A composite wall is made of two layers of 0.3 m and 0.15 m thickness with surfaces held at 600°C and 20°C respectively. If the conductivities are 20 and 50 W/mK, determine the heat conducted. In order to restrict the heat loss to 5 kW/m2 another layer of 0.15 m thickness is proposed. Determine the thermal conductivity of the material required

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Netta00 Netta00 · Answer 1 · 2019-09-09T08:29:00+02:00

Answer:

[tex]Q=32.22\dfrac{KW}{m^2}[/tex]

[tex]K_3=1.5\ \frac{W}{m.K}[/tex]

Explanation:

At initial condition

As we know that thermal resistance

[tex]R_{th}=\dfrac{L}{KA}[/tex]

So the total thermal resistance

[tex]R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}[/tex]

Now by putting the values

[tex]R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K-2A}[/tex]\

[tex]R_{th}=\dfrac{0.3}{20A}+\dfrac{0.15}{50A}[/tex]

[tex]R_{th}=\dfrac{0.018}{A}\ \frac{K}{W}[/tex]

So the heat conduction

[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]

[tex]Q=\dfrac{600-20}{0.018}[/tex]

[tex]Q=32.22\dfrac{KW}{m^2}[/tex]

At final condition another layer is added

[tex]Given\ that\ heat\ flux\ is\ 5\ \frac{KW}{m^2}[/tex]

So the total thermal resistance

[tex]R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}[/tex]

[tex]R_{th}=\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}\ \frac{K}{W}[/tex]

[tex]\dfrac{0.018}{A}+\dfrac{0.15}{K_3}=\dfrac{\Delta T}{q}[/tex]

[tex]\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}=\dfrac{580}{5000A}[/tex]

[tex]K_3=1.5\ \frac{W}{m.K}[/tex]