Odisha CHSE +2 Vocational Textile Designing study materials
Odisha CHSE +2 Vocational Textile Designing study resources

Odisha CHSE +2 Vocational Textile Designing: Textile Fibres, Yarns and Fabric Structure

Odisha CHSE +2 Vocational Textile Designing · Work through your exam, one topic at a time.

All guide sections

Textile Fibres, Yarns and Fabric Structure

What this domain tests

This domain covers fibre classification and properties, yarn structure and count, and the three fabric-forming systems. Questions often ask you to link fibre to yarn to fabric performance, or to calculate a count conversion.

Core rules

  • Fibres are natural (cotton, linen, wool, silk) or man-made (regenerated such as viscose; synthetic such as polyester, nylon, acrylic). Staple fibres are short and spun; filament fibres are continuous.
  • Key properties: cotton absorbs and breathes but wrinkles; polyester is strong and quick-drying but hydrophobic; wool insulates and felts; silk has lustre and low creep.
  • Yarn count: indirect Ne (higher number = finer) versus direct Tex (g/1000 m) and denier (g/9000 m). For cotton, Tex = 590.5 ÷ Ne; denier = 9 × Tex.
  • Twist direction (S or Z) and twist factor affect handle, strength and surface.
  • Fabric structure: woven (warp and weft interlacing), knitted (loops), nonwoven (bonded web). Woven weaves include plain, twill and satin; cover factor and sett (EPI/PPI) control density and GSM.

Worked example

A cotton yarn is 30s Ne. Convert it to Tex and denier. Tex = 590.5 ÷ 30 = 19.7 tex. Denier = 9 × 19.7 = 177 denier. Reasoning: Ne is indirect, so a higher Ne means a finer yarn; 30s cotton is finer than 20s. The Tex value tells you 19.7 g per 1000 m, which is useful for costing and for predicting fabric weight.

Traps

Mixing direct and indirect count systems; using 840 yd/lb for worsted or metric counts; assuming satin has no interlacing points (it has long floats, not none); ignoring moisture regain when comparing oven-dry weights; treating all silk as mulberry silk.