GATE (TF) Textile 2017 Question Paper Solution | GATE/2017/TF/44

Question 44 (Textile Engineering & Fibre Science)

Let a cheese of 160 mm traverse length be wound on a rotary traverse machine having a drum of 75 mm diameter and 2.5 crossings. If the drum rotates at 3250 rpm, then the coil angle, in degrees, accurate to one decimal place, is __73.5 to 76.0__.

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Solution:

Given in the question–
Traverse length of cheese(L)=160mm
=160 x 10^-3 meter

Dia of drum(D_d)=75 mm
=75 x 10^-3 meter

No. of crossings=2.5

Drum rotational speed(N_d)=3250rpm

Coil angle( $\phi$ )=?

Formula:

$tan(\phi)=\frac{V_s}{V_t}$

Where, Vt=Traverse speed , Vs=Package Surface speed and $\phi$ =coil angle

$V_t=2LN_t$

Where, N_t is traverse rotational speed

Traverse ratio=2 x No. of crossings

Traverse ratio=2 x 2.5

Traverse ratio=5

$Traverse ratio(T.R.)=\frac{N_p}{N_t}$

$5=\frac{3250}{N_t}$

$N_t=\frac{3250}{5}$

$N_t=650 rpm$

Now,
$V_t=2LN_t$

$V_t=2\times160\times10^-3\times650$

Traverse speed(V_t)=208 m/min

And

$V_s=\pi \times D_d \times N_d$ (Package surface speed=Drum surface speed)

$V_s=3.14 \times 75\times10^-3 \times 3250$

V_s=765.4 m/min

$tan(\phi)=\frac{765.4}{208}$

$tan(\phi)=3.68$

$\phi=tan^-1(3.68)$

$Coil angle(\phi)=74.79^o$ (Answer)

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