Krogh Cylinder

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

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

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

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

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

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

7. Capillary Transport What are appropriate reaction rates and boundary conditions? Constant rate of consumption (determined by tissue metabolism, not O2 concentration) Not metabolic (no consumption) Louisiana Tech University Ruston, LA 71272

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

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

10. Constant Reaction: Roadmap The roadmap for solving the steady state problem is as follows: • Integrate the differential equation once to obtain a solution for flux as a function of r. • Use the zero flux boundary condition at Rk 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. Louisiana Tech University Ruston, LA 71272

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 Rk. Louisiana Tech University Ruston, LA 71272

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

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

14. Boundary Condition at Capillary Wall From the problem statement (Slide 9) c(Rc) = c0. Evaluate the previous solution forc(r)atr=Rc. Solve forb. b Louisiana Tech University Ruston, LA 71272

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

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

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

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

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

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 = rc is not homogeneous because it is in the form: Louisiana Tech University Ruston, LA 71272

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

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: Why are these forms obvious? The boundary conditions become: The initial condition becomes: Louisiana Tech University Ruston, LA 71272

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

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

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

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

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. Louisiana Tech University Ruston, LA 71272

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

29. Change to Homogeneous 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. Diffusion equation: Let: Then: Louisiana Tech University Ruston, LA 71272

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. Louisiana Tech University Ruston, LA 71272

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

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

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

34. Separation of Variables Homogeneous diffusion equation: Function of t only Function of h only Louisiana Tech University Ruston, LA 71272

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

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

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

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

39. Bessel Functions Louisiana Tech University Ruston, LA 71272

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

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 hk. I.e. The only way these terms can go to zero for all t is if: For every value of l. Louisiana Tech University Ruston, LA 71272

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

43. Characteristic Values We conclude that the allowable values of l are those which satisfy: This is not as simple as previous cases, where the Y0 term became zero, but it is possible to find these values from MatLab or Excel, given a value for hk. Louisiana Tech University Ruston, LA 71272

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

45. To Calculate (and Plot) in Excel Louisiana Tech University Ruston, LA 71272

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 l=1.2. With the value 1.2 in cell A1 and the formula in cell A2: “OK” will change the value of cell A1 to the 4th root, 0.110266. Goal Seek A2 Set cell: To value: 0 By changing cell: A1 Cancel OK Louisiana Tech University Ruston, LA 71272

47. First Six Roots This process gives the following value for the first 6 roots: Louisiana Tech University Ruston, LA 71272

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

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

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