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)

Thermodynamics of Energy Conversion

1. Vapour Power Cycles

Power cycle is a process where devices are used to continuously produce power. A typical vapor power cycle is the steam power plant shown below:

In this system:

• the boiler receives heat from a high temperature source and converts liquid water into high pressure and high-temperature steam.
• This steam is then fed to a turbine that drives an electric generator. The steam expands through the turbine and exhausts at a low pressure. This expansion process occurs adiabatically and is nearly as reversible as possible.
• The exhaust from the turbine is sent to a condenser where cooling water is used. A pump is used to increase the pressure of the liquid condensate. A small fraction of the work obtained from the turbine is used to operate the pump.

The most efficient cycle that can operate between two constant temperature reservoirs, T_H and T_C is the Carnot engine and its thermal efficiency is given by:

                           

Let us consider a typical nuclear power plant that has a capacity of 750,000 kW. Steam is generated from a boiler at around 300°C and 7 MPa. The condenser operates at around 40°C and 7 kPa. The thermal efficiency of a Carnot engine operating between these two temperatures is:

This means that, in the best-case scenario the power plant can only convert 45% of heat received by the boiler into work while the other 55% is discarded to the surroundings. 

In practice, the efficiency of an actual power plant will be much less than that of a Carnot engine. If we assume that η = 0.3 then:

The heat of vaporization of water is approximately 2260 kJ/kg. Therefore, the steam circulation rate is:

The specific volume of steam at 300°C and 7 MPa is 0.02947 m³/kg and a reasonable velocity for high-pressure steam in a pipe is 25 m/s. Hence, the diameter of the pipe is:

The representation of a Carnot heat engine on a T-S diagram is:

Shaft Work in a Reversible Steady-State Process

In the analysis of a power cycle, it is necessary to calculate the work required for the pump to increase the pressure of the exit stream from the condenser to the boiler pressure, i.e., process 1 → 2. In a reversible steady-state process, the equation to calculate the shaft work is derived as follows.

The first law of thermodynamics for a closed system is given by:

Note that equations 3 and 6 consist of only properties and their differential changes.

These properties and their changes are state functions and are not dependent on the path or process involved. Therefore, both equations hold for all reversible and irreversible processes and for a change of state in either a steady-state flow system or a closed system.

Now consider the first law of thermodynamics for a steady-state flow system

Thus, the reversible shaft work in a steady-flow process with negligible changes in kinetic and potential energies is given by:

The power required can be obtained by multiplying equation 12 by the mass flow rate, ṁ = Q̇ρ. The result is:

2. Rankine Cycle

Named after William J.M. Rankine (1820-1872), the Rankine cycle is the ideal cycle for a simple power plant. As was shown in the T-S diagram of a Carnot cycle from the previous section, the fluid at the exit of a condenser i.e., state 1 is a mixture of liquid and vapour. Practically, it is much easier to pump a pure liquid rather than a two-phase mixture. Thus, for a Rankine cycle, state 1 is a saturated liquid.

The Rankine cycle can be represented on a T-S diagram as follows:

In a Rankine cycle with superheat, the steam at the exit of the boiler is superheated as shown:

In analyzing the Rankine cycle, it is helpful to think of the efficiency depending on the average temperature at which heat is supplied and the average temperature at which heat is rejected. 

Thus, if the boiler and condenser pressures are the same, the efficiency of a Rankine cycle with superheat is greater than the efficiency of a Rankine cycle.

Fluid Statics and Pressure Measurement

Let us determine the pressure at any point in a fluid a rest. For this, let us consider a wedge-shaped particle that is exposed on all sides to a fluid. This is presented as follows:

The cross section of the particle can be used to sketch a free-body diagram, as shown below:

The dimensions Δx, Δy, and Δz are small and approach zero as the particle shrinks to a point. Pressure and gravity are the only forces considered to be acting on the particle. Applying Newton’s second law in the x- and z-directions respectively, we obtain the following equations;

where:

• p_x, p_z, and p_s are average pressures acting on the three corresponding faces
• a_x and a_z are the accelerations
• ρ is the particle density

The net force equals zero in a static fluid. After simplification, with a_x = a_z = 0, these two equations become:

p_xΔz – p_sΔs(sinθ) = 0                              and

p_zΔx – p_sΔs(cosθ) – (ρg/2)ΔxΔz = 0

(ρg/2)ΔxΔz can be neglected as it is a higher-order term containing ΔxΔz, which is very small in comparison to the other terms. From the geometry of the wedge, we find that:

                                      Δz = Δs sinθ              and            Δx = Δs cosθ      

Since θ is chosen arbitrarily chosen, substitution into the pressure equations we obtain:

This shows that pressure at a point is the same in all directions and is applicable to both 2 dimension and 3-dimension cases.  

Let us consider an element of a fluid at rest, as illustrated below:

The element chosen has a volume, dx dy dz, and is sketched on a coordinate system where the positive z-direction is downward, coincident with the direction of the gravity force.

The following figure is a view of the element looking in the positive y-direction; the force acting on the right face is pdydz and that on the left face is (p + (∂p/∂x)dx)dydz, both normal to their respective surfaces. 

Summing forces in the x-direction, we have the following for a static fluid:

this means that pressure does not vary with respect to x. Similarly, for the y-direction we can assume the same as x-direction, hence:

These two equations show that there is no pressure variation in any lateral direction.

If we create a free body diagram for the z-direction and add up the forces, we obtain:

This shows that pressure does vary in a static fluid in the z-direction i.e., pressure increases with depth. This is mathematical shown by integrating both sides of the above equation:

where point 1 is a reference point such as the free surface of a liquid and point 2 is a point of interest. Since the density is a constant for incompressible fluids, we get: 

p₂ – p₁ = ρg(z₂ – z₁) = ρgΔz

where Δz is the depth below the liquid surface. This relationship is the basic equation of hydrostatics and is often written as:

Δp = ρgz