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Lieberkuhn
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Homework Statement
A furnace wall is made up of 3 layers:
1. 4mm layer of material with thermal conductivity of 52 W/m.K
2. 2mm layer of material with thermal conductivity of 20 W/m.K
3. 1mm layer of material with thermal conductivity of 3 W/m.K
There is an air gap between layers 1&2 with a thermal resistance of 0.16 K/W.
The temperature at the inner surface of layer 1 is 873.15 K
The ambient temperature outside layer 3 is 343.15 K
The heat transfer coefficient from outside surface to surroundings is 17 W/m^2 .K
Find the following:
1. Rate of heat loss per square metre of outside surface (heat flux)
2. Temperature at each interface of wall, including outside surface temperature.
Homework Equations
h=q/ΔT[/B]
h = Heat transfer coefficient (in W/m^2 .K)
q = Heat flux (in W/m^2)
ΔT = Change in temperature (in K)
q=(kΔT)/L
q = Heat flux (in W/m^2)
k = Thermal conductivity (in W/mK)
ΔT = Change in temperature (in K)
L = Thickness of material (in m)
R=L/kA
R = Thermal resistance (in K/W)
L = Thickness of material (in m)
k = Thermal conductivity (in W/mK)
A = Area (in m^2)
q=ΔT/[(L1/k1)+(L2/k2)+(L3/k3)]
q = Heat flux (in W/m^2)
ΔT = Change in temperature between outer surfaces (in K)
L1 = Thickness of layer 1 (in m)
L2 = Thickness of layer 2 (in m)
L3 = Thickness of layer 3 (in m)
k1 = Thermal conductivity of layer 1 (in W/mK)
k2 = Thermal conductivity of layer 2 (in W/mK)
k3 = Thermal conductivity of layer 3 (in W/mK)
The Attempt at a Solution
Found the following equation but do not know how to remove A (area) to find q (heat flux).
3147.064q=28111200-8486.4A
Any useful hints would be much appreciated!