Production Calculation of Circular Knitting Machine
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 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/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
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)
