**Introduction:**

It has been expounded in the sections above that the output of a circular knitting machine depends on a series of different influencing variables. A wealth of machine data and data on the fabric to be produced is required for calculating production capacity. In article we will discuss about the circular knitting machine production calculation.

In this respect the cylinder diameter d in inch, the gauge E, the system count S, the machine rpm n, and the efficiency level q of the circular knitting machine must be known. The following data on the fabric to be produced must also be available:

- The construction(e.g. single-jersey, rib, purl etc.)
- The course density or courses /cm, and
- The weight per unit area in gm /m
^{2}.

**Circular Knitting Machine Production Calculation**

**Machine output:**

The machine capacity or performance in running m/hr is calculated in accordance with the following equation:

Machine capacity, L

Speed of machine in rpm X No. of system or feeders on the machine

X efficiency X 60minutes

= ……………………………………………………………………………………………………………………….m/hr

…………………….No. of feeders or systems per course X courses per cm. X 100

**Example-1:**

Calculate the length in meters of a plain, single sided or single- jersey fabric knitted at 20 courses/cm. on a 30” diameter 22-gauge circular machine having108 feeds. The machine operates for 8 hours at 36 rpm at 87% efficiency.

**Solution:**

Machine capacity i.e. the total length of the fabric in meters,

Speed of machine in rpm X No. of system or feeders on the machine

X efficiency X 60minutes

= ………………………………………………………………………………………………………………………….m/hr

…………………No. of feeders or systems per course X courses per cm. X 100

….36 X 108 X 87 X 60 X 8

= ……………………………………

……..1 X 20 x 100 X 100

= 811.82 meters (ANS)

**Example-2:**

Calculate the length in meters of a plain, single sided or single-jersey fabric knitted at 16 courses/cm. on a 26” diameter 28 gauge circular machine having 104feeds. The machine operates for 8 hours at 29rpm at 95% efficiency.

Machine capacity i.e. the total length of the fabric in meters,

Speed of machine in rpm X No. of system or feeders on the machine

X efficiency X 60minutes

= ……………………………………………………………………………………………………………………..m/hr

…………………No. of feeders or systems per course X courses per cm. X 100

….29 X 104 X 95 X 60 X8

= …………………………………..

………1 X 16 X 100 X 100

= 859.56 meters (ANS)

Author of this Aricle:Raju TexKnitting Technologist, B.J.Group, Mawna, Gazipur Chittagong Textile Engineering College Facebook: https://www.facebook.com/ERZRAZU