Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 36 Chp09:FluidStatics Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. The Fish Feels More “Compressed” The Deeper it goes – An example of HydroStatic Pressure Fluid Statics • Definition of Pressure • Pressure is defined as the amount of force exerted on a unit area of a surface:

3. Direction of fluid pressure on boundaries Furnace duct Pipe or tube Heat exchanger • Pressure is a Normal Force • i.e., it acts perpendicularto surfaces • It is also called a Surface Force Dam

4. Absolute and Gage Pressure • Absolute Pressure, p: The pressure of a fluid is expressed relative to that of a vacuum (absence of any substance) • Gage Pressure, pg: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere (the BaseLine Pressure) which has an Absolute pressure, p0. • p0≡ BaseLine Pressure:

5. Pressure Units

6. Consider a Vertical Column of NonMoving Fluid with density, ρ, and Geometry at Right Static Fluid Pressure Distribution • Taking a slice of fluid at vertical position-z note that by Equilibrium:The sum of the z-directed forces acting on the fluid-slice must equal zero

7. Consider the Fluid Slice at z under the influence of gravity S is the Top & Bot Slice-Surface Area The z-Direction Fluid Forces z • Let pzand pz+Δz denote the pressures at the base and top of the slice, where the elevations are z and z+Δz respectively y x

8. A Force Balance in the z-Dir The z-Direction Fluid Forces z y x • Solving for Δp/Δz and Taking the limit Δz→0

9. For Constant ρ Constant Density Case • Next Consider the typical Case of a “Free Surface” at Atmospheric Pressure, p0, and total Fluid Depth, H • If h is the DEPTH, then the z↔hreln:

10. Then Constant Density Case • Sub into Δp Eqn • Now if at h = 0, p = p0, and Δh = h2 – 0 = h • In this Case

11. The Gage Pressure,pg, is that pressure which is Above the p0 Baseline Constant Density Case • Thus the typical Formulation for Liquids at atmospheric pressure • The Gage Pressure is used for Liquid-Tanks or Dams

12. Since the acceleration of gravity, g, is almost always regarded as constant, the ρliqg product is often called the Specific Weight Specific Weight • Then the pressure eqns

13. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces A floating body displaces its own weight in the fluid in which it floats Bouyancy F1 Free liq surf h1 Δh h2 F2

14. The upper surface of the body is subjected to a smaller force than the lower surface Thus the net bouyant force acts UPwards Bouyancy F1 Free liq surf h1 Δh h2 F2 • The net force due to pressure in the vertical direction FB = F2- F1 = (pbottom - ptop)(ΔxΔy) = Δp(ΔxΔy)

15. The pressure difference Bouyancy F1 Free liq surf h1 pbottom – ptop = ρg(h2-h1) = ρgΔh Δh h2 Δp = ρgΔh • Subbing for Δp in the FB Eqn F2 • So Finally the Bouyant Force Eqn FB = (ρgΔh)(ΔxΔy) • But ΔhΔxΔy is the Volume, V, of the fluid element

17. Consider Fluid Pressure acting on a submerged FLAT Surface with any azimuthal angle θ Equivalent Point Load • In this situation the Net Force acting on the Flat Surface is the pressure at the Flat Surface’s GEOMETRIC Centroid times the Surface area • i.e., Fliq = the AVERAGE Pressure times the Area

18. Fliq is NOT Positioned at the Geometric Centroid; Instead it is located at the CENTER of PRESSURE Equivalent Point Load • Calculation of the Center of Pressure Requires Moment Analysis as was described last lecture

19. Pressure acts as a function of depth Force on a Submerged Surface 0 y Δy Resultant, R • Then the Magnitude of the Resultant Force is Equal to the Area under the curve d d • Face Width of Structure INTO the Screen = b

20. The location of the Resultant of the pressure dist. is found the by same technique as used on the horizontal beam Force on a Submerged Surface 0 y S Resultant, R d d • Thus the Location of the Resultant Hydrostatic force Passes Thru the Pressure-Area Centroid • The Distance OS is Also Called the Center of Pressure

21. How does one find the forces on a SUBMERGED surface (red) at an angle? Construct the FBD Angled Submerged Surface • Detach The Triangular Fluid Volume from the bulk fluid to Reveal Forces: • Weight of the Volume • Pressure at NORMAL Surfaces • Can also ATTACH the Fluid to the Body if desired

22. Angled SubMerged Surface • The resulting force distribution without the weight of the water would show Equal to the Trapezoidal Area Under the Pressure Curve Times the Width b

23. Consider The P-distribution on a non-linear surface. Liquid Generates Resultant, R on The Surface NonLinear Submerged Surface • Use Detached Fluid Volume as F.B.D. • The Force Exerted by the Surface is Simply −R

24. HydroStatic Free Body Diagram • Examine Support Structure for Non-Rectilinear Surfaces • Detach From Surrounding Liquid, and And from the Structure, a LIQUID VOLUME that Exposes Flat X&Y Surfaces exposed to Liquid Pressure • Pressure is • Constant at X-Surfaces (Horizontal) • Triangular or Trapezoidal at Y-Surfaces (Vertical)

25. Centroids of Parabolas Sector Spandrel

26. WhtBd Example: P9-122 • Determine the resultant horizontal and vertical force components that the water exerts on the side of the dam. Find for R the PoA on the Dam-Face • The dam is 25 ft long • SW for Water = 62.4 lb/cu-ft

28. Resultant Location

29. Solve By MATLAB Deqn = 1.0e+03 * 0.1218 0.2600 -5.0333 x = -7.5856 5.4500 y = 7.4257 xD= 5.4500 yD= 7.4257 % Bruce Mayer, PE % ENGR36 * 03Dec12 % ENGR36_Dam_Face_P9_122_1212.m % W = 260; % kip P = 487;.5 % kip % Deqn = [P/4 W -(25*P/3+30*W/8)] x = roots(Deqn) y = (x(2))^2/4 xp = linspace(0,10, 500); yp1 = polyval(Deqn, xp); yp = xp.^2/4; xD = x(2) yD = y plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('P9-122 Dam-Face Force Location')

30. Given Fresh Water dam with Geometry as shown Example  Small Dam • If the Dam is 24ft wide (into screen) Find the Resultant of the pressure forces on the Dam face BC • Assume • Detach the Parabolic Sector of Water

31. Create FBD for Detached Water-Chunk Example: Small Dam • The gage Pressure at the Bottom

32. The Force dF18 at the base of the Dam with Face Width, b (into paper), consider a vertical Distance dh located at 18ft Deep Example: Small Dam

33. Now to Covert dF to w (lb/vertical-Foot) Simply Divide by the vertical distance that generated dF Example: Small Dam • In this case b = 1ft

34. For a Submerged Flat surface, with Face Width, b, in a constant density Fluid the Load per Unit-Length profile value, w(h) can found as Example: Small Dam Digression • In Units of lb/ft or lb/in or N/m

35. The Load Profile is a TriAngle → A = ½BH B = 1123 lb/ft H = 18 ft Then the Total Water Push, P, is the Area under the Load Profile Example: Small Dam • Previous Slide

36. Then the Volume of the Detached Water Example: Small Dam • And the Weight, W4, of the Attached Water

37. And the Location of the of P is the Center of Pressure which is located at the Centroid of the LOAD PROFILE In this case the CP is 6ft above the bottom Example: Small Dam • Also W4 is applied at the CG of the parabolic sector • From “Parabola” Slide

38. Now the Dam also pushes on the Detached Water For Equilibrium the Push by the Dam on the Water-Chunk must be the negative of Resultant of the Load ON the Dam The Applied Loads are P and W4 Example: Small Dam • Then the FBD for the Water-Chunk

39. The Water Chunk FBD Example: Small Dam • Notice that This is a 3-Force Body • Thus the Forces are CONCURRENT and no Moments are Generated • The Force Triangle Must CLOSE • Use to Find Mag & Dir for R

40. The Force Triangle: Example: Small Dam • The the Load per Unit Width of the Dam • 12580 lb/ft-Width • Directed 36.5° below the Horizontal and to the Left relative to the drawing

41. Resultant Location on Dam • Determine the CoOrdinates, Horizontally & Vertically, for the Point of Application of the 12 580 lb • Take the ΣMC=0 about the upper-left Corner where the water-surface touches the dam

43. W = 7488 P = 10109 k = 0.0309 Deqn = 1.0e+05 * 0.0023 0.1011 -1.6624 x = -56.4769 12.7360 y = 5.0064 xD = 12.7360 yD = 5.0064 MATLAB Results xcanNOT be Negative in this Physical Circumstance

44. PoA for R on Dam Face

45. % Bruce Mayer, PE % ENGR36 * 23Jul12 % ENGR36_Dam_Face_1207.m % W = 7488 P = 10109 k = 10/18^2 Deqn = [k*W P -(6*W+12*P)] x = roots(Deqn) y = k*(x(2))^2 xp = linspace(0,18, 500); yp = polyval(Deqn, xp); yp = k*xp.^2; xD = x(2) yD = y plot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('Dam Face Force Location‘) MATLAB Code

46. WhiteBoard Work Barrel DamProblem • A 55-gallon, 23-inch diameter DRUM is placed on its side to act as a DAM in a 30-in wide freshwater channel. The drum is anchored to the sides of the channel. Determine the resultant of the pressure forces acting on the drum and anchors. • ATTACH some Water Segments to the Drum