Question

Problem 2: A hot metal ball 0.00245 W/m°C; α-0151* 10-s m2/s; d 18 cm) initially at a unifom temperature of 310°C is cooled by placing it in air at 29 C is flowing at a constant 4 m/s. In order to continue on in the manufacturing process, the temperature at the center of the ball must be 70°C. (μ. = 1.872*10-5 kg m-s). The air You mav assume that the properties of air may bem odeled as V = 1.608 x 10-5 m2/s Pr = 0.7282 μ. = 2.934-10-s kg/m-s Given this infomation, determine: a. The average heat transfer coefficient b. Te tirne required for the temperature ofthe center ofthe sphere to reach 70°C-state and justfy any assumptions The surface temperature of the sphere when the center is 70°C. The percent of total heat transferred to the surroundings when the center temperature is 70°C c. d.

Answer 1

k = 0.00245 w/m degree C alpha = 0.0151 times 10%-6 m^2/s d = 0.18m initial temp = 310 degree C T_infinity 29 degree C T_infinity = 29 degree C v = 1.608 times 10^-5 m^2/s v = 4 m/s (a) Reynolds number = vd/upsilon = 4 times 0.018/1.608 times 10^-5 = 0.44 times 10^-5 = 4.4 times 10^4 Nussle number is given by, N_4 = h_D/k = 2 + [0.4Re^1/2 + 0.06 Re^2/3] P_r^0.4 [mu_infinity_0/mu_s]^1/4 = 2 + [0.4 (4 times 10^4)^1/2 + 0.06 (4.4 times 10^4)^2/3] (0.7282)^0.4 (1.872/2.934) = 135 h = k/D = 0.00245/0.18 times 135 h = 1.83 w/m degree c (b) Here, Biot number = hr/L = 1.83 times 0.09/0.00245 = 67.22 > 0.1 1/B_i = 0.014 theta_0 = T_0 - T_infinity/T_i - T_infinity = 70.29/310 - 29 41/281 = 0.14 From graph, tau = alpha t/r^2 = 0.25 t = tau (r^2)/alpha = 0.25 times (0.09)^2/0.15 times 10^-6 m^2/s = 13500 sec = 3.75 hrs \r\n (c) Here, Q = mc_p (T_2 - T_1)