A 60-problem set built on the chemical specification: process balances, reactors, columns and exchangers rather than circuits or beams. It is our set, not the NCEES one; NCEES sells its own, 2 × 50 questions, $50 each.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
practice problems60
on the real exam110
real exam time5 h 20 min
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How the set is weighted
Fifty problems follow the chemical areas in proportion to their item ranges: six on balances, the largest area at 10–15 questions, and five each on fluids, thermodynamics, heat transfer and mass transfer. Reactions, process design and chemistry take four each, control and safety three, and solids handling, materials and engineering sciences two.
The last ten come from the four areas every FE discipline shares: four mathematics, two statistics, two economics and two ethics. Their full sets are on the shared FE practice test and engineering economics.
60 FE Chemical practice problems
0 / 60 answered · 0 correct
Question 1 of 60 · Materials Science · 4–6 items
Which polymer has the highest resistance to concentrated sulfuric acid?
Solution
Fluoropolymers (PTFE, PFA) are the standard lining and gasket materials for concentrated sulfuric acid; polyolefins, PVC and polyamides are limited by concentration and temperature.
PTFE's carbon-fluorine backbone is inert to virtually all acids, including concentrated sulfuric acid, across a wide temperature range.
HDPE handles dilute and moderate sulfuric acid, but strong oxidizing concentrated acid attacks it, especially when warm.
PVC resists many dilute acids, but concentrated sulfuric acid degrades it, and its temperature limit is low.
Polyamides hydrolyze in strong acids; nylon has poor resistance to concentrated sulfuric acid.
Question 2 of 60 · Materials Science · 4–6 items
A carbon steel process vessel is monitored using linear polarization resistance. At the operating conditions, the measured polarization resistance is 25 Ω · cm2. The anodic and cathodic Tafel slopes are both determined to be 0.120 V/decade. Assuming the uniform corrosion reaction is the oxidation of iron to Fe2+, what is the expected corrosion penetration rate?
Given data:
Atomic weight of Fe = 55.85 g/mol
Density of Fe = 7.87 g cm−3
Faraday's constant = 96,485 C/mol
One dichromate accepts only 6 electrons, half of what ethanol releases, so charge (+12 vs +6) cannot balance.
Ethanol gives 12 electrons (C: −2 → +4, two carbons), so two Cr2O72− (6 e− each) are needed; H, O and charge (+12) then balance with 16 H+ and 11 H2O.
With 2 dichromates the hydrogens do not balance (12 + 6 on the left, 16 on the right), and the charge is +8 against +12.
This reuses the dichromate half-reaction's water but not its 14 H+, and the electron count is still half of what ethanol gives.
Question 4 of 60 · Chemistry and Biology · 7–11 items
Metal
Measured Diluted Sample (mg/L)
Method Detection Limit (mg/L)
EPA Discharge Limit (mg/L)
Chromium
0.18
0.05
2.0
Copper
0.26
0.02
2.0
Nickel
0.08
0.10
1.5
Zinc
0.35
0.05
5.0
A chemical plant analyzes its wastewater effluent for heavy metals before discharge. The analytical laboratory dilutes the original sample by a factor of 10 prior to analysis to prevent matrix interference. The laboratory reports the measured concentrations of the diluted sample alongside the Method Detection Limit (MDL) and the EPA discharge limits. Based on the provided table, which metal legally exceeds its EPA discharge limit when properly accounting for the dilution factor and detection limits?
original = measured × dilution factor: Cr 1.8, Cu 2.6, Zn 3.5 mg/L; Ni < MDL → < 1.0 mg/L
Only copper (2.6 > 2.0 mg/L) exceeds its limit.
Zinc in the original sample is 0.35 × 10 = 3.5 mg/L, below its 5.0 mg/L limit.
Chromium is 0.18 × 10 = 1.8 mg/L, just under the 2.0 mg/L limit.
Only copper exceeds its limit once each reading is multiplied by 10; without the correction none would.
Corrected for the 10× dilution, copper is 0.26 × 10 = 2.6 mg/L, above its 2.0 mg/L limit; its reading is well above the 0.02 mg/L detection limit.
The nickel reading (0.08) is below its 0.10 mg/L detection limit, so it is reported as not detected (< 1.0 mg/L in the original), under the 1.5 mg/L limit.
Question 5 of 60 · Chemistry and Biology · 7–11 items
During a routine quality control analysis at a metallurgical processing facility, a chemist performs a redox titration to determine the concentration of dissolved iron. A 25.0 mL wastewater sample containing unknown quantities of Fe2+ requires exactly 18.4 mL of a standardized 0.020 M KMnO4 solution to reach the faint pink visual endpoint in a highly acidic medium. Based on the balanced redox half-reactions, determine both the true normality of the KMnO4 titrant solution and the molarity of the Fe2+ in the original wastewater sample.
Given
Sample volume = 25.0 mL
Titrant volume = 18.4 mL
KMnO4 molarity = 0.020 M
Find
Normality of KMnO4 and molarity of Fe2+
Solution
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O → N = 5 × 0.020 M = 0.100 N
N1V1 = N2V2: NFe = 0.100 × 18.4/25.0 = 0.0736 N = 0.0736 M (1 e− per Fe2+)
0.004 N divides the molarity by 5 instead of multiplying; MnO4− → Mn2+ gains 5 electrons per ion.
0.020 N treats permanganate as one equivalent per mole; in acid it accepts 5 electrons.
N = 5 × 0.020 = 0.100 N; Fe2+ → Fe3+ is one electron, so MFe = 0.100 × 18.4/25.0 = 0.0736 M.
The normality is right, but the iron is then computed with the 0.020 M molarity instead of the normality.
0.3680 multiplies normality by titrant volume and divides by 5 mL, mixing up the sample volume.
Question 6 of 60 · Chemistry and Biology · 7–11 items
Environmental engineers at a chemical manufacturing plant are evaluating a newly generated aqueous waste stream to determine appropriate disposal protocols. The stream consists primarily of a 0.50 M chloroacetic acid solution (Ka = 1.38 × 10−3). To verify compliance with RCRA regulations, calculate the exact pH of this waste stream and determine if it exhibits the characteristic of corrosivity. According to standard EPA regulatory definitions, an aqueous liquid is classified as a corrosive hazard if its pH is less than or equal to 2.0.
Given
Chloroacetic acid concentration = 0.50 M
Ka = 1.38 × 10−3
Corrosivity limit pH ≤ 2.0
Find
pH and corrosivity classification
Solution
Ka = x2/(C − x): x2 + 1.38 × 10−3x − 6.9 × 10−4 = 0 → x = [H+] = 0.0256 M
pH 0.30 treats chloroacetic acid as a strong acid ([H+] = 0.50 M); it is weak (Ka = 1.38 × 10−3).
Solving x2/(0.50 − x) = 1.38 × 10−3 gives [H+] = 0.0256 M, so pH = 1.59 ≤ 2.0: the waste is RCRA-corrosive (D002).
The pH is right, but the rule is pH ≤ 2.0, so 1.59 is corrosive.
2.86 is the pKa; it equals the pH only in a 1:1 buffer, not in a solution of the acid alone.
3.16 = −log(KaC) takes [H+] = KaC instead of √(KaC).
Question 7 of 60 · Thermodynamics · 8–12 items
When two substances are mixed at the same temperature to form an ideal solution, what is the overall enthalpy change of mixing?
Solution
ideal solution: ΔHmix = 0, ΔVmix = 0, ΔSmix = −RΣxiln xi > 0, ΔGmix = RTΣxiln xi < 0
At the same temperature there is no sensible heating to account for; heat capacities do not enter the mixing enthalpy.
Heat release on mixing (ΔHmix < 0) signals stronger unlike-pair attraction, i.e. a non-ideal solution with γ < 1.
In an ideal solution like and unlike interactions are equal, so ΔHmix = 0 (and ΔVmix = 0); only the entropy changes.
An endothermic mix (ΔHmix > 0) means weaker unlike interactions, positive deviation from Raoult's law.
Question 8 of 60 · Thermodynamics · 8–12 items
Parameter
Value
ΔG°rxn at 298 K
5.0 kJ/mol
Current PX
2.0 atm
Current PY
0.4 atm
A gas-phase reaction X ⇌ 2Y occurs at 298 K. Based on the standard Gibbs free energy of reaction and the current partial pressures provided, evaluate the reaction quotient Q, the equilibrium constant K, and the immediate direction of spontaneity.
Q = PY2/PX = 0.16/2.0 = 0.080 and K = e−5,000/(8.314 × 298) = 0.133; Q < K, so ΔG < 0 and the reaction shifts forward.
Q = 0.200 drops the exponent on PY; with that error Q > K would point the reaction backward.
K = 7.52 drops the minus sign in K = e−ΔG°/RT; a positive ΔG° gives K < 1.
Q = 0.200 forgets the stoichiometric exponent: PY must be squared for X ⇌ 2Y.
Q and K are right, but Q < K means too little product, so the reaction runs forward.
Question 9 of 60 · Material/Energy Balances · 10–15 items
Stream
Total flow (kg/h)
Inert content (wt%)
Fresh feed
1,000
5
Product
Unknown
0
Purge
Unknown
20
A separation process consists of a fresh feed, a pure product stream, and a purge stream. The system operates at steady state. Based on the overall process flow data provided in the table, what is the calculated mass flow rate of the purge stream?
1000.0 kg/h is the fresh feed; the purge is only part of what leaves.
750.0 kg/h is the product stream (1,000 − 250), not the purge.
All inert leaves in the purge: 0.05 × 1,000 = 0.20 × P, so P = 250.0 kg/h.
200.0 kg/h applies the purge's 20 % to the fresh feed; the balance is on the inert, which must all leave in the purge.
50.0 kg/h is the inert mass flow itself (0.05 × 1,000); the purge carries it at only 20 % concentration.
Question 10 of 60 · Material/Energy Balances · 10–15 items
Stream
Solids (wt%)
Water (wt%)
Fresh Feed
10
90
Main Unit Product
50
50
Main Unit Vapor
0
100
Mixed Final Product
20
80
A 100 kg/h fresh juice feed is processed in a separation network. An evaporator concentrates the juice by removing a stream of pure water. To preserve flavor, a portion of the fresh feed bypasses the evaporator and mixes with the main unit product downstream. Using the steady-state composition data provided, calculate the mass flow rate of the bypass stream.
Given
Fresh feed = 100 kg/h at 10 wt% solids
Main unit product = 50 wt% solids
Vapor = pure water
Mixed final product = 20 wt% solids
Find
Bypass flow rate (kg/h)
Solution
overall solids: 0.10 × 100 = 0.20·P → P = 50 kg/h (water evaporated W = 50 kg/h)
evaporator: Fe → 0.5 solids product of 0.2Fe, vapor 0.8Fe = 50 → Fe = 62.5 kg/h
bypass = 100 − 62.5 = 37.50 kg/h
87.50 kg/h subtracts the evaporator product from the fresh feed; that difference is the evaporator vapor plus bypass, not the bypass alone.
62.50 kg/h is the feed sent to the evaporator, the complement of the bypass.
50.00 kg/h is the final product (and also the water evaporated), not the bypass.
Overall solids: 10 = 0.20P → P = 50 kg/h; the bypass b and the 0.2-fraction concentrate from the evaporator feed (100 − b) add up to 50 kg/h, so b = 37.50 kg/h.
12.50 kg/h is the concentrate leaving the evaporator (0.1 × 62.5/0.5), not the bypass.
Question 11 of 60 · Material/Energy Balances · 10–15 items
A tank initially holds 500 kg of a 5% salt solution. A 20% salt stream flows in at 10 kg/min, and nothing leaves the tank. How long does it take for the tank solution to reach 10% salt?
Given
Initial tank mass = 500 kg
Initial salt fraction = 5 %
Inlet salt fraction = 20 %
Inlet rate = 10 kg/min
Target salt fraction = 10 %
Find
Time t (min)
Solution
salt: ms = 25 + 0.20 × 10t · total: M = 500 + 10t
(25 + 2t)/(500 + 10t) = 0.10 → 25 + 2t = 50 + t → t = 25.0 min
50.0 min forgets the 25 kg of salt already in the tank.
Salt: 25 + 2t; mass: 500 + 10t; (25 + 2t)/(500 + 10t) = 0.10 gives t = 25.0 min.
20.3 min uses the overflowing (constant-mass) tank solution; this tank has no outflow and fills up.
12.5 min adds salt at 2 kg/min until 50 kg, treating the tank mass as fixed at 500 kg; the tank grows by 10 kg/min.
Question 12 of 60 · Material/Energy Balances · 10–15 items
A process separates a feed stream (100 kg/h) into a distillate (25 kg/h) and bottoms (75 kg/h). If the feed contains 40% component A, and the distillate contains 90% component A, what is the composition of component A in the bottoms?
40.0 % is the feed composition; separation moves A into the distillate, so the bottoms must be leaner.
Question 13 of 60 · Material/Energy Balances · 10–15 items
A reactor processes 200 kg/h of feed containing 30% reactant A. If the conversion is 75%, what is the mass flow rate of product B if each kg of A produces 0.8 kg of B?
Given
Feed = 200 kg/h
Reactant A in feed = 30%
Conversion = 75%
Yield = 0.8 kg B per kg A reacted
Find
Product B flow (kg/h)
Solution
B = F × xA × X × yield = 200 × 0.30 × 0.75 × 0.8 = 36 kg/h
120 kg/h treats the whole 200 kg/h feed as reactant A.
48 kg/h converts all 60 kg/h of A, ignoring the 75 % conversion.
45 kg/h is the A reacted; each kilogram of A gives only 0.8 kg of B.
A fed = 0.30 × 200 = 60 kg/h; A reacted = 0.75 × 60 = 45 kg/h; B = 0.8 × 45 = 36 kg/h.
Question 14 of 60 · Material/Energy Balances · 10–15 items
In an adiabatic process, 5 kg/s of water at 25°C is mixed with 3 kg/s of water at 75°C. What is the final temperature of the mixture? Assume water has a constant specific heat capacity of 4.18 kJ/(kg·°C).
Given
Stream 1 flow m1 = 5 kg/s
Stream 1 temperature T1 = 25 °C
Stream 2 flow m2 = 3 kg/s
Stream 2 temperature T2 = 75 °C
Specific heat cp = 4.18 kJ/(kg·°C)
Find
Mixture temperature (°C)
Solution
Σṁcp(T − Ti) = 0 → T = (5 × 25 + 3 × 75)/8 = 43.75 °C
T = (5 × 25 + 3 × 75)/(5 + 3) = 350/8 = 43.75 °C (cp cancels).
50.00 °C is the plain average; the colder stream is larger, so the mixture is cooler than that.
55.00 °C weights the 50 °C difference by 3/5 instead of 3/8 (it divides by one stream instead of the total).
56.25 °C swaps the flows, giving the larger weight to the hot stream.
Question 15 of 60 · Heat Transfer · 8–12 items
Temperature (°C)
Pressure (kPa)
hf (kJ/kg)
hfg (kJ/kg)
hg (kJ/kg)
100
101.3
419.0
2257.0
2676.0
A condenser must condense a saturated steam mass flow of 2.0 kg/s at a saturation temperature of 100 °C. Cooling water enters at 20 °C and leaves at 35 °C. Using the provided thermodynamic table, calculate the minimum required mass flow rate of the cooling water. The specific heat of the cooling water is 4.18 kJ/(kg·K).
Given
Steam flow ṁ = 2.0 kg/s
Saturation temperature = 100°C
hfg at 100°C = 2257.0 kJ/kg
Cooling water inlet = 20°C
Cooling water outlet = 35°C
Cooling water cp = 4.18 kJ/(kg·K)
Find
Cooling water flow (kg/s)
Solution
q = ṁshfg = 2.0 kg/s × 2,257 kJ/kg = 4,514 kW
ṁw = q/(cpΔT) = 4,514/(4.18 × 15) = 72.0 kg/s
1,079.9 kg/s leaves the 15 °C temperature rise out of the denominator.
30.9 kg/s divides by the outlet temperature (35 °C) instead of the 15 °C rise.
13.4 kg/s uses hf (419 kJ/kg) as the heat released; the condensing steam gives up hfg.
Question 16 of 60 · Mass Transfer and Separation · 8–12 items
In the design of a continuous packed column for gas absorption, the stage-equivalent height (HETP) is frequently analyzed relative to the continuous transfer-unit height (HTU). Under what specific operational condition are the HETP and the overall gas-phase HTU mathematically identical?
Solution
HETP = HTUOG·ln λ/(λ − 1), λ = mG/L
λ → 1 (operating line parallel to the equilibrium line): HETP = HTUOG, NOG = number of theoretical stages.
Which film controls changes the HTU values but not the HETP/HTU ratio, which depends only on λ.
Minimum reflux is a distillation operating limit; it does not make HETP equal to HTUOG.
For A = L/(mG) > 1, λ < 1 and HETP < HTUOG; they are equal only at A = 1.
HETP = HTUOG·ln λ/(λ − 1) with λ = mG/L; parallel lines mean λ = 1, where the ratio → 1.
Feed thermal condition belongs to distillation McCabe-Thiele analysis; it does not link HETP and HTU.
Question 17 of 60 · Mass Transfer and Separation · 8–12 items
An aqueous feed stream flows at 150 L/h containing a solute concentration of 12.0 g L−1. A single-stage extraction process using a pure immiscible organic solvent is designed to reduce the aqueous raffinate solute concentration to 3.0 g/L. The solute distribution coefficient between the organic and aqueous phases is KD = 5.0 (concentration basis). Determine the required volumetric flow rate of the organic solvent.
Given
Feed flow rate = 150 L/h
Feed solute concentration = 12.0 g/L
Raffinate solute concentration = 3.0 g/L
Distribution coefficient KD = 5.0
Find
Organic solvent flow rate (L/h)
Solution
F(cF − cR) = S·yE, yE = KDcR = 5.0 × 3.0 = 15 g/L
S = 150 × 9.0/15 = 90.0 L/h
450.0 L/h divides the 1,350 g/h removed by the raffinate concentration (3.0 g/L), forgetting KD.
150.0 L/h simply matches the feed flow; the solvent rate follows from the solute balance.
120.0 L/h divides all the solute fed (1,800 g/h) by 15 g/L, as if none stayed in the raffinate.
112.5 L/h assumes the extract leaves at the feed concentration (12 g/L).
Question 18 of 60 · Mass Transfer and Separation · 8–12 items
A binary distillation column processes a saturated liquid feed (q = 1) with a light key mole fraction of 0.50. The desired distillate composition is 0.90 mole fraction of the light key. The relative volatility is constant at 3.0. Determine the minimum reflux ratio required for this separation.
0.75 is the equilibrium vapor composition y′ at the pinch, not the reflux ratio.
0.90 is 1.5 × Rmin, a typical operating reflux, not the minimum.
1.60 uses (xD − x′) instead of (xD − y′) in the numerator.
Question 19 of 60 · Mass Transfer and Separation · 8–12 items
A two-stage reverse osmosis system is operated continuously. The retentate from Stage 1 serves directly as the feed for Stage 2. Operating specifications for the system are provided in the table below. If the fresh raw water feed has a salt concentration of 1000 mg/L, determine the expected salt concentration in the final retentate stream leaving Stage 2.
Parameter
Stage 1
Stage 2
Feed Source
Raw Water
Stage 1 Retentate
Fresh Feed Flow (m3/h)
100
N/A
Stage Cut (Recovery)
0.50
0.50
Salt Rejection
0.96
0.96
Given
Fresh feed flow = 100 m3/h
Feed salt concentration = 1000 mg/L
Stage cut (each stage) = 0.50
Salt rejection (each stage) = 0.96
Stage 2 feed = Stage 1 retentate
Find
Final retentate concentration (mg/L)
Solution
stage: salt in = QfCf; permeate Qf/2 at 0.04Cf; retentate Qf/2 at Cf(1 − 0.02)/0.5 = 1.96Cf
Cr2 = 1,000 × 1.96 × 1.96 = 3,841.6 mg/L
4000.0 mg/L assumes complete rejection in both stages (1,000 × 2 × 2).
3920.0 mg/L applies 96 % rejection in Stage 1 but treats Stage 2 as rejecting all salt (1,960 × 2).
Each stage: Cp = 0.04Cf, and the retentate (half the flow) holds the rest: Cr = Cf(1 − 0.5 × 0.04)/0.5 = 1.96Cf; two stages: 1,000 × 1.962 = 3,841.6 mg/L.
2000.0 mg/L is Stage 1 with 100 % rejection; at 96 % some salt passes into the permeate.
1960.0 mg/L is the Stage 1 retentate; Stage 2 concentrates it again.
Question 20 of 60 · Mass Transfer and Separation · 8–12 items
In an absorption tower, an unreactive gas is absorbed into a liquid solvent. The system operates with linear equilibrium where the mole fraction in the gas phase (y) is related to the mole fraction in the liquid phase (x) by the relationship y = 1.5x.
The individual gas-film mass transfer coefficient is ky = 0.04 kmol/(m2 · h · Δy) and the individual liquid-film coefficient is kx = 0.60 kmol/(m2 · h · Δx).
Determine the overall gas-phase mass transfer coefficient (Ky) for this process.
1/Ky = 1/ky + m/kx = 25 + 1.5/0.60 = 27.5, so Ky = 0.0364 kmol/(m2·h).
0.0383 uses 1/(m·kx) for the liquid term; the slope multiplies the liquid resistance, m/kx.
0.0400 is ky alone; the liquid film adds about 9 % to the resistance.
0.640 adds the two coefficients; resistances in series add, not coefficients.
27.5 is the total resistance 1/Ky; it must be inverted.
Question 21 of 60 · Solids Handling · 3–5 items
A granular catalyst is recirculated through a screw conveyor, and samples are taken before the first pass and after the twentieth. Which of the following describes the change in median particle size (d50) and true (skeletal) particle density caused solely by attrition?
Solution
Attrition = abrasion and chipping: more fines, smaller d50, rounder particles. True (skeletal) density is a material property and does not change; bulk density may go either way depending on how the fines pack.
Breaking a particle does not lower the density of the material it is made of.
Attrition only creates fines and rounds particles: the size distribution shifts down, while the true (skeletal) density of the solid stays the same.
Repeated screw passes chip and abrade particles, so the median size falls, and the solid itself is not densified.
Attrition breaks particles into smaller pieces of the same solid; the skeletal density of the material does not change.
Attrition produces fines and smaller fragments, so the median size goes down, not up.
Question 22 of 60 · Chemical Reaction Engineering · 7–11 items
A PFR and a CSTR of equal volumes are processing the same feed for a reaction with positive reaction order. Which statement is true?
Solution
Levenspiel: V/FA0 = ∫dX/(−rA) (PFR, area under curve) vs X/(−rA)exit (CSTR, rectangle). For n > 0, 1/(−rA) rises with X, so the rectangle is larger: PFR needs less volume.
A CSTR beats a PFR only for negative-order or autocatalytic regions, where the rate rises with conversion.
For any positive order the PFR wins; the kinetics only change by how much.
They coincide only for zero-order kinetics, which is excluded by «positive order».
For n > 0 the rate falls with conversion; a CSTR runs entirely at the low exit rate, while a PFR integrates through higher rates, so equal volumes give the PFR more conversion.
Question 23 of 60 · Chemical Reaction Engineering · 7–11 items
A gas-phase reaction A → 3B occurs isothermally and isobarically in a plug-flow reactor. Pure A is fed at FA0 = 10 mol/s with an initial concentration of CA0 = 0.5 mol L−1. The reaction is first-order with respect to A, with a rate constant of k = 0.2 s−1. Determine the reactor volume required to achieve 80% conversion.
400.0 L is the CSTR volume without expansion, υ0X/[k(1 − X)]; the reactor is a PFR.
482.8 L drops the −εX term of the integrated PFR equation.
642.8 L adds εX instead of subtracting it.
1040.0 L is the CSTR volume with expansion, υ0X(1 + εX)/[k(1 − X)]; the reactor is a PFR.
Question 24 of 60 · Process Design · 7–11 items
In process design, what does the term 'turndown ratio' refer to?
Solution
Turndown ratio = Qmax/Qmin at acceptable performance (trays: weeping vs flooding; valves: rangeability; burners: stable flame).
Startup and shutdown durations are not described by turndown.
Actual over theoretical yield is the percent yield of a reaction.
Turndown = maximum/minimum controllable throughput, e.g. 4:1 for a tray column or a control valve's range.
Cost ratios belong to estimating (installation factors), not to operability.
Question 25 of 60 · Process Design · 7–11 items
Location
Hot stream temperature (°C)
Cold stream temperature (°C)
Exchanger 1 (hot end)
140
110
Exchanger 1 (cold end)
80
50
Exchanger 2 (hot end)
80
70
Exchanger 2 (cold end)
50
25
A process engineer is retrofitting a heat exchanger network to maximize energy recovery. Based on the terminal temperatures of the matched process streams provided in the table, and given a design minimum approach temperature (ΔTmin) of 10 °C, at which location does the thermodynamic pinch point occur?
Given
ΔTmin = 10°C
Exchanger 1 hot end: 140°C / 110°C
Exchanger 1 cold end: 80°C / 50°C
Exchanger 2 hot end: 80°C / 70°C
Exchanger 2 cold end: 50°C / 25°C
Find
Location of the pinch
Solution
approach ΔT = Thot − Tcold: E1 hot 30, E1 cold 30, E2 hot 10, E2 cold 25 °C
The pinch is where ΔT = ΔTmin = 10 °C: Exchanger 2, hot end. No heat should cross the pinch.
Exchanger 1's cold end has a 30 °C approach (80 vs 50 °C), well above ΔTmin.
Both ends of Exchanger 1 sit at 30 °C, far from the 10 °C minimum approach.
The approach there is 80 − 70 = 10 °C, equal to ΔTmin; the pinch is where the hot and cold curves come closest.
The hot end of Exchanger 1 has a 30 °C approach (140 vs 110 °C).
The 25 °C approach (50 vs 25 °C) at Exchanger 2's cold end is not the closest point.
Question 26 of 60 · Process Control · 4–6 items
Which of the following is NOT typically a manipulated variable in a chemical process control system?
Solution
Controlled variable: what is measured and held at set point (purity, temperature, level). Manipulated variable: what the controller moves (valve position → flow, steam, feed).
Steam flow is manipulated through its valve to control temperature.
Valve position is the most direct manipulated variable; controller outputs move valves.
Feed rate is set through a valve and is commonly manipulated, for example to control level or throughput.
Purity is a controlled (measured) variable; the controller adjusts reflux, steam or feed to hold it.
Question 27 of 60 · Process Control · 4–6 items
For a process with dead time, which control strategy would be most appropriate to improve control performance?
Solution
Smith predictor: model Gm(s)e−θs in parallel; the controller sees the predicted undelayed output, so it can be tuned as if θ = 0 (needs an accurate model).
A Smith predictor uses a process model to predict the undelayed response and feeds that back, removing the dead time from the loop's characteristic equation.
On-off control cycles continuously and gets worse with dead time, because the measurement lags every switch.
Derivative action alone has no steady-state action, and dead time makes derivative of the delayed signal useless.
P-only control must be detuned for dead time and leaves offset; it does not address the delay.
Question 28 of 60 · Process Control · 4–6 items
For a process control strategy that integrates feedforward control with conventional feedback in a distillation column, which factor is most critical for the success of the feedforward action?
Solution
Feedforward: Gff(s) = −Gd(s)/Gp(s) acting on the measured disturbance. It needs a timely, accurate disturbance measurement and a decent model; feedback trims what it misses.
A long integral time slows the feedback trim; it does not make feedforward work.
Feedforward acts on the measured disturbance before it upsets the column; if that measurement is wrong or late, the correction is wrong or late.
A slower sensor delays the feedforward correction, which defeats its purpose.
Feedback gain matters for trim, but feedforward's success rests on measuring the disturbance and having a good process model.
Question 29 of 60 · Safety, Health, and Environment · 5–8 items
Which approach represents an inherently safer design principle?
Inspection is a procedural (administrative) safeguard; it manages the hazard rather than removing it.
Substitution is one of the inherently safer strategies (minimize, substitute, moderate, simplify): it removes the hazard instead of guarding it.
Extra relief devices are engineered add-on protection, not an inherent change to the process.
Fire suppression mitigates consequences after release; it is an active add-on, not inherently safer design.
Question 30 of 60 · Safety, Health, and Environment · 5–8 items
A chemical processing facility keeps the inventories below in its process (reactor feed and charge vessels, not atmospheric storage tanks). Under OSHA Process Safety Management (29 CFR 1910.119), a process is covered if it involves an Appendix A highly hazardous chemical at or above its threshold quantity, or 10,000 lb or more of a flammable liquid. Which chemicals make this process subject to PSM? (1 lb = 0.4536 kg)
Covered: chlorine (Appendix A) and toluene (flammable liquid ≥ 10,000 lb in a process; atmospheric storage would be exempt), 29 CFR 1910.119(a)(1).
Ammonia does not reach its threshold once 4,000 kg is converted to 8,818 lb.
Ammonia is under its threshold: 4,000 kg = 8,818 lb < 10,000 lb.
Comparing kilograms directly with thresholds in pounds misses that chlorine and toluene exceed theirs.
Chlorine is covered (800 kg = 1,764 lb > 1,500 lb), but so is toluene: 5,000 kg = 11,023 lb of a flammable liquid held in process vessels exceeds 10,000 lb.
In pounds: chlorine 1,764 lb ≥ 1,500 lb (Appendix A) and toluene 11,023 lb ≥ 10,000 lb (flammable liquid in process) are covered; ammonia at 8,818 lb is below 10,000 lb.
Question 31 of 60 · Safety, Health, and Environment · 5–8 items
What does the term "LC50" represent in toxicology?
Solution
LC50: median lethal concentration (inhalation/aquatic, with exposure time); LD50: median lethal dose (oral/dermal, mg/kg). Lower values = more toxic.
A median effect concentration for a non-lethal response is an EC50; LC50 refers to death.
Environmental decay is described by a half-life, not by LC50.
LC50 = lethal concentration for 50 % of test animals over a specified exposure time (e.g. 4 h inhalation).
The lowest level with an observed effect is the LOAEL, not a lethal median.
Question 32 of 60 · Engineering Sciences · 4–6 items
A 15 kg object initially at rest is subjected to a constant force of 45 N. What is the object's speed after 5 seconds?
Given
Mass m = 15 kg
Force F = 45 N
Time t = 5 s
Find
Speed after 5 s (m/s)
Solution
a = F/m = 45 N/15 kg = 3.0 m/s2
v = v0 + at = 0 + 3.0 × 5 = 15.0 m/s
225.0 multiplies force by time (45 × 5) and never divides by the 15 kg mass.
75.0 multiplies the acceleration by t2 instead of t.
37.5 is ½at2, the distance traveled in meters, not the speed.
a = F/m = 45/15 = 3.0 m/s2; from rest, v = at = 3.0 × 5 = 15.0 m/s.
Question 33 of 60 · Engineering Sciences · 4–6 items
A standard Wheatstone bridge circuit is modeled with three clockwise mesh currents. Mesh 1 (i1) contains the driving voltage source Vs and the two bottom arms (R3 and R4). Mesh 2 (i2) is the top-left loop containing R1, the middle galvanometer equivalent resistance R5, and R3. Mesh 3 (i3) is the top-right loop containing R2, R4, and R5. Identify the correctly formulated Kirchhoff's Voltage Law (KVL) equation for Mesh 2 using the passive sign convention.
Solution
KVL, mesh 2: R1i2 + R5(i2 − i3) + R3(i2 − i1) = 0
→ −R3i1 + (R1 + R3 + R5)i2 − R5i3 = 0
The coupling terms are right, but the source Vs sits in mesh 1, so mesh 2's equation equals 0.
Both mutual terms have the wrong sign and pair with the wrong resistors.
The mutual terms are wrong: mesh 2 shares R3 with mesh 1 and R5 with mesh 3, not R1 and R3.
With all mesh currents clockwise the shared-branch terms enter with a minus sign, and Vs is not in this mesh.
Mesh 2 sums its own resistors times i2 and subtracts each shared resistor times the neighboring mesh current (R3 with mesh 1, R5 with mesh 3); no source lies in mesh 2, so the right side is 0.
Question 34 of 60 · Fluid Mechanics/Dynamics · 8–12 items
Which of the following best describes the characteristic of laminar flow?
Solution
Laminar: Re < ≈ 2,100 in pipes, viscous forces dominate, layers do not mix, ΔP ∝ V (Hagen-Poiseuille).
Re above about 4,000 is turbulent flow; laminar pipe flow is below about 2,100.
Eddies and mixing across streamlines are the mark of turbulence.
ΔP ∝ V2 is the fully turbulent (rough-pipe) behavior; in laminar flow ΔP ∝ V.
In laminar flow layers slide past one another with no cross-stream mixing; viscous forces dominate.
Question 35 of 60 · Fluid Mechanics/Dynamics · 8–12 items
Water (ρ = 1,000 kg/m3) enters a sudden pipe expansion at 5 m/s, and the square of the inlet-to-outlet area ratio is (A1/A2)2 = 0.0625. Neglecting losses, what is the ideal pressure rise across the expansion?
With the real (Borda-Carnot) loss the rise is only ρv2(v1 − v2) = 4,690 Pa.
12,500 Pa is the whole inlet velocity head ½ρv12; the outlet still carries some kinetic energy.
Bernoulli without losses: ΔP = ½ρv12[1 − (A1/A2)2] = 0.5 × 1,000 × 25 × 0.9375 = 11,700 Pa rise.
9,380 Pa uses A1/A2 = 0.25 instead of its square in the bracket.
4,690 Pa is the real rise from a momentum (Borda-Carnot) balance, ρv2(v1 − v2); the question asks for the frictionless (ideal) value.
Question 36 of 60 · Fluid Mechanics/Dynamics · 8–12 items
In the Bernoulli equation, which of the following terms represents the velocity head?
Solution
P/ρg + v2/2g + z = constant (Bernoulli) — pressure, velocity and elevation heads
The velocity head is the kinetic energy per unit weight, v2/2g.
hL is the head loss to friction, which appears in the energy equation, not in ideal Bernoulli.
P/ρg is the pressure head.
z is the elevation (potential) head.
Question 37 of 60 · Fluid Mechanics/Dynamics · 8–12 items
For laminar flow over a flat plate the boundary layer thickness is approximately δ = 5x/√Rex, with Rex = ρUx/μ. Water (ρ = 1,000 kg/m3, μ = 0.001 Pa·s) flows at U = 0.2 m/s along a flat plate. What is the boundary layer thickness at x = 2 m from the leading edge?
0.0250 mm forgets the square root: 5x/Rex instead of 5x/√Rex.
7.91 mm drops the x in the numerator (5/√Rex is the ratio δ/x, not δ).
Rex = 1,000 × 0.2 × 2/0.001 = 4 × 105 (laminar, below 5 × 105); δ = 5x/√Rex = 10/632 = 0.0158 m = 15.8 mm.
56.1 mm applies the turbulent correlation 0.37x/Rex0.2; at Rex = 4 × 105 the boundary layer is still laminar.
Question 38 of 60 · Fluid Mechanics/Dynamics · 8–12 items
A chemical process uses a pipe with an internal diameter of 0.10 m and a total length of 200 m. The fluid flows at a velocity of 3.0 m/s. The literature provides a Fanning friction factor of fF = 0.005 for this specific fluid and pipe material under current conditions. Calculate the frictional head loss.
FE Reference Handbook, Fluid Mechanics: Darcy-Weisbach (Darcy f = 4 × Fanning f).
180.00 m omits g: 4fF(L/D)V2/2 has units of m2/s2, not meters of head.
73.39 m applies the factor of 4 twice.
hf = 4fF(L/D)(V2/2g) = 4 × 0.005 × 2,000 × 0.459 m = 18.35 m.
4.59 m plugs the Fanning factor straight into the Darcy-Weisbach equation, which needs fD = 4fF = 0.020.
1.15 m divides by 4 instead of multiplying: the Darcy factor is four times the Fanning factor.
Question 39 of 60 · Thermodynamics · 8–12 items
Which equation correctly represents the first law of thermodynamics for a closed system undergoing a process?
Solution
closed system: ΔU (+ ΔKE + ΔPE) = Q − W, W = work done by the system
FE Reference Handbook, Thermodynamics: first law (closed systems).
This is the definition of Gibbs energy at constant T, not the first law.
ΔH = Q applies to a constant-pressure process with only boundary work; the general closed-system balance is written for U.
For a closed system the energy change is heat added minus work done by the system (with W positive when the system does work).
ΔS = Qrev/T defines entropy change for a reversible process; it is a second-law statement.
Question 40 of 60 · Heat Transfer · 8–12 items
For laminar forced convection over a flat plate the local coefficient follows hx = C·Rex1/2·Pr1/3. If the local Reynolds number doubles while the Prandtl number stays constant, what is the percent increase in hx?
Given
Correlation hx = C·Rex1/2·Pr1/3
Rex doubles, Pr constant
Find
Percent increase in hx
Solution
h2/h1 = (Re2/Re1)1/2 = 20.5 = 1.414 → +41.4 %
300.0 % squares the Reynolds ratio (22 = 4, an increase of 300 %); the laminar exponent is ½.
100.0 % assumes hx grows linearly with Rex; the laminar exponent is ½.
50.0 % takes the exponent ½ as half of a 100 % increase; powers do not scale that way.
hx ∝ Rex1/2: doubling Rex multiplies hx by √2 = 1.414, an increase of 41.4 %.
Question 41 of 60 · Heat Transfer · 8–12 items
Which flow configuration in a shell and tube heat exchanger typically gives the highest heat transfer rate for the same surface area?
Solution
For the same U, A and inlet temperatures, counterflow gives the highest effectiveness: εcounter ≥ εcross ≥ εparallel. Multipass shell-and-tube units approach it with F < 1.
Cross flow sits between parallel and counter-current; its correction factor F is below 1.
Counter-current flow keeps the largest mean temperature difference along the exchanger (highest LMTD), so it transfers the most heat for a given area.
Parallel flow has the largest ΔT at the inlet and the smallest at the outlet, so it gives the lowest LMTD for the same terminal temperatures.
Mixing a stream evens out its temperature and lowers the effective driving force compared with pure counterflow.
Question 42 of 60 · Heat Transfer · 8–12 items
Property
Value
Thermal conductivity, k
40.0 W/(m·K)
Density, ρ
8000 kg/m3
Specific heat, cp
500 J/(kg·K)
Characteristic length, Lc
0.05 m
A transient conduction analysis of a solid uses the Fourier number Fo = αt/Lc2, with α = k/(ρcp). What is the Fourier number for a time of 120 s?
0.12 uses twice the characteristic length (0.10 m) in Lc2.
0.024 divides by Lc instead of Lc2.
0.0080 puts the time in minutes (2 min); α is in m2/s, so t must be in seconds.
4.8 × 10−5 takes Lc as 5 (centimeters) instead of 0.05 m.
Question 43 of 60 · Mathematics · 6–9 items
An engineer needs to estimate the density of an organic solvent mixture at a specific process temperature using experimental data. The available data set contains density values measured at discrete temperatures ranging from 20°C to 80°C. The current operating process is at 95°C. The engineer applies a linear interpolation formula to the available data points to estimate the density at the operating condition. Which of the following statements best evaluates this approach?
Solution
Interpolation estimates between known points (20–80 °C); a target at 95 °C is extrapolation, valid only if the trend is known to continue (e.g. from a physical model).
A good fit inside 20–80 °C says nothing about behavior outside it; correlation does not license extrapolation.
Using the two points nearest 95 °C (70 and 80 °C) still projects the line past the data, which is still extrapolation.
Density data can be fitted empirically without an equation of state; the problem is the range, not the method.
95 °C lies outside the measured 20–80 °C range, so this is extrapolation, not interpolation, and the error is unbounded.
A higher-order polynomial usually extrapolates worse; nothing requires one inside the data range.
Question 44 of 60 · Mathematics · 6–9 items
Using Taylor series expansion, what is the approximate value of e0.1 to three terms?
Given
Value x = 0.1
Number of terms = 3
Find
Approximate value of e0.1
Solution
ex ≈ 1 + x + x2/2! → e0.1 ≈ 1 + 0.1 + 0.005 = 1.105 (exact 1.10517)
FE Reference Handbook, Mathematics: Taylor series.
1.150 uses x/2 = 0.05 as the third term instead of x2/2.
1.110 drops the factorial and uses x2 = 0.01 as the third term instead of x2/2.
ex ≈ 1 + x + x2/2! = 1 + 0.1 + 0.005 = 1.105.
1.100 stops after two terms (1 + x); the third term x2/2 = 0.005 is missing.
Question 45 of 60 · Mathematics · 6–9 items
For the function f(x) = ln(x) + √x on the interval [1, 4], what is the average rate of change of f(x) over the interval, the value the Mean Value Theorem guarantees f′(c) takes somewhere in it?
Question 47 of 60 · Probability and Statistics · 4–6 items
A measurement of 15 samples of chemical reactor output gave a mean conversion of 87.5% with a standard deviation of 2.1%. A new catalyst is tested on 12 samples, giving a mean conversion of 89.3% with a standard deviation of 1.9%. Using α = 0.05, what is the critical t-value for testing if the new catalyst gives a statistically significant improvement?
FE Reference Handbook, Probability and Statistics: Student's t-distribution table.
1.65 is the one-tailed z value; with 15 + 12 small samples and estimated variances the t distribution (df = 25) applies.
An improvement test is one-tailed; df = n1 + n2 − 2 = 25, and t0.05, 25 = 1.708 ≈ 1.71.
1.96 is the two-tailed z value at α = 0.05; the question tests one direction with small samples.
2.06 is t0.025, 25, the two-tailed critical value; 'improvement' is a one-sided alternative.
Question 48 of 60 · Probability and Statistics · 4–6 items
A chemical engineer measured viscosity (cP) at several temperatures (°C): (20, 8.5), (30, 7.2), (40, 6.1), (50, 5.3), (60, 4.6). The least-squares line through these data is y = −0.097x + 10.22, with a total sum of squares SST = 9.57 and an error sum of squares SSE = 0.163. What is the coefficient of determination R2?
Given
Data (T °C, viscosity cP) = (20, 8.5), (30, 7.2), (40, 6.1), (50, 5.3), (60, 4.6)
Regression line y = −0.097x + 10.22
SST = 9.57
SSE = 0.163
Find
Coefficient of determination R2
Solution
R2 = 1 − SSE/SST = 1 − 0.163/9.57 = 0.983
FE Reference Handbook, Probability and Statistics: least squares and the coefficient of determination.
R2 = 1 − SSE/SST = 1 − 0.163/9.57 = 0.983.
0.966 squares R2 again (0.9832); R2 is already the squared correlation.
0.017 is SSE/SST, the unexplained fraction; R2 is one minus it.
−0.097 is the slope of the regression line; R2 lies between 0 and 1.
Question 49 of 60 · Ethics and Professional Practice · 3–5 items
A chemical engineer develops a highly efficient continuous polymerization process. Without seeking legal review, the engineer presents the initial research findings, including yield improvements and general process efficiencies, at a public engineering conference. The presentation does not reveal the specific catalyst composition or the exact reactor operating parameters. Under standard intellectual property principles, how does this public presentation most likely impact the company's future strategy?
Solution
Disclosure before filing becomes prior art (novelty); most countries have no grace period, the US gives one year for the inventor's own disclosure. Trade-secret protection survives for information not disclosed and kept confidential.
Priority goes to the first inventor to file (first-to-file since 2013); a conference talk is not a filing.
Presenting work does not transfer ownership; employment agreements normally assign inventions to the employer.
Only what was disclosed loses secrecy; the catalyst composition and reactor settings were not presented and can remain trade secrets.
The talk is prior art against patenting what was shown, while the catalyst composition and reactor settings, never disclosed, remain trade secrets if kept confidential.
A US patent term runs 20 years from the filing date of the application, not from a public presentation.
Question 50 of 60 · Ethics and Professional Practice · 3–5 items
A client verbally requests an increase in a pump's capacity halfway through a design project. The engineer complies without updating the formal contract. The modified system subsequently fails. What is the primary risk exposure for the engineer in this scenario?
Solution
Professional liability for engineering services rests on negligence (failure to meet the standard of care). Scope changes should be documented by contract amendment and the design rechecked for the new capacity.
Fraud requires an intentional false statement; complying with a verbal request is not deception.
Changing the design on an undocumented verbal request, without updating scope and checking the change, exposes the engineer to a negligence claim for failing the standard of care.
Indemnification clauses allocate losses between parties; they are not the source of the engineer's exposure here.
Courts generally do not imply a warranty of results for professional services; the claim against a design professional is negligence.
Strict liability applies to products; design services are judged by the professional standard of care.
Question 51 of 60 · Economics · 4–6 items
Parameter
Defender
Challenger
Current book value
$50,000
N/A
Current market value
$30,000
N/A
Purchase price
N/A
$100,000
A plant manager is performing a replacement study comparing an existing centrifuge (Defender) with a new model (Challenger). The corporate tax rate is 20%. Calculate the net initial cash investment (Year 0 outlay) required to undertake the replacement.
Given
Tax rate = 20%
Defender book value = $50,000
Defender market value = $30,000
Challenger purchase price = $100,000
Find
Net initial cash investment ($)
Solution
loss on sale = book − market = $50,000 − $30,000 = $20,000 → tax saving = 0.20 × $20,000 = $4,000
net outlay = $100,000 − $30,000 − $4,000 = $66,000
$100,000 is the purchase price alone; the defender's $30,000 sale and the $4,000 tax saving reduce the outlay.
$80,000 subtracts only the $20,000 book loss from the purchase price instead of the $30,000 cash received.
$74,000 treats the tax effect of the loss as a cost; a loss on sale reduces taxes.
$70,000 nets the sale proceeds but ignores the $4,000 tax saving on the loss.
Selling below book creates a $20,000 loss; at 20 % it saves $4,000 of tax, so outlay = $100,000 − $30,000 − $4,000 = $66,000.
Question 52 of 60 · Economics · 4–6 items
Year
5-Year MACRS Rate
Straight-Line Rate
1
20.00%
20.00%
2
32.00%
20.00%
3
19.20%
20.00%
4
11.52%
20.00%
5
11.52%
20.00%
6
5.76%
0.00%
A profitable chemical manufacturing plant is evaluating the depreciation method for a newly purchased reactor. The company has a positive minimum attractive rate of return (MARR). Which statement best describes why adopting the MACRS depreciation schedule results in a higher Net Present Value (NPV) for the project compared to the Straight-Line schedule?
Solution
PW(tax savings) = Σ t·Dj·(P/F, i, j)
Total D is the same (100 %); MACRS moves it forward (20 % + 32 % in years 1–2), so at i > 0 its present worth is larger.
Depreciation method does not exempt salvage value from tax; gains over book value are taxed either way.
MACRS front-loads deductions (52 % in years 1–2 versus 40 %); at a positive MARR earlier tax savings have a higher present worth.
Both schedules write off 100 % of the cost, so the undiscounted tax savings are equal; only their timing differs.
Both schedules total 100 % of the basis; MACRS changes the timing, not the total.
Depreciation is a non-cash deduction; it does not change revenue.
Question 53 of 60 · Thermodynamics · 8–12 items
A gas expands from 0.1 m³ to 0.3 m³ against a constant external pressure of 100 kPa. How much work is done by the gas?
Given
Initial volume V1 = 0.1 m3
Final volume V2 = 0.3 m3
Constant external pressure P = 100 kPa
Find
Work done by the gas W (kJ)
Solution
Wb = P(V2 − V1) = 100 kPa × (0.3 − 0.1) m3 = 20 kJ (done by the gas)
30 kJ multiplies the pressure by the final volume (0.3 m3) instead of the volume change.
W = Pext(V2 − V1) = 100 kPa × 0.2 m3 = 20 kJ done by the gas.
10 kJ multiplies the pressure by the initial volume (0.1 m3) instead of the volume change.
−20 kJ is the work done on the gas; the question asks for the work done by the gas, which is positive during expansion.
Question 54 of 60 · Thermodynamics · 8–12 items
What is the Joule-Thomson coefficient μJT = (∂T/∂P)h of an ideal gas?
Energy rises as product size falls, so it cannot be proportional to the size squared.
No comminution law uses the cube of the reduction ratio; Bond uses inverse square roots of size.
Rittinger: E/m = KR(1/x2 − 1/x1), i.e. energy goes with the new surface created; it suits fine grinding.
Energy proportional to ln(x1/x2) is Kick's law, used for coarse crushing.
Question 57 of 60 · Chemical Reaction Engineering · 7–11 items
A decomposition reaction follows second-order kinetics with k = 0.05 L/(mol·min). If the initial concentration is 2.0 mol/L, how long are the first half-life (2.0 → 1.0 mol/L) and the second half-life (1.0 → 0.5 mol/L)?
Given
Rate constant k = 0.05 L/(mol·min)
Initial concentration C0 = 2.0 mol/L
Find
First and second half-lives (min)
Solution
second order: 1/C = 1/C0 + kt → t1/2 = 1/(kC0)
first: 1/(0.05 × 2.0) = 10 min · second: 1/(0.05 × 1.0) = 20 min
Second order: t1/2 = 1/(kC0). From 2.0 mol/L: 1/(0.05 × 2.0) = 10 min; from 1.0 mol/L: 1/(0.05 × 1.0) = 20 min.
The values are swapped: a second-order half-life grows as the concentration falls, so the second one is the longer.
Equal successive half-lives are the mark of first-order kinetics; here t1/2 depends on the starting concentration.
20 and 40 min use the starting concentrations one step too low (1.0 and 0.5 mol/L); the first half-life starts at 2.0 mol/L.
14 min = ln 2/k treats k as a first-order constant; the units L/(mol·min) mark a second-order reaction.
Question 58 of 60 · Chemical Reaction Engineering · 7–11 items
A liquid-phase exothermic reaction runs in a jacketed, non-isothermal CSTR. At steady state the heat generated, G(T) = (−ΔHR)kCAV, equals the heat removed, R(T) = UA(T − Tc) + ṁcp(T − T0). At the design point G = 5,000 kW and T = 350 K, with ṁcp = 40 kW/K, feed temperature T0 = 300 K and jacket temperature Tc = 320 K. What UA is required to hold this steady state?
Given
Heat generation G = 5,000 kW
Reactor temperature T = 350 K
ṁcp = 40 kW/K
Feed temperature T0 = 300 K
Jacket temperature Tc = 320 K
Find
Required UA (kW/K)
Solution
G = UA(T − Tc) + ṁcp(T − T0)
5,000 = UA × 30 + 40 × 50 → UA = 3,000/30 = 100.0 kW/K
233.3 kW/K adds the feed term to the generation instead of subtracting it.
166.7 kW/K makes the jacket remove all 5,000 kW and ignores the heat taken up by the cold feed.
150.0 kW/K uses Tc − T0 = 20 K as the jacket driving force; the jacket removes heat at T − Tc = 30 K.
126.7 kW/K computes the feed term with the jacket difference (40 × 30 = 1,200 kW).
The feed absorbs ṁcp(T − T0) = 40 × 50 = 2,000 kW, so the jacket removes 3,000 kW: UA = 3,000/30 = 100.0 kW/K.
Question 59 of 60 · Process Design · 7–11 items
Which of the following is NOT typically included in a piping and instrumentation diagram (P&ID)?
Solution
P&ID: all process equipment (tagged), piping with line numbers and sizes, valves, instruments, control and safety loops. PFD: major equipment, main streams, heat and material balance. Neither shows civil details.
Instrument bubbles and their signal and process connections are what the «I» in P&ID stands for.
Foundations are civil/structural drawings; a P&ID shows process equipment, piping, valves, instruments and control loops.
Every vessel, pump and exchanger on a P&ID carries its equipment tag number.
Control valves, with their fail positions, are a core element of every P&ID.
Question 60 of 60 · Process Design · 7–11 items
When sizing a centrifugal pump for a process service, which parameter must be established first?
Solution
Pump sizing order: design flow → system head at that flow (system curve) → pump selection on the curve (best efficiency point) → NPSHA vs NPSHR → power, materials, seals.
Required flow (with design margin) comes first; head is read at that flow on the pump curve, then NPSH, materials and drive follow.
Materials are chosen for the fluid after the hydraulic duty is set.
Seal selection depends on the fluid and pressure once the pump is chosen; it is a later detail.
Motor size and efficiency follow from the brake power, which needs flow and head first.
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