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 ^{—} Zn_{2} 
H_{2} 
1.0 
1  3z_{N}, 
NH, 
 
2z_{n}, 
Total 
2.0 
22z_{n}, 
Thus, the equilibrium moles of N_{2} can be written as:
Obviously, (initial moles  equilibrium moles) represents the amount of N_{2} converted, and this is equal to Zn_{2}
If Zn_{2} moles of N_{2} are converted, then 3zn_{2} moles of H_{2} are converted.
Thus, at equilibrium, there will be 1 3zn_{2} moles of H_{2}.
On the same basis, if Zn_{2} moles of N, are converted, then 2zn_{2} moles of ammonia must be produced.
If we write K_{p} 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) 
n_{2} 
1.0 
1 0.189 = 0.811 
H_{2} 
1.0 
1 3x0.189 = 0.433 
NH, 
 
2x0.189 = 0.378 
Obviously, H_{2} being a limiting component, we can check what happens if H_{2} is present in stoichiometry proportions. We find that Zn_{2} becomes 0.57, and hence, the amount of ammonia produced is 1.14 kmol.
Q2 Solution:
The reaction involved is the methanesteam 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) 
CH_{4} 
1.0 
1 ^{} Zch_{4} 
H_{2}0 
6.0 
6 — Zch_{4} 
CO 
 
Zch_{4} 
H, 
 
3Zch_{4} 
Total 
7 
7 + 2 z 
At the same time, we recognise that amount of CH_{4} converted must be 1 x = Zch_{4 }Using this and the reaction stoichiometry, the moles H_{2}0 converted must be Zch_{4 }and the moles of H_{2}0 at equilibrium must be 6Zch_{4}
Again taking the reaction stoichiometry, the amount of CO produced must be Zch_{4 }and the amount of H_{2} produced must be 3zch_{4}
Taking the sum of the equilibrium moles, we find an expression: 7 + 2zch_{4} If we express K_{p} in terms of component partial pressure 'p and write p, = y_{f} 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 Zch_{4} = 0.5, we find a value of K_{p} = 9.59 X 10~^{3}.
We know In K_{r} = 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 Zch_{4} 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 Zch_{4} and taking the products from this reaction as the starting amounts for Reaction 2, does an analysis in terms of z_{C}o
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 CH_{4} and H_{2}0.
Initial moles of CH_{4} = 1 kmol and H_{2}0 = 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, K_{pl} for Reaction 1 can be calculated as 1.352 and K_{p2} 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 z_{C}m and Zco^{:}
Essentially to answer the question, we need to solve for Zch_{4} 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 K_{Pl} was multiplied by K_{P2}/K_{Pl}. The resulting equation was then subtracted from equation (6.2) and the values of Zch_{4} 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 Zch_{4} and Zco It is also important to constrain 0 < Zch_{4} < 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) 
Catoms 
Composition 
Amount (kmol) 
Catoms 

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 

СН_{4} 
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 = H_{2} in output gas + H_{2}0 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 + H_{2}0 x 2 in output gas
Thus, H_{2} in output gas:
86.8 = 28.53 + 13.94 + 2H_{2}0 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 10^{6}kJ.
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 N_{2} = 1  Zn_{2}
Draw up a mass balance table at equilibrium:
Component 
Initial (kmol) 
Equilibrium (kmol) 
N, 
1.0 
1  Zn, 
H, 
3.0 
3 — 3ZN2 
NH, 
 
2z_{n}, 
Total 
4.0 
4  2zn_{2} 
Take compositions in mole fractions:
Thus:
b. Total pressure = 150 bar
Take yj P = p,, where p_{t} 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 K_{p}, 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 — Zn_{2} 
H, 
3.0 
3 — 3zn_{2} 
NH, 
 
2zn, 
Total 
4.0 
4  2zn, 
At T = 454°C (727 K), use the equation for K_{p}: 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 lefthand side to two volumes on the righthand side, a higher pressure pushes the equilibrium to the right. Cases 13 indicate this happens.
In terms of the quantitative statement, it is easy to show that K_{p} increases as T reduces  Zn_{2} will increase as T reduces.
Inspection of equation for K_{p} in terms Zn_{2} 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) 
n_{2} 
1.0 
1 ^{—} Zn_{2} 
h_{2} 
3.0 
3  3Zn_{2} 
NH_{3} 
 
2zn_{2} 
Inert 
0.5 

Total 
4.5 
4.5  2zn_{2} 
The K_{p} can be written as:
This can be solved using Goal Seek or an appropriate method. It can be shown that Zn_{2} =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, T^{2} 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.