Question

Problem 4 Consider a thin-walled cylindrical vessel of unit length, containing volatile solid reactants (specific gravity of 1.5, specific heat capacity of 2200 J/kg-C), which are initially at a temperature of 25° C. The reactants are suddenly stirred thoroughly, and they react endothermically, with a heat generation rate of of 101 W/m3. The inner and outer radius of the thin annulus, made of a material whose thermal conductivity is 12 W/mK, are 0.2-m and 0.215-m, respectively. The vessel is surrounded by ambient fluid at 25° C and a convection coefficient of 5 W/m2K. . With a neat schematic of the problem, show that temperature gradient across the thin-walled shell will be negligible by evaluating the Biot number Set up the differential equation to compute reactant temperature as a function of time, and compute the thermal time constant. . What is the temperature of the reactants after 2 hours have elapsed?

Answer 1

q = heat generation rate = 10 W/m^3 k = thermal conductivity of material ( material) = 12 W/mk r_i = 0.2 m r_0 = 0.215 m Biot Number = conductive resistance by annulus/convective Resistance to air = ln (r_2/r)/2 pi k l/1/h2 pi r_2 l = h ln (r_2/r_1 middot r_2/K = 0.0065 * since Biot No. is less than 0.01 so we can neglect temp gradient across the glinder well Applying the energy balance equation to the control volume shown by dotted lines E^dot_m - E^dot_out = (dE/dt)_c.v - 1) 0 - hA (T - T_alpha) - q^dot times volume = delta VC_p dT/d tau - 2) [where dE/dt]_cv = delta VC_p dp/dl] \r\n Temp after 2 hours tau = 3600 times 2 seconds In eq 3 as Initial tau_i = 25 degree C, tau_alpha = 25 degree C there is no convective neat transfer & after some time it is very less. so eq^3 becomes d tau/dt = -3.03 times 10^-6 implies tau_f - 298 = -3.03 times 10^-6 times 7200 y^bar = 298 - 0.021816