Chapter 2 conduction

CHAPTER CONDUCTION Conduction Conduction is the mode of energy transfer as heat due to temperature difference in a solid or any phase of material where the mass is contiguous and in thermal contact Microscopically this mode of energy transfer is attributed to free electron flow from higher to lower energy levels, lattice vibration and molecular collision However no macroscopic mass movement is involved The temperature in the body will be a function of location and time In the popular cartesian coordinates : T = T(x, y, z, τ) How temperature varies with position within the body : The temperature field is obtained by deriving and solving the differential equation based on energy balance relations for the volume Conduction T2 < T1 T1 q insulator x A qx the conducted heat rate in the x-direction has phenomenologically been found to be adequately represented by: Temperature gradient q x dT   kA dx Conductivity Fourier’s Law T2 < T1 T1 q insulator x A qx dT qx ''   k A dx Baron Jean Baptiste Joseph Fourier (1768-1830)   T  T  T  q ''  k T  k  i j k  y z   x del operator  Thermal conductivity The equation of heat conduction in one direction qx dT qx ''   k A dx The constant of proportionality k is the thermal conductivity of material, which is a measure of the ability of a material to conduct heat Thermal conductivity k of a material (W/mK) can be defined as the rate of heat transfer Jun/sec through a unit thickness of the material (m) per unit area (1m2 )per unit temperature difference (1K)    Thermal Conductivity of Gases Thermal Conductivity of Liquids Thermal Conductivity of Solids + Metals and alloys; + Solid dielectrics (non-metals); Conductivity Thermal conductivity • Solid > liquid > gas • Pure metals > alloys > nonmetallic solids > insulation systems • Thermal conductivity of liquid increases with decreasing liquid molecular weight • Liquid metal > nonmetallic liquid • Thermal conductivity of gas increases with decreasing gas molecular weight Conductivity Thermal Diffusivity Thermal diffusivity represents how fast heat diffuses through a materia Heat conducted k    Heat stored  Cp (m / s) Specific heat Cp (J/kg.0C) represent the heat storage capability of a material; expresses heat storage capability per unit mass The heat capacity ρCp represent the heat storage capability of a materia per unit volume The thermal conductivity k represents how well a material conducts heat, the heat capacity ρCp represents how much energy a material stores per unit volume →The larger the thermal diffusivity is the faster the propagation of heat into the medium A small value of thermal diffusivity means that heat is mostly absorbed by the material and a small amount of heat will be conducted further Example A thick–walled tube of stainless steel = 19 ⁄ ℃ with inner diameter (ID) and outer diameter (OD) is covered with a layer of asbestos insulation = 0.2 ⁄ ℃ If the inside wall temperature of the pipe and outside temperature of asbestos are maintained at 600℃ and 100℃, respectively a) Calculate the heat loss per meter of length b) Calculate the tube – insulation interface temperature = 600℃ = 100℃ Example • The heat loss per meter of length − = ln + = 680 ⁄ ln • The tube – insulation interface temperature − = ln = 680 ⁄ = 595.8 ℃ Critical thickness of insulation q q T1  T2 T2  T3   r2   r3  1   ln  ln  2k A  r1  2k B  r2  T1  T3 T T  r   r  RA  RB 1 ln   ln  2k A  r1  2k B  r2  Sphere • Heat flux =− ⁄ : heat flux per surface area = = • Boundary condition = • Temperature profile =0 − − − − = − − Steady conduction in one direction WITH HEAT SOURCES T1 H T2 q L δ plane wall cylindrical wall spherical wall Plane wall Boundary condition = = Plane wall Temperature profile = = + − − = = − 2 = = + − + =0 − − 1− 1− − + − − − − Plane wall =− Heat flux = − 1+ − − Cylinder Boundary condition = = Cylinder + Temperature profile = − + − − = − − − − − ln + − ln − ⁄ −1 + 1− − =0 − ⁄ ln ln − 1− ln ⁄ ln ⁄ Cylinder Heat flux = =− + − − ln − Sphere Boundary condition = = Sphere Temperature profile − − Heat flux − = − − − + 1− =0 − − + − = + − − − 6 − + How to solve the conduction problem ? Try to simplify the problem to one–dimension Select a simple shape factor model represent the physical situation Seek some simple analytical solutions, if they are too complicated, move to the numerical techniques In practice, recognize that convection and radiation boundary conditions are subject to large uncertainties CHAPTER CONDUCTION