1. Louisiana Tech University Ruston, LA 71272 Slide 1 Krogh Cylinder Steven A. Jones BIEN 501 Wednesday, May 7, 2008

2. Louisiana Tech University Ruston, LA 71272 Slide 2 Announcements 1.All homeworks have been assigned. 2.Final exam will be taken from parts of the homework. 3.No homework on Krogh cylinder. 4.Krogh cylinder will not be on the exam. 5.Friday – finish Krogh and do Comparmental Models 6.Monday – Review and Course Evaluations 7.Wednesday - Exam 8.Friday – No class, will be available for questions. 9.Today – Office Hours will start at 10:30.

3. Louisiana Tech University Ruston, LA 71272 Slide 3 Energy Balance Major Learning Objectives: 1.Learn a simple model of capillary transport.

4. Louisiana Tech University Ruston, LA 71272 Slide 4 The Krogh Cylinder Capillary Tissue

5. Louisiana Tech University Ruston, LA 71272 Slide 5 Assumptions The geometry follows the Krogh cylinder configuration –Radial symmetry –Transport from capillary –Capillary influences a region of radius R k. Reactions are continuously distributed There is a radial location at which there is no flux

6. Louisiana Tech University Ruston, LA 71272 Slide 6 Capillary Transport Consider the following simple model for capillary transport: Reactive Tissue Capillary Interior Matrix

7. Louisiana Tech University Ruston, LA 71272 Slide 7 Capillary Transport What are appropriate reaction rates and boundary conditions? Constant rate of consumption (determined by tissue metabolism, not O 2 concentration) Not metabolic (no consumption)

8. Louisiana Tech University Ruston, LA 71272 Slide 8 Diffusion Equation For steady state:

9. Louisiana Tech University Ruston, LA 71272 Slide 9 Constant Rate of Reaction Assume the rate of reaction, r x, is constant: And that the concentration is constant at the capillary wall and zero at the edge of the Krogh cylinder ( M will be numerially negative since the substance is being consumed).

10. Louisiana Tech University Ruston, LA 71272 Slide 10 Constant Reaction: Roadmap Integrate the differential equation once to obtain a solution for flux as a function of r. Use the zero flux boundary condition at R k to determine the first constant of integration. Substitute the constant back into the previously- integrated differential equation. Integrate again to obtain the form for concentration. Use the constant concentration boundary condition to determine the second constant of integration. The roadmap for solving the steady state problem is as follows:

11. Louisiana Tech University Ruston, LA 71272 Slide 11 First Integration Because there is only one independent variable, : Integrate once: Write in terms of flux: We will use this form to satisfy the no-flux boundary condition at R k.

12. Louisiana Tech University Ruston, LA 71272 Slide 12 Flux Boundary Condition at R k Since flux is 0 at the edge of the cylinder ( R k ), Substitute back into the (once-integrated) differential equation (boxed equation in slide 10):

13. Louisiana Tech University Ruston, LA 71272 Slide 13 Second Integration for Concentration With the previous differential equation: Divide by r : Integrate:

14. Louisiana Tech University Ruston, LA 71272 Slide 14 Boundary Condition at Capillary Wall From the problem statement (Slide 9) c(R c ) = c 0. Evaluate the previous solution for c(r) at r=R c. b Solve for b.

15. Louisiana Tech University Ruston, LA 71272 Slide 15 Simplify Combine like terms, recalling that : Or, in terms of partial pressures:

16. Louisiana Tech University Ruston, LA 71272 Slide 16 Efffect of Different M Values

17. Louisiana Tech University Ruston, LA 71272 Slide 17 Plot of the Solution Note that the solution is not valid beyond R k.

18. Louisiana Tech University Ruston, LA 71272 Slide 18 Finding R k The steady state equation is a function of R k, but an important question is, “What is R k, given a certain metabolic rate?” Non-starvation: Halfway between capillaries. Starvation: Is the solution still valid?

19. Louisiana Tech University Ruston, LA 71272 Slide 19 Non Steady State Diffusion equation: Initial Condition: Boundary Conditions:

20. Louisiana Tech University Ruston, LA 71272 Slide 20 Homogeneous Boundary Conditions The problem will be easier to solve if we can make the boundary conditions homogeneous, i.e. of the form: Our boundary condition at r = r c is not homogeneous because it is in the form:

21. Louisiana Tech University Ruston, LA 71272 Slide 21 Homogeneous Boundary Conditions However, if we define the following new variable: The boundary condition at r c becomes: And the boundary condition at r k is still homogeneous:

22. Louisiana Tech University Ruston, LA 71272 Slide 22 Non-Dimensionalization The new concentration variable also has the advantage of being dimensionless. We can non- dimensionalize the rest of the problem as follows: Let: The boundary conditions become: The initial condition becomes: Why are these forms obvious?

23. Louisiana Tech University Ruston, LA 71272 Slide 23 Non-Dimensionalization The diffusion equation can now be non-dimensionalized: Use: So that:

24. Louisiana Tech University Ruston, LA 71272 Slide 24 Non-Dimensionalization (Continued) Use: To determine that:

25. Louisiana Tech University Ruston, LA 71272 Slide 25 Non-Dimensionalization (Continued) Now apply: To: To get:

26. Louisiana Tech University Ruston, LA 71272 Slide 26 Non-Dimensionalization (Cont) Simplify Multiply by :

27. Louisiana Tech University Ruston, LA 71272 Slide 27 Non-Dimensionalization (Cont) Examine The term on the right hand side must be non-dimensional because the left hand side of the equation is non- dimensional. Thus, we have found the correct non- dimensionalization for the reaction rate.

28. Louisiana Tech University Ruston, LA 71272 Slide 28 The Mathematical Problem The problem reduces mathematically to: Differential Equation Boundary Conditions Initial Condition

29. Louisiana Tech University Ruston, LA 71272 Slide 29 Change to Homogeneous Diffusion equation: Let: Then: Follow the approach of section 3.4.1 (rectangular channel) to change the non-homogeneous equation to a homogeneous equation and a simpler non-homogeneous equation.

30. Louisiana Tech University Ruston, LA 71272 Slide 30 Divide the Equation The equation is solved if we solve both of the following equations: In other words, we look for a time-dependent part and a time independent (steady state) solution. Note that since the reaction term does not depend on time, it can be satisfied completely by the time-independent term.

31. Louisiana Tech University Ruston, LA 71272 Slide 31 Solution to the Spatial Part We already know that the solution to: Is: And that this form satisfies the boundary conditions at r c and r k. It will do so for all time (because it does not depend on time).

32. Louisiana Tech University Ruston, LA 71272 Slide 32 Non-Dimensionalize the Spatial Part In terms of the non-dimensional variables: And this form also becomes zero at the two boundaries.

33. Louisiana Tech University Ruston, LA 71272 Slide 33 Transient Part We therefore require that: With Boundary Conditions And the Initial Condition

34. Louisiana Tech University Ruston, LA 71272 Slide 34 Separation of Variables Homogeneous diffusion equation: Function of  only Function of  only

35. Louisiana Tech University Ruston, LA 71272 Slide 35 The Two ODEs Solutions: What does this equation remind you of?

36. Louisiana Tech University Ruston, LA 71272 Slide 36 Radial Dependence Could it perhaps be a zero-order Bessel Function?

37. Louisiana Tech University Ruston, LA 71272 Slide 37 Radial Dependence Differential equation for radial dependence becomes: So the solution is:

38. Louisiana Tech University Ruston, LA 71272 Slide 38 Radial Dependence This is Bessel’s equation, so the solution is:

39. Louisiana Tech University Ruston, LA 71272 Slide 39 Bessel Functions

40. Louisiana Tech University Ruston, LA 71272 Slide 40 Derivatives of Bessel Functions The derivatives of Bessel functions can be obtained from the general relations: Specifically:

41. Louisiana Tech University Ruston, LA 71272 Slide 41 Flux Boundary Condition In the solution we will have terms like: We will be requiring the gradient of these terms to go to zero at  k. I.e. The only way these terms can go to zero for all  is if: For every value of.

42. Louisiana Tech University Ruston, LA 71272 Slide 42 Relationship between A and B In other words: And for the boundary condition at the capillary wall:

43. Louisiana Tech University Ruston, LA 71272 Slide 43 Characteristic Values We conclude that the allowable values of are those which satisfy: This is not as simple as previous cases, where the Y 0 term became zero, but it is possible to find these values from MatLab or Excel, given a value for  k.

44. Louisiana Tech University Ruston, LA 71272 Slide 44 The Characteristic Function (  k = 10)

45. Louisiana Tech University Ruston, LA 71272 Slide 45 To Calculate (and Plot) in Excel 0.1 =BESSELY(A1*etak,1)*BESSELJ(A1,0) - BESSELJ(A1*etak,1)*BESSELY(A1,0) 0.15 =BESSELY(A2*etak,1)*BESSELJ(A2,0) - BESSELJ(A2*etak,1)*BESSELY(A2,0) 0.2 =BESSELY(A3*etak,1)*BESSELJ(A3,0) - BESSELJ(A3*etak,1)*BESSELY(A3,0) 0.25 =BESSELY(A4*etak,1)*BESSELJ(A4,0) - BESSELJ(A4*etak,1)*BESSELY(A4,0)

46. Louisiana Tech University Ruston, LA 71272 Slide 46 Finding the Roots Use “Tools | Goal Seek” to find the roots of the equation. For example, the plot indicates that one root is near =1.2. With the value 1.2 in cell A1 and the formula in cell A2: Set cell: A2 To value: 0 By changing cell: A1 Goal Seek OK Cancel “OK” will change the value of cell A1 to the 4 th root, 0.110266.

47. Louisiana Tech University Ruston, LA 71272 Slide 47 First Six Roots 0.110266 0.497895 0.855445 1.208701 1.560322 1.8 This process gives the following value for the first 6 roots: 1 2 3 4 5 6

48. Louisiana Tech University Ruston, LA 71272 Slide 48 Complete Solution The complete solution has the form: Where the values of n are obtained as described above.

49. Louisiana Tech University Ruston, LA 71272 Slide 49 A Prettier Form We obtain the somewhat more aesthetically pleasing form: If we define:

50. Louisiana Tech University Ruston, LA 71272 Slide 50 Initial Condition Now apply the initial condition: To: So:

51. Louisiana Tech University Ruston, LA 71272 Slide 51 Orthogonality For simplicity, define: These are called “Cylindrical Functions.” They satisfy the radial Laplacian operator, are zero for  = 1, and have zero derivative for  = . According to the Sturm-Liouville Theorem, the following orthogonality relationship holds for these functions:

52. Louisiana Tech University Ruston, LA 71272 Slide 52 Sturm-Liouville Compare: To: Note that  is not the first derivative of  2, but if we divide by  : and 1 is the 1 st derivative of 

53. Louisiana Tech University Ruston, LA 71272 Slide 53 Sturm-Liouville Compare: To: The w(  ) is particularly important because it tells us the weighting factor.

54. Louisiana Tech University Ruston, LA 71272 Slide 54 Initial Condition With the initial condition: Multiply both sides by and integrate:

55. Louisiana Tech University Ruston, LA 71272 Slide 55 Initial Condition Now use the orthogonality condition: These integrals can be found in tables and worked out to determine the values for the  n ’s, from which the final answer is obtained.

56. Louisiana Tech University Ruston, LA 71272 Slide 56 Fourier Bessel Series For Example:

57. Louisiana Tech University Ruston, LA 71272 Slide 57 Integral of Cylindrical Functions To obtain the previous result, take Bessel’s Equation, substituting in the cylindrical function, and multiply by and then integrate.

58. Louisiana Tech University Ruston, LA 71272 Slide 58 Norm of Cylindrical Functions For the first term: Integrate by parts with:

59. Louisiana Tech University Ruston, LA 71272 Slide 59 Integration by Parts, First Term

60. Louisiana Tech University Ruston, LA 71272 Slide 60 Integration by Parts, Second Term The second integral of Slide 54 is: But: So the second integral of Slide 54 is cancels with the last term of I 1, leaving:

61. Louisiana Tech University Ruston, LA 71272 Slide 61 Last Term For the last term:

62. Louisiana Tech University Ruston, LA 71272 Slide 62 Last Term

63. Louisiana Tech University Ruston, LA 71272 Slide 63 Combine the Three Integrals The expression will simplify further, given that we will be working with problems for which either the function or its derivative is zero at each boundary.

64. Louisiana Tech University Ruston, LA 71272 Slide 64 What about the similarity solution As it turns out, we did not need to abandon the similarity solution. We could have done the same thing we did with the separation of variables solution. I.e. we could have said that the complete solution is the sum of the particular solution and a sum of similarity solutions.

65. Louisiana Tech University Ruston, LA 71272 Slide 65

66. Louisiana Tech University Ruston, LA 71272 Slide 66 Similarity Solution? We could attempt a similarity solution: Which transforms the equation to:

67. Louisiana Tech University Ruston, LA 71272 Slide 67 Transform Variables From Previous: Multiply by r 2 /D

68. Louisiana Tech University Ruston, LA 71272 Slide 68 The Problem From Previous: If this had been a “no reaction” problem, the method would work. Unfortunately, the reaction term prevents the similarity solution from working, so we need to take another approach.