Page 469 - Modelling in Transport Phenomena A Conceptual Approach

10.2. ENERGY TRANSPORT 449

Substitution of Eq. (10.2-84) into Eq. (10.285) yields

Q
-=E- -2 sin2 A,
X, X, + sin X, cos X, [I - exp (- x;T)] (10.2-86)
QO n=O
where Qo is the amount of heat transferred to the slab when the driving force is
constant and equal to its greatest (or? initial) value, i.e.,
Qo = 2LWHpCp(Tm -To) (10.287)

Example 10.3 A cake baked at 175 "C for half an hour is taken out of the oven
and inverted on a rack to cool. The kitchen temperature is 2OoC ana! the average
heat transfer coefficient is 12 W/ m2. K. If the thickness of the cake is 6 cm, esti-
mate the time it takes for the center to reach 40°C. Take k = 0.18Wjm.K and
a = 1.2 x m2/s for the cake.
Solution

The Biot number is

- (12)(0.03) =2
-
(0.18)
From Table 10.2 XI = 1.077. Considering only the first tern of the series in Eq.
(10.2-82), the temperature at the center, T,, is

Substitution of the values into Eq. (2) gives

20 - 40 2 sin61.7
-
20 - 175 - 1.077 + sin61.7cos61.7 exp [- (1.077)~~] (3)
in which 1.077rad = 61.7". Solving for T yields

T = 1.907 (4)
Therefore, the time is

- (1.907) (0.03)2 = 14,303s N 4h
-
1.2 x 10-7
Comment: The actual cooling time is obviously less than 4 h as a result of the
heat loss from the edges as well as the heat transfer to the rack by conduction.