**Production Calculation of Circular Knitting Machine**

**Raju Tex**

Knitting Technologist,

B.J.Group, Mawna, Gazipur

Chittagong Textile Engineering College

**Facebook:**https://www.facebook.com/ERZRAZU

**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.

Circular knitting machine |

- The construction(e.g. single-jersey, rib, purl etc.)
- The course density or courses/cm, and
- The weight per unit area in gm/m2.

**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

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

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

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)**

## 0 comments:

## Post a Comment