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Has to be done on Matlab The conservation of heat can be used to develop a heat balance for a long, thin rod (Fig. 27.2).

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Has to be done on Matlab

The conservation of heat can be used to develop a heat balance for a long, thin rod (Fig. 27.2). If the rod is not insulated along its length and the system is at a steady state, the equation that results is dx2d2T+h(TaT)=0 where h is a heat transfer coefficient (m2) that parameterizes the rate of heat dissipation to the surrounding air and Ta is the temperature of the surrounding air (C). To obtain a solution for Eq. (27.1), there must be appropriate boundary conditions. A simple case is where the temperatures at the ends of the bar are held at fixed values. These can be expressed mathematically as T(0)T(L)=T1=T2 With these conditions. Eq. (27.1) can be solved analytically using calculus. For a 10-m rod with Ta=20,T1=40,T2=200, and h=0.01, the solution is T=73.4523e0.1x53.4523e0.1x+20 FIGURE 27.2 A noninsulated uniform rod positioned between two bodies of constant but different temperature. For this case T1>T2 and T2>Ta. Use the information above to find the solution of the governing equation using finite difference method and compare it with the analytical solution above by plotting both solutions on the same graph

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