1. Louisiana Tech University Ruston, LA 71272 Momentum Balance Steven A. Jones BIEN 501/CMEN 513 Friday, March 17, 2006

2. Louisiana Tech University Ruston, LA 71272 Momentum Balance Learning Objectives: 1.State the motivation for curvilinear coordinates. 2.State the meanings of terms in the Transport Theorem 3.Differentiate between momentum as a property to be transported and velocity as the transporting agent. 4.Show the relationship between the total time derivative in the Transport Theorem and Newton’s second law. 5.Apply the Transport Theorem to a simple case (Poiseuille flow). 6.Identify the types of forces in fluid mechanics. 7.Explain the need for a shear stress model in fluid mechanics. The Stress Tensor. Appendix A.5 Show components of the stress tensor in Cartesian and cylindrical coordinates. Vectors and GeometryVectors and Geometry

3. Louisiana Tech University Ruston, LA 71272 Motivation for Curvilinear Coordinates Fully developed pipe flow (Poiseuille) Flow around a small particle (Stokes Flow) Applications: How fast does a blood cell settle? What is the motion of a catalyzing particle? Application: What is the flow stress on an endothelial cell?

4. Louisiana Tech University Ruston, LA 71272 Cylindrical coordinates are simpler because of the boundary conditions: Cylindrical Coordinates: Examples Fully developed pipe flow (Poiseuille) In cartesian coordinates, there are three velocity components to worry about. In spherical coordinates, one of these components is zero (u  ). r  

5. Louisiana Tech University Ruston, LA 71272 Stokes (Creeping Flow) In cartesian coordinates, there are three velocity components to worry about. To confirm the three components, consider the point ( x, y, z ) = (1, 1, 1). Slice parallel to the equator (say the equator is in the xz plane): This velocity vector has an x and z component (visible above) and a y component (visible to the left). Top View x z x z x y x z

6. Louisiana Tech University Ruston, LA 71272 Momentum Balance Consider flow entering a control volume: The rate at which momentum is generated in a chunk of fluid that is entering the control volume is governed by the Reynolds Transport Theorem

7. Louisiana Tech University Ruston, LA 71272 Momentum Balance Consider flow entering a control volume: The property  in this case is momentum per unit volume,  =  v. Both  and v are bold (vectors).

8. Louisiana Tech University Ruston, LA 71272 Momentum Balance It is useful to recall the meanings of the terms. Rate at which the momentum of the fluid passing through the sample volume increases (production of momentum). Rate at which the momentum increases inside the sample volume (partial derivative) Flux of momentum through the surface of the control volume.

9. Louisiana Tech University Ruston, LA 71272 Say What? The momentum of the car passing through the location of measurement is increasing. The momentum at the location of measurement is not increasing. Location of Measurement 25 mph 40 mph Rate at which the momentum of the fluid passing through the sample volume increases (production of momentum).

10. Louisiana Tech University Ruston, LA 71272 Momentum Balance & Newton The momentum balance is a statement of Newton’s second law. Production of Momentum (Force per unit volume). Eulerian form of the time derivative of momentum (i.e. ma per unit volume).

11. Louisiana Tech University Ruston, LA 71272 Momentum Balance & Newton Lagrangian time derivative Eulerian form of the time derivative of momentum (i.e. ma per unit volume). Eulerian time derivative

12. Louisiana Tech University Ruston, LA 71272 Momentum Balance It is also useful to note that this is three equations, one for each velocity component. For example, the v 1 component of this equation is: But note that the full vector v remains in the last integral.

13. Louisiana Tech University Ruston, LA 71272 Momentum Surface Flux The roles of the velocity components differ, depending on which surface is under consideration. The momentum being carried through the surface. The velocity vector that carries momentum through the surface. v1v1 v2v2 In the figure to the left: The velocity component perpendicular to the plane (v 2 ) carries momentum (  v 1 ) through the plane.

14. Louisiana Tech University Ruston, LA 71272 Momentum Shell Balance Fully developed pipe flow (Poiseuille) dr dz Assumptions: 1.Steady, incompressible flow (no changes with time) 2.Fully developed flow 3.Velocity is a function of r only ( v=v ( r )) 4.No radial or circumferential velocity components. 5.Pressure changes linearly with z and is independent of r. Note: 3, 4 and 5 follow from 1 and 2, but it takes a while to demonstrate the connection. vrvr

15. Louisiana Tech University Ruston, LA 71272 The Control Volume The control volume is an annular region dz long and dr thick. We will be concerned with 4 surfaces: rr  rr  rz Outer Cylinder

16. Louisiana Tech University Ruston, LA 71272 The Control Volume Inner Cylinder rr  rr  rz

17. Louisiana Tech University Ruston, LA 71272 The Control Volume zz  zr  zz Left Annulus

18. Louisiana Tech University Ruston, LA 71272 zz  zr  zz The Control Volume Right Annulus

19. Louisiana Tech University Ruston, LA 71272 Continuity The mass entering the annular region = the mass exiting. dr dz Thus: This equation is automatically satisfied by assumption 3 (velocity does not depend on z).

20. Louisiana Tech University Ruston, LA 71272 Momentum in Poiseuille Flow The momentum entering the annular region - the momentum leaving= momentum destruction. (Newton’s 2 nd law – F=ma ) dr dz In fluid mechanics, we talk about momentum per unit volume and force per unit volume. For example, the force per unit volume caused by gravity is  g since F=mg. (Units are g cm/s 2 ).

21. Louisiana Tech University Ruston, LA 71272 Momentum Rate of momentum flow into the annulus is: Again, because velocity does not change with z, these two terms cancel one another. Rate of momentum flow out is: dr dz

22. Louisiana Tech University Ruston, LA 71272 Shearing Force Denote the shearing force at the cylindrical surface at r as  ( r ). The combined shearing force on the outer and inner cylinders is: dr dz Note the signs of the two terms above.

23. Louisiana Tech University Ruston, LA 71272 Pressure Force The only force remaining is that cause by pressure on the two surfaces at r and r+dr. dr dz This force must balance the shearing force:

24. Louisiana Tech University Ruston, LA 71272 Force Balance Divide by 2  dr dz :

25. Louisiana Tech University Ruston, LA 71272 Force Balance Take the limit as Now we need a model that describes the relationship between the shear rate and the stress. From the previous slide:

26. Louisiana Tech University Ruston, LA 71272 Shear Stress Model The Newtonian model relating stress and strain rate is: Thus, In our case,

27. Louisiana Tech University Ruston, LA 71272 Differential Equation So, with: The equation is: and

28. Louisiana Tech University Ruston, LA 71272 Differential Equation If viscosity is constant, By symmetry, Since the pressure gradient is constant (assumption 4), we can integrate once:, so C 1 = 0.

29. Louisiana Tech University Ruston, LA 71272 Differential Equation If viscosity is constant, By symmetry, Since the pressure gradient is constant (assumption 4), we can integrate once:, so C 1 = 0.

30. Louisiana Tech University Ruston, LA 71272 Differential Equation Integrate again. The no-slip condition at r=R is u z =0, so With

31. Louisiana Tech University Ruston, LA 71272 Review, Poiseuille Flow 1.Use a shell balance to relate velocity to the forces. 2.Use a model for stress to write it in terms of velocity gradients. 3.Integrate 4.Use symmetry and no-slip conditions to evaluate the constants of integration. If you have had any course in fluid mechanics before, you have almost certainly used this procedure already.

32. Louisiana Tech University Ruston, LA 71272 Moment of Momentum Balance It is useful to recall the meanings of the terms. Rate at which the moment of momentum of the fluid passing through the sample volume increases (production of momentum). Rate at which the moment of momentum increases inside the sample volume (partial derivative) Flux of moment of momentum through the surface of the control volume.

33. Louisiana Tech University Ruston, LA 71272 Types of Forces 1.External Forces (gravity, electrostatic) 2. Mutual forces (arise from within the body) a.Intermolecular b.electrostatic 3. Interfacial Forces (act on surfaces)

34. Louisiana Tech University Ruston, LA 71272 Types of Forces 1. Body Forces (Three Dimensional) Gravity Magnetism 2. Surface Forces (Two Dimensional) a.Pressure x Area – normal to a surface b.Shear stresses x Area – Tangential to the surface 3. Interfacial Forces (One Dimensional) e.g. surface tension x length)

35. Louisiana Tech University Ruston, LA 71272 Types of Forces 4. Tension (Zero Dimensional) The tension in a guitar string. OK, really this is 2 dimensional, but it is treated as zero-dimensional in the equations for the vibrating string.

36. Louisiana Tech University Ruston, LA 71272 The Stress Tensor The first subscript is the face on which the stress is imposed. The second subscript is the direction in which the stress is imposed.

37. Louisiana Tech University Ruston, LA 71272 The Stress Tensor The diagonal terms (normal stresses) are often denoted by  i.

38. Louisiana Tech University Ruston, LA 71272 Exercise For a general case, what is the momentum balance in the  -direction on the differential element shown (in cylindrical coordinates)? dr dd dz

39. Louisiana Tech University Ruston, LA 71272 Look at the  r term Divide by dr, d , dz

40. Louisiana Tech University Ruston, LA 71272 Contact Forces Diagonal elements are often denoted as  B-P P t(z,P)t(z,P) Stress Principle: Regardless of how we define P, we can find t( z,n ) n

41. Louisiana Tech University Ruston, LA 71272 Contact Forces Diagonal elements are often denoted as  B-P t(z,P)t(z,P) Stress Principle: Regardless of how we define P, we can find t(z,n) n P

42. Louisiana Tech University Ruston, LA 71272 Cauchy’s Lemma Stress exerted by B-P on P is equal and opposite to the force exerted by P on B-P. BPBP t(z,n)t(z,n) n P BPBP t(z, n)t(z, n) P nn

43. Louisiana Tech University Ruston, LA 71272 Finding t(z,n) If we know t(z,n) for some surface normal n, how does it change as the orientation of the surface changes? BPBP t(z,n)t(z,n) n P

44. Louisiana Tech University Ruston, LA 71272 The Tetrahedron We can find the dependence of t on n from a momentum balance on the tetrahedron below. Assume that we know the surface forces on the sides parallel to the cartesian basis vectors. We can then solve for the stress on the fourth surface. A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2

45. Louisiana Tech University Ruston, LA 71272 The Stresses on A i Must distinguish between the normals to the surfaces A i and the directions of the stresses. In this derivation, stresses on each surface can point in arbitrary directions. t i is a vector, not a component. A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 t1t1

46. Louisiana Tech University Ruston, LA 71272 Components of t i Recall the stress tensor: A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 t1t1 Surface (row) Direction of Force (Column)

47. Louisiana Tech University Ruston, LA 71272 Total Derivative Let t 1, t 2, and t 3 be the stresses on the three A i A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 Value is constant if region is small. Volume of the tetrahedron.

48. Louisiana Tech University Ruston, LA 71272 Body Forces Similarly, A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 Value is constant if region is small. Volume of the tetrahedron.

49. Louisiana Tech University Ruston, LA 71272 Surface Forces A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 Area of the surface. t does not vary for differential volume.

50. Louisiana Tech University Ruston, LA 71272 Find A i A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 Each of the A i is the projection of A on the coordinate plane. Note that A projected on the z 1 z 2 plane is just (i.e. dot n with the  coordinate).

51. Louisiana Tech University Ruston, LA 71272 Infinitessimal Momentum Balance A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 Thus: Divide by A Take the limit as the tetrahedron becomes infinitessimally small (h  0)

52. Louisiana Tech University Ruston, LA 71272 Meaning of This Relationship A3A3 A2A2 A1A1 n z1z1 z3z3 z2z2 If we examine the stress at a point, and we wish to determine how it changes with the direction of the chosen normal vector (i.e. with the orientation of the surface of the body), we find that: Where are the stresses on the surfaces perpendicular to the coordinate directions.

53. Louisiana Tech University Ruston, LA 71272 Mohr’s Circle Those students familiar with solid mechanics will recall the Mohr’s Circle, which is a statement of the previous relationship for 2-dimensions in solids.

54. Louisiana Tech University Ruston, LA 71272 Symmetry The stress tensor is symmetric. I.e.  ij=  ji

55. Louisiana Tech University Ruston, LA 71272 Kroneker Delta The Kroneker delta is defined as: It can be thought of as a compact notation for the identity matrix:

56. Louisiana Tech University Ruston, LA 71272 Permutation Tensor The permutation tensor is defined as: It is a sparse tensor, so the only components (of 27 possible) that are not zero are: 1 2 3 1 3 2 Positive Negative

57. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta A well known result is: This expression is a 2 nd order tensor, each component of which is the sum of 9 terms. For example, with m =1 and n =2. (note sums over j and k ) j ’s are the same k ’s are the same

58. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta Consider the mn component of If m=n=1, then there are only 2 possibilities for j and k that do not lead to zero values of the permutation tensor. They can be 2 and 3, for if either is 1, then the value is zero. The same result occurs for m=n= 2 and m=n= 3. I.e. if m=n, then the value us 2.

59. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta If m  n, then the first  is nonzero only if its j and k indices are not m. But in that case, since n must be one of these other two values and the second  must be zero. I.e., the expression is zero when m  n. The two results combine as follows: This expression is valuable because it allows us to relate something that looks complicated in terms of something that is more readily understandable.

60. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta Consider another expression: This expression is frightening because it is a 4 th order tensor. It has 81 components, each of which is made of 3 terms. Yet, all terms for which i=j or m=n or will be zero. Let i= 1, then j=2, k=3 or j=3, k=2 give nonzero results. If k=3, then there are only two nonzero values of m and n. Overall, the “important” values of the subscripts are: Gives +1 Gives  1

61. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta Consider another expression: Gives +1 Gives  1 Only 6 terms are non-zero (those for which i  j and either i=m, j=n or i=n and j=m. Two of these are:

62. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta Thus,

63. Louisiana Tech University Ruston, LA 71272 Permutation Tensor and Delta It can be shown, through similar enumeration, that the delta expression: Gives the same results and that therefore: