Chemical Engineering - MCQ Practice Questions

Understanding:

We must identify the correct expression for the effective mean temperature difference used in shell-and-tube heat exchanger design when the arrangement is not purely countercurrent.

Formula:

The standard design equation for a heat exchanger is:

where the corrected mean temperature difference is:

Step 1: Role of the LMTD

The Log Mean Temperature Difference (LMTD) is first calculated assuming purely countercurrent flow between the same terminal temperatures.

Step 2: Role of the correction factor $F$

For multi-pass arrangements (e.g., 1-2, 2-4 exchangers), the actual mean driving force is less than the countercurrent LMTD. The dimensionless correction factor $F$ ($0<F\le 1$) accounts for the departure from pure countercurrent flow. It is obtained from standard charts as a function of two dimensionless ratios $P$ and $R$.

Step 3: Corrected mean temperature difference

The effective temperature difference used in the design equation is:

The LMTD is always calculated on the countercurrent basis regardless of the actual flow arrangement; $F$ then corrects it downward.

Answer:

The corrected mean temperature difference is $F$ multiplied by the countercurrent LMTD.

Quick Tip:

A value of $F<0.75$ is generally considered thermodynamically inefficient; the designer should reconsider the number of shell passes or flow arrangement.

Understanding:

We seek the condition that defines the optimum insulation thickness by minimising total annual cost.

Formula:

At the minimum of total cost:

Step 1: Differentiate total cost

Step 2: Interpret the condition

Since $C_i$ increases with $x$, $\frac{d{C}_i}{dx}>0$, and since $C_h$ decreases with $x$, $\frac{d{C}_h}{dx}<0$. At the optimum, these two slopes are equal in magnitude but opposite in sign, so their sum is zero. This is NOT the same as $C_i=C_h$ (which is a common misconception).

Step 3: Confirm it is a minimum

The second derivative test on $C_T$ confirms a minimum when the curvature is positive, which is the case for the typical cost curves in insulation problems.

Answer:

The optimum thickness satisfies the condition that the sum of the derivatives equals zero.

Quick Tip:

The condition $C_i=C_h$ gives the intersection of the two cost curves, not the minimum of their sum. Always differentiate $C_T$ to find the true optimum.

Understanding:

We must find the minimum height of a cylindrical tank that holds a given volume.

Formula:

Volume of a vertical cylinder:

Step 1: Solve for $H$

Step 2: Round to nearest 0.1 m

Verification:

Answer:

The minimum height of the cylindrical storage tank is approximately 7.9 m.

Quick Tip:

A common error is using diameter instead of radius in $\pi r^2$, which would give $H=100/(\pi \times 16)\approx 1.99\text{m}$ — always halve the diameter first.

Understanding:

We derive how the optimum economic pipe diameter scales with flow rate.

Formula:

For turbulent flow, pumping cost $\propto Q^a{D}^{-m}$ and capital cost $\propto D^n$. Total cost:

Step 1: Minimise with respect to $D$

Step 2: Solve for $D^{\ast }$

For turbulent flow in a smooth pipe, the friction factor gives $a=2$ in the pumping cost exponent (since power $\propto Q\cdot \Delta P\propto Q^3{D}^{-5}$, and capital cost $\propto D^n$), yielding:

Step 3: Interpretation

This is the standard result in plant design for economic pipe sizing. For typical values $m\approx 5$ and $n\approx 1$, $D^{\ast }\propto Q^{1/3}$, consistent with the widely used rule $D_{\text{opt}}\propto Q^{0.35\text{–}0.45}$.

Answer:

The optimum economic pipe diameter scales with flow rate as:

Quick Tip:

The Peters and Timmerhaus correlation for optimum economic diameter ($D_{\text{opt}}\approx 0.363Q^{0.45}{\rho }^{0.13}$ in SI) is derived from exactly this minimisation.

Understanding:

We must identify the correct thin-wall formula for hoop (circumferential) stress in a cylindrical pressure vessel.

Formula:

From equilibrium of a half-cylinder of unit length:

Step 1: Derive by equilibrium

Consider a free-body diagram of a unit length of a half-cylinder. The bursting force due to pressure acting on the projected area $2r\times 1$ is:

This is resisted by two wall cross-sections, each of area $t\times 1$:

Step 2: Equate forces

Step 3: Contrast with longitudinal stress

The longitudinal (axial) stress is half the hoop stress:

Hoop stress is therefore the governing stress for cylindrical vessel design, which is why the wall thickness is sized against it.

Answer:

The hoop stress in a thin-walled cylindrical pressure vessel is:

Quick Tip:

A common trap is confusing hoop stress $Pr/t$ with longitudinal stress $Pr/2t$. Remember: hoop stress is always twice the longitudinal stress in a cylinder, making it the design-limiting stress.

Understanding:

We must state the six-tenths rule for scaling equipment capital cost with capacity in plant design.

Formula:

The six-tenths (power-law or economy-of-scale) rule:

where the exponent $n=0.6$ for most chemical plant equipment.

Step 1: Origin of the exponent

Capital cost scales roughly as surface area (for vessels) while capacity scales as volume. Since area $\propto V^{2/3}\approx V^{0.667}$, the empirical exponent rounds to 0.6. Extensive plant data confirm this value.

Step 2: Apply the rule

For a capacity ratio $S_2/S_1$:

Step 3: Validity

The six-tenths rule is most reliable when $0.1\le S_2/S_1\le 10$. Outside this range, more detailed cost correlations (e.g., Guthrie or Ulrich method) are recommended.

Answer:

The six-tenths rule gives the scaled equipment cost as:

Quick Tip:

For gas compressors the exponent is closer to 0.82, and for furnaces it is about 0.77. The default exponent of 0.6 applies broadly to vessels, heat exchangers, and distillation columns.

Understanding:

We must identify the slope and orientation of the $q$-line on the McCabe–Thiele diagram for a saturated liquid (bubble-point) feed.

Formula:

The $q$-line equation is:

where $q$ is the liquid fraction parameter and $z_F$ is the feed composition.

Step 1: Value of $q$ for saturated liquid feed

For a feed that is a saturated liquid (bubble-point liquid), all the feed enters as liquid:

Step 2: Slope of the $q$-line

An infinite slope corresponds to a vertical line on the $x$–$y$ diagram.

Step 3: Summary of $q$ values

Answer:

For a saturated liquid feed, $q=1$, giving the $q$-line an infinite slope, i.e., the line is vertical passing through $x=z_F$.

Quick Tip:

Memorising the five feed conditions with their $q$ values and corresponding $q$-line orientations is essential for McCabe–Thiele problems in GATE Chemical Engineering.

Understanding:

We must calculate the payback period for a plant investment.

Formula:

Step 1: Calculate the denominator

Step 2: Calculate PBP

Verification:

An investment of \5\text{ M}$recoveredat$\$1\text{ M/yr}$ clearly takes exactly 5 years. $✓$

Answer:

The payback period for this plant is 5 years.

Quick Tip:

Depreciation is added back to net profit because it is a non-cash expense — the cash is still available to recover the investment. Using net profit alone (giving PBP $=6.25$ yr) is a very common exam trap.

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).