Completed solution of the exercises book Incropera

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PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law, qq q A = - k dT dx A=k A TT Lcond xx 12== ′′ ⋅⋅ − . Solving for T2 gives TT qL kA21 cond=− . Substituting numerical values, find TC - 3000W 0.025m 0.2W / m K 10m2 2= × ⋅×415/G24 T C - 37.5 C2 = 415/G24/G24 TC . 2 = 378/G24 < COMMENTS: Note direction of heat flow and fact that T2 must be less than T1. PROBLEM 1.2 KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air. ANALYSIS: From Fourier’s law, it is evident that the gradient, xdT dx q k ′′=− , is a constant, and hence the temperature distribution is linear, if xq′′ and k are each constant. The heat flux must be constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C are ( ) 212x 25 C 15 CdT T Tq k k 1W m K 133.3W mdx L 0.30 m −−−′′ =− = = ⋅ = /G24/G24 . (1) 22xxq q A 133.3W m 20 m 2667 W′′=× = × = . (2) < Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 ≤ T2 ≤ 38°C, with different wall thermal…