Heat flux through the wall

between two semi-infinite solids Interfacial contact between two semi-infinite solids TA,i kA, rA, cA Ts kB, rB, cB TB,i Boundary and interfacial conditions

The interface temperature is not function of time. Ex) A: man, B: wood (pine) or steel (AISI 1302) Assume wood: steel:

Objects with Constant Surface Temperatures Utilization of solution to convection boundary condition As Bi → ∞, Ts,2 → T∞

Semi-Infinite Solid In dimensionless form Plane wall, Cylinder, and Sphere for Bi → ∞

Summary of transient heat transfer results for constant surface temperature cases

Objects with Constant Surface Heat Fluxes Summary of transient heat transfer results for constant surface heat flux cases

Example 5.7 Cancer treatment by laser heating using nanoshells 1) Prior to treatment, antibodies are attached to the nanoscale particles. 2) Doped particles are then injected into the patient’s bloodstream and distributed throughout the body. 3) The antibodies are attracted to malignant sites, and therefore carry and adhere the nanoshells only to cancerous tissue. 4) A laser beam penetrates through the tissue between the skin and the cancer, is absorbed by the nanoshells, and, in turn, heats and destroys the cancerous tissues.

Known: Size of a small sphere Thermal conductivity (k), reflectivity (r), and extinction coefficient (k) of tissue Depth of sphere below the surface of the skin

laser heat flux Find: Heat transfer rate from the tumor to the surrounding healthy tissue for a steady-state treatment temperature of Tt,ss = 55ºC at the surface of the tumor. Laser power needed to sustain the tumor surface temperature at Tt,ss = 55ºC. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding tissue is neglected. Water property can be used. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding is considered and the thermal mass of the tumor is neglected.

laser heat flux Assumptions: 1D conduction in the radial direction. Constant properties. Healthy tissue can be treated as an infinite medium. The treated tumor absorbs all irradiation incident from the laser. Lumped capacitance behavior for the tumor. Neglect potential nanoscale heat transfer effects. Neglect the effect of perfusion.

q (Case 12 of Table 4.1) 1. Steady-state heat loss q from the tumor

laser heat flux projected area of the tumor: 2. Laser power Pl, Energy balance : heat transfer rate from tumor = absorbed laser energy

laser heat flux 3. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding tissue is neglected.

4. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding is considered and thermal mass of the tumor is neglected. Heat transfer between a sphere and an exterior infinite medium subjected to constant heat flux By trial and error,

Periodic Heating Oscillating surface temperature thermal penetration depth (reduction of temperature amplitude by 90% relative to that of surface) Quasi-steady state temperature distribution Surface heat flux

Sinusoidal heating by a strip C1: depends on thermal contact resistance at interface between heater strip and underlying material

Finite Difference Method Numerical Method Finite Difference Method Dt p-1 p p+1 in time Dx Dy (m,n) (m,n+1) (m-1,n) (m+1,n) (m,n-1) in space truncation error: O(Dt) first order accuracy in time

Explicit Method (Euler Method) : forward difference stability criterion: If the system is one-dimensional in x, or or stability criterion:

one-dimensional in x, t p = 5 p = 4 p = 3 p = 2 p = 1 p = 0 x m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Boundary node subjected to convection 1 1 2 3 qconv,in qcond,out See Table 5.3 (p. 306) stability criterion: or

Example 5.9 m + 1 m - 1 m qcond,in qcond,out L = 10 mm qcond,in Fuel element: Steady operation Sudden change to Symmetry adiabat Coolant 1 2 3 4 5 m + 1 m - 1 m 5 4 qcond,in qcond,out L = 10 mm qcond,in qconv,out Dx = 2 mm Find: Temperature distribution at 1.5 s after a change in operating power by using the explicit finite difference method

m + 1 m - 1 m qcond,in qcond,out Dx Thus, Symmetry adiabat 1 2 3 4 5 For node 0, set 5 4 qcond,in qconv,out Dx/2 For node 5, or

Dt: stability criterion or Thus, choose Dt = 0.3 s Then,

nodal equations Initial distribution: steady-state solution with

Calculated nodal temperatures

Comments: Expanding the finite difference solution, the new steady-state condition may be determined.

Implicit Method (fully) : backward difference stability criterion : no restriction qconv,in qcond,out 1 If the system is one-dimensional in x, Boundary node subjected to convection

Explicit Method one-dimensional in x, t p = 5 p = 4 p = 3 p = 2 p = 1 p = 0 x m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Implicit Method (fully) one-dimensional in x, t p = 5 p = 4 p = 3 p = 2 p = 1 p = 0 x m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Crank – Nicolson Method Second order accuracy in time, but serious stability problem

backward difference : forward difference: averaging: stability criterion: or explicit method:

Example 5.10 very thick slab of copper 1 2 3 4 m + 1 m - 1 m 1 2 3 4 m + 1 m - 1 m Dx = 75 mm 1 Dx/2 qcond,in qrad,in qcond,out qcond,out T(x,0) = 20°C Find: Using the explicit FDM, determine temperature at the surface and 150 mm from the surface after 2 min, T(0, 2 min), T(150 mm, 2 min) Repeat the calculations using the implicit FDM. Determine the same temperatures analytically.

Determination of nodal points T(x,2 min) 20 x 150 300 Table A.1, copper (300 K) :

2 min → p = 5 Explicit FDM m + 1 m - 1 m qcond,in qcond,out Dx = 75 mm qrad,in qcond,out Dx/2 node 0: or interior nodes: time step: stability criterion Table A.1, copper (300 K) : 2 min → p = 5

finite-difference equations and After 2 min, and

Improvement of the accuracy domain length: 600 mm After 2 min, and When Dt = 24 s, and

Implicit FDM node 0: or Arbitrarily choosing, interior nodes: A set of nine equations must be solved simultaneously for each time increment. The equations are in the form [A][T]=[C].

[A][T] = [C]

After 2 min, and

Analytical Solution Comparison

Implicit method with Dx = 18 Implicit method with Dx = 18.75 mm (37 nodalpoints) and Dt = 6 s (Fo = 2.0) exact:

Conduction (Chap.2 – Chap5.) 2. Introduction Fourier law, Newton’s law of cooling Heat diffusion equation & boundary conditions 3. One-Dimensional Steady-State Conduction Without heat generation: electric network analogy With heat generation Extended surfaces 4. Two-Dimensional Steady-State Conduction Analytical method: Separation of variables Conduction shape factor Numerical method 5. Transient Conduction Lumped capacitance method: Bi and Fo numbers Analytical method: separation of variables Semi-infinite solid: similarity solution