GATE (TF) Textile 2006 Question Paper Solution | GATE/2006/TF/66

Question 66 (Textile Engineering & Fibre Science)

The following data refers to a square fabric
– Fabric construction: 2/2 Twill
– Count of yarn: 60 tex
– Crimp in yarn: 5%

Applying Brierley’s formula of maximum set and using the values of constant for cotton namely, m=0.39 and k=200, the number of yarns/dm works out to be

(A)	180
(B)	210
(C)	230
(D)	260

[Show Answer]

Option C is correct.
Given in the question-

Fabric construction: 2/2 Twill

i.e., Ends(E)=4

Interlacement(I)=2

Count of yarn(C): 60 tex
C=590.5/60
C=9.84
Crimp in yarn(c): 5%

yarns/dm=?
m=0.39

k=200,

Brierley’s frormula-

$Fabric sett(T)=f^m \times (K \times C)^\frac{1}{2}$

Where, $f=\frac{E}{I}$ is the average float.

$f=\frac{E}{I}$

$f=\frac{4}{2}$

f=2

C is count and m & K are constant

$Fabric sett(T)=2^(0.39) \times (200 \times 9.84)^\frac{1}{2}$

$Fabric sett(T)=1.31 \times (1968)^\frac{1}{2}$

$Fabric sett(T)=1.31 \times 44.36$

Fabric sett(T)=58.11

Fabric sett(T)=58 threads per inch

T=58/2.54 threads per cm

T=22.84

T=2284 threads per m

T=228.4 yarns per dm (Ans)

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