Introduction
The process of applying a protective adhesive (PVC, CMC, PVA etc.) coating upon the yarn (warp) surface is called sizing. This is the most important operation to attain maximum weaving efficiency specially for blended and filament yarns. Due to sizing increases elasticity of yarn, yarn strength, smoothness, weight of the yarn and frictional resistance and decreases yarn hairiness, flexibility, absorbency and static electricity. So, calculations play an important role in sizing machine. Some important production calculation (Actual production, Calculated production, Time required, Length of warp, Weight of sized warp, Weight of unsized warp, Weight of size, Size%, Count of sized yarn, No. of weavers beam, No of loom etc.) of Sizing machine are discussed below.
Some Important Conversions for Sizing Machine Production Calculation:
- 36 inch = 3 Feet = 1 Yard;
- 1 Meter = 1.0936 Yards;
- 1 Pound(lb) = 453.6 Gram = 16 Ounce (Oz);
- 1 Lea = 120 Yards;
- 840 Yard = 7 Lea = 1 Hank;
- 1 Pound (lb) = 7000 Grains;
- 1 Meter = 39.37 Inch;
- 1 inch = 2.54 cm;
- 1 cm = 10 mm;
- 1 Kg = 2.205 Pound (lb)
Sizing Machine Production Calculation in Weaving Mill
1. The actual production per day of 8 hrs = Calculated production per day of 8 hrs x Efficiency
2. Total Length of yarn sized = Total length of warp x No of ends
………………………………………………Total length of warp in yds
3.Total wt. of sized warp in lbs = —————————————————— + Size%
…………………………………………………….840 x Yarn count
Example: The calculated production of a high-speed slasher is 108 yds per min. If the efficiency of the machine is 78%. Calculate the followings-
- The actual production per day of 8 hrs.
- The total length of yarn if the total ends is 3280.
- The total wt. of the sized warp if it is sized to 12% and the count of unsized are 40’s.
Solution:
Here, Calculated production = 108 yds/min
Efficiency = 78% = 0.78
Calculated production per day of 8 hrs = 108 x 60 x 8 yds/8 hrs
= 51840 yds/8 hrs
a) Actual production per day of 8 hrs = Calculated production per day of 8 hrs x Efficiency
= 51840 x 0.78 yds
= 40435.2 yds (Ans)
Here we calculated, Total length of warp = 40435.2 yds
And given, No of ends = 3280
b) Total length of yarn sized = Total length of warp x No of ends
= 40435.2 x 3280 yds
= 132627456 yds (Ans)
Here we calculated, Total length of warp in yds = 132627456 yds
Given, Yarn count = 40’s
Size% = 10%
…………………………………………………….Total length of warp in yds
c) Total wt. of sized warp in lbs = ——————————————– + Size%
…………………………………………………………..840 x Yarn count
….132627456
= ——————– + 10%
…..840 x 40
= 3947.25 + 10%
= 4341.97 lbs (Ans)
………………………………….Length of warp in yds
4. Wt. of unsized warp = ———————————————- x No of ends
……………………………………840 x Yarn count
5. The wt. of the size to be put on the warp = Wt. of unsized warp x % of size required to be put on warp
6. Wt. of sized warp = Wt. of unsized warp + Wt. of size on warp
…………………………………………………..Length of warp in yds
7. The count of the sized warp = ——————————————————- x No of ends
……………………………………………..840 x Wt. of sized warp in lbs
……………………………………………………………………………….100
Or, Count of sized warp = Count of unsized warp x —————————-
…………………………………………………………………………..100 + Size%
Example: A warp containing 2850 ends is required to be sized to 22%. The length of the sized warp on the beam is required to be 1180 yds. If the counts of the yarn 40’s. Find out-
- Wt. of unsized warp.
- The wt. of the size to be put on the warp of the given length.
- The wt. of sized warp.
- The count of the sized warp.
Solution:
Here given, The length of warp = 1180 yds
No of ends = 2850
Yarn count = 40’s
Size% = 22%
……………………………………….Length of warp in yds
a) Wt. of unsized warp = —————————————- x No of ends
…………………………………………..840 x Yarn count
……1180
= ——————— x 2850 lbs
….840 x 40
= 100.09 lbs (Ans)
b) The wt. of the size to be put on the warp = Wt. of unsized warp x % of size required to be put on warp
= 100.09 x 22%
= 22.02 lbs (Ans)
Here we calculated, Wt. of unsized warp = 100.09 lbs
And Wt. of size on warp = 22.02 lbs
c) Wt. of sized warp = Wt. of unsized warp + Wt. of size on warp
= 100.09 + 22.02 lbs
= 122.11 lbs (Ans)
…………………………………………………………..Length of warp in yds
d) The count of the sized warp = ————————————————- x No of ends
……………………………………………………..840 x Wt. of sized warp in lbs
…………1180
= ———————– x 2850
….840 x 122.11
= 32.78 ≈ 33’s (Ans)
……………………………………………………………………………………….100
Or, Count of sized warp = Count of unsized warp x ———————————–
………………………………………………………………………………….100 + Size%
……………..100
= 40 x —————————-
…………100 + 22
= 32.78 ≈ 33’s (Ans)
8. Wt. of size on the yarn = Wt. of sized warp – Wt. of unsized warp
……………………………………………………………………….Wt. of size
9. The percentage of size to be put on warp = ————————————– x 100%
…………………………………………………………………Wt. of unsized warp
Example: The wt. of sized yarn on a beam was found to be 83.6 lb. The beam contains 1080 yds of warp whose count before sizing was 40’s. If the no of ends in warp 2900. Calculate-
- The wt. of size on the yarn.
- The percentage of size put on the warp.
Solution:
Here, Length of warp = 1080 yds
Yarn count before sizing = 50’s
Wt. of sized warp = 83.6 lb
No of ends = 2900
………………………………..Length of warp in yds
Wt. of unsized warp = ———————————————- x No of ends
…………………………………..840 x Yarn count
…….1080
= —————- x 2900 lbs
….840 x 50
= 74.57 lbs
a) Wt. of size on the yarn = Wt. of sized warp – Wt. of unsized warp
= 83.6 – 74.57 lbs
= 9.03 lbs (Ans)
…………………………………………………………………………………Wt. of size
b) The percentage of size put on the warp = ————————————- x 100%
…………………………………………………………………………Wt of unsized warp
…..9.03
= —————- x 100
….74.57
= 12.1% (Ans)
10. Total length of sized warp = Wt. of sized warp x Count of sized warp
……………………………………..Total length of warp
11. Length of sized warp = ————————————–
…………………………………………..No of ends
Example: A warp containing 2400 ends and sized to 10%. If the sized warp wt 120 lbs of 40’s. Calculate the length of the sized warp and total length of sized yarn.
Solution:
Here, Count of sized yarn = 40’s
Size% = 10%
No of ends = 2400
Wt. of sized warp = 120 lbs
Total length of sized warp = Wt. of sized warp x Count of sized warp
= 120 x 40 hanks
= 4800 hanks (Ans)
……………………………………..Total length of warp
Length of sized warp = ————————————————–
…………………………………………..No of ends
….4800
= ————- = 2 hanks (Ans)
….2400
…………………………………..(Warp length on a back beam – Waste) x (elongation% + 100)/100
12. No of weavers beam/set = ————————————————————————————
………………………………………………….Length of warp in yds on a weavers beam
Example: A set of 8 beams each containing 32000 yds of warp is to be used for producing of weavers beam on a high-speed slasher. If the % of elongation is 0.5% and waste of warp is 60 yds. Calculate the no of beams. The length of sized warp on a weaver is 1200 yds.
Solution:
Here given, Length of warp on a back beam = 32000 yds
Waste of warp = 60 yds
Length of warp in yds on a weavers beam = 1200 yds
% of elongation = 0.5%
……………………………..(Warp length on a back beam – Waste) x (% of elongation + 100)/100
No of weavers beam per set = —————————————————————————————–
…………………………………………………….Length of warp in yds on a weavers beam
….(32000-60) x (0.5 + 100)/100
= —————————————————-
………………….1200
= 26.75 ≈ 27 (Ans)
………………………..(Length of warp on a back beam – Waste) x (% of elongation + 100)/100
13. Time required = —————————————————————————————————–
………………………………………………Actual production in yds per hr of slasher
Example: The actual production of a high-speed slasher is 1500 yds/hr. If % of elongation is 0.5%, waste of warp = 45 yds. Time that would be required to size set of 6 back beams containing 36226 yds of warp each.
Solution:
Here given, Length of warp on a back beam = 36226 yds
Waste of warp = 45 yds
Actual production in yds per hr of slasher = 1500 yds
% of elongation = 0.5%
……………………………………(Warp length on a back beam – Waste) x (elongation% + 100)/100
No of weavers beam per set = ——————————————————————————————
………………………………………………………Actual production in yds per hr of slasher
….(36226 – 45) x (0.5 + 100)/100
= —————————————————-
…………………..4500
= 8.08 hrs. (Ans)
………………………..Actual Production
14. Efficiency = ——————————————– x 100%
…………………….Calculated Production
Example: The actual production of a high-speed slasher is 32400 yds per day of 8 hrs. If calculated speed is 90 yds per min. find efficiency.
Solution:
Here given, actual production = 32400 yds/8hr
Actual Production = 32400/8 = 4050 yds/hr
Calculated production = 90 yds/min = 90 x 60 = 5400 yds/hr
……………………….Actual Production
Efficiency = ———————————————— x 100%
……………………Calculated Production
….4050
= ————— x 100
….5400
= 75% (Ans)
Average length of warp in yds consumed per loom per hr,
……………………………………………..100 + warp regain% 100 + warp waste%
= Average production (actual) x ——————————- x ——————————–
………………………………………………………….100 100
…………………………………………………….Production (actual) of slasher in yds/hr
15. No of looms per slasher = ——————————————————————————–
……………………………………..Avg length of warp consumed per loom (actual) in yds + Waste
Example: The average production (actual) per loom of a weaving shed is 6.1 yds of cloth per hour. The up-take of warp in weaving was found to be 8%. If the production (actual) of the slasher, which is required to supply loom beam to this weaving shed is 29128 yds per day of 8 hrs, calculate the no of looms it can keep running. Allow 0.75% as waste of warp.
Solution:
Here, Average production (actual) = 6.1 yds/hr
Warp regain% = 8%
Warp waste% = 0.75%
Actual production of slasher per 8 hrs = 29128 yds
Average length of warp in yds consumed per loom per hr,
………………………………………………100 + warp regain% 100 + warp waste%
= Average production (actual) x ——————————– x ———————————
………………………………………………………….100 100
…………100 + 8 100+ 0.75
= 6.1 x ————– x ——————
…………..100 100
= 6.637 yds
Production (actual) of slasher per hour = 29128/8 = 3641 yds
……………………………………………………..Production (actual) of slasher in yds/hr
No of looms per slasher = ———————————————————————————————
……………………………………Avg length of warp consumed per loom (actual) in yds + Waste
…..3641
= ———– = 548 (Ans)
….6.637
Conclusion
I have tried to explain/show all the production calculation of sizing machine in weaving mill with so many examples through this article. Hope this article will be helpful for those who are new to the weaving mill or those who are students.
Author of this Article:
Md. Imran Hossain
B.Sc. in Textile Engineering
Shahid Abdur Rab Serniabat Textile Engineering College, Barisal.
Email: [email protected]
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