Question

A composite rod is made from stainless steel and iron and has a length of 0.498 m. The cross section of this composite rod is shown in the drawing below and consists of a square within a circle.

The square cross section of the steel is 1.14 cm on a side. The temperature at one end of the rod is 75.9 °C, while it is 17.9 °C at the other end. Assuming that no heat exits through the cylindrical outer surface, find the total amount of heat conducted through the rod in three minutes.

Answer 1

Assuming that square has its maximum extension, i.e. its corners lie on the circular circumference of the cross section, the diameter of the circle equals the diagonal length of the square. The diagonal length of square is square-root od 2 times its side length.

So the total area of the composite corss section:

A_total = (π/4)*D^2 = (π/4)*(√2*1.14*10^-2)^2 = (π/2)*(1.14*10^-2)^2 = 2.04*10^-4 m^2

The area of the steel cross section:

A_steel = (1.14*10^-2)^2 = 1.29*10^-4 m^2

So the area of the iron section:

A_iron = A_total - A_steel = 2.04*10^-4 - 1.29*10^-4 = 7.5*10^-5 m^2

The heat flow rate through each section can be found from integral form of fourier's law for stationary 1-D heat flow:

H = ∆Q/∆t = k*A*∆T / ∆x

The thermal conductivities are:

k_iron = 80 Wm^-2°C^-1

k_steel = 16 Wm^-2°C^-1

H_iron = k_iron*A_iron*∆T / ∆x

= 80*( 7.5*10^-5)*(75.9 - 17.9) / 0.498   = 0.698 W

H_steel = 16*(1.29*10^-4)*(75.9 - 17.9) / 0.498 = 0.240 W

The total heat flow rate:

H_total = H_iron + H_steel = 0.698 + 0.240 = 0.938 W

So the required heat conducted through the rod in 3 minutes (180 seconds):

  ∆Q = H_total*∆t = (0.938)*180 = 168.84 J