IChO 1968 · Theory problem 3

Combustion of blast-furnace gas

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The gas escaping from a blast furnace has the following composition:

12.0 volume % of \(\ce{CO2}\) 28.0 volume % of \(\ce{CO}\)
3.0 volume % of \(\ce{H2}\) 0.6 volume % of \(\ce{CH4}\)
0.2 volume % of \(\ce{C2H4}\) 56.2 volume % of \(\ce{N2}\)

Problems:

3.1 Calculate the theoretical consumption of air (in \(\mathrm{m^3}\)) which is necessary for a total combustion of \(200\ \mathrm{m^3}\) of the above gas if both the gas and air are measured at the same temperature. (Oxygen content in the air is about 20 % by volume).

3.2 Determine the composition of combustion products if the gas is burned in a 20 % excess of air.

Official English text of the problem.

Answers

  • 3.1: The theoretical air requirement is \(\boxed{173\ \mathrm{m^3}}\).
  • 3.2: With 20% excess air, the product compositions are approximately:
Product: name, formula, SMILES Wet basis (volume %) Dry basis (volume %)
Carbon dioxide, \(\ce{CO2}\), O=C=O 21.8 22.3
Water, \(\ce{H2O}\), O 2.4 Removed
Oxygen, \(\ce{O2}\), O=O 1.8 1.9
Nitrogen, \(\ce{N2}\), N#N 74.0 75.8

The wet composition assumes all water remains as vapour; the dry composition excludes water.

Solution

3.1

At the same temperature and pressure, ideal-gas volumes are proportional to amounts in moles. Thus, balanced equations also give the reacting gas-volume ratios.

The combustible constituents are carbon monoxide (\(\ce{CO}\), [C-]#[O+]), hydrogen (\(\ce{H2}\), [H][H]), methane (\(\ce{CH4}\), C), and ethene (\(\ce{C2H4}\), C=C). Their complete combustion reactions are:

\[ \ce{2CO + O2 -> 2CO2} \]

\[ \ce{2H2 + O2 -> 2H2O} \]

\[ \ce{CH4 + 2O2 -> CO2 + 2H2O} \]

\[ \ce{C2H4 + 3O2 -> 2CO2 + 2H2O} \]

The carbon dioxide already present requires no oxygen, and nitrogen is treated as inert.

For \(200\ \mathrm{m^3}\) of blast-furnace gas:

Combustible gas Initial volume (\(\mathrm{m^3}\)) Oxygen required per unit gas volume Required oxygen (\(\mathrm{m^3}\))
\(\ce{CO}\) \(200(0.280)=56.0\) \(1/2\) 28.0
\(\ce{H2}\) \(200(0.030)=6.0\) \(1/2\) 3.0
\(\ce{CH4}\) \(200(0.006)=1.2\) 2 2.4
\(\ce{C2H4}\) \(200(0.002)=0.4\) 3 1.2

Hence,

\[ V_{\ce{O2},\mathrm{required}} =28.0+3.0+2.4+1.2 =34.6\ \mathrm{m^3}. \]

Since air contains 20% oxygen by volume,

\[ \boxed{ V_{\mathrm{air,theoretical}} =\frac{34.6}{0.20} =173\ \mathrm{m^3} }. \]

3.2

A 20% excess means supplying 1.20 times the theoretical air requirement:

\[ V_{\mathrm{air,supplied}} =1.20(173)=207.6\ \mathrm{m^3}. \]

Taking the remaining 80% of air as nitrogen, this supplies

\[ V_{\ce{O2},\mathrm{supplied}} =0.20(207.6)=41.52\ \mathrm{m^3}, \]

\[ V_{\ce{N2},\mathrm{air}} =0.80(207.6)=166.08\ \mathrm{m^3}. \]

All following product volumes are expressed at the same reference temperature and pressure, with water treated as vapour.

Carbon dioxide consists of the original carbon dioxide plus that formed from carbon monoxide, methane, and ethene:

\[ \begin{aligned} V_{\ce{CO2}} &=200(0.120)+56.0+1.2+2(0.4)\\ &=\boxed{82.0\ \mathrm{m^3}}. \end{aligned} \]

Water is formed from hydrogen, methane, and ethene:

\[ V_{\ce{H2O}} =6.0+2(1.2)+2(0.4) =\boxed{9.2\ \mathrm{m^3}}. \]

The unconsumed oxygen is

\[ V_{\ce{O2}} =41.52-34.6 =\boxed{6.92\ \mathrm{m^3}}. \]

Nitrogen comes from both the fuel gas and the supplied air:

\[ V_{\ce{N2}} =200(0.562)+166.08 =\boxed{278.48\ \mathrm{m^3}}. \]

Thus, the total wet product volume is

\[ V_{\mathrm{wet}} =82.0+9.2+6.92+278.48 =376.6\ \mathrm{m^3}. \]

Each wet volume percentage is calculated from

\[ \text{Volume percentage of species }i =\frac{V_i}{376.6\ \mathrm{m^3}}\times100\%. \]

For example,

\[ \text{Wet }\ce{CO2}\text{ percentage} =\frac{82.0}{376.6}\times100\% \approx21.8\%. \]

For the dry composition, exclude the water:

\[ V_{\mathrm{dry}} =376.6-9.2 =367.4\ \mathrm{m^3}. \]

Dividing the volumes of carbon dioxide, oxygen, and nitrogen by \(367.4\ \mathrm{m^3}\) gives the dry percentages reported above.

The solution was checked against the official solution.