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Question 4:
Solution:
Subject: Chemical Reaction Engineering
Topic: Non-isothermal reactors
Given,
The reaction A → B is a first order reaction with a rate constant k =0.1 min−1 at initial
temperature T i=300 ° C , and heat of reaction ∆ H =−50 kJ /mol. The reaction takes place in
an adiabatic batch reactor with an initial concentration of A as C A 0=2 mol /l and proceeds till
80% conversion of A. So, we can say that the conversion of A is X A =0.8 .
Step 1: Write down the first order rate equation and other relevant equations
Explanation:
The first order rate equation is given as:
−kt
C A=C A 0 e
Where C A is the concentration at time t.
Also, the rate constant term depends on temperature of the reaction mixture and since the
conditions are adiabatic and the reaction is an exothermic reaction, the temperature within the
reactor will change as the reaction proceeds.
Dependence of rate constant on temperature is given by the Arrhenius equation:
ln =
[
k 2 Ea 1
−
1
k1 R T 1 T 2 ]
Where k 2, k 1 are the rate constants at temperatures T 2, T 1. E a is the activation energy required
for the reaction and R is the gas constant.
Step 2: Calculate the rate constant at the final temperature (k 2) when the reaction has
reached 80% completion.
Explanation:
On observing the Arrhenius equation, we observe that we do have data for the final reaction
temperature T 2 and the activation energy of the reaction E a. So, it is not possible to proceed
further without knowledge of these parameters. Neither do we have the heat capacity data of
the reactants and products from which we can calculate the final temperature T 2.
Final Answer:
The calculation for reaction time requires additional data particularly the activation energy of
the reaction and the final temperature of the reaction mixture. So, there is insufficient data in
the problem.
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Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 4:
Solution:
Subject: Chemical Reaction Engineering
Topic: Non-isothermal reactors
Given,
The reaction A → B is a first order reaction with a rate constant k =0.1 min−1 at initial
temperature T i=300 ° C , and heat of reaction ∆ H =−50 kJ /mol. The reaction takes place in
an adiabatic batch reactor with an initial concentration of A as C A 0=2 mol /l and proceeds till
80% conversion of A. So, we can say that the conversion of A is X A =0.8 .
Step 1: Write down the first order rate equation and other relevant equations
Explanation:
The first order rate equation is given as:
−kt
C A=C A 0 e
Where C A is the concentration at time t.
Also, the rate constant term depends on temperature of the reaction mixture and since the
conditions are adiabatic and the reaction is an exothermic reaction, the temperature within the
reactor will change as the reaction proceeds.
Dependence of rate constant on temperature is given by the Arrhenius equation:
ln =
[
k 2 Ea 1
−
1
k1 R T 1 T 2 ]
Where k 2, k 1 are the rate constants at temperatures T 2, T 1. E a is the activation energy required
for the reaction and R is the gas constant.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question-1
Solution 1
Subject: Mass Transfer
Topic: Distillation
Given,
Feed mole percentage of A = 40%, B = 60%, Distillate composition of A = 80%. The
equilibrium relation for A is y=1.5 x . Total reflux conditions exist in the column.
Assumption: Assuming that a feed rate of F=100 mol /min is fed into the reactor for a total
time of 1 min after which the equilibrium is allowed to establish and feed is cut off.
Step 1: Preliminary Calculations
Explanation:
Convert the respective mole percentages to mole fractions by dividing the value by 100. So,
the feed mole fraction of A ( z Af ) is given by:
40
z Af = =0.4
100
Similarly, the mole fraction of B in feed will be z Bf =0.6, and the mole fraction of A in
distillate will be x D =0.8 .
Step 2: Calculate the molar mass of A in the feed
Explanation:
Molar flow rate of A in the feed stream ( F A) is:
F A=F × z Af
F A=100 × 0.4=40 mol /min
Step 3: Calculate molar flow of A in the distillate (D) and the bottoms(B)
Explanation:
Since the conditions within the column are that of Total Reflux, the molar flow rate of liquid
and vapour streams will be equal. So, the flow will be half of the feed flow rate at any given
time within the reactor for each of the phases:
100
D= =50 mol /min
2
100
B= =50 mol /min
2
Now, since we know x D =0.8, we can calculate the molar flow of A in the distillate stream:
D A =50 ×0.8=40 mol /min
We observe here that the flow rate of A in distillate stream ( D A ) is equal to the flow rate of A
in the feed stream ( F A) which is 40 mol/min . So, according to mass balance:
F a=D A + B A
On substituting the respective values:
40=40+B A
So, B A =40−40=0
This means that there is no A in the bottoms. This situation is both theoretically and
practically impossible and would require an infinite number of trays in the column. No
matter, how many trays are there in the distillation column, practically there will still be a
trace amount of A in the bottoms.
Final Answer:
The operation requires an infinite number of theoretical stages or trays in the column.
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Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 5 is not available for Chemical Engineering.
Question 2:
Solution:
Subject: Heat Transfer
Topic: Heat Exchanger
Step 1: Identify the information given in the problem
Explanation:
Flow rate of water is given as m=5 kg /min. The initial and final temperatures of water are
T i=25 °C and T f =75° C respectively. Specific heat capacity of water is given as
kJ
C p=4.18 . ° C .
kg
Also convert the mass flow rate of water from kg/min to kg/s as:
5
m=5 kg /min= kg /s
60
Step 2: Calculate the heat duty required for the operation
Explanation:
Heat duty for the operation Q is given by:
Q=m×C p ( T f −T i )
Substitute the information into the equation:
5
Q= ×4.18 × ( 75−25 ) kW
60
Q=17.4167 kW
Final Answer:
The heat duty required is 17.4167 kW.
