Production Related Formulas, Problems and Its Solutions of Knitting
Introduction:
The production of a circular knitting machine is done by both length and weight. There are several types of formulas for the determination of production. Some of them are given below –
Formulas for the determination of production in length –
Course/Minute
1. Fabric length = …………………………………
Course/cm
2. Course/Min = No. of feeder X Cylinder speed
3. Stitch density = Course/cm X Wales/cm
No. of Wales
4. Fabric width = ………………………….
Wales/cm
No. of Needle Ļ X Cylinder Dia X Gauge
= …………………………… = ………………………………………..
Wales/cm Wales/cm
Formulas for the determination of production in weight –
Course/min X Yarn length per course
1. Fabric weight = ………………………………………………………………
Yarn count
2. Yarn length per course = Total needle no. of machine X Loop length
3. Total needle no. m/c = Ļ X Cylinder dia X Gauge
Some mathematical problems and its solutions
1. Calculate the production per day in kg of a plain single jersey knitted at 30 inch diameter, 24 gauge circular machine having 96 feeders and 0.25 cm stitch length produced by 30/1’s. The machine operates at 25 rpm at 70% efficiency.
Solution:
Here data given,
Machine dia = 30”
Machine gauge = 24
No. of feeders = 96
Stitch length = 0.25 cm
Yarn count = 30/1’
Machine rpm = 25
Efficiency = 70%
Now, length of yarn in a loop = 0.25 cm
So, Length of yarn in full course = 0.25 X Ļ X G X d cm
= 0.25 X Ļ X 24 X 30 cm
So, Length of yarn used in a minute for producing course,
= 0.25 X Ļ X 24 X 30 X 96 X 25 cm
We get, Production per day at 70% efficiency,
0.25 X Ļ X 24 X 30 X 96 X 25 X 60 X 24 X 70
= ……………………………………………………………………….
2.54 X 36 X 840 X 30 X 100 X 2.2
= 269.85 kg (ANS)
Or,
RPM of cylinder X No. of feeder X Ļ X Cylinder dia (inch)
Production in weight = ……………………………………………………………………
1000 X 1000
Gauge X Loop length (mm) X tex X 60 X 24 X Efficiency
X …………………………………………………………………………………….
1000
25 X 96 X Ļ X 30 X 24 X 0.25 X 10 X 590.5/30 X 60 X 24 X 0.70
= ………………………………………………………………………………………
1000 X 1000 X 1000
= 269 kg/day (ANS)
2. Calculate the production of a single jersey circular knitting m/c per shift from the following data –
Here,
Cylinder dai = 30”
Cylinder speed = 20 rpm
No. of feed = 36
No. of course per inch = 30
Machine eff = 80%
Solution:
Here the given data,
Cylinder dai = 30”
Cylinder speed = 20 rpm
No. of feed = 36
No. of course per inch = 30
Machine eff = 80%
Production/8 hr =?
We know,
No. of course per min = No. of feeder X Cylinder speed
= 36 X 20 = 720
Again,
No. of course per min
Production = ………………………………….. X Machine efficiency
No. of course per inch
= 720/30 X 80/100 inch/min
= 720/30 X (60 X 8)/36 X 80/100 gauge/8hr
= 256 gauge/shift (ANS)
3. Determine the no. of course per cm of a fabric from the following data. This fabric is produced 1152 m per shift in a single jersey circular knitting machine.
Here,
Knitting m/c speed = 20 rpm
No. of feeder = 48
Machine efficiency = 75%
Solution:
Given,
Knitting m/c speed = 20 rpm
No. of feeder = 48
Machine efficiency = 75%
Length of produced fabric per shift = 1152 m
No. of course per cm =?
We know,
Fabric production per shift,
Course per min
= …………………………. X 60 X 8 X Efficiency
Course per cm
No. of feeder X Machine speed X 60 X 8
= ………………………………………………………………….. X Efficiency
Course per cm
48 X 20 X 60 X 8 75
Or, 1152 X 100 = ………………………………... X ……………
Course/cm 100
48 X 20 X 60 X 8 X 75
Or, Course/cm = ………………………………………. = 3 (ANS)
1152 X 100 X 100
