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Automobile Emission Control Questions and Answers

31.

A canister holds \( 1.5 \) kg of carbon at a working capacity of \( 0.08 \) kg of vapour per kg. Its vapour capacity is

Answer: C

Correct answer: 0.12 kg

Multiply the carbon mass by its working capacity:

\[ 1.5 \times 0.08 = 0.12\ \mathrm{kg} \]

Working capacity is deliberately less than the total adsorption capacity, since the canister is never purged completely dry. The figure must cover the vapour generated over a hot day's parking plus refuelling losses where the system handles them - and once it saturates, further vapour escapes to atmosphere.

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32.

Ten per cent of a fleet emits five times as much per vehicle as the other ninety per cent. That decile contributes

Answer: D

The decile emits \( 0.10 \times 5 = 0.5 \) units against the rest's \( 0.90 \times 1 = 0.9 \), so its share is \( 0.5/1.4 = 0.357 \). A small population of badly maintained vehicles carrying so large a share of the total is exactly why inspection and maintenance programmes are worth running.

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33.

A catalyst reaches light-off at \( 300 \) degrees, \( 90 \) s after a cold start. Over a \( 600 \) s test, that period is

Answer: A

Correct answer: 15 per cent of the test

Express the pre-light-off period as a fraction of the cycle:

\[ \frac{90}{600} \times 100 = 15\% \]

Fifteen per cent of the duration accounts for a far larger share of the emissions, since the catalyst is converting almost nothing during it. That disproportion is what drives close-coupled converters, low thermal mass substrates and electrically heated catalysts - all measures to shorten those ninety seconds.

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34.

Lambda readings of \( 0.97 \), \( 1.00 \) and \( 1.03 \) are compared. Which lies within a typical three-way catalyst window of plus or minus one per cent?

Answer: B

A window of one per cent admits values from 0.99 to 1.01, so only the stoichiometric reading qualifies. It is a remarkably narrow band, and holding the engine within it is precisely what the closed loop fuelling system exists to do.

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35.

Statement I: A three-way catalyst requires closed loop fuelling control.

Statement II: It can oxidise and reduce simultaneously only within about one per cent of stoichiometric.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. The three-way catalyst must oxidise carbon monoxide and hydrocarbons while simultaneously reducing nitrogen oxides, and those requirements conflict: oxidation wants oxygen, reduction wants none.

Only within a narrow window around stoichiometric does the exhaust suit both. Holding the mixture there needs a lambda sensor and closed loop control, oscillating rapidly about the target - which is why the catalyst and the sensor arrived together.

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36.

Statement I: Oxides of nitrogen peak when the mixture is deeply rich.

Statement II: Thermal formation requires both high flame temperature and available oxygen.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

The second statement gives the mechanism correctly, and it refutes the first. Thermal formation of nitrogen oxides needs high temperature and available oxygen together.

A deeply rich mixture is hot but oxygen-starved, so formation falls. The peak occurs slightly lean of stoichiometric, where the flame is still hot and oxygen is in surplus. That is why the emission curves for nitrogen oxides and for carbon monoxide peak on opposite sides of stoichiometric.

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37.

Statement I: A diesel engine cannot use a three-way catalyst.

Statement II: A diesel runs lean, so the exhaust contains ample oxygen throughout.

In the light of the above statements, choose the correct answer.

Answer: A

Both hold and the second is the reason: reduction of nitrogen oxides is impossible in an oxygen-rich exhaust. Separate devices are therefore needed, an oxidation catalyst for the one duty and a reduction system for the other.

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38.

Statement I: Exhaust gas recirculation lowers nitrogen oxide emissions.

Statement II: Recirculated exhaust absorbs combustion heat without contributing any of its own.

In the light of the above statements, choose the correct answer.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. Recirculated exhaust is inert: it takes up heat as the charge burns without releasing any, so the peak flame temperature falls.

Since thermal formation of nitrogen oxides rises steeply with temperature, a modest reduction cuts the emission substantially. The cost is combustion quality - too much recirculation makes the burn slow and unstable - which is why the rate is scheduled against load and speed rather than applied continuously.

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39.

Statement I: Cold start emissions can be ignored on a long journey.

Statement II: The catalyst does almost nothing until it reaches its light-off temperature.

In the light of the above statements, choose the correct answer.

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

The second statement is true and it is exactly why the first fails. Until light-off the catalyst converts almost nothing, so the first minute or two passes essentially untreated exhaust.

That fixed quantity is emitted whatever the journey length, so it cannot be ignored - it simply becomes a smaller share of a longer trip while remaining the same absolute amount. Averaged over a fleet of short urban journeys it dominates, which is why cold start is a regulated test condition.

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40.

Statement I: A carbon canister purge must be scheduled carefully by the engine controller.

Statement II: An uncontrolled purge delivers a slug of fuel vapour that drives the mixture rich.

In the light of the above statements, choose the correct answer.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the care needed. The canister holds adsorbed fuel vapour, and purging it admits an uncontrolled quantity of fuel to the intake.

An unscheduled purge therefore drives the mixture rich, upsetting the closed loop control and the catalyst's operating window. The controller purges when conditions allow - warm engine, steady load - and trims the injection to compensate, learning the vapour concentration as it goes.

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