# Textile Calculation

Problems and solutions of textile

## Mathematical Problems and Solutions of Combing

Mathematical Problems and Solutions of Combing
Md. Abu Sayed
Founder of Textile Apex
Pabna Textile engineering college, Pabna, Bangladesh
Cell : +8801745214773

What is Combing?
Combing is the process in yarn spinning by which straightening and parallelizing of fibres and the removal of short fibres 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. Some mathematical problems and solutions of combing are as follows –

1. Ribbon/feed lap weight per yd 580 grain. 6 laps are feeded 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).