Sustainability and Stability - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 806

where C A is in moles of aromatics per cubic centimeter. The oil is to be fed to the reactor at
a constant rate of 0.5 cm 3 /min. The temperature of the feed stream is increased very slowly
from 250 to 375 C and then decreased very slowly from 375 to 250 C.
Assume that the density of the oil is constant at 0.85 g/cm 3 and that the hydrogenation of
the aromatics is the only reaction. The pseudosteady-state approximation can be made for the
slow changes in feed temperature, i.e., it can be assumed that steady state is achieved at all
feed temperatures since the changes in feed temperature occur very slowly.
Prepare a plot of fractional conversion of the aromatics as a function of the feed tempera-
ture and comment on the system behavior.
Solution. The reaction under concern is
Aþ H 2 /

r 0 ¼ kC A products

Steady mole balance on A over the CSTR leads to
QðC A0 C A Þ¼m cat r 0 (E16-3.1)
where m cat is the mass of the catalyst in the reactor. The left-hand side of Eqn (E16-3.1) repre-
sents the mass supplying rate of A (MS A ), while the right-hand side represents the mass
consumption rates of A (MC A ). The general energy balance:
X

N S

F j0 ðH j H j0 ÞþV X

N R

r i DH Ri þ X

N S

d T
d t

d ð pV Þ
d t ¼ Q W s

C Pj n j (3.133)

j¼1 i¼1 j¼1

can be reduced for a steady CSTR with negligible stirrer work input and adiabatic
operation to

C P ðT T 0 Þ¼m cat ðDH R Þr 0 (E16-3.2)
The left-hand side of Eqn (E16-3.2) is the heat of removal (H R ) from the reactor, while the
right-hand side is the heat of generation in the reactor (H G ).
To obtain the steady-state solutions, one is required to solve both the mass balance Eqn
(E16-3.1) and energy balance Eqn (E16-3.2) for a nonisothermal reactor operation:
MS A ¼ MC A and H R ¼ H G . Eqn (E16-3.2) O (E16-3.1) yields
T T 0

Q r

C A0 C A ¼ DH R

(E16-3.3)

r C P

which can be rearranged to give

T ¼ T 0 þ DH R
r

C A0 f A (E16-3.4)

C P

Substituting the rate expression and conversion in Eqn (E16-3.1) , we obtain
QC A0 f A ¼ m cat k 0 C A0 ð1 f A Þexp

E
RT

(E16-3.5)

That is

E = R

ln f A
1 f A

T ¼ (E16-3.6)

Q
m cat k 0