Page 295 - Modelling in Transport Phenomena A Conceptual Approach

8.2. ENERGY TRANSPORT WITHOUT CONVECTION 275

Figure 8.22 Rate of heat loss as a function of insulation thickness.

Example 8.11 Consider a hollow steel sphere of inside radius R1 = lOcm and
outside radius RZ = 20 cm. The inside surface is maintained at a constant temper-
ature of 18OOC and the outside surface dissipates heat to ambient temperature at
20 OC by convection with an average heat transfer coeficient of 11 W/ m2. K. To
reduce the rate of heat loss, it is proposed to cover the outer surface of the sphere
by two types of insulating materials X and Y in series. Each insulating material
has a thickness of 3cm. The thermal conductivities of the insulating materials X
and Y are 0.04 and 0.12W/ m. K, respectively. One of your fiends claim that
the order in which the two insdating materials are put around the sphere does not
make a difference in the rate of heat loss. As an engineer, do you agree?
Solution
Physical properties

For steel: k = 45 W/ m. K
Analysis
The rate of heat loss can be determined from Eq. (8.2-57). If the surface is first
coved by X and then Y, the rate of heat loss is
4~ (180 - 20)
Q = 0.1 0.03 0.03 1
(45)(0.1)(0.2) + (0.04)(0.2)(0.23) (0.12)(0.23)(0.26) (0.26)2(11)
= 91.6 W

On the other hand, covering the surface first by Y and then X gives the rate of
heat loss as
4~ (180 - 20)
Q=
0.1 0.03 0.03 1
(45) (0.1) (0.2) (0.12) (0.2) (0.23) -I- (0.04) (0.23) (0.26) ( 0.26)2 ( 1 1)
= 103.5W
Therefore, the order of the layers with different thermal conductivities does make
a difference.