GATE (TF) Textile 2009 Question Paper Solution | GATE/2009/TF/52

Common Data Questions

The feed rate of a card is 100 kg/hour and the delivery rate is 400 m/min. Licker-in droppings and flat strips are 3% and 1% respectively.

Question 52 (Textile Engineering & Fibre Science)

If the total draft in the card is decreased by 10%, the sliver linear density in ktex will be

(A)6.4
(B)5.4
(C)4.4
(D)3.4
[Show Answer]

Option C is correct
Given in the question-

Given in the question-

Feed rate of card=100 kg/hr
=100 x 1000/60 gm/min
=1666.66 gm/min

Delivery rate=400 m/min

Licker-in droppings=3%

Flat strips=1%

The count of card sliver(Ne)=?

Firstly we have to find gm/m

\frac{g}{m}=\frac{Feed rate of card}{Delivery rate}

\frac{g}{m}=\frac{1666.66}{400}

\frac{g}{m}=4.16665

Now, tex is the weight of 1000 meter length

Tex=4166.65

Tex after total waste removed-

Total waste=Licker-in dropping+Flat stripe

Total waste=3%+1%

Total waste=4%

Tex=4166.65 x (1-Total waste%)

Tex=4166.65 x (1-4%)

Tex=4166.65 x (1-0.04)

Tex=4166.65 x 0.96

Tex=3999.984

Relationship between tex and Ne-

Ne=590.5/tex

Ne=590.5/3999.984

Ne_2=0.1476

Delivered card sliver count(Ne)=0.15

Feed count(tex)=\frac{Feed rate of card}{Delivery rate} \times 1000

Feed count(tex)=\frac{1666.66}{400} \times 1000

Feed count(tex)=4.16665 x 1000

Feed count(tex)=4166.65

Ne_1=590.5/4166.65

Ne_1=0.14

Now,

Draft=\frac{delivered count(Ne_2)}{Feed count (Ne_1)}

Draft=\frac{0.15}{0.14}

Total Draft=1.071

If total draft is decreased by 10%

Then, Draft=Total draft-Total draft x 10%

Draft=1.071-1.071 x 10%

Draft=1.071-1.071 x 0.1

New Draft=0.9639

Now,
New Draft=\frac{feed count}{delivered count} , (In tex system)

0.9639=\frac{4166.65}{delivered count}

Delivered count=4166.65/0.9639

Delivered count=4322.69 tex

Sliver linear density in ktex=4.32 (Ans)

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