Advanced Chemical Engineering interview questions

125 questions with answers (CHE-301 to CHE-425). Try each one aloud before reading the answer.

Material & Energy Balances

CHE-301 · A reaction A → B has 25% single-pass conversion. Unreacted A is fully separated and recycled. For 100 mol/h fresh A, find the reactor feed and recycle rates.

Overall conversion is 100%, so 100 mol/h A reacts. Reactor feed F gives 0.25F = 100, so F = 400 mol/h. Recycle = 400 − 100 = 300 mol/h, a recycle ratio of 3.

CHE-302 · Fresh feed of 100 mol/h contains 0.5 mol% inert. If purge inert concentration must not exceed 5 mol%, what purge flow is needed?

At steady state, all inert leaves in the purge: 0.5 mol/h. Purge flow = 0.5/0.05 = 10 mol/h. The remaining reactants are lost with it, so purge composition is an economic trade-off.

CHE-303 · How would you calculate the adiabatic flame temperature of a fuel with excess air?

Write the combustion balance to get product moles. Set reactant enthalpy at feed temperature plus heat of combustion equal to product sensible enthalpy: Σnᵢ∫cₚ,ᵢdT from 298 K to T_ad. Solve iteratively with temperature-dependent cₚ, including dissociation at very high temperatures.

CHE-304 · A reactor converts 10 mol/s of A with ΔHᵣ = −100 kJ/mol. Feed and product temperatures are equal. What cooling duty is needed?

With equal inlet and outlet temperatures and negligible other effects, Q = ξ·ΔHᵣ = 10 × (−100) = −1000 kJ/s. About 1 MW must be removed.

CHE-305 · Propane is burned with 25% excess air. Find the flue gas composition per mol of propane.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. O₂ fed = 6.25 mol; N₂ = 6.25 × 79/21 = 23.5 mol. Flue gas: CO₂ 3, H₂O 4, O₂ 1.25, N₂ 23.5; total 31.75 mol (about 9.4% CO₂ wet).

CHE-306 · What is pinch analysis?

A heat-integration method that plots hot and cold composite curves, with a chosen minimum approach ΔT_min. The pinch is where the curves are closest, and the method sets minimum hot and cold utility targets before network design.

CHE-307 · What are the golden rules of pinch design?

Do not transfer heat across the pinch, do not use cold utility above the pinch, and do not use hot utility below it. Violating any rule raises both hot and cold utility by the amount of the violation.

CHE-308 · Plant flow and composition data do not satisfy mass balances. How do you handle it?

Use data reconciliation: adjust measurements within their uncertainties using weighted least squares subject to balance constraints. Then use gross-error detection to flag faulty meters, leaks or unmeasured streams before using the data for accounting or optimization.

CHE-309 · A well-mixed tank of volume V initially contains salt at C₀ and is fed pure water at flow v. How does concentration change?

Unsteady balance: V·dC/dt = −vC, so C(t) = C₀·e^(−t/τ), with τ = V/v. After one residence time, 36.8% of the initial concentration remains.

CHE-310 · A solid with 8% water (1000 kg/h) must be supplied at 2% water. A dryer removes all water from the stream it treats. What fraction must bypass the dryer?

Bypass fraction b carries 80b kg water. Product mass = 920 + 80b. 80b/(920 + 80b) = 0.02 gives 78.4b = 18.4, so b ≈ 0.235. About 23.5% bypasses the dryer.

CHE-311 · How do sequential-modular and equation-oriented process simulation differ?

Sequential-modular simulators (most commercial flowsheets) solve units one at a time and converge recycle loops by tearing streams. Equation-oriented simulators solve all equations simultaneously, which is faster for optimization and specification-heavy problems but harder to initialize and debug.

Chemical Engineering Thermodynamics

CHE-312 · How do you select a thermodynamic property model for a simulation?

Use cubic equations of state (Peng–Robinson, SRK) for hydrocarbons, gases and high pressure. Use activity models (NRTL, UNIQUAC, Wilson) for polar, non-ideal liquids at low to moderate pressure. Use electrolyte models for ionic systems and steam tables for water. Validate against experimental VLE data.

CHE-313 · A reaction at 298 K has ΔG° = −20 kJ/mol. What is its equilibrium constant?

K = exp(−ΔG°/RT) = exp(20,000/(8.314 × 298)) = exp(8.07) ≈ 3.2 × 10³. Products are strongly favored.

CHE-314 · How can Rankine cycle efficiency be improved?

Raise boiler pressure and superheat temperature, lower condenser pressure, use reheat between turbine stages (also reducing wet steam at the turbine exit) and use regenerative feedwater heating with bled steam.

CHE-315 · What is the Carnot COP of a refrigerator operating between −10 °C and 35 °C?

COP = T_c/(T_h − T_c) = 263/(308 − 263) = 263/45 ≈ 5.8. Real vapor-compression units achieve much less because of irreversibilities.

CHE-316 · Calculate the reversible adiabatic work to compress air (γ = 1.4) from 1 bar, 300 K to 4 bar.

W = [γ/(γ − 1)]·RT₁[(P₂/P₁)^((γ−1)/γ) − 1] = 3.5 × 8.314 × 300 × (4^0.2857 − 1) = 8730 × 0.486 ≈ 4.24 kJ/mol. Outlet temperature ≈ 300 × 1.486 ≈ 446 K.

CHE-317 · How is bubble pressure calculated with modified Raoult's law?

yᵢP = xᵢγᵢPᵢˢᵃᵗ. Summing yᵢ = 1 gives P_bubble = Σxᵢγᵢ(T, x)Pᵢˢᵃᵗ(T), then yᵢ = xᵢγᵢPᵢˢᵃᵗ/P. Bubble temperature requires iteration on T.

CHE-318 · How do Wilson, NRTL and UNIQUAC models differ?

All express excess Gibbs energy using binary interaction parameters fitted to data. Wilson cannot predict liquid–liquid immiscibility. NRTL adds a non-randomness parameter and handles LLE. UNIQUAC uses molecular size and surface area terms, and its group-contribution version UNIFAC can predict without data.

CHE-319 · How does pressure affect equilibrium conversion in a gas-phase reaction?

K = K_y·P^Δν (ideal gases). K is fixed by temperature, so if Δν < 0 (fewer moles of products, as in NH₃ synthesis), raising P raises K_y and conversion. If Δν > 0 (steam reforming), lower pressure favors conversion.

CHE-320 · What is exergy analysis and why is it useful?

Exergy is the maximum useful work obtainable as a stream comes to equilibrium with the environment. Exergy analysis locates where irreversibilities destroy work potential, such as in large ΔT heat transfer, throttling or combustion, pointing to the best targets for efficiency improvement.

CHE-321 · How is an isothermal flash calculation solved?

Using the Rachford–Rice equation Σzᵢ(Kᵢ − 1)/[1 + V/F·(Kᵢ − 1)] = 0 for the vapor fraction V/F, then xᵢ = zᵢ/[1 + V/F(Kᵢ − 1)] and yᵢ = Kᵢxᵢ. K-values are updated iteratively if they depend on composition.

CHE-322 · What is retrograde condensation?

In multicomponent mixtures near the critical region, lowering pressure at constant temperature can cause liquid to condense rather than vaporize. It occurs between the critical point and the cricondentherm and matters in gas-condensate reservoirs and pipelines.

Fluid Mechanics & Mechanical Operations

CHE-323 · Water flows at 2 m/s through a 100 m long, 0.1 m diameter pipe with Darcy friction factor 0.02. What is the pressure drop?

ΔP = f(L/D)(ρV²/2) = 0.02 × 1000 × (1000 × 4/2) = 40,000 Pa = 0.4 bar.

CHE-324 · What power is needed to pump 0.01 m³/s of water against a 50 m head with 70% efficiency?

P = ρgQH/η = 1000 × 9.81 × 0.01 × 50/0.7 ≈ 7.0 kW.

CHE-325 · Calculate NPSHa for water at 25 °C from an open tank at 101.3 kPa, with the liquid level 3 m above the pump and 1 m suction friction loss.

NPSHa = (P_atm − P_v)/(ρg) + z − h_f = (101.3 − 3.17) × 1000/9810 + 3 − 1 ≈ 10.0 + 2 = 12.0 m.

CHE-326 · When would you run centrifugal pumps in series or in parallel?

Series adds heads at the same flow, suiting high-head, steep system curves. Parallel adds flows at the same head, suiting flat system curves or variable demand. Pumps in parallel should have similar curves to avoid one being dead-headed.

CHE-327 · What are the main flow regimes in horizontal gas–liquid pipe flow?

Stratified, wavy, plug, slug, annular and dispersed bubble or mist flow, depending on gas and liquid superficial velocities and properties. Slug flow causes large pressure fluctuations and mechanical loads, so it is often avoided in design.

CHE-328 · What is water hammer and how is it estimated?

A pressure surge caused by sudden valve closure or pump trip. The Joukowsky equation gives ΔP = ρ·c·ΔV, where c is the pressure-wave speed (about 1000–1400 m/s in water-filled pipes). Stopping 2 m/s flow can add over 20 bar. Slow-closing valves and surge vessels mitigate it.

CHE-329 · What pressure drop occurs across a 1 m fluidized bed with voidage 0.45 and particle density 2500 kg/m³ in air?

At fluidization, ΔP = (1 − ε)(ρₚ − ρ_f)gL ≈ 0.55 × 2500 × 9.81 × 1 ≈ 13.5 kPa. This stays roughly constant as gas velocity increases beyond minimum fluidization.

CHE-330 · How do you scale up a constant-pressure filter from lab data?

Plot t/V against V from lab tests to get specific cake resistance α and medium resistance. Account for cake compressibility (α = α₀ΔPˢ), then size area for the plant cycle time, including washing and discharge time, and confirm with pilot tests on representative slurry.

CHE-331 · How is a control valve sized for liquid service?

Using the flow coefficient: Q = C_v√(ΔP/SG), with Q in US gpm and ΔP in psi (or K_v in metric units). Select a valve operating at about 20–80% open at design flow, and check for choked flow, cavitation and flashing.

CHE-332 · What criteria are used to scale up agitated vessels?

Keep geometric similarity, then choose a criterion based on the process: constant power per volume for mass transfer and dispersion, constant tip speed for shear-sensitive systems, constant blend time for fast mixing, or just-suspended speed for solids. Not all can be held at once.

CHE-333 · How would you select equipment to separate fine solids from a liquid?

Consider particle size, density difference, solids concentration, cake properties and required dryness. Options include thickeners for coarse dense solids, centrifuges such as decanter or disc-stack for fine or small density difference, filter presses for dry cake, and membranes for submicron particles. Lab settling and filtration tests guide the choice.

Heat Transfer

CHE-334 · A heat exchanger must transfer 500 kW with U = 500 W/m²·K and corrected mean temperature difference 40 K. What area is needed?

A = Q/(U·ΔT_m) = 500,000/(500 × 40) = 25 m². A margin is then added for fouling and uncertainty.

CHE-335 · What is the effectiveness of a condenser with NTU = 2?

For a condensing stream, C_r = 0, so ε = 1 − e^(−NTU) = 1 − e^(−2) ≈ 0.865. The cooling water reaches 86.5% of the maximum possible temperature rise.

CHE-336 · How do the Kern and Bell–Delaware methods differ for shell-side design?

Kern's method uses a simple equivalent diameter correlation and is quick but approximate. Bell–Delaware corrects an ideal crossflow coefficient for baffle leakage, bundle bypass, window flow and baffle spacing effects, giving much better accuracy. Stream analysis software is more rigorous still.

CHE-337 · A heat exchanger's outlet temperature has slowly drifted over months. How do you troubleshoot?

Calculate the current U from flows and temperatures and compare with clean and design values to quantify fouling. Check pressure drop for plugging, verify instrument accuracy and flows, and look for bypassing or air or non-condensables. Then plan cleaning, adjust velocity, or use antifoulants or water treatment.

CHE-338 · What causes flow-induced tube vibration in shell-and-tube exchangers?

Vortex shedding, turbulent buffeting, fluid-elastic instability (most damaging) and acoustic resonance in gas service. Excessive crossflow velocity and long unsupported spans are the main causes. They are prevented by more support plates, lower velocity, no-tubes-in-window designs or rod baffles.

CHE-339 · How do non-condensable gases affect a condenser?

They accumulate at the condensing surface, forming a gas film through which vapor must diffuse, sharply lowering the coefficient. A few percent of air can greatly reduce performance. Continuous venting and designing for vapor to sweep toward the vent are essential.

CHE-340 · What factors must be considered in designing a multiple-effect evaporator?

Boiling point elevation and hydrostatic head, which reduce available ΔT in each effect, feed arrangement (forward, backward, mixed) depending on viscosity and feed temperature, vapor–liquid separation, fouling, and the trade-off between steam economy and capital as effects are added.

CHE-341 · What net radiation is exchanged between 1 m² of a black surface at 1000 K and black surroundings at 300 K?

q = σA(T₁⁴ − T₂⁴) = 5.67 × 10⁻⁸ × (10¹² − 8.1 × 10⁹) ≈ 56 kW.

CHE-342 · How is a fired heater arranged?

The radiant section is where process tubes receive heat mainly by flame radiation, absorbing most of the duty. The convection section above recovers heat from flue gases by convection, often with finned tubes. The shield tubes, stack draft, burner arrangement and tube-skin temperature limits are key design issues.

CHE-343 · When would you choose an air cooler over a water cooler?

Air coolers suit sites where water is scarce or expensive, avoid water treatment and plume issues, and suit process outlet temperatures well above ambient. Water coolers are more compact and reach lower outlet temperatures. Air coolers are limited by summer dry-bulb temperature.

CHE-344 · How long does a 10 mm steel ball (ρ = 7800 kg/m³, c = 470 J/kg·K, k = 45 W/m·K) take to reach 63% of its temperature change in air with h = 100 W/m²·K?

V/A = D/6 = 1.67 mm, so Bi = 100 × 0.00167/45 ≈ 0.004 and lumped analysis applies. τ = ρcV/(hA) = 7800 × 470 × 0.00167/100 ≈ 61 s.

Mass Transfer

CHE-345 · A packed absorber has H_OG = 0.5 m and N_OG = 8. What packing height is needed?

Z = H_OG × N_OG = 0.5 × 8 = 4 m, plus allowances for distributors, redistribution and a safety margin.

CHE-346 · For dilute absorption with a very large absorption factor (negligible back pressure), what N_OG is required for 99% removal?

When mG/L → 0, N_OG = ln(y₁/y₂) = ln(100) ≈ 4.6 transfer units.

CHE-347 · How is constant-rate drying time calculated?

t_c = (L_s/(A·N_c))·(X₁ − X_c), where L_s is mass of dry solid, A drying area, N_c the constant drying flux, X₁ initial and X_c critical moisture content on a dry basis.

CHE-348 · A material with L_s/A = 20 kg/m² dries at a constant rate of 2 kg/m²·h from X = 0.40 to X_c = 0.15. How long does the constant-rate period last?

t = (20/2) × (0.40 − 0.15) = 10 × 0.25 = 2.5 h.

CHE-349 · When does Knudsen diffusion dominate over molecular diffusion in pores?

When the gas mean free path is larger than the pore diameter, at small pores or low pressure, so molecules collide more with walls than each other. Knudsen diffusivity D_K ∝ d_pore·√(T/M), independent of pressure.

CHE-350 · What are breakthrough curves and the mass transfer zone in fixed-bed adsorption?

The mass transfer zone is the region where adsorbent is partly loaded and moves through the bed. The breakthrough curve is the outlet concentration against time. Breakthrough occurs when the outlet reaches a set limit, and a narrow MTZ means better bed utilization.

CHE-351 · How do PSA and TSA differ?

Pressure Swing Adsorption adsorbs at high pressure and regenerates by lowering pressure, with short cycles of minutes (H₂ purification, N₂ or O₂ from air). Temperature Swing Adsorption regenerates by heating, with long cycles of hours, suiting strongly adsorbed species like water in gas drying.

CHE-352 · What trade-off limits polymer gas separation membranes?

Permeability and selectivity: more permeable polymers are usually less selective. This upper-bound trade-off (the Robeson plot) drives membrane material development and staged process designs.

CHE-353 · What is the osmotic pressure of 0.5 M NaCl solution at 25 °C?

van't Hoff: π = iCRT = 2 × 500 mol/m³ × 8.314 × 298 ≈ 2.48 × 10⁶ Pa ≈ 25 bar. Reverse osmosis must apply pressure well above this.

CHE-354 · How is a cooling tower designed?

Using Merkel's method: required tower characteristic KaV/L = ∫ dT/(h_s − h_a), integrated over the water range, where h_s is saturated-air enthalpy at water temperature and h_a is bulk air enthalpy. This is matched to fill performance data at the chosen L/G, approach and design wet-bulb.

CHE-355 · A solution of 60 kg solute in 100 kg water is cooled to where solubility is 30 kg per 100 kg water, without evaporation. What yield of anhydrous crystals is obtained?

Crystals = 60 − 30 = 30 kg, a 50% yield. Hydrated crystals or evaporation would change the calculation.

Chemical Reaction Engineering

CHE-356 · If a reaction rate doubles from 25 °C to 35 °C, what is its activation energy?

Arrhenius k = A·e^(−Ea/RT). ln 2 = (Ea/R)(1/298 − 1/308), so Ea = 0.693 × 8.314/(1.09 × 10⁻⁴) ≈ 53 kJ/mol.

CHE-357 · What are the main challenges in reactor scale-up?

Surface-to-volume ratio falls, so heat removal becomes limiting. Mixing and blend time worsen, mass transfer changes, residence time distribution changes, and safety margins shrink. Hot spots, selectivity loss and runaway risk require pilot data and dimensionless analysis.

CHE-358 · How is a Levenspiel plot used for reactor selection?

Plot F_A0/(−r_A) against X. A PFR volume is the area under the curve; a CSTR volume is the rectangle up to exit X. For a curve with a minimum (autocatalytic), a CSTR up to the minimum followed by a PFR minimizes total volume.

CHE-359 · For a second-order reaction at 80% conversion, compare CSTR and PFR sizes.

CSTR: kC_A0τ = X/(1 − X)² = 0.8/0.04 = 20. PFR: kC_A0τ = X/(1 − X) = 4. The CSTR needs five times the volume.

CHE-360 · Why can an exothermic CSTR have multiple steady states?

Heat generation against temperature is S-shaped (Arrhenius kinetics with conversion limit), while heat removal is a straight line. They can intersect at three points: low and high steady states are stable, the middle one is unstable. Ignition and extinction behavior and hysteresis follow.

CHE-361 · How does recycle affect a PFR's behavior?

A recycle ratio R makes the PFR behave between plug flow (R = 0) and a CSTR (R → ∞). Recycle is used with autocatalytic reactions, to dilute feed for temperature control or to improve wetting, at the cost of lower average reactant concentration.

CHE-362 · How is the tanks-in-series model fitted to RTD data?

From a tracer test, compute mean residence time t̄ and variance σ². The equivalent number of ideal CSTRs is N = t̄²/σ². N = 1 is a CSTR, and large N approaches plug flow. Conversion can then be predicted from N tanks.

CHE-363 · What reactor would you choose for a highly exothermic gas-phase catalytic oxidation?

A multitubular fixed-bed reactor cooled by boiling water or molten salt, with small tube diameters to limit hot spots, or a fluidized bed for near-isothermal operation. Diluted catalyst and controlled feed concentration below flammable limits are also used.

CHE-364 · How do you determine whether a catalytic reaction is mass-transfer limited?

Vary flow velocity at constant space time: if rate changes, external film diffusion matters. Vary particle size: if rate changes, internal pore diffusion matters. Also, a low apparent activation energy (around half the true value) suggests strong pore diffusion limitation. The Weisz–Prater criterion quantifies it.

CHE-365 · What is optimal temperature progression for a reversible exothermic reaction?

At low conversion, high temperature maximizes rate. As conversion rises, equilibrium limits, so temperature must fall. Multi-bed adiabatic reactors with interstage cooling, as in SO₂ oxidation and ammonia synthesis, approximate this optimal falling profile.

Separation Processes

CHE-366 · What is the McCabe–Thiele method?

A graphical method for binary distillation. It assumes constant molar overflow and steps off theoretical stages between the equilibrium curve and the rectifying and stripping operating lines, which meet on the q-line. It gives stage count, feed stage and minimum reflux.

CHE-367 · Using Fenske, find the minimum stages for α = 2.5, x_D = 0.95 and x_B = 0.05.

N_min = ln[(0.95/0.05)(0.95/0.05)]/ln 2.5 = ln(361)/0.916 = 5.89/0.916 ≈ 6.4 theoretical stages, including the reboiler.

CHE-368 · What is the Fenske–Underwood–Gilliland shortcut method?

Fenske gives minimum stages at total reflux, Underwood gives minimum reflux, and the Gilliland correlation relates actual stages to actual reflux. Kirkbride's equation then estimates feed location. It is used for preliminary multicomponent column design.

CHE-369 · Find the minimum reflux for a saturated liquid feed with z = 0.5, α = 2.5 and x_D = 0.95.

The pinch is at x = 0.5 on the q-line: y* = 2.5 × 0.5/(1 + 1.5 × 0.5) = 0.714. R_min = (x_D − y*)/(y* − x*) = (0.95 − 0.714)/(0.714 − 0.5) ≈ 1.10.

CHE-370 · How can distillation energy use be reduced?

Through heat integration such as feed–bottoms exchange, vapor recompression or heat pumps for close boilers, multi-effect (pressure-cascaded) columns, dividing-wall columns for three-product separations, side reboilers and condensers, and optimizing reflux and feed condition.

CHE-371 · What heuristics guide distillation column sequencing?

Remove corrosive or hazardous components first, do the easiest splits (largest α) first and the hardest last, remove the most plentiful component early, favor direct sequences (lightest overhead), and favor equimolar splits between top and bottom.

CHE-372 · Using Kremser, how many ideal stages are needed for 95% absorption with absorption factor A = 1.5 and solute-free solvent?

N = ln[(y₁/y₂)(1 − 1/A) + 1/A]/ln A = ln[20 × 0.333 + 0.667]/ln 1.5 = ln(7.33)/0.405 ≈ 4.9 stages.

CHE-373 · Column differential pressure is rising and product purity is falling. What do you suspect and do?

Likely incipient flooding, foaming or fouling of trays or packing. Check vapor and liquid loads against capacity, run a gamma scan, look for foaming agents, and verify reboiler duty and feed rate. Short-term, reduce boilup or feed; long-term, clean trays, add antifoam or revamp internals.

CHE-374 · How do you select a liquid–liquid extraction contactor?

Mixer-settlers suit few stages, high throughput and difficult phase separation. Agitated columns (rotating disc, Karr, Scheibel) give many stages compactly. Pulsed columns suit radioactive or corrosive service. Centrifugal extractors suit short contact time, small density difference or emulsions.

CHE-375 · Why is supercritical CO₂ used for extraction?

CO₂ becomes supercritical above about 31 °C and 73.8 bar, so solvent power is tunable with pressure and temperature. It is non-toxic and leaves no solvent residue, making it ideal for decaffeination, flavors and natural products. Solute is recovered simply by depressurizing.

Process Control

CHE-376 · A P-only controller with K_c = 4 controls a process with gain K_p = 0.5. What is the offset for a unit step setpoint change?

Offset = 1/(1 + K_cK_p) = 1/(1 + 2) ≈ 0.33, or 33% of the setpoint change. Adding integral action removes it.

CHE-377 · How does IMC or lambda tuning work for a first-order plus dead-time process?

The closed-loop time constant λ is chosen as the tuning knob. For a FOPDT process (K, τ, θ), a common PI form is K_c = τ/[K(λ + θ)] and τ_I = τ. Larger λ gives slower, more robust control.

CHE-378 · What is a Smith predictor?

A dead-time compensator in which a process model without delay predicts the undelayed output for the controller, while the delayed model removes the predicted effect from the measurement. It allows tighter tuning for dead-time-dominant loops but needs an accurate model.

CHE-379 · What is Model Predictive Control and where is it used?

MPC uses a dynamic process model to predict future outputs and solves an optimization at each interval to find manipulated-variable moves that track targets while respecting constraints. It suits multivariable, interacting, constrained units like distillation, FCC and crude units.

CHE-380 · A flow loop oscillates continuously. How do you tell valve stiction from poor tuning?

Put the loop in manual: if oscillation stops, it is a control problem. Stiction shows a sawtooth controller output and square-wave process variable, and step tests show sticky small moves. Tuning-caused oscillation is sinusoidal. Fix stiction with valve maintenance or a positioner, not detuning.

CHE-381 · How would you control temperature in an exothermic batch reactor?

Use cascade control: the reactor temperature master resets the jacket temperature slave, which manipulates heating and cooling via split range. Add feed-rate limits for semi-batch addition, high-temperature interlocks, emergency cooling and quench or dump systems.

CHE-382 · What is the Relative Gain Array?

λᵢⱼ = (open-loop gain with other loops open) ÷ (gain with other loops closed). Pair variables with relative gains close to 1. Avoid negative pairings, and expect strong interaction when values are large.

CHE-383 · What is the snowball effect in plantwide control?

In reactor–recycle systems, if all fresh feeds are flow-controlled and the reactor is fixed, a small feed or disturbance change can cause very large recycle flow changes. It is prevented by fixing a flow somewhere in the recycle loop and adjusting fresh feed to inventory.

CHE-384 · How is integral windup prevented?

By stopping integration when the output saturates (conditional integration), back-calculation that resets the integral to match the limited output, external reset feedback from the actual valve position in cascade or override schemes, and correct output clamping.

CHE-385 · Why does a valve's installed characteristic differ from its inherent characteristic?

As flow rises, line and equipment pressure drops increase and valve ΔP falls. An inherently linear valve then behaves like quick-opening, while an equal-percentage valve becomes nearly linear. That is why equal-percentage valves are chosen when the valve takes a small share of system ΔP.

Process Equipment Design

CHE-386 · What are the main steps in designing a shell-and-tube heat exchanger?

Define duty and properties; choose TEMA type and fluid allocation; estimate U and calculate area with F-corrected LMTD; choose tube size, length, pitch, passes and baffles; calculate film coefficients and pressure drops; iterate until U and ΔP meet limits; then check vibration and mechanical design.

CHE-387 · An exchanger needs 50 m² area using 25 mm OD tubes 4.88 m long. How many tubes are needed?

Area per tube = π × 0.025 × 4.88 ≈ 0.383 m². Tubes = 50/0.383 ≈ 131, rounded up to the nearest tube count from layout tables for the chosen shell and passes.

CHE-388 · How is a vessel under vacuum (external pressure) designed?

It fails by elastic buckling rather than yielding, so thickness depends on the length-to-diameter and diameter-to-thickness ratios, using code charts. Stiffening rings reduce effective length and allow a thinner shell. Full vacuum should be designed for whenever steam-out or cooling can cause it.

CHE-389 · Size the maximum vapor velocity in a vertical knock-out drum using K = 0.07 m/s, ρ_L = 800 kg/m³ and ρ_V = 5 kg/m³.

Souders–Brown: u_max = K√((ρ_L − ρ_V)/ρ_V) = 0.07 × √(795/5) ≈ 0.07 × 12.6 ≈ 0.88 m/s. Drum area = vapor volumetric flow ÷ u_max, and length is set by liquid holdup time.

CHE-390 · How do you choose a reboiler type?

Vertical thermosyphon for clean, moderate-duty services; horizontal thermosyphon for larger duties or low headroom; kettle for wide boiling ranges, vacuum, high vaporization fraction or when it should act as a stage; forced circulation for viscous or fouling fluids; internal bundle for small duties.

CHE-391 · What must be considered in relief valve sizing and selection?

Identify the governing relief scenario and load, phase of relieving fluid (gas, liquid or two-phase), set pressure and allowable overpressure, back pressure (conventional versus balanced bellows or pilot valves), and inlet and outlet line losses. API 520 and API 521 are the standard references.

CHE-392 · How is differential thermal expansion handled in fixed tubesheet exchangers?

Calculate the shell-to-tube expansion difference from metal temperatures. If tube or shell stresses or tube-to-tubesheet joint loads are excessive, add a shell expansion bellows or select a U-tube or floating head design.

CHE-393 · How do jackets, internal coils and external loops compare for reactor heat removal?

Jackets are simple and clean but have limited area, which shrinks relative to volume on scale-up. Internal coils add area but complicate cleaning and mixing. External pumped-loop exchangers give unlimited area and scalability but add hold-up outside the reactor and pump risks.

CHE-394 · What are key design considerations for large atmospheric storage tanks?

Shell course thickness by the one-foot or variable-design-point method for hydrostatic head, roof type, bottom and foundation settlement, wind and seismic stability, venting for filling, emptying and fire cases, corrosion, and secondary containment (bunds). API 650 is the common code for welded steel tanks.

CHE-395 · Why is piping flexibility analysis performed?

Hot pipes expand and can overload nozzles on pumps, compressors and vessels or overstress the pipe itself. Stress analysis checks sustained, thermal and occasional loads and adds loops, expansion joints, guides and spring supports to keep loads within allowable limits.

Process Safety & HAZOP

CHE-396 · What is HAZOP?

A Hazard and Operability study: a multidisciplinary team systematically reviews each P&ID node by applying guidewords (No, More, Less, Reverse, As well as, Part of, Other than) to process parameters to find deviations, their causes, consequences and safeguards, and to record actions.

CHE-397 · An initiating event occurs 0.5 times per year. The BPCS (PFD 0.1) and an alarm with operator response (PFD 0.1) are credited. With a target of 10⁻⁵ per year, what SIL is needed?

Mitigated frequency without SIF = 0.5 × 0.1 × 0.1 = 5 × 10⁻³/yr. Required SIF PFD = 10⁻⁵/(5 × 10⁻³) = 2 × 10⁻³, which falls in SIL 2 (10⁻³ to 10⁻²).

CHE-398 · What are the principles of inherently safer design?

Minimize (reduce hazardous inventories), Substitute (use less hazardous materials), Moderate (use milder conditions or dilution), and Simplify (design out opportunities for error). They remove hazards rather than adding safeguards.

CHE-399 · What lessons came from the Flixborough explosion?

In 1974, a temporary bypass pipe installed between cyclohexane oxidation reactors without proper design or review ruptured, and the vapor cloud explosion killed 28 people. Key lessons: rigorous management of change, engineering design for temporary modifications, minimized inventories and occupied buildings away from hazards.

CHE-400 · Estimate the LFL of a 50:50 mixture of methane (LFL 5%) and propane (LFL 2.1%).

Le Chatelier's rule: LFL_mix = 1/Σ(yᵢ/LFLᵢ) = 1/(0.5/5 + 0.5/2.1) = 1/(0.10 + 0.238) ≈ 3.0 vol%.

CHE-401 · How is the thermal runaway hazard of a reaction assessed?

Screen with DSC for decomposition energy and onset. Use reaction calorimetry for heat of reaction, heat release rate and accumulation, and adiabatic calorimetry (ARC, VSP) for adiabatic temperature rise, time to maximum rate and gas generation. Then compare maximum temperature of synthesis reaction with decomposition onset and boiling point.

CHE-402 · In a HAZOP of a pump discharge line, analyze the deviation "No Flow."

Causes: pump trip, closed or failed control valve, blocked strainer, empty suction vessel. Consequences: pump overheating or seal damage, downstream starvation, reverse flow. Safeguards: low-flow alarm and trip, minimum-flow recycle, check valve, low-level trip on suction. Actions are recorded if safeguards are inadequate.

CHE-403 · What are individual risk and societal risk in QRA?

Individual risk is the annual probability of fatality for a person at a given location, shown as contours. Societal risk relates frequency to number of fatalities, shown as an F–N curve, and reflects aversion to large-scale accidents.

CHE-404 · Why is two-phase relief sizing important for reactive systems?

In runaway or foaming reactions, the vessel contents can swell and vent as a two-phase mixture, which needs a much larger relief area than vapor-only flow. DIERS methods, using calorimetry data, size vents for tempered, gassy or hybrid systems.

CHE-405 · What are leading and lagging process safety indicators?

Lagging indicators measure past failures, such as loss of containment events and process safety incidents. Leading indicators measure the health of barriers before incidents, such as overdue inspections, bypassed trips, demand rates on safety systems and MOC compliance.

Industrial Chemical Processes & Petrochemicals

CHE-406 · What conditions are used in the Haber–Bosch process and why?

High pressure (about 150–300 bar) favors fewer gas moles (N₂ + 3H₂ ⇌ 2NH₃). Moderate temperature (about 400–500 °C) balances rate against an equilibrium that worsens with temperature. A promoted iron catalyst is used, with ammonia condensed out and unreacted gas recycled.

CHE-407 · What are typical steam methane reforming conditions?

Nickel catalyst in fired tubes at about 800–900 °C outlet and 20–30 bar, with a steam-to-carbon ratio around 3 to suppress coking. The strongly endothermic reaction is favored by high temperature and low pressure.

CHE-408 · Why is the FCC unit described as heat-balanced?

Burning coke from spent catalyst in the regenerator supplies heat that the hot regenerated catalyst carries to the riser to vaporize and crack the feed. Coke yield adjusts until heat produced equals heat required, so feed quality and operating changes shift the balance.

CHE-409 · What are key design features of an ethylene cracking furnace?

High coil outlet temperature (around 800–850 °C), very short residence time (fractions of a second) and low hydrocarbon partial pressure via steam dilution to maximize olefins. Rapid quenching in transfer-line exchangers stops secondary reactions. Coke build-up requires periodic decoking.

CHE-410 · How does a refinery manage sulfur?

Sulfur compounds are converted to H₂S by hydrotreating over Co–Mo or Ni–Mo catalysts. Amine units absorb H₂S from gas streams and sour water strippers remove it from water. The concentrated H₂S goes to Claus and tail-gas units to produce elemental sulfur.

CHE-411 · How much hydrogen is needed to make 1000 kg of ammonia?

N₂ + 3H₂ → 2NH₃. 1000/17.03 ≈ 58.7 kmol NH₃ needs 88.1 kmol H₂ ≈ 178 kg H₂ (and about 822 kg N₂), before purge and loss.

CHE-412 · How do mercury, diaphragm and membrane chlor-alkali cells compare?

Mercury cells give high-purity caustic but release mercury and are being phased out. Diaphragm cells use asbestos or polymer diaphragms and produce weak, salty caustic that needs evaporation. Membrane cells use ion-exchange membranes, giving high-purity caustic with the lowest energy use, and are now standard.

CHE-413 · What conditions favor NO₂ absorption in nitric acid production?

NO oxidation to NO₂ is favored by high pressure and low temperature, unusually for gas reactions, and absorption is exothermic. Absorption towers therefore run at elevated pressure with intensive cooling. Tail gas NOₓ is reduced by catalytic abatement.

CHE-414 · What makes bulk free-radical polymerization reactors difficult to design?

Viscosity rises by orders of magnitude, so mixing and heat transfer deteriorate. Polymerization is highly exothermic, and the gel (Trommsdorff) effect sharply accelerates rate as termination slows. Solution, suspension or emulsion processes and limited conversion are used to keep control.

CHE-415 · How does hydrogen production by water electrolysis compare with steam methane reforming?

SMR is cheaper at scale but emits CO₂ unless combined with carbon capture. Electrolysis (alkaline, PEM, solid oxide) produces zero-direct-emission hydrogen with renewable power. Theoretical energy is about 39.4 kWh/kg H₂ (HHV), with real systems needing more, so electricity cost dominates economics.

Process Economics & Plant Design

CHE-416 · A project costs $1000 and returns $300 a year for 5 years. What is its NPV at 10%?

Annuity factor = [1 − 1.1⁻⁵]/0.1 ≈ 3.791. PV = 300 × 3.791 ≈ $1137. NPV ≈ 1137 − 1000 = $137, so the project is acceptable.

CHE-417 · How is the economic optimum pipe diameter determined?

Express annual cost as capital charges (pipe, fittings, installation, which rise with D) plus pumping energy (which falls roughly with D⁻⁵ for turbulent flow). Differentiate total cost with respect to D or use optimum-velocity correlations, then round to standard pipe size.

CHE-418 · How is the optimum number of effects in an evaporator chosen?

Each added effect cuts steam use roughly in proportion but adds capital cost, and available ΔT per effect shrinks, increasing area. The optimum minimizes total annual cost, with higher steam cost and larger capacity favoring more effects or mechanical vapor recompression.

CHE-419 · How do Lang, Hand and module factor methods differ?

Lang applies one factor to total equipment cost. Hand applies separate factors by equipment type. Guthrie's module method applies factors for each equipment item to account for installation, piping, instruments and materials, then adds contingency and fees, giving better accuracy at the study stage.

CHE-420 · What is discounted payback period?

The time for cumulative discounted cash flows to recover the investment. It is longer than simple payback and accounts for time value but still ignores cash flows after payback.

CHE-421 · Project A has higher NPV but lower IRR than project B. Which should you choose if they are mutually exclusive?

Choose A, provided the discount rate reflects the true cost of capital. NPV measures absolute value created. IRR can mislead for projects of different scale or timing, and when cash flows change sign more than once.

CHE-422 · What is the capital recovery factor for 10% interest over 10 years?

CRF = i(1 + i)ⁿ/[(1 + i)ⁿ − 1] = 0.1 × 2.594/1.594 ≈ 0.163. A $1 million investment is equivalent to about $163,000 per year.

CHE-423 · How is safety integrated into plant layout?

By using consequence modeling to set separation distances between hazardous units, control rooms and boundaries; grouping by hazard; placing ignition sources upwind of likely releases; preventing domino effects; and providing access, escape routes, drainage and blast-resistant buildings where needed.

CHE-424 · How would you evaluate a debottlenecking project economically?

Identify the limiting equipment by test runs or simulation, estimate the capacity gain and incremental cost for each option, and calculate incremental NPV and IRR using extra margin on additional sales. Account for downtime for tie-ins, the next bottleneck and market capacity to absorb product.

CHE-425 · How do Monte Carlo analysis and tornado diagrams improve project evaluation?

A tornado diagram ranks variables by their effect on NPV when each is varied across its range. Monte Carlo samples all uncertain inputs from probability distributions simultaneously, producing an NPV distribution and probability of loss instead of a single deterministic number.

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