fundamentals of heat and mass transfer solutions manual phần 4

BIẾT: máy bay bức tường, ban đầu có một phân bố nhiệt độ tuyến tính trạng thái ổn định với các ranh giới duy trì ở mức T (0, t) = T1 và T (L, t) = T2, đột nhiên kinh nghiệm một hệ thống nhất thể tích nhiệt do dòng điện. Ranh giới điều kiện T1 và T2 vẫn cố định theo thời gian. | PROBLEM KNOWN Plane wall initially having a linear steady-state temperature distribution with boundaries maintained at T 0 t T1 and T L t T2 suddenly experiences a uniform volumetric heat generation due to the electrical current. Boundary conditions T1 and T2 remain fixed with time. FIND a On T-x coordinates sketch the temperature distributions for the following cases initial conditions t 0 steady-state conditions t assuming the maximum temperature exceeds T2 and two intermediate times label important features b For the three-nodal network shown derive the finite-difference equation using either the implicit or explicit method c With a time increment of At 5 s obtain values of Tm for the first 45s of elapsed time determine the corresponding heat fluxes at the boundaries and d Determine the effect of mesh size by repeating the foregoing analysis using grids of 5 and 11 nodal points. SCHEMATIC E1 o c L 20 mm Tn 100 C q 2x107W m3 ASSUMPTIONS 1 Two-dimensional transient conduction 2 Uniform volumetric heat generation for t 0 3 Constant properties. PROPERTIES Wall Given p 4000 kg m3 c 500 J kg-K k 10 W m-K. ANALYSIS a The temperature distribution on T-x coordinates for the requested cases are shown below. Note the following key features 1 linear initial temperature distribution 2 non-symmetrical parabolic steady-state temperature distribution 3 gradient at x L is first positive then zero and becomes negative and 4 gradient at x 0 is always positive. b Performing an energy balance on the control volume about node m above for unit area find Ein - Eout Eg Est Tn - T T T 1 - tP k 1 12 À lm k 1 11 À lm q 1 Ax p 1 cAx im im Ax Ax At Fo T1 T2 - 2Tm -ầA Tm 1 - Tm pcp For the Tm term in brackets use p for explicit and p 1 for implicit form Explicit Implicit Tm 1 Fo tP T2p 1 - 2Fo Tm qAt pcp Tp Tm 1 Fo tP 1 TP 1 q Ay pcp 1 2 Continued. PROBLEM Cont. c With a time increment At 5s the FDEs Eqs. 1 and 2 become Explicit Implicit where Tn 1 75 Tm 1 TP 75 3 4 .