**Calculation of Jute Carding**

**Engr Avijit Dey**

Production Officer

Fabian Group, Chittagong

**Mobile:**01836124490

**E-mail:**avijitdey43@yahoo.com

**Introduction:**

**Jute carding**is a combining operation where jute reeds are splitted and extraneous matters are removed. Jute fibres are formed into 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.

**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 3m/min find the card production in an 8 hr day when it runs at 80 percent efficiency and the deliver sliver count when there is moisture and waste loss of 6 percent of the input weight. The sliver is delivered by a roll former which has a 4percent lead over the delivery rollers of the card.

**Answer:****Draft on the card**= 500 ÷ 28 = 17.85

**Delivery speed of card**= 3×17.85 = 53.5m/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

= 121.5ktex

121.5×21400

**Daily production**=

**…………………………**= 2600kg

**(ANS).**

1000

**Problem: 2**

It is necessary to produce breaker card sliver at a count of 18lb/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,**

24.5×7×3.1416yd

=

**……………………………………**=14.95 yd

36

That is the clock length is 14.95 yd.

**Again,**

Dollop × 100

=

**………………………………………**= Delivery sliver (lb/100yd)

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 = 41lb

**(ANS).**

**Problem: 3**

The cylinder of a finisher card rotates at180 r.p.m. Find its linear speed when its radius is24m. 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 = 11.45ft/min

Similarly, doffer surface speed = 57.3ft/min

Therefore, delivery speed = 57.3×2 =114.6ft/min.

Machine draft = 114.6÷11.45 = 10

**(ANS).**

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