2.Winding

NumericalProblem

1.Theemptydiameterofaspindledrivencylindricalpackageis5cm.Thespindlespeedis2000r.p.m.andtraversevelocityis100m/min.Determine

a)Windingspeedandangleofwindatthestart

b)Windingspeedandangleofwindwhenpackagediameterbecomesdouble

Solution:

a)Here,Spindler.p.m.(n)=2000

Traversespeed(Vt)=100m/min

d=diameterofcylindricalpackage=5cm

So,windingspeedandangleofwindatthestartare329.69m/minand17.66respectively.

(b)Whenpackagediameterbecomesdoublei.ed=10cm

So,windingspeedandangleofwindatthestartare628.32m/minand9.04respectively.

2.Acheeseof150mmtraverselengthiswoundonarotarytraversemachineequippedwith75mmdiameterdrumsof2.5crossings.Calculatethewindingspeedandcoilangleifthedrumrotatesat3250r.p.m..

Solution:

3.Thediameterandlengthofthedrumofarandomwinderis100mmand300mmrespectively.Ifthedrum is rotating at 4000 r.p.m. and crossing on drum is 3 then calculate the slippage% ifwindingspeedis1200m/min.

Solution:

4.Aconehavingnegligibleconicityandmaximummeandiameterof200mmisbeingbuiltusinga50mmdiameter cylindricalwinding drum rotating at 3000 r.p.m. The traverse perminute is 320. Theconewasrunningat960r.p.m.whenthemeanpackagediameterwas150mm.Whatistheslippage% between drum and package? Assuming no change in contact point between the cone and drumduringpackagebuiltup,calculatetheconediametersatwhichpatterningmayoccur.

Solution:

Therpmofthecone(n)is960whenpackagediameter(d)is150mm=15cm

Now,consideringx=1,thevalueofdbecomes45cmwhichisnotacceptable.Because,45>20cm(maximummeanpackagediameter)

Thevaluesofdatwhichpatterningmayoccurare15cm,11.25cm,9cm,7.5cmetc(correspondingvaluesofxare3,4,5and6).

5.Whatisthenearestvalueoftraverseratioto3topreventpatterninginacheesewhenthediameteris5cm?Theyarnismadeupofcottonfibreandthecountis20s.

Solution:

6.a)Foraspindledrivenwinderwhenthepackagediameteris10cmthewindangleis20.Determinetheangleofwindwhenpackagediameteris15cm.

b)Inadrumdrivenwindertheangleofwindis30.Thedrumhaving5cmdiametermakes5revolutionsforonedoubletraverse.Calculatethelengthofthewindingdrum.

Solution

a)Forspindledrivenwinders,

b)Fordrumdrivenwinders,

7.Aprecisionwinderwithtraverselengthof20cmisoperatingatconstantwindingspeedof1000m/min.Thespindler.p.m.is3000whenthepackagediameteris10cm.Whenthepackagediameterincreasesto20cm,whatwillbethespindler.p.m.?

Solution

Constantwindingspeed=w=1000m/min

Thespindlespeedhastobereduced,withtheincreaseinpackagediameter,tomaintainconstantwindingspeed.Thetraversespeed(R)willalsobereducedasitisconnectedwithspindledrive.

Spindler.p.m.(n1)=3000whenpackagediameter(d1)=10cm=0.1m

Traverselength=L=20cm=0.2m

Increasedpackagediameter(d2)=20cm

8.Acheeseof20cmtraverseisbeingwoundonaprecisionwinderwithtraverseratioof5/2.Whenthecheesediameteris10cm,whatshouldbethetraverserationearestto5/2forthepreventionofpatterning?Theyarnismadeofcottonfibrehavingpackingfractionis0.6andcountis4Ne.

Solution

9.Aprecisionwinderwithconstantspindlespeedisbeingusedfortheproductionofcheese.Thetraverselength150mmandthetraverseratiois6.Iftheminimumangleofwindis80,determinethemaximumpackagediameter.

Solution

10.Acheesehaving150mmheight,50mmcorediameterand125mmmaximumdiameterisbeingwoundontwodifferentmachinesasfollows:a)Agroovedrumrotarytraversewinderwherethedrumdiameteris75mmandthedrumismaking5revolutionsperdoubletraverse.b)Aconstantspindlespeedprecisionwinderonwhichthespindlerotates6timesperdoubletraverse.

Formachine(a)determinethetraverseratioatpackagediameter50mmand100mmformachine(b)findtheangleofwindatthesamediameters.

Solution

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