hw7-solution

# hw7-solution - 1111117 1 Filamﬂr FXch/Lso as warm Jm...

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Unformatted text preview: 1111117 1, Filamﬂr FXch/Lso as warm Jm +£><1looo|< m NLJﬂ’t the mash Q What is We now. aka/mgr, ® 2. :34 44.6 Modal TszvoJluvc x1 @ Wrween Exavm‘sz 4% AMA ’rYZSMT @ ® © rum/UH 7 Solution: If we number the triangles starting from the right angle [[5.2.4];[5.6.2];[1.2.3];[1.4.2]] the element wise conductivity matrix does not to be recomputed and it is still 1 .8000 -0.9000 -0.9000 -0.9000 0.9000 0 -0.9000 0 0.9000 as in exercise 47. The heat load vector due to internal generation rate changes: nodes 3,6, get contribution from just one element, nodes 1,4,5 get contribution from two elements, and node 2 gets contribution from four elements. The surface heat transfer matrix and load vector do not change and may be reused. The solutIon Is Solution from T = exercise 48 32.61 32.0881 £391) 31.0920 [T] = 25'10 34.8659 _ 24.84 257471 Relative change of temperature at Node 2: 21.88 245402 >> (31 .0920-30.8094)l30.8094*100 ‘ 21.8391 ans = 0.9173 ”2., Solvc +2! 41% Tun/UPLra/fmvt in ’l’ht. +n‘m3vJ—W/ dvlmdun smallest Swvge/‘l' I)": ‘qu Fl-rr‘mgwm'ovx oh ’11“, w‘jl/ct. triﬂing +an %n=ﬂL(T‘Ta1 r—- D .. 3 “3.0 w/m/oK/,e,=5~w/W3/°K/ la=9C, Q—Ioow/am Solution: Since the geometry, the material properties, the internal heat generation rate, and the boundary conditions are all symmetric with respect to three symmetry planes, the smallest subset of the mesh that one can use is a single T3 element in the interior, and a single L2 element on the boundary. Node Locations-- [[0,0];[b12,0];[0,bl2*sin(30/180*pi)]] Degree of freedom numbers: 1,2,3 Solution 17.8571 7.1429 36.6071 in degrees Celsius 3- Rbfmt EXeru'SL “r51 Ming ‘itvvis «mama/tan, rid—L Tim‘s Ira—Mme rth is SOML‘HMQS mum ”mam-Q”. WWW“ ”HUM in‘ltﬁtmﬁvu WICHVIDA ’ﬁwm Ckwtrl'tr lo wombL use MS demrt rug}. Solution: The integral [(5:2)] = [3(2) cV[N][N]T Az dS is evaluated at three quadrature points. At each ofthose three points one of the basis functions assumes value 1, while the other two functions are identically zero there. Therefore the capacity matrix is evaluated as a diagonal matrix with entries equal to CVAZIS(E) 3. Element wise diagonal capacity matrices would be preferred by the explicit Eulertime integration method.This method is only conditionally stable, which means that the time step is typically very short. Therefore to coverthe time interval of integration one would need to take a lot of steps, and therefore making each step very efficient is critical. Since the system matrix on the left is only the capacity matrix for the explicit Euler method, the solution step can be made very quickly. No benefit derives from using a diagonal capacity matrix for the implicit methods discussed in Chapter 10. ...
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