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Mathematical Problems of Spiral Drawing Frame

Mathematical Problems of Spiral Drawing Frame
Engr Avijit Dey
Production Officer
Fabian Group, Chittagong
Mobile: 01836124490
E-mail: avijitdey43@yahoo.com




Introduction:
In this method of faller- bar propulsion there are two spiral screws on each side, one set directly above the other. The ends of each faller –bar are cut to fit into the groves on the spiral so that as the screws rotate they drive the faller – bar along. As each faller comes to the end of the top screw it is knocked down on to the bottom one by a cam on the top screw, springs holding it steady as as it falls into the grooves of the bottom screw. The bottom spiral is more coarsely pitched than the top one so that the faller – bars are returned quickly to the back of the machine ready to be lifted by cams on the bottom screw up into the spirals of the top screw. By having a coarse spiral on the bottom fewer bars are needed to complete the gill sheet.

Problem : 1
On a spiral D/F, Drawing roller dia 2 ½”, & its speed 78.5 ft/min. Draft = 6; The faller speed 320 faller drops/min; Faller lead 2%, Find the pitch of screw; Find r.p.m of 2” dia retaining or back roller .

Solution :

S.Speed of Drawing roller = 78.5 ft/min.

Draft = 6

So S.speed of ret roller or back roller = 78.5÷6 


= 13.08 ft/min
=13.08×12inches/min
=157 inches/min

Dia of ret roller = 2”

                                   157
So,r.p.m. of ret R/r = ………. = 24.98 = 25 (Ans).
                                  11×2

Again,

Faller bar S.speed = Ret R/r S.Speed + Ret R/r S.Speed ×lead%

=157+157× (2÷100) 

= 160.14 inches/min

Faller drops/min× Pitch

            Faller bar s.speed         160.14
Pitch = ……………………… = ………… = 0.5 inches. (ANS)
             Faller drops/min            320

Problem-2:

For1st D/F

  • 2 Heads
  • 2 del/heads.
  • Del speed 127 ft/min.
  • Delivered sliver 8lbs/100 yds.
For 2nd D/F

  • Draft 6
  • Eff = 90%
  • Loss =1%
  • And head = 5
  • 2del/head
  • Doubling 3:1
S.speed of the back roller =?

Prod/hr of 2nd D/F =?

Solution :

Total length of delivery of 1st D/F = 127×2×2 ft/min. = 508 ft/min.

                                                                508
S.Speed of 2nd drawing back roller = …………ft/min.
                                                              5×2×3

     508×100
= ……………… ft/min [At 90%eff]
   5×2×3×80

= 18.82ft/min.
(ANS) 
 
Delivery roller surface speed = 6×18.82 ft/min.

So Total length of delivered sliver of 2nd d/f = 6×18.82×5×2 ft/min [At 100% eff]

   6×18.82×5×2×90
= ……………………….. [At 90% eff]
            100

  6×18.82×5×2×90
= …………………………. [yds/hr]
         100×3

Lbs/100 yds of 2nd D/F delivered sliver = 8×(3÷6) × (99÷100) = 3.96 lbs/100 yds.

                                     6×18.82×5×2×90×60×3.96
So prod/hr of 2nd D/F = …………………………………
                                               100×3×1000

= 804.89
= 805lbs. (ANS) 

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