CH-08 (523178), страница 3
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Superscripts (k + 1) and (k) referto the improved and previous distributions, respectively. The order of sweep willaffect how the temperatures should be upgraded. For example, if the temperaturesare to be re-averaged from top to bottom and left to right, referring to Figure 1, thenEquation 3 is to be modified as:Ti(,kj +1) =(1 ( k +1)T+ Ti(+k1), j + Ti(,kj−+11) + Ti(,kj+)14 i −1, j)(14)Notice that the neighboring temperatures in the row above, i–1, and in the columnto the left, j–1, have already been upgraded while those in the row below, i + 1, andin the column to the right, j + 1, are yet to be upgraded.
Similar modifications areto be made to Equations 6 to 12 during relaxation.© 2001 by CRC Press LLCThe program Relaxatn is developed according to the relaxation methoddescribed above. For solving the problem shown in Figure 3, both FORTRAN andQuickBASIC versions are made available for interactively specifying the tolerance.Sample results are presented below.FORTRAN VERSION© 2001 by CRC Press LLCSample ResultsThe program Relaxatn is first applied for an interactively entered value of equalto 100.
Only one relaxation needs to be implemented as shown below. The temperature distribution for Sweep #1 is actually the initial assumed distribution. Onecannot assess how accurate this distribution is. The second run specifies that be© 2001 by CRC Press LLCequal to 1. The results show that 136 relaxation steps are required. For giving moreinsight on how the relaxation has proceeded, Sweeps #10, #30, #50, #100, and #137are presented for interested readers. It clearly indicates that a tolerance of equal to100 is definitely inadequate.© 2001 by CRC Press LLCQUICKBASIC VERSION© 2001 by CRC Press LLCSample ResultsIrregular BoundariesPractically, there are cases where the domain of heat conduction have boundarieswhich are quite irregular geometrically as illustrated in Figure 4.
For such cases, theequation derived based on the relaxation method, Equation 3, which states that thetemperature at any point has the average value of those at its four neighboring pointsif they are equally apart, has to be modified. The modified equation can be derivedusing a simple argument applied in both x and y directions. For example, considerthe temperature at the point G, TG, in Figure 4.
First, let us investigate the horizontal,y direction (for convenience of associating x and y with the row and column indicesi and j, respectively as in Figure 2). We observe that TG is affected more by thetemperature at the point C, TC, than by that at the point I, TI because the point C iscloser to the point G than the point I. Since the closer the point, the greater theinfluence, based on linear variation of the temperature we can then write:TG =∆yβ′∆y1β′TC +TI =TC +T∆y + β′∆y∆y + β∆y1 + β′1 + β′ I(15)where the increment from point I to point G is the regular increment y while thatbetween G and C is less and equal to y with having a value between 0 and1. Similarly, along the vertical, x direction and considering the points B, G, and Hand a regular increment x, we can have:TG =© 2001 by CRC Press LLC∆xα ′∆x1α′TB +TH =TB +T∆x + α ′∆x∆x + α ′∆x1 + α′1 + α′ H(16)FIGURE 4.
There are cases where the domain of heat conduction have boundaries whichare quite irregular geometrically.where like , has a value between 0 and 1. As often is the case, the regularincrements x and y are taken to be equal to each other for the simplicity ofcomputation. Equations 15 and 16 can then be combined and by taking both x andy directions into consideration, an averaging approach leads to:TG =1 1α′1β′TB +TH +TC +TI 2 1 + α′1 + α′1 + β′1 + β′ (17)For every group of five points such as B, C, G, H, and I in Figure 4 situated atany irregular boundary, the values of and have to be measured and Equation17 is to be used during the relaxation process if the boundary temperatures areknown.If some points along an irregular boundary are insulated such as the points Band C in Figure 4, we need to derive new formula to replace Equation 6 or Equation10.
The insulated condition along BC requires Tn = 0 where n is the directionnormal to the cord BC when the arc BC is approximated linearly. If the values of ‘and ‘ are known, we can replace the condition T/n = 0 with∂T ∂TdX ∂TdY ∂T∂T∂T∂T=+=+ β′=0sin θ +cos θ = α ′∂n ∂Xdn ∂Ydn ∂X∂Y∂X∂Y(18)The remainder of derivation is left as a homework problem.MATLAB APPLICATIONA Relaxatn.m file can be created to perform interactive MATLAB operationsand generate plots of the temperature distributions during the course of relaxation.This file may be prepared as follows:© 2001 by CRC Press LLCThis file can be applied to solve the sample problem run by first specifying theboundary temperatures described in Equation 4 to obtain an initial distribution byentering the MATLAB instructions:© 2001 by CRC Press LLCThe fprintf command enables a label be added, in which the format %3.0frequests 3 columns be provided without the decimal point for printing the value ofNR, and \n requests that next printout should be started on a new line.
The Relaxatn.mcan now be utilized to perform the relaxations. Let first perform one relaxation byentering>> NR = NR + 1; fprintf(‘Sweep # %3.0f \n’,NR), [D,T] = feval(A:Relaxatn’,T]The resulting display of the error defined in Equation 13 and the second temperature distribution is:© 2001 by CRC Press LLCIn case that we need to have the 30th temperature distribution by performing29 consecutive relaxations, we enter:>> for NR = 3:30; [D,T] = feval(A:Relaxatn',T]; end>> fprintf('Sweep # %3.0f \n',NR),D,TThe resulting display of the error defined in Equation 13 and the 30th temperaturedistribution is:To obtain a plot of this temperature distribution after the initial temperaturedistribution has been relaxed 29 times, with gridwork and title as shown in Figure 5a,the interactive MATLAB instructions entered are:>> V = 0:1:50; contour(T,V’), grid, title(‘* After 30 relaxations *’)Notice that 51 contours having values 0 through 50 with an increment of 1defined in the row matrix V.
In Figure 5a, the contour having a value equal to 0 is© 2001 by CRC Press LLCFIGURE 5. After 38 relaxations (a) and after 137 relaxations (b).© 2001 by CRC Press LLCalong the upper edge (Y = 10) and right edge (X = 20), the first curved contour ofthe right has a value equal to 1, and the values of the contours are increased fromright to left until the point marked “5” which has a temperature equal to 50 isreached. It should be noted that along the left edge, the uppermost point marked“10” has a temperature equal to zero and the temperatures are increased linearly (asfor the initial conditions) to 50 at the point marked with “5”, and from that pointdown to the point marked “1” the entire lower portion of the left edge is insulated.For obtaining the 137th sweep, we can continue to call the service Relaxatn.mby similarly applying the MATLAB instructions as follows:>> for NR = 31:137; [D,T] = feval(A:Relaxatn',T]; end>> fprintf('Sweep # %3.0f \n',NR),D,TThe resulting display of the error defined in Equation 13 and the 137th temperature distribution is:© 2001 by CRC Press LLCFigure 5b shows the 137th temperature distribution when the interactive MATLAB instructions entered are:>> V = 0:1:50; contour(T,V’), grid, title(‘* After 137 relaxations *’)Notice that area near the insulated boundaries at the right-lower corner has finallyreached a steady-state temperature distribution, i.e., changes of the entire temperaturedistribution will be insignificant if more relaxations were pursued.MATHEMATICA APPLICATIONSTo apply the relaxation method for finding the steady-state temperature distribution of the heated plate already solved by the FORTRAN, QuickBASIC, andMATLAB versions, here we make use of the Do, If, and While commands ofMathematica to generate similar results through the following interactive operations:In[1]: = t = Table[0,{10},{20}]; eps = 100; nr = 0; d = eps + 1;In[2]: = Do[t[[i,1]] = (I–1)*10,{i,2,6}];In[3]: = (While[d>eps, d = 0;nr = nr + 1;Do[Do[ts = t[[i,j]]; t[[i,j]] = .25*(t[[I–1,j]] + t[[I + 1,j]]+ t[[i,j–1]] + t[[i,j + 1]]);d = Abs[ts-t[[i,j]]] + d;,{j,2,19}],{i,2,6}];Do[ts = t[[i,1]]; t[[i,1]] = .25*(t[[I–1,1]] + t[[I + 1,1]]+ 2*t[[i,2]]); d = Abs[ts-t[[i,1]]] + d;Do[ts = t[[i,j]]; t[[i,j]] = .25*(t[[I–1,j]] + t[[I + 1,j]]+ t[[i,j–1]] + t[[i,j + 1]]);d = Abs[ts-t[[i,j]]] + d;,{j,2,19}];If[i = = 7,Continue, ts = t[[i,20]];t[[i,20]] = .25*(t[[I–1,20]] + t[[I + 1,20]] + 2*t[[i,19]]);d = Abs[ts-t[[i,20]]] + d;],{i,7,9}];ts = t[[10,1]]; t[[10,1]] = .5*(t[[9,1]] + t[[10,2]]);d = Abs[ts-t[[10,1]]] + d;Do[ts = t[[10,j]]; t[[10,j]] = .25*(2*t[[9,j]] + t[[10,j–1]]+ t[[10,j + 1]]); d = Abs[ts-t[[10,j]]] + d;,{j,2,19}];ts = t[[10,20]]; t[[10,20]] = .5*(t[[9,20]] + t[[10,19]]);d = Abs[ts-t[[10,20]]] + d;])In[4]: = Print[“Sweep #”,nr]; Round[N[t,2]]Out[4] = Sweep #2{{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{10, 4, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{20, 9, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},© 2001 by CRC Press LLC{30, 13, 5, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{40, 18, 7, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{50, 20, 8, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{17, 11, 5, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{ 6, 5, 3, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{ 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},{ 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}Notice that In[2] initializes the boundary temperatures, nr keeps the count ofhow many sweeps have been performed, and the function Round is employed inIn[4] to round the temperature value to a two-digit integer.
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