Ideal Reactors - Solved Examples

 Example 1

The following reaction is irreversible and first order:

The reaction is carried out in a PFR with 80 tubes. Each tube has a diameter of 5cm and a height of 1m. The feed consists of reactant A and 30% inerts flowing at 200kg/h with a pressure of 50bar and temperature of 600°C (873K). The output conversion is 90%. What is the average residence time? (The molecular weight of A is 60g/mol & R = 0.0821 L • atm/ mol • K)

Solution

The space time for a first order reaction in a PFR is given by the following equation:

The mass flow in each tube is: (200/80) = 1.25kg/h

In molar basis:

Thus, the initial concentration of A is given by:

We can now calculate the initial volumetric flow:

The volume of each tube can be calculated as follows:

The residence time can be calculated as follows:

The expansion parameter, ε_A can be determined as follows:

The rate constant, k, can now be solved using equation 1:

Thus, the average residence time can now be solved using:

Example 2

The following second order irreversible reaction is carried out in gas phase in a PFR reactor:

The PFR feed consists of 50% by weight reactant A (MW = 40) and the rest is inert (MW_inert = 20). The reaction occurs at a constant temperature of 70C and pressure of 5.25atm. The rate constant is 400m³/(kmol.ks).

For a production rate of R set at 30kmol/h with a 40% conversion, what should be the volume of the reactor?

Solution

For a second-order reaction, the rate can be expressed as:

Substituting the PFR design equation into the above equation and integrating we obtain:

With a mass basis of 1.0g, we can calculate ε_A. The molar values for each feed components are 0.5/40 = 0.0125 mol of A and 0.5/20 = 0.025 mol of inert. Thus, the volume balance is as shown below:

The expansion parameter, ε_A is = (0.05 - 0.0373)/0.0373 = 0.340. The initial concentration of A is:

Equation 1 can now be solved: 

Given that the output molar flow F_R = 30kmol/h, the volumetric flow rate, v_o is:

Thus, the reactor volume is:

Reactor Sizing Examples

 Reactor sizing was discussed in a previous blog entry (Click here to view).

Example 1

Consider a liquid phase reaction occurring in a PFR with the following data and Equation

X	0	0.4	0.8
-rA (mol/dm3s)	0.01	0.008	0.002
1/(-rA) (dm3s/mol)	100	125	500

If the molar feed to the PFR is 2 mol/s, what is the volume of the PFR needed to achieve 80% conversion.

Answer

For a PFR, 

Using the data provided in the table, we can plot a Levenspiel Plot:

Using Simpson's three point rule:

Thus, to achieve an 80% conversion, a PFR of  293.3 dm³ is needed

Example 2

A second order irreversible reaction is carried out in the gas phase inside a PFR. 

A reactant of molecular weight 40 and 50% by weight, and the rest with an inert of molecular weight of 20 are fed to the reactor. The reaction is carried out at constant temperature of 70°C and constant pressure of 5.25 atm. The rate constant is 400m³/(kmol.ks).

Calculate the volume of the PFR needed to achieve a 40% conversion of A to produce 30kmol/hr of product R.

Answer

For a second order reaction:

When substituting into the PFR equation and integrating the equation we get:

            (1) 

To calculate ε_A, let us consider a 1.0g sample of reagent mixture which has 0.5/40 = 0.0125 mol of A and 0.5/20 = 0.025 mol of inert. The following volume balance is then obtained. 

The expansion parameter, ε_A, = (0.05 - 0.0373)/0.0373 = 0.0340

The initial concentration of A, C_A0, is given by:

Substituting into equation 1,

Given the product flow rate (F_R) is 30 kmol/hr, we can obtain the volumetric flow rate: 

The reactor volume can now be found using the following:

Adiabatic Tubular Reactors

 

Steady-State PFR with Heat Exchange

 

CSTR - Non-Adiabatic Case

The Effect of Reaction Heat on a Non-Adiabatic CSTR

Non-Isothermal Reactor Design