Suppose you have a 05m times 05m surface that needs a fin to

Suppose you have a 0.5[m] times 0.5[m] surface that needs a fin to enhance heat transfer. You have 100 cubic centimeters of material with which to cast a straight fin rectangular fin and weld it to the surface. None of the three dimensions of the fin may be smaller than 2 centimeters, and the fin should not extend more than 20 centimeters from the surface. The material has a conductivity of and k = 5 [W/m degree C] and the prevailing convection heat-transfer coefficient is h = 60[W/m^2 degree C]. Whatever your design, you will need to weld it to the surface and the weld creates a contact resistance of 2 times 10^-4 [m^2 degree C/W]. What fin dimensions design maximize the overall heat transfer? What, in [W/degree C], is the overall heat-transfer coefficient of your fin?

Solution

Fin heat loss = h ( Perimeter of finx length) x (temp differ)

equating heat source to convection loss = -h P (temp diff) / cross section of fin

the general equation is given by dT² /dX² - hP/A (temp diff) = 0 solving this and using fourier law for heat transfer we get  = -kA dt/dx = -kA dtheta/dx

which gives q = sqrt ( hPkA (temp diff) = sqrt[ ( 60x 1,04x 5 x 1.385 )Temp diff] = 432.12 ( tempdiff) W

This temp difference is the temperature between fin and the ambient.

Since there is resistance at the root weld, the temperature will be tranmitted to the fin depending on the contact resistance given as 2x 10^-4 m²degC/W

ie. 2x 10^-4 x 1.385 (temp diff) /W = 2.77x 10^-4 temp diff /W

The area should be max for heat transfer and hence the area of rectangular plate is taken as 20cm high and length 0.5 meter with thickness as 2 cm