Daryl Jan Padilla
CSTR-Continuous Stirred tank reactor
General mole balance equation
In – Out + Generation = Accumulation
𝑑𝐹𝐴𝑂 𝑑𝐹𝐴 𝑑𝑁𝐴
− + ∫ 𝑟𝐴 𝑑𝑉 =
𝑑𝑉 𝑑𝑉 𝑑𝑡
At steady state:
𝑑𝑁𝐴
=0
𝑑𝑡
Well mixed therefore 𝑟𝐴 would be the same throughout dV would be from 0 to V value:
𝑉
∫ 𝑟𝐴 𝑑𝑉 = 𝑟𝐴 𝑉
0
𝐹𝐴𝑂 − 𝐹𝐴 + 𝑟𝐴 𝑉 = 0
(𝐹𝐴𝑂 − 𝐹𝐴 )
𝑉=
−𝑟𝐴
Conversion for single reaction:
(𝐹𝐴𝑂 − 𝐹𝐴 )
𝑥=
𝐹𝐴𝑂
…………………………………………………………………………………………………………………………………………………………………
(𝐹𝐴𝑂 − 𝐹𝐴 )
𝑉=
−𝑟𝐴
𝐹𝐴 = 𝐹𝐴𝑂 − 𝐹𝐴𝑂 𝑥
(𝐹𝐴𝑂 𝑥)
𝑉=
−𝑟𝐴
For reactor in series:
Tank 1: conversion from 0 to x1
(𝐹𝐴𝑂 𝑥1 )
𝑉1 =
−𝑟𝐴
Tank 2: conversion from x1 to x2
(𝐹𝐴𝑂 (𝑥2 − 𝑥1 ))
𝑉2 =
−𝑟𝐴
PFR-Plug flow reactors
Daryl Jan Padilla
General mole balance equation
In – Out + Generation = Accumulation
𝑑𝐹𝐴𝑂 𝑑𝐹𝐴 𝑑𝑁𝐴
− + ∫ 𝑟𝐴 𝑑𝑉 =
𝑑𝑉 𝑑𝑉 𝑑𝑡
At steady state:
𝑑𝑁𝐴
=0
𝑑𝑡
Differentiating:
𝑑𝐹𝐴
0− + 𝑟𝐴 = 0
𝑑𝑉
𝑑𝐹𝐴
= 𝑟𝐴
𝑑𝑉
Conversion for single reaction:
(𝐹𝐴𝑂 − 𝐹𝐴 )
𝑥=
𝐹𝐴𝑂
…………………………………………………………………………………………………………………………………………………………………
𝑑𝐹𝐴
= 𝑟𝐴
𝑑𝑉
𝐹𝐴 = 𝐹𝐴𝑂 − 𝐹𝐴𝑂 𝑥
Differential:
𝑑𝐹𝐴 = 0 − 𝐹𝐴𝑂 𝑑𝑥 (substituting values of 𝑑𝐹𝐴 )
𝑑𝑥
𝐹𝐴𝑂 = 𝑟𝐴
𝑑𝑉
Integration:
𝐹𝐴𝑂 𝑑𝑥
= 𝑑𝑉
−𝑟𝐴
At t=0 and x=0
𝑥
𝑑𝑥
𝑉 = 𝐹𝐴𝑂 ∫
0 −𝑟𝐴
Use Simpson’s rule calculating any Area below the curve
Daryl Jan Padilla
Example 2.2
The reaction described by A B is to be carried out in a flow reactor. Species A enters the reactor at a
molar flowrate of 0.4 mole/s, using the data provided:
1. Calculate the volume necessary to achieve 80% conversion in CSTR.
2. Shade the area that would correspond to the necessary volume.
3. Calculate the volume necessary to achieve 80% conversion in PFR.
4. Shade the area that would correspond to the necessary volume.
5. For two CSTR in series, 40% conversion is achieved in the first reactor. What is the volume of
each of the two reactor necessary to achieve 80% overall conversion of the entering species A?
6. For two PFR in series, 40% conversion is achieved in the first reactor. What is the volume of each
of the two reactor necessary to achieve 80% overall conversion of the entering species A?
Given:
𝑚𝑜𝑙𝑒
𝐹𝐴𝑂 = 0.4
𝑠
X -rA 1/-rA
3 3
(mol/m × s) (m × s / mol)
0.0 0.45 2.22
0.1 0.37 2.70
0.2 0.30 3.33
0.4 0.195 5.13
0.6 0.113 8.85
0.7 0.079 12.66
0.8 0.05 20.00
Daryl Jan Padilla
Graphing:
1-4) 1 would likely to shade everything under the curve but I don’t know how to operate in Microsoft
word
Plot 80% conversion
0.5
0.45
0.4
0.35
0.3
-rA
0.25
0.2
0.15
0.1
0.05
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
x
Plot 80% conversion
25
20
15
1/-rA
10
0
0 0.2 0.4 0.6 0.8 1
x
Daryl Jan Padilla
Assumed at steady state:
𝑑𝑁𝐴
=0
𝑑𝑡
Volume at 80% conversion in a CSTR
(𝐹𝐴𝑂 𝑥)
𝑉=
−𝑟𝐴
(0.4 × 0.80)
𝑉= = 6.4 𝑚3
0.05
Daryl Jan Padilla
Volume at 80% conversion in a PFR
Simpson’s rule
𝑥1
ℎ
∫ 𝑓(𝑥)𝑑𝑥 = (𝑓 + 4𝑓1 + 2𝑓2 + 4𝑓3 + 𝑓4 )
𝑥0 3 0
𝑥4 − 𝑥0
ℎ=
𝑛
𝑥
𝑑𝑥
𝑉 = 𝐹𝐴𝑂 ∫
0 −𝑟𝐴
∆𝑋 1 4 2 4 1
𝑉 = 𝐹𝐴𝑂 [ + + + + ]
3 −𝑟𝐴 −𝑟𝐴1 −𝑟𝐴2 −𝑟𝐴3 −𝑟𝐴4
𝑥2 − 𝑥1
∆𝑋 =
𝑛 = 𝑠𝑢𝑏𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
0.8 − 0 1 4 2 4 1
𝑉 = 0.4 [ + + + + ]
3(4) 0.45 0.3 0.195 0.113 0.05
𝑉 = 2.165605224 𝑚3
𝑉 = 2.1656 𝑚3
Daryl Jan Padilla
5-6)
For series reactor:
𝐹𝐴𝑂
𝑉1 , 𝑟𝐴1 𝐹𝐴1 , 𝑥1
𝐹𝐴2 , 𝑥2
𝑉2 , 𝑟𝐴2
For two CSTR, first tank 40% and the second tank is 80% conversion
(0.4(0.4))
𝑉1 = = 0.8205128205 𝑚3
0.195
𝑉1 = 0.8205 𝑚3
(0.4(0.8 − 0.4))
𝑉2 = = 3.2 𝑚3
0.05
Daryl Jan Padilla
𝑉1 , 𝑟𝐴1 𝑉2 , 𝑟𝐴2
𝐹𝐴𝑂 𝐹𝐴1 , 𝑥1 𝐹𝐴2 , 𝑥2
For two PFR, first tank 40% and the second tank is 80% conversion
𝑥
𝑑𝑥
𝑉1 = 𝐹𝐴𝑂 ∫
0 −𝑟𝐴
Simpson’s rule (refer to previous equation):
0.4 − 0 1 4 1
𝑉1 = 0.4 [ + ++ ]
3(2) 0.45 0.3 0.195
𝑉1 = 0.551566516 𝑚3
𝑉1 = 0.5516 𝑚3
0.8 − 0.4 1 4 1
𝑉2 = 0.4 [ + ++ ]
3(2) 0.195 0.113 0.05
𝑉2 = 1.614038272 𝑚3
𝑉2 = 1.6140 𝑚3
