GATE (TF) Textile 2010 Question Paper Solution | GATE/2010/TF/48

A speed frame with all the drafting rollers of 25.4 mm diameter has the following process parameters: Roving count – 1.5s N_e; Front drafting roller speed – 220 rpm; Roving diameter – 3.0 m; Roving Twist Multiplier – 1.2 tpi Ne^-1/2, Empty bobbin diameter – 76.2 mm.

Question 48 (Textile Engineering & Fibre Science)

Spindle speed will be

(A)	824 rpm
(B)	968 rpm
(C)	1016 rpm
(D)	1106 rpm

[Show Answer]

Given in the question-

Diameter of all drafting rollers(D)=25.4 mm
$D=25.4 \times 10^-3$ meter

Roving count=1.5s Ne

Roving diameter=3.00 mm

$Roving TM=1.2 \times tpi \timesNe^\frac{1}{2}$

Front roller speed(N)=220 rpm

Empty bobbin diameter=76.2 mm

Spindle speed=?

Calculation-

$FRD=\pi \times D \times N$

Where ,
D=Dia of roller

N=front roll speed

$FRD=3.14 \times 25.4 \times 10^-3 \times 220$

FRD=17.55 m/min

FRD=17.55 X 39.34 inch/min

FRD=690.42 inch/min

And

$TPI=TM \times \sqrt Count$

$TPI=1.2 \times \sqrt 1.5$

$TPI=1.2 \times 1.225$

TPI=1.47 turns per inch

Now,by formula-

$TPI=\frac{Spindle speed}{FRD}$

$1.47=\frac{Spindle speed}{690.42}$

$1.47 \times 690.42=Spindle speed$

Spindle speed=1014.91
=1015 rpm (Ans)

Option C is approx correct.

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