REACTORS: IDEALIZED CHEMICAL REACTORS - Biomass as a Sustainable Energy Source for the Future - page 178

Example 6.5 Conversion in a tubular reactor using the RTD

The following liquid-phase reaction is performed in a tubular reactor:

2min −1

A ! b B+ c C+ − R A =kc A with k = 0 :

A pulse of tracer is introduced in the inlet of a reactor, and the tracer concentration
as a function of time is measured at the reactor outlet (Table 6.2). What is the con-
version of reactant A?
The area under the curve is Equation (6.44)
Area = X c i Δ L −1

t i = 1+4+6+3+1 ð Þ × Δ t= 1+4+6+3+1 ð Þ 2 = 30 g min

The average time that the reactant spends in the reactor is Equation (6.45)

X

t i c i Δ t i

ð 2×1+4×4+6×6+8×3+1×10 Þ ×2

i

t= X

:

= =5 9 min

ð 1+4+6+3+1 Þ ×2

c i Δ t i

i

The E curve can be obtained from Equation (6.46) and is shown in Table 6.3.
In a plug flow tubular reactor, all the molecules spend exactly 5.9 min in the
reactor. According to Equation (6.29),

c A
c A0 = exp

ðÞ − kt = exp ð − 0 : 2 5 : 9 Þ ≈ 0 : 31

The conversion of A is X A =1 − 0.31 ≈ 0.69.

However, according to the E curve, ≈ 27% of the molecules spend more than

6 min in the reactor, and 40% of the molecules spend between 4 and 6 min in the

reactor. To estimate the total c A /c A0 , we have to analyze each fraction of molecules
independently, according to their residence time (Table 6.4).
Thus, the conversion in the real reactor is slightly lower than the conversion in
the ideal PFR.

≈

TABLE 6.2 Concentration of tracer at the exit of the reactor
Time t (min)

0 2 4 6 8 10 12

Concentration of tracer (gL −1 ) 0 1 4 6 3 1 0

TABLE 6.3 E curve obtained from the concentrations shown in Table 6.2
Time t (min)

0 2 4 6 8 10 12

E = c/area (min −1 ) 0 0.033 0.13 0.20 0.10 0.033 0

E Δ t 0 0.067 0.27 0.40 0.20 0.067 0 =1