Question

The inner pipe of a double-pipe heat exchanger has an OD of 1.9 in. and contains 36 rectangular fins of the type shown in the sketch below. The fins are made of steel (k 34.9 Btu/h ft. "F) and are 0.5 in. high and 0.035 in. thi temperature is 250°F and the fluid in the annulus surrounding the fins is at 150°F with a heat-transfer c ck. The pipe wall ft2 、 c, F. Calculate: (a) The fin efficiency (b) The rate of heat transfer from one fin per foot of pipe length. (c) The prime surface area per foot of pipe length. (d) The rate of heat transfer from the prime surface per foot of pipe length. (e) The weighted efficiency of the finned surface.

Help with a-g, answers given. Show work

Answer 1

D = 1.9 in 36 Fins k = 34.9 Btu/h. ft . F t_0 = 250 degree F T_ambent = 150 degree F h = 30 Btu/h. Ft^2 . F degree h = heat transfer coefficient = 30 Btu/h - Ft^2 F p = parameter of Fin = 2 times (0.035 + L) = 2 times (0.035 + 12) in 20058 K = Thermal conductivity of Fin A_c = Area of cross section = t times L = 0.035 times 12 in^2 = 0.42 in^2 = 2.9166 times 10^-3 Ft^2 1) Fin efficiency therefore Find m = squareroot h p/k A_c m = squareroot 30 times 2.0058/34.9 times 2.9166 times 10^-3 m = 24.31 Efficiency of Actual heat Fin = transferred by Fin/maximum heat transfer by Fin If entire Fin area is at Fin base = theta Fin/theta_max = squareroot h p k A_c tan h (mL) (t_0 - t_a)/h A (t_0 - t_a) Here, theta_Fin formula used is for insulated Fin tip But in our cure Fin is of finite length a converting heat at tip. so, using corrective length i.e, L_e = L + A_c/p = 0.5 + 0.42/24.07