Page 111 - The engineering of chemical reactions

Conversion in a Constant-Density PFTR 95
This gives
ln-=-kZ
CA(z)
CA0 LI
Solving for CA, we obtain

at z and

CA(L) = CA& -kLh = cAoe-k’

at2 = L.

:::c Example 3-3 The reaction A + B, Y = kCA occurs in PFTR with 90% conversion. If
T3 k = 0.5 min-l, C~~ = 2 moles/liter, and v = 4 liters/min, what residence time and reactor
;;I volume will be required?
From the previous equation we have
v 1 CA(L)
t=--=----In-=
V k CAo
= 2 In 10 = 4.61 min
so that
V = vt = 4 x 4.61 = 18.4 liters
Note that this answer is totally different than for this reaction and conversion in a
ns CSTR and that the residence time and reactor volume required are considerably
g smaller in a PFTR than in a CSTR.
Il”,n
;s d
;‘iB How long a 2 cm diameter tube would be required for this conversion and what would be
i~ji: the fluid velocity?
1^1
The length is the volume divided by the cross-sectional area or
4 v 4 x 18.4 x lo3
L=----= = 5860 cm = 58.6 m
J-CD2 rrx2x2
and
L 5860
- = 1270 cm/min = 21 crnkec
’ = t = 4.61
Thus, while the PFTR reactor volume is much smaller than the CSTR for this
conversion, the PFTR tube length may become impractical, particularly when
pumping costs are considered.

Ts’ / Example 3-4 The reaction A + B, r = kc; occurs in PFTR with 90% conversion. If
‘$, k = 0.5 liter mole-l min-1, C A0 = 2 moles/liter, and v = 4 liter/min, what residence time
“:-’ and reactor volume will be required?