Question

(a) The heat flux through the faces at the ends of bar is found to be proportional to un au/an at the ends. If the bar is perfectly insulated, also at the ends x 0 and x L are adiabatic conditions, Q1 ux(0, t) = 0 0 (2'7)*n prove that the solution of the heat transfer problem above (adiabatic conditions at both ends) gives as, 2 an: nnx u(x, t) Ao t An cos n-1 where Ao and An are an arbitrary constant The heat equation is: au at ax2 (17 marks) (b) If L and a 1 for the solution of heat transfer problem in Q1 (a), find the temperature in the bar with the initial temperature, f(x)= k = constant. (3 marks) A half-range expansions given as the following function Q2 2k for 0

Answer 1

Heat eq^n is partial differential u/partial differential t = alpha^2 partial differential^2 u/partial differential x^2: u_x (0, t) = 0 u_y (x, t) = 0 - (1) By variable separable (a) u(x, t) = X (x) T (t) (let) - (2) Putting value in eq^n (1) XT = alpha^2 X T implies X/X = T/alpha^2 T = K (Say(=) implies X/X = k implies X - KX = 0 Let K = -mu^2 therefore X (x) = c_1 cos mu x + c_2 sin mu x - (*) Again sol^n T, is T = c_3 e^-alpha^2 mu^2 t - (**) Thus eq^n (2) becomes, u (x, t) = (c_1 cos mu x + c_2 sin mu x) c_3 e^-alpha^2 mu^2 t generalized sol^n, u (x, t) = sigma^infinity_n = 0(A_n cos mu x + B_N sin mu x) e^-alpha^2 mu^2 t initial cond^n u_x (0, t) = 0 therefore 0 = mu B_n middot e^-alpha^2 mu^2 t implies B_n = 0 (since mu notequalto0 therefore u 9x, t) = sigma^infinity_n = 0 A_n cos mu x middot e^alpha^2 mu^2 t