1/22/05ME 2591 ME 259 Heat Transfer Lecture Slides II Dr. Gregory A. Kallio Dept. of Mechanical Engineering, Mechatronic Engineering & Manufacturing Technology California State University, Chico

1/22/05ME 2592 Steady-State Conduction Heat Transfer u Incropera & DeWitt coverage: –Chapter 2: General Concepts of Heat Conduction –Chapter 3: One-Dimensional, Steady-State Conduction –Chapter 4: Two-Dimensional, Steady-State Conduction

1/22/05ME 2593 General Concepts of Heat Conduction Reading: Incropera & DeWitt Chapter 2

1/22/05ME 2594 Generalized Heat Conduction u Fourier’s law, 1-D form: u Fourier’s law, general form: - q” is the heat flux vector, which has three components; in Cartesian coordinates: (magnitude)

1/22/05ME 2595 The Temperature Gradient u  T is the temperature gradient, which is: –a vector quantity that points in direction of maximum temperature increase –always perpendicular to constant temperature surfaces, or isotherms (Cartesian) (Cylindrical) (Spherical)

1/22/05ME 2596 Thermal Conductivity u k is the thermal conductivity of the material undergoing conduction, which is a tensor quantity in the most general case: –most materials are homogeneous, isotropic, and their structure is time-independent; hence: which is a scalar and usually assumed to be a constant if evaluated at the average temperature of the material

1/22/05ME 2597 Total Heat Rate u Total heat rate ( q ) is found by integrating the heat flux over the appropriate area: u k and  T must be known in order to calculate q” from Fourier’s law – k is usually obtained from material property tables –to find  T, another equation is required; this additional equation is derived by applying the conservation of energy principle to a differential control volume undergoing conduction heat transfer; this yields the general Heat Diffusion (Conduction) Equation

1/22/05ME 2598 Heat Diffusion (Conduction) Equation u For a homogeneous, isotropic solid material undergoing heat conduction: u Cylindrical and spherical coordinate system forms given in text (p ) u This is a second-order, partial differential equation (PDE); its solution yields the temperature field, T(x,y,z,t), within a given solid material

1/22/05ME 2599 Heat Diffusion (Conduction) Equation u For constant thermal conductivity ( k ): u For k = constant, steady-state conditions, and no internal heat generation –this is known as Laplace’s equation, which appears in other branches of engineering science (e.g., fluids, electrostatics, and solid mechanics)

1/22/05ME Boundary Conditions and Initial Condition u Boundary Conditions: known conditions at solution domain boundaries u Initial Condition: known condition at t = 0 u Number of boundary conditions required to solve the heat diffusion equation is equal to the number of spatial dimensions multiplied by two u There is only one initial condition, which takes the form –where T i may be a constant or a function of x,y, and z

1/22/05ME Types of Boundary Conditions for Conduction Problems u Specified surface temperature, e.g., u Specified surface heat flux, e.g., u Specified convection ( h, T  given), e.g., u Specified radiation ( , T sur given), e.g.,

1/22/05ME Solving the Heat Diffusion Equation u Choose a coordinate system that best fits the problem geometry. u Identify the independent variables ( x,y,z,t ), e,g, is it a S-S problem? Is conduction 1-D, 2-D, or 3-D? Justify assumptions. u Determine if k can be treated as constant and if u Write the general heat conduction equation using the chosen coordinates. u Reduce equation to simplest form based upon assumptions. u Write boundary conditions and initial condition (if applicable). u Obtain a general solution for T(x,y,z,t) by some method; if impossible, resort to numerical methods.

1/22/05ME Solving the Heat Diffusion Equation, cont. u Solve for the constants in the general solution by applying the boundary conditions and initial condition to obtain a particular solution. u Check solution for correctness (e.g., at boundaries or limits such as x = 0, t = 0, t  , etc.) u Calculate heat flux or total heat rate using Fourier’s law, if required. u Optional: rearrange solution into a nondimensional form

1/22/05ME Example: u GIVEN : Rectangular copper bar of dimensions L x W x H is insulated on the bottom and initially at T i throughout. Suddenly, the ends are subjected and maintained at temperatures T 1 and T 2, respectively, and the other three sides are exposed to forced convection with known h, T . u FIND: Governing heat equation, BCs, and initial condition

1/22/05ME One-Dimensional, Steady- State Heat Conduction Reading: Incropera & DeWitt, Chapter 3

1/22/05ME D, S-S Conduction in Simple Geometries w/o Heat Generation u Plane Wall –if k = constant, general heat diffusion equation reduces to –separating variables and integrating yields –where T(x) is the general solution ; C 1 and C 2 are integration constants that are determined from boundary conditions x L

1/22/05ME D, S-S Conduction in Simple Geometries w/o Heat Generation u Plane Wall, cont. –suppose the boundary conditions are –integration constants are then found to be –the particular solution for the temperature distribution in the plane wall is now

1/22/05ME D, S-S Conduction in Simple Geometries w/o Heat Generation u Plane wall, cont. –The conduction heat rate is found from Fourier’s law: –If k were not constant, e.g., k = k(T), the analysis would yield »note that the temperature distribution would be nonlinear, in general

1/22/05ME D, S-S Conduction in Simple Geometries w/o Heat Generation u Electric Circuit Analogy –heat rate in plane wall can be written as –in electrical circuits we have Ohm’s law: –analogy:

1/22/05ME Thermal Circuits for Plane Walls u Series Systems u Parallel Systems

1/22/05ME Thermal Circuits for Plane Walls, cont. u Complex Systems

1/22/05ME Thermal Resistances for Other Geometries Due to Conduction u Cylindrical Wall u Spherical Wall l r1r1 r2r2 r2r2 r1r1

1/22/05ME Convective & Radiative Thermal Resistance u Convection u Radiation

1/22/05ME Critical Radius Concept u Since the surface areas of cylinders and spheres increase with r, there exist competing heat transfer effects with the addition of insulation under convective boundary conditions (see Example 3.4) u A critical radius ( r cr ) exists for radial systems, where: –adding insulation up to this radius will increase heat transfer – adding insulation beyond this radius will decrease heat transfer u For cylindrical systems, r cr = k ins /h u For spherical systems, r cr = 2k ins /h

1/22/05ME Thermal Contact Resistance u Thermal contact resistance exists at solid-solid interfaces due to surface roughness, creating gaps of air or other material: A B q

1/22/05ME Thermal Contact Resistance u R” t,c is usually experimentally measured and depends upon –thermal conductivity of solids A and B –surface finish & cleanliness –contact pressure –gap material –temperature at contact plane u See Tables 3.1, 3.2 for typical values

1/22/05ME EXAMPLE u Given: two, 1cm thick plates of milled, cold- rolled steel, 3.18  m roughness, clean, in air under 1 MPa contact pressure u Find: Thermal circuit and compare thermal resistances

1/22/05ME D, S-S Conduction in Simple Geometries with Heat Generation u Thermal energy can be generated within a material due to conversion from some other energy form: –Electrical –Nuclear –Chemical u Governing heat diffusion equation if k = constant:

1/22/05ME S-S Heat Transfer from Extended Surfaces (i.e., fins) u Consider plane wall exposed to convection where T s >T  : u How could you enhance q ? –increase h –decrease T  –increase A s (attach fins)

1/22/05ME Fin Nomenclature u x = longitudinal direction of fin u L = fin length (base to tip) u L c = fin length corrected for tip area u W = fin width (parallel to base) u t = fin thickness at base u A f = fin surface area exposed to fluid u A c = fin cross-sectional area, normal to heat flow u Ap = fin (side) profile area u P = fin perimeter that encompasses A c u D = pin fin diameter u T b = temperature at base of fin

1/22/05ME D Conduction Model for Thin Fins u If L >> t and k/L >> h, then the temperature gradient in the longitudinal direction (x) is much greater than that in the transverse direction (y); therefore u Another way of viewing fin heat transfer is to imagine 1-D conduction with a negative heat generation rate along its length due to convection

1/22/05ME Fin Performance u Fin Effectiveness u Fin Efficiency –for a straight fin of uniform cross-section: –where L c = L + t / 2 (corrected fin length)

1/22/05ME Calculating Single Fin Heat Rate from Fin Efficiency u Calculate corrected fin length, L c u Calculate profile area, A p u Evaluate parameter u Determine fin efficiency  f from Figure 3.18, 3.19, or Table 3.5 u Calculate maximum heat transfer rate from fin: u Calculate actual heat rate:

1/22/05ME Maximum Heat Rate for Fins of Given Volume u Analysis: u “Optimal” design results:

1/22/05ME Fin Thermal Resistance u Fin heat rate: u Define fin thermal resistance: u Single fin thermal circuit:

1/22/05ME Analysis of Fin Arrays u Total heat transfer = heat transfer from N fins + heat transfer from exposed base Thermal circuit: –where

1/22/05ME Analysis of Fin Arrays, cont. u Overall thermal resistance:

1/22/05ME Example u Given: Annular array of 10 aluminum fins, spaced 4mm apart C-C, with inner and outer radii of 1.35 and 2.6 cm, and thickness of 1 mm. Temperature difference between base and ambient air is 180°C with a convection coefficient of 125 W/m 2 -K. Contact resistance of 2.75x10 -4 m 2 -K/W exists at base. u Find: a) Total heat rate w/o and with fins b) Effect of R” t,c on heat rate

1/22/05ME Two-Dimensional, Steady- State Heat Conduction Reading: Incropera & DeWitt Chapter 4

1/22/05ME Governing Equation u Heat Diffusion Equation reduces to: u Solving the HDE for 2-D, S-S heat conduction by exact analysis is impossible for all but the most simple geometries with simple boundary conditions.

1/22/05ME Solution Methods u Analytical Methods –Separation of variables (see section 4.2) –Laplace transform –Similarity technique –Conformal mapping u Graphical Methods –Plot isotherms & heat flux lines u Numerical Methods –Finite-difference method (FDM) –Finite-element method (FEM)

1/22/05ME Conduction Shape Factor u The heat rate in some 2-D geometries that contain two isothermal boundaries ( T 1, T 2 ) with k = constant can be expressed as –where S = conduction shape factor (see Table 4.1) u Define 2-D thermal resistance:

1/22/05ME Conduction Shape Factor, cont. u Practical applications: –Heat loss from underground spherical tanks: Case 1 –Heat loss from underground pipes and cables: Case 2, Case 4 –Heat loss from an edge or corner of an object: Case 8, Case 9 –Heat loss from electronic components mounted on a thick substrate: Case 10