Introduction to Heat and Mass Transfer Week 7
Example
Solution Technique Using either finite difference method or finite volume method, we end up with a set of simultaneous algebraic equations in terms of nodal temperatures C T S where,» C = Coefficient Matrix (N N matrix)» T = Solution Matrix (Column Vector)» S = Load Matrix (Column Vector)
Solution Technique (contd.) Direct Method: involves inversion of coefficient matrix and multiplication of inverted matrix with load matrix to obtain the solution matrix Many standard methods of matrix inversion» Gauss Elimination» Gauss Jordan Elimination» LU Factorization For diagonally dominant systems, more efficient methods such as Thomas Algorithm etc. available Computationally matrix inversion is an expensive process not preferred in practical problems with large meshes
Solution Technique (contd.) For example:
Solution Technique (contd.) Iterative Method: involves initial guess at certain nodes to begin the process of solution followed by successive refinement to obtain converged solution Many standard methods of matrix inversion» Jacobi Iteration» Gauss Seidel Iteration» Successive Over (Under) Relaxation Iterative methods are computationally much more efficient compared to direct methods almost always employed in practical problems with large meshes
Solution Technique (contd.) For example:
Solution Technique (contd.)
HW # 4 prob. 1 Write the discretized equation for a typical interior node for the case of three dimensional, steady state heat conduction with constant properties and no thermal energy generation by using (a) finite difference method; (b) finite volume method.
HW # 4 prob. 2 Consider steady heat transfer in a solid whose cross section is shown below. Heat is generated at a rate of 2 10 6 W/m 3. The left surface of the body is insulated and the bottom surface is maintained at an uniform temperature of 90C. The entire top surface is subjected to convection to ambient air at 25C with a convection coefficient of 80 W/m 2 C. An uniform heat flux of 5000 W/m 2 is applied to the right surface. Consider an uniform mesh with Dx = Dy = 1.2 cm.» Obtain discretized equations for appropriate nodes» Determine the nodal temperatures
Problem Figure Insulated h = 880 W/m 2 C T = 25 C k = 15 W/m C q = 2 10 6 W/m 3 2 2.4 cm 1.2 cm q = 5000 W/m 2 90 C 6 cm
HW # 4 will due on 11/14, right before the class! Late HW will not be accepted!!
Project #2 Project topic: Numerical Method Individual project report (no page limit)» Find the separate discretized equations of the given domain equations in terms of nodal temperatures (by FDM or FVM)» Determine the nodal temperatures by programming and find the solutions of the problems» Please also attach your program (no computer language restriction) at the end Due on Tuesday(11/21) before the class, late project report is not accepted
Problem: Cold Plate Analysis of cold plate used to thermally control IBM multi-chip, thermal conduction module. Features: Heat dissipated in the chips is transferred by conduction through spring-loaded aluminum pistons to an aluminum cold plate. Nominal operating conditions may be assumed to provide a uniformly distributed heat flux of q o 10 W/m at the base of the cold plate. Heat is transferred from the cold plate by water flowing through channels in the cold plate. 5 2
Problem: Cold Plate (cont.) Schematic: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties.
Problems (a) Cold plate temperature distribution for the prescribed conditions. (b) Options for operating at larger power levels while remaining within a maximum cold plate temperature of 60 C. (c) Follow (b), options for operating at larger power levels while using a copper cold plate with k = 400 W/m-K. (d) Options for operating at larger power levels while increasing the h = 10000 W/m 2 -K (a practical upper limit) and with k = 400 W/m-K.
Example Consider a thermal conduction module (TCM) with 100 chips, each dissipating 3 W of power. The module is cooled by water at 25 C flowing through the cold plate on top of the module. The thermal resistances in the path of heat flow are Rchip = 1 C/W between the junction and the surface chip, Rint = 8 C/W between the surface of the chip and the outer surface of the thermal conduction module, and Rext = 6 C/W between the outer surface of the module and the cooling water. Determine the junction temperature of the chip.
Example
Closure Coverage thus far..» talked about finite difference method and finite volume method as representative numerical methods for heat transfer problems» discussed the uses and limitations of direct and iterative solution techniques
Closure (contd.) Finite difference formulation of heat conduction problems using central difference approximation and discretized equations for interior nodes Finite volume formulation of heat conduction problems using conservation of energy and discretized equations for interior nodes Treatment of boundary nodes and the related discretized equations corresponding to common BCs Use of direct and iterative techniques for solving the resulting set of algebraic equations
Transient Analysis Strategy For a general transient conduction problem, we can follow the following procedure:» Compute Biot number and check whether Bi < 0.1» If Bi < 0.1, then we can usually implement lumped system analysis with reasonable accuracy» Otherwise we compute required constants as function of Biot number and implement approximate 1D analytical solutions for large plane walls (small thickness) or infinitely long cylinders (l/r o >10) or spheres (We can implement Heisler-Grober charts for the these standard shapes)» If multidimensional effects are present, then we consider the given solid as product of relevant 1D analytical solutions for standard shapes
Next Topic Transient Conduction» Lumped Capacitance Method Validity Simplified Transient Analysis
Lumped Capacitance Method The most general case of heat transfer through a solid involves variation of temperature in three dimensions and dependence of temperature on time during some transient Heat diffusion equation must be solved to yield a general solution T(x,y,z,t) Under specific conditions, the temperature through the body can be assumed to be uniform i.e. no temperature gradients When can we have negligible temperature gradient?
Lumped Capacitance Method (contd.) Consider plane wall subjected to convection T 1 k solid T T L k A R 1 2 solid T T 1/ h A R t, cond 2 conv t, conv T 2 q cond q conv T T DT T T DT 1 2 solid 2 solid fluid h,t R R t, cond t, conv L Condition for Lumped Capacitance Validity
Lumped Capacitance Method (contd.) Validity for lumped capacitance approach is governed by the dimensionless Biot number (Bi) given by: where, Bi h L conv c k» h conv = Heat transfer coefficient (W/m 2 -K)» k solid = Thermal conductivity of solid (W/m-K)» L c = Characteristics length of the solid (m) = V/A s Theoretically, lumped capacitance valid for Bi = 0; but practically lumped capacitance valid for Bi < 0.1 solid
Lumped Capacitance Method (contd.)
Example Consider two identical hot solids and surrounding air. One solid is cooled by a fan while the other is allowed to cool naturally in air. For which solid lumped analysis is more valid?
HW # 5 prob.1 Consider the same two solids now one dropped in large container filled with water and the other allowed to cool in air. For which solid lumped system analysis is more valid?
Simplified Transient Analysis Consider a solid with initial temperature T i subjected to convection T T h A conv s exp t T T VC i i p i t exp t VC p t C t, solid R t, conv ha s
Simplified Transient Analysis (contd.) Considering the transient temperature distribution, we can write: conv c exp 2 i solid c Dimensionless time is given by Fourier number (Fo): c where,» a = Thermal diffusivity (m 2 /s) h L at k L at Fo 2 L
HW # 5 prob.2 A thermal energy storage system involves a packed bed of 75-mm diameter aluminum spheres. During charging process, hot gas enters increasing thermal energy stored in the colder spheres. In the discharging process, cold gas flows taking away stored thermal energy from the warmer spheres. In a particular charging process, hot gas enters the system at 300C. The initial temperature of aluminum spheres is 25C and convection coefficient is 75 W/m 2 -K.» How long does it take a sphere near the inlet of the system to accumulate 80% of maximum possible thermal energy?» What is the temperature at the center of the sphere?
HW # 5 prob.3 A 300 mm diameter solid steel sphere is coated with a dielectric material layer of thickness 2 mm and thermal conductivity 0.04 W/m-K. The coated sphere is initially at a uniform temperature of 500C and is suddenly quenched in a large oil bath which has a temperature of 100C and convection heat transfer coefficient of 3300 W/m 2 -K.» Estimate the time required for the coated sphere to reach a temperature of 140C
Closure In this lecture, we» introduced the notion of lumped capacitance and conditions for its validity» talked about the simplified transient conduction analysis when lumped capacitance criterion applies
Closure (contd.) Biot number and its significance for transient analysis Bi R R t, cond conv c t, conv h L k solid Simplified transient analysis and non-dimensional temperature distribution for lumped systems T T h A exp exp conv s t Bi Fo T T VC i i p