TwoDimensional Conduction: Shape Factors and Dimensionless Conduction Heat Rates


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1 TwoDimensional Conduction: Shape Factors and Dimensionless Conduction Heat Rates Chapter 4 Sections 4.1 and 4.3 make use of commercial FEA program to look at this. D Conduction General Considerations Twodimensional conduction: There are many real situations where the heat transfer is not onedimensional and is two or threedimensional. In these cases we may apply a number of different approaches depending on specifics of the problem. These include analytical, graphical and numerical approaches
2 D Conduction General Considerations Twodimensional conduction: Temperature distribution is characterized by two spatial coordinates, e.g., T (x,y). Heat flux vector is characterized by two directional components, e.g., q and q. x y Assuming steadystate, twodimensional conduction in a rectangular domain with constant thermal conductivity and heat generation, the heat equation is: T T q ( x, y) + + = 0 x y k Assuming steadystate, twodimensional conduction in a rectangular domain with constant thermal conductivity and no heat generation, the heat equation is: T x T y + = 0 Solution Methods The Heat Equation and Methods of Solution Solution Methods: Exact/Analytical: Separation of Variables (Section 4.) Limited to simple geometries and boundary conditions. i ( ) Approximate/Graphical q = 0 : Flux Plotting (Section 4 S.1) Of limited value for quantitative considerations but a quick aid to establishing physical insights. Approximate/Numerical: FiniteDifference, Finite Element or Boundary Element Method. Most useful approach and adaptable to any level of complexity.
3 D Conduction General Considerations Consider heat transfer in a long, prismatic solid with two isothermal surfaces and two insulated surfaces: Note the shapes of lines of constant temperature (isotherms) and heat flow lines (adiabats). What is the relationship between isotherms and heat flow lines? Perpendicular! Exact/Analytical: Separation of Variables (Section 4.) Exact mathematical solutions to heat conduction equation Only limited cases can be solved this way Simple D case is illustrated in Section 4. of Text for a rectangular plate with three sides of the plate held at a constant temperature T 1 and the fourth held at T. letting ( T T) 1 θ =, then the H.C. eqn is ( T T) 1 θ θ + = 0 and θ has value between 0 and 1 x y The boundary conditions are then 0 or 1 as shown
4 Exact/Analytical: Separation of Variables (Section 4.) The solution uses separation of variables, i.e., letting θ ( xy, ) = Xx ( ) Y( y) which by substitution (and dividing by XY) gives 1 X 1 Y = and θ has value between 0 and 1 X x Y y which can be reduced to ordinary differential eqns X x X y + λ X = λ Y = 0 0 for which a solution exists (although quite complicated) See Eqn 4.19 Exact/Analytical: Superposition principle In certain cases Superposition can be used for the solution of linear, homogeneous PDE s any superposition of solutions of linear, homogeneous PDE s is again a solution, and the particular solutions may then be combined to obtain more general solutions. the superposition principle, also known as superposition property, states that, for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually. So that if input A produces response X and input B produces response Y then input (A + B) produces response (X + Y). Mathematically, for a linear system, F, defined by F(x) = y, where x is some sort of stimulus (input) and y is some sort of response (output), the superposition (i.e., sum) of stimuli yields a superposition of the respective responses: F(x 1 +x )=F(x 1 )+F(x ) The superposition principle holds because, by definition, a linear system must be additive. Laplace s equation for heat conduction is a linear differential equation
5 Exact/Analytical: Superposition principle Laplace s equation for heat conduction is a linear differential equation θ x θ y + = 0 Using superposition it is possible to solve a problem with complicated boundary conditions by adding the solutions for problems with simpler boundary conditions Example would be a rectangular plate where the boundaries are functions of x and y respectively. Then T(x,y)= T 1 (x,y) + T (x,y) This is simply illustrated Superposition  Simple Examples In electrical engineering combined waveform wave 1 wave Two waves in phase Two waves 180 out of phase Temperature fields and heat flow
6 Flux Plots Flux Plotting : Approximate/Graphical Method Utility: Requires delineation of isotherms and heat flow lines. Provides a i quick means of estimating the rate of heat flow. q = 0 Procedure: Systematic construction of nearly perpendicular isotherms and heat flow lines to achieve a network of curvilinear squares. Rules: On a schematic of the twodimensional conduction domain, identify all lines of symmetry, which are equivalent to adiabats and hence heat flow lines. Sketch approximately uniformly spaced isotherms on the schematic, choosing a small to moderate number in accordance with the desired fineness of the network and rendering them approximately perpendicular to all adiabats at points of intersection. Draw heat flow lines in accordance with requirements for a network of curvilinear squares. See Supplemental Section 4 S.1. ( ) Flux Plots (cont.) Example: Square channel with isothermal inner and outer surfaces. Note simplification achieved by identifying lines of symmetry. Requirements for curvilinear squares: Intersection of isotherms and heat flow lines at right angles Approximate equivalence of sums of opposite sides ab + cd ac + bd Δx Δ y (4 S.1) Determination of heat rate: M i i= 1 q= q Mq, where M is the number of lanes i
7 Flux Plots (cont.) Example: Square channel with isothermal inner and outer surfaces. Determination of heat rate: M ΔT ΔT j i i M k( y ) j M ka Δ, i i= 1 q= q Mq 1 M q kδt N, 1 Δ x where l is depth of channel normal to page, N j= 1 Δ T = Δ T = N ΔT, j j Δx where N is the number of temperature increments in each lane M q kδt,for Δx Δy 1 N The Conduction Shape Factor Determination of heat rate: The previous method and expression can be used to define the shape factor, S q = S k Δ T, 1 M where S =, N Note for k = constant, q = 0 and boundary surfaces are isothermal 1 R t, cond ( D = ) S k Exact and approximate results for common two and threedimensional systems are provided in Table 4.1(a). For example, S
8 Shape Factor The Conduction Shape Factor Twoorthree dimensional heat transfer in a medium bounded by two isothermal surfaces at T 1 and T may be represented in terms of a conduction shape factor S. ( ) q= Sk T T (4.0) 1 Case 6. Long (L>>w) circular cylinder centered in square solid of equal length π L S = 1n 1.08 / ( w D) Twodimensional conduction resistance: R cond ( ) ( Sk) 1 D = (4.1)
9 Dimensionless Heat Rate The Dimensionless Conduction Heat Rate For isothermal objects of surface area A s and temperature T 1 embedded in an infinite medium at temperature T, heat transfer may be represented by a dimensionless conduction heat rate q. ( ) q q ka T T / L * ss * = ss s 1 c (4.3) where the object s characteristic length L c is ( 4 ) 1 / L A / π c s (4.) Exact and approximate results for common systems are provided in Table 4.1 (b). For example, Case 1. Isothermal sphere q = 1 * ss * For any object, q 1. ss
10 Problem: Shape Factor Problem 4.9 Heatgenerating radioactive waste in a buried container of known size and shape. Find container surface temperature. Schematic: ASSUMPTIONS: (1) Steadystate conditions, () Soil is a homogeneous medium of known and constant properties, (3) Negligible contact resistance. PROPERTIES: Table A.3, Soil (300 K): k=0.5 W/m K. ANALYSIS: From an energy balance on the container, π D q= k T T 1 D/ 4z ( ) 1 q= i Eg and from case 1 of Table 4.1(a), Problem: Shape Factor (cont.) Hence, ( 1 4 ) q D / z T1 = T + k π D 500W ( 1m/40m) o o =0 C+ =9.7 C 0.5 W/m K π(m) COMMENTS: (1) If the canister is buried within an infinite medium of temperature T =0 C, we may use Case 1 of Table 4.1(b). With Eqns. (4.) and (4.3) yield * As = π D / and q ss = 1, q 500W T1 = T + = kπ D 0.5W/m K π(m) o o 0 C+ =96.5 C Does this value make sense to you? () Using case 1 of Table 4.1(a) with z, evaluate T 1. Does this value make sense to you?
11 Problem: Thermal Circuit Problem 4.4: Attachment of a long aluminum pin fin (D=5mm) to a base material of aluminum or stainless steel. Determine the fin heat rate and the junction temperature (a) without and (b) with a junction resistance. Schematic: The heat flow lines shown in the figure presume a fin effectiveness of ε > 1. How would the lines look f for ε < 1? f The heat flow lines shown in the figure presume a fin effectiveness of ε > 1. How would the lines look f for ε < 1? f ε <1 ε >1
12 Problem: Thermal Circuit (cont) ASSUMPTIONS: (1) Steadystate conditions, () Constant properties, (3) Large base material, (4) Infinite fin. PROPERTIES: Aluminum alloy, k = 40 W/m K, Stainless steel, k = 15 W/m K. ANALYSIS: (a,b) From the thermal circuit with the junction resistance, the heat rate and junction temperature are Tb T Tb T q f = = (1) Rtot Rb + Rt, j + Rf (, ) Tj = T + qf Rf + Rt j () From Case 10 of Table 4.1, S=D. Hence, from Eq. (4.1) Rb = 1 Skb = 1 ( Dkb) = ( 0.005m kb) 1 K/W With A c = π D, the junction resistance is 4 5 Rt, j= Rt, j Ac= 3 10 m K W π ( 0.005m) 4 = 1.58K W 1/ With qf = ( hpkac) θb for an infinite fin (Table 3.4) and P= πd, 1/ 3 1/ R f = ( hpkac) = 50 W m Kπ ( 0.005m) 40 W m K 4 = 16.4K W Problem: Thermal Circuit (cont.) Without R t,j With R t,j Base R b (K/W) q f (W) T j ( C) q f (W) T j ( C) Al alloy St. steel COMMENTS: (1) Why is the effect of the base material on the heat rate and the junction temperature substantial for the stainless steel and not for the aluminum? () Why is the relative effect of the contact resistance on the heat rate and the junction temperature more pronounced for the aluminum alloy base than for the stainless steel?
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