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 Ne; 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=?


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


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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