Introduction:
Jute drawing frames are divided into two types depending on their mechanism used to proper the faller bars. In this article, we will know about the mathematical problems of push bar drawing in Jute Spinning.
- Push-bar.
- Spiral.
Push-bar drawing frame:
In this type of machine, the fallers have especially cranked ends that 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.
Mathematical Problems of Push Bar Drawing in Jute Spinning
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, the pitch of the faller bar = ½”, faller bar lead = 10%, calculate the faller bar drops/min.
Solution:
…………………………………..100×100
Length of production = ………………. yds/hr
………………………………………..8
….600×100×3
= ………………….ft/min
………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
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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
Author of this Article: Engr Avijit Dey Production Officer Fabian Group, Chittagong Mobile: +88-01836124490 E-mail: [email protected]