GATE Textile Engineering Textile Fibres Flashcards

24 sample cards from the Textile Fibres deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. How do crystalline and amorphous regions each influence fibre properties?

   Crystalline regions give strength, stiffness (modulus), dimensional stability, and resistance to solvents/dyes/moisture. Amorphous regions provide flexibility, elasticity, moisture absorption, dyeability and accessibility to chemicals. Higher crystallinity raises strength but lowers extensibility and dye uptake.

2. Define the glass transition temperature ($T_g$) of a polymer fibre.

   $T_g$ is the temperature at which the amorphous regions change from a hard, rigid, glassy state to a soft, rubbery state. It is a second-order transition: below $T_g$ chain segments are frozen; above $T_g$ they gain large-scale segmental mobility. It strongly affects setting, dyeing and dimensional stability.

3. Define crystallization temperature ($T_c$) and where it lies relative to $T_g$ and $T_m$.

   $T_c$ is the temperature at which a polymer's mobile amorphous chains, on cooling from the melt (or heating from the glass), arrange into crystals (cold crystallization). It lies between $T_g$ and $T_m$: $$T_g < T_c < T_m$$

4. Define the melting temperature ($T_m$) of a fibre and its thermodynamic nature.

   $T_m$ is the temperature at which the ordered crystalline regions melt into a disordered liquid; it is a first-order transition occurring over a range. At $T_m$ the free energy change $\Delta G = 0$, so $$T_m = \frac{\Delta H_m}{\Delta S_m}$$ where $\Delta H_m$ and $\Delta S_m$ are the enthalpy and entropy of fusion.

5. What is the approximate empirical relationship between $T_g$ and $T_m$ (in kelvin)?

   For many polymers the ratio holds approximately: $$\frac{T_g}{T_m} \approx \frac{1}{2}\ \text{(symmetrical)} \quad\text{to}\quad \frac{2}{3}\ \text{(unsymmetrical)}$$ with $T_g$ and $T_m$ in kelvin (Boyer–Beaman rule).

6. List the molecular factors that increase the melting temperature ($T_m$) of a fibre polymer.

   $T_m$ increases with: chain stiffness/rigidity (aromatic rings), strong intermolecular forces (H-bonding, polarity), high chain symmetry and regularity, high molecular weight, and crosslinking. Higher $\Delta H_m$ (cohesive energy) and lower $\Delta S_m$ (more rigid chains) both raise $T_m$.

7. What molecular factors lower or raise $T_g$ of a fibre?

   $T_g$ is raised by chain stiffness, bulky/polar side groups, strong intermolecular forces, crosslinking and high molecular weight. $T_g$ is lowered by chain flexibility, plasticizers, moisture (e.g., water plasticizes cellulose/nylon), flexible spacer groups and bulky but flexible pendant groups.

8. How does moisture act as a plasticizer for hygroscopic fibres?

   Absorbed water molecules penetrate amorphous regions and break interchain hydrogen bonds, increasing chain mobility and free volume. This lowers the effective $T_g$ (e.g., wet nylon, wool and cellulose set or relax at lower temperatures than when dry).

9. What is addition (chain-growth) polymerization, and name a fibre made by it?

   Addition polymerization links unsaturated monomers (containing $\ce{C=C}$) without loss of any small molecule, via initiation, propagation and termination of free radicals/ions. Example fibres: polyacrylonitrile (acrylic), polypropylene, and polyethylene. e.g. $$\ce{n CH2=CHX -> [-CH2-CHX-]_n}$$

10. What is condensation (step-growth) polymerization, and name two fibres made by it?

    Condensation polymerization joins bifunctional monomers with elimination of a small molecule (water, HCl, methanol) at each step. Examples: nylon-6,6 and polyester (PET) — both eliminate water; the molecular weight builds slowly throughout the reaction.

11. Write the reaction for nylon-6,6 formation and state the by-product.

    Nylon-6,6 forms by condensation of hexamethylene diamine and adipic acid: $$\ce{n\,H2N(CH2)6NH2 + n\,HOOC(CH2)4COOH -> [-NH(CH2)6NH-CO(CH2)4CO-]_n + 2n\,H2O}$$ The by-product is water; the repeat unit has $6+6$ carbons.

12. How is Nylon-6 produced and from which monomer?

    Nylon-6 is made by ring-opening polymerization of the single monomer $\varepsilon$-caprolactam (a cyclic lactam with $6$ carbons). It is heated with a small amount of water (hydrolytic polymerization); no small molecule is eliminated, so it is technically a chain-growth ring-opening (addition) polymerization.

13. Write the repeat unit of Nylon-6 and compare its structure to Nylon-6,6.

    Nylon-6 repeat unit: $$\ce{[-NH-(CH2)5-CO-]_n}$$ It is made from one monomer ($\varepsilon$-caprolactam) and all amide groups point the same direction. Nylon-6,6 uses two monomers and amide groups alternate direction. This gives Nylon-6,6 a higher $T_m$ ($\approx 265\,^{\circ}\mathrm{C}$) than Nylon-6 ($\approx 215$–$220\,^{\circ}\mathrm{C}$).

14. Why does Nylon-6,6 have a higher melting point than Nylon-6 despite identical chemistry?

    In Nylon-6,6 the symmetrical arrangement (two different monomers, amide dipoles alternating) allows better chain packing and more effective hydrogen bonding between adjacent chains, raising crystallinity and $T_m$ ($\approx 265\,^{\circ}\mathrm{C}$) above Nylon-6 ($\approx 220\,^{\circ}\mathrm{C}$).

15. What is moisture regain and write its formula?

    Moisture regain is the mass of water in a fibre expressed as a percentage of its oven-dry mass: $$R=\frac{W-D}{D}\times 100$$ where $W$ is conditioned (moist) mass and $D$ is oven-dry mass. (Moisture content uses $W$ in the denominator instead.)

16. Rank cotton, wool, silk, nylon and polyester by approximate standard moisture regain.

    Approximate regain at standard conditions ($65\%$ RH, $20\,^{\circ}\mathrm{C}$): wool $\approx 14$–$16\%$ (highest), silk $\approx 11\%$, cotton $\approx 7$–$8.5\%$, nylon $\approx 4\%$, polyester $\approx 0.4\%$ (lowest). Hygroscopicity tracks the amorphous content and polar groups.

17. What is the role of the lumen in cotton and how does it relate to maturity?

    The lumen is the central canal left by the dried-out protoplast. In mature cotton the secondary wall is thick and the lumen is small (high strength, good dyeing). In immature cotton the wall is thin and the lumen large/collapsed, giving weak, dye-resistant 'dead' fibres and neps.

18. Define tenacity and the unit used for fibres.

    Tenacity is the specific strength of a fibre — the breaking force per unit linear density, independent of cross-sectional area. Units are grams-force per denier ($\mathrm{gf/den}$) or cN/tex (centinewton per tex). It is used because fibre cross-sections are irregular.

19. What is the difference between denier and tex as units of linear density?

    Both express mass per unit length. Denier $=$ mass in grams of $9000\ \mathrm{m}$ of fibre; tex $=$ mass in grams of $1000\ \mathrm{m}$. Conversion: $$\text{denier} = 9 \times \text{tex}$$ Higher value means coarser fibre.

20. What are the principal uses of the major natural fibres?

    Cotton: apparel, home textiles, medical. Wool: warm clothing, suiting, carpets, blankets (warmth, resilience). Silk: luxury apparel, scarves, ties (lustre, strength). Jute/bast: sacking, hessian, ropes, geotextiles (cheap, strong, biodegradable).

21. Why are wool and silk damaged by alkalis but relatively resistant to mild acids?

    Wool and silk are proteins whose peptide and (in wool) cystine bonds are hydrolyzed/cleaved by alkalis, dissolving the fibre. Mild acids are less harmful because the amide linkages are more resistant and the protein is stable at lower pH — the reverse of cellulosic fibres, which resist alkali but are degraded by acids.

22. What is birefringence in fibres and what does it indicate?

    Birefringence is the difference between the refractive indices parallel and perpendicular to the fibre axis: $$\Delta n = n_{\parallel} - n_{\perp}$$ A high positive value indicates strong molecular orientation along the fibre axis; it is used in identification and to assess orientation/draw.

23. What structural feature gives meta-aramid (Nomex) its flame resistance rather than high strength?

    In meta-aramid the amide links attach to the benzene ring in the meta (1,3) positions, giving kinked, less-oriented chains — so lower strength/modulus than para-aramid, but excellent thermal stability and inherent flame resistance (high $T_g$, chars rather than melts), used in protective and firefighter clothing.

24. Summarize the effect of drawing (orientation) on fibre $T_g$, crystallinity and tenacity.

    Drawing stretches the filament, aligning chains along the fibre axis. This increases molecular orientation and crystallinity, raises tenacity and modulus, reduces extensibility, and slightly raises the effective $T_g$ (oriented amorphous chains are more constrained). It is essential in melt-spun synthetics like nylon and polyester.

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