1. 1 MECH 221 FLUID MECHANICS (Fall 06/07) Chapter 2: FLUID STATICS Instructor: Professor C. T. HSU

2. MECH 221 – Chapter 2 2 2.1. Hydrostatic Pressure  Fluid mechanics is the study of fluids in macroscopic motion. For a special static case: No Motion at All  Recall that by definition, a fluid moves and deforms when subjected to shear stress and, conversely, a fluid that is static (at rest) is not subjected to any shear stress. Otherwise it will move.  No shear stress, i.e., Normal stress only

3. MECH 221 – Chapter 2 3 2.1. Hydrostatic Pressure  The force/stress on any given surface immersed in a fluid at rest, is always perpendicular (normal) to the surface. This normal stress is called “pressure”  Fluid statics is to determine the pressure field  At any given point in a fluid at rest, the normal stress is the same in all directions (hydrostatic pressure)

4. MECH 221 – Chapter 2 4 2.1. Hydrostatic Pressure  Proof: Take a small, arbitrary, wedged shaped element of fluid Fluid is in equilibrium, so ∑F = 0 Let the fluid element be sufficiently small so that we can assume that the pressure is constant on any surface (uniformly distributed).

5. MECH 221 – Chapter 2 5 2.1. Hydrostatic Pressure  F 1 =p 1  A 1  F 2 =p 2  A 2  F 3 =p 3  A 3   m =  V  Fluid Density :   Fluid Volume :  V=  x  y  z/2

6. MECH 221 – Chapter 2 6 2.1. Hydrostatic Pressure  Look at the side view ∑F x = 0 : F 1 cos - F 2 = 0 p 1  A 1 cos - p 2  A 2 = 0 Since  A 1 cos =  A 2 =  y  z  p 1 = p 2 ∑F z = 0 : F 1 sin +  m.g = F 3 p 1  A 1 sin +   V g = p 3  A 3 p 1 (  x/sin )  y sin +  g  x  y  z/2 = p 3  x  y p 1 +  g  z/2 = p 3

7. MECH 221 – Chapter 2 7 2.1. Hydrostatic Pressure  Shrink the element down to an infinitesimal point, so that  z0, then p 1 = p 3  p 1 = p 2 = p 3  Notes: Normal stress at any point in a fluid in equilibrium is the same in all directions. This stress is called hydrostatic pressure. Pressure has units of force per unit area. P = F/A [N/m 2 ] The objective of hydrostatics is to find the pressure field (distribution) in a given body of fluid at rest.

8. MECH 221 – Chapter 2 8 2.2. Vertical Pressure Variation  Take a fluid element of small control volume in a tank at rest  Force balance: (P+  P) A +  g A  y = P A  P/  y = -  g  Negative sign indicates that P decreases as y increases A

9. MECH 221 – Chapter 2 9 2.2. Vertical Pressure Variation  For a constant density fluid, we can integrate for any 2 vertical points in the fluid (1) & (2). P 2 - P 1 = -  g (y 2 - y 1 ) If  = (y), then: ∫dP = -g ∫ (y)dy If  = (p,y) such as for ideal gas P =  RT where T=T (y) ∫dP/P = -(g/R) ∫dy/T(y) The integration at the right hand side depends on the distribution of T(y).

10. MECH 221 – Chapter 2 10 2.3. Horizontal Pressure Variation  Take a fluid element of small control volume  Force balance: P 1 A = P 2 A P 1 = P 2  Static pressure is constant in any horizontal plane.  Having the vertical & horizontal variations, it is possible to determine the pressure at any point in a fluid at rest.

11. MECH 221 – Chapter 2 11 2.3. Horizontal Pressure Variation  Absolute Pressure v.s. Gage Pressure Absolute pressure:  Measured from absolute zero Gage pressure:  Measured from atmospheric pressure If negative, it is called vacuum pressure  P abs = P atm + P gage

12. MECH 221 – Chapter 2 12 2.4. Forces on Immersed Surfaces  For constant density fluid: *The pressure varies with depth, P= gh. *The pressure acts perpendicularly to an immersed surface 2.4.1. Plane Surface Let the surface be infinitely thin, i.e. NO volume Plate has arbitrary plan form, and is set at an arbitrary angle, , with the horizontal.

13. MECH 221 – Chapter 2 13 2.4. Forces on Immersed Surfaces  Looking at the top plate surface only, the pressure acting on the plate at any given h is: P = P atm + gh  So, the pressure distribution on the surface is,

14. MECH 221 – Chapter 2 14 2.4. Forces on Immersed Surfaces  To find the total force on the top surface, integrate P over the area of the plate, F = ∫P dA = P atm A + g ∫h dA  Note that h = y sin, therefore: F = P atm A + g sin ∫y dA  Recall that the location of c.g.( center of gravity ) in y is: y c.g. = (1/A) ∫y dA

15. MECH 221 – Chapter 2 15 2.4. Forces on Immersed Surfaces  So, F = P atm A + g sin y c.g. A  Or, F = P atm A + gh c.g. A = (P atm + gh c.g )A  If P c.g =P atm + gh c.g, then the pressure acting at c.g. is: F = P c.g. A  In a fluid of uniform density, the force on a submerge plane surface is equal to the pressure at the c.g. of the plane multiplied by the area of the plane.  F is independent of .  The shape of the plate is not important

16. MECH 221 – Chapter 2 16 2.4. Forces on Immersed Surfaces  Where does the total/resultant force act? Similar to c.g., the point on the surface where the resultant force is applied is called the Center of Pressure, c.p.

17. MECH 221 – Chapter 2 17 2.4. Forces on Immersed Surfaces  The moment of the resultant force about the x- axis should equal the moment of the original distributed pressure about the x-axis  y c.p. F = ∫y dF =  g sin  ∫y 2 dA +P atm ∫y dA  Recall that the moment of inertia about the x-axis, I ox, is by definition: I ox = ∫y 2 dA = y 2 c.g. A + I c.g. x

18. MECH 221 – Chapter 2 18 2.4. Forces on Immersed Surfaces  I c.g. x - moment of inertia about the x-axis at c.g.  y c.p. F = g sin I ox +P atm y c.g. A = g sin (y 2 c.g. A + I c.g. x) +P atm y c.g. A =(g sin y c.g. A + P atm A) y c.g. + g sin I c.g. x  y c.p. = y c.g. + (g sin I c.g. x) / (P c.g. A)

19. MECH 221 – Chapter 2 19 2.4. Forces on Immersed Surfaces  Similarly, x c.p. = x c.g. + (g sin I c.g. y) / (P c.g. A) I c.g. y - moment of inertia about the y-axis at c.g. *Tables of I c.g. for common shapes are available *For simple pressure distribution profiles, the c.p. is usually at "c.g." of the profile

20. MECH 221 – Chapter 2 20 2.4. Forces on Immersed Surfaces 2.4.2. Curved Surface  Suppose a warped plate is submerged in water, what is the resulting force on it? The problem can be simplified by examining the horizontal and vertical components separately.

21. MECH 221 – Chapter 2 21 2.4. Forces on Immersed Surfaces 2.4.2.1. Horizontal Force  Zoom on an arbitrary point 'a'.  Locally, it is like a flat plate P a is the pressure acting at 'a', and it is normal to the surface.  The force due to the pressure at 'a' is: F a = P a A a, which acts along the same direction as P a  Its horizontal component is: F a H = F a sin = P a.A a sin

22. MECH 221 – Chapter 2 22 2.4. Forces on Immersed Surfaces  But, A a sin is the vertical projection of 'a', so that the horizontal force at 'a' due to pressure is equal to the force that would be exerted on a plane, vertical projection of 'a'. This can be generalized for the entire plane  The horizontal force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane  The line of action on a curved surface is the same as the line of action on a projected plane

23. MECH 221 – Chapter 2 23 2.4. Forces on Immersed Surfaces  This is true because for every point on the vertical projection there is a corresponding point on the warped plate that has the same pressure.

24. MECH 221 – Chapter 2 24 2.4. Forces on Immersed Surfaces 2.4.2.2. Vertical Force  Similar to the previous approach, F a V = F a cos = P a A a cos   A a cos is the horizontal projection of 'a', but this is only at a point!  Notice that if one looks at the entire plate, the pressures on the horizontal projection are not equal to the pressures on the plate

25. MECH 221 – Chapter 2 25 2.4. Forces on Immersed Surfaces  Note : P a = gh a  F a V = gh a A a cos  In general, P a ≠ P a' Consequently, one needs to integrate along the curved plate This is not difficult if the shape of the plate is given in a functional form

26. MECH 221 – Chapter 2 26 2.4. Forces on Immersed Surfaces  The ultimate result is: The vertical component of the force on a curved surface is equal to the total weight of the volume of fluid above it  The line of action is through the c.g. of the volume  If the lower side of a surface is exposed while the upper side is not, the resulting vertical force is equal to the weight of the fluid that would be above the surface

27. MECH 221 – Chapter 2 27 2.4. Forces on Immersed Surfaces  So far, only surfaces (not volumes) have been discussed  In fact, only one side of the surface has been considered  Note that for a surface to be in equilibrium, there has to be an equal and opposite force on the other side

28. MECH 221 – Chapter 2 28 2.5. Bodies with Volume (Buoyancy)  The volume can be constructed from two curved surfaces put together, and thus utilize the previous results.  Since the vertical projections of both plates are the same, FH ab = FH cd, Where FV ab =  g (vol. 1-a-b-2-1), FV cd =  g (vol. 1'-d-c-2'-1')  *Note that this is true regardless of whetherthere is or there isn't any fluid above c-d.

29. MECH 221 – Chapter 2 29 2.5. Bodies with Volume (Buoyancy)  Join the two plates together  Total force: F B =FV cd -FV ab =  g(vol. a-b-c-d)  This force F B is called Buoyancy Force

30. MECH 221 – Chapter 2 30 2.6. Archimedes' Principle  The net vertical force on an immersed body of arbitrary shape due to the pressure forces acting on the surfaces of the body is equal to the weight of the displaced fluid * The line of action is through the center of the mass of the displaced fluid volume * Direction of buoyant force is upward  If a body immersed in a fluid is in equilibrium, then: W = F B  W is the weight of the body.

31. MECH 221 – Chapter 2 31 2.6. Archimedes' Principle  For a body in a fluid of varying density, e.g. ocean, the body will sink or rise until it is at a height where its density is equal to the density of the fluid  For a body in a constant density fluid, the body will float at a level such that the weight of the volume of fluid it displaces is equal to its own weight

32. MECH 221 – Chapter 2 32 2.7. Pressure Variation with Rigid-Body Motion  The variation of pressure with distance is balanced by the total accelerations that may be due to gravitational acceleration g, constant linear acceleration a l and constant rotational acceleration a r. Generally, a = -(g + a l + a r )  For g in the vertical y direction, g = g j  For linear acceleration in the x and y directions, a l = ax i + ay j  For fluid rotates rigidly at a constant angular velocity ω, the acceleration a r is in the radial r direction, i.e., a r = -rω 2 e r where e r is the unit vector in r direction