1. 2.9 RELATIVE EQUILIBRIUM Fluid statics: no shear stresses  the variation of pressure is simple to compute 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. 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

2. Uniform Linear Acceleration Fig. 2.31: a liquid in an open vessel is given a uniform linear acceleration a 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  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 Fig. 2.31b: 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

3. Figure 2.31 Acceleration with free surface

4. 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,

5. To evaluate the constant of integration c: let x = 0, y = 0, p = p 0 ; then c = p 0 and When the accelerated incompressible fluid has a free surface, its equation is given by setting p = 0 in the above eq. Solving it for y gives 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

6. Example 2.15 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. Ft 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,

7. Figure 2.32 Tank completely filled with liquid

8. Example 2.16 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 s shown in the figure. When the origin is taken at 0, Eq. (2.9.2) becomes Then, for y = 0, along the bottom,

9. Figure 2.33 Uniform linear acceleration of container

10. Uniform Rotation about a Vertical Axis Forced-vortex motion: rotation of a fluid, moving as a solid, about an axis  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. No shear stresses exist in the liquid, and the only acceleration that occurs is directed radially 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: (2.2.5)

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

12. For constant angular velocity w, any particle of fluid P has an acceleration w 2 r directed radially inward (a = -iw 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 p is a function of y and r only: For a liquid (γ ≈ const) integration yields c is the constant of integration

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

14. When a free surface occurs in a container that is being rotated, the fluid volume underneath the paraboloid of revolution is the original fluid volume The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig. 2.35). The rise of liquid from its vertex to the wall of the cylinder is w 2 r 0 2 /rg (Eq. (2.9.6)), for a circular cylinder rotating about its axis. Since a paraboloid 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 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 w, r 0, and depth before rotation are given

15. Figure 2.35 Rotation of circular cylinder about its axis

16. Example 2.17 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 w = 200 x 2π/60 = 20.95 rad/s, γ = 1.2 x 9806 = 11.767 N/m3, rA = 1 m, and rB = 1.5 m. When the second equation is subtracted from the first and the values are substituted, Hence

17. Example 2.18 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 vortical axis through its midpoint 6.73 rad/s. Draw the paraboloid of zero pressure, and determine the pressure at the bottom and midpoint of the tube. In Fig. 2.36, the zero-pressure paraboloid 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, =.289 m and

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

19. Fluid Pressure Forces in Relative Equilibrium 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.

20. 과제물 스캐닝된 파일을 텍스트 파일로 만들 것. (2,3,4,5 장, 자습서 ) 교과서의 모든 예제는 반드시 제출하여야 함. 정지유체의 한점에 작용하는 압력은 모든 방향에 대해서 같음을 증명하라. 힘의 평형으로부터 오일러 방정식과 압력의 식을 유도하라. 기체에 대하여 온도가 변하지 않는 경우와 변하는 경우에 대하여 압력의 식을 유도하라. 액주계에 있어서 일반적으로 압력을 계산하는 절차를 설명하라. ( 단순 액주계 와 차분 액주계의 예를 포함하여 ). 면적이 다른 경우와 경사진 경우의 미차액주계에서 압력을 계산하는 식을 유 도하라. 1 차 2 차 모멘트에 대하여 설명하라. 부록에 나와있는 도형에 대하여 도심축과 x 축에 대한 2 차모멘트에 대한 식을 유도하라. 부록 B 의 수치해석 기법을 설명하라. 프로그램밍 포함. ( 교과서와 다른 책의 내 용 참조 ). 예제 2.7 의 문제를 Bisection 과 Newton-Raphson Method 를 사용하여 컴퓨터 프 로그램을 작성하여 풀어라.

21. 과제물 Runge-Kutta Method 를 설명하고 수치해석 책에 나와있는 문제를 풀어라 ( 프로 그램밍 포함 : C 및 Fortran).