Three areas share this page: Process Design with 7–11 questions, Process Control with 4–6 and Materials Science with 4–6. Together they ask whether you can turn a flowsheet into equipment that is sized, priced, controlled and built from the right metal.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam7–11
problems on this page18
of 17 knowledge areas14
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What the specification covers
Process Design lists five subtopics. In our words:
Process flow diagrams and P&IDs: reading streams, equipment and instrument symbols.
Equipment selection: sizing and scale-up.
Cost estimation: cost indices and equipment costing.
Design and optimization: efficiency, sustainability, green engineering, inherently safer design, evaluating specifications.
Design standards: regulatory, ASTM, ISO, OSHA.
18 design, control and materials problems
0 / 18 answered · 0 correct
Question 1 of 18 · Process Control · 4–6 items
In the design of a cascade control system, what is the most critical characteristic of the secondary (slave) loop variable relative to the primary variable?
Solution
Cascade: primary (master) controller output = set point of the secondary (slave) loop. The secondary must be faster (typically ≥ 3–5 × the primary's speed) and catch disturbances entering it early (e.g. steam pressure in a jacket-temperature loop).
Low noise helps any loop, but it is not what makes cascade work; speed of the inner loop is.
A slower secondary inverts the requirement: the inner loop could not settle before the primary controller moves its set point.
Measurement cost is a practical concern, but a cheap slow secondary would make cascade worse than single-loop control.
The inner loop must respond several times faster than the outer one, so it rejects its own disturbances before they reach the primary variable.
Question 2 of 18 · Process Control · 4–6 items
A chemical reactor's temperature control loop becomes highly unstable after a piping modification relocates the sensor downstream. A new step test reveals a transport delay of 45 s and a process time constant of 20 s. The original PI controller settings, which modeled the response as a simple first-order lag without delay, were retained. What is the fundamental cause of this new instability?
Solution
FOPDT: G(s) = Ke−θs/(τs + 1). With θ/τ = 45/20 = 2.25 the delay dominates; dead time adds phase lag −ωθ, so the ultimate gain falls and the old PI gain exceeds it. Retune for θ (e.g. IMC or Ziegler-Nichols on the new step test).
The measured time constant (20 s) is not the issue; the new element is the 45 s transport delay the model lacks.
The 45 s transport delay now exceeds the 20 s time constant; PI settings tuned for zero dead time are far too aggressive and the phase lag drives the loop unstable.
Drift produces a slow offset, not loop instability; the step test shows a delay, not a biased signal.
A long integral time means weaker integral action, which makes a loop more stable, not less.
Too little gain gives a sluggish, stable loop; instability comes from too much gain for the dead time present.
Question 3 of 18 · Process Control · 4–6 items
A flow loop uses a series (ideal) PI controller governed by the equation Gc(s) = Kc (1 + 1/(τI s)). The required tuning parameters were determined to be proportional gain Kc = 2.0 and integral time τI = 0.5 minutes. However, the Distributed Control System (DCS) for this unit employs a parallel PI algorithm defined as GDCS(s) = Kp + Ki/s, and requires the integral gain Ki to be entered in units of s−1. What is the correct value of Ki that must be entered into the DCS to achieve identical dynamic performance?
60.0 multiplies Kc·τI by 60, two errors at once; integral gain is a gain divided by a time.
4.00 is Kc/τI in min−1; the DCS needs it per second (÷ 60).
1.00 multiplies Kc by τI instead of dividing.
Ki = Kc/τI = 2.0/(0.5 min × 60 s/min) = 2.0/30 s = 0.0667 s−1.
0.0333 s−1 is 1/τI alone; the parallel integral gain also carries Kc.
Question 4 of 18 · Process Control · 4–6 items
A process engineer is tuning a temperature control loop and analyzes open-loop step response data to identify a First-Order Plus Dead Time (FOPDT) model. At t = 0 minutes, the manipulated variable (MV) undergoes a unit step change of 10%. The initial steady-state process variable (PV) is 100.0°C. The PV begins to respond exactly at t = 2.0 minutes and reaches 106.32°C at t = 7.0 minutes. The final steady-state PV reached after a long time is 110.0°C. Using the identified FOPDT model, what is the predicted PV at exactly t = 10.0 minutes?
Given
MV step = 10%
Initial PV = 100.0°C
PV starts responding at t = 2.0 min
PV = 106.32°C at t = 7.0 min
Final PV = 110.0°C
Find
PV at t = 10.0 min (°C)
Solution
K = ΔPV/ΔMV = (110.0 − 100.0)/10 % = 1.0 °C/% · θ = 2.0 min
(106.32 − 100)/10 = 0.632 at t − θ = 5.0 min → τ = 5.0 min
108.65 °C ignores the dead time and uses 10/5 in the exponent.
K = 10 °C/10 % = 1 °C/%, θ = 2 min, and 6.32/10 = 63.2 % at 5 min after the delay gives τ = 5 min; PV(10) = 100 + 10(1 − e−8/5) = 107.98 °C.
107.60 °C lumps the dead time into a 7-minute time constant with no delay (1 − e−10/7).
106.99 °C subtracts the dead time twice, using (10 − 4)/5 in the exponent.
106.81 °C reads the time constant as 7 min (the 63.2 % point measured from t = 0); the dead time must be subtracted first, τ = 7 − 2 = 5 min.
Question 5 of 18 · Materials Science · 4–6 items
An engineer selects an amorphous polymer piping rated with a dry glass transition temperature (Tg) of 95 °C for a process stream operating continuously at 85 °C. The fluid is a mixture of water and methyl ethyl ketone (MEK). Which statement best evaluates the long-term reliability of this design choice?
Solution
Rated Tg = 95 °C applies to the dry polymer. Absorbed solvent (MEK) acts as a plasticizer and lowers Tg; at 85 °C the wet polymer may be near or above its effective Tg and lose stiffness.
Hydrolysis attacks particular chemistries (polyesters, polyamides), not every amorphous polymer; the ketone is the real hazard.
Amorphous thermoplastics are typically the most vulnerable class to ketones, not the most resistant.
MEK swells and plasticizes many amorphous polymers, which lowers the effective Tg, possibly below 85 °C, so the pipe can soften and creep.
Solvents can induce some crystallization in a few polymers, but the general and immediate threat here is plasticization.
The 95 °C rating is the dry Tg; the solvent lowers it, so the 10 °C margin is not safe.
Question 6 of 18 · Materials Science · 4–6 items
A shell-and-tube heat exchanger will cool a process stream at about 30 °C that contains elevated concentrations of aqueous chlorides and dissolved oxygen. Which of the following materials is the most appropriate to specify to limit localized pitting corrosion?
Solution
Pitting resistance rises with Cr, Mo and N (PREN = %Cr + 3.3·%Mo + 16·%N). 316 (≈ 2–3 % Mo) > 304 > 430/410 > carbon steel. At high temperature, chloride stress-corrosion cracking of austenitics becomes the governing risk instead.
The 2–3 % molybdenum in 316 raises its pitting resistance well above 304, which makes it the best of these options for aerated chloride water.
Martensitic grades (e.g. 410) have low chromium and no molybdenum and pit readily in chlorides.
Plain 430 ferritic steel has no molybdenum and pits more easily than 304 in aerated chlorides.
Carbon steel corrodes generally in aerated water; it has no passive film to protect.
304 has no molybdenum and is prone to chloride pitting; 316 is the usual upgrade.
Question 7 of 18 · Materials Science · 4–6 items
A process engineering team explicitly rejects the use of 316L Stainless Steel for a new process line operating at 150°C carrying a concentrated aqueous chloride solution. What is the most accurate primary technical rationale for discarding this material option in this specific service scenario?
Solution
Chloride SCC needs a susceptible alloy (austenitic 300 series), tensile stress and hot chloride solution (> ≈ 60 °C). Duplex, super-austenitic or nickel alloys are used instead.
316L's passive film gives low uniform corrosion; chlorides attack it locally (pits, cracks), not uniformly.
Austenitic steel does expand more than carbon steel, but that is handled by flexibility design and is not why 316L is rejected here.
Austenitic stainless steels such as 316L crack under tensile stress in hot chloride solutions, typically above about 60 °C; 150 °C concentrated chloride is a classic SCC service.
Creep matters in the several-hundred-degree range; 150 °C is far too low for 316L to creep.
Austenitic (fcc) steels are among the least susceptible to hydrogen embrittlement.
Question 8 of 18 · Materials Science · 4–6 items
An engineer must select the primary piping material for a reactor effluent line operating at 650 °C that transports a highly concentrated, corrosive chloride gas. Utilizing the material properties data below, determine which alloy satisfies both the mechanical and chemical constraints of this service.
Alloy material
Maximum operating temperature
Chloride pitting resistance
Carbon steel
400 °C
Poor
Type 304 stainless steel
870 °C
Poor
Inconel 625
980 °C
Excellent
Monel 400
480 °C
Good
Solution
Screen both constraints: Tmax > 650 °C → 304 SS, Inconel 625; chloride resistance → Inconel 625 (excellent), Monel 400 (good, but 480 °C). Only Inconel 625 passes both.
304 is rated to 870 °C, but its chloride pitting resistance is poor.
Only Inconel 625 combines a rating above 650 °C (980 °C) with excellent chloride resistance.
Monel resists chlorides well but is rated only to 480 °C, below the 650 °C service.
Carbon steel fails both tests: its limit is 400 °C and its chloride resistance is poor.
Question 9 of 18 · Process Design · 7–11 items
NPS Designator
Internal Diameter (mm)
NPS 6
154.1
NPS 8
202.7
NPS 10
254.5
NPS 12
304.8
A chemical plant is designing a new pipeline to transport liquid water at a mass flow rate of 150,000 kg/hr. The fluid density is 1000 kg m−3. To minimize erosion and pressure drop, the engineering guidelines require the superficial velocity to not exceed 1.5 m/s. Using the provided piping dimensional data, determine the smallest nominal pipe size that satisfies this design velocity constraint.
Dmin = √(4Q/(πvmax)) = √(4 × 0.0417/(π × 1.5)) = 0.188 m → smallest ID above it: NPS 8 (202.7 mm, v = 1.29 m/s)
Q = 150,000/3,600/1,000 = 0.0417 m3/s; in NPS 8 (202.7 mm) v = 0.0417/0.0323 = 1.29 m/s ≤ 1.5 m/s, while NPS 6 gives 2.23 m/s.
NPS 12 (0.57 m/s) satisfies the limit, but it is not the smallest size that does.
NPS 10 (0.82 m/s) works but is one size larger than needed.
NPS 6 (154.1 mm) gives 2.23 m/s, above the 1.5 m/s limit.
Question 10 of 18 · Process Design · 7–11 items
Select the correct fundamental formulation for determining the required design wall thickness t of a cylindrical thin-wall pressure vessel, given the internal pressure P, internal radius r, allowable material stress σallow, and a specified linear corrosion allowance CA.
Solution
σhoop = P·r/t → t = P·r/σallow + CA
(Design codes add joint efficiency and a pressure correction, e.g. t = PR/(SE − 0.6P) + CA.) FE Reference Handbook, Mechanics of Materials: thin-walled cylinders.
Subtracting the allowance leaves the vessel thinner than the pressure requires once corrosion has occurred.
Multiplying by CA makes the thickness depend on the allowance's units and vanish when CA = 0; the allowance is additive.
Adding 1 to a length and multiplying mixes a dimensionless factor with a thickness; CA is a length added to t.
Hoop stress σ = Pr/t gives the pressure thickness Pr/σallow; the corrosion allowance is metal that will be lost, so it is added on top.
Using the diameter without the factor ½ doubles the hoop-stress thickness (t = PD/2σ).
Question 11 of 18 · Process Design · 7–11 items
A chemical plant installed in 2010 cost $4.5 million. Using the Chemical Engineering Plant Cost Index (CEPCI) values of 550.8 for 2010 and 657.5 for 2022, what would be the approximate cost to build the same plant in 2022?
Given
Plant cost in 2010 = $4.5 million
CEPCI 2010 = 550.8
CEPCI 2022 = 657.5
Find
Plant cost in 2022 ($ million)
Solution
C2 = C1·(I2/I1) = $4.5 million × (657.5/550.8) = $5.37 million
$5.00 million applies the six-tenths exponent to the index ratio (4.5 × 1.1940.6); that exponent is for capacity, not for time.
$4.50 million ignores inflation between 2010 and 2022.
$3.77 million inverts the index ratio (550.8/657.5), de-escalating the cost instead of bringing it forward.
Question 12 of 18 · Process Design · 7–11 items
A process facility audits its pipe sizing methodologies to ensure line velocities fall within appropriate limits for specific fluid phases. Based on the operational data below, which process stream is most likely over-sized due to incorrectly applying a liquid velocity constraint to a highly compressible fluid?
Stream Identifier
Fluid Phase
Current Velocity (m/s)
Stream I
Cooling Water
1.5
Stream II
Superheated Steam
1.2
Stream III
Two-Phase Hydrocarbon
18.0
Stream IV
Natural Gas
22.0
Solution
Typical design velocities: liquids ≈ 1–3 m/s, gases ≈ 15–30 m/s, steam ≈ 20–40 m/s. Steam at 1.2 m/s points to a liquid criterion applied to a compressible fluid (an oversized, costly line).
1.5 m/s is a normal velocity for cooling water, a liquid.
18 m/s is a reasonable velocity for a two-phase hydrocarbon line; it does not look like a liquid limit misapplied.
22 m/s is within the usual range for a gas line, so it shows no sign of a liquid criterion.
1.2 m/s is a liquid-line velocity; steam lines are normally sized for roughly 20–40 m/s, so this line is grossly oversized.
Question 13 of 18 · Process Design · 7–11 items
A newly installed high-alloy piping system operating at 300°C experiences a sudden loss of containment at a flanged joint. Post-incident analysis confirms that both the main pipe and the flange metallurgy were fully compatible with the process fluid and correctly rated for the extreme temperature and pressure. Which of the following is the most likely root cause of the containment failure?
Solution
Materials selection covers every wetted and load-bearing component: gaskets, seals, packing and bolting must also suit the temperature and fluid. Elastomer or fiber gaskets have temperature limits well below metal ratings.
With pipe and flanges rated and compatible, the gasket is the component left; a gasket not rated for 300 °C degrades and loses its seal.
The metallurgy was confirmed compatible with the fluid, and the leak is at the flanged joint, not at a weld.
The pipe was rated for the temperature; creep would show as deformation of the pipe, not a sudden flange leak.
Galvanic attack needs dissimilar metals in an electrolyte; the metallurgy was confirmed compatible.
Nothing in the case describes cycling, and fatigue cracks the metal over time rather than causing a sudden joint leak.
Question 14 of 18 · Process Design · 7–11 items
What is the purpose of a Lang factor in process design?
Control stability is a dynamics question (gain and phase margins), unrelated to the Lang factor.
Lang factor × delivered cost of major equipment ≈ fixed capital investment (about 3.1 solids, 3.6 solid-fluid, 4.7 fluid plants).
Exchanger thermal performance is effectiveness or LMTD; the Lang factor is a cost multiplier.
Wall thickness comes from pressure design (hoop stress plus corrosion allowance), not from cost factors.
Question 15 of 18 · Process Design · 7–11 items
In utilizing the six-tenths rule for equipment cost estimation during process design, what underlying assumption about economies of scale is being made?
A constant cost would be an exponent of 0; larger equipment does cost more.
The relation is a power law with exponent 0.6, not an exponential function.
C2/C1 = (Q2/Q1)0.6: doubling capacity raises cost by only about 52 %, an economy of scale.
Linear scaling would be an exponent of 1; the six-tenths rule uses 0.6.
Question 16 of 18 · Process Design · 7–11 items
A process engineer is responsible for sizing a continuous stirred-tank reactor (CSTR) and its associated temporary storage system. The engineer sizes the total volume of all vessels precisely to match the nominal liquid holdup volume dictated by the steady-state process kinetics, without incorporating standard freeboard space. Which of the following is the most likely consequence of neglecting this ullage space in the dynamic system design?
Solution
Freeboard (ullage) absorbs level swings during dynamic operation, foaming, thermal expansion and vapor disengagement; vessels are sized above the working holdup (often 10–20 % extra volume).
Lack of vapor space makes pressure rise faster on thermal expansion or gas evolution, not fall.
Vessel volume above the liquid holds no liquid, so it has no effect on reactant concentration.
Vessel freeboard does not set the suction-line velocity, which depends on flow and pipe size.
Missing headspace leaves no room for vapor to disengage; it does not cause early separation.
With no freeboard, any level surge, foaming or upset fills the vessel and liquid is carried into the vent or relief system.
Question 17 of 18 · Process Design · 7–11 items
When designing a chemical process for a batch reactor, what is the primary advantage compared to a continuous stirred tank reactor?
Solution
Batch: flexible, multi-product, high conversion per volume, but higher labor cost and down time. Continuous (CSTR/PFR): high throughput, steady quality, low labor per unit.
Batch operation needs more labor per unit of product (charging, emptying, cleaning), so labor cost is higher, not lower.
Batch heat transfer is limited by jacket area and changes during the cycle; it is not better than a CSTR.
A batch reactor can be cleaned and switched between recipes, which suits small-volume, multi-product production (pharmaceuticals, specialty chemicals).
Down time between batches lowers throughput; continuous reactors give higher production rates.
Question 18 of 18 · Process Control · 4–6 items
A sensitive thermal process is approximated as a first-order-plus-dead-time (FOPDT) system with time constant τ = 20 min and dead time θ = 2 min. To keep operation safe, the control engineer uses lambda tuning instead of Ziegler-Nichols, choosing a closed-loop time constant λ = 20 min. What is the main advantage of lambda tuning here?
Solution
lambda (IMC) PI for G = Ke−θs/(τs + 1): Kc = τ/[K(λ + θ)], τI = τ
The closed loop then follows a first-order response with time constant λ (no overshoot); larger λ = more robust, slower. Ziegler-Nichols targets quarter-amplitude decay, which overshoots.
Zero offset comes from the integral action in any PI controller; lambda tuning does not decouple τ from θ.
With λ equal to the open-loop time constant the loop is deliberately no faster than the process; Ziegler-Nichols settings are the aggressive ones.
Lambda tuning of a FOPDT process gives a PI controller; it neither uses aggressive derivative action nor removes dead time (only a predictor can).
Minimum-IAE tunings are more aggressive; lambda tuning trades speed for robustness and a predictable, non-oscillatory response.
Lambda (IMC) tuning sets Kc and τI so the closed loop behaves like a first-order lag with time constant λ; with λ = τ = 20 min the response is smooth and does not overshoot.
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Common traps
Time and capacity in one step. A cost index moves a price through time; a capacity exponent moves it across sizes. When both differ, apply both, one after the other.
Scale-up by volume alone. Geometric similarity keeps shape ratios; heat-transfer area per unit volume falls as a vessel grows, which is why a jacket that cooled the pilot reactor may not cool the plant one.
Symbols read as words. A P&ID shows instrument function by letter code (first letter the measured variable, the next ones the function); read the code before the line type.
Sources: Peters, Timmerhaus and West, Plant Design and Economics for Chemical Engineers; Turton et al., Analysis, Synthesis, and Design of Chemical Processes.
Process control
Process Control is area 15 with 4–6 questions. It shares this page because a control loop is drawn on the same P&ID as the equipment. The outline covers process dynamics (first- and second-order responses, gains, time constants, stability, damping, transfer functions), control strategies (feedback, feedforward, cascade, ratio, PID tuning, alarms and safety equipment) and loop hardware (pairing measured and manipulated variables, sensors, control valves, DCS and PLC programming, interlocks).
Gain with units and sign. A process gain is output change over input change, in the units of each; a reverse-acting process has a negative gain, and the controller action must match it.
Fail-safe position. A valve fails to the position that is safe for the process: cooling water fails open, fuel gas fails closed. Decide from the hazard, not from habit.
Feedforward without feedback. Feedforward cancels a measured disturbance but cannot correct model error; in practice it is paired with a feedback trim.
Sources: Seborg, Edgar, Mellichamp and Doyle, Process Dynamics and Control. Block diagrams, Bode plots and Routh stability are covered once for the site on FE Electrical and Computer control systems.
Materials of construction
Materials Science is area 04 with 4–6 questions, and on this exam it means choosing what the plant is built from: properties under temperature and stress, compatibility of metals, polymers, ceramics and composites with the process fluid, and corrosion and its control.
A corrosion allowance for the wrong mechanism. Extra wall thickness covers uniform corrosion; it does nothing against pitting or cracking.
Room-temperature strength at service temperature. Allowable stress falls as temperature rises; read it at the design temperature.
Sources: Perry’s Chemical Engineers’ Handbook, section on materials of construction. Crystal structures, phase diagrams and heat treatment are covered once for the site on FE Mechanical material properties.
How it connects
Time-value-of-money calculations for a project choice are covered once for all FE disciplines on engineering economics. Equipment duties come from heat transfer, fluids and reactions; inherently safer design continues on safety.