Winding. Numerical Problems:

Transcription

1 Winding Numerical Problem: 1) The empty diameter of a pindle driven cylindrical package i 5 cm. The pindle peed i 000 rpm and travere velocity i 100 m/min. Determine a) Winding peed and angle of wind at the tart b) Winding peed and angle of wind when package diameter become double (a) Here, Spindle rpm (n) = 000 Travere peed (V t ) = 100 m/min d= diameter of cylindrical package = 5 cm Surface peed (V ) = dn = = m/min Vt tan1 = 100 V = 0.318, where θ 1 = angle of wind or, θ 1 = Winding peed= t V ( V ) = ((100) (314.16) ) =39.69 m/min. (b) When package diameter become double i.e d = 10 cm Surface peed (V ) = dn ' = m/min. = 68.3 m/min

2 Vt 100 tan = V 68.3 = or, θ = ) A cheee of 150 mm travere length i wound on a rotary travere machine equipped with 75 mm diameter drum of.5 croing. Calculate the winding peed and coil angle if the drum rotate at 350 rpm. Surface peed of drum (V ) = DN where D= diameter of drum = 75 mm = 7.5 cm N = rpm of the drum = 350 V = m/min. 100 Travere peed (V t ) = LN S where L= travere length = 150 mm = 15 cm V t = 195 m/min 5100 Now, S (no. of drum revolution for double travere ) = croing =.5 = 5 So, winding peed = ( DN) LN S = ((765.76) (195) ) = 790. m/min

3 Vt 195 tan = 0.55 V So, θ (angle of wind) = Therefore, coil angle = ) The diameter and length of the drum of a random winder i 100 mm and 300 mm repectively. If the drum i rotating at 4000 rpm and croing on drum i 3 then calculate the lippage% if winding peed i 100 m/min. The drum diameter (D) = 100mm= 10 cm drum length (L) = 300mm= 30 cm drum rpm (N) = 4000 No. of revolution of drum per double travere (S) = 3 =6 Let, the tranlation ratio between package urface peed and drum urface peed i x. Therefore, Package urface peed Drum urface peed x Conidering the lippage, the expreion of winding peed will be a follow. Winding peed = ( DN. x) LN S x = or, x= or, x = 0.9 So, lippage i 0.1 or 10% = 100 m/min.

4 4) A cone having negligible conicity and maximum mean diameter of 00 mm i being built uing a 50 mm diameter cylindrical winding drum rotating at 3000 rpm. The travere cam i rotating at 30 rpm. Find the diameter at which patterning can occur if the cone wa running at 960 rpm when the mean package diameter wa 150 mm and auming no movement of point of drive during package built up. What i the lippage % in the above example? The rpm of the cone (n) i 960 when package diameter (d) i 150mm = 15 cm dn Therefore, the lippage i =1 DN where d = mean package (cone) diameter, D = drum diameter, N = rpm of the drum Slippage = = 0.4 i.e. 4% Maximum cone diameter (avg.) = 00mm n Patterning will occur when travere ratio ( ) will be integer. N t N t i the rpm of the travere cam (travere/min) = 30 So, value of n will be = 30.x where x number) N (x belong to the et of natural Now, conidering that the lippage between the drum and cone i contant,

5 n d = = contant So, 30. xd= Now, conidering x = 1, the value of d become 45 cm which i not acceptable. Becaue, 45> 0 cm (maximum mean package diameter) The value of d at which patterning may occur are 15 cm, 11.5 cm, 9 cm, 7.5 cm etc (correponding value of n are 3,4, 5 and 6). 5) What i the nearet value of travere ratio to 3 to prevent patterning in a cheee of 5 cm diameter? The yarn i made up of cotton fibre and the count i 0. Diameter of cotton yarn =.54 8 Ne cm So, linear gain i 0.0 cm. = cm = 0.0 cm Revolution gain = linear gain d = = So, nearet value of travere ratio i 3± ) a) For a pindle driven winder when the package diameter i 10 cm the wind angle i 0 0. Determine the angle of wind when package diameter i 15 cm. b) In a drum driven winder the angle of wind i The croll i 5 and drum diameter i 5 cm. Calculate the length of the winding drum. a) For pindle driven winder,

6 Travere peed tan = Surface peed R = no. of double travere/min. d = package diameter n = package (pindle rpm) A, R n Vt LR Contant V dn πdn and L are contant, o d tan So, 1 1 tan 10 tan 0 d 15 d tan o = contant or, θ = o So, the angle of wind will be o when package diameter i 15 cm. b) For Drum driven winder, LN Vt L tan S V DN DS where L = Length of Drum N = rpm of Drum S = croll of Drum So, O tan30 L.5.5 or, L =.67 cm. So, the Length of winding drum i.67 cm. 7) A preciion wider with travere length of 0 cm i operating at contant winding peed of 1000 m/min. The pindle rpm i 3000 when the package diameter i 10 cm. What will be the pindle rpm when the package diameter increae to 0 cm? Contant winding peed = w = 1000 m/min

7 The pindle peed ha to be reduced, with the increae in package diameter, to maintain contant winding peed. The travere peed (R) will alo be reduced. Spindle rpm (n 1 ) = 3000 when package diameter (d 1 ) = 10 cm = 0.1 m Travere length = L = 0 cm = 0. m Increaed package diameter (d ) = 0 cm Let R n k or, R = nk (contant) Now, w ( LR) ( d n ) So, or, 1 1 ( Ln k) ( d n ) w n (4 L k d ) (4.04 k.01) or, k In cae : w n (4 L k d ) 1000 n (4 L k d ) n ( ) So, n = rpm The pindle rpm will be rpm. (An)

8 8) A cheee of 0 cm travere i being wound on a preciion winder with travere ratio of 5/. When the cheee diameter wa 10 cm, what hould be the travere ratio nearet to 5/ for the prevention of patterning? The yarn i made of cotton fibre and count i 4 (Ne), packing fraction i 0.6. The path of yarn ha been hown in the figure above. wind/min 5 Travere ratio=.5 double travere/min Therefore, wind/travere = coil occupy 0 cm of package width. Therefore, one coil will occupy 16 cm of package width. package width occupied by one coil 16 tan package circumference.10 So, ϴ = 7 o Yarn diameter (cm) = d.54 d g in in 7 o g d Revolution gain 0.003

9 Revolution gain So, for avoiding patterning, travere ratio hould be.503 or.497 9) A preciion winder with contant pindle peed i ued for the production of cheee if travere length 150 mm and the nominal wind per double travere of 6. If the minimum angle of wind i 8 0, determine the maximum package diameter. [11.33 cm] Travere length of cheee (L) = 150 mm= 15cm Wind per double travere = Package rpm ( n) Double travere per min ( R ) = 6 In cae of preciion winder, the angle of wind (θ) reduce a the package diameter (d) increae. Therefore, θ min correpond to d max tan min or, tan 8 0 = Vt LR 15R (for pindle driven machine) V dn d n 30 1 d 6 max max = 5 d 5 or, d max = = 11.33cm 0 tan8 So, maximum cheee diameter i cm max

10 10) A cheee of 150 mm travere length and 50 mm core diameter and 15 mm maximum diameter i wound on two machine a follow: a) A groove drum rotary travere winder where the drum diameter i 75 mm and the drum making 5 revolution per double travere. b) A contant pindle peed preciion winder on which the pindle rotate 6 time per double travere. For machine (a) determine the wind per double travere at package diameter 50 mm and 100 mm; for machine (b) find the angle of wind at the ame diameter. Travere length (L) = 150 mm =15 cm Minimum diameter of the cheee (d min ) = 5 cm Maximum diameter of the cheee (d max ) = 1.5 cm a) Drum diameter (D) = 75 mm = 7.5 cm Revolution of drum per double travere (S) = 5 D Wind per double travere = S d When d= 5cm, 7.5 Wind per double travere = 5 = When d= 10cm, D 7.5 Wind per double travere = S 5 = 3.75 d 10 So, the value of wind per double travere are 7.5 and Vt b) tan1 ( pindle rotate at 6 time per double travere) V LR or, tan 1 dn, here n R 6 or, θ 1 = = 15 1 = Vt ' 15 1 Again, tan V ' 10 6 or, θ = So, the value of angle of wind are and