Question

Copper is a very good conductor of heat and thus not a very good insulator. It has a thermal conductivity of 398 W/(m-°C). A square Cu plate with cross-sectional area 0.15 ×10-2 m2 is sandwiched between two isothermal plates, each of which is at a different temperature. The Cu plate is fully insulated along its exposed edges so that the only significant heat flux is directed through the thickness of the plate. One side of the Cu plate is maintained at 190.8°C by an electric energy supply of 1,086 watts. Assuming steady-state heat flow conditions, how thick, in mm, must this Cu plate be so that the temperature on the other side of the Cu plate does not exceed 24°C ?

Answer 1

Answer:

The Cu plate must be at least 91.69 mm thick

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, we must solve the equation for d, assuming that the maximum possible temperature of the other side of the plate is 24°C:

$Q=\frac{T_1-T_0}{d}kA\\d=\frac{T_1-T_0}{Q}kA\\d=0.0917 m =91.69 mm$

$T_0: Temperature \ on \ the \other \ side \ of \ the \ plate\\T_1: Temperature \ at \ the \ first \ side \ of \ the \ plate\\k: thermal \ conductivity \\d: Cu \ plate \ thickness \\ Q: heat \ flow\\ A: cross-sectional \ area$