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

`Q=32.22(KW)/(m^2)`

`K_3=1.5\ (W)/(m.K)`

Step-by-step explanation:

At initial condition

As we know that thermal resistance

`R_(th)=(L)/(KA)`

So the total thermal resistance

`R_(th)=(L_1)/(K_1A)+(L_2)/(K_2A)`

Now by putting the values

`R_(th)=(L_1)/(K_1A)+(L_2)/(K-2A)`\

`R_(th)=(0.3)/(20A)+(0.15)/(50A)`

`R_(th)=(0.018)/(A)\ (K)/(W)`

So the heat conduction

`Q=(\Delta T)/(R_(th))`

`Q=(600-20)/(0.018)`

`Q=32.22(KW)/(m^2)`

At final condition another layer is added

`Given\ that\ heat\ flux\ is\ 5\ (KW)/(m^2)`

So the total thermal resistance

`R_(th)=(L_1)/(K_1A)+(L_2)/(K_2A)+(L_3)/(K_3A)`

`R_(th)=(0.018)/(A)+(0.15)/(K_3A)\ (K)/(W)`

`(0.018)/(A)+(0.15)/(K_3)=(\Delta T)/(q)`

`(0.018)/(A)+(0.15)/(K_3A)=(580)/(5000A)`

`K_3=1.5\ (W)/(m.K)`