Question

A spherical stainless steel (? = 7,800 kgm-3; c = 460 Jkg-1K-1; k = 17 Wm-1K-1; ? = 0.9) reactor of 1.0-m inner diameter and 1.0-cm wall thickness is filled with reactants (? = 1,100 kgm-3; c = 4,200 Jkg-1K-1) that are initially at 25 ?C. An exothermic reaction within the reactor generates heat uniformly at a temperature-dependent volumetric rate of Qoexp(-A/T), where Qo = 6 kWm-3, A = 85 K and T is the reactive mixture temperature in K. The exterior surface of the reactor is exposed to 25 ?C ambient air with a convective heat transfer coefficient of 20 Wm-2K-1, and also exchanges radiation with surrounding surfaces that are at 35 ?C. The reactant mixture is well stirred with a convective heat transfer coefficient of 250 Wm-2K-1 so that the temperature within the reactor is nearly uniform. You may neglect the thermal capacitance of the reactor wall and consider the instantaneous heat transfer rate through the reactor wall to be equal to the instantaneous rate of heat loss to the air and surroundings.

a) What is the temperature of the reactant mixture after 5 hours of process time?

b) What is the temperature at the outer surface of the reactor at that time?

c) What is the steady-state temperature of the outer surface of the reactor?

Answer 1

Solving the DE on Polymath, we got the following profile for the tempaerature of the inside contents:

Here, Y-axis: Temperature in Kelvin; X-axis= Time in seconds.

a) From the report generated on polymath, Temperature inside after 5 hours: 314 K

Now, radiation resistance = (T_in - 308)/(T_in⁴ - 308⁴)* (5.67 x 10 ^-8)*0.9*A = 0.05184 K/W

Total resistance = 0.017235 + 0.05184 = 0.069075

Conduction resistance = 4.6347 x 10^-5

Fraction of conduction resistance = 4.6347 x 10^-5 / 0.069075 = 6.7 x 10^-4

which is a very small number.

Hence, inside temp = Outer surface temperature in both cases. (Outer wall will be slightly less)

c) We see steady state temp = 358 K