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

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Presentation on theme: "Louisiana Tech University Ruston, LA 71272 Slide 1 Krogh Cylinder Steven A. Jones BIEN 501 Friday, April 20, 2007."— Presentation transcript:

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

2 Louisiana Tech University Ruston, LA Slide 2 Energy Balance Major Learning Objectives: 1.Learn a simple model of capillary transport.

3 Louisiana Tech University Ruston, LA Slide 3 The Krogh Cylinder

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

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

6 Louisiana Tech University Ruston, LA Slide 6 Capillary Transport What are appropriate reaction rates and boundary conditions? Constant rate of consumption (determined by tissue metabolism, not O 2 concentration No reaction

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

8 Louisiana Tech University Ruston, LA Slide 8 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).

9 Louisiana Tech University Ruston, LA Slide 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:

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

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

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

13 Louisiana Tech University Ruston, LA Slide 13 Boundary Condition at Capillary Wall From the problem statement (Slide 9) c c (R c ) = c 0 b

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

15 Louisiana Tech University Ruston, LA Slide 15 Plot of the Solution

16 Louisiana Tech University Ruston, LA Slide 16 Plot of the Solution Note that the solution is not valid beyond r k.

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

18 Louisiana Tech University Ruston, LA Slide 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 = r c is not homogeneous because it is in the form:

19 Louisiana Tech University Ruston, LA Slide 19 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:

20 Louisiana Tech University Ruston, LA Slide 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:

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

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

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

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

25 Louisiana Tech University Ruston, LA Slide 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.

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

27 Louisiana Tech University Ruston, LA Slide 27 Change to Homogeneous Diffusion equation: Let: Then: Follow the approach of section (rectangular channel) to change the non-homogeneous equation to a homogeneous equation and a simpler non-homogeneous equation.

28 Louisiana Tech University Ruston, LA Slide 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.

29 Louisiana Tech University Ruston, LA Slide 29 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).

30 Louisiana Tech University Ruston, LA Slide 30 Non-Dimensionalize the Spatial Part In terms of the non-dimensional variables: And this form also becomes zero at the two boundaries.

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

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

33 Louisiana Tech University Ruston, LA Slide 33 Boundary conditions for g Initial Condition: Boundary Conditions: 0 c0c0

34 Louisiana Tech University Ruston, LA Slide 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:

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

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

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

38 Louisiana Tech University Ruston, LA Slide 38 Radial Dependence

39 Louisiana Tech University Ruston, LA Slide 39 Radial Dependence

40 Louisiana Tech University Ruston, LA Slide 40 Radial Dependence So since: Note: When we used Bessel’s equation in Womersley flow, we did not use the Y 0 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.

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

42 Louisiana Tech University Ruston, LA Slide 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.

43 Louisiana Tech University Ruston, LA Slide 43 Bessel Functions

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

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

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

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

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

49 Louisiana Tech University Ruston, LA Slide 49 Boundary Condition at r = r c

50 Louisiana Tech University Ruston, LA Slide 50 Boundary Condition at r = r c

51 Louisiana Tech University Ruston, LA Slide 51 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.

52 Louisiana Tech University Ruston, LA Slide 52 Counter-Current Exchange Conservation of Mass Membrane Diffusion:

53 Louisiana Tech University Ruston, LA Slide 53 Counter-Current Exchange Conservation of Mass might better be written as: And Membrane Diffusion as: Where:

54 Louisiana Tech University Ruston, LA Slide 54 Combine the Two Equations Solution: Where:

55 Louisiana Tech University Ruston, LA Slide 55 Integrate w.r.t. z

56 Louisiana Tech University Ruston, LA Slide 56

57 Louisiana Tech University Ruston, LA Slide 57

58 Louisiana Tech University Ruston, LA Slide 58 Differential Form Describes how much a volume of material will increase in temperature with a given amount of heat input. I.e., if I add x number of Joules to a volume 1 cm 3, it will increase in temperature by T degrees. T Joules in

59 Louisiana Tech University Ruston, LA Slide 59 Thermal Conductivity k – defines the rate at which heat “flows” through a material. Fourier’s law (A hot cup of coffee will become cold). Fourier’s law is strictly analogous to Fick’s law for diffusion. q is flux, i.e. the amount of heat passing through a surface per unit area. Gradient drives the flux.

60 Louisiana Tech University Ruston, LA Slide 60 Heat Production, Example Consider the case of a resistor: V1V1 V2V2 The resistor dissipates power according to, (where I is the current) or, equivalently The volume of the resistor is. Therefore, at any spatial location within the resistor, it is generating: Joules/s/cm 3

61 Louisiana Tech University Ruston, LA Slide 61 Boundary Conditions Constant temperature (T=T 0 ) Constant flux Heat transfer: More general: –Surface Condition or on the closed surface. –Initial condition within the volume.

62 Louisiana Tech University Ruston, LA Slide 62 Boundary Conditions Constant temperature (T=T 0 ) Constant flux Heat transfer: More general: –Surface Condition or on the closed surface. –Initial condition within the volume.

63 Louisiana Tech University Ruston, LA Slide 63 Semi-Infinite Slab (of marble) Flow of Heat T=T1T=T1 T=T 0 uniform How does temperature change with time? Initial Temperature Profile

64 Louisiana Tech University Ruston, LA Slide 64 Semi-Infinite Slab (of marble) Differential Equation (no source term): Boundary Conditions:

65 Louisiana Tech University Ruston, LA Slide 65 Semi-Infinite Slab (of marble) For 1-dimensional geometry and constant k : This problem is mathematically identical to the fluid flow near an infinitely long plate that is suddenly set in motion.

66 Louisiana Tech University Ruston, LA Slide 66 Semi-Infinite Slab (of marble) Dimensionless Temperature: Dimensionless temperature describes the difference between temperature and a reference temperature with respect to some fixed temperature difference, in this case T 1 – T 0.

67 Louisiana Tech University Ruston, LA Slide 67 Semi-Infinite Slab (of marble) Equations in Terms of Dimensionless Temperature: Boundary Conditions: T * is valuable because it nondimensionalizes the equations and simplifies the boundary conditions.

68 Louisiana Tech University Ruston, LA Slide 68 Semi-Infinite Slab (of marble) Assume a similarity solution, and define:. We make use of the following relationships:

69 Louisiana Tech University Ruston, LA Slide 69 Semi-Infinite Slab (of marble) With these substitutions, the differential equation becomes: This can be divided by to yield: The combination can now be replaced with to give:.

70 Louisiana Tech University Ruston, LA Slide 70 Semi-Infinite Slab (of marble) Instead of trying to solve directly for T *, try to solve for the first derivative:. The equation is rewritten as: which is separated as:. Thus:. So:

71 Louisiana Tech University Ruston, LA Slide 71 Semi-Infinite Slab (of marble) Integrate:.. To obtain:. This integral cannot be evaluated in closed form by standard methods. However, it is tabulated in handbooks and it can be evaluated under standard software. The integral is called the Error function (because of it’s origins in probability theory, where Gaussian functions are important).

72 Louisiana Tech University Ruston, LA Slide 72 Newton’s Law of Cooling Often we must evaluate the heat transfer in a body that is in contact with a fluid (e.g. heat dissipation from a jet engine, cooling of an engine by a radiator system, heat loss from a cannonball that is shot through the air). The boundary between the solid and fluid conforms neither to a constant temperature, nor to a constant flux. We make the assumption that the rate of heat loss per unit area is governed by:

73 Louisiana Tech University Ruston, LA Slide 73 Newton’s Law of Cooling The heat transfer coefficent, h, is a function of the velocity of the cannonball. I.e. the higher the velocity, the more rapidly heat is extracted from the cannonball. One generally assumes that the heat transfer coefficient does not depend on temperature. However, in free convection problems, it can be a strong function of temperature because the velocity of the fluid depends on the fluid viscosity, which depends on temperature.

74 Louisiana Tech University Ruston, LA Slide 74 Free Convection In free convection, the fluid moves as a result of heating of the fluid near the body in question. Fluid becomes less dense near the body. Bouyency causes it to move up, enhancing transfer of heat.

75 Louisiana Tech University Ruston, LA Slide 75 Forced Convection In forced convection (vs. free convection), the velocity is better controlled because it does not depend strongly on the heat flow itself. A fan, for example, controls the velocity of the fluid.

76 Louisiana Tech University Ruston, LA Slide 76 Convection vs. Conduction Convection enhances flow of heat by increasing the temperature difference across the boundary. Because “hot” fluid is removed from near the body, fluid near the body is colder, therefore the temperature gradient is higher and heat transfer is higher, by Fourier’s law. In other words, Fourier’s law still holds at the boundary. Small gradient, small heat transfer. Large gradient, large heat transfer.

77 Louisiana Tech University Ruston, LA Slide 77

78 Louisiana Tech University Ruston, LA Slide 78 Similarity Solution? We could attempt a similarity solution: Which transforms the equation to:

79 Louisiana Tech University Ruston, LA Slide 79 Transform Variables From Previous: Multiply by r 2 /D

80 Louisiana Tech University Ruston, LA Slide 80 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.


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