Fuels and Combustion Questions and Answers

Answer: C

Correct answer: exactly sufficient for complete combustion

The theoretical or stoichiometric air is the amount calculated from the fuel’s composition to convert all its carbon to carbon dioxide, all its hydrogen to water and all its sulphur to sulphur dioxide, with nothing left over. It is a computed reference quantity, not what is actually supplied - every real furnace supplies more, because perfect mixing is unattainable.

Answer: D

Nitrogen and unused oxygen must be heated to flue gas temperature and then discarded, and the larger gas volume shortens the residence time. Too little excess air is worse still, since unburnt fuel wastes its whole calorific value.

Answer: A

Correct answer: carbon monoxide

The Orsat apparatus absorbs the constituents in sequence in three reagents - potassium hydroxide for carbon dioxide, alkaline pyrogallol for oxygen, and ammoniacal cuprous chloride for carbon monoxide - measuring the volume contraction after each. The remainder is taken as nitrogen. The analysis is on a dry basis, because the sample is collected over water, so water vapour is not among the determinations.

Answer: B

Burning carbon to monoxide gives about 2430 kcal per kilogram against 8080 for complete combustion, so some seventy per cent of the heat leaves the furnace still chemically locked in the gas. It is why even a few tenths of a per cent of monoxide is a serious loss.

Answer: C

Correct answer: poor mixing of fuel and air

Oxygen and carbon monoxide together are contradictory on their face: oxygen says air was in surplus, carbon monoxide says some fuel did not find any. Both can only be true if the two never met, so the furnace as a whole has enough air while some regions run rich and others lean. The remedy is better burner mixing or improved distribution, not more air - which would only raise the stack loss.

Answer: D

Correct answer: begins to burn without further external heating

Ignition temperature is the temperature at which combustion becomes self-sustaining: the heat released by the reaction is enough to keep the fuel at reaction temperature, so the external source can be withdrawn. Below it the reaction stops when the heating stops. It is a practical rather than a thermodynamic constant, depending on pressure, on the fuel’s state of division and on the test method.

Answer: A

Correct answer: concentrations in air within which it will propagate a flame

Every flammable gas has a lower and an upper limit, and only between them will a flame started at one point travel through the mixture. Below the lower limit the mixture is too lean to release enough heat to ignite the next layer; above the upper limit there is too little oxygen. Hydrogen’s very wide range, roughly 4 to 75 per cent, is what makes it so hazardous to handle.

Answer: B

Correct answer: dissociation and heat loss to the surroundings

The theoretical flame temperature assumes complete combustion in an adiabatic vessel. Two things spoil it: at high temperature carbon dioxide and water dissociate back to carbon monoxide, hydrogen and oxygen, absorbing heat, and no real furnace is adiabatic - radiation and conduction remove heat as fast as it is released. Excess air lowers the temperature too, but the theoretical figure is defined without it.

Answer: C

Correct answer: 2.67 kg

The reaction is \( \mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2} \), so 12 kg of carbon takes 32 kg of oxygen and \[ \frac{32}{12} = 2.67\ \text{kg of oxygen per kg of carbon.} \] The figure 8 belongs to hydrogen and 1 to sulphur; together the three are the whole of Dulong’s oxygen calculation.

Answer: D

Correct answer: 8 kg

The reaction is \( 2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O} \), so 4 kg of hydrogen takes 32 kg of oxygen and \[ \frac{32}{4} = 8\ \text{kg of oxygen per kg of hydrogen.} \] Hydrogen is by far the most oxygen-hungry of the three combustible elements, which is why the hydrogen term dominates the correction for combined oxygen in Dulong’s formula.