Calculation of Jute Carding:
Introduction: Jute carding is a combining operation where jute reeds are split and extraneous matters are removed. Jute fibers are formed into a ribbon called “sliver”. There are three different carding sections: (i) breaker carding (ii) inner carding (iii) and finisher carding These are of a simpler nature than those required for the spreader, being confined chiefly to draft, count, and speed. The following three are typical of those met with in practice. in this article, we will discuss the calculation of jute carding.
Calculation of Jute Carding
Problem: 1
Eight ends of 300 ktex spreader sliver are fed to a breaker with a draft constant of 500 which has a draft change pinion of 28 fitted. If the feed speed is 3 m/min find the card production in an 8 hr day when it runs at 80 percent efficiency and the deliver sliver count when there are moisture and waste loss of 6 percent of the input weight. The sliver is delivered by a roll former which has a 4 percent lead over the delivery rollers of the card.
Answer:
Draft on the card = 500 ÷ 28 = 17.85
Delivery speed of card = 3×17.85 = 53.5 m/min
Therefore, roll former delivery speed = 53.5× 1. 04 = 55.6 m/min
Daily delivery = 55.6 × 60 × 8 × 0.8 = 21400 m
………………………………………….300×8×0.94
Sliver count at roll former =……………………….
…………………………………………..17.85× 1. 04
= 121.5k tex
……………………………….121.5×21400
Daily production = ………………………… = 2600 kg (ANS).
………………………………………1000
Problem: 2
It is necessary to produce breaker card sliver at a count of 18 lb/100 yd.
What dollop weight must be used to meet the following conditions?
Draft constant = 440
Draft pinion = 30
Feed sheet roller 7 in diameter
Feed sheet roller 24.5 revolutions per clock revolution
Moisture and waste loss 3.5 percent.
Answer:
In a revolution of the clock the feed sheet travels,
……24.5×7×3.1416 yd
= ……………………………… =14.95 yd
…………………36
That is the clock length is 14.95 yd.
Again,
………..Dollop × 100
= ……………………………… = Delivery sliver (lb/100 yd)
….Clock length × draft
Draft = 440÷30 = 14.7
……………………………………………………………………………………18×15×14.7
Hence, ignoring loss for the moment, Dollop weight = …………..………………
…………………………………………………………………………………………..100
= 39.7 lb
To allow for losses this must be increased by 1. 035,
Correct dollop weight = 41 lb (ANS).
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Problem: 3
The cylinder of a finisher card rotates at 180 r.p.m. Find its linear speed when its radius is 24m. The feed roller travels at (1÷200) of the cylinder speed, the doffer at (1÷40) of the cylinder speed and at half the speed of the delivery rollers. Find the card draft.
Answer:
………………………………………180 × 2×3.1416×24
Cylinder surface speed = ………………………………….
……………………………………………………12
= 2290 ft/min.
Therefore, feed roller surface speed= 2290 ÷ 200 = 11.45 ft/min
Similarly, doffer surface speed = 57.3 ft/min
Therefore, delivery speed = 57.3×2 =114.6 ft/min.
Machine draft = 114.6÷11.45 = 10 (ANS).
Author of this Article: Engr Avijit Dey Production Officer Fabian Group, Chittagong Mobile: 01836124490 E-mail: [email protected]