Find Out the Production of Interlock Circular Knitting Machine
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.
The following data was
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)
