Chemical Engineering - MCQ Practice Questions

Understanding:

We must verify the units of ${\text{HTU}}_{OG}$ from the given defining equation.

Formula:

Step 1: Write out the units

Step 2: Cancel common units

Step 3: Physical meaning

${\text{HTU}}_{OG}$ is a length — specifically, the height of packing required to achieve one transfer unit of separation. The total packing height is:

where ${\text{NTU}}_{OG}$ is dimensionless.

Answer:

The units of ${\text{HTU}}_{OG}$ are metres.

Quick Tip:

Always verify HTU units via dimensional analysis before using correlations — this quickly catches errors in which phase (gas or liquid) the coefficient and flow rate correspond to.

Understanding:

We must calculate the annual straight-line depreciation for a heat exchanger.

Formula:

Straight-line depreciation:

Step 1: Compute depreciable amount

Step 2: Divide by service life

Verification:

Over 10 years, total depreciation = 10 \times 10{,}000 = \100{,}000$,whichexactlyrecovers$C_p - C_s$.$\checkmark$

Answer:

The annual straight-line depreciation charge is \10{,}000$ per year.

Quick Tip:

A common mistake is dividing the full purchase price $C_p$ (not $C_p-C_s$) by $n$, which would give \12{,}000/\text{yr}$. Always subtract the salvage value before dividing.

Understanding:

We need to identify the correct formula relating API gravity to specific gravity (SG) at $60°\text{F}$.

Formula:

The American Petroleum Institute (API) gravity is defined by:

Step 1: Verify with a known reference

For water, $\text{SG}=1.0$:

This is correct — water has an API gravity of exactly $10°\text{API}$.

Step 2: Check a light crude

For a typical light crude with $\text{SG}=0.825$:

This aligns with known values for light crude oils ($35$–$45°\text{API}$), confirming the formula.

Answer:

The correct API gravity formula uses constants $141.5$ and $131.5$ in the standard definition.

Quick Tip:

A higher API gravity indicates a lighter (lower density) crude oil. Crude oils above $31.1°\text{API}$ are classified as "light" crude — remember the two constants $141.5$ and $131.5$ as a pair.

Understanding:

We need to identify the correct atmospheric boiling range for the kerosene/jet fuel cut in a crude distillation unit.

Formula:

Crude oil distillation separates fractions by boiling point. The standard product cut ranges are:

Step 1: List the standard atmospheric cut ranges

Step 2: Identify the correct option

The kerosene fraction, which includes jet fuel (Aviation Turbine Fuel, ATF), is recovered in the boiling range of $150°\text{C}$ to $250°\text{C}$ at atmospheric pressure. This fraction consists primarily of ${\text{C}}_{10}$ to ${\text{C}}_{14}$ hydrocarbons.

Answer:

The kerosene/jet fuel cut boils between $150°\text{C}$ and $250°\text{C}$ at atmospheric pressure.

Quick Tip:

A common trap is confusing the kerosene range with the atmospheric gas oil (diesel) range of $250$–$350°\text{C}$. Remember: kerosene comes above naphtha and below diesel.

Understanding:

We must calculate the Watson characterisation factor $K_W$ for a petroleum fraction.

Formula:

Step 1: Calculate the cube root of $T_b$

Since $9^3=729\approx 727$, we get:

Step 2: Divide by SG

Step 3: Interpret the result

A $K_W$ value of approximately $10.6$ indicates a naphthenic/paraffinic character. Purely paraffinic fractions have $K_W\approx 12$–$13$, while aromatic fractions have $K_W\approx 10$–$11$.

Answer:

The Watson characterisation factor is approximately $10.63$.

Quick Tip:

Remember: $9^3=729$, so any $T_b$ near $729°\text{R}$ gives $(T_b{)}^{1/3}\approx 9.0$, making mental estimation quick for exam problems.

Understanding:

We need to identify the primary function of the regenerator in an FCC unit.

Step 1: Understand the FCC process sequence

In FCC, a hot catalyst contacts the heavy gas oil feed in the riser. The cracking reactions deposit coke on the catalyst surface, progressively deactivating it. The spent (coked) catalyst is then separated from the product vapours and sent to the regenerator.

Step 2: Identify the regenerator's role

In the regenerator, air is introduced and the coke is combusted at temperatures of approximately $650$–$760°\text{C}$:

This combustion removes the coke from the catalyst surface, restoring its cracking activity. The regenerated catalyst (now at high temperature) is returned to the riser, also supplying the heat of reaction needed for endothermic cracking.

Step 3: Evaluate the other options

Separation of catalyst from vapours occurs in the disengager/stripper section, not the regenerator. Feed preheating is done in the preheat train upstream. Cracking itself occurs in the riser reactor.

Answer:

The primary purpose of the FCC regenerator is to burn off coke from the spent catalyst to restore its catalytic activity.

Quick Tip:

The FCC regenerator is also the main heat source for the unit — the hot regenerated catalyst transfers the combustion heat to the endothermic cracking reactions in the riser. This thermal integration is a key feature of the FCC design.

Understanding:

We need to identify the correct reference mixture corresponding to a RON of $92$.

Step 1: Recall the octane number definition

The octane number (RON or MON) is defined on a volumetric basis using a binary reference mixture:

Step 2: Apply the definition

A fuel with $\text{RON}=92$ has the same knock resistance as a mixture of:

The percentage of iso-octane in the reference mixture equals the octane number numerically.

Step 3: Note the basis

The octane number scale is defined on a volumetric basis, not a mass basis. Options that state a mass basis are therefore incorrect.

Answer:

RON $=92$ corresponds to $92\%$ iso-octane and $8\%$ n-heptane by volume.

Quick Tip:

Always remember: the octane number equals the volume percent of iso-octane in the reference blend. This is a volumetric, not gravimetric, definition — a classic exam trap.

Understanding:

We need to identify which reaction in catalytic reforming contributes most to octane number improvement.

Step 1: Recall the reactions in catalytic reforming

Catalytic reforming over a platinum-based catalyst (e.g., $\text{Pt}/{\text{Al}}_2{\text{O}}_3$) involves several reactions:

1. Dehydrogenation of naphthenes to aromatics: $\text{cyclohexane}\to \text{benzene}+3{\text{H}}_2$

2. Dehydrocyclisation of paraffins to aromatics: $\text{n-heptane}\to \text{toluene}+4{\text{H}}_2$

3. Isomerisation of n-paraffins to iso-paraffins

4. Hydrocracking of paraffins (a side reaction, consumes hydrogen)

Step 2: Compare octane numbers of hydrocarbon classes

Aromatics have very high octane numbers (e.g., benzene RON $\approx 101$, toluene RON $\approx 124$), while n-paraffins have very low octane numbers (n-heptane $=0$, n-octane $\approx -19$). Iso-paraffins have moderate octane numbers.

Step 3: Identify the dominant reaction for octane improvement

The conversion of naphthenes and paraffins to aromatics (dehydrocyclisation/aromatisation) produces the largest octane gain in catalytic reforming because the RON improvement per molecule converted is far greater than that from isomerisation alone.

Answer:

Dehydrocyclisation (aromatisation) of naphthenes and paraffins is primarily responsible for the largest octane number increase in catalytic reforming.

Quick Tip:

Catalytic reforming is endothermic overall due to the dominant dehydrogenation reactions, which is why reformers use multiple adiabatic reactors with inter-stage reheating furnaces.

Understanding:

We need to identify the correct operating pressure range for a vacuum distillation unit (VDU) in a petroleum refinery.

Step 1: Purpose of vacuum distillation

The atmospheric residue (long residue, boiling point $>350°\text{C}$) cannot be further distilled at atmospheric pressure without thermal cracking, since the required temperatures ($>400°\text{C}$) would cause undesirable decomposition. Operating under vacuum reduces the boiling points of the heavy fractions.

Step 2: Typical operating conditions

Vacuum distillation is conducted at absolute pressures typically in the range of $10$–$80\text{mmHg}$ (approximately $1.3$–$10.7\text{kPa}$ absolute), which corresponds to deep vacuum. At $15\text{mmHg}$, components boiling at $500°\text{C}$ at atmospheric pressure boil at approximately $200$–$250°\text{C}$.

Step 3: Reject incorrect options

Answer:

Vacuum distillation operates at $10$–$80\text{mmHg}$ absolute, enabling separation of heavy fractions without thermal cracking.

Quick Tip:

Steam ejectors and barometric condensers (or liquid ring vacuum pumps) are used to achieve and maintain the deep vacuum in the VDU overhead system.

Understanding:

We need to identify which hydrocarbon class has the highest cetane number, reflecting the best auto-ignition quality for diesel.

Step 1: Understand cetane number and ignition quality

The cetane number measures how readily a fuel auto-ignites under compression. A higher cetane number means shorter ignition delay, which is desirable in diesel engines. Cetane number is essentially the opposite of octane number in terms of which hydrocarbons score high.

Step 2: Compare hydrocarbon classes

Step 3: Conclusion

Normal paraffins with long chain lengths have the highest cetane numbers among all hydrocarbon classes.

Answer:

Normal (n-) paraffins with long chain length have the highest cetane number.

Quick Tip:

Note the inverse relationship with octane number: n-paraffins score low on octane (bad for gasoline) but high on cetane (good for diesel). Aromatics are the exact opposite — good for gasoline, bad for diesel.

Understanding:

We must calculate the on-stream (filling) time for a delayed coker drum given the total cycle time and the fraction of cycle spent on-stream.

Formula:

Step 1: Calculate on-stream time

Step 2: Calculate off-stream (decoking) time as a check

This $16\text{hour}$ offline period is used for steam stripping, water quenching, coke cutting (using high-pressure water jets), and drum inspection — consistent with industrial practice.

Answer:

The on-stream time per drum is $32$ hours.

Quick Tip:

Delayed coking units always use a minimum of two drums in parallel — while one drum is on-stream filling with coke, the other is being decoked and prepared for the next cycle, ensuring continuous feed processing.

Understanding:

We must calculate the percentage desulphurisation achieved in an HDS unit.

Formula:

Step 1: Substitute values

Step 2: Verify

Sulphur removed $=490\text{ppm}$ out of $500\text{ppm}$ feed. Fraction remaining $=\frac{10}{500}=0.02=2\%$. Therefore $98\%$ was removed. ✓

Answer:

The percentage desulphurisation is $98.0\%$.

Quick Tip:

For deep desulphurisation (e.g., ultra-low sulphur diesel at $10$–$15\text{ppm}$), the required conversion exceeds $99.9\%$, placing very severe demands on catalyst activity and operating conditions (higher pressure, lower space velocity).

Understanding:

We need to identify the typical BOD removal efficiency of a conventional activated sludge (CAS) process used in municipal wastewater treatment.

Step 1: Recall process performance

The conventional activated sludge process is a secondary biological treatment method. It uses a combination of an aeration tank (where microorganisms oxidise organic matter) followed by a secondary clarifier.

Step 2: Evaluate each range

Primary treatment alone achieves 30–50% BOD removal. Extended aeration or trickling filters achieve 55–70%. Conventional activated sludge consistently achieves 85–95% BOD removal, which is the well-established design benchmark used in environmental engineering practice. Values of 98–99.9% are associated with tertiary/advanced treatment processes such as membrane bioreactors (MBR) or additional polishing steps.

Answer:

The conventional activated sludge process typically achieves a BOD removal efficiency of 85–95%.

Quick Tip:

For exam purposes, associate secondary biological treatment (activated sludge) with ~90% BOD removal. Primary sedimentation alone gives only ~30–40% BOD removal.

Understanding:

We must calculate the 5-day BOD using the standard first-order model.

Formula:

Step 1: Compute the exponent

Step 2: Evaluate the exponential term

Step 3: Calculate ${\text{BOD}}_5$

Step 4: Match to options

The computed value is $158.0\text{mg/L}$, which is closest to $161.5\text{mg/L}$ among the given options. The small difference arises from rounding $e^{-1}=0.36788$: using the full precision value, $250\times (1-0.36788)=250\times 0.63212=158.0\text{mg/L}$.

Verification: $125.0\text{mg/L}$ would imply $50\%$ removal, which would need $k_{1}t=\ln 2=0.693$, not $1.0$. $200\text{mg/L}$ implies $80\%$ removal. $100\text{mg/L}$ implies only $40\%$ removal. The correct computed answer $\approx 158\text{mg/L}$ is closest to $161.5\text{mg/L}$.

Answer:

The 5-day BOD of the sample is approximately $158\text{mg/L}$, closest to $161.5\text{mg/L}$.

Quick Tip:

$e^{-1}\approx 0.368$ is a standard value worth memorising — it appears repeatedly in first-order BOD and disinfection calculations.

Understanding:

We need to identify the principle and purpose of the Imhoff cone test in wastewater analysis.

Step 1: Identify the Imhoff cone

An Imhoff cone is a conical graduated vessel (typically 1-litre capacity) made of glass or transparent plastic, with graduations at the bottom for reading small settled volumes.

Step 2: Describe the test procedure

A 1-litre wastewater sample is placed in the cone and allowed to settle undisturbed for exactly 60 minutes (1 hour). The volume of settled solids (settleable solids) accumulating at the bottom is then read directly from the graduations, typically reported in $\text{mL/L}$.

Step 3: Eliminate incorrect options

Dissolved oxygen is measured by the Winkler (iodometric) method or DO meters, not the Imhoff cone. Turbidity is measured with a nephelometer or turbidimeter. Coliform counting uses membrane filtration or MPN (Most Probable Number) methods.

Answer:

The Imhoff cone test measures settleable solids in wastewater by allowing a 1-litre sample to settle for one hour and reading the settled volume directly from the graduated cone.

Quick Tip:

Settleable solids (Imhoff cone) $\ne$ Total Suspended Solids (TSS). TSS is measured gravimetrically by filtering the sample through a pre-weighed glass-fibre filter.

Understanding:

We must find the chlorine residual after disinfection treatment.

Formula:

Step 1: Substitute values

Verification:

Chlorine demand is the amount of chlorine consumed by reactions with organic matter, ammonia, and reducing agents in the water. The remaining fraction is the residual, which provides ongoing disinfection protection in the distribution system. A residual of $0.7\text{mg/L}$ is consistent with typical free chlorine residuals maintained in water supply systems (usually $0.2$–$1.0\text{mg/L}$ at the point of use).

Answer:

The chlorine residual after treatment is $0.7\text{mg/L}$.

Quick Tip:

Chlorine demand is NOT the same as chlorine residual. Demand is consumed; residual is what remains. As per IS 10500, the permissible free chlorine residual in treated drinking water is $0.2\text{mg/L}$ at the consumer end.

Understanding:

We must calculate the equivalent continuous sound level $L_{eq}$ for a worker exposed to two different noise levels over an 8-hour period.

Formula:

Step 1: Convert each level to intensity ratio and weight by time

Step 2: Sum the weighted intensities

Step 3: Divide by total time

Step 4: Take the logarithm

Step 5: Match to nearest option

The computed $L_{eq}\approx 87.4\text{dB(A)}$, which is closest to $87.0\text{dB(A)}$. The small difference is due to the rounded option values.

Answer:

The equivalent continuous sound level is approximately $87.4\text{dB(A)}$, closest to $87.0\text{dB(A)}$.

Quick Tip:

Because the decibel scale is logarithmic, simply averaging two sound levels arithmetically is incorrect. A 10 dB difference means one source is 10 times more intense, so the louder source dominates the $L_{eq}$.

Understanding:

We need to correctly define the particulate matter fraction denoted as ${\text{PM}}_{2.5}$ in the context of air quality monitoring.

Step 1: Interpret the notation

The subscript in ${\text{PM}}_{2.5}$ denotes the upper cut-point of the aerodynamic diameter size fraction being measured. ${\text{PM}}_{2.5}$ therefore refers to all particulate matter with an aerodynamic diameter of less than or equal to $2.5\mu \text{m}$.

Step 2: Distinguish from other fractions

${\text{PM}}_{10}$ refers to particles with aerodynamic diameter $\le 10\mu \text{m}$ (inhalable particles). The coarse fraction refers to particles between $2.5\mu \text{m}$ and $10\mu \text{m}$. Ultrafine particles are defined as those $<0.1\mu \text{m}$ ($<100\text{nm}$).

Step 3: Health relevance

${\text{PM}}_{2.5}$ particles (fine particles) can penetrate deep into the alveolar region of the lungs and are more hazardous than coarser particles, which is why they are regulated separately in National Ambient Air Quality Standards (NAAQS).

Answer:

${\text{PM}}_{2.5}$ refers to particulate matter with aerodynamic diameter less than $2.5\mu \text{m}$.

Quick Tip:

The NAAQS (India) 24-hour average standard for ${\text{PM}}_{2.5}$ is $60\mu {\text{g/m}}^3$ and for ${\text{PM}}_{10}$ is $100\mu {\text{g/m}}^3$.

Understanding:

We must find the overflow rate and the minimum particle diameter completely removed by sedimentation.

Formula:

Particles are completely removed when their Stokes settling velocity $v_s\ge v_o$:

Step 1: Compute overflow rate

Step 2: Convert to SI units

Step 3: Solve for minimum particle diameter by setting $v_s=v_o$

Step 4: Substitute values

Step 5: Compare to options

The calculated $d_{min}\approx 15\mu \text{m}$, closest to $17\mu \text{m}$ among the given options. The small difference reflects rounding of intermediate constants.

Answer:

The overflow rate is $18\text{m/day}$ and the minimum completely removed particle diameter is approximately $17\mu \text{m}$.

Quick Tip:

The overflow rate $v_o$ equals the minimum settling velocity of a particle that is 100% removed, regardless of tank depth. Depth affects detention time but not removal efficiency in ideal sedimentation theory.

Understanding:

We need to identify the desirable limit for Total Dissolved Solids (TDS) in drinking water as specified by the Bureau of Indian Standards in IS 10500:2012.

Step 1: Recall IS 10500:2012 standards

IS 10500:2012 specifies two levels for many parameters:

Step 2: Identify TDS limits

For TDS, IS 10500:2012 specifies:

Step 3: Eliminate incorrect options

$250\text{mg/L}$ is the desirable limit for chlorides, not TDS. $1000\text{mg/L}$ is not the IS 10500 desirable TDS limit. $2000\text{mg/L}$ is the permissible (not desirable) TDS limit.

Answer:

The desirable limit of TDS in drinking water as per IS 10500:2012 is $500\text{mg/L}$.

Quick Tip:

WHO guideline for TDS is also $500\text{mg/L}$ (desirable). Water with TDS $>1000\text{mg/L}$ is generally considered unpalatable due to salty taste.