Transport Phenomena - Heat Transfer Problem :
Minimum thickness for a composite furnace wall
Problem.
The wall of a furnace comprises three layers as shown in the figure. The first layer is refractory (whose maximum allowable temperature is 1400oC) while the second layer is insulation (whose maximum allowable temperature is 1093oC). The third layer is a plate of 6.35 mm thickness of steel [thermal conductivity = 45 W/(m K)]. Assume the layers to be in very good thermal contact.
Figure. Layers in a composite furnace wall.
The temperature T0 on the inside of the refractory is 1370oC, while the temperature T3 on the outside of the steel plate is 37.8oC. The heat loss through the furnace wall is expected to be 15800 W/m2. Determine the thickness of refractory and insulation that results in the minimum total thickness of the wall.
Given thermal conductivities in W/(m K):
Layer
k at 37.8oC
k at 1093oC
Refractory
3.12
6.23
Insulation
1.56
3.12
Solution.
Click here for stepwise solution
Step. Thermal resistance representation for composite furnace wall
In general, the heat flow is given by Q =
DT/Rth and
the thermal resistance for a rectangular slab is Rth
= Dx/(kA), where
DT is the temperature driving force
(thermal potential difference), Dx is the
slab thickness, k is the thermal conductivity, and A is the cross-sectional area of the slab.
The thermal resistances for the three layers are in series as shown in the figure below.
Figure. Thermal resistance representation of composite furnace wall.
Based on the thermal resistance representation for the composite furnace wall, the heat flux q is
(1)
In the refractory and insulation, the thermal conductivity k varies with temperature. If a linear variation is assumed, then the arithmetic mean is to be used for the thermal conductivity.
Step. Temperature at insulation - steel interface
The temperature T2 at the interface between the insulation and steel layers is given by
(2)
The above expression can be used to calculate temperature T2 (as done later).
Step. Thickness of refractory and insulation
The thickness of the refractory (x1 - x0) and insulation (x2 - x1) are
(3)
On adding the above two equations, we get
(4)
Since q, T0 and T2 are known in the above expression, the first two terms on the right-hand side are nearly fixed. The last term is negative as k12 is less than k01 (i.e., thermal conductivity of insulation is less than that of refractory). Since the aim is to minimize (x2 - x0), the temperature T1 must be maximized.
Step. Substitution of numerical values
The temperature at the insulation - steel interface is
(5)
Next, the temperature T1 is set to 1093oC (i.e., the maximum allowable temperature for the insulation). By linear extrapolation, the thermal conductivity of the refractory at 1370oC is 7.05 W/(m K). Substitution of numerical values gives the thickness of refractory as
(6)
and the thickness of insulation as
(7)
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