Question

A wall is constructed of a section of stainless steel (k = 16 W/m -°C) 40 mm thick with identical layers of plastic on both sides of the steel The overall heat-transfer coefficient, considering convection on both sides of the plastic, is 120 W/m2 °C If the overall temperature difference across the arrangement is 60°C, calculate the temperature difference across the stainless steel The area of the wall is one meter square

Answer 1

The temperature difference across the stainless steel is 18°C

Step-by-step explanation:

The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.

The heat flux q/A can be expressed as:

`q/A = h*\Delta T`

It can also be expressed as

`q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)`

being Δx/k*(T1-T2) the conductive heat flux through the steel.

If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:

`(k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^(\circ) C`