Chapter 2 Pressure and Fluid Statics

1. Chapter 2 Pressure and Fluid Statics

2. introduction • Many fluid problems do not involve motion. They concern the pressure distribution in a static fluid and its effect on solid surfaces and on floating and submerged bodies. • When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

3. Introduction • Assuming a known fluid in a given gravity field, the pressure may easily be calculated by integration. • Important applications: • pressure distribution in the atmosphere and the oceans,(Dams, Gates,Tanks) • the design of manometer pressure instruments, • Forces on submerged flat and curved surfaces, • Buoyancy on a submerged body, and • the behavior of floating bodies.(ships)

4. Pressure • Pressure is defined as a normal force exerted by a fluid per unit area. • Units of pressure are N/m2, which is called a pascal (Pa). • Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kPa = 103 Pa) and megapascal (1 MPa = 106 Pa) are commonly used. • Other units include bar, atm, kgf/cm2, lbf/in2=psi.

5. Absolute, gage, and vacuum pressures • Actual pressure at a given point is called the absolute pressure. • Most pressure-measuring devices are calibrated to read zero in the atmosphere, and therefore indicate gage pressure, Pgage=Pabs - Patm. • Pressure below atmospheric pressure are called vacuum pressure, Pvac=Patm - Pabs.

6. Absolute, gage, and vacuum pressures Pgage=Pabs - Patm

7. Pressure at a Point • Pressure at any point in a fluid is the same in all directions. • Pressure has a magnitude, but not a specific direction, and thus it is a scalar quantity.

8. Variation of Pressure with Depth • In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers. • To obtain a relation for the variation of pressure with depth, consider rectangular element • Force balance in z-direction gives • Dividing by Dx and rearranging gives

9. Variation of Pressure with Depth • Pressure in a fluid at rest is independent of the shape of the container. • Pressure is the same at all points on a horizontal plane in a given fluid.

10. Example • El BroloceLake, a sea water lake near El Broloce , Kafr El Shaikh, has a maximum depth of 60 m in some areas, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum depth. • Using Table 2.1 ϒ = 10050N/m3 , z=-60m, pa= 91kPa p=pa-ϒz P=91x103 -10050*(-60)=91x103 + 603000=694kPa Why

11. Example • El Noubaria Lake, a water lake near Kafr El Dawar, Bahira, has a maximum depth of 60 m in some areas, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum depth. • Using Table 2.1 ϒ = 9790N/m3 , z=-60m, pa= 91kPa p=pa-ϒz P=91x103 -9790*(-60)=91x103 + 587400=678.4kPa

12. Scuba Diving and Hydrostatic Pressure

13. Scuba Diving and Hydrostatic Pressure • Pressure on diver at 100 ft? • Danger of emergency ascent? 1 100 ft 2 Boyle’s law If you hold your breath on ascent, your lung volume would increase by a factor of 4, which would result in embolism and/or death.

14. Pascal’s Law • Pressure applied to a confined fluid increases the pressure throughout by the same amount. • In picture, pistons are at same height: • Ratio A2/A1 is called ideal mechanical advantage

15. The Manometer • An elevation change of Dz in a fluid at rest corresponds to DP/rg. • A device based on this is called a manometer. • A manometer consists of a U-tube containing one or more fluids such as mercury, water, alcohol, or oil. • Heavy fluids such as mercury are used if large pressure differences are anticipated.

16. Mutlifluid Manometer • For multi-fluid systems • Pressure change across a fluid column of height h is DP = rgh. • Pressure increases downward, and decreases upward. • Two points at the same elevation in a continuous fluid are at the same pressure. • Pressure can be determined by adding and subtracting rgh terms.

17. Example Patm Patm • In opposite Fig. a tank contains water and immiscible oil at 20°C. What is h in centimeters if the density of the oil is 898 kg/m3?

18. Example-Solution For water take the density = 998 kg/m3. Apply the hydrostatic relation from the oil surface to the water surface, skipping the 8-cm part: Patm + (898)(g)(h+0.12) - (998)(g)(0.06+0.12)=Patm h ≈ 0.08 m ≈ 8.0 cm

19. Example All fluids in the Figure are at 20°C. If atmospheric pressure = 101.33 kPa and the bottom pressure is 242 kPa absolute, what is the specific gravity of fluid X?

20. Example –Solution Simply apply the hydrostatic formula from top to bottom: 242000 = 101330 + (8720)(1.0) + (9790)(2.0) +γ (3.0) + (133100)(0.5)

21. Measuring Pressure Drops • Manometers are well--suited to measure pressure drops across valves, pipes, heat exchangers, etc. • Relation for pressure drop P1-P2 is obtained by starting at point 1 and adding or subtracting rgh terms until we reach point 2. • If fluid in pipe is a gas, r2>>r1 and P1-P2= rgh

22. The Barometer • Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure. • PC can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to Patm. • Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance.

23. The Barometer Two "H" shaped resonators are patterned on the sensor, each operating at a high frequency output. As pressure is applied, the bridges are simultaneously stressed, one in compression and one in tension. The resulting change in resonant frequency produces a high differential output (kHz) directly proportional to the applied pressure. This simple time-based function is managed by a microprocessor.

24. Burdon tube In category 2, elastic deformation instruments, a popular, inexpensive, and reliable device is the bourdon tube, sketched in Fig. When pressurized internally, a curved tube with flattened cross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflection can be used to drive electric-output sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to drive another sensor.

25. Fluid Statics • Fluid Statics deals with problems associated with fluids at rest. • In fluid statics, there is no relative motion between adjacent fluid layers. • Therefore, there is no shear stress in the fluid trying to deform it. • The only stress in fluid statics is normal stress • Normal stress is due to pressure • Variation of pressure is due only to the weight of the fluid → fluid statics is only relevant in presence of gravity fields. • Applications: Floating or submerged bodies, water dams and gates, liquid storage tanks, etc.

26. High Dam

27. High Dam

28. High Dam

29. High Dam • Example of elevation head z converted to velocity head V2/2g. We'll be discussed this in more detail later (Bernoulli equation).

30. Hydrostatic Forces on Plane Surfaces • On a plane surface, the hydrostatic forces form a system of parallel forces • For many applications, magnitude and location of application, which is called center of pressure, must be determined. • Atmospheric pressure Patm can be neglected when it acts on both sides of the surface.

31. Resultant Force The magnitude of FRacting on a plane surface of a completely submerged plate in a homogenous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface

32. Center of Pressure • Line of action of resultant force FR=PCA does not pass through the centroid of the surface. In general, it lies underneath where the pressure is higher. • Vertical location of Center of Pressure is determined by equation the moment of the resultant force to the moment of the distributed pressure force. • $Ixx,C is tabulated for simple geometries.

33. Hydrostatic Forces on Curved Surfaces • FR 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 FH and FV separately.

34. Hydrostatic Forces on Curved Surfaces • Horizontal force component on curved surface: FH=Fx. Line of action on vertical plane gives y coordinate of center of pressure on curved surface. • Vertical force component on curved surface: FV=Fy+W, where W is the weight of the liquid in the enclosed block W=rgV. x coordinate of the center of pressure is a combination of line of action on horizontal plane (centroid of area) and line of action through volume (centroid of volume). • Magnitude of force FR=(FH2+FV2)1/2 • Angle of force is a = tan-1(FV/FH)

35. Centroid moments ofinertia for various cross sections

36. Centroid moments ofinertia for various cross sections In most cases the ambient pressure pa is neglected because it acts on both sides of the plate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam. In this case pCG ,hCG, and the center of pressure becomes independent of specific weight

37. Example The gate in the figureis 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute the force on the gate due to seawater pressure, the horizontal force P exerted by the wall at point A, and the reactions at the hinge B.

38. Example – solution (a) By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B. The depth hCG is thus 15 - 3 = 12 ft. The gate area is 5(10) = 50 ft2. Neglect pa as acting on both sides of the gate. the hydrostatic force on the gate is F= pCGA=ϒhCGA=(64 lbf/ft3)(12 ft)(50 ft2)= 38,400 lbf

39. Example – solution (b) First we must find the center of pressure of F. A free-body diagram of the gate is shown in the figure. The gate is a rectangle, hence since pa is neglected.

40. Example – solution The distance from point B to force F is thus 10 - l - 5 = 4.583 ft. Summing moments counterclockwise about B gives ( c ) With F and P known, the reactions Bxand Bzare found by summing forces on the gate

41. Example – solution

42. Example A tank of oil has a right-triangular panel near the bottom, as in the figure . Omitting pa, find the • hydrostatic force and • (b) CP on the panel.

43. Example – solution • The triangle has properties given in the figure. The centroid is one-third up (4 m) and one-third over (2 m) from the lower left corner, as shown. The area is 1/2(6 m)(12 m) = 36 m2 The moments of inertia are

44. Example – solution The depth to the centroid is hCG= 5 + 4= 9 m; thus the hydrostatic force (b) The CP position is given by

45. Example forces FHand FVon the dam and the position CP where they act. The width of the dam is 50 ft.

46. Example – solution The vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with its centroid halfway down, or hCG= 12 ft. The force FHis thus The line of action of FHis below the centroid by an amount Thus FHis 12 4 16 ft

47. Example – solution The vertical component FV equals the weight of the parabolic portion of fluid above the curved surface. The geometric properties of a parabola are shown in Fig. The weight of this amount of water is This acts downward on the surface at a distance 3x0/8 = 3.75 ft over from the origin of coordinates. Note that the vertical distance 3z0/5 in Fig. is irrelevant.

48. Example – solution The total resultant force acting on the dam is As seen in Fig., this force acts down and to the right at an angle of 29°= The force F passes through the point (x, z) (3.75 ft, 8 ft). If we move down along the 29° line until we strike the dam, we find an equivalent center of pressure on the dam at xCP = 5.43 ftzCP= 7.07 ft

49. Buoyancy and Stability • Buoyancy is due to the fluid displaced by a body. FB=rfgV. • Archimedes principal : The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.

50. Buoyancy and Stability • Buoyancy force FB is equal only to the displaced volume rfgVdisplaced. • Three scenarios possible • rbody<rfluid: Floating body • rbody=rfluid: Neutrally buoyant • rbody>rfluid: Sinking body