SME
Chemical Reaction Engineering
Batch, CSTR and PFR design
Use this for “A first-order liquid reaction with k = 0.2 per min is fed at 10 L/min. What CSTR and PFR volumes reach 80% conversion?”
Chart
This is an example. Type your own values to start a new problem.
τCSTR (min)
=
X
÷
(
k (min⁻¹)
×
(1
−
X
))
τPFR (min)
=
−ln(1
−
X
)
÷
k (min⁻¹)
VCSTR (L)
=
v₀ (L/min)
×
τCSTR (min)
,
VPFR (L)
=
v₀ (L/min)
×
τPFR (min)
Formulas
τCSTR
=
X
k
× (1 −
X
)
20
=
0.8
0.2 × (1 − 0.8)
τPFR
=
−
ln(1 −
X
)
k
8.047
=
−
ln(1 − 0.8)
0.2
VCSTR
=
v₀
×
τCSTR
200 = 10 × 20
VPFR
=
v₀
×
τPFR
80.47 = 10 × 8.047
Assumptions
•
First order, −rA = kCA, isothermal, liquid phase (constant density).
•
Steady state; the CSTR is well mixed, so it runs at the low exit concentration.
•
The PFR has no mixing along its length; space time τ = V ÷ v₀.
Step-by-step
WE KNOW
k = 0.2 min⁻¹, X = 0.8, v₀ = 10 L/min
FIND
CSTR space time (τCSTR), PFR space time (τPFR), CSTR volume (VCSTR), PFR volume (VPFR)
Step 1 · Find CSTR space time (τCSTR)
Use τCSTR = X ÷ (k × (1 − X))
The
CSTR
balance
FA0X
=
(−rA)V
holds
at
the
exit
rate
kCA0(1
−
X
);
divide
by
v₀
for
τ.
τCSTR
=
0.8
0.2 × (1 − 0.8)
τCSTR
=
0.8
0.2 × 0.2
τCSTR
=
0.8
0.04
τCSTR = 20 min
Step 2 · Find PFR space time (τPFR)
Use τPFR = −ln(1 − X) ÷ k
Each
slice
of
the
PFR
works
at
its
own
rate,
so
τ
=
∫
dX
k
(1 −
X
)
from
0
to
X
,
which
is
−
ln(1 −
X
)
k
.
τPFR
=
−
ln(1 − 0.8)
0.2
τPFR
=
−
ln(0.2)
0.2
τPFR
=
−
−1.6094
0.2
τPFR
=
1.6094
0.2
τPFR = 8.047 min
Step 3 · Find CSTR volume (VCSTR)
Use VCSTR = v₀ × τCSTR
The volume the feed fills in one space time.
VCSTR = 10 × 20
VCSTR = 200 L
Step 4 · Find PFR volume (VPFR)
Use VPFR = v₀ × τPFR
The volume the feed fills in one space time.
VPFR = 10 × 8.047
VPFR = 80.47 L
Check
Put the numbers back in. Both sides should match.
20
=
0.8
0.2 × (1 − 0.8)
✓
8.047
=
−
ln(1 − 0.8)
0.2
✓
200 = 10 × 20 ✓
80.47 = 10 × 8.047 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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