Page 111 - Numerical Methods for Chemical Engineering

100 2 Nonlinear algebraic systems

Table 2.2 Drag coefficient for flow around a
smooth sphere at Reynolds’ numbers between
350 000 and 1 000 000. From Table 5–22 of Perry
& Green (1984)

Re C D
350 000 0.396
400 000 0.0891
500 000 0.0799
700 000 0.0945
1 000 000 0.110

Between Reynolds’ numbers of 350 000 and 1 000 000, there is a transient drop in the
drag constant as the boundary layer first becomes turbulent, delaying separation on the
downstream side. In this regime, we may use interpolation (MATLAB function interp1)of
the data in Table 2.2.
3
Estimate the terminal velocity of a smooth spherical iron particle (density 7850 kg/m )
3
in water (density 1000 kg/m , viscosity 0.001 Pa s) as a function of the particle diameter
(in meters) over the range 0.1 mm to 10 cm.
2.B.1 Consider a CSTR, with the single gas-phase elementary reaction

A + B → C (2.164)
that has a rate constant k of 0.05 l/(mol/s). The feed stream to the reactor, at a volumetric
flow rate υ 0 in liters per second, is at a total pressure P 0 = 5 atm. It comprises reactants
A and B, and a nonreacting diluent gas. The partial pressures of the reactants are P A0 =
0.5 atm and P B0 = 0.5 atm. Assume that the ideal gas law holds, and that the reactor is
operated isothermally at 350 K. Assume that the inlet stream is at the same temperature and
pressure as the reactor contents. For a reactor volume of 1000 l, plot the conversion of A as
a function of υ 0 . Make sure to account for the fact that the total number of moles in the gas
phase decreases with increasing conversion, so that the outlet volumetric flow rate will be
somewhat smaller than υ 0 . For further discussion, consult Fogler (1999).
2.B.2. Consider a 1000-l CSTR in which the following reactions are taking place:

A + B → C r R1 = k 1 c A c B
(2.165)
C + B → D r R2 = k 2 c B c C
A → E r R3 = k 3 c A
We have the following kinetic data
l −2 l
−2
k 1 (298 K) = 2.1 × 10 k 1 (315 K) = 3.6 × 10
mol s mol s
l −2 l
−2
k 2 (298 K) = 1.5 × 10 k 2 (315 K) = 4.5 × 10 (2.166)
mol s mol s
k 3 (298 K) = 0.00012 s −1 k 3 (315 K) = 0.00026 s −1