How to calculate Knit fabric production?

Knitting is one of the most used fabric production techniques, where yarns are interlooped to form the flat fabric. Two types of knitting exist, namely, warp knitting and weft knitting. A weft-knitted fabric can be produced using one single strand of yarn Figure-1(a). while a warp-knitted fabric or a woven fabric requires the same number of yarns as required through the width of the fabric Figure-1(b). Knit fabric production calculation has been explained in detail as below.
How to calculate knit fabric production

How to calculate knit fabric production

How to calculate knit fabric structure

Knit Fabric Production Calculation:
👉1. Production in length (meter/hour):

= \frac{{Course/hour}}{{Course/meter*Feeder/course}}

= \frac{{Machine RPM*60*No.of.feeders*Efficiency\% }}{{Course/inch*39.37*Feeder/course}}

= \frac{{Machine.RPM*No.of.feeders*Eff\% }}{{Course/inch*Feeder/course}}x1.524

= \frac{{Machine.RPM*No.of.feeders*Eff\% }}{{Course/cm*2.54*Feeder/course}}x1.524

= \frac{{Machine.RPM*No.of.feeders*Eff\% }}{{Course/cm*Feeder/course}}x0.60

Example: a circular knitting machine has 30” diameter, 24 gauge and 96 feeders. It is producing a single jersey fabric with 15 course per cm. If the machine runs at 22 rpm and 85% efficiency, calculate the length of the fabric produced in one hour.

Solution:
Here, Machine RPM= 22, Course/cm= 15, No of feeders= 96, Feeder/course for single jersey fabric= 1, Efficiency= 85%
Therefore,
\Pr oduction = \frac{{Machine.RPM*No.of.feeders*Eff\% }}{{Course/cm*Feeder/course}}x0.60
= \frac{{22*96*0.85}}{{15*1}}x0.60
= 71.81meter/hour

👉2. Fabric width (meter):

= \frac{{\pi *Machine.dia.in.inch*Machine.gauge}}{{Wales/cm*100}}


Example: a circular knitting machine has 30” diameter, 28 gauge and 96 feeders. It is producing a single jersey fabric with 15 course per cm and 20 wales per cm. If the machine runs at 20 rpm and 85% efficiency, calculate the width and length of the fabric produced in one hour.

Solution: Here,
Machine RPM= 20, Course/cm= 15, Machine dia= 30 inch, Wales/cm= 20, Machine gauge= 28, Feeder/course for single jersey fabric= 1, No of feeders= 96, and Efficiency= 85%.

Therefore,

(i).Fabric.production = \frac{{Machine.RPM*no.of.feeders*Eff\% }}{{Course/cm*feeder/course}}0.60meter/hour
= \frac{{20*96*0.85}}{{15*1}}0.60meter/hour
= 65.28meter/hour

(ii).Fabric.width = \frac{{\pi *Machine.dia.in.inch*Machine.gauge}}{{Wales/cm*100}}meter
= \frac{{3.1412*30*28}}{{20*100}}meter
= 1.32meter

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