LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE

1. LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Outward normal vector: consider an arbitrarily shaped simply-connected volume. I have drawn a cylinder below because that is what I could draw in Excel, but the volume could be any shape. The direction in which any location on the volume is facing is characterized in terms of a unit outward normal vector . The direction of depends upon the location on the surface of the volume: for example here or here or here In index notation the unit normal vector can be written simply as ni (or nj or np etc.)

2. dA dV LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE The divergence theorem is a simple example of a more general integral theorem that comes in many other flavors. It relates an integral defined over the surface of a control volume to another over the internal volume of the control volume Using the velocity vector as an example, the standard divergence theorem takes the form where S denotes the surface of the volume, dA denotes an element of surface area, V denotes the volume of the control volume, and dV denotes an element of volume. Note that the unit outward normal vector is denoted as ni in index notation.

3. LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE The following integral theorems which are related to the divergence theorem will also prove useful in this course: The first of these holds for any scalar (zeroth-order tensor) A and the second holds for any matrix (second-order tensor) Aij, but the examples of pressure and the stress tensor have been chosen for a reason that will become apparent before long.

4. z+z x y y+y z x+x LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE All of these theorems represent 3-dimensional analogs of the simple statement that integration and differentiation are reciprocal operations, or more specifically Here we prove the following theorem for a an infinitesimal rectangular control volume: The control volume has dimensions x, y and z

5. The contribution to from these two faces is thus z+z x y y+y z x+x LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Now consider the two faces in the y direction. On the left face and on the right face

6. But z+z x y y+y z x+x LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Similarly computing the contribution from the other two faces, it is found that so establishing the theorem for an infinitesimal body.

7. LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE The stress tensor contains the normal stresses 11, 22 and 33 and the shear stresses 12, 21, 13, 31, 23 and 32. In fluids, the normal stresses 11, 22 and 33 have components associated with pressure that are all equal to each other. Since a normal stress is defined as positive for the case of tension, and since fluids are generally in compression, it makes sense to change the sign convention and define these pressure components as so that p is positive when the fluid is in compression.

8. LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE A fluid can be defined as a state of matter which cannot withstand a sustained shear stress without flowing. The flow continues until application of the shear stress is halted. Thus for a fluid to be at rest in static equilibrium, all the shear stresses 12, 13, 23, 21, 32 and 31 must be vanishing, so that the stress tensor takes the form Now why should the three normal stresses be equal, or equivalently, why should pressure p be isotropic(the same in every direction)?

9. LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE To show why this is the case, we first consider a fluid at rest (fluid statics, or hydrostatics in the case of water). Gravity pulls the water down. But the water does not move because it is in static balance. Static balance is obtained when the pressure distribution is just sufficient to balance gravity. Consider the illustrated control volume with height z in the vertical direction and area A in the horizontal plane. For simplicity, the fluid has a free surface, the location of which is located by the inverted triangle. (The existence of a free surface is not necessary for the derivation. In addition, the fluid is assumed to be incompressible, so that density  is (to a high degree of approximation) constant. A z z+z z

10. or or LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE The downward force of gravity acting on the control volume is given as gzA. The pressure force acting on the bottom face is p(z)A, and it acts upward. (Remember that a positive pressure always acts to compress) The pressure force acting on the top face is p(z+z)A, and it acts downward. For static equilibrium, then, the force balance is A p(z+z) z z+z z p(z)

11. A p(z+z) z z+z z p(z) p LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE The relation or Is the foundation of the study of fluid statics of incompressible fluids. It says that pressure increases linearly downward in the fluid. But this relation in and of itself is insufficient to establish that pressure is isotropic, as it only considers the vertical direction.

12.  LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Now consider the static fluid in the illustrated volume with a right-triangular cross-section and angle . The volume has dimensions x, y and z where z denotes the upward vertical direction. p3 z The pressures acting on the bottom face, left side face and diagonal face are denoted as p1, p2 and p3, respectively. Note that each pressure acts inward normal to the face in question.  p2 y The volume is (1/2)xyz and the weight is g(1/2)xyz (which acts only downward.) x p1 The area A of the diagonal face is given as The pressure force on the diagonal face has components p3Acos acting downward and p3Asin acting to the left in the horizontal plane.

13. But so the equation reduces to  p3 p3 p1 p1 LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Force balance in the vertical direction: p3 z  p2 In the limit as x, y and z all  0 we obtain the following result at a point: y x p1 

14. Force balance in horizontal direction: But so the equation reduces to  In the limit as x, y and z all  0 we obtain the following result at a point: p3 p3 p2 p2 LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE p3 z  p2 y x p1 

15.  p p or p LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Now the angle  was not specified, so the result holds for any angle. The result, then, is that the pressure p at a point is isotropic, or the same in any direction. Strictly speaking, the result holds only for static fluids. But it easily generalizes to moving fluids. For example, let the fluid be accelerating in the vertical direction at rate az. The force balance then generalizes to or in the limit as z  0,

16. It is easy to see that such a buoyant force should exist. Since pressure increases linearly downward according to the relation the upward-acting pressure on the bottom of the body must be larger than the downward-acting pressure on the top of the body. p LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Archimedes’ theorem states that a body submerged in a static fluid is buoyed upward by a force equal to the weight of fluid it displaces. Thus if the body has volume V and the upward vertical coordinate is z = x3, the buoyant force FBi acting on the body is

17. The total pressure force is obtained by integrating over the whole body: But using the divergence theorem and the relation for static pressure it is found that LECTURE 2: DIVERGENCE THEOREM, PRESSURE, ARCHIMEDES PRINCIPLE Now let’s prove Archimedes’ theorem using the divergence theorem. Consider an elemental area dA on the surface of the body with unit outward normal vector ni. Since pressure pushes inward, the pressure force on the elemental area is ni dA p so establishing the desired result.