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English Pages 1072 [2088] Year 2019
PROBLEM 1.1 KNOWN: Temperature distribution in wall of Example 1.1. FIND: Heat fluxes and heat rates at x = 0 and x = L. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through the wall, (2) constant thermal conductivity, (3) no internal thermal energy generation within the wall. PROPERTIES: Thermal conductivity of wall (given): k = 1.7 W/m·K. ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
q′′x = −k
dT dx
(1)
Since the temperature distribution is T(x) = a + bx, the temperature gradient is
dT =b dx
(2)
Hence, the heat flux is constant throughout the wall, and is
q′′x =−k
dT =−kb =−1.7 W/m ⋅ K × ( −1000 K/m ) =1700 W/m 2 dx
0.1 Hence, the lumped capacitance method is inappropriate. Using the one-term series approximation, Eqs. 5.52 with Table 5.1, θ* = C1 exp −ζ12 Fo J o ζ1r* r* = r ro = 1
(
T ( ro , t ) − T∞ = θ* = Ti − T∞
= Bi hr = o k 1.84
) ( )
− 350 ) C (175 = ( 25 − 350 ) C
0.54
= ζ1 1.546 rad
= C1 1.318
Continued...
PROBLEM 7.52 (Cont.) 0.54 = 1.318exp[-(1.546rad)2Fo]J o (1.546 × 1) Using Table B.4 to evaluate J o (1.546) = 0.4859, find Fo = 0.0725 where
= Fo
5.68 ×10−7 m 2 s × t o = = 5.68 ×10−3 t o 2 2 ro ( 0.010 m )
α to
(6)
3.06. Therefore, we specify N = 4.