Circular copper rods of diameter D= 1.75 mm and length L= 25 mm are used to enhance heat transfer from a surface that is maintained at Ts,1= 100°C. One end of the rod is attached to this surface (at x= 0), while the other end (x= 25 mm) is joined to a second surface, which is maintained at Ts,2= 0°C. Air flowing between the surfaces (and over the rods) is also at a temperature of [infinity] T[infinity]= 0°C, and a convection coefficient of h= 125 W/m2·K is maintained. (1) What is the rate of heat transfer by convection from a single copper rod to the air, in W? (2) What is the total rate of heat transfer from a 1 m × 1 m section of the surface at 100°C, if a bundle of the rods is installed on 4-mm centers, in W?

a. Rate of heat transfer = 1.273 W

b. Total rate of heat transfer = 285,933.8 W

Explanation:

Given

a. Applying the conservation of energy

The general heat conduction equation for the temperature distribution is given as

Using boundary conditions, we find the constant values

Substituting the first boundary condition in the temperature distribution, we get

Using the second boundary conditions

Substituting the second boundary condition in the temperature distribution, we get

Since

We therefore substitute to find

Substituting into the heat conduction equation, we obtain

We proceed to differentiate the temperature distribution with respect to x

The thermal conductivity of copper from the table A.1, "Thermophysical properties of selected metallic solids" at an average temperature of 323K is 400W/m.K

Solving for m

Using Fourier's law, we solve for the heat transfer rate by conduction

Calculating the conductive heat transfer rate at x = 0, we get

Calculating the conductive heat transfer rate at x = 0.025 m, we get

b. The rods are placed 4 mm apart on a 1 m by 1 m section

To find the number of rods on each side with 1 m being equal to 1000 mm,

We therefore calculate the total number of rods, N in 1 m by 1 m section

To calculate the bare surface area, A

Total rate of heat transfer can be calculated thus