Lap Former & Comber Machine Production Calculations with Example

Introduction
Combing is the removing process of short fibers, neps & impurities of carded sliver and to make the fiber more parallel, uniform & straight by using a comb machine or comb associated by knives, brushes and rollers. Combing is necessary for better yarn appearance and regularity. Longer fibers are finer than shorter fibers. After combing higher count (uniform & finer sliver) is possible to keep minimum no of fiber in yarn diameter. So, calculations play an important role in combing. Some important calculations (Actual Draft & Mechanical Draft, Draft Constant, DCP, Comber Noil%, Machine Production, Production Constant, Bett Tension, sliver Hank etc.) of Comber machine are discussed below.

Comber Machine Production Calculations

Some Important Conversion Factors for Comber Machine Production Calculations:

  • 36 inch = 3 Feet = 1 Yard;
  • 1 Meter = 1.0936 Yards;
  • 1 Pound(lb) = 453.6 Gram = 16 Ounce (Oz);
  • 1 Lea = 120 Yards;
  • 840 Yard = 7 Lea = 1 Hank;
  • 1 Pound (lb) = 7000 Grains;
  • 1 Meter = 39.37 Inch;
  • 1 inch = 2.54 cm;
  • 1 cm = 10 mm;
  • 1 Kg = 2.205 Pound (lb)
  • Count means the number of 840 yards length that weight exactly 1 pound (lb)
  • KTex = Number of kilograms per 1000 meters.

Lap Former and Comber Machine Production Calculations with Formula:

………………………..Delivery Count x No of Doubling
1. Actual Draft = ————————————————————————-
…………………………..Feed Count x No of Delivery

……………………………………………………(100 – Noil Extraction %)
2. Mechanical Draft = Actual Draft x ———————————————-
…………………………………………………………………100

3. Draft Constant = Mechanical Draft x DCP

Example: Find out required DCP to produce 3.4 Ktex sliver from 64 Ktex comber lap, the comber machine particulars are as the following- no of head = 8, no of delivery = 2, noil extraction = 15%, draft constant = 1698.

Solution:

Here given,

Feed Count = 3.4 Ktex
Delivery Count = 64 Ktex
No of Doubling = 8
No of Head = 2
Noil Extraction% = 15
Draft Constant = 1698

……………………..Delivery Count x No of Doubling
Actual Draft = ——————————————————————–
………………………..Feed Count x No of Delivery

….64 x 8
= ————– = 75.3
3.4 x 2

………………………………………………….100 – Noil%
Mechanical Draft = Actual Draft x ——————–
………………………………………………………100

…………..100 – 15
= 75.3 x —————–
……………..100

= 64

Draft Constant = Mechanical Draft x DCP

…………………Draft Constant
Or, DCP = —————————————–
………………..Mechanical Draft

1698
= ————— = 26.53≈ 27T (Ans)
…..64

4. Production per Day = Nip/Min x Feed/Nip x No of Head x Feed Lap Wt. x Efficiency x Noil% x 60 x 24

………………………………………….Required Production
5. No of Comber Required = ——————————————————–
…………………………………………..Actual Production

Example: Calculate the production per day of a combing machine running with following particulars – feed lap wt = 1100 grains/yard, nips/min = 400, no of combing head = 8, Noil extraction = 17%, feed/nip = 5 mm, efficiency = 90%. Also find out the no of modern comber m/c to produce 10000 lb/day.

Solution:

Nip/Min = 400

……………………………………….5
Feed/Nip = 5 mm = ————————-= 0.005468 yard
……………………………..10 x 2.54 x 36

No of Head = 8

Feed Lap Wt = 1100 grains/yard
= 1100/7000 lbs/yard
= 0.1571 lbs/yard

Efficiency = 90% = 90/100 = 0.9

Noil% = 17% = 100-17/100 = 0.83

Production per Day = Nip/Min x Feed/Nip x No of Head x Feed Lap Wt. x Efficiency x Noil% x 60 x 24

= 400 x 0.005468 x 8 x 0.1571 x 0.9 x 0.83 x 60 x 24lbs/day/comber

= 2956.9lbs/day/comber (Ans)

Here, Required Production = 10000 lbs/day

Actual Production = 2956.9 lbs/day

No of Comber M/C Required = 10000/2956.9 = 3.38 ≈ 3 (Ans)

…………………Nips/Min x Feed/Nip x No of Head x Lap Wt. x Effi. x Noil% x 60
6. Production/hr = ———————————————————————————————
……………………………………………………1000 x 1000

Example: The cylinder of a 6 head comber is running at a speed of 150 nips per minute and each nip feeds 0.185″ lap. The hank of lap is 0.0166. Calculate the production of comber in hours at 75% efficiency and 12% waste.

Solution:

Here Given,

Nips/Min = 150

Nip/Feed = 0.185”
= 0.185 x 2.54 x 10 = 4.699 mm

Lap Hank = 0.0166
= 32.53 gm/yard
= 29.745 gm/m (Lap Wt.)

Efficiency = 75% = 0.75
Waste/Noil% = 12% = 100-12/100 = 0.88
No of Comber Head = 6

……………………….Nips/Min x Feed/Nip x No of Head x Lap Wt. x Effi. x Noil% x 60
Production/hr = ————————————————————————————————–
………………………………………………………………….1000 x 1000

150 x 4.699 x 6 x 29.745 x 0.75 x 0.88 x 60
= ————————————————————— kg
……………………..1000 x 1000

= 4.98 kg (Ans)

7. Production per Shift = Feed/Min x Lap Wt x Efficiency x noil% x No of Head x 60 x Shift x Bett Tension

Example: Calculate the production of comber machine per shift. Using the following particulars-

Lap Hank = 70 KTex
Nips/Min = 350
Feed/Nip = 5 mm
Bett Tension = 12 %
Noil % = 18 %
Efficiency (%) = 90 %
No. Of Head = 8
Shift Time = 8 Hours

Solution:

Here Given,

Lap Hank = 70 KTex

70 x 1000
= —————– = 70 gm/m
…….1000

So, Lap Wt. = 70 gm/m

Bett Tension % = 12

…………………………..Bett Tension in%
Bett tension = 1 + ———————————
…………………………………..100

…………..12
= 1 + ————- = 1.12
…………100

Here given,

Feed/Nip = 5mm
Nips/Min = 350

Feed/Min = Feed/Nip x Nips/Min
= 5 x 350 mm
= 1750 mm
= 1.75 Meter

Efficiency = 90% = 0.9

Noil% = 18% = (100-18)/100 = 0.82

Also Given,

No of Head = 8
Shift Time = 8 hrs.

Production per Shift = Feed/Min x Lap Wt x Efficiency x noil% x No of Head x 60 x Shift x Bett Tension

= 1.75 x 70 x 0.9 x 0.82 x 8 x 60 x 8 x 1.12
= 388813.824 grams
= 388.813 kgs (Ans)

8. Production constant = π x Roller Dia (D) x Coiler Calendar Roller RPM (N)

9. Production per Shift = Product Constant x Sliver Wt. x Efficiency x 60 x 8

Example: Find out the production constant and production/shift in a comber machine as per following parameters:

Coiler calendar roller speed = 120 rpm
Roller radius = 1.25″
Delivery sliver wt. = 72 grains/yd
Efficiency (%) = 90%

Solution:

Here Given,

Coiler Calendar Roller Speed = 120 RPM
Roller Radius, r = 1.25”
Delivery Sliver Weight = 72 Grains/yard = 72/7000 = 0.0102857 lb/yard
Efficiency = 90% = 0.9

Production Constant = πDN = π x 2r x N
= 2 x 3.1416 x 1.25 x 120 inch/min
= (2 x 3.1416 x 1.25 x 120)/36 yards/min
= 26.18 yards/min (Ans)

Production per Shift = Product Constant x Sliver Wt. x Efficiency x 60 x 8
= 26.18 x 0.0102857 x 0.9 x 60 x 8 lbs
= 116.33 lbs (Ans)

………………………………………………………………….π x CCR RPM x CCR Dia
10. Surface Speed of Coiler Calendar Roller = ———————————————-
………………………………………………………………………………39.37

……………………………………….7000
11. Sliver Hank = ——————————————————-
……………………….Sliver wt. (grains/yd) x 840

……………………….SS of CCR x 0.59 x No of Head x Efficiency x 60 x 8
12. Production = ————————————————————————————
………………………………….Sliver Hank (Ne) x 1000 x 100

Example: Calculate the m/c production per shift with the help of following particulars-

  • Coiler Calendar Roller RPM = 125
  • Coiler Calendar Roller Dia = 2.25”
  • Delivery Sliver wt. = 70 Grains/yard
  • No of Heads = 8
  • Efficiency = 90%

Solution:

………………………………………………………………π x CCR RPM x CCR Dia
Surface Speed of Coiler Calendar Roller = ———————————————–
…………………………………………………………………………….39.37

3.1416 x 125 x 2.25
= ——————————- meters/min
………….39.37

= 22.44 meters/min

………………………………..7000
Sliver Hank = ————————————————
…………………Sliver wt. (grains/yd) x 840

……7000
= —————– = 0.119 Ne
70 x 840

…………………………………….SS of CCR x 0.59 x No of Head x Efficiency x 60 x 8
Production per shift = ———————————————————————————-
……………………………………………………Sliver Hank (Ne) x 1000 x 100

22.44 x 0.59 x 8 x 90 60 x 8
= ———————————————- kgs
…….0.119 x 1000 x 100

= 384.5 kgs (Ans)

………………………….Wt. of Noil
13. Noil% = ————————————————– x 100
………………..Wt. of Noil + Wt. of Sliver

Example: Find out the Noil%, If the sliver weight is 70 grains/yard and the wt. of noil is 0.89 gram/meter.

Solution:

Here given,

Wt. of Noil = 0.89 gram/meter
Wt. of Sliver = 70 grains/yard

Sliver Wt. = 70 grains/yard
= 70/7000 lb/yard
= 70 x 453.6/7000 gram/yard

70 x 453.6 x 39.37
= —————————- gram/meter
……….7000 x 36

= 4.96 gram/meter

……………………….Wt. of Noil
Noil% = —————————————— x 100
…………..Wt. of Noil + Wt. of Sliver

……….0.89
= ——————– x 100
0.89 + 4.96

= 15.21% (Ans)

Conclusion
I have tried to explain & solve the machine production calculation and other calculations formulas with examples of Comber Machine (Lap Former) through this article. Hope this article will be helpful for those who are new to the spinning mill or those who are students.

Author of this Article:
Md. Imran Hossain
B.Sc. in Textile Engineering
Shahid Abdur Rab Serniabat Textile Engineering College, Barisal.
Email: mdimranhossain.te@gmail.com

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