Mathematical Problems of Push Bar Drawing in Jute Spinning

Mathematical Problems of Push Bar Drawing in Jute Spinning
Engr Avijit Dey
Production Officer
Fabian Group, Chittagong
Mobile: +88-01836124490
E-mail: avijitdey43@yahoo.com




Introduction :
Jute drawing frames are divided into two types depending on their mechanism used to proper the faller bars. 
 
  1. Push-bar.
  2. Spiral.
Push-bar drawing frame :
In this type of machines, the fallers have specially cranked ends which run in slides on the machine frame. The fallers are driven by a large carrier wheel at the back of the machine. The earlier models had collars on each faller- bar which bore against each other but in modern frames the bars bear across the full width, the bar behind pushing the in front- hence the name.
Push-bar drawing frame
Problem :1
A high speed push bar D/F with 2 heads, 2 delivery/head, doubling 2:1, Produces 600 lbs/hr of 8 lbs/100 yds sliver. Draft 4, pitch of the faller bar = ½”, faller bar lead = 10%, calculate the faller bar drops/min .

Solution:

                                     100×100                  600×100×3
Length of production = …………… yds/hr = …………………ft/min
                                           8                           8×60

   600×100×3×12
= ………………….. = 4500 inches/min.
          8×60

                                                                         4500
Sliver length production per delivery roller = …………… inches/min.
                                                                          2×2

= 1125 inches/min.

                                                                            10
Surface speed of faller sheet = 281.25+281.25×………… = 300.375
                                                                           100

We know, 
Faller bar surface speed = Faller drops/min× pitch.

                                     Faller bar S.S.
Or, Faller drops/min =…………………
                                         Pitch

   309.375
= …………… = 618.75 (ANS).
       ½

Problem : 2
A high speed type push bar 1st d/f 2heads, 2del per head, doubling 2:1 is being fed with 16 lbs/100 yds sliver. If d/f draft is 4 and the back roller surface speed of d/f is 23.44 ft/min.

Calculate,
a. lbs/100 yds from the d/f.

Solution:

                                                2
lbs/100 yds from the d/f =16 ×…… = 8
                                                4

We know,

              Delivery roller S.Speed
Draft = ……………………………………
             Retaining roller S.Speed.

Or, Delivery roller S.Speed = draft× Ret.R/R S.Speed.

= 4×23.44×1 ft/min( for one sliver)

= 4×23.44×2×2( for 2head each of ft/min.2delivery sliver )

  4×23.44×2×2
= …………………. yds/hr.
           3

  4×23.44×2×2×60
= ………………..…..yds/hr
             3

Now, wt of 100 yds sliver =8 lbs.

                             4×23.44×2×2×60×8
So, Delivery wt = …………………………… = 600.64lbs/hr. (ANS)
                                    3×100

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