Incropera

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,

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.

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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, dT dx = − 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
(2)

(

)

(1)

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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.
3500

2500 Heat loss, qx (W)

1500

500

-500

-1500 -20 -10 0 10 20 30 40

Ambient air temperature, T2 (C) Wall