Question

A large stainless steel (AISI 316) plate having a thickness of 5 mm is initially at a temperature of 70°C, and both sides of the plate are suddenly exposed to a convection environment at 20°C with the heat transfer coefficient of h= 800 W/m 2 oC. What will be the center and surface temperatures of the plate after 10 s? For stainless steel (AISI 316), α = 348x10-6 m2/s and k = 13.4 W/m K.

Answer 1

(a) Center of the plate: Given that, here time: T = 10S Also thermal diffusivity of stairtess steel (AISI 316) = alpha = 3.48^* 10^-6 m^2/s and thermal conductivity = 13.4 w/mk thickness = 2 L = 5mm Therefore L = 25 mm therefore Fourier no = F_0 = alpha middot T/L^2 = 3.48 times 10^-6 times 10/(0.0025)^2 = 3.568 now Biot no = h. middot L/k \r\n = 800 times 0.0025/13.8 = 0.15 also initial temp = T_i = 70^degree C surrounding temp = T_infinity = 20C Therefore fro Heisler chart at F_0 = 5.568 and B_i = 0.15 (T_0 - T_infinity)/(T_i - T_infinity) = 0.53 times (T_0 = center temp) Therefore t_0 = 0.53 times (70 - 20) + 20 = 46.5 C \r\n (b) Surface temp (T_1) Here we have to find temp, at x = L Therefore X/L = L/L = 1 Therefore At X/L = 1 and B_i = 0.15 from heistress chart (T_L - T_infinity)/(T_0 - T_infinity) = 093