calculations in weft knitting machines
TRANSCRIPT
CALCULATIONS IN WEFT KNITTING MACHINE
A.MURALIKRISHNAN
LECTURER – TEXTILES
P.A.C.RAMASAMYRAJA POLYTECHNIC COLLGE,
RAJAPALAYAM, TAMILNADU, INDIA
Email: [email protected]
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Courses Courses : It is a horizontal rows of loops across the width of Fabric produced by the adjacent needles
- No of courses determines the length of fabrics
- Measured in Courses / cm or Courses/Inch
• No. of courses in 1 machine revolution = number of feeders
• No. of loops in course = no. of needles in machine
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Courses
• No of courses produced in 1 rev. = No. of feeders
Example 1: A knitting machine is running at a speed of 20 rpm with 30 feeders. Calculate the total no. of courses produced in 1 hr.
Ans:
Total number of courses = 20 x 30 x 60
= 36000 courses/hr
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Wales • It is a vertical column of loops produced by the Same
needles.
– Number of wales determines the width of fabric and
– measured in wales /cm or wales/Inch
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Machine parametersGauge: Number of needles per 1 inch.
Ex: 18 , 20, 22 ….
M/c Diameter : Diameter of cylinder Ex: 18 “, 20 “…..
Total no. of needles in a machine :3.142 x machine cylinder diameter x gauge
Example 2: Find out the total number of needles in a machine having cylinder diameter of 20” and 26 gauge
Ans: Total number of needles = 3.142 x 20 x 26 = 1634 needles
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Circumference of fabric
• Total no. of wales in a fabric
= total number of needles
Wales / inch = Machine gauge/ (1 - Shrinkage/100)
Example 3: Find the approximate circumference of fabric and WPI in a knitted fabric having 24 gauge, 20 inch diameter (assume fabric shrinkage is 35%)
Ans: Circumference of fabric = 3.142 x 20x (1-0.35) = 40.8 “
WPI = machine gauge / (1-shrinkage%) = 24/(1-0.35) = 36.92
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Stitch length or Loop length: Length of yarn in loop
• ( ranges from 0.28 “ to 0.35” or 7 to 10 mm)
Course length: Length of yarn consumed for full course
during knitting.
= Total no. of loops in 1 course x Loop length
(No. of loops in course = no. of needles in m/c )
• Total no. of courses for 1 rev = Total no. of feeder
Total no. of courses for n rev = n x feeders
Total no. of loops for n rev = n x feeders x total no. of needles.
Example 4: A knitting machine having following specifications: Machine Speed– 32 rpm, 24 gauge, 30 “ cylinder diameter, 40 feeders, 85% efficiency, CPI – 45, Loop lenth – 0.28” and yarn tex – 30. Calculate
1. Total no. of courses produced per shift
2. Total length of fabric produced per shift
3. Total weight of yarn required per shift.
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
• Loop shape factor: CPI / WPI =1.3
• Stitch density : Total no. of loops in a given area
S.D = CPI x WPI or CPCM x WPCM
Example 5: Calculate stitch density for a knitting machine having 24 gauge, 20 “ cylinder diameter, shrinkage – 30%, Loop shape factor – 1.3
Ans: WPI = 24/0.75 = 34.28
CPI = WPI x 1.3 = 44.57
Stitch density = 1527.9 loops / sq. inch
Loop shape factor and stitch density
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Tightness factor
• It is a ratio of area covered by yarn in one loop to the area occupied by the loop.
• Indicate the relative tightness or looseness of knitted fabric.
• T.F = √Tex / l where l = loop length in cm
Example 6: Find out the tightness factor of knitted fabric made of 36s count with loop length of 0.12”.
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Important formulas based on constants• CPI = Course constant / LL (Kc / LL)
• WPI = Wale constant / LL (Kw / LL)
• S.D = CPI x WPI or (Course const x wale const)/ LL2
= Kc x Kw / LL2
• Fabric GSM = CPCM x WPCM x LL in cm x Tex / 10
– Or Ks x Tex / 10 x LL in cm
Example 7: Given data : Kc – 5.5 , Kw – 4.2, 32 Ne, Tightnessfactor =15, Loop length- 0.28”.
Calculate 1. CPCM , WPCM
2. Stitch density
3. GSM
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Example 8: A knitting machine is running at 25 rpm with 60feeders and the fabric density is measured with CPI = 36.
WPI = 27 (14.2 c/cm; 10.6 w/cm) and the fabric weight is 173g/M2 and width -70”.
Calculate 1. Production in length per hour
2. Production in weight per hour
[1] Production in length per hour :
= (60 feeders X 25 rpm X 60 min)/(36 CPI)
= 2500 “/hr or (63.4 m/hr)
[2] The production in weight per hour :
given the finished width of fabric is 70 inches (1.78 meter)
= (63.4 m/hr X 1. 76 m X 173 g/m2 )/(1000g)
= 19.5 kg/hr (43 lb/hr)
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
Example 9:
• A knitting machine have 14 gauge, 20” dia, 40feeders operating at 25 rpm with 90% efficiency.Calculate the length, width and weight per hour forthe following fabrics using two fold resultant yarncount of 12s Ne.
• 1x1 rib, loop length – 0.2” , CPI = 27, WPI = 20
• Interlock, loop length – 0.12” , CPI = 37, WPI = 30
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India
References
• Spencer D J, “Knitting Technology”, Pergamon Press, UK, 1998.
• Anbumani N , “Knitting- Fundamentals, Machines, Structures and Developments”, New Age International Private Limited, India, 2006
Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India