Chemical Engineering Tutorials: Ideal Reactors

Wednesday, 18 June 2025

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: