###### 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)