Question

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

Answer 1

Answer:

$Q=32.22\dfrac{KW}{m^2}$

$K_3=1.5\ \frac{W}{m.K}$

Explanation:

At initial condition

As we know that  thermal resistance

$R_{th}=\dfrac{L}{KA}$

So the total thermal resistance

$R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}$

Now by putting the values

$R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K-2A}$\

$R_{th}=\dfrac{0.3}{20A}+\dfrac{0.15}{50A}$

$R_{th}=\dfrac{0.018}{A}\ \frac{K}{W}$

So the heat conduction

$Q=\dfrac{\Delta T}{R_{th}}$

$Q=\dfrac{600-20}{0.018}$

$Q=32.22\dfrac{KW}{m^2}$

At final condition another layer is added

$Given\ that\ heat\ flux\ is\ 5\ \frac{KW}{m^2}$

So the total thermal resistance

$R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}$

$R_{th}=\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}\ \frac{K}{W}$

$\dfrac{0.018}{A}+\dfrac{0.15}{K_3}=\dfrac{\Delta T}{q}$

$\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}=\dfrac{580}{5000A}$

$K_3=1.5\ \frac{W}{m.K}$