This document provides the calculations to determine the daily production rate of a knitting machine given various machine specifications and operating parameters. It first calculates the production rate of a single machine as 4.97 kg/day with 54 feeders. It then scales this up to the total production rate across all feeders of 228.123 kg/day at 85% efficiency. It also provides an alternative calculation method reaching the same result and examples for calculating production rates using different machine specifications.

1. Find out the production per day of 24hrs in kg at 85% efficiency when machine gauge = 24, machine diameter
= 30″ yarn count = 30Ne; RPM = 30; stitch length = 2.60 mm. Number of feeder = 54
Solution:
Here data given, Efficiency = 85%
Machine diameter = 30″
M/c gauge = 24 needles per inch
Yarn count = 30Ne
M/c rpm = 30
Stitch length = 2.60 mm
Number of feeder = 54
Machine circumference = π × d
Number of needle = π × d × G
Length of yarn in a course = π × d× G× Stitch length
Length of yarn used to produce courses in a minute = π × d× G× Stitch length × m/c rpm
= (π × 30× 24 × 2.6 × 30) mm/minute
= 176342.4 mm/minute
= 176.342 m/minute
= 176.342 × 1.09 yards /minute
= 192.21 yard/minute
=
30x840
192.21
pounds/minute
= 0.0076 pounds/minute
= 0.0076 × 60 × 24 pounds/day
= 10.944 pounds/day
=
2.2
10.944
kg /day
= 4.97 kg /day
Now with 54 feeders, the production will be, (4.97 × 54) = 268.38 kg/day
Production/m/c/day at 85% efficiency = 268.38 × 0.85 kg /day
= 228.123 kg /day
Alternate solution:
Here data given, Efficiency = 85% Machine diameter = 30″
M/c gauge = 24 needles per inch Yarn count = 30Ne
M/c rpm = 30 Stitch length = 2.60 mm
Production in Kg/day =
2.2(Ne)xCountx840x1000
Efficiency×feederofNumber×1.09×24×60×rpm×(mm)lengthStitch×G×(inch)d×π
=
2.2x30x840x1000
0.85×54×1.09×24×60×30×2.6×24×30×π
= 228.123 kg /day

2. For Practice…………….
1. Find out the production per hour in pounds at 70% efficiency when machine gauge = 32, machine diameter
= 30″ yarn count = 40 Ne; RPM = 24; stitch length = 2.20 mm. [Number of feeder = 90]
2. What will be the production per shift (8 hrs) in kg at 95% efficiency when machine gauge = 24, machine
diameter = 30″ yarn count = 30 Ne; RPM = 30; stitch length = 2.45 mm. [Number of feeder = 50]
3. Find out the stitch length of a fabric if that is produced from a machine 5 kg per day at 80% efficiency. And
machine gauge = 24, machine diameter = 28″ yarn count = 27Ne; and machine RPM = 26. [Number of feeder =
60]
4. In a knitting factory, Fabric will be produced tomorrow from 28 Ne Yarn. It has 50 machines, all of them
have 24 gauge, 30″ diameter, 60 feeders and stitch length should be = 3.00 mm. But 25 machines have an rpm
of 24, and other 25 machines have an rpm of 28. What will be the production of that factory tomorrow?