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’slaw,
q cond = q x = q ′′ ⋅ A = -k x
Solving for T2 gives
T −T dT ⋅ A = kA 1 2 . dx L
T2 = T1 −
q cond L . kA
Substituting numerical values, find
T2 = 415 C -
3000W × 0.025m 0.2W / m ⋅ K × 10m2
T2 = 415 C - 37.5 C T2 = 378 C.
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.
KNOWN: Inner surface temperature and thermalconductivity 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, dTdx = − q′′ k , is a constant, and x hence the temperature distribution is linear, if q′′ and k are each constant. The heat flux must be x 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
25 C − −15 C dT T1 − T2 q′′ = − k =k = 1W m ⋅ K= 133.3W m 2 . x dx L 0.30 m q x = q′′ × A = 133.3 W m 2 × 20 m 2 = 2667 W . x
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 conductivities, k.
2500 Heat loss, qx (W)
-1500 -20 -10 0 10 20 30 40
Ambient air temperature, T2 (C) Wallthermal 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 temperaturedistribution in a plane wall would not be linear.
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
T−T 7°C q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
( ∆t ) =
4312 W × $0.01/ MJ 0.9 ×106 J / MJ
( 24 h / d × 3600s / h ) = $4.14 / d
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and theconcrete.
PROBLEM 1.4 KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness. FIND: Thermal conductivity, k, of the wood. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’slaw, Eq. 1.2. Rearranging, k=q′′ x L W = 40 T − T2 m2 1
( 40-20 ) C
k = 0.10 W / m ⋅ K.
COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference.
PROBLEM 1.5 KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions. FIND: Heat loss through window. SCHEMATIC:
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