QUESTION 4 (continued): (i) (2 marks) From discretised heat equation, write the general update equation for an interior node (sz : ) This corresponds to the equation for the value of T1 at time level m + 1 from the discrete solution values at time level mi (ii) (3 marks) At each time step, the boundary node values can be found from the discretised boundary conditions as follows: 5T TX\" 7 T37" 7 Tom=Tca * w 0' 31/ FL Ay Use these boundary conditions in the general update equation to Write down the update equation for each interior solution node, yl : Ay, yz : 2Ay and ya : 3A1}, (iii) (3 marks) To march the solution in time, we will use a xed time step of At = $1): Use this to simplify your update equations for the interior nodes as much as possible QUESTION 4 (continued): (iv) (6 marks) Use the update equations derived in part (iii) to march the discrete solution forward in time for 3 time steps from the initial condition of T : TH at to : 0. The nal set of temperatures calculated should he at time t3 : 3At. QUESTION 4 (20 MARKS): A copper bar with length L = 1 m is extracted from an oven where it has been annealing. It has a uniform initial temperature T(0, y) 2 TH = 1000 K. To cool, it is placed standing on its end upon a highly conductive plate at room temperature TE, 0) = To = 300 K. This situation is illustrated in the gure below. Assume the heat loss is dominated by conduction to the plate, and heat losses from all other surfaces can be neglected. Thus 6; _ at the top end of the rod 33,; ly=L 0. insulated V g q=0 n. E c: :74 (\\l :- 3'3 0 2 i l The temperature in the rod is governed by the 1D heat equation: ST 821" 6% 5y Here1 D = 1 X 104 mQ/s is the thermal diffusivity. To compute an approximate evolution of the temperature prole in the rod, we discretise it into to nodes with spacing A3,: = 0.25 m, as shown on the right in the gure. We then approximate the second order space derivative with the central difference expression, and the time derivative with a forward difference expression. This gives the following discrete equation: Trad-1 _ Tm Tm _ 2Tm Tn. t ( 3+1 1 + 31). (4) At 2 D A92 QUESTION 4 (continued): (V) (4 marks) How many time steps are required before the solution at y = L changes? How does this compare to the rate of information propagation expected in the heat equation? Suggest an alternative numerical scheme that would increase the accu racy of the solution. (vi) (2 marks) In order to obtain the solution at a given time more quickly, one possi bility is increasing the time step. For the numerical scheme considered here, would it be appropriate to do this? Be sure to justify you answer. Some formulas Lagrange polynomial of degree n - 1 Pn-1(2) = > In-1,i(x)yi i=0 where the basis functions are: n-1 In-1,j (2) = II C - Ci i-0,i/j Cj - Ci Gaussian Quadrature ( f(x) da ~ [wif(Ii) # terms values of ri weight wi 2 -0.57735027 1.0 +0.57735027 1.0 -0.77459667 0.5555555 0.0 0.8888888 +0.77459667 0.5555555 Finite-difference formulas f'(20) = JI - fo h " + O(h) (forward) f'(xo) = fo - J-1 + O(h) (backward) h f'(20) = f1 - f-1 + 0(h2) (central) 2h f"(20) = f2 - 2f1+ fo + O(h) h2 f"(20) = f1 - 2fo + f-1 + 0(h2) h2 Gauss-Seidel Method :-1 n-1 k+1 Caijack , 0 siEn ari -0 j=i+1