Answers to Problems in Chapter 5

Answers to Questions Related to Syngas and Ammonia Gas Production

Q1 Solution:

We are given the stoichiometric equation:

The text uses a fractional conversion, and this is consistently used to work out the reaction mass balance. As this approach is usually adopted, draw up an equilibrium balance table:

Basis: 1 kmol

Component

Initial (kmol)

Equilibrium (kmol)

N,

1.0

1 Zn2

H2

1.0

1 - 3zN,

NH,

-

2zn,

Total

2.0

2-2zn,

Thus, the equilibrium moles of N2 can be written as:

Obviously, (initial moles - equilibrium moles) represents the amount of N2 converted, and this is equal to Zn2

If Zn2 moles of N2 are converted, then 3zn2 moles of H2 are converted.

Thus, at equilibrium, there will be 1 -3zn2 moles of H2.

On the same basis, if Zn2 moles of N, are converted, then 2zn2 moles of ammonia must be produced.

If we write Kp as:

as explained in the main text, we can write:

where /?, is the partial pressure of component T, y, is the gas/vapour mole fraction of T and P is the total pressure.

Thus, using the data from equilibrium mass balance and dropping the subscripts:

You will note that the expression (2 - 2z)2 appears because each of the component moles has to be divided by the total equilibrium moles. This equation can be solved graphically, but it is better to use systems such as MATLAB or Excel.

Using Excel, the equation was written as:

Using SOLVER, the value of z can be calculated as 0.189.

This gives the final table:

Component

Initial (kmol)

Equilibrium (mol)

n2

1.0

1 -0.189 = 0.811

H2

1.0

1 -3x0.189 = 0.433

NH,

-

2x0.189 = 0.378

Obviously, H2 being a limiting component, we can check what happens if H2 is present in stoichiometry proportions. We find that Zn2 becomes 0.57, and hence, the amount of ammonia produced is 1.14 kmol.

Q2 Solution:

The reaction involved is the methane-steam reaction:

If we draw up a simple equilibrium balance taking 1.0 kmol of methane as basis, then we can say:

Component

Initial (kmol)

Equilibrium (kmol)

CH4

1.0

1 - Zch4

H20

6.0

6 — Zch4

CO

-

Zch4

H,

-

3Zch4

Total

7

7 + 2 z

At the same time, we recognise that amount of CH4 converted must be 1 -x = Zch4 Using this and the reaction stoichiometry, the moles H20 converted must be Zch4 and the moles of H20 at equilibrium must be 6-Zch4-

Again taking the reaction stoichiometry, the amount of CO produced must be Zch4 and the amount of H2 produced must be 3zch4-

Taking the sum of the equilibrium moles, we find an expression: 7 + 2zch4- If we express Kp in terms of component partial pressure 'p- and write p, = yf P, where p, is the partial pressure, y, the component mole fraction and P the total pressure:

Writing each mole fraction as component moles/total moles:

For a value of Zch4 = 0.5, we find a value of Kp = 9.59 X 10~3.

We know In Kr = ln(9.59 x 10~3) = -4.641.

Q3 Solution:

The important parts of the book text to access are the К values and the equations in Appendix 3.

The other important table to access is the table that gives the primary reformer analysis in terms of two fractional conversions Zch4 for reaction and Zco for Reaction 2. There is an explanation of the primary reformer analysis given in Appendix 2.

The analysis carried out, in effect, assumes that the thermodynamics is dependent on the initial and final states only and not the path taken.

The analysis finds the equilibrium products for Reaction 1 in terms of Zch4 and taking the products from this reaction as the starting amounts for Reaction 2, does an analysis in terms of zCo-

Taking the book text table that analyses the primary reformer, we can follow its logic and adopt the figures appropriate to this problem.

Assume equal moles of CH4 and H20.

Initial moles of CH4 = 1 kmol and H20 = 3.5 kmol.

On this basis, we can write the equilibrium mass balance table:

Component

Initial Moles

Moles after Reaction 1

Moles after Reaction 2

CH,

1.0

CO

H,0

1.0

CO,

-

H,

-

Total

2.0

At the temperature stated, Kpl for Reaction 1 can be calculated as 1.352 and Kp2 for Reaction 2 as 2.280, thus writing:

using the fact that P = 1.

The appropriate values of у can be written in terms of zCm and Zco:

Essentially to answer the question, we need to solve for Zch4 and Zco, two equations and two unknowns in this case.

This was done using SOLVER in Excel, to find a function to set to zero.

The equation for KPl was multiplied by KP2/KPl. The resulting equation was then subtracted from equation (6.2) and the values of Zch4 and Zco calculated using SOLVER to find the z values that made this function zero.

SOLVER gave a solution, and it should be noted that the solution is sensitive to the initial guessed values of Zch4 and Zco- It is also important to constrain 0 < Zch4 < 1 and 0 < Zco < 1 •

Values of 0.892 and 0.596 are obtained as specified in the question.

Q4 Solution:

Take a basis of 100 kmol fuel gas.

We are given input and output compositions.

A mass balance must be carried out; no reaction equations are specified.

This is a classic situation where atom balances must be carried out.

An input/output table can be drawn up:

Component

Input Fuel Gas

Air

Output Gas (Dry) W

Composition

Amount

(kmol)

C-atoms

Composition

Amount

(kmol)

C-atoms

СО

21.0

21.0

21.0

4.3

0.043 W

0.043 W

СО,

7.1

7.1

7.1

15.0

0.15 W

0.15 W

н.

28.6

28.6

8.0

0.08 W

СН4

7.4

7.4

7.4

2.1

0.021 W

0.021 W

N,

35.7

35.7

79.0

63.9

0.639 W

о.

21.0

Н,0

Total

100

35.5

0.214 W

The following balances can be carried out:

Carbon atom balance:

Carbon atoms in = Carbon atoms out 35.5 = 0.214 W

This can be solved to give W, the kmol of the output stream on a dry basis. Nitrogen atom balance:

Nitrogen in fuel gas + Nitrogen in air = Nitrogen in output gas

35.7 + Nitrogen in air = 0.699 x 165.89

Nitrogen in air = 80.26 kmol

Thus, O, in air = 80.26 x (21/79) = 21.33 kmol

Hydrogen atom balance.

H, in fuel gas = H2 in output gas + H20 in output gas 28.6 X 2 + 7.4 x 4’= 0.086 x 165.89 X 2 + 0.021 x 165.89 x 4 + H20 x 2 in output gas

Thus, H2 in output gas:

86.8 = 28.53 + 13.94 + 2H20 in output gas H,0 in output gas = 22.17 kmol.

It is now possible to write a summary:

With 165.89 kmol of output gas CO = 165.89 x 0.043 = 7.13 kmol CO, = 165.89 x 0.15 = 24.88 kmol H, = 165.89 x 0.086 = 14.27 kmol CH, = 165.89 x 0.021 = 3.48 kmol N, = 165.89 x 0.699 = 115.96 kmol Total = 165.72 kmol

In the input air, the N, has been calculated as 80.26 kmol and O, as 21.33 kmol.

Thus, the ratio of fuel gas to air must be --—-- = 0.984.

e (80.26 + 21.33)

b. As indicated in the question, finding the flame temperature requires a statement of solution of the adiabatic energy balance.

For the input, all the values of m,• have already been calculated.

The temperature 7} is specified (298 K), so the input energy can be calculated using the polynomial equations.

This should give a total energy input of 4.201 x 106kJ.

In the output stream, the output temperature is the unknown.

All the values of m„ have been found in the equilibrium mass balance.

The individual enthalpies can be found from the polynomials.

SOLVER was used to set a function

Input energy - Output energy to zero by altering the value of T„

The value of T„ = 1725.9 К was found.

Q5 Solution:

a. Basis: 1 kmol

Therefore, equilibrium moles N2 = 1 - Zn2

Draw up a mass balance table at equilibrium:

Component

Initial (kmol)

Equilibrium (kmol)

N,

1.0

1 - Zn,

H,

3.0

3 — 3ZN2

NH,

-

2zn,

Total

4.0

4 - 2zn2

Take compositions in mole fractions:

Thus:

b. Total pressure = 150 bar

Take yj P = p,, where pt is the partial pressure of component ‘i P is the total pressure and y, is the mole fraction of component T.

We know P = 150 bar.

Thus:

We can write:

To find a temperature, use the given equation:

Substituting in Kp, the value of temperature can be found to be 800.3 К = 527.3°C. Q6 Solution:

For T = 454°C (727 K) and P = 200 bar:

Component

Initial (kmol)

Equilibrium (kmol)

N,

1.0

1 — Zn2

H,

3.0

3 — 3zn2

NH,

-

2zn,

Total

4.0

4 - 2zn,

At T = 454°C (727 K), use the equation for Kp: Thus, at 727 K:

We can write this as:

We can use SOLVER or Goal Seek in Excel.

Find z = 0.387

For the same stoichiometry and temp P = 500 bar

Find К = 4.08 x 10~5 using Goal Seek z = 0.56

For the same stoichiometry but temperature = 399°C (672 K) with P = 200 bar Find z = 0. 522 a. Summarising:

T

P

Z

1

727

200

0.387

2

727

500

0.560

3

672

200

0.522

For Cases 1 and 2 with temperature constant Pressure increases, and conversion increases Increasing P increases z

Cases 1 and 3 with pressure constant As temperature reduces, conversion increases

b. These are consistent with Le Chatelier’s principle. For exothermic reactions, any condition allows energy transfer from system to surroundings and pushes equilibrium to the right, and a reduced temperature achieves this.

For a reaction where we go from four volumes on the left-hand side to two volumes on the right-hand side, a higher pressure pushes the equilibrium to the right. Cases 1-3 indicate this happens.

In terms of the quantitative statement, it is easy to show that Kp increases as T reduces - Zn2 will increase as T reduces.

Inspection of equation for Kp in terms Zn2 indicates that if the pressure term is taken over to the other side of the equation, then z will increase as P increases.

c. For the conditions stated, T = 454°C (727 K) and P = 200 atm. It has been calculated that К = 4.08 x 10~5. The equilibrium mass balance can be written: Basis: 1 kmol

Component

Initial (kmol)

Equilibrium (kmol)

n2

1.0

1 Zn2

h2

3.0

3 - 3Zn2

NH3

-

2zn2

Inert

0.5

Total

4.5

4.5 - 2zn2

The Kp can be written as:

This can be solved using Goal Seek or an appropriate method. It can be shown that Zn2 =0.361. There has been reduction in z. Numerically, the equilibrium mixture is now (4.5 - 2z).

Mathematically, this reduces the value of z.

This dilution effect can sometimes used in reactions where the equilibrium reaction mixture volume has increased. Theoretically, an increased pressure would reduce z and a decrease in pressure is often not practical or desirable. This ‘diluent’ effect can sometimes be used in such cases.

Answers to Problems Relating to Syngas Production Not Included in the Main Text

Q1 Solution:

It is necessary to find the component enthalpies for the flue gas at the stated input and output conditions. The enthalpy polynomials are used and the following figures found:

Component

kmol

Input Stream Enthalpy

Output Stream Enthalpy

k)

kj

СО,

56.7!

-1.91E + 07

-I.93E+07

Water

110.57

-2.14E + 07

-2.17E+07

Nitrogen

485.28

1.96E + 07

1.85E+07

Oxygen

16.83

6.60E + 05

6.25E+05

-2.03E + 07

-2.19E+07

Total

-20,322.487.73

-21.885,745.40

In the problem, it states that the enthalpy of vaporisation (latent heat) is:

Taking the energy being transferred from the flue gas to the process stream as the adjusted value for a realistic 2% energy loss, we obtain the figure:

Using the appropriate latent heat of the steam, we can calculate the kg of steam that can be generated:

Q2 Solution:

To fix the flue gas outlet temperature, the process energy balance is carried out. Energy Balance:

Consider the steam stream fed into the superheater. The entering stream is saturated at a pressure of 62 bar. From steam tables, the corresponding saturation pressure can be read (interpolated) as 278°C. This steam passes through the superheater where energy from the flue gas transfers to the steam and exits in the outlet stream at a pressure of 58.5 bar. Steam tables indicate that the temperature corresponding to this at saturated stream conditions is 273.9°C. Thus,

420 - 273.9 = 146.1 degrees of superheat are present.

From steam tables, the specific enthalpy of the saturated steam at inlet is read as 2782 kJ/kg.

The superheated steam at outlet at 420°C can be read as 3229 kJ/kg.

As stated, 9704 kg of steam enter the superheater; thus, the amount of energy required to raise the superheated steam at 420°C is:

As has been demonstrated in previous calculations, we have to carry out an iterative process to fix the outlet temperature.

The inlet temperature for the flue gas is known to be 892°C. Using the enthalpy polynomials in Appendix 3, the component enthalpies can be calculated at 1165 K, and using the relevant component amounts, the stream enthalpies in kJ can be calculated as follows:

Composition

In

CO,

56.7!

-1.93E+07

Water

110.57

-2.17E+07

Nitrogen

485.28

1.85E+07

Oxygen

16.83

6.25E+05

-2.19E+07

The energy required for the steam superheat stream is known. The energy balance can be written as:

The energy out term for the flue gas can be written in terms of the enthalpy polynomials as:

where H, is the component enthalpy represented by the relevant component enthalpies, m, is the component moles and N is the number of components.

The relevant constants are found, as usual, in Appendix 3.

Each of the polynomial expressions carried terms in T, T2 and 7"’ where it is was then necessary to calculate the value of T.

As in a number of problems solved here, SOLVER in an Excel spreadsheet was set to produce an answer by setting a function of the form:

T, the temperature, is the unknown in the polynomial expressions. A value of 726.8°C is found.

 
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