This Demonstration calculates temperature profiles for steady-state conduction through multiple walls in series. The physical situation is a composite wall with the end temperatures fixed (as a result of convection or radiation). Therefore, the flux is equal to the difference of the two end temperatures over the resistance. The total thickness of the walls is 10 cm. You can vary the temperature of the left wall; the temperature of the right wall is fixed at 45 °C. You can vary the thicknesses of the first three walls and the conductivity of the fourth wall. The heat flux, which is the same through each wall, is calculated, and the temperature distribution through each wall is plotted.
The total heat-transfer resistance through four walls in series was used to calculate the steady-state heat flux through the walls. This flux value will be the same for the entire system since there is no heat generation, so the flux equation (with modified resistance and temperature terms) can be applied to each part of the composite wall. The heat flux, conductivities, wall thicknesses, and the left wall temperature were used to determine the temperatures at the interfaces between walls.
to are the endpoints of the walls and to are their conductivities.