How To Calculate Fabric Production of Jacquard Circular Knitting Machine
Jacquard circular knitting machine
A jacquard circular knitting machine is disclosed provided with a needle cylinder driving mechanism capable of optionally and readily controlling the rotary motion of the needle cylinder according to the pattern to be knitted.
A two-color jacquard fabric is to be produced, and the following machine and fabric data were assumed:
Example-1:
Values of circular knitting machine:
- Machine diameter = 26”
- Gauge E = 20
- Number of feeders = 60
- Machine speed = 27rpm
- Machine efficiency = 80%
Values of article:
- Structure: two colour jacquard (two part
- Courses for one course)
- Yarn: cotton yarn Nm = 50/1(80%), polyester filament yarn dtex = 67f14(20%)
- Course density = 10.5courses/cm.
- Wales density = 16wales/cm.
- Fabric weight = 90gm/m2
We know that,
n X S X 60 X ŋ
Machine performance L in meter per hour = ………………………………………..
feeders/course X courses/cm X 100
27 X 60 X 60 X 0.80
= …………………………
2 X 10.5 X l00
= 37 m/hr.
D X π X E
Fabric width, WB in meter = …………………
wpcm X 100
26 X 3.14 X 20
= ……………………
16 X 100
= 1.02m
L X WB X Weight in GSM
Machine performance in Kg per hour = ……………………………
1000
37 X 1.02 X 90
= …………………
1000
= 3.396 Kg/hr
Example-2:
Values of circular knitting machine:
- Machine diameter = 30”
- Gauge E = 20
- Number of feeders = 96
- Machine speed = 23rpm
- Machine efficiency = 80%
Values of article:
- Structure: three colour jacquard
- Yarn: polyester dtex = 150/1
- Course density = 12courses/cm.
- Wales density = 11wales/cm.
- Fabric weight = 180gm/m2
n X S X 60 X ŋ
Machine performance L in meter per hour = ………………………………………..
feeders/course X courses/cm X 100
23 X 96 X 60 X 0.80
= ……………………………
3 X 12 X 100
= 29.44 m/hr.
D X π X E
Fabric width, WB in meter = …………………
wpcm X 100
30 X 3.14 X 20
= ……………………
11 X l00
= 1.7m
L X WB X Weight in GSM
Machine performance in Kg per hour = ……………………………
1000
29.44 X 1.7 X 18
= ……………………
1000
= 9 Kg/h
Example-3:
Calculate the production of a single-jersey circular knitting machine in kg/hr from the following data:
Values of circular knitting machine:
- Machine diameter = 30”
- Gauge E = 28
- Number of feeders = 96
- Machine speed = 35rpm
- Machine efficiency = 80%
Values of article:
- Structure: plain(Single jersey)
- Yarn: cotton Nm 50/1(Ne29.6/1)
- Stitch length = 0.25cm
n X S X(π X d X E X stitch length in cm)X 60 X ŋ X 0.4536
= …………………………………………………………………………
100 x l00 X Ne x 840 x 0.9144
n X S X(d X E X stitch length in cm.) X ŋ
= ……………………………………...........…… X 0.00001112598
Ne
35 X 96 x (30 X 28 X 0.25) X 80
= ……………………………….....…. X 0.00001112598
29.6
=21.22 Kg. (ANS)
