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Find Out the Production of Interlock Circular Knitting Machine

Find Out the Production of Interlock Circular Knitting Machine
Raju Tex
Knitting Technologist,
B.J.Group, Mawna, Gazipur
Chittagong Textile Engineering College
Facebook: https://www.facebook.com/ERZRAZU




Introduction:
An interlock fabric comprising, in the simplest case, two part courses. These part courses complement each other to make a full course, and therefore two systems or feeders are required for producing one course.
Interlock Circular Knitting Machine
The following data were assumed for the interlock fabric production:

Example-1:
Values of circular knitting machine:
  • Machine diameter = 30”
  • Gauge E = 28
  • Number of feeders = 96
  • Machine speed = 31rpm
  • Machine efficiency = 85%
Values of article:
  • Structure: plain interlock
  • Yarn: polyester dtex = 76/1
  • Course density = 17courses/cm.
  • Wales density = 14 wales/cm.
  • Fabric weight = 100 gm/m2
                                                                              N X S X 60 X ŋ
Machine performance L in meter per hour = ………………………………………..
                                                                  feeders/course X courses/cm X 100


       31 X 96 X 60 X 0.85
= ……………………………
            2 X 17 X 100

= 44.64m/hr

                                              D X π X E
Fabric width, WB in meter = …………………
                                             wpcm X 100


    30 X 3.14 X 28
= ……………………
        14 X 100

= 1.88 m

                                                            L X WB X Weight in GSM
Machine performance in Kg per hour = ……………………………
                                                                          1000


    44.64 X 1.88 X 10
= ………………………
             1000

= 8.39 Kg/hr (ANS)

Elxample-2:
Values of circular knitting machine:
  • Machine diameter = 30”
  • Gauge E = 42
  • Number of feeders = 108
  • Machine speed = 31rpm
  • Machine efficiency = 87%
Values of article:
  • Structure: plain interlock
  • Yarn: polyester filament yarn dtex 50 of 88/1
  • Course density = 19 courses/cm.
  • Wales density = 23wales/cm.
  • Fabric weight = 100gm/m2
                                                                           N X S X 60 X ŋ
Machine performance L in meter per hour = ………………………………………..
                                                                  feeders/course X courses/cm X 100


    31 X 108 X 60 X 0.87
= …………………………
           2 x 19 x 10

= 45.99m/hr

                                             D X π X E
Fabric width, WB in meter = ……………………
                                            wpcm X 100


    30 X 3.14 X 4
= …………………
        23 x 100

= 1.72 m

                                                            L X WB X Weight in GSM
Machine performance in Kg per hour = …………………………………
                                                                           1000


    44.99 X 1.72 X 10O
= …………………………
              1000

= 7.91Kg/hr. (ANS)

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