COMSATS University Islamabad, Lahore Campus
Assignment 1 – Chemical Reaction Engineering SP-2022
Course Title:
Chemical Reaction Engineering
Course Code:
Course Instructor
Dr. Asad Ullah Khan/ Dr. Arif Hussain
Semester:
6th
Programme
Name:
A
Batch:
Submission Date: 16-03-2022
Student’s Name:
Shafaq Imtiaz
FA19-CHE
Section:
A
Total Marks
Reg. No.
CHE-331
Credit
Hours:
BS Chemical Engineering
Assignment Date:
4(3,1)
03-03-2022
10
FA19-CHE-019(A)
Important Instructions / Guidelines: Copying other student’s assignments are strictly prohibited. In addition, offenses will be
reported to the HOD office, which may result in further disciplinary action, including suspension or expulsion from the program.
Penalties will be given without discussion or warning. Note that you are responsible for not leaving copies of your assignments
lying around and for protecting your files accordingly.
Question 1: - (CLO1, C2, PLO1)
(4)
Referring to the text material and the reference book on Chemical Reaction Engineering, discuss in detail on the
following industrial reactors (with photos). Your discussion must include the characteristics, kind of phases,
their specific use, advantages, and disadvantages.
1.
2.
3.
4.
Semi-batch
CSTRs in series
Fluidized bed reactor
Trickle bed reactor
Question 2: - (CLO1, C2, PLO1)
(4)
Now that you have developed a little knowledge of application of Chemical Reaction Engineering, find an article
on the web or in a journal that deals with the application of CRE, or that covers a topic that interests you.
Give a brief synopsis of the article and describe anything new you learned from the article and/or how it relates to
what you have learned in class. (Maybe you could keep a journal, too, because you never know when an
interesting concept will come in handy later in life.)
Question 3: - (CLO2, C3, PLO1)
(2)
The reaction A B is to be carried out isothermally in a continuous flow reactor. Apply the fundamental design
equations to find the volume of CSTR and PFR necessary to consume 99% of A (i.e., CA = 0.01CAo). When the
entering molar flow rate is 5 mol/hr and υo = 10dm3/h, CA0 = 0.5mol/dm3, assuming the reaction rate –rA as
a) –rA = k
b) –rA = kCA
c) –rA = kCA 2
k = 0.05mol/h.dm3 * (n+x/n+y)
k = 0.0001s-1 * (n+x/n+y)
k = 3dm3/mol.h * (n+x/n+y)
Repeat (a), (b) and (c) to calculate the time necessary to consume 99.9% of species A in a 1000dm3 constant volume
batch reactor. (*where n is numerical part of your roll number and values of x and y are according to the following:
x = 5 and y = 3 for n = 1 to 20; x = 3 and y = 4 for n = 21 to 40 and x = 6, and y = 5 for n > 40).
Page 1 of 9
Solution 01:
1. Semi-batch:
A semi-batch reactor is a variation of a batch reactor in which one
reactant may be added intermittently or continuously to another
reactant contained in a vessel as reaction proceeds.
Characteristics:


Run on steady-state or unsteady-state with respect to the flowing phase
semi-batch reactor gives benefit of anything by changing the contacting pattern.
Phases:
 The reaction may be single-phase or multi-phase.
 For a multiphase reaction in which one phase flows continuously through a vessel containing a
batch of another phase.
Specific Use:

Semi-batch reactors are required in case of chemical reactions with a high heat effect, so one of
the reagents is slowly fed to the other component(s), which is already in the reactor.
Advantages:


Good temperature control.
Unwanted side-reactions minimized.
Disadvantages:
 High labor cost per unit product.
 Large scale production difficult to achieve.
 Reactor
operations
difficult
to
analyze.
2. CSTRs in series:
The CSTR is always operating on lowest concentration, the exit concentration. When say two
CSTRs are in series, the first operates at a higher
concentration, therefore the rate is greater, therefore the
conversion is greater. The second reactor in series builds
on the conversion in the first reactor.
Characteristics:
 Exit stream has the same composition as in the tank.
 The feed assumes a uniform composition throughout the
reactor.
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 Run at steady state with continuous flow of reactants and products.
Phases:
 Liquid phase
 Gas-liquid reactions
 Solid-liquid reactions
Specific use:
 When agitation is required and series configurations for different concentration streams.
Advantages:
 Easily adapts to two phase runs.
 Good control.
 Easy to clean.
Disadvantages:
 Lowest conversion per unit volume.
 By passing and channeling possible with poor agitation.
3. Fluidized bed Reactor:
In fluidized bed, the fluidization of the particles in the reactor leads
to the surface of the particles being continuously turned over.
Characteristics:
 The distinguishing feature of a fluidized bed reactor is that the bed
of solid particles or catalyst is supported by an up flow of gas.
 Thermal
uniformity,
offered
by fluidized systems.
Phases:
Fluidized beds are typically categorized as either being:
 2 phase system which are not aerated
 3 phase system which is aerated by sparging
Specific use:
 Used in the petrochemical for catalytic cracking.
Advantages:
 A high concentration of cells can be immobilized in the reactor due to the larger surface area for
cell immobilization is available
 Mass transfer rates are higher due to the larger surface area and the higher levels of mixing in the
reactor.
Disadvantages:
Fluidized bed reactors are more difficult to design. Design considerations include:
Page 3 of 9


Setting the flow rate to achieve fluidization
Ensuring that bubble size remains small during the fermentation.
4. Trickle-bed reactor:
A trickle-bed reactor (TBR) is a chemical reactor that uses the downward
movement of a liquid and the downward (co-current) or upward (countercurrent) movement of gas over a packed bed of (catalyst) particles.
Characteristics:
 Trickle bed reactors are solid-liquid-gas contacting devices.
 The fluid flows over catalyst particles and forms fine films, rivulets, or droplets.
Phases:
 Liquid phase
 Gas-liquid reactions
 Solid-liquid reactions
Specific use:
 In the treatment of wastewater trickle bed reactors are used where the required biomass resides
on the packed bed surface.
Advantages:
 Lower total energy consumption since solids are stagnant, not suspended in slurry
 Simple to operate under high temperatures and pressures.
Disadvantages:
 Difficult to scale up due to dependence on fluid dynamics of system
 Channeling may occur, leading to inefficiencies
Solution 02:
The Journal selected is “Application of Chemical Reaction Engineering principles to biological
systems” by Jong Hwan Sung, available in PMC.
In this Journal, the author explains that Living cells can be described by the principles of chemical
reaction engineering. In principle, biological systems can be considered as chemical reactors (a single
reactor or combination of reactors) at various length and time scales. The application of chemical
engineering principles to understand and describe cellular systems is well known in the bioprocess
engineering literature. Single cells can be described mathematically using the principles of chemical
reaction engineering, and their changes in physiology can be linked to changes in external parameters
such as concentrations of nutrients, chemical signaling factors, and physical parameters such as pH,
temperature, and shear stress.
Concept that a living cell is basically a chemical reactor:
This concept that a living cell is basically a chemical reactor can be extended further to biological systems
of a wide range of length scale. Considering the fact that the human body has hierarchical structures,
Page 4 of 9
with appropriate simplification and segmentation, the part or the whole of the human body can be
considered as a ‘living reactor’, with inputs and outputs, and reactions occurring inside. The ‘living
reactor’ can refer to a part or the whole of an organism, ranging from a single molecule (DNA or protein)
to cells, tissues, organs, and the whole body.
some illustrative examples of biological systems that can be modeled using principles of chemical
reaction engineering are
PBPK modeling:
 Physiologically based pharmacokinetic models are basically a mass balance on a substance in the
body, treating the body as a ‘reactor’.
Enzyme reactions:
 The knowledge about the important parameters related to the reaction kinetics of enzymes and
transport of molecules within the reactor helps one to optimize reactor design and operating
conditions.
Solution 03:
Given data:
A

B
Isothermal reaction
99% of A consumption, conversion = 0.99
CA = 0.01 CA0
FA0 = 5mol/hr
υo = 10dm3/hr
CA0 = 0.5 mol/dm3
Continuous flow reactor  FA
a) –rA = k
k = 0.05mol/h.dm3 * (n+x/n+y)
Calculation of K using n = 19(roll no) & x = 5 & y=3
K = 0.0545 mol/h.dm3
- CSTR Volume:
𝑽=
𝑭𝑨𝟎−𝑭𝑨
−𝒓𝑨
Rewriting equation in terms of concentration:
𝑽=
𝑪𝑨𝟎 𝑽𝟎 − 𝑪𝑨 𝑽𝟎
−𝒓𝑨
Substituting values into equation:
Given that -rA = k, therefore: 
-rA = 1.0909 mol/h.dm3
Page 5 of 9
𝑽=
𝒎𝒐𝒍
𝒅𝒎𝟑
𝒎𝒐𝒍
𝒅𝒎𝟑
)(𝟏𝟎
) −(𝟎.𝟎𝟎𝟓
)(𝟏𝟎
)
𝒅𝒎𝟑
𝒉𝒓
𝒅𝒎𝟑
𝒉𝒓
𝒎𝒐𝒍
0.0545
𝒉.𝒅𝒎𝟑
(𝟎.𝟓
V = 90.825 dm3
- PFR Volume:
𝒅𝑭𝑨
𝒅𝑽
= 𝒓𝑨
𝒅𝑪𝒂 𝑽𝟎
=-K
𝒅𝑽
Rearranging equation:
𝑪𝒂𝟎
𝒗𝟎
∫ 𝒅𝑽 =
∫ 𝒅𝑪𝒂
𝒌
𝑪𝒂
𝑽 =
𝒗𝟎
𝒌
(CA0 -Ca)
Substituting values into equation:
𝑽 = (𝟏𝟎
𝒅𝒎𝟑
𝒉𝒓
𝒎𝒐𝒍
𝒎𝒐𝒍
𝒎𝒐𝒍
) /(0.0545 𝒉.𝒅𝒎𝟑) (𝟎. 𝟓 𝒅𝒎𝟑 − 𝟎. 𝟎𝟎𝟓 𝒅𝒎𝟑)
V = 90.825 dm3
b) –rA = kCA
k = 0.0001s-1 * (n+x/n+y)
- Calculation of K using n = 19 & x = 5 & y=3
K = 0.00010909 s-1
- CSTR Volume:
𝑽=
𝑭𝑨𝟎−𝑭𝑨
−𝒓𝑨
Rewriting equation in terms of concentration:
𝑽=
𝑪𝑨𝟎 𝑽𝟎 − 𝑪𝑨 𝑽𝟎
Substituting values into equation:
Given that -rA = kCa, therefore: 
mol/dm3
𝒌𝑪𝒂
-rA = 0.00010909 s-1 * 3600 s/hour * 0.005
Page 6 of 9
𝑽=
𝒎𝒐𝒍
𝒅𝒎𝟑
𝒎𝒐𝒍
𝒅𝒎𝟑
)(𝟏𝟎
) −(𝟎.𝟎𝟎𝟓
)(𝟏𝟎
)
𝒅𝒎𝟑
𝒉𝒓
𝒅𝒎𝟑
𝒉𝒓
𝟏
𝟑𝟔𝟎𝟎𝟎𝒔
𝒎𝒐𝒍
0.00010909 ( ) ∗
∗ 𝟎.𝟎𝟎𝟓
𝒔
𝟏𝒉𝒓
𝒅𝒎𝟑
(𝟎.𝟓
V = 2520.85 dm3
- PFR Volume:
𝒅𝑭𝑨
= 𝒓𝑨
𝒅𝑽
𝒅𝑪𝒂 𝑽𝟎
= - KCa
𝒅𝑽
Rearranging equation:
𝑪𝒂𝟎
𝑽𝟎
𝒅𝑪𝒂
∫ 𝒅𝑽 =
∫
𝒌
𝑪𝒂
𝑪𝒂
𝑽=
𝐕=
𝒗𝟎
𝒌
𝐯𝟎
𝐤
(lnCA0 - lnCa)
CA0
(ln( CA ))
Substituting values:
𝐕=
𝐝𝐦𝟑
)
𝐡𝐫
𝟏
𝟑𝟔𝟎𝟎𝟎𝒔
0.00010909 ( ) ∗
𝒔
𝟏𝒉𝒓
𝟏𝟎(
0.5
(ln(0.005))
V = 117.262 dm3
c) –rA = kCA 2
k = 3dm3/mol.h * (n+x/n+y)
- Calculation of K using n = 19 & x = 5 & y=3
K = 3.2727 dm3/mol.h
- CSTR Volume:
𝑽=
𝑭𝑨𝟎−𝑭𝑨
−𝒓𝑨
Rewriting equation in terms of concentration:
𝑽=
𝑪𝑨𝟎 𝑽𝟎 − 𝑪𝑨 𝑽𝟎
𝒌𝑪𝒂𝟐
Page 7 of 9
Substituting values into equation:
Given that -rA = kCa2, therefore: 
-rA = 3.2727 dm3/mol.h * (0.005 mol/dm3)2
𝒎𝒐𝒍
𝒅𝒎𝟑
𝒎𝒐𝒍
𝒅𝒎𝟑
)(𝟏𝟎
) −(𝟎.𝟎𝟎𝟓
)(𝟏𝟎
)
𝒅𝒎𝟑
𝒉𝒓
𝒅𝒎𝟑
𝒉𝒓
𝒅𝒎𝟑
𝒎𝒐𝒍 ^𝟐
3.2727
∗ 𝟎.𝟎𝟎𝟎𝟎𝟐𝟓
𝒎𝒐𝒍.𝒉
𝒅𝒎𝟔
(𝟎.𝟓
𝑽=
V = 60500.50 dm3
- PFR Volume:
𝒅𝑭𝑨
𝒅𝑽
= 𝒓𝑨
𝒅𝑪𝒂 𝑽𝟎
= - KCa2
𝒅𝑽
Rearranging equation:
𝑪𝒂𝟎
𝑽𝟎
𝒅𝑪𝒂
∫ 𝒅𝑽 =
∫
𝒌
𝑪𝒂𝟐
𝑪𝒂
Integrating Equation:
𝑽=
𝑽=
𝟏𝟎 𝒅𝒎𝟑/𝒉𝒓
𝒗𝟎
𝒌
1
( 𝐶𝐴-
1
𝐶𝐴0
)
1
1
( 0.005𝑚𝑜𝑙/𝑑𝑚3 - 0.5𝑚𝑜𝑙/𝑑𝑚3)
𝒅𝒎𝟑
𝟑.𝟐𝟕𝟐𝟕
𝒎𝒐𝒍.𝒉𝒓
V = 605 dm3
Time required for 99.9% conversion of A using a batch reactor:
Where:
V = 1000 dm3
CA = 0.01 CAO
a) –rA = k.
K = 0.0545 mol/h.dm3
𝑵𝑨𝑶 𝒅𝑵
t=∫𝑵𝑨
−𝒓𝑨𝑽
Page 8 of 9
Constant Volume V=V0 ,
CA= NA/V
𝑪𝑨𝑶 𝒅𝑪𝑨
t=∫𝑪𝑨
−𝒓𝑨
𝑪𝑨𝑶 𝒅𝑪𝑨
t=∫𝑪𝑨
𝑲
1
= 𝐾 (𝐶𝐴0 − 𝐶𝐴)
Zero order:
1
1
t= 𝐾 (𝐶𝐴0 − 0.01𝐶𝐴𝑂)= 0.545 (𝐶𝐴0 − 0.01𝐶𝐴𝑂)
t=9.08hr
b) –rA = kCA
K = 0.00010909 s-1
First Order:
𝟏
CA0
𝐭 = 𝐤(ln( CA ))
𝟏
CA0
𝐭 = 0.00010909(ln(0.01CAo))=42214.41s
t= 11.7hr
c) –rA = kCA 2
K = 3.2727 dm3/mol.hr
Second order:
𝟏
1
𝒕 = 𝒌( 𝐶𝐴𝟏
1
𝐶𝐴0
1
)
𝒕 = 𝟑.𝟐𝟕𝟐𝟕( 0.01(0.5)-
1
0.5
)
t= 60.5hr
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