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Unformatted text preview: STEADY-STATE CONDUCTION— MULTIPLE DIMENSIONS I 3-1 INTRODUCTION In Chap. 2 steady-state heat transfer Was calculated in systems in which the temperature gradient and area could be expressed in terms of one space co- ordinate. We now wish to analyze the more general case of two-dimensional heat ﬂow. For steady state, the Laplace equation applies. 62T 627‘ 6x2 + 5y: : (3-1) assuming constant thermal conductivity. The solution to this equation may be obtained by analytical, numerical, or graphical techniques. The objective of any heat-transfer analysis is usually to predict heat flow or the temperature which results from a certain heat ﬂow. The solution to Eq. (3— 1) will give the temperature in a two-dimensional body as a function of the two independent space coordinates x and y. Then the heat flow in the x and y directions may be calculated from the Fourier equations 6T ‘Y = —kA,— 3—2 (I. .ax ( l _ {3T (1,. : kkAv— (3‘3) , , 6}, These heat-ﬂow quantites are directed either in the x direction or in the y direction. The total heat ﬂow at any point in the material is the resultant of the it}, and Y], at that point. Thus the total heat-flow vector is directed so that it is perpendicular to the lines of constant temperature in the material, as shown in Fig. 3-]. So if the temperature distribution in the material is knOWn. we may easily establish the heat flow. Fig- 3'1 Sketch showing the heat flow in two dimenstons. I 3-5 NUMERICAL METHOD OF ANALYSIS An immense number of analytical solutions for conduction heat-transfer prob- lems have been accumulated in the literature over the past 100 years. Even so, in many practical situations the geometry or boundary conditions are such that an analytical solution has not been obtained at all, or if the solution has been developed, it involves such a complex series solution that numerical evaluation becomes exccedingly dif cult. For such situations the most fruitful approach to the problem is one based on ﬁnite-difference techniques, the basic principles of which we shall outline in this section. Consider a two—dimensional body which is to be divided into equal incre- ments in both the .r and y directions, as shown in Fig. 3-5. The nodal points are designated as showu, the in locations indicating the x increment and the n locations indicating the y increment. We wish to establish the temperatures at any of these nodal points within the body, using Eq. (3-1) as a governing Condition. Finite differences are used to approximate differential increments in the temperature and space coordinates; and the smaller we choose these approximated. The temperature gradients may be Written as follows: Fig. 3-5 Sketch illustrating nomenclature used in two; dimensional numerical analysis of heat conduction. m,” 8T Tm+l.n __ T _ a —____ 5X m+ti‘2.n Ax E W Tum ~ Tm—I,n ~ ——__ mil/2.0 A .X' x E __ Tm.“+1 — Tmm 0)’ marl-1.12 a Ay 6T N Tn“, — TIME, Ejmm—llz ~ ﬂy a] 91" 62]" N 6x m+lf2.n — (kin—1Q,” TmHJI + Tmﬁlgn _ 233nm N 3T“ : W M ,_ (my)2 Thus the ﬁnite—difference approximation for Eq. (3-1} becomes 621‘ ﬂy rim-rm 6y m.nil."2 Tm"+1 + Tmnil _ 2T)“ __ : —-‘___—___ , . ,r: 6y2 mm: I‘In'l-LH + Tut—Ln _ ZTIPJJI + Tm.n+] + TmJi—l _ 27171.” 0 ‘L———_+_— _—*——‘———_—__" : (AX)l (Ay }* 1rm+[,i'1 + Tm—ln + Tnmi+l + Tnmi—l _ 4THLH : 0 state conditions. ' peratures at the various nodes. A very simple example is shown in Fig. 3-6. and the four equations for nodes 1, 2, 3, and 4 would be 100+500+T2+T3~4T,=0 T1+500+100+T4~4T230 100+T1+T4+100—4T3=0 T3+T2+100+100—4T4=0 These equations have the solution 1“, = T; = 250°C T3 = T4 = 150°C Of course, we could recognize from symmetry that T, = T; and T3 2 T4 and would then only need two nodal equations. 100+T,+100—3T3=0 Once the temperatures are determined, the heat ﬂow may be calculated from _ AT = k —~— q 2 A): M Where the ATis taken at the boundaries. In the example the heat ﬂow may be calculated at either the 500°C face or the three 100°C faces. If a sufﬁciently ﬁne grid is used, the two values should be very nearly the same. T= 500“C T: IOODC T: 100%: Fig. 3-6 Four~node problem. The Gauss-Seidel method is one, An old method suitable for hand calculations with a small number of nodes is called the re- laxation method. In this technique the nodal equation is set equal to some residual j? and the following calculation procedure followed: Whit 1. Values of the nodal temperatures are assumed. 2. The value of the residual for each node is calculated from the respective equation and the assumed temperatures. 3. The residuals are “relaxed” to zero by changing the assumptions of the nodal temperatures. The largest residuals are usually relaxed ﬁrst. 4. As each nodal temperature is changed, a new residual must be calculated for connecting nodes. 5. The procedure is continued until the residuals are sufﬁciently close to zero. In Table 33 a relaxation solution for the system of Fig. 3-6 is shown. For the most part, the relaxation method would be employed as an expedient vehicle only when a computer was not readily available. Other methods of solution include a transient analysis carried through to steady state (see Chap. 4), direct elimination (Gauss elimination 1, or more sophisticated iterative techniques Table 3—3 Relaxation Table for System of Fig. 3-6 '1'. RI T: R3 T3 _ 7 133 T4 _ R4 300 n 100 300 — 100 200 — 100 200 — 100 275 0 ~ 125 — 125 — 30 270 a 5 a 130 — 45 — 165 MED 30 a 70 160 — 5 — 10 255 10 A 65 — 25 0 261) a 25 a 20 a 5 155 a 5 — 25 a 15 250 15 — 35 5 — 15 150 5 — 20 150 5 0 [\J J! | O O O O When the solid is exposed to some convection boundary condition, the temperatures at the surface must be computed differently from the method given above. Consider the boundary shown in Fig. 3-7. The energy balance on node (at, n) is Tm u _ m 7 l H Ax Tm.” a Tm n + | Ax Tm.” _ Tm n+ I ' I ' ' k ' k — _—' A A} Ax 2 fly 2 Ay : IIA})(THLH — TI) If Ax : Ay. the boundary temperature is expressed in the equation A I A. l Tm,” + _ 1k r Tm — §(2T1H—l.u + Tm.n+l + IHJI—l) : 0 An equation ofthis type must be written for each node along the surface shown in Fig. 3—7. So when a convection boundary condition is present, an equation like (\$25) is used at the boundary and an equation like (3-24) is used for the interior points. Fig. 3—7 Nomenclature for nodal equation Wlll’t con- vective boundary condition. course, more equations mean more cumbersome solutions. Fortunately, com- puters and even programmable calculators have the capability to obtain these The nodal equations may be written as 0I1T1+ 0|sz “t ' " + HINT” = C: (tuft + (IEQTZ + ' ' ' : C3 amt". + - - - 2 C3 (3—27) anilTl + (11:17:) + + “unfit ‘ Cu where T., T3, . . . . T” are the unknown nodal temperatures. By using the matrix notation (1')] (232 _Cz T2 [A] : as. [C] : [7] = a,” a”; an” '0! T" Eq. (3-27) can be expressed as [ANTI = [C] (3-28) and the problem is to ﬁnd the inverse of [A] such that [TJ = [AF’lC] (3-29) Designating [24]"I by b” b]; b.” [AW — ."gf . . . . . . . . _ b,” b,,2 bm, the ﬁnal solutions for the unknown temperatures are written in expanded form as #4 1 0 l 0 0 0 0 0 T. —600 l —4 l 0 l O 0 0 0 T2 #500 O 2 —4.67 0 0 I 0 0 0 T3 —567 l D 0 —4 l 0 l 0 0 T4 * 100 0 l 0 1 *4 I 0 I 0 T5 : 0 0 0 l 0 2 —4.67 0 0 1 T6 —67 0 0 (l 2 0 0 ~46? I 0 I7 — 167 0 0 0 0 2 0 1 k4.67 1 T8 —67 0 0 O 0 O I O 1 *2 67 T9 767 ...
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