GATE (TF) Textile 2005 Question Paper Solution | GATE/2005/TF/44

Question 44 (Textile Engineering & Fibre Science)

A bobbin leading speed frame running at 1000 rpm spindle speed is producing a roving with 2.5 turns per inch. The diameter of empty bobbin is 5 cm. When the bobbin attains 10 cm diameter, the percentage change in the bobbin speed will be approximately

(A)	1 %
(B)	3 %
(C)	5 %
(D)	7 %

[Show Answer]

Option B is correct.

Given-

Spindle speed=1000 rpm

TPI=2.5

Dia of empty bobbin(d1)=5cm

Final bobbin dia(d2)=10cm

% change in bobbin speed=?

TPI=Spindle speed/FRD

2.5=1000/FRD

FRD=1000/2.5

FRD=400 inch/min

FRD=400 x 2.5 cm/min

Front roll delivery(FRD)=1000 cm/min

In Bobbin lead speed frame-

Bobbin speed-flyer speed(spindle speed)= $\frac{FRD}{\pi \times dia of bobbin}$

Bobbin speed=flyer speed(spindle speed)+ $\frac{FRD}{\pi \times dia of bobbin}$

Bobbin speed=1000+ $\frac{1000}{3.14 \times dia of bobbin}$

At empty bobbin dia,

Bobbin speed 1=1000+ $\frac{1000}{3.14 \times 5}$

At 10 cm bobbin dia-

Bobbin speed 2=1000+ $\frac{1000}{3.14 \times 10}$

% change in bobbin speed= $\frac{Bobbin speed 1-Bobbin speed 2}{Bobbin speed 1} \times 100$

% change in bobbin speed= $\frac{(1000+\frac{1000}{3.14 \times 5})-(1000+\frac{1000}{3.14 \times 10})}{1000+\frac{1000}{3.14 \times 5}} \times 100$

% change in bobbin speed= $\frac{(\frac{1000}{3.14 \times 5})-(\frac{1000}{3.14 \times 10})}{1000+\frac{1000}{3.14 \times 5}} \times 100$

% change in bobbin speed= $\frac{31.84}{1063.69} \times 100$

% change in bobbin speed=2.99 (Ans)

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