A metal circular rectangular fin has a base diameter of 20 m

A metal circular rectangular fin has a base diameter of 20 mm pipe, a length of 27 mm, k= 20 W/m.k and a thickness of 2mm. The base temperature is 217 C, h= 30 W/m^2.K and T_a = air temperature = 27C and fin spacing (pitch) is 20mm. Find the fin efficiency Area per pitch for A_0 (original_base area before fin added), Af (fin per pitch), A_exp(exposed_base per pitch_after_fin_added) Find q_0 (original pipe without fin), qf (one fin), qt (of pipe plus fin) and ratio of qt (with fin) to q_0(withoutfin). (all per pitch) If the fin given is a linear rectangular fin what is the fin efficiency? Is it tanh(mL)?

Solution

a)The fin efficiency = qf/q =

A0 = 27x 2 xpi x 10 = 282.6mm²

Af = 27x (10.46-2) = 28.42mm²

q0 = k (T2-T1) x A = 20 x (180) x (10.46 x27) = 1016712

qf = sqrt ( hx Perimeter of fin) x k X cross section of fin x temp at base

   = Sqrt ( 30x 44x 40x 217 = 1513.7

Hence qf/q0 can be found.

Generally the heat tranfer equation can be written as d2(theta)/dx2 -m2 (theta) = 0

substituting theta as e ^alpha x and assuming the fin is long, we get

q = sqrt( h p k A Theta )

for insulated tip solving the equation we get q = -KA dt/dx = -KA Dtheta/dx

this gives q= sqrt ( hPAk Theta ) tanhL