Ideal Flow Reactors - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 181

+ d V VV

F j e

F j 0

dV

FIGURE 5.3 A schematic diagram of the PFR showing a differential control volume ( V , V þ d V ).

In general,

X

N S

r ¼ C j M j (5.10)

j¼1

In a general case, Eqn (5.7) needs to be written each for every component (species) except one
involved. Thus, we would have N s partial differential equations to solve even for an
isothermal isobaric reactor. If temperature changes, the energy balance equation must be
solved simultaneously as well. Pressure drop or momentum balance makes another addition
to the equations.
In industrial operations, we would prefer to run the reactor at maximum capacity all the
time to maximize the throughput and minimize the unit cost to produce the product. There-
fore, often times steady-state operations are desired. When steady state is reached, every
quantified measure in the reactor is independent on time, i.e. time derivatives are zero. Equa-
tion (5.7) is reduced to

d F j
d V

r j ¼ (5.11)

For a single reaction that is carried out in a steady PFR, there is only one independent
concentration or reaction mixture content variable and all other concentrations can be related
through stoichiometry as shown in Chapter 3. Without lose of generosity, let us use compo-
nent A as the key component of consideration. Equation (5.11) applied to species A gives

d F A
d V

r A ¼ (5.12)

an ordinary differential equation.
In general, the rate of reaction is a function of concentration and temperature. From
Chapter 3, we learned that the stoichiometry can be applied to relate the amount of every
species in the reaction mixture. The amount change of a component participating in the reac-
tion divided by its stoichiometric coefficient is the universal extent of reaction for a single
reaction. The stoichiometry can be written in a flow reactor as
d F j
n j ¼

d F A
n A ¼ r d V

(5.13)

When there is no side feed or outlet stream, Eqn (5.13) can be integrated to give

F j F j0
n j

F A F A 0
n A

F A 0

¼ ¼

n A f A

(5.14)