ISOTHERMAL REACTOR DESIGN
Goal:
Determine volume necessary to achieve a specified conversion, X, or
the conversion that can be achieved for a specified reactor volume, V.
CHEMICAL REACTION ENGINEERING ALGORITHM
Step 1: Mole balance/Design equation
Step 2: Rate law
Identify rate for reaction you wish to carry out.
Example:
The reaction,
has the rate law
Step 3: Stoichiometry
1) Specify whether reaction is carried out in liquid phase or in gas phase
2) Express the concentration as a function of conversion
Step 4: Combine
Combine rate law and stoichiometry to express rate law as a function of
conversion
and subsequently combine with mole balance to express differential
/algebraic equation as a function of conversion
Step 5: Evaluate Equation
Determine V required to achieve X, or X that can be achieved for given V.
SUMMARY
EXAMPLE: The elementary liquid phase reaction
The liquid phase reaction
2𝐴 ⟶ 𝐵
is carried out isothermally in a CSTR. Pure A enters at a volumetric flow
rate of 25 dm3/s and at a concentration of 0.2 mol/dm3.
What CSTR volume is necessary to achieve a 90% conversion when k =
10 dm3/(mol.s)?
Step 1: Mole balance/Design equation
CSTR design equation in terms of conversion:
𝑉=
𝐹𝐴0 𝑋
−𝑟𝐴
Step 2: Rate law
−𝑟𝐴 = 𝑘𝐶𝐴 2
Step 3: Stoichiometry
Phase?...
𝐹𝐴 = 𝐹𝐴0 (1 − 𝑋)
From 𝐶𝐴 =
𝐹𝐴
𝑣
, it follows that
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣
and 𝑣 = 𝑣0 … why?
So,
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣0
→ 𝐶𝐴 = 𝐶𝐴0 (1 − 𝑋)
Therefore, obtained concentration in terms of X.
Step 4: Combine
−𝑟𝐴 = 𝑘𝐶𝐴 2 and 𝐶𝐴 = 𝐶𝐴0 (1 − 𝑋)
…
−𝑟𝐴 = 𝑘 [𝐶𝐴0 (1 − 𝑋)]2
−𝑟𝐴 = 𝑘𝐶𝐴0 2 (1 − 𝑋)2
Therefore, obtained −𝑟𝐴 in terms of X.
Then sub into,
𝑉=
𝑉=
𝐹𝐴0 𝑋
−𝑟𝐴
𝐹𝐴0 𝑋
𝑘𝐶𝐴0 2 (1 − 𝑋)2
Therefore, obtained mole balance equation as function of X.
Step 5: Evaluate Equation
𝑉=
(25)(0.9)
10(0.2)2 (1 − 0.9)2
∴ 𝑉 = 1125 𝑑𝑚3
Determine the reactor space time, τ
𝜏=
𝑉
1125
=
= 45 𝑠
𝑣0
25
Homework Problem 1
EXAMPLE: The elementary gas phase reaction
The gas phase reaction
2𝐴 ⟶ 𝐵
is carried out isothermally in a CSTR. Pure A enters at a volumetric flow
rate of 2.5 dm3/s and a concentration of 0.2 mol/dm3.
What CSTR and PFR volume is necessary to achieve a 90% conversion
when k = 0.5 dm3/(mol.s)?
Additional information: 𝑇0 = 500 𝐾, 𝑃0 = 8.2 𝑎𝑡𝑚
FOR CSTR CASE
Step 1: Mole balance/Design equation
𝐹𝐴0 𝑋
𝑉=
−𝑟𝐴
Step 2: Rate law
−𝑟𝐴 = 𝑘𝐶𝐴 2
Step 3: Stoichiometry
Phase?...
𝐹𝐴 = 𝐹𝐴0 (1 − 𝑋)
From 𝐶𝐴 =
𝐹𝐴
𝑣
, it follows that
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣
But now 𝑣 ≠ 𝑣0 … why?
𝑃 𝑇0
𝑣 = 𝑣0 (1 + 𝜀𝑋) ( ) ( )
𝑃0 𝑇
where
𝜀 = 𝑦𝐴0 𝛿
𝛿=
𝑐 𝑑 𝑏 𝑎
+ − −
𝑎 𝑎 𝑎 𝑎
∴𝛿=
𝑐 𝑑 𝑏
+ − −1
𝑎 𝑎 𝑎
where 𝑎, 𝑏, 𝑐, 𝑑 are the stochiometric coefficients in the reaction for the
general reaction A + B → C + D.
BUT, in this example, A is the reactant, and B is the product so,
𝛿=
=
𝑏 𝑎
−
𝑎 𝑎
𝑏
−1
𝑎
Also, isothermal reactor so 𝑇0 = 𝑇, i.e., 𝑇0 ⁄𝑇 = 1
Pressure drop normally only considered for PBRs only unless otherwise
stated, so in this case
𝑃0 = 𝑃, i.e., 𝑃0 ⁄𝑃 = 1
Hence,
𝑣 = 𝑣0 (1 + 𝜀𝑋)
So,
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣0 (1 + 𝜀𝑋)
(1−𝑋)
→ 𝐶𝐴 = 𝐶𝐴0 (1+𝜀𝑋)
1
and 𝜀 = (1) ( − 1) = −0.5
2
(1−𝑋)
→ 𝐶𝐴 = 𝐶𝐴0 (1−0.5𝑋)
Therefore, obtained concentration in terms of X.
Step 4: Combine
(1−𝑋)
−𝑟𝐴 = 𝑘𝐶𝐴 2 and 𝐶𝐴 = 𝐶𝐴0 (1−0.5𝑋)
…
(1−𝑋) 2
−𝑟𝐴 = 𝑘 [𝐶𝐴0 (1−0.5𝑋)]
(1 − 𝑋)2
−𝑟𝐴 = 𝑘𝐶𝐴0
(1 − 0.5𝑋)2
2
Therefore, obtained −𝑟𝐴 in terms of X.
Then sub into,
𝑉=
𝐹𝐴0 𝑋
−𝑟𝐴
𝐶𝐴0 𝑣0 𝑋
𝑣0 𝑋 (1 − 0.5𝑋)2
𝑉=
2 = 𝑘𝐶
(1 − 𝑋)2
2 (1 − 𝑋)
𝐴0
𝑘𝐶𝐴0
(1 − 0.5𝑋)2
Therefore, obtained mole balance equation as function of X.
Step 5: Evaluate Equation
2.5
(0.9)(1 − 0.5 × 0.9)2
𝑉=
(0.5)(0.2)
(1 − 0.9)2
∴ 𝑉 = 680.6 𝑑𝑚3
Determine the reactor space time, τ
𝑉
680.6
𝜏=
=
= 272.3 𝑠
𝑣0
2.5
FOR PFR CASE
Step 1: Mole balance/Design equation
𝑋
𝑑𝑋
0 −𝑟𝐴
𝑑𝑉 = 𝐹𝐴0 ∫
Step 2: Rate law
−𝑟𝐴 = 𝑘𝐶𝐴 2
Step 3: Stoichiometry
Phase?...
𝐹𝐴 = 𝐹𝐴0 (1 − 𝑋)
From 𝐶𝐴 =
𝐹𝐴
𝑣
, it follows that
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣
and 𝑣 ≠ 𝑣0
𝑃 𝑇0
𝑣 = 𝑣0 (1 + 𝜀𝑋) ( ) ( )
𝑃0 𝑇
But isothermal reactor so 𝑇0 = 𝑇, i.e., 𝑇0 ⁄𝑇 = 1
and
𝑃0 = 𝑃, i.e., 𝑃0 ⁄𝑃 = 1
Hence,
𝑣 = 𝑣0 (1 + 𝜀𝑋)
So,
𝐹𝐴 𝐹𝐴0 (1 − 𝑋)
=
𝑣
𝑣0 (1 + 𝜀𝑋)
(1−𝑋)
→ 𝐶𝐴 = 𝐶𝐴0 (1+𝜀𝑋)
1
and 𝜀 = (1) ( − 1) = −0.5
2
(1−𝑋)
→ 𝐶𝐴 = 𝐶𝐴0 (1−0.5𝑋)
Therefore, obtained concentration in terms of X.
Step 4: Combine
(1−𝑋)
−𝑟𝐴 = 𝑘𝐶𝐴 2 and 𝐶𝐴 = 𝐶𝐴0 (1−0.5𝑋)
…
(1−𝑋)
−𝑟𝐴 = 𝑘 [𝐶𝐴0 (1−0.5𝑋)]
2
(1 − 𝑋)2
−𝑟𝐴 = 𝑘𝐶𝐴0
(1 − 0.5𝑋)2
2
Therefore, obtained −𝑟𝐴 in terms of X.
Then sub into,
𝑋
𝑑𝑋
0 −𝑟𝐴
𝑑𝑉 = 𝐹𝐴0 ∫
Substituting and integrating from 0 to V,
𝑋(
𝑣0
1 − 0.5𝑋)2
𝑉=
∫
𝑑𝑋
𝑘𝐶𝐴0 0 (1 − 𝑋)2
Therefore, obtained mole balance equation as function of X.
Step 5: Evaluate Equation
𝑋(
𝑣0
1 − 0.5𝑋)2
𝑉=
∫
𝑑𝑋
𝑘𝐶𝐴0 0 (1 − 𝑋)2
Appendix A in textbook (Fogler, 6th Ed.),
Therefore,
∫
𝑋 (1
0
(1 + (−0.5))2 𝑋
− 0.5𝑋)2
2
𝑑𝑋 = 2(−0.5)(1 + (−0.5)) ln(1 − (−0.5)) + (−0.5) 𝑋 +
(1 − 𝑋)2
1−𝑋
So,
2
𝑣0
(1 + (−0.5)) 𝑋
𝑉=
[2(−0.5)(1 + (−0.5)) ln(1 − (−0.5)) + (−0.5)2 𝑋 +
]
𝑘𝐶𝐴0
1−𝑋
∴ 𝑉 = 90.7 𝑑𝑚3
Determine the reactor space time, τ
𝜏=
𝑉
90.7
=
= 36.3 𝑠
𝑣0
2.5
Homework Problem 2
For the gas phase reaction occurring in a PFR,
(a) Determine the space time if the reactor is operated isothermally.
(b) If 𝑣0 = 10 𝑑𝑚3 /𝑠 what would the required volume of the reactor be?