Chapter 1 - PROBLEM 1.1 KNOWN: Heat rate, q, through...

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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, 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 = - k dT dx A=kA TT L cond xx 12 == ′′ ⋅⋅ - . Solving for T 2 gives qL kA 21 cond =- Substituting numerical values, find TC - 3000W 0.025m 0.2W / m K 10m 2 2 = × ⋅× 415 T C-37.5 C 2 = 415 . 2 = 378 < COMMENTS: Note direction of heat flow and fact that T 2 must be less than T 1 .
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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 x q 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 ( 29 2 12 x 25 C 15 C dT T T q k k 1W m K 133.3W m dx L 0.30m -- - ′′ =- = = = . (1) 22 xx q q A 133.3W m 20m 2667 W = × = . (2) < Combining Eqs. (1) and (2), the heat rate q x can be determined for the range of ambient temperature, -15 T 2 38 ° C, with different wall thermal conductivities, k. -20 -10 0 10 20 30 40 Ambient air temperature, T2 (C) -1500 -500 500 1500 2500 3500 Heat 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.
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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 (29 12 TT 7 C q k LW 1.4W / m K 11m 8m 4312 W t 0.20m == × = < The daily cost of natural gas that must be combusted to compensate for the heat loss is g d 6 f qC 4312W $0.01/ MJ C t 24h / d 3600s / h $4.14/ d 0.9 10 J / MJ η × =∆ = × = × < 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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PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness.
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This note was uploaded on 10/20/2008 for the course ME 415 taught by Professor Biddle during the Fall '08 term at Cal Poly Pomona.

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Chapter 1 - PROBLEM 1.1 KNOWN: Heat rate, q, through...

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