(Two steady state problems) Let u(x, t) denote the temperature in a metal rod of length...

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(Two steady state problems) Let u(x, t) denote the temperature in a metal rod of length L, with thermal conductivity Ko> 0 and thermal diffusivity k > 0. The rod has uniform cross-sectional area, and is insulated on its sides. The rod experiences non-uniform internal heating of the form Q(x) = Mx/L, for 0 < x < L, where M> 0 is a positive constant. The governing equation for u(x, t), the temperature of the rod, is the heat equation du J²u Ət əx² (*) The rod is subject to boundary conditions at each end of the rod. Two sets of possible boundary conditions are: = k +M² 0<r<L, t> 0. L' (i) u(0, t) = 30°C, du(L, t) = 0 and (ii) au(0, t) = 0, u(L, t) = 30°C. In both cases the initial temperature (at t= 0) is constant (in z) at 30°C. (a) Recall that a steady-state solution is a solution that does not depend on time, that is, u(x, t) = u(x). For each set of boundary conditions given above, find the solution of the steady-state problem (and only the steady-state problem, do not solve the entire time dependent problem). (b) Find the steady-state heat flux, (x), for each of these problems. Which way does the heat flow in each case? Justify your answer. (c) Problems (i) and (ii) differ only in terms of which boundary condition goes at which end. But the solution of one of the problems has a significantly larger change of temperature from one end of the rod to the other. Calculate the temperature difference from one end of the rod to the other for each case, between (i) and (ii), which problem has the bigger change of temperature from one end of the rod to the other? Justify. What is the physical mechanism that causes this difference? Explain. (d) For (i) and for (ii), sketch the steady-state solutions, as functions of z. (Assume k = 1, M = 6 and L = 1 for this sketch only.) Identify which sketch goes with (i) vs. (ii). (Two steady state problems) Let u(x, t) denote the temperature in a metal rod of length L, with thermal conductivity Ko> 0 and thermal diffusivity k > 0. The rod has uniform cross-sectional area, and is insulated on its sides. The rod experiences non-uniform internal heating of the form Q(x) = Mx/L, for 0 < x < L, where M> 0 is a positive constant. The governing equation for u(x, t), the temperature of the rod, is the heat equation du J²u Ət əx² (*) The rod is subject to boundary conditions at each end of the rod. Two sets of possible boundary conditions are: = k +M² 0<r<L, t> 0. L' (i) u(0, t) = 30°C, du(L, t) = 0 and (ii) au(0, t) = 0, u(L, t) = 30°C. In both cases the initial temperature (at t= 0) is constant (in z) at 30°C. (a) Recall that a steady-state solution is a solution that does not depend on time, that is, u(x, t) = u(x). For each set of boundary conditions given above, find the solution of the steady-state problem (and only the steady-state problem, do not solve the entire time dependent problem). (b) Find the steady-state heat flux, (x), for each of these problems. Which way does the heat flow in each case? Justify your answer. (c) Problems (i) and (ii) differ only in terms of which boundary condition goes at which end. But the solution of one of the problems has a significantly larger change of temperature from one end of the rod to the other. Calculate the temperature difference from one end of the rod to the other for each case, between (i) and (ii), which problem has the bigger change of temperature from one end of the rod to the other? Justify. What is the physical mechanism that causes this difference? Explain. (d) For (i) and for (ii), sketch the steady-state solutions, as functions of z. (Assume k = 1, M = 6 and L = 1 for this sketch only.) Identify which sketch goes with (i) vs. (ii).