Question

50 70 100 180 100 70 50 20 20 20 10mm 20 40 60 90 140 90 60 40 Figure 1: A rectangular plate with temperature defined on the boundaries 4 Figure 1 shows a rectangular plate made of an homogeneous isotropic material The temperature distribution in this plate satisfies the indicated boundary condi- tions (given in degrees centigrade) and has reached a steady-state condition so that the temperature is described by Laplace's equation a2T a2T (i) Draw a sketch of the solution domain showing clearly the line of symme- try for the temperature distribution and indicating which of the unknown temperatures are equal to each other (ii) Use the finite difference formulae on the formulae sheet to formulate the finite difference equations required to find estimates of the nodal temper- atures, TA. TB. Tc and TD

Answer 1

partial differential^2 T/ partial differential x^2 + partial differential^2 T/partial differential y^2 = 0 partial differential^2 T/partial differential x^2 = T_i + 1, j - 2 T_i, j + T_i - 1, j/(delta x)^2 partial differential^2 T/ partial differential y^2 = T_i, j + 1 - 2 T_i, j + T_i, j - 1/(delta y)^2 so partial differential^2T/ partial differential x^2 = partial differential^2 T/ partial differential y^2 implies T_i + 1, j - 2 T_i, j + T_i - 1, j/(delta x)^2 = -[T_i, j + 1 - 2 T_i, j + T_i, j - 1/(delta y)^2] In our problem delta x = delta y = 10mm so T_i + 1, j + T_i, j + 1 + T_i - 1, j + T_i, j- 1/delta = T_i, j This is the equation now From the figure if clear that T_4 = T_G, T_8 = T_F T_c = T_E \r\n We can write the following equations T_pi = 1/4 [20 + 50 + T_B + 40] T_A = 1/4 = [1/4 [110 + T_B] implies 4T_A - T_B = 110 T_8 = 1/4 [T_A + 0T + T_c + 60] implies T_A - 4T_B + T_C = -130 T_c = 1/4 [T_3 + 100 + T_D + 90]