1. SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING

2. GUIDED BY PROF. ANERI CHAVAN NO. NAMEENROLLMENT NO 1PANCHAL JIGAR M.130450106022 2PANDYA VASISTH P.130450106023 3PARMAR HARSHAD130450106025 4KALPESH130450106026 5VISHAL130450106027 TOPIC:- Fluids In a Relative Equilibrium

3. T OPICS :- Fluid Statics What is Relative Equilibrium? Liquid in a uniform linear acceleration. Liquid Containers subjected to constant Vertical acceleration Liquid Containers subjected to constant Horizontal acceleration. Liquid Containers Subjected to constant Rotation.

4. Fluid statics deals with situations of "static equilibrium." In static equilibrium: No part of a fluid is in motion relative to another part of the fluid - no shear stress is present in the fluid - only normal isotropic stresses (i.e.. pressure) exist. The fluid could simply be at rest, or it could be moving. F LUID S TATICS

5. If motion of the fluid is involved, the only motions possible in static equilibrium are a solid body translation, a solid body rotation, or a combination of the two (ex. a body of fluid may be rotating about its axis at the same time that it is being translated along a particular direction). The terminology "solid body" implies that the movement of the fluid is equivalent to that of a body in which no part is capable of movement relative to any other part, as in a solid.

6. The science of fluid statics will be treat in two parts: the study of pressure and its variation throughout a fluid; the study of pressure forces on finite surfaces. Special cases of fluids moving as solids are included in the treatment of statics because of the similarity of forces involved.

7. Since there is no motion of a fluid layer relative to an adjacent layer, there are no shear stresses in the fluid. Hence, all free bodies in fluid statics have only normal pressure forces acting on their surfaces.

8. RELATIVE EQUILIBRIUM In fluid statics the variation of pressure is simple to compute, thanks to the absence of shear stresses. For fluid motion such that no layer moves relative to an adjacent layer, the shear stress is also zero throughout the fluid. A fluid with a translation at uniform velocity still follows the laws of static variation of pressure.

9. When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid), no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body. Two cases are of interest, a uniform linear acceleration and a uniform rotation about a vertical axis. When moving thus, the fluid is said to be in Relative Equilibrium.

10. F LUID P RESSURE F ORCES IN R ELATIVE E QUILIBRIUM The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface. When the pressure varies linearly over the plane surface (linear acceleration), the magnitude of force is given by the product of pressure at the centroid and area, since the volume of the pressure prism is given by p G A. For nonlinear distributions the magnitude and line of action can be found by integration.

11. U NIFORM L INEAR A CCELERATION A liquid in an open vessel is given a uniform linear acceleration a as in Fig. 2.31. After some time the liquid adjusts to the acceleration so that it moves as a solid; i.e., the distance between any two fluid particles remains fixed, and no shear stresses occur. By selecting a Cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig. 2.31a), the z axis is normal to a and there is no acceleration component in that direction. Eq`n (2.2.5) applies to this situation,

12. Figure 2.33 Figure 2.33 Uniform linear acceleration of container

13. Fig. 2.31b shows the pressure gradient ∇ p is then the vector sum of -ρa and -jγ. Since ∇ p is in the direction of maximum change in p (the gradient), at right angles to ∇ p there is no change in p. Surfaces of constant pressure, including the free surface, must therefore be normal to ∇ p.

14. Figure 2.31 Figure 2.31 Acceleration with free surface

15. To obtain a convenient algebraic expression for variation of p with x, y, and z, that is, p = p(x, y, z), Eq. (2.2.5) is written in component form: Since p is a function of position (x, y, z), its total differential is Substituting for the partial differentials gives which can be integrated for an incompressible fluid,

16. To evaluate the constant of integration c, let x = 0, y = 0, p = p 0 ; then c = p 0 and (2.9.2) When the accelerated incompressible fluid has a free surface, its equation is given by setting p = 0 in the above Eq. (2.9.2). Solving it for y gives (2.9.3) The lines of constant pressure, p = const, have the slope and are parallel to the free surface. The y intercept of the free surface is

17. Example 2.14 The tank in Fig. 2.32 is filled with oil, relative density 0.8, and accelerated as shown. There is a small opening in the rank at A. Determine the pressure at B and C; and the acceleration a x required to make the pressure at B zero. By selecting point A as origin and by applying Eq. (2.9.2) for a y = 0 At B, x = 1.8 m, y = - 1.2 m, and p = 2.35 kPa. At C, x = -0.15 m, y = -1.35 m, and p = 11.18 kPa. For zero pressure at B, from Eq. (2.9.2) with origin at A,

18. Figure 2.32 Figure 2.32 Tank completely filled with liquid

19. Example 2.15 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig. 2.33). It is given a constant linear acceleration a x = g/2, a y = -g/4. Develop an equation for variation of pressure along its base. The free surface has the slope: hence, the free surface is located as shown in the figure. When the origin is taken at 0, Eq. (2.9.2) becomes Then, for y = 0, along the bottom,

20. U NIFORM R OTATION ABOUT A V ERTICAL A XIS Rotation of a fluid, moving as a solid, about an axis is called forced-vertex motion. Every particle of fluid has the same angular velocity. This motion is to be distinguished from free-vortex motion, in which each particle moves in a circular path with a speed varying inversely as the distance from the center. A liquid in a container, when rotated about a vertical axis at constant angular velocity, moves like a solid alter some time interval. (2.2.5)

21. No shear stresses exist in the liquid, and the only acceleration that occurs is directed racially inward toward the axis of rotation. By selecting a coordinate system (Fig. 2.34a) with the unit vector I in the r direction and j in the vertical upward direction with y the axis of rotation, the following equation may be applied to determine pressure variation throughout the fluid:

22. Figure 2.34 Figure 2.34 Rotation of a fluid about a vertical axis

23. For constant angular velocity ω, any particle of fluid P has an acceleration ω 2 r directed racially inward (a = -Iω 2 r). Vector addition of -jγ and -ρa (Fig. 2.34b) yields ∇ p, the pressure gradient. The pressure does not vary normal to this line at a point  if P is taken at the surface, the free surface is normal to ∇ p. Expanding Eq. (2.2.5) k is the unit vector along the z axis (or tangential direction). Then

24. Since p is a function of y and r, the total differential dp is For a liquid (γ ≈ const) integration yields in which c is the constant of integration.

25. If the value of pressure at the origin (r = 0, y = 0) is p 0, then c = p 0 and (2.9.5) When the particular horizontal plane (y = 0) for which p 0 = 0 is selected and the above equation is divided by γ, (2.9.6) which shows that the head, or vertical depth, varies as the square of the radius. The surfaces of equal pressure are parabolic of revolution.

26. When a free surface occurs in a container that is being rotated, the fluid volume underneath the parabolic of revolution is the original fluid volume. The shape of the parabolic depends only upon the angular velocity ω. For a circular cylinder rotating about its axis (Fig. 2.35) the rise of liquid from its vertex to the wall of the cylinder is ω 2 r 0 2 /2g (from Eq. (2.9.6)). Since a parabolic of revolution has a volume equal to one-half its circumscribing cylinder, the volume of the liquid above the horizontal plane through the vertex is

27. When the liquid is at rest, this liquid is also above the plane through the vertex to a uniform depth of Hence, the liquid rises along the walls the same amount as the center drops, thereby permitting the vertex to be located when ω, r 0, and depth before rotation are given.

28. Figure 2.35 Rotation of circular cylinder about its axis

29. Example 2.16 A liquid, relative density 1.2, is rotated at 200 rpm about a vertical axis. At one point A in the fluid 1 m from the axis, the pressure is 70 kPa. What is the pressure at a point B which is 2 m higher than A and 1.5 m from the axis? When Eq. (2.9.5) is written for the two points, Then ω = 200 × 2π/60 = 20.95 rad/s, γ = 1.2 × 9806 = 11.767 N/m 3, r A = 1 m, and r B = 1.5 m. When the second equation is subtracted from the first and the values are substituted, Hence

30. Example 2.17 A straight tube 2 m long, closed at the bottom and filled with water, is inclined 30 o with the vertical and rotated about a vertical axis through its midpoint 6.73 rad/s. Draw the parabolic of zero pressure, and determine the pressure at the bottom and midpoint of the tube. In Fig. 2.36, the zero-pressure parabolic passes through point A. If the origin is taken at the vertex, that is, p 0 = 0, Eq. (2.9.6) becomes which locates the vertex at O, 0.577 m below A. The pressure at the bottom of the tube is or At the midpoint, = 0.289 m and

31. Figure 2.36 Rotation of inclined tube of liquid about a vertical axis

32. P RESSURE D ISTRIBUTION IN A L IQUID S UBJECT TO H ORIZONTAL A CCELERATION Figure shows a liquid contained in a tank which has an acceleration a. A particle of mass m on the free surface at O will have the same acceleration as the tank and will be subjected to an accelerating force F. From Newton’s second law,

33. F=ma (2.30) Also, F is the resultant of the fluid pressure force R, acting normally to the free surface at O, and the weight of the particle mg, acting vertically. Therefore, F = mgTanθ (2.31) Comparing Eqs. (2.30) and (2.31) Tanθ = a/g and is constant for all points on the free surface. Thus, the free surface is a plane inclined at a constant angle θ to the horizontal. Since the acceleration is horizontal, vertical forces are not changed and the pressure at any depth h below the surface will be γh. Planes of equal pressure lie parallel to the free surface.