Automobile Emission Control Questions and Answers

Answer: A

Lambda is the actual ratio divided by the stoichiometric one, \( 13.2/14.7 = 0.898 \). A value this far below one is well outside the three-way catalyst's window and would give high carbon monoxide and hydrocarbon emissions.

Answer: B

Correct answer: 60 per cent

The excess is measured against the limit:

\[ \frac{1.6 - 1.0}{1.0} \times 100 = 60\% \]

The distractor 37.5 per cent divides by the measured value instead of the limit, which answers a different question - what fraction of the emission is excess. Compliance is judged against the standard, so the standard is the denominator.

Answer: C

Correct answer: 0.48 g/km

A 96 per cent conversion leaves 4 per cent:

\[ 12 \times 0.04 = 0.48\ \mathrm{g/km} \]

The distractor 11.52 is the amount converted rather than the amount remaining. The figure shows how much a catalyst does: without it the vehicle would emit twelve grams a kilometre, an order of magnitude above any modern limit. It also shows why light-off matters - before the catalyst is hot, the full 12 g/km passes through.

Answer: D

The product is \( 200000 \times 12000 \times 0.48 = 1.152\times10^9 \) grams, or 1152 tonnes. Fleet totals of this kind are what an urban inventory is built from, and they scale directly with both the fleet size and the annual mileage.

Answer: A

Fifteen per cent of four hundred milligrams is sixty. That inert fraction absorbs combustion heat without releasing any, which is exactly how it lowers the peak temperature and the thermal nitrogen oxide formation with it.

Answer: B

Five per cent of thirty litres is 1.5 litres a hundred kilometres. Tank sizing follows from this figure and the required service interval, and running out of the solution now derates the engine on most modern vehicles.

Answer: C

Correct answer: 50

Divide the annual distance by the regeneration interval:

\[ \frac{25\,000}{500} = 50 \]

Each regeneration raises the exhaust temperature enough to burn off the soot, usually by injecting extra fuel, so fifty of them a year carry a real fuel penalty. The interval also depends on how the vehicle is used: heavy urban duty loads the filter faster and, worse, may never allow regeneration to complete.

Answer: D

Correct answer: 40 per cent of the total

The remainder is simply what the first 2 km does not account for:

\[ 100 - 60 = 40\% \]

So a fifth of the distance produces three fifths of the hydrocarbons. The reason is the cold catalyst, which converts almost nothing until light-off, together with the rich mixture the engine needs to run smoothly when cold. It is why a short journey emits far more per kilometre than a long one.

Answer: A

Correct answer: 33.6 km/h

Convert the time to hours before dividing:

\[ \frac{11}{1180/3600} = \frac{11 \times 3600}{1180} = 33.6\ \mathrm{km/h} \]

An average of about 34 km/h characterises a mixed urban and extra-urban cycle, and the figure matters because emissions are strongly load dependent. A cycle whose average speed is far below real driving will understate the emissions a vehicle produces in service - the criticism that led to the newer test procedures.

Answer: B

Correct answer: 97.1 per cent

The reduction is measured against the original level:

\[ \frac{350 - 10}{350} \times 100 = 97.1\% \]

Removing all but a fortieth of the sulphur was an expensive refinery investment, and it was undertaken because the aftertreatment could not work otherwise. The distractor 2.9 per cent is the fraction remaining, which is a useful check - the two must sum to 100.