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ME 259 Heat Transfer Lecture Slides II

1/22/05. ME 259. 2. Steady-State Conduction Heat Transfer. Incropera

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ME 259 Heat Transfer Lecture Slides II

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

    2. 1/22/05 ME 259 2 Steady-State Conduction Heat Transfer Incropera & DeWitt coverage: Chapter 2: General Concepts of Heat Conduction Chapter 3: One-Dimensional, Steady-State Conduction Chapter 4: Two-Dimensional, Steady-State Conduction

    3. 1/22/05 ME 259 3 General Concepts of Heat Conduction Reading: Incropera & DeWitt Chapter 2

    4. 1/22/05 ME 259 4 Generalized Heat Conduction Fourier’s law, 1-D form: Fourier’s law, general form: q” is the heat flux vector, which has three components; in Cartesian coordinates: (magnitude)

    5. 1/22/05 ME 259 5 The Temperature Gradient ?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)

    6. 1/22/05 ME 259 6 Thermal Conductivity 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

    7. 1/22/05 ME 259 7 Total Heat Rate Total heat rate (q) is found by integrating the heat flux over the appropriate area: 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

    8. 1/22/05 ME 259 8 Heat Diffusion (Conduction) Equation For a homogeneous, isotropic solid material undergoing heat conduction: Cylindrical and spherical coordinate system forms given in text (p. 64-65) 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

    9. 1/22/05 ME 259 9 Heat Diffusion (Conduction) Equation For constant thermal conductivity (k): 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)

    10. 1/22/05 ME 259 10 Boundary Conditions and Initial Condition Boundary Conditions: known conditions at solution domain boundaries Initial Condition: known condition at t = 0 Number of boundary conditions required to solve the heat diffusion equation is equal to the number of spatial dimensions multiplied by two There is only one initial condition, which takes the form where Ti may be a constant or a function of x,y, and z

    11. 1/22/05 ME 259 11 Types of Boundary Conditions for Conduction Problems Specified surface temperature, e.g., Specified surface heat flux, e.g., Specified convection (h, T? given), e.g., Specified radiation (?, Tsur given), e.g.,

    12. 1/22/05 ME 259 12 Solving the Heat Diffusion Equation Choose a coordinate system that best fits the problem geometry. 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. Determine if k can be treated as constant and if Write the general heat conduction equation using the chosen coordinates. Reduce equation to simplest form based upon assumptions. Write boundary conditions and initial condition (if applicable). Obtain a general solution for T(x,y,z,t) by some method; if impossible, resort to numerical methods.

    13. 1/22/05 ME 259 13 Solving the Heat Diffusion Equation, cont. Solve for the constants in the general solution by applying the boundary conditions and initial condition to obtain a particular solution. Check solution for correctness (e.g., at boundaries or limits such as x = 0, t = 0, t ? ? , etc.) Calculate heat flux or total heat rate using Fourier’s law, if required. Optional: rearrange solution into a nondimensional form

    14. 1/22/05 ME 259 14 Example: GIVEN: Rectangular copper bar of dimensions L x W x H is insulated on the bottom and initially at Ti throughout . Suddenly, the ends are subjected and maintained at temperatures T1 and T2 , respectively, and the other three sides are exposed to forced convection with known h, T?. FIND: Governing heat equation, BCs, and initial condition

    15. 1/22/05 ME 259 15 One-Dimensional, Steady-State Heat Conduction Reading: Incropera & DeWitt, Chapter 3

    16. 1/22/05 ME 259 16 1-D, S-S Conduction in Simple Geometries w/o Heat Generation Plane Wall if k = constant, general heat diffusion equation reduces to separating variables and integrating yields where T(x) is the general solution; C1 and C2 are integration constants that are determined from boundary conditions

    17. 1/22/05 ME 259 17 1-D, S-S Conduction in Simple Geometries w/o Heat Generation 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

    18. 1/22/05 ME 259 18 1-D, S-S Conduction in Simple Geometries w/o Heat Generation 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

    19. 1/22/05 ME 259 19 1-D, S-S Conduction in Simple Geometries w/o Heat Generation Electric Circuit Analogy heat rate in plane wall can be written as in electrical circuits we have Ohm’s law: analogy:

    20. 1/22/05 ME 259 20 Thermal Circuits for Plane Walls Series Systems Parallel Systems

    21. 1/22/05 ME 259 21 Thermal Circuits for Plane Walls, cont. Complex Systems

    22. 1/22/05 ME 259 22 Thermal Resistances for Other Geometries Due to Conduction Cylindrical Wall Spherical Wall

    23. 1/22/05 ME 259 23 Convective & Radiative Thermal Resistance Convection Radiation

    24. 1/22/05 ME 259 24 Critical Radius Concept 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) A critical radius (rcr) exists for radial systems, where: adding insulation up to this radius will increase heat transfer adding insulation beyond this radius will decrease heat transfer For cylindrical systems, rcr = kins/h For spherical systems, rcr = 2kins/h

    25. 1/22/05 ME 259 25 Thermal Contact Resistance Thermal contact resistance exists at solid-solid interfaces due to surface roughness, creating gaps of air or other material:

    26. 1/22/05 ME 259 26 Thermal Contact Resistance 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 See Tables 3.1, 3.2 for typical values

    27. 1/22/05 ME 259 27 EXAMPLE Given: two, 1cm thick plates of milled, cold-rolled steel, 3.18?m roughness, clean, in air under 1 MPa contact pressure Find: Thermal circuit and compare thermal resistances

    28. 1/22/05 ME 259 28 1-D, S-S Conduction in Simple Geometries with Heat Generation Thermal energy can be generated within a material due to conversion from some other energy form: Electrical Nuclear Chemical Governing heat diffusion equation if k = constant:

    29. 1/22/05 ME 259 29 S-S Heat Transfer from Extended Surfaces (i.e., fins) Consider plane wall exposed to convection where Ts>T?: How could you enhance q ? increase h decrease T? increase As (attach fins)

    30. 1/22/05 ME 259 30 Fin Nomenclature x = longitudinal direction of fin L = fin length (base to tip) Lc = fin length corrected for tip area W = fin width (parallel to base) t = fin thickness at base Af = fin surface area exposed to fluid Ac = fin cross-sectional area, normal to heat flow Ap = fin (side) profile area P = fin perimeter that encompasses Ac D = pin fin diameter Tb = temperature at base of fin

    31. 1/22/05 ME 259 31 1-D Conduction Model for Thin Fins 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 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

    32. 1/22/05 ME 259 32 Fin Performance Fin Effectiveness Fin Efficiency for a straight fin of uniform cross-section: where Lc = L + t / 2 (corrected fin length)

    33. 1/22/05 ME 259 33 Calculating Single Fin Heat Rate from Fin Efficiency Calculate corrected fin length, Lc Calculate profile area, Ap Evaluate parameter Determine fin efficiency ?f from Figure 3.18, 3.19, or Table 3.5 Calculate maximum heat transfer rate from fin: Calculate actual heat rate:

    34. 1/22/05 ME 259 34 Maximum Heat Rate for Fins of Given Volume Analysis: “Optimal” design results:

    35. 1/22/05 ME 259 35 Fin Thermal Resistance Fin heat rate: Define fin thermal resistance: Single fin thermal circuit:

    36. 1/22/05 ME 259 36 Analysis of Fin Arrays Total heat transfer = heat transfer from N fins + heat transfer from exposed base Thermal circuit: where

    37. 1/22/05 ME 259 37 Analysis of Fin Arrays, cont. Overall thermal resistance:

    38. 1/22/05 ME 259 38 Example 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/m2-K. Contact resistance of 2.75x10-4 m2-K/W exists at base. Find: a) Total heat rate w/o and with fins b) Effect of R”t,c on heat rate

    39. 1/22/05 ME 259 39 Two-Dimensional, Steady-State Heat Conduction Reading: Incropera & DeWitt Chapter 4

    40. 1/22/05 ME 259 40 Governing Equation Heat Diffusion Equation reduces to: 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.

    41. 1/22/05 ME 259 41 Solution Methods Analytical Methods Separation of variables (see section 4.2) Laplace transform Similarity technique Conformal mapping Graphical Methods Plot isotherms & heat flux lines Numerical Methods Finite-difference method (FDM) Finite-element method (FEM)

    42. 1/22/05 ME 259 42 Conduction Shape Factor The heat rate in some 2-D geometries that contain two isothermal boundaries (T1, T2) with k = constant can be expressed as where S = conduction shape factor (see Table 4.1) Define 2-D thermal resistance:

    43. 1/22/05 ME 259 43 Conduction Shape Factor, cont. 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

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