Size Reduction, Material Handling and Mechanical Separations; Processing of Produce and Storage Systems
2. Screening, cleaning and grading; fluidization and conveying
Screening separates by size. No screen is perfect: some fines are carried over with the oversize and some oversize passes through torn or worn meshes. The effectiveness of a screen for recovering fines is the product of the recovery of fines in the underflow and the rejection of coarse material from it. Worked: a feed of 100 kg/h containing 40 kg of fines gives 30 kg/h of underflow that is 95 per cent fines (28.5 kg) and 70 kg/h of overflow. The recovery of fines is 28.5/40 = 0.7125; the coarse material in the feed is 60 kg, of which 1.5 kg reaches the underflow, so the rejection is 58.5/60 = 0.975; the effectiveness is 0.7125 × 0.975 = 69.5 per cent.
| Machine | Separates by |
|---|---|
| Air-screen cleaner | size (screens) and terminal velocity (air blast) |
| Aspirator | terminal velocity — light chaff and shrivelled seed lifted out |
| Specific-gravity separator | density, on an inclined, vibrating, air-fluidised deck |
| Indented cylinder | length — short seeds lifted in the indents, long ones left |
| Spiral separator | shape (roundness) — round seeds roll faster and fly outward |
| Magnetic separator | surface texture, via iron powder that sticks to rough or sticky seed coats |
| Colour (optical) sorter | colour, rejected by air jets |
Fluidization. When air flows up through a bed of granular solids, the pressure drop rises with velocity until it equals the bed’s buoyant weight per unit area, ΔP = (1 − ε)(ρ_p − ρ_f)gL, at the minimum fluidization velocity; beyond it the bed expands and behaves like a boiling liquid while ΔP stays constant. A 0.5 m bed of particles of density 1200 kg/m³ at voidage 0.4 fluidises at a pressure drop of 0.6 × 1200 × 9.81 × 0.5 = 3532 Pa (neglecting air density). Fluid beds dry, freeze and roast granular foods uniformly. Conveying: pneumatic conveyors carry grain in an air stream (dilute or dense phase), flexibly but at a high energy cost; bucket elevators lift grain vertically, discharging by centrifugal action at the head pulley when the bucket speed makes v²/r ≥ g — for 60 rev/min, a pulley radius of g/ω² = 9.81/(2π)² = 0.248 m — or by gravity (continuous and positive discharge) at lower speeds; screw (auger) conveyors, whose capacity is ≈ (π/4)(D² − d²) × pitch × N × bulk density × a loading factor, mix as they convey and suit short and inclined runs; and belt conveyors carry large tonnages over long distances with the least power per tonne.
3. Centrifugal separation, homogenization, filtration and membrane separation
A particle settling through a fluid under gravity reaches the Stokes terminal velocity v = D²(ρ_p − ρ)g/(18μ), valid in the laminar regime (particle Reynolds number below about 1). A 10 µm particle 100 kg/m³ denser than water (μ = 10⁻³ Pa·s) settles at only 5.45 µm/s — which is why fine separations are done in a centrifuge, where g is replaced by ω²r. The relative centrifugal force RCF = ω²r/g measures the gain: at 3000 rev/min and r = 0.1 m, RCF ≈ 1006. The cream separator is a disc-stack centrifuge that splits milk into skim and cream by density; decanters separate solids from liquids; cyclones separate dust or powder from air (spray-dryer outlets, grain-handling exhausts) by a vortex that throws particles to the wall.
Homogenization forces milk through a narrow valve at high pressure, breaking fat globules into much smaller ones so that — by Stokes’ law, since the rise velocity goes as D² — they no longer cream; a second stage at lower pressure breaks up the clusters the first forms. Filtration passes a slurry through a medium that holds the solids as a cake; at constant pressure the filtrate volume V and time t follow the Ruth equation t/V = (μαc/(2A²ΔP))V + μR_m/(AΔP), a straight line from which the cake and medium resistances are found; filter aids such as diatomaceous earth keep compressible cakes porous. Membrane separation is pressure-driven filtration through a semi-permeable membrane, graded by pore size: microfiltration (MF) removes bacteria and fat, ultrafiltration (UF) retains proteins while passing lactose and salts (whey protein concentrate, cheese milk), nanofiltration (NF) retains divalent ions and sugars while passing most monovalent ions, and reverse osmosis (RO) retains almost everything but water, at pressures that must exceed the solution’s osmotic pressure — used to pre-concentrate juice and milk without heat.
4. Processing of seeds, spices, fruits and vegetables; value addition
- Seeds: cleaning and upgrading on the machines above (removing weed seed, inert matter and light or broken seed), drying to a safe moisture, grading for size, treating with fungicide or insecticide, and packing in moisture-proof bags, with germination and purity tested.
- Spices: cleaning, blanching or curing where needed (turmeric is boiled, pepper may be dipped in hot water), drying (sun or mechanical), polishing, and grinding — cryogenic grinding with liquid nitrogen keeps the volatile oils that ambient grinding drives off as heat; oleoresins and essential oils are value-added extracts.
- Fruits and vegetables: washing, sorting and grading, peeling (mechanical, steam, lye), cutting, blanching, pulping and juice extraction, clarification, concentration, and preservation by canning, freezing, drying, pickling, or conversion to jams, jellies, squashes, ketchups and dehydrated products.
- Value addition raises the worth of produce by processing it nearer the farm — grading and packing, primary processing (cleaning, drying, milling), and secondary products — cutting post-harvest losses and adding employment; by-products (husk, bran, peel, pomace) become fuel, oil, feed or pectin.
5. Storage systems: controlled and modified atmospheres, perishables, godowns, bins and silos, packaging
Fresh produce keeps respiring after harvest, consuming its own sugars and releasing heat, CO₂ and water; storage life is extended by slowing that respiration. Controlled-atmosphere (CA) storage holds the produce in a gas-tight cold room with reduced O₂ (a few per cent) and raised CO₂, maintained by nitrogen generators and CO₂ scrubbers, with ethylene removed; apples are the classic CA crop. Modified-atmosphere packaging (MAP) creates a similar atmosphere passively inside a package whose film permeability is matched to the produce’s respiration. Perishable storage starts with pre-cooling to remove field heat quickly — hydro-cooling, forced-air cooling, vacuum cooling for leafy vegetables, and package icing — followed by cold storage at the right temperature and high relative humidity, avoiding chilling injury in tropical produce such as bananas and mangoes.
Grain storage keeps dry grain cool and away from insects, moulds, rodents and moisture. Godowns (warehouses) store bagged grain in stacks on dunnage, fumigated under covers; CAP (cover-and-plinth) storage stacks bags on raised plinths under polythene covers in the open. Bins and silos store grain in bulk and allow mechanised handling, aeration and fumigation. In a tall (deep) bin, wall friction carries part of the grain’s weight, so the pressures do not grow linearly with depth as they would in a liquid: Janssen’s equation gives the lateral pressure p = (γR/μ′)[1 − e^(−μ′kh/R)], with R the hydraulic radius of the bin, μ′ the grain-on-wall friction coefficient and k the ratio of lateral to vertical pressure — rising towards the limit γR/μ′ with depth. A bin is classed deep when the plane of rupture of the grain meets the opposite wall below the surface, and shallow when it cuts the grain surface, for which Rankine’s theory applies.
Packaging protects against moisture, gases, light, microbes and handling damage. Materials range from jute and paper to plastics — LDPE (a good moisture barrier but gas-permeable), HDPE, polypropylene, polyester, and multilayer laminates combining a barrier layer (aluminium foil, EVOH) with a sealable one — judged by their water-vapour and oxygen transmission rates. Machines include form-fill-seal machines (vertical for powders and granules, horizontal for solid items), vacuum and gas-flush packers, aseptic fillers, and can seamers.
Key takeaways
- Rittinger E = K_R(1/L₂ − 1/L₁) for fine grinding, Kick E = K_K ln(L₁/L₂) for coarse crushing, Bond E = W_i(10/√P − 10/√F) between.
- Screen effectiveness = recovery of the wanted fraction × rejection of the unwanted one.
- Each cleaner exploits one property: aspirator — terminal velocity, indented cylinder — length, spiral — roundness, gravity table — density.
- At fluidization ΔP = (1 − ε)(ρ_p − ρ_f)gL; a bucket elevator discharges centrifugally when ω²r ≥ g.
- Stokes v ∝ D²; RCF = ω²r/g; membranes rank MF > UF > NF > RO in pore size.
- CA storage lowers O₂ and raises CO₂ to slow respiration; Janssen’s equation gives the lateral pressure in a deep bin.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Grinding a material from 6 mm to 2 mm takes 10 kJ/kg. By Rittinger’s law, how much energy does it take to grind the same material from 6 mm to 1 mm, in kJ/kg?
Numerical answer — type the value.
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Answer: 25
K_R(1/2 − 1/6) = K_R/3 = 10, so K_R = 30 kJ·mm/kg; E = 30(1/1 − 1/6) = 30 × 5/6 = 25 kJ/kg. Scaling by the size reduction in millimetres (5/4 × 10 = 12.5) ignores that Rittinger’s energy follows new surface, which grows as 1/L.Grinding a material from 6 mm to 2 mm takes 10 kJ/kg. By Kick’s law, how much energy does it take to grind it from 6 mm to 1 mm, in kJ/kg (to one decimal place)?
Numerical answer — type the value.
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Answer: 16.3
K_K ln(6/2) = K_K ln 3 = 10, so K_K = 9.102 kJ/kg; E = 9.102 ln 6 = 9.102 × 1.792 = 16.3 kJ/kg. Rittinger’s law gives 25 kJ/kg for the same data — the two laws diverge as the product gets finer, and Kick’s is meant for coarse sizes.A material with a Bond work index of 12 kWh/t is ground from a feed 80-per-cent-passing size of 4000 µm to a product 80-per-cent-passing size of 400 µm. What energy is required, in kWh/t (to one decimal place)?
Numerical answer — type the value.
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Answer: 4.1
E = W_i(10/√P − 10/√F) = 12(10/√400 − 10/√4000) = 12(0.5 − 0.1581) = 12 × 0.3419 = 4.10 kWh/t. Swapping P and F gives a negative energy, and entering sizes in millimetres breaks the constant 10, which assumes micrometres.In a sieve analysis of ground maize, the cumulative percentages retained on the standard sieves add up to 350. What is the fineness modulus?
Numerical answer — type the value.
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Answer: 3.5
FM = Σ(cumulative percentage retained)/100 = 350/100 = 3.5. Summing the individual (non-cumulative) percentages always gives 100 and a meaningless FM of 1.A screen receives 100 kg/h of feed containing 40 per cent fines. The underflow is 30 kg/h and contains 95 per cent fines. Defining effectiveness as the recovery of fines in the underflow multiplied by the rejection of coarse material from it, what is the screen effectiveness, in per cent (to one decimal place)?
Numerical answer — type the value.
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Answer: 69.5
Fines in feed = 40 kg; in underflow = 0.95 × 30 = 28.5 kg; recovery = 28.5/40 = 0.7125. Coarse in feed = 60 kg; coarse in underflow = 1.5 kg; rejection = 58.5/60 = 0.975. E = 0.7125 × 0.975 = 0.695 = 69.5 per cent. Reporting the recovery alone (71.3 per cent) ignores the coarse contamination.A bed of granular solids 0.5 m deep has a voidage of 0.4 and a particle density of 1200 kg/m³. Neglecting the density of air, what is the pressure drop across the bed at incipient fluidization, in Pa (to the nearest whole number)?
Numerical answer — type the value.
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Answer: 3532
ΔP = (1 − ε)(ρ_p − ρ_f)gL = 0.6 × 1200 × 9.81 × 0.5 = 3531.6 ≈ 3532 Pa — the buoyant weight of the solids per unit area. Using ε instead of (1 − ε) gives 2354 Pa, the weight of air-filled space rather than of particles.A particle of diameter 10 µm is 100 kg/m³ denser than water, whose viscosity is 1 × 10⁻³ Pa·s. What is its Stokes settling velocity under gravity, in µm/s (to two decimal places)? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 5.45
v = D²Δρg/(18μ) = (10 × 10⁻⁶)² × 100 × 9.81/(18 × 10⁻³) = 9.81 × 10⁻⁸/0.018 = 5.45 × 10⁻⁶ m/s = 5.45 µm/s. Forgetting to square the diameter gives a velocity a hundred thousand times too large.A centrifuge bowl rotates at 3000 rev/min with an effective radius of 0.1 m. What is the relative centrifugal force (the ratio of centrifugal to gravitational acceleration), to the nearest whole number? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 1006
ω = 2π × 3000/60 = 314.16 rad/s; ω²r = 98 696 × 0.1 = 9870 m/s²; RCF = 9870/9.81 = 1006. Using N in rev/min directly (3000² × 0.1/9.81) gives over 90 000, forgetting to convert to rad/s.The head pulley of a bucket elevator turns at 60 rev/min. What is the smallest pulley radius (to the bucket centre) at which grain is thrown out by centrifugal discharge, in m (to three decimal places)? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 0.248
Centrifugal discharge needs ω²r ≥ g at the top of the pulley. ω = 2π rad/s, so r = g/ω² = 9.81/39.48 = 0.248 m. Using ω = 60 rad/s (rev/min mistaken for rad/s) gives an impossibly small 2.7 mm.Controlled-atmosphere storage extends the life of fresh apples mainly by:
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Answer: A — lowering O₂ and raising CO₂ around the fruit to slow its respiration
Respiration consumes O₂ and releases CO₂, so a low-O₂, high-CO₂ atmosphere at low temperature slows it and delays ripening and softening. Raising O₂ does the opposite, dry air would shrivel the fruit, and freezing would destroy its texture.Which of the following statements about membrane separations are correct?
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Answer: A — Reverse osmosis retains dissolved salts and sugars and passes almost only water; B — Ultrafiltration of whey retains proteins while passing lactose and salts; C — Microfiltration has the largest pores of the four processes
Pore size falls from MF through UF and NF to RO. RO passes essentially only water; UF holds the large protein molecules and passes lactose and minerals; MF has the widest pores. NF does the reverse of the fourth statement — it retains divalent ions and passes most monovalent ones, which is how it demineralises whey partially.Janssen’s equation is used in the design of grain silos to find:
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Answer: A — the lateral pressure on the walls of a deep bin, which approaches a limit with depth because of wall friction
Janssen balanced the weight of a slice of grain against the friction on the wall around it, which is why pressure in a deep bin levels off towards γR/μ′ instead of growing like a liquid’s. The linear, liquid-like pressure of a shallow bin is Rankine’s case.Match each cleaning machine with the property it uses. Which of the following pairings are correct?
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Answer: A — Aspirator — terminal velocity; B — Indented cylinder — length; D — Specific-gravity separator — density
The aspirator lifts light material in a rising air stream; the indented cylinder picks short seeds out of long ones; the gravity table stratifies by density. The spiral separator sorts by roundness — round seeds roll faster and are flung outward — and colour sorting is done by optical sorters, so the third pairing is wrong.