1. 2016/5/30 林再興教授編 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.

2. 2016/5/30 林再興教授編 2 §2.1 Presure at a point At rest, on any plane Gas P 2 Shear stress = 0 A or Normal stress Liquid P 1 P 3 =P 1 =P 2 =P 3 =P 4 =P =Pressure   P 4 ( positive for compression ) θ

3. 2016/5/30 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 ?

4. 2016/5/30 林再興教授編 4 Equation of motion ( or equilibrium equation) In y-direction, Σ F y = 0 P y δx δz – ( P s δx δs)sin θ = 0 P y δx δz – P s δxδz = 0 P y – P s = 0 or P y = P s

5. 2016/5/30 林再興教授編 5 In z-direction ΣF z = 0 P z δx δy – ( P s δx δs)cosθ –ρg δx δy δz(1/2) = 0 Since δ s cos θ = δ y  P z δx δy – P s δx δy –1/2 ρ g δx δy δz = 0 P z = P s + 1/2 ρ gδz For δx δy δz → 0 P z = P s =>P y = P s = P z

6. Pascal's law Since the angle θ was arbitrarily chosen, we can obtain P y = P x = P z  P y = P x = P z = P S ~ Pascal's law 2016/5/30 林再興教授編 6

7. 2016/5/30 林再興教授編 7 §2-2 Basic Equation for Pressure Field pressure net force pressure gradientnet force Let P = the pressure at the center of the element. (x,y,z) pressure force in y direction on a fluid element

8. 2016/5/30 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

9. 2016/5/30 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 )

10. 2016/5/30 林再興教授編 10 Surface forces (1) pressure gradient (2) viscous stress Body force ( gravitational potential)

11. 2016/5/30 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

12. 2016/5/30 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

13. 2016/5/30 林再興教授編 13 (2.3) …(2.4) Pressure does not change in a horizontal plane ( or x-y plane) z p 2 ─z 2 p 1 ─z 1

14. 2016/5/30 林再興教授編 14 §2.3.1 Incompressible Fluid Incompressible fluid, ρ=const. & assuming g = Const. z p 2 ─z 2 p 1 ─z 1  h= where h is called pressure head Hydrostatic pressure distribution ─ the pressure varies linearly with depth For H 2 Oh = 23.1 ft for ΔP = 10 psia For Hg h = 518mm for ΔP = 10 psia

15. 2016/5/30 林再興教授編 15 P = P 0 + ρgh where P 0 : pressure at free surface; h: Depth below the free surface) same h  same P

16. 2016/5/30 林再興教授編 16 Example 2.1 Given: As figure on the right SG(gasoline)=0.68 h 1 = 17 ft (Gasoline) h 2 = 3 ft (water) Find: pressure at point (1) & (2) in units of lb/ft 2, lb/in 2, and as a pressure head in feet of water

17. 2016/5/30 林再興教授編 17 Solution: (a)P 1 = P 0 + ρgh 1 (b)P 1 = P 0 + ρ H2O gh H2O h H2O = (c)P 2 = P 1 + ρ H2O gh 2 = 722.13(lbf/ft 2 ) + 1.94(slug/ft 3 )  32.2(ft/s 2 )  3(ft) = 722.13 + 187.4 = 909.53

18. 2016/5/30 林再興教授編 18 (d) P 2 = P 0 + ρ H 2 O gh H 2 O  h H 2 O = Transmission of fluid pressure since p 1 = p 2

19. 2016/5/30 林再興教授編 19 §2.3.2 Compressible Fluid --- Compressible fluid air, oxygen(O 2 ), hydrogen(N 2 ) (ρg) gas = f(p,T) Eq(2.4)  dp = -ρgdz (ρg) air = 0.076 lbf/ft 3 at p=14.7psia & T=60 ℉ (ρg) H 2 O = 62.4 lbf/ft 3 at p=14.7 psia &T=60 ℉

20. 2016/5/30 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

21. 2016/5/30 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 (1)p 2 (at top) / p 1 (at bottom) = ? if air is compressible fluid at T = 59 ℉ (2) p 2 / p 1 = ? if air is assumed to be incompressible fluid (ρg) air = 0.076 lbf/ft 3 at 14.7psia

22. 2016/5/30 林再興教授編 22 Solution: (1)From Eq (2.10) (2) If air is incompressible fluid dp = -ρgdz  p 2 –p 1 = -ρg(z 2 – z 1 )

23. 2016/5/30 林再興教授編 23 §2.4 Standard Atmosphere From Eq.(2.9) where T a = 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. 0 R --- (2.10)

24. 2016/5/30 林再興教授編 24 Fig 2.6 P.51 (fig. 2.6)

25. 2016/5/30 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 ─ P Abs > 0 psia ; Patm ~ 14.7 psia (atmosphere pressure) Gage pressure ─ P gage = 0 psig (atmosphere pressure) P =P gage +14.7 psia [=] psia P gage > 0  P > P atm P gage < 0  P < P atm

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27. 2016/5/30 林再興教授編 27 20.83 psia 13.54 psia 20.83 psia 8.33 psia 7.29 psig - 5.21 psig 5.21 psi vacuum 13.54 psi vacuum 0 psia -13.54 psig

28. 2016/5/30 林再興教授編 28 Mercury barometer P atm = ρgh + P vapor ρgh because P vapor = 0.000023 psia at T=68 ℉ P = 14.7 psia  h=760mmHg(abs.) =29.9in Hg(abs.) =34ft H 2 O(abs.)

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31. 2016/5/30 林再興教授編 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

32. 2016/5/30 林再興教授編 32 §2.6.1 Piezometer tube P = P A = ρgh + P o [=] psia (absolute pressure) P = P A = ρgh [=] psig (gage pressure) Disadvantages 1. P A > P atm 2. h 1 should be reasonable 3. Fluid in container must be a pressure at point(A) liquid is desired

33. 2016/5/30 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.

34. 2016/5/30 林再興教授編 34 Method(A) P A -P 2 = -ρ 1 g(z 1 - z 2 ) +) P 2 -P 4 = -ρ 2 g(z 3 - z 4 ) P A -P 4 = -ρ 1 g(z 1 -z 2 )-ρ 2 g(z 3 -z 4 ) => P A = -ρ 1 gh 1 -ρ 2 g(-h 2 ) or P A = -ρ 1 gh 1 +ρ 2 gh 2 => P A +ρ 1 gh 1 -ρ 2 gh 2 = 0 => P A =ρ 2 gh 2 -ρ 1 gh 1 Method(B) 0 P A +ρ 1 gh 1 -ρ 2 gh 2 =P 4 =>P A +ρ 1 gh 1 -ρ 2 gh 2 =0 =P 2 =P 3 =P 4 =>P A =ρ 2 gh 2 -ρ 1 gh 1 (4)

35. 2016/5/30 林再興教授編 35 Example 2.4 Give Right figure SG(oil) = 0.9 SG(Hg) = 13.6 h 1 = 36 inch h 2 = 6 inch h 3 = 9 inch Find ： P A = ？ Solution P A +ρ 0 g(h 1 +h 2 ) -ρ Hg gh 3 = p gage p A = -ρ 0 gh(h 1 +h 2 ) +ρ Hg gh 3 = -0.9 * 1.94 slug/ft 3 * 32.2 ft/s 2 *(36+6)in * 1ft / 12inch + 13.6 * 1.94 * 32.2 * 9/12 = 440lb f / ft 2 *(1ft 2 / 144in )= 3.06psig

36. 2016/5/30 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. P A + ρ 1 gh 1 - ρ 2 gh 2 - ρ 3 gh 3 =P B P A -P B =- ρ 1 gh 1 + ρ 2 gh 2 + ρ 3 gh 3

37. 2016/5/30 林再興教授編 37 Example2.5 Given: Right figure Find: (1) P A -P B =f(γ 1, γ 2,h 1,h 2 ) (2) P A -P B =? if γ 1 =9.8 kN / m 3, γ 2 =15.6 kN / m 3 h 1 =1.0m and h 2 =0.5m Solution: (1) P A - ρ 1 gh 1 - ρ 2 gh 2 + ρ 1 g(h 2 +h 1 )=P B P A -P B = ρ 1 gh 1 + ρ 2 gh 2 - ρ 1 g(h 2 +h 1 )=h 2 ( ρ 2 g- ρ 1 g) (2) P A -P B =h 2 ( ρ 2 g- ρ 1 g) =0.5(15.6*10 3 -9.8*10 3 )=2.9*10 3 N/m 2 =2.9kpa

38. 2016/5/30 林再興教授編 38 §2.6.3 Inclined-Tube Manometer —To measure Small pressure changes P A + ρ 1 gh 1 - ρ 2 g(l 2 sinθ)– ρ 3 gh 3 =P B P A -P B = ρ 2 gl 2 sinθ+ ρ 3 gh 3 – ρ 1 gh 1 If fluid in ρ 1 & ρ 3 is gas, then ρ 1 gh 1 →0, ρ 3 gh 3 →0,and P A -P B = ρ 2 g l 2 sinθ  l 2 ↗ as sinθ ↘ 0

39. 2016/5/30 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

40. 2016/5/30 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

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42. 2016/5/30 林再興教授編 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

43. 2016/5/30 林再興教授編 43 §2.8 Hydrostatic Force on a Plane Surface Assume ρ=Const (incompressible fluid) => P=P a +ρgh

44. 2016/5/30 林再興教授編 44 The total hydrostatic force (F R ) (2.17) for ρ=Const,θ=Const The first moment of the area with respect to the X-axis = => F R =P 0 A+ρgsinθy CG A or F R =P 0 A+ρgh CG A ─ (2.18) F R ┴Surface ﹝ F R =(P atm +ρgh C )A ﹞ or F R =P CG A, where P CG =P atm + ρgh c

45. 2016/5/30 林再興教授編 45 To find the center pressure or resultant force(x R,y R ) : Base on moment of the resultant force=moment of the distributed pressure force To find y R  F R y R = = = =ρg sinθ  {ρg sinθy CG A}y R =ρg sinθ  y CG A y R =  y R =  y R = where = Second moment of the area(moment of inertia) = I X

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47. 2016/5/30 林再興教授編 47 The parallel axis theorem(to express I X ) I X =I XC +Ay CG 2 where I XC =the second moment of the area with respect to an axis passing through its centroid and parallel to the X-axis. y R = + y CG ------- (2.19) Note: (1) I XC /y CG A > 0 => y R > y CG (2) Resultant force does not pass through the centroid,but is always below it.

48. 2016/5/30 林再興教授編 48 To find x R => F R x R = = => ρg sinθ y CG A x R = ρg sinθ => x R = I xyc = the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area

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50. 2016/5/30 林再興教授編 50 Example2.6 Given ： figure on right Diameter of circular gate=4m ρg = 9.8 KN/m 3 D shaft =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

51. 2016/5/30 林再興教授編 51 Solution ： (a )F R =P CG A=ρgh CG A =9.8*10 3 (N/m 3 )*(10m)*(π·2 2 )m 3 =1231*10 3 N=1231KN =1.23MN y R = + y CG ---- (2.19) = + y CG = + y CG (fig. 2.18 P.64) = + = 0.0866+11.5 = 11.58m

52. 2016/5/30 林再興教授編 52 y R -y CG =0.0866m below the shaft x R = + x CG = +x CG = 0 + x CG = x CG (b)Center of rotation at c Σr · F = 0 or ΣM = 0 M - F R ·(y R -y CG ) = 0 M = F R ·(y R -y CG ) = 1231*10 3 *0.0866 =1.07*10 5 N-M