A driving cycle in emissions testing is
Answer: A
It is meant to represent typical use so that vehicles can be compared on a common basis. How well the cycle reflects real driving has been a long-running controversy, and real driving emissions testing was introduced to address it.
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Cold start emissions are disproportionately important because
Answer: B
Until the bed reaches roughly three hundred degrees the conversion efficiency is very low, and a large share of a short trip's total emission occurs in those first minutes. Close-coupled converters and electrical heating both attack this directly.
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An inspection and maintenance programme controls emissions by
Answer: C
New vehicle standards do nothing about a fleet whose control equipment has degraded or been tampered with. A small proportion of badly maintained vehicles typically accounts for a large share of a fleet's total emission.
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A lean burn engine reduces nitrogen oxides by
Answer: D
Excess air dilutes and cools the charge, and the emission falls once the mixture is lean enough. The difficulty is that a three-way catalyst cannot reduce nitrogen oxides in the resulting oxygen-rich exhaust, so a lean NOx trap or reduction catalyst is needed.
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An engine runs at an air-fuel ratio of \( 13.2 \) against a stoichiometric \( 14.7 \). Its lambda value is
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.
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A vehicle emits \( 1.6 \) g/km of carbon monoxide against a limit of \( 1.0 \) g/km. It exceeds the limit by
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.
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A catalyst converts \( 96 \) per cent of the carbon monoxide entering at \( 12 \) g/km. The tailpipe emission is
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.
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A fleet of \( 200000 \) vehicles each travelling \( 12000 \) km a year at \( 0.48 \) g/km emits annually
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.
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Exhaust gas recirculation at \( 15 \) per cent replaces that fraction of the fresh charge. Of \( 400 \) mg of charge per cycle, the recirculated mass is
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.
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A urea dosing system uses solution at \( 5 \) per cent of diesel consumption. A vehicle burning \( 30 \) litres per \( 100 \) km uses
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.
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