# Tema31 - CHAPTER 31 31.1 The equation to be solved is d 2T...

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CHAPTER 31 31.1 The equation to be solved is d T dx 2 2 20 = - Assume a solution of the form T = ax 2 + bx + c which can be differentiated twice to give T ” = 2 a . Substituting this result into the differential equation gives a = - 10. The boundary conditions can be used to evaluate the remaining coefficients. For the first condition at x = 0, 50 10 0 0 2 = - + + ( ) ( ) b c or c = 50. Similarly, for the second condition. 100 10 10 10 50 2 = - + + ( ) ( ) b which can be solved for b = 105. Therefore, the final solution is T x x = - + + 10 105 50 2 The results are plotted in Fig. 31.5. 0 100 200 300 0 5 10 31.2 The heat source term in the first row of Eq. (31.26) can be evaluated by substituting Eq. (31.3) and integrating to give 20 2 5 2 5 25 0 2 5 . . . - = x dx Similarly, Eq. (31.4) can be substituted into the heat source term of the second row of Eq. (31.26), which can also be integrated to yield 20 0 2 5 25 0 2 5 x dx - = . . These results along with the other parameter values can be substituted into Eq. (31.26) to give 0 4 0 4 25 1 2 1 . . ( ) T T dT dx x - = - + and

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- + = + 0 4 0 4 25 1 2 2 . . ( ) T T dT dx x 31.3 In a manner similar to Fig. 31.7, the equations can be assembled for the total system, 0 4 0 4 0 4 0 8 0 4 0 4 0 8 0 4 0 4 08 0 4 0 4 0 4 25 50 50 50 25 1 2 3 4 5 1 1 . . . . . . . . . . . . . ( ) / ( ) / - - - - - - - - = - + + T T T T T dT x dx dT x dx The unknown end temperatures can be substituted to give 1 0 4 0 8 0 4 0 4 0 8 0 4 0 4 0 8 0 4 1 5 70 50 90 15 1 2 3 4 5 - - - - - - - - = - . . . . . . . . . ( ) / ( ) / dT x dx T T T dT x dx These equations can be solved for dT x dx T T T dT x dx ( ) / ( ) / 1 2 3 4 5 105 250 325 275 95 - = - The solution, along with the analytical solution (dashed line) is shown below: 0 100 200 300 0 5 10 31.4 0 2 2 = - - D d c dx U dc dx kc R D d c dx U dc dx kc = - - 2 2 ~ ~ ~ D d c dx U dc dx kc N dx x x i 2 2 1 2 ~ ~ ~ - -
D d c dx N x dx i x x 2 2 1 2 ~ ( ) (1) - U dc dx N x dx i x x ~ ( ) 1 2 (2) - k cN x dx i x x ~ ( ) 1 2 (3) Term (1): D d c dx N x dx D dc dx x c c x x dc dx x c c x x i x x 2 2 1 1 2 2 1 2 2 1 2 1 1 2 ~ ( ) ( ) ( ) = - - - - - - - Term (2): dc dx N x dx c c x x N x dx i x x i x x ~ ( ) ( ) 1 2 1 2 2 1 2 1 ∫ ∫ = - - N x dx x x i x x ( ) 1 2 2 1 2 = - = - dc dx N x dx c c i x x ~ ( ) 1 2 2 1 2 - = - - - U dc dx N

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Tema31 - CHAPTER 31 31.1 The equation to be solved is d 2T...

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