Textile Fibres: Classification, Natural and High-Performance Fibres, Identification and Blends
2. Essential requirements of a fibre-forming polymer
Few polymers make good fibres, because fibre use needs strength along the axis, modest extension and stability in wash and wear. A fibre-forming polymer must satisfy the following.
- Linear, long chain with few or no branches or cross-links, so chains can be pulled parallel to the fibre axis. A cross-linked network cannot melt or dissolve, so it cannot be spun.
- Sufficient molecular weight (degree of polymerisation of the order of a hundred or more) so that entanglement and chain overlap give strength; but too high a molecular weight makes the melt or solution too viscous to extrude.
- Regular, symmetrical chain that can pack into crystals and be oriented by drawing. This is why isotactic polypropylene is a fibre and atactic is a soft wax-like solid.
- Inter-chain attraction: hydrogen bonds (amide, urethane, hydroxyl) or strong dipoles (nitrile, ester) hold oriented chains together; a purely van der Waals polymer needs very high molecular weight (polyethylene, polypropylene).
- Processability and thermal window: it must either melt without decomposing or dissolve in a practical solvent. Its Tm must be above ironing temperature (so the fabric does not melt) yet below the decomposition temperature (so it can be melt-spun), and its Tg should be above the temperature of use for dimensional stability.
- Further needs: adequate resistance to light, chemicals and moisture, and an acceptable cost.
3. Gross and fine structure of cotton, wool and silk
Cotton is a seed hair that is about 90% cellulose (cellulose I crystal form), with wax, pectin and protein in the outer layers. Gross structure: a flattened, twisted ribbon with convolutions that reverse in sense, a kidney-bean cross-section when mature, and a collapsed central canal, the lumen. Fine structure, from outside in: the cuticle (wax and pectin), the primary wall (a thin net of cellulose fibrils), the secondary wall (the bulk, deposited as daily growth layers in which fibrils spiral about the axis and reverse at points along the length) and the lumen. Mature fibres have a thick secondary wall; immature fibres have a thin wall, are ribbon-like, take dye poorly and form neps; dead fibres are the extreme case. Hence fineness, maturity and length decide spinning quality.
Wool is the protein keratin, rich in the sulphur amino acid cystine whose disulphide cross-links give stability and resilience. Gross structure: a crimped, round to oval fibre with an outer cuticle of overlapping scales (points toward the tip, causing a directional frictional effect and hence felting), a cortex of spindle-shaped cells forming the bulk, and in coarse fibres a hollow medulla. The cortex is bilateral: ortho- and para-cortex differ in cystine content and swelling, which produces crimp. Fine structure: cortical cells contain macrofibrils built of microfibrils (α-helical coiled-coil keratin) embedded in an amorphous, sulphur-rich matrix. Stretching converts α-helix toward the extended β-form, which is why wool extends greatly and recovers when moist.
Silk from Bombyx mori is a continuous filament of the protein fibroin (rich in glycine and alanine, forming β-pleated-sheet crystallites), spun as two triangular-section filaments (brins) held together by the gummy protein sericin. Degumming in hot soap or alkali removes sericin, roughly a quarter of the raw weight, giving the soft lustrous fibre. Wild (tussah) silk has a flatter, more fibrillar structure and is coarser.
4. Bast fibres, and properties and uses of natural and man-made fibres
Bast fibres (flax, jute, hemp, ramie) come from the phloem of the stem. Each technical fibre is a bundle of short ultimate cells cemented by pectin and, in jute and hemp, lignin. Retting (microbial, water or dew) breaks down the cement so the bundles can be separated. Flax and ramie are strong, lustrous, low in extension and used for linen and fine fabrics; jute is lignified, coarse and brittle, the cheap fibre for hessian, sacking and carpet backing; hemp is coarse and durable (ropes, canvas). All are largely cellulose, so they have high regain and lose strength less on wetting than viscose.
| Fibre | Density, g/cm³ | Regain, % | Notable property and use |
|---|---|---|---|
| Cotton | 1.52 | about 8 | Comfort, stronger wet; apparel, home textiles |
| Wool | 1.31 | 13 to 16 | Warmth, resilience, flame-resistant; suiting, knitwear |
| Silk | 1.33 | about 11 | Lustre, drape, high strength; luxury apparel |
| Viscose | 1.52 | 12 to 13 | Soft, absorbent, weak when wet; apparel, hygiene |
| Nylon 6, 66 | 1.14 | about 4 | Toughness, abrasion resistance; hosiery, carpets, ropes |
| Polyester (PET) | 1.38 | about 0.4 | Crease resistance, easy care; apparel, blends, bottles-to-fibre |
| Acrylic | 1.17 | 1 to 2 | Wool-like bulk, light stable; knitwear, blankets |
| Polypropylene | 0.91 | about 0 | Lightest, hydrophobic; geotextiles, carpets, nonwovens |
5. Carbon, aramid and UHMWPE fibres
These high-performance fibres reach strength and modulus far above conventional textile fibres by using chains that are rigid or fully extended and highly oriented.
- Carbon fibre: made by heating a precursor, mostly polyacrylonitrile (PAN). Steps: stabilisation in air at about 200 to 300 °C (cyclisation and oxidation into a ladder structure that no longer melts), carbonisation in inert gas at about 1000 °C or more (non-carbon atoms leave), and optionally graphitisation at higher temperature for higher modulus. Result: very high modulus and strength, low density (about 1.8 g/cm³), electrical conductivity, but brittle with low breaking extension. Used mainly in composites (aerospace, sporting goods).
- Aramid (aromatic polyamide). Para-aramid (poly-p-phenylene terephthalamide, e.g. Kevlar-type) has rigid rod chains that form a liquid-crystalline solution in concentrated sulphuric acid, is dry-jet-wet spun, and gives very high strength and modulus with density about 1.44 g/cm³; it does not melt but decomposes at high temperature; poor UV resistance. Uses: ballistic armour, tyre cord, ropes, composites. Meta-aramid (Nomex-type) has kinked chains, lower strength, and is valued for heat and flame resistance (protective clothing, filter felts).
- UHMWPE (ultra-high molecular weight polyethylene, molar mass of millions): gel spun and ultra-drawn to nearly fully extended chains. Very high specific strength, density about 0.97 g/cm³ (floats on water), excellent abrasion and chemical resistance, near-zero moisture regain; but low melting point (about 150 °C) and marked creep. Uses: ropes, ballistic panels, cut-resistant gloves, sails.
6. Fibre identification and blend analysis
Identification proceeds from simple to specific tests. Physical: microscopy (longitudinal view and cross-section: cotton twisted ribbon, wool scales, viscose striations, silk smooth triangles), density in a density-gradient column, and melting point. Chemical: burning (cellulosics burn quickly with a paper smell and soft ash; wool and silk burn with a burnt-hair smell and leave a crushable bead; nylon and polyester melt and leave a hard bead; polypropylene melts and burns with a waxy smell), solubility, and staining. Instrumental methods (IR spectra, DSC melting) settle doubtful cases.
| Component removed | Reagent (typical) | Residue |
|---|---|---|
| Wool or silk | Hot dilute sodium hydroxide (wool); concentrated hydrochloric acid (silk) | Cotton, polyester |
| Cotton or viscose | About 70% sulphuric acid at room temperature | Polyester, acrylic, wool partly |
| Nylon | Formic acid (about 85%) | Cotton, polyester, wool |
| Acrylic | Dimethylformamide, hot | Cotton, wool, polyester |
Blend analysis by selective dissolution. Take a dry, cleaned sample of mass m (oven-dry). Remove one component with the reagent, wash, dry and weigh the residue r. Because the reagent attacks the undissolved fibre slightly, multiply by a correction factor d (a number above 1, the ratio of original to treated dry mass of that fibre alone). Then the percentage of the undissolved component is P = 100 × r × d / m, and the rest is the dissolved component. Worked example: 2.000 g oven-dry wool–cotton sample; after boiling with alkali the dry cotton residue is 1.230 g, d = 1.02. Cotton = 1.230 × 1.02 = 1.2546 g = 62.73%; wool = 37.27%. When blends are bought by conditioned mass, convert each dry fraction with its commercial regain: for 67:33 polyester:cotton (dry) with regains 0.4% and 8.5%, 100 kg dry stock weighs 67 × 1.004 + 33 × 1.085 = 103.07 kg at commercial regain.
Key takeaways
- Fibres are natural (vegetable, animal, mineral) or man-made (regenerated, synthetic, inorganic); staple versus filament; 1 tex = 9 denier = 10 dtex, and a larger number means a coarser fibre.
- A fibre-forming polymer is linear, of adequate molecular weight, regular enough to crystallise and orient, has inter-chain attraction, and has Tm between ironing and decomposition temperatures; PAN and cellulose decompose before melting and are solution-spun.
- Cotton: cuticle, primary wall, spiral secondary wall, lumen, with maturity governing quality. Wool: keratin with disulphide cross-links, scaly cuticle, bilateral cortex. Silk: fibroin brins glued by sericin, removed by degumming.
- Bast fibres are stem fibres freed by retting; jute is lignified and coarse, flax and ramie fine and strong. Moisture regain is on dry mass: M = 100R/(100 + R).
- Carbon (PAN, stabilise then carbonise, brittle, stiff), para-aramid (liquid-crystal rod chains, heat resistant, strong) and UHMWPE (gel-spun, density below 1, low melting, creeps) are the three high-performance fibres.
- Blend analysis: dissolve one component, dry the residue, apply the correction factor, P = 100 r d/m; buying by conditioned mass needs the mass-weighted commercial regain.
Practice questions (17)
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.
Which of the following is a regenerated cellulosic fibre, in which cellulose is dissolved and then re-formed into filaments?
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Answer: B — Viscose, because cellulose is dissolved as xanthate and regenerated by acid
Viscose is cellulose that has been dissolved (as cellulose xanthate) and regenerated into filaments by an acid bath, so the product is cellulose again. Nylon 6 is a synthetic polyamide, flax is a natural bast fibre, and degumming only purifies natural silk; none of them involves dissolving and re-forming cellulose.Isotactic polypropylene is a good fibre-forming polymer whereas atactic polypropylene of the same molecular weight is not. The main reason is that the atactic form
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Answer: C — has irregular side-group placement, so chains cannot pack into crystals or orient well
Tacticity controls chain regularity. Isotactic chains pack into crystals and orient on drawing, giving strength; atactic chains cannot crystallise and give a soft, weak material. Both forms are linear, have only van der Waals forces (no hydrogen bonds), and are not cross-linked, which rules out the other statements.In the wool fibre, crimp arises mainly because the cortex is bilateral. The two halves differ in
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Answer: A — cystine content and swelling behaviour (ortho- and para-cortex)
Ortho-cortex and para-cortex differ in cystine content, packing and swelling, so on wetting or drying the two sides change length unequally and the fibre curls into crimp. Scale direction explains felting, not crimp; the medulla is a central canal in coarse wool; sericin belongs to silk, not wool.Degumming of raw silk removes
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Answer: D — sericin, leaving the fibroin filaments
Sericin is the gummy protein cementing the two fibroin brins; hot soap or alkali dissolves it and leaves soft, lustrous fibroin. Removing fibroin would destroy the fibre. Silk has no significant disulphide cross-linking (that is wool), and wax and pectin are cotton constituents.A polyester–cotton fabric is to be analysed by selective dissolution so that the polyester can be weighed as the residue. Which treatment is suitable?
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Answer: B — Treatment with about 70% sulphuric acid at room temperature, which dissolves cotton only
Cellulose dissolves in about 70% sulphuric acid at room temperature while polyester is unaffected, so the polyester remains as the weighable residue. Dilute alkali does not dissolve cotton (it is the reagent for wool), formic acid dissolves nylon rather than cotton, and heating would degrade both fibres.Which statement about UHMWPE fibre is correct?
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Answer: A — It floats on water, softens near 150 °C and creeps
UHMWPE has density about 0.97 g/cm³ (floats), a melting point near 150 °C and noticeable creep. It is gel spun, has essentially zero regain, and melts rather than decomposing; carbonising PAN is the route to carbon fibre.Para-aramid and meta-aramid fibres are both aromatic polyamides, yet para-aramid is chosen for ballistic armour and meta-aramid for flame-protective clothing. The reason is that
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Answer: D — para chains are straight rigid rods and orient to very high strength, kinked meta chains do not
Para linkage gives straight, rigid, rod-like chains that form a liquid-crystalline dope and orient almost perfectly, hence very high strength and modulus. Meta linkage kinks the chain, lowering strength though retaining good thermal stability. Both contain amide groups with hydrogen bonding, and neither melts conveniently, so both are solution-spun.In making carbon fibre from PAN, the precursor is first heated in air at about 200 to 300 °C. The purpose of this stabilisation step is to
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Answer: A — make an infusible cyclised ladder structure
Heating in air cyclises the nitrile groups and introduces oxygen, giving an infusible ladder polymer; without it the fibre would melt or fuse when taken to carbonisation temperature. Graphitisation, not stabilisation, raises the graphitic order, and there is no sericin in PAN.A cotton sample contains many immature fibres. Compared with mature cotton these fibres have
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Answer: B — a thin wall, ribbon form and a tendency to form neps
Immature cotton has a thin secondary wall, so the fibre collapses to a ribbon, is weaker, entangles into neps and takes dye unevenly. A thick wall characterises mature fibre, lignin is a bast-fibre component, and scales belong to wool.Which of the following are essential requirements of a polymer that is to be made into textile fibre?
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Answer: A — The chains must be essentially linear and regular enough to be oriented and, preferably, crystallised; C — It must be of adequate molecular weight, and be able to melt without decomposing or dissolve in a usable solvent
A fibre polymer must be linear and regular (so chains orient and crystallise), of adequate molecular weight, and processable by melting or dissolving. A cross-linked network can neither melt nor dissolve, so it cannot be spun, and an oligomer of very low degree of polymerisation lacks entanglement and gives a weak, brittle product.Which of the following are bast fibres?
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Answer: A — Flax; B — Ramie; D — Jute
Flax, ramie and jute are obtained from the phloem of plant stems and are freed by retting. Sisal is a leaf fibre taken from the leaves of the agave plant, so it is not a bast fibre.Which statements about blend analysis by selective dissolution are correct?
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Answer: A — The residue is dried to constant mass before weighing so that moisture is not counted as fibre; B — A correction factor above 1 compensates for a small loss of the undissolved fibre in the reagent
Constant dry mass and the correction factor d (above 1) are standard practice. Hot alkali dissolves wool, not cotton, so the reagent choice stated is reversed; and the dry-mass percentages must be converted with commercial regain if conditioned-mass proportions are wanted.A 2.000 g oven-dry wool–cotton sample is boiled in dilute alkali to dissolve the wool. The dried cotton residue is 1.230 g. The correction factor for loss of cotton in the reagent is d = 1.02. Percentage of wool by dry mass, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 37.3
Corrected cotton = 1.230 × 1.02 = 1.2546 g, so cotton = 1.2546/2.000 = 62.73%. Wool = 100 − 62.73 = 37.27%, which is 37.3% to one decimal place. Omitting d would give 38.5%.A polyester–cotton blend is 67% polyester and 33% cotton by oven-dry mass. Use commercial regains of 0.4% for polyester and 8.5% for cotton. The mass at commercial regain of a lot whose oven-dry mass is 100 kg, in kg to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 103.07
Each dry component is multiplied by (1 + R/100). Polyester: 67 × 1.004 = 67.268 kg. Cotton: 33 × 1.085 = 35.805 kg. Total = 67.268 + 35.805 = 103.073, i.e. 103.07 kg. Equivalently the blend regain is 3.07%.A polyester filament of circular cross-section is 1.5 denier. Take the polyester density as 1.38 g/cm³ and 1 tex = 9 denier. Its diameter in micrometres, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 12.4
Linear density = 1.5/9 = 0.1667 tex = 0.1667 g per 1000 m = 1.667 × 10⁻⁷ kg/m. Area = (mass per length)/density = 1.667 × 10⁻⁷ / 1380 = 1.208 × 10⁻¹⁰ m². Diameter = √(4A/π) = 1.240 × 10⁻⁵ m = 12.4 µm.A fibre sample weighs 5.40 g when conditioned and 4.98 g after oven-drying. Its moisture regain, as a percentage of oven-dry mass to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 8.43
Regain = (W − W₀)/W₀ × 100 = (5.40 − 4.98)/4.98 × 100 = 0.42/4.98 × 100 = 8.43%. The moisture content (on conditioned mass) would be 7.78%, the common wrong answer.A continuous-filament polyester yarn is labelled 75 denier with 36 filaments, all of equal fineness. The linear density of one filament in dtex (1 dtex = 1 g per 10,000 m; 1 denier = 1 g per 9000 m), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.31
Yarn dtex = 75 × 10/9 = 83.33. Per filament = 83.33/36 = 2.315, i.e. 2.31 dtex (denier per filament is 75/36 = 2.08). Dividing without converting units, 2.08, is the usual slip.