Krogh Cylinder

1. Krogh Cylinder Steven A. Jones BIEN 501 Friday, April 20, 2007 Louisiana Tech University Ruston, LA 71272

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

3. The Krogh Cylinder Louisiana Tech University Ruston, LA 71272

4. Assumptions • The geometry follows the Krogh cylinder configuration • Reactions are continuously distributed • There is a radial location at which there is no flux Louisiana Tech University Ruston, LA 71272

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

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

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

8. 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

9. Constant Rate of Reaction Because there is only one independent variable: (Note the change from partial to total derivative) Integrate once: Divide by r and integrate again: Louisiana Tech University Ruston, LA 71272

10. Constant Rate of Reaction Because there is only one independent variable, : Integrate once: Write in terms of flux: Louisiana Tech University Ruston, LA 71272

11. Flux Boundary Condition at Rk Since flux is 0 at the edge of the cylinder (Rk), Substitute back into the differential equation: Louisiana Tech University Ruston, LA 71272

12. Solution for Concentration Substitute back into the differential equation: Divide by r: Integrate: Louisiana Tech University Ruston, LA 71272

13. Boundary Condition at Capillary Wall From the problem statement (Slide 9) cc(Rc) = c0 b Louisiana Tech University Ruston, LA 71272

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

15. Plot of the Solution Louisiana Tech University Ruston, LA 71272

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

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

18. 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

19. 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

20. 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: Louisiana Tech University Ruston, LA 71272

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

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

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

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

25. 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

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

27. 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

28. 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

29. 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

30. 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

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

32. Boundary conditions Recall what we did: And that: Initial Condition: Boundary Conditions: Louisiana Tech University Ruston, LA 71272

33. Boundary conditions for g Initial Condition: Boundary Conditions: c0 0 Louisiana Tech University Ruston, LA 71272

34. It follows that we must solve The equation is solved if we solve both of the following equations: With the following initial/boundary conditions: Louisiana Tech University Ruston, LA 71272

35. Separation of Variables Homogeneous diffusion equation: Louisiana Tech University Ruston, LA 71272

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

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

38. Radial Dependence Louisiana Tech University Ruston, LA 71272

39. Radial Dependence Louisiana Tech University Ruston, LA 71272

40. Radial Dependence So since: Note: When we used Bessel’s equation in Womersley flow, we did not use the Y0 term because it goes to At z=0. However, in this case, we do not need to go to r=0, so we will keep it in the solution. 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 rk. I.e. The only way these terms can go to zero for all t is if: For every value of s. Louisiana Tech University Ruston, LA 71272

42. Relationship between A and B In other words: In contrast to problems we have seen before, in which the separation variable could take on only discrete values, here we can satisfy the boundary condition for any value of s. Louisiana Tech University Ruston, LA 71272

43. Bessel Functions Louisiana Tech University Ruston, LA 71272

44. s = 0 We must also consider the case for s = 0. But this expression is already a part of the steady state solution. Louisiana Tech University Ruston, LA 71272

45. Complete Solution If s could take on only discrete values, the complete solution would be: Louisiana Tech University Ruston, LA 71272

46. Complete Solution However, since s can take on any value, the complete solution must be: Louisiana Tech University Ruston, LA 71272

47. Complete Solution And since: Louisiana Tech University Ruston, LA 71272

48. Initial Condition The initial condition is: Louisiana Tech University Ruston, LA 71272

49. Boundary Condition at r = rc Louisiana Tech University Ruston, LA 71272

50. Boundary Condition at r = rc Louisiana Tech University Ruston, LA 71272