23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2017
Question 2 of 6: Coal Combustion — Air, Flue-Gas Mass and Volume
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
16-Chem-A1, May 2017 — three-hour, open-book examination. The paper is in two parts: Part A (Questions 1–3, process mass & energy balances) and Part B (Questions 4–6, chemical thermodynamics). Candidates answer two questions from each part; all six are worked below as a complete study resource.
Reference texts (this subject). Felder & Rousseau, Elementary Principles of Chemical Processes, 4th ed. (mass & energy balances, combustion, recycle/purge); Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics, 8th ed. (fugacity, activity coefficients, generalized correlations, equations of state); property data from the paper's appended Smith–Van Ness tables (App. B critical constants; App. E Lee/Kesler charts).
Question 2: Coal Combustion — Air, Flue-Gas Mass and Volume (Part A)
Given. Ultimate analysis (mass fractions on 1 lb coal) and the boiler-exit temperature $600\ ^\circ\text{F} = 1059.7\ ^\circ\text{R}$ at 1 atm.
C
H₂
O₂
N₂
S
Ash
0.8339
0.0456
0.0505
0.0103
0.0064
0.0533
Find. (a) theoretical air; (b) flue-gas mass; (c) flue-gas volume at 600 °F; (d) air at 20% excess; (e) flue-gas volume at 20% excess; (f) % CO₂ dry and wet.
Approach. Work on a 1 lb-coal basis: sum the element oxygen demands, credit the fuel oxygen, convert to air through the 23.2 wt% O₂ content, and carry nitrogen as the inert tie element to close mass and molar (Orsat) bases.
Stoichiometric oxygen and air (a). C→CO₂ needs $\tfrac{32}{12}$, H₂→H₂O needs 8, S→SO₂ needs 1 lb O₂ per lb; credit the fuel oxygen:
$$O_2^{\text{req}} = 0.8339(2.667) + 0.0456(8) + 0.0064(1) = 2.595,\quad O_2^{\text{net}} = 2.595 - 0.0505 = 2.544\ \text{lb}$$
$$m_{\text{air}} = \frac{2.544}{0.2319} = \boxed{10.97\ \text{lb air / lb coal}}$$
Flue-gas mass (b). By overall mass balance, gas = coal + air − retained ash:
$$m_{\text{gas}} = 1 + 10.97 - 0.0533 = 11.91\ \text{lb} \;\;(\text{check by products: } 3.058_{\text{CO}_2}+0.410_{\text{H}_2\text{O}}+0.013_{\text{SO}_2}+8.434_{\text{N}_2}=11.91)$$
Flue-gas volume at 600 °F (c). Convert products to moles (all O₂ consumed at stoichiometric) and apply the ideal-gas law:
$$n_{\text{wet}} = 0.0695_{\text{CO}_2}+0.0226_{\text{H}_2\text{O}}+0.0002_{\text{SO}_2}+0.2995_{\text{N}_2} = 0.3918\ \text{lbmol}$$
$$V = \frac{nRT}{P} = \frac{0.3918(10.732)(1059.7)}{14.7} = 303\ \text{ft}^3/\text{lb coal}$$
Air and volume at 20% excess (d, e). Excess air scales the air and adds unreacted O₂ with its N₂ to the gas:
$$m_{\text{air}}^{20\%} = 1.2(10.97) = 13.16\ \text{lb}, \qquad n_{\text{wet}}^{20\%} = 0.3918 + 0.2\,n_{\text{O}_2}\!\left(1+\tfrac{79}{21}\right) = 0.4675\ \text{lbmol}$$
$$V^{20\%} = \frac{0.4675(10.732)(1059.7)}{14.7} = 362\ \text{ft}^3/\text{lb coal}$$
CO₂ percentage (f). At exactly stoichiometric air the CO₂ fraction is the Orsat maximum:
$$\%\text{CO}_2^{\text{wet}} = \frac{0.0695}{0.3918} = 17.7\%, \qquad \%\text{CO}_2^{\text{dry}} = \frac{0.0695}{0.3918 - 0.0226} = 18.8\%$$
Adding excess air dilutes these values, so the numbers above are the peak attainable.
