Question

An alloy fin with a thermal conductivity of 200 W/ሺm ∙ Kሻ has a length of 2.5 cm and a thickness of 3.5 mm. The base of the fin is held at 400o C and the ambient is at 20o C. Assuming that heat loss from the tip is negligible and that the convective heat transfer coefficient is 10 W/ሺmଶ ∙ Kሻ and the fin to be 1m wide, Find the heat transfer flux.

Answer 1

Answer: heat flux into the fun is 21.714 mW/m^2

Step-by-step explanation:

Heat flux Q = q/A

q = heat transfer rate W

A = area m^2

q = area * conductivity * temperature gradient

Temperature gradient = difference in temperature of the metal faces divided by the thickness.

Therefore Q = k * ( temp. gradient)

Q = 200 * ((400-20)/3.5*10^-2)

Q = 21714285.71 = 21.714 mW/m^2

Answer 2: convective heat transfer flux between fin and air

is 3800W/m^2

Explanation :

q = hA*(Ts-Ta)

h = convective heat transfer coefficient

Ts = temperature of fin

Ta = temperature of air

Q = q/A

Q = h(Ts-Ta)

Q = 10(400 - 20)

Q = 3800 W/m^2