# Textile Calculation

Problems and solutions of textile

## Some Essential Calculations for Knitting

Some Essential Calculations for Knitting Md. Abu Sayed
Founder of Textile Apex
Pabna Textile engineering college, Pabna, Bangladesh
Cell : +8801745214773

Introduction:
Knitting is completely calculation based production. Before production any item knitter must calculate first. If calculation is done perfectly then production efficiency will increase. Now I’m giving some problems of knitting for helping purpose novice knitting muster. In previous article I have written about different necessary formula for knitting. You can read it to understand following problems.

Calculations for Knitting
Problem-1: Calculate the production per hour (in gauge) of a circular knitting machine from the following data –
• Cylinder dia – 30”
• Cylinder speed – 20 rpm
• No. of feeder – 36
• Efficiency – 80%
• Course/inch – 30
• Course/rpm – 36
Solution

Given,
• Cylinder dia – 30”
• Cylinder speed – 20 rpm
• No. of feeder – 36
• Efficiency – 80%
• Course/inch – 30
• Course/rpm – 36
• Production/8 hour =?
We know that,

Course/min = No. of feeder X Cylinder speed

= 36 X 20

= 720

Again,

Course/min
Production = ……………………… X  Efficiency
Course/inch

= (720/30) X (80/100) inch/min

= (720/30) X (60/30) X (80/100) gauge/hr

= 32 gauge (ANS)

Problem-2: Calculate the production per 8hr (in meter) of a circular knitting machine from the following data –
• No. of feeder – 48
• Fabric width – 264 cm
• Stitch density – 15
• Machine speed – 20 rpm
• Gauge – 14
• Efficiency – 75%
Solution:

Given,
• No. of feeder – 48
• Fabric width – 264 cm
• Stitch density – 15
• Machine speed – 20 rpm
• Gauge – 14
• Efficiency – 75%
• Production/8hr =?
We know that,

No. of Wales in fabric                           Gauge X π X Dia
Wales/cm = …………………………………... = ……………………………….
Width of fabric                                   Width of fabric

= (14 X 3.14 X 30)/264 = 4.995 = 5

Again,

Course/cm = (Stitch density)/(Wales/cm) = 15/5 = 3

Course/min                         No. of feeder X m/c speed
Production = ……………………. X Efficiency = ………………………… X Efficiency
Course/cm                                     Course/cm

= (48 X 20)/3 X .75 cm/min

= (48 X 20)/3 X (60 X 8)/100 X .75 m/8hr

= 1152 m/8hr (ANS)

Problem-3: Calculate the production per day (in gauge) of a circular knitting machine from the following data –
• Machine speed – 25 rpm
• Machine gauge – 28
• Machine dia – 24”
• No. of feeder – 96
• Course/cm – 16
• Efficiency – 90%
Solution:

Given,
• Machine speed – 25 rpm
• Machine gauge – 28
• Machine dia – 24”
• No. of feeder – 96
• Course/cm – 16
• Efficiency – 90%
• Production/24hr =?
We know,

Course/min = No. of feeder X Machine speed = 96 X 25 = 2400

And, Course/inch = Course/cm X 2.54 = 16 X 2.54 [2.54 cm = 1 inch]

= 40.64

Again,

Course/min
Production = …………………….. X Efficiency
Course/inch

= (2400/40.64) X 0.90 inch/min

= (2400/40.64) X (60 X 24)/36 X 0/90 Gauge/24 hr

= 2126 gauge/24 hr (ANS)

Problem-4: Calculate the production per shift (in kg) of a single jersey circular knitting machine from the following data –
• Cylinder speed – 35 rpm
• No. of feeder – 96
• Cylinder dia – 30 inch
• Machine gauge- 28
• Efficiency – 85%
• Course density – 18 course/ cm
• Wales density – 13 Wales/cm
• Fabric wt. – 125 gm/m2
We know that,

Fabric length per shift

No. of feeder X Machine speed
= …………………………………………………….. X 60 X 8 X (85/100)
Course/cm

= (96 X 35)/18 X 60 X 8 X (85/100) m/shift

= 761.6 m/shift

Total amount of walse
Fabric width = …………………………………….
Wales/cm

Π X Machine dia X Machine gauge
= …………………………………………………
Wales/cm

= (3.14 X 30 X 28)/13 = 202.89 cm

= 2.03 m

Total wt. of fabric per shift

= Total length of fabric X fabric width X Wt. of fabric

= 761.6 X 2.03 X 125 gm/m2

= (761.6 X 2.03 X 125)/1000 kg/m2

= 193.26 kg/m2 (ANS)

Problem-5: In a circular knitting machine, machine speed = 135 rpm, no. of feeder = 18, Produced fabrics course/cm = 28, Wales/cm = 24. Calculate the production per 8 hr and fabric density.

Solution:

Given,
• Machine speed = 135 rpm
• No. of feeder = 18
• Course/cm = 28
• Wales/cm = 24
• Production/8 hr =?
• Stitch density =?
We know that,

Stitch density = Course/cm X Wales/cm = 28 X 24 = 672 (ANS)

No. of feeder X Machine speed
Production = ……………………………………………………
Course/cm

= (18 X 135)/28 cm/min

= (18 X 135)/28 X (60 X 8)/100 m/8hr

= 416.57 m/8hr (ANS)