Page 496 - Elements of Chemical Reaction Engineering Ebook

Sec. 8.3 Nonisothermal Continuous-Flow Reactors 467

5. Energy balance. Equation (8-56):

For the acetone reaction system,

PFR with heat 6. Parameter evaluation:
exchange a. Mole balance. On a per tube basis, uo = 0.002 m31s. The concentration of
acetone is 18.8 mol/ m3 , so the entering molar flow rate is

mol
FAo = CAouo = (188 $1 (2 X
S
The value of k at 1035 X is 3.58 s-l ; consequently, we have

b. Energy balance. From the adiabatic case above we already have AC, , CpA, cxA ,
PA , 'yA , Aa, AP, and Ay . The heat-transfer area per unit volume of pipe is

U = 110 J/m2.s.K
Combining the overall heat-transfer coefficient with the area yields
Vu = 16,500 J/m3-s.K
We now use Equations (E8-7.1) through (E8-7.6), and Equations (E&-7.10) and
(E8-7.11) along with the POLYMATH program (Table E8-7.3), to determine the con-
version and temperature profiles shown in Figure E8-7.3.
The corresponding variables in the POLYMATH program are
tl = T, dh = AHRx(T), dcp = AC,, cpa = C,,, ua = Ua

- TABLE E8-7.3. POLYMATH PROGRAM FOR PFR WITH HEAT EXCHANGE -
Equations: Initial Values:
-___ _.
d(t)Jd(v)=(uaX(ta-t)+raXdh)/(faO%(cpa+xXdcp)) 1035
d<x>jd<v>=-ra/faO 0
f aO= .O376
ua=lG500
ta=1150
cpa=26.6+. l.833t-t-. 000@459XtXt
dcp=~.8-.0115~t-.OOOOO38l%t~t
caO=i8.8
t0=1035
term=-.0000@127%(tX~3-298%X3)
dh-80770+6.8X (it-298) -. 00575% ( t#%2-298%%2) +term
ra=-c:aOY3.58X~xp~34222%~l~tO-l/t~~~~l-x~X~tO/t~~~l+x~
vo = 0, Uf = 0.001