Jute Carding:
Jute carding is the process of separating and aligning the jute fibers in preparation for spinning. Jute fibers are extracted from the jute plant, which is grown mainly in India and Bangladesh. The fibers are long, soft, and lustrous, but they are also tough and difficult to spin directly.
In jute spinning, carding is the major area for the jute yarn production process. In the jute carding machine, the primary function of jute carding is done by the action of worker and cylinder and the cleaning and the cleaning actions are done between worker and stripper. For jute yarn production, some jute carding calculation formula is given below.
Some Important Jute Carding Calculation Formulas
Jute carding calculation formula is as follows:
……………………………………Surface speed (SS) of cylinder roller – SS of worker roller
1. Carding efficiency = …………………………………………………………………………………………
………………………………………………Surface speed (SS) of cylinder
………………………………SS of cylinder
2. Carding Ratio = ……………………………
………………………………SS of worker
………………………………………………Dollop wt.
3. Wt. of feed sliver/100 yd = ………………………. X 100%
……………………………………………Clock length
………………………………………………………………………………Doubling
4. Wt. of delivery/100 yd = wt. of feed sliver/100yd X …………………..
…………………………………………………………………………………Draft
……Dollop wt. ………………… Doubling
= …………………… X 100 X …………………..
……Clock length…………………Draft
…………………………………SS of delivery roller
5. Mechanical Draft = ……………………………..
……………………………………SS of feed roller
……Calculated feed sliver wt. /100 yd
= …………………………………………………………
…Calculated delivered sliver wt. /100 yd
……………………………Wt. /100 yd of feed material
6. Actual draft = ………………………………………………….. X Doubling X Waste%
…………………………Wt. /100 yd of delivered material
= Mechanical draft X Waste%
7. Draft constant = DCP X Draft
8. SS of feed roller = π X Dia in inch X Speed
………Length of feed material
= …………………………………………
………………Doubling
………SS of delivery roller
= ……………………………………
……………Draft
9. SS of delivery roller = π X Dia in inch X Speed
= SS of delivery roller X Lead%
10. Finisher card sliver wt. /100 yd
………B/C sliver wt. /100 yd
= ………………………………………. X Doubling X Waste%
……………F/C draft
……Dollop wt.…………………B/C doubling……F/C doubling
= …………………… X 100 X ……………………. X ………………….. X Waste%
……Clock length…………………B/C draft…………F/C draft
11. Production /hr = SS of delivery roller in inch /min X 60/36 X wt. of sliver/yd X Eff% X Waste%
…………………………………………………………………Dollop wt.
= SS of feed roller in inch/min X 60/36 X ……………………… X Eff% X Waste%
………………………………………………………………Clock length
Some Problems and Solutions:
1. The surface speed of the cylinder is 2400 feet/min. If the carding efficiency 90%, then what will be the SS of worker roller.
Solution:
………………………………SS of cylinder – SS of worker
Carding efficiency = …………………………………………………….. X 100
…………………………………………SS of cylinder
=> 90 = (2400 – x) / 2400 X 100 (Let, SS of worker roller be x)
=> 90 X 2400 = 240000 – 100x
=> 100x = 240000 – 216000
=> X = 2400/100 = 240 ft/min (ANS)
2. Find out the wt/100yd of BC sliver when,
- Dollop wt. = 30 lb
- Clock length = 12.23
- Draft = 11.23
- Evaporation and short fibre loss = 4%
Solution:
We know, Breaker card sliver wt. /100yd = (Dollop wt.) / (Clock length) X 100 X Doubling / Draft X Waste%
= 30/12.9 X 100 X 1/11.23 X (100 – 4)/100
= 19.88 lb/100yd. (ANS)
3. If B/C draft is 12, dollop wt. is 38 lb and Clock length (CL) is 14.4 yd, Then find –
- Feed sliver wt./100yd,
- Delivered sliver wt./100yd.
Solution:
Feed sliver wt./100 yd = DW/CL X 100
= 38/14.4 X 100
= 263.89 lb.
Delivered sliver wt./100yd = 263.89/Draft
= 263.89/12
= 21.99 lb. (ANS)
4. A B/C is working with 14.4 yds CL having FR speed of 6.94 rpm and dia 9.75 inches. What will be the time reqd to feed one dollop of 36 lb?
Solution:
Given,
- CL = 14.4 yd
- FR speed = 6.94 rpm
- FR dia = 9.75 inch
- Dollop wt. = 36 lb
- Time to finish one dollop =?
Distance traversed by the feed roller in one min = πdn = π X 9.75 X 6.94 inch/min
= 3.14 X 9.75/36 X 6.94 yd/min
= 5.90 yd/min.
The time required to feed one dollop = (CL in yd)/SS of FR in yd
= 14.4/5.9 min = 2.44 min (ANS)
5. A B/C is running with a CL 12.9 yd and DW 30 lb. having SS of the feed roller 18 feet/min. If the efficiency of B/C is 80% and evaporation loss is 4%, then what will be the production/shift in a metric ton? [1 MT = 2204 lb]
Solution:
Given,
- CL = 12.9 yd
- DW = 30 lb
- SS of feed roller = 18 feet/min
- Evaporation loss = 4%
- Efficiency =80%
- Production/shift =?
We know,
Production/shift = SS of feed roller in inch/min X 60 X DW/Cl X Eff% X Waste%
= 18/3 X 30/12.9 X 80/100 X (100 – 4)/100 X 60 X 8 X 1/2204
= 2.33 (ANS)