1. Pharos Univ. ME 259 Fluid Mechanics Static Forces on Inclined and Curved Surfaces

2. Main Topics The Basic Equations of Fluid Statics Pressure Variation in a Static Fluid Hydrostatic Force on Submerged Surfaces Buoyancy

3. The Basic Equations of Fluid Statics Body Force

4. The Basic Equations of Fluid Statics Surface Force

5. The Basic Equations of Fluid Statics Surface Force

6. The Basic Equations of Fluid Statics Surface Force

7. The Basic Equations of Fluid Statics Total Force

8. The Basic Equations of Fluid Statics Newton’s Second Law

9. The Basic Equations of Fluid Statics Pressure-Height Relation

10. 2.3.1 Pressure and head  In a liquid with a free surface the pressure at any depth h measured from the free surface can be found by applying equation (2.3) to the figure.  From equation (2.3): P 1 – P 2 =  g (y a -y) But y a -y = h, and P 2 = P atm (atmospheric pressure since it is at free surface). Thus, P 1 – P atm =  gh or P 1 = P atm +  gh (abs) (2.4) or in terms of gauge pressure (P atm = 0),: P 1 =  gh =  h (2.5) h P1P1 P 2 = P atm y yaya Free surface

11. Pressure Variation in a Static Fluid Incompressible Fluid: Manometers

12. Pressure Variation in a Static Fluid Compressible Fluid: Ideal Gas Need additional information, e.g., T(z) for atmosphere

13. Differential Manometer  The liquids in manometer will rise or fall as the pressure at either end changes. P 1 = P A +  1 ga P 2 = P B +  1 g(b-h) +  man gh P 1 = P 2 (same level) P A +  1 ga = P B +  1 g(b-h) +  man gh or P A - P B =  1 g(b-h) +  man gh -  1 ga P A - P B =  1 g(b-a) + gh(  man -  1 ) Figure 2.13: Differential manometer

14. Hydrostatic Force on Submerged Surfaces Plane Submerged Surface

15. Center of Pressure Line of action of resultant force F R =P C A lies underneath where the pressure is higher. Location of Center of Pressure is determined by the moment. I xx,C is tabulated for simple geometries.

16. Hydrostatic Force on Submerged Surfaces Plane Submerged Surface We can find F R, and y´ and x´, by integrating, or …

17. Hydrostatic Force on Submerged Surfaces Plane Submerged Surface Algebraic Equations – Total Pressure Force

18. Hydrostatic Force on Submerged Surfaces Plane Submerged Surface Algebraic Equations – Net Pressure Force

19. Table 2.1 Second Moments of Area GG h b G G h h/3 G d G Rectangle Triangle  d 4 /64  d 2 /4 Circle bh 3 /36bh/2 bh 3 /12bh IgIg Area Shape G G h h/3 G h G b G G d b

20.  A 6-m deep tank contains 4 m of water and 2-m of oil as shown in the diagram below. Determine the pressure at point A and at the tank bottom. Draw the pressure diag. Pressure at oil water interface (P A ) P A = P atm + P oil (due to 2 m of oil) = 0 +  oil gh oil = 0 + 0.98 x 1000 x 9.81 x 2 = 15696 Pa PA = 15.7 kPa (gauge) Pressure at the bottom of the tank; P B = P A +  water gh water P B = 15.7x1000 + 1000 x 9.81 x 4 = 54940 Pa P B = 54.9 kPa (gauge)  water = 1000 kg/m 3 SG of oil = 0.98 P atm = 0 4 m 2 m PAPA P A =15.7 kPa B A oil water P B = 54.9 kPA Pressure Diagram

21. Hydrostatic Force on Submerged Surfaces Curved Submerged Surface

22. Hydrostatic Force on Submerged Surfaces Curved Submerged Surface Horizontal Force = Equivalent Vertical Plane Force Vertical Force = Weight of Fluid Directly Above (+ Free Surface Pressure Force)

23. Hydrostatic Forces on Curved Surfaces F R on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface. Easiest approach: determine horizontal and vertical components F H and F V separately.

24. Forces on Curved Surfaces h1h1 h2h2

25. Submerged Curved Surface Resultant force:Horizontal and vertical components Horizontal component: F H =  s*w*(h + s/2), Where, h = Depth to the top of rectangle (beginning of curve surface) s = projected rectangle height w = projected rectangle length or width Center of pressure h p = h C + I C /(h C A) h C = h + s/2 Vertical Component F V =  Volume  A*w Where, A = entire area of fluid w = projected rectangle length or width

26. Hydrostatic Buoyant Force Archimedes’ principle  When a body is submerged or floating, the resultant force by the fluid is called the buoyancy force. This buoyancy force is acting vertically upward  The buoyancy force is equal to the weight of the fluid displaced by body.  The buoyancy force acts at the centroid of the displaced volume of fluid.  A floating body displaces a volume of fluid whose weight - body weight For equilibrium: + ΣF y = 0 F b – W = 0 or F b = W Therefore we can write ;  F b = weight of fluid displaced by the body Or F b = W = mg =  g  Where F b = buoyant force  = displaced volume of fluid W = weight of fluid W = mg F b = W GBGB GBGB W = mg Volume of displaced fluid

27. Buoyancy

28. For example, for a hot air balloon

29. Buoyancy and Stability Buoyancy is due to the fluid displaced by a body. F B =  f gV. Archimedes principal : The buoyant force = Weight of the fluid displaced by the body, and it acts through the centroid of the displaced volume.

30. Buoyancy and Stability Buoyancy force F B is equal only to the displaced volume  f gV displaced. Three scenarios possible  body <  fluid : Floating body  body =  fluid : Neutrally buoyant  body >  fluid : Sinking body

31. Stability of Immersed Bodies Rotational stability of immersed bodies depends upon relative location of center of gravity G and center of buoyancy B. G below B: stable G above B: unstable G coincides with B: neutrally stable.

32. Stability of Floating Bodies If body is bottom heavy (G lower than B), it is always stable. Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment. Measure of stability is the metacentric height GM. If GM>1, ship is stable.