Incropera's Principle of Heat and Mass Transfer [Solutions] [8 ed.] 1119660378, 9781119660378

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Incropera's Principle of Heat and Mass Transfer [Solutions] [8 ed.]
 1119660378, 9781119660378

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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.