psig 林再興教授編

  36. The U-tube manometer is also widely used to measure the difference in pressure between two containers or two points in a given system. PA+ ρ1gh1- ρ2gh2- ρ3gh3=PB PA-PB=- ρ1gh1+ ρ2gh2+ ρ3gh3 林再興教授編

  37. Example2.5 Given: Right figure Find:(1)PA-PB=f(γ1, γ2,h1,h2) (2)PA-PB=? if γ1=9.8kN/m3, γ2=15.6kN/m3 h1=1.0m and h2=0.5m Solution: (1) PA- ρ1gh1- ρ2gh2+ ρ1g(h2+h1)=PB PA-PB= ρ1gh1+ ρ2gh2- ρ1g(h2+h1)=h2(ρ2g- ρ1g) (2) PA-PB=h2(ρ2g- ρ1g) =0.5(15.6*103-9.8*103)=2.9*103 N/m2 =2.9kpa 林再興教授編

  38. §2.6.3 Inclined-Tube Manometer —To measure Small pressure changes PA+ ρ1gh1- ρ2g(l2sinθ)–ρ3gh3=PB PA-PB= ρ2gl2sinθ+ ρ3gh3–ρ1gh1 If fluid in ρ1 &ρ3 is gas, then ρ1gh1→0, ρ3gh3→0,and PA-PB= ρ2g l2sinθ l2↗ as sinθ↘0 林再興教授編

  39. §2.7 Mechanical and Electronic pressure Measuring Devices Disadvantage of manometers (1) not well suited for measuring very high pressure. (2) not well suited for measuring pressure that are changing rapidly with time. (3) require the measurement of one or more column heights.  time consuming 林再興教授編

  40. Other types of pressure measuring instruments. Idea: pressure acts on an elastic structure, the structure will deform,and then deformation can be related to the magnitude of the pressure. (1) Bourdon pressure gage (gage pressure) (2) The aneroid barometer(Bourdon type) — for measuring atmospheric pressure. (absolute pressure) (3) pressure transducer — pressure → electrical output 林再興教授編

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  42. Example: (a) Bourdon tube is connected to a linear Variable differential transformer (LVDT.) (b) Thin,elastic diaphragm which is in contact with the fluid. — pressure changesdiaphragm deflectselectrical voltage (i) strain gage (ii)piezoelectric crystal 林再興教授編

  43. §2.8 Hydrostatic Force on a Plane Surface Assume ρ=Const (incompressible fluid) => P=Pa+ρgh 林再興教授編

  44. The total hydrostatic force (FR) (2.17) for ρ=Const,θ=Const The first moment of the area with respect to the X-axis = => FR=P0A+ρgsinθyCGA or FR=P0A+ρghCGA ─ (2.18) FR┴Surface ﹝FR=(Patm+ρghC)A﹞ or FR=PCGA , where PCG=Patm+ρghc 林再興教授編

  45. To find the center pressure or resultant force(xR,yR) : Base on moment of the resultant force=moment of the distributed pressure force To find yR  FRyR = = = =ρg sinθ  {ρg sinθyCGA}yR=ρg sinθ  yCG A yR=  yR =  yR = where = Second moment of the area(moment of inertia) = IX 林再興教授編

  46. 林再興教授編

  47. The parallel axis theorem(to express IX) IX=IXC+AyCG2where IXC=the second moment of the area with respect to an axis passing through its centroid and parallel to the X-axis. yR= + yCG ------- (2.19) Note: (1) IXC /yCG A > 0 => yR > yCG (2) Resultant force does not pass through the centroid,but is always below it. 林再興教授編

  48. To find xR => FRxR = = => ρg sinθ yCGA xR = ρg sinθ => xR = => xR = Ixyc = the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area 林再興教授編

  49. 林再興教授編

  50. Example2.6 Given：figure on right Diameter of circular gate=4m ρg = 9.8 KN/m3 Dshaft=10m Determine： (a)the magnitude and location of the resultant force exerted on gate by the water (b)the moment that would have to be applied to the shaft to open the gate 林再興教授編

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