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Exam (elaborations) TEST BANK FOR Introduction to Heat 4th Edition and Fundamentals of Heat 5th Edition By Incropera F.P. and Dewitt D.P. (Student Study Guide and Solution Manual) $15.49
Exam (elaborations) TEST BANK FOR Introduction to Heat 4th Edition and Fundamentals of Heat 5th Edition By Incropera F.P. and Dewitt D.P. (Student Study Guide and Solution Manual)
Exam (elaborations) TEST BANK FOR Introduction to Heat 4th Edition and Fundamentals of Heat 5th Edition By Incropera F.P. and Dewitt D.P. (Student Study Guide and Solution Manual)
PROBLEM 1.1
KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inn...
PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T 1. FIND: The outer temperature of the wall, T 2. 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 = - kdT
dxA=kATT
Lcond xx12==′′⋅⋅−. Solving for T 2 gives TTqL
kA21cond=− . 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 T 2 must be less than T 1. 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, x dT 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 T 2 = -15°C are ()2 12x25 C 15 CdT T Tq k k 1W m K 133.3W mdx L 0.30m−−−′′=− = = ⋅ =/G24/G24
. (1) 22xxq q A 133.3W m 20m 2667W′′=× = × = . (2) < Combining Eqs. (1) and (2), the heat rate q x can be determined for the range of ambient temperature, -15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k. -20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)-1500-500500150025003500Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K For the concrete wall, k = 1 W/m ⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity. COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear. PROBLEM 1.3 KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas. FIND: Daily cost of heat loss. SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties. ANALYSIS: The rate of heat loss by conduction through the slab is () ()12TT 7 Cq k LW 1.4W/m K 11m 8m 4312 Wt 0.20m−°== ⋅ × = < The daily cost of natural gas that must be combusted to compensate for the heat loss is () ()g
d6fqC 4312W $0.01/MJC t 24h/d 3600s/h $4.14/d
0.9 10 J/MJ η×=Δ = × =
× < COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete.
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