Combing is the process in yarn spinning by which straightening and parallelizing of fibres and the removal of short fibers and impurities by using a comb on combs assisted by brushes and rollers is called combing. To produce a uniform sliver of required weight per unit length combing is necessary. Here goes the need to know the mathematical solutions.
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Combing Mathematical Problems and Solutions
Some mathematical problems and solutions of combing are as follows –
1. Ribbon/ feed lap weight per yd 580 grain. 6 laps are fed in a combing machine. If delivered sliver weight per yds is 46.60 grain, then what will be the ribbon lap hank and draft?
Solution:
Here given,
- Ribbon lap wt./yd = 580 grain
- Doubling = 6
- Delivered sliver wt./yd = 46.60 grain
- Ribbon lap hank =?
- Draft =?
We know that,
…………………………….Ribbon lap length/Length unit
Ribbon lap hank =…………………………………………………
…………………………….Ribbon lap weight/Weight unit
= Ribbon lap length/ Length unit X Weight unit/ Ribbon lap weight
= 1/840 X 7000/580
= 0.0144 (ANS).
……………………….Feed sliver weight
Actual draft = ……………………………….. X Doubling
……………………Delivered sliver weight
= 580/46.60 X 6
= 74.68 (ANS).
2. In a combing machine coiler calendar roller speed is 118 rpm, radius 1”, Delivered sliver weight per yd is 46.6 grain. Calculate the production constant. If efficiency is 90% then also calculate the production per shift.
Solution:
Here given,
- Coiler calendar roller speed (N) = 118 rpm
- Coiler calendar roller radius (r) = 1”
- Delivered sliver weight per yd = 46.6 grain
- Production constant =?
- Production/shift at 90% eff. =?
We know that,
Production constant = Circumference speed of coiler calendar roller / 36
= (π X 2r X N) / 36
= (3.14 X 2 X 1 X 118) / 36
= 20.58
……………………………………………………………….Sliver weight
Production/shift = Production constant X ……………………… X 60 X shift X Eff.
……………………………………………………………………..7000
= 20.584 X 46.6/7000 X 60 X 8 X 90/100 lb/shift
= 59.20 lb/shift (ANS).
3. If in a combing machine draft constant is 2780 and DCP is 46, then what will be the mechanical draft of the machine? If waste is 18%, then what will be the actual draft?
Solution:
Here given,
- Draft constant = 2780
- DCP = 46
- Waste = 18%
- Mechanical draft =?
- Actual draft =?
We know,
Mechanical draft = Draft constant/DCP
= 2780/46
= 60.43 (ANS).
Again,
Actual draft = (Mechanical draft X 100) / (100 – Waste%)
= (60.43 X 100)/(100 – 18)
= 6043 / 82
= 73.70 (ANS).
4. Calculate the DCP of a combing machine from these parameters. Feed lap hank = 64 K Tex, Delivered single sliver hank = 3.40 K Tex, Doubling = 6, Waste = 15%, and draft constant = 1696. The machine is giving twine sliver delivery.
Solution:
Here given,
- Feed lap hank = 64 K Tex
- Delivered single sliver hank = 3.40 K Tex
- Doubling = 6
- Waste = 15%
- Draft constant = 1696
- DCP =?
We know that,
Delivered twin sliver hank = 3.14 X 2 = 6.80 (Direct system)
Again, Actual draft = Feed lap hank / Delivered sliver hank X Doubling
= (64 X 6) X 6.8
= 56.47
And, Mechanical draft = Actual draft X (100 – Waste%) / 100
= 56.47 X 85/100
= 48
Now, DCP = Draft constant / Mechanical draft
= 1696 / 48
= 35.33
= 35T (ANS).
5. Calculate the production for 8 hrs of a combing machine from the following data –
- Feed/Nip = 6.7 mm
- Waste = 16%
- Nip/min = 120
- Efficiency = 90%
- No. of head = 6 Feed lap hank = 68 K Tex
Solution:
Here given,
- Feed/Nip = 6.7 mm
- Waste = 16%
- Nip/min = 120
- Efficiency = 90%
- No. of head = 6 Feed lap hank = 68 K Tex
- Production/8 hrs =?
We know that,
Production = Nip/min X Feed/Nip X No. of head X Waste% X Eff.%
= 120 X 6.7 X 6 X {(100 – 16) / 100} X 90/100 mm/min
= 120 X 60 X 8 X 6.7/1000 X 68/1000 X 6 X 84/100 X 90/100
= 119.04 kg/8hrs (ANS).
Author of this Article: Md. Abu Sayed Founder of Textile Apex Pabna Textile engineering college, Pabna, Bangladesh Email: [email protected] Cell : +8801745214773