Separations carry 8–12 questions, from Fick’s law to a full column. Expect graphical stage work, mass-transfer coefficients and the drying and humidification calculations that the outline files under this area.
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ExamFE Chemical
questions on the exam8–12
problems on this page16
of 17 knowledge areas10
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What the specification covers
The outline lists six subtopics. In our words:
Molecular diffusion, steady and unsteady, and estimating diffusivities.
Convective mass transfer: film and overall coefficients, eddy diffusion.
Continuous-contact methods: transfer units, HTU and HETP, packed height.
Humidification, drying and evaporation as unit operations.
16 separation problems
0 / 16 answered · 0 correct
Question 1 of 16 · Mass Transfer and Separation · 8–12 items
A reverse osmosis process operates with a solute flux modeled by Js = B(Cfeed − Cpermeate). An operator notes that doubling the applied trans-membrane pressure significantly increases pure water recovery (stage-cut) but yields only a minimal change in the total mass of salt crossing the membrane per unit area. Which statement best explains this transport behavior?
Solution
Jw = A(ΔP − Δπ) · Js = B(Cfeed − Cpermeate)
Raising ΔP raises water flux but not salt flux, so more water carries the same salt: rejection improves with pressure (solution-diffusion model).
In the solution-diffusion model Js = B(Cfeed − Cpermeate) has no pressure term, so salt passage stays nearly constant while water flux rises with ΔP.
More water does dilute the permeate, but that concerns permeate concentration, not the salt mass flux, which the observation says barely changed.
RO membranes are dense, not porous sieves, and salt keeps crossing by diffusion at any recovery.
Water flux follows Jw = A(ΔP − Δπ), so it is driven by the applied pressure; that is why recovery rose when the pressure doubled.
The opposite holds: solute flux hardly depends on pressure, so higher pressure improves apparent rejection.
Question 2 of 16 · Mass Transfer and Separation · 8–12 items
Equipment section
Efficiency
Partial condenser
100 %
Column trays (overall)
60 %
Partial reboiler
100 %
A binary distillation column is designed to separate a chemical mixture using a partial condenser and a partial reboiler. A graphical McCabe-Thiele analysis determines that exactly 14 total theoretical equilibrium stages are required to achieve the specified distillate and bottoms compositions. Using the provided equipment efficiency parameters, determine the actual number of trays required inside the distillation column.
FE Reference Handbook, Chemical Engineering: distillation (overall column efficiency).
8 multiplies the 12 theoretical trays by the 60 % efficiency (7.2, rounded up) instead of dividing.
12 is the number of theoretical trays inside the column (14 − 2); it has not been divided by the 60 % tray efficiency.
13 subtracts only the partial condenser (14 − 1) and applies no tray efficiency.
14 counts the total theoretical stages, including the two that the partial condenser and partial reboiler provide, with no efficiency applied.
The partial condenser and the partial reboiler each act as one equilibrium stage, so 14 − 2 = 12 theoretical trays; 12/0.60 = 20 actual trays.
Question 3 of 16 · Mass Transfer and Separation · 8–12 items
A porous solid is dried in a batch dryer under constant air conditions, and the drying rate R is plotted against the free moisture content X. At first dR/dX = 0. Below a critical moisture content, dR/dX > 0. Which of the following best describes the mechanism that governs the process when dR/dX > 0?
Solution
Constant-rate period (dR/dX = 0): the surface stays wet and external heat and mass transfer control. Below the critical moisture content (falling-rate period, R falls with X) internal movement of moisture controls.
If capillary flow kept the surface wet at a constant rate, the drying rate would stay constant rather than fall with X.
External (gas-film) resistance controls the constant-rate period; in the falling-rate period the limit has moved inside the solid.
Surface heat and mass transfer set the constant drying rate; once the rate depends on X, the interior controls.
Below the critical moisture content the surface is no longer wet, and the rate falls with X because moisture must diffuse or wick from the interior.
Free evaporation of unbound surface water is the constant-rate period, where dR/dX = 0.
Question 4 of 16 · Mass Transfer and Separation · 8–12 items
A distillation column operates at total reflux. The desired distillate composition is xD = 0.95. The vapor-liquid equilibrium data for the mixture is provided in the table below. How many theoretical equilibrium stages are required to reach a bottoms mole fraction of 0.50?
Two theoretical stages (McCabe-Thiele stepping at total reflux).
One step from y = 0.95 only reaches x = 0.80; the bottoms target of 0.50 needs a second step.
At total reflux the operating line is y = x: from y = 0.95 the curve gives x = 0.80, then y = 0.80 gives x = 0.50, so two stages reach the bottoms target.
Three stages counts the total condenser as an extra equilibrium stage; a total condenser changes no composition.
Four stages adds both the condenser and a separate reboiler stage to the two steps; the two steps already include the reboiler.
Five stages counts each row of the VLE table as a stage instead of stepping between the curve and the diagonal.
Question 5 of 16 · Mass Transfer and Separation · 8–12 items
A distillation column is operating with a saturated liquid feed. If the feed is changed to saturated vapor while the feed flow rate, feed composition, distillate flow rate and reflux ratio stay the same, how does this change affect the internal vapor and liquid flows in the stripping section?
Solution
L̄ = L + qF · V̄ = V − (1 − q)F
With R and D fixed, L and V are unchanged; q: 1 → 0 gives L̄ = L (down from L + F) and V̄ = V − F (down from V).
Stripping liquid does fall, but vapor falls too: with saturated vapor feed the feed adds nothing to the reboiler vapor.
L̄ = L + qF and V̄ = V − (1 − q)F; changing q from 1 to 0 lowers L̄ by F and lowers V̄ by F.
Liquid in the stripping section falls, because the feed no longer adds liquid there.
With a fixed reflux ratio the rectifying vapor is unchanged, so the stripping vapor V̄ = V − F drops by the feed rate.
Removing liquid from the feed can only reduce stripping liquid, and the feed vapor now joins the rectifying section instead of the stripping one.
Question 6 of 16 · Mass Transfer and Separation · 8–12 items
A reverse osmosis membrane is used to desalinate seawater. The feed water operates at a hydraulic pressure of 65 bar and has an osmotic pressure of 32 bar. The permeate stream is collected at a hydraulic pressure of 2 bar and has an osmotic pressure of 1 bar. If the membrane has a water permeability constant of 1.5 L/(m2 · h · bar), determine the theoretical steady-state volumetric water flux across the membrane.
49.5 uses only the feed side (65 − 32 bar) and ignores both permeate pressures.
94.5 counts the hydraulic difference only (63 bar) and forgets that osmotic pressure opposes the flow.
Question 7 of 16 · Mass Transfer and Separation · 8–12 items
A solid feed enters a dryer with an initial moisture content of 0.60 kg water/kg dry solid. After passing through the drying process, the product leaves with a moisture content of 15% measured on a wet basis. Calculate the mass of water removed by the dryer per 100 kg of completely dry solid.
water removed = 100 kg × (0.60 − 0.1765) = 42.35 kg
Product water = 100 × 0.15/0.85 = 17.65 kg per 100 kg dry solid; removed = 60.00 − 17.65 = 42.35 kg.
36.00 kg takes 15 % of the total initial mass (160 kg) as the water left, mixing the wet basis with the feed instead of the product.
26.47 kg takes 100 kg of wet feed (62.5 kg dry solid) as the basis instead of 100 kg of dry solid.
17.65 kg is the water still in the product (100 × 0.15/0.85), not the water removed.
9.00 kg is 15 % of the initial 60 kg of water, a misreading of the wet-basis moisture as a fraction removed.
Question 8 of 16 · Mass Transfer and Separation · 8–12 items
A countercurrent liquid-liquid extraction system is being designed to remove a solute from a feed stream using a pure, solute-free solvent. The process requires a fractional solute recovery of 95%. If the operating flow rates and equilibrium conditions dictate an extraction factor (E = mV/L) of 1.5, determine the number of theoretical ideal stages required using the Kremser equation.
FE Reference Handbook, Chemical Engineering: Kremser equation for countercurrent stages.
19.00 is the ratio of recovered to unrecovered solute (0.95/0.05), not a stage count.
8.39 adds one to the large-E approximation; the approximation itself already overstates N.
7.39 = ln(1/φ)/ln E uses the large-E approximation, which is poor at E = 1.5.
5.91 is N + 1 from the Kremser expression; one still has to be subtracted.
With fraction unextracted φ = 0.05: EN+1 = 1 + (E − 1)/φ = 11, so N + 1 = ln 11/ln 1.5 = 5.91 and N = 4.91.
Question 9 of 16 · Mass Transfer and Separation · 8–12 items
A multi-component distillation column is designed to separate a mixture characterized by a constant relative volatility of 2.0 between the light and heavy keys. The required distillate composition is 95 mol% light key and 5 mol% heavy key. The bottoms composition must be 5 mol% light key and 95 mol% heavy key. Using shortcut relations, determine the minimum number of theoretical stages required at total reflux.
17.00 squares the separation factor a second time (ln 3612).
Question 10 of 16 · Mass Transfer and Separation · 8–12 items
A chemical plant is selecting an extraction solvent for a process operating continuously at 40°C. High mass transfer efficiency is desired, but strict corporate safety protocols prohibit utilizing any solvent capable of forming a flammable vapor-air mixture under normal operating temperatures (i.e., the process temperature must remain strictly below the flash point). Based on the properties in the table, which solvent is the optimal choice?
Solvent candidate
Selectivity
Distribution ratio
Flash point (°C)
Hexane
15.0
2.5
−22
Toluene
18.0
3.0
4
Cyclohexanol
12.0
1.8
68
Propylene carbonate
5.0
0.5
132
Solution
Screen first: flash point > 40 °C leaves cyclohexanol (68 °C) and propylene carbonate (132 °C). Then rank by selectivity and distribution ratio: cyclohexanol (12.0, 1.8) beats propylene carbonate (5.0, 0.5).
Two solvents, cyclohexanol (68 °C) and propylene carbonate (132 °C), have flash points above the 40 °C process temperature.
Toluene has the best selectivity and distribution ratio, but its 4 °C flash point breaks the safety rule.
Of the two solvents whose flash points exceed 40 °C, cyclohexanol has the higher selectivity (12.0) and distribution ratio (1.8), so it extracts best.
Hexane's flash point is −22 °C, far below 40 °C, so it forms flammable vapor at operating temperature.
Propylene carbonate passes the flash-point screen, but its selectivity of 5.0 and distribution ratio of 0.5 make it the weakest extractant that qualifies.
Question 11 of 16 · Mass Transfer and Separation · 8–12 items
System
kGa (mol/m³·s·atm)
kL a (1/s)
H (atm·m³/mol)
Ammonia-Water
10.0
0.50
0.01
Oxygen-Water
5.0
0.02
10.0
Carbon Dioxide-Water
2.0
0.10
1.0
Hydrogen Sulfide-Water
20.0
0.05
5.0
Based on the overall mass transfer parameters provided in the table, which absorption system is primarily limited by gas-phase mass transfer resistance?
Given
Ammonia-water: kGa = 10.0, kLa = 0.50, H = 0.01
Oxygen-water: kGa = 5.0, kLa = 0.02, H = 10.0
CO2-water: kGa = 2.0, kLa = 0.10, H = 1.0
H2S-water: kGa = 20.0, kLa = 0.05, H = 5.0
Find
Gas-phase-controlled system
Solution
1/KGa = 1/kGa + H/kLa
NH3: 0.10 + 0.01/0.50 = 0.10 + 0.02 → gas film 83 % of the total
O2, CO2 and H2S all have H/kLa ≫ 1/kGa (liquid-film control). FE Reference Handbook, Chemical Engineering: two-film theory.
H/kLa = 5.0/0.05 = 100 against 1/kGa = 0.05, so H2S here is strongly liquid-film controlled.
Oxygen is sparingly soluble (H = 10): H/kLa = 500 dwarfs 1/kGa = 0.2, the classic liquid-film-controlled case.
H/kLa = 1.0/0.10 = 10 versus 1/kGa = 0.5: the liquid film carries about 95 % of the resistance.
Ammonia-water breaks the pattern: its gas-film term (0.10) is five times its liquid-film term (0.02).
1/kGa = 0.10 and H/kLa = 0.01/0.50 = 0.02: the gas film holds 83 % of the resistance, as expected for a very soluble gas.
Question 12 of 16 · Mass Transfer and Separation · 8–12 items
A spherical particle of radius R releases a solute by steady-state diffusion into a large stagnant fluid, with no reaction. The concentration at the particle surface is held at Cs, and the concentration far from the particle is 0. Which boundary conditions apply to the diffusion equation in spherical coordinates?
Thickness sets the flux of every gas through the membrane equally, so it changes productivity, not selectivity.
Pressure ratio affects how much of the ideal selectivity is realized, but it does not characterize the membrane itself.
The ideal selectivity of a dense membrane is αA/B = PA/PB, the ratio of the two gases' permeabilities.
Porosity matters for porous filters; gas-separation membranes are dense, and selectivity comes from solubility and diffusivity differences.
Question 15 of 16 · Mass Transfer and Separation · 8–12 items
In a liquid-liquid extraction process, the distribution coefficient (K) is defined as:
Solution
K = yextract/xraffinate at equilibrium
The extraction factor E = KS/F combines K with the solvent-to-feed ratio.
The minimum solvent ratio is a design limit read from the operating and equilibrium lines, not a definition of the distribution coefficient.
K = y/x, the solute concentration in the extract over that in the raffinate once the phases are at equilibrium.
Recovery depends on K, the solvent ratio and the number of stages; it is a result, not the definition of K.
That is a flow ratio set by the designer; K is an equilibrium property of the solute between the two phases.
Question 16 of 16 · Mass Transfer and Separation · 8–12 items
A liquid-liquid extraction system separates into two distinct phases in thermodynamic equilibrium. The compositions at the endpoints of the tie-line are provided in the table below. If the overall two-phase mixture contains 52.0 wt% solvent and 11.0 wt% solute, determine the mass percentage of the total mixture that constitutes the extract phase.
73.3 % divides the overall solute by the extract solute (11/15) and ignores the raffinate end of the tie line (5 % solute).
65.0 % divides the mixture's solvent by the extract's (52/80), forgetting that the raffinate also holds 10 % solvent.
Lever rule on solvent: fE = (52 − 10)/(80 − 10) = 0.60; the solute balance gives the same, (11 − 5)/(15 − 5) = 0.60.
52.0 % is the overall solvent content of the mixture, not a phase fraction.
40.0 % is the raffinate share, (80 − 52)/(80 − 10); the lever arm opposite the extract end gives the extract.
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Common traps
The q-line for the feed. A saturated-liquid feed (q = 1) draws a vertical q-line, a saturated vapor (q = 0) a horizontal one. In general:
y = (q/(q − 1))·x − xF/(q − 1)
Minimum reflux used as the design reflux. Minimum reflux needs infinite stages; a design always runs above it.
HTU·NTU against HETP. Packed height is Z = HTU·NTU for continuous contact and Z = N·HETP when stages were counted. The two are not interchangeable without the operating-to-equilibrium slope ratio.
Mole and mass fractions on one diagram. Equilibrium curves are in mole fraction; convert mass compositions before plotting.
Axes swapped. On an x-y diagram the liquid composition is x on the horizontal axis; reading a vapor value off it shifts every stage.
Sources: McCabe, Smith and Harriott, Unit Operations of Chemical Engineering; Seader and Henley, Separation Process Principles.
The outline names graphical equilibrium-stage methods and McCabe-Thiele, so expect to read or step off stages on a given diagram rather than plot one from scratch.
Where does drying sit in the outline?
Yes. Humidification, drying and evaporation are listed under this area, not under thermodynamics.
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