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:

Solved Example for Thermodynamics of Energy Conversion

For the following examples, Appendix B for steam values that I have referred to in these questions was obtained from:

M.D Koretsky, Engineering and Chemical Thermodynamics, Wiley, 2004.

Example 

Consider a power plant operating on a Rankine cycle using steam as the working fluid. The boiler pressure is 2.5MPa and the steam leaving the boiler is superheated to a temperature 126°C above its saturation temperature. The condenser temperature is 50°C and it discharges saturated liquid. The efficiency of the turbine is 0.90 and of the pump 0.80 as compared to reversible and adiabatic machines operating at the same pressure ranges.

a) Sketch the cycle on a T-S diagram.

b) Calculate the thermal efficiency of the cycle.

c) Compare the thermal efficiency of this Rankine cycle with the thermal efficiency of a Carnot cycle receiving heat at the saturation temperature of steam at 2.5MPa and discarding it at 50°C.

Solution

a) 

b) In order to calculate the thermal efficiency, we need to calculate enthalpy values at states 1, 2, 3 and 4.

c)

Examples of Unsteady-State Applications

Example 1

An insulated rigid tank of volume 0.3m³ is connected to a large pipeline carrying air at 1400 kPa and 300°C. The valve between the pipeline and the tank is opened and the tank fills with air until the pressure is 1400 kPa and then the valve is closed. Determine the final temperature of the air in the tank if:

a) The tank is initially empty,
b) The tank initially contains air at 350 kPa and 139°C.

Solution

Assume the system is the contents of the tank. 

Example 2

A rectangular steel tank having an internal volume of 1m³ contains air at 2.5MPa and 20°C. A relief valve is opened slightly allowing air to escape to the atmosphere. The valve is closed when the pressure in the tank reaches 350 kPa:

a) Calculate the amount of heat that must be added so as to keep the tank contents at 20°C throughout the process.

b) Calculate the final temperature if the process takes place adiabatically.

Solution

Note:

The result of equation 14 informs us that the gas that remains in the tank undergoes a reversible adiabatic expansion. Hence, the problem can be solved by choosing the contents of the tank in the final state as the system. The same amount of gas occupies less volume at the initial state as shown:

This is a closed system and since the gas on one side of the imaginary boundary has the same temperature as the gas on the other side we can assume the system is adiabatic as no heat is exchanged across the boundary. 

Furthermore, with the exception of the region around the valve - which is outside our chosen system - the gas in the cylinder is undergoing a uniform expansion thus there is no pressure, velocity or temperature gradients within the cylinder. Thus it can be assumed that the changes occurring in the system are reversible. 

Solved Example for First Law of Thermodynamics for a Closed System #2

Question 1 was solved in a previous blog entry (Click here).

Note: The Appendix B for steam values that I have referred to in these questions was obtained from: 

M.D Koretsky, Engineering and Chemical Thermodynamics, Wiley, 2004.

However, any steam tables from other reputable publishers can also work.

Question 2

An insulated rigid tank contains 0.2 kg of wet steam at 65°C. The steam is stirred with a paddle wheel until the pressure reaches 500 kPa. If the volume of the tank is 0.15 m3, determine:

a) The final temperature in the tank,

b) The work done by the paddle wheel.

Answer: 

Question 3

A rigid tank containing saturated steam at 600 kPa is cooled until the quality reaches to 50%. If the volume of the tank is 0.3 m³, determine:

a) The final temperature and pressure in the tank,

b) The amount of heat that must be removed.

Answer:

Solved Example for First Law of Thermodynamics for a Closed System #1

Question 1 

A horizontal, perfectly insulated piston-cylinder device contains 1.6mol of helium at 800 kPa. Its volume is 0.005m³ and the ambient pressure is 100 kPa.

a) Calculate the work done when the gas is expanded reversibly until the internal pressure is equal to the ambient pressure. Also sketch the process on a P - V diagram.

b) Calculate the work done when the gas is expanded very suddenly until the internal pressure is equal to the ambient pressure.

Answer:

Homogeneous Liquid Phase Second Order Irreversible Reaction