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-
Where, is the average float.
f=2
C is count and m & K are constant
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)