Sizing:
The process of applying a proactive adhesive coating upon the surface of the yarn is called sizing. This is the most important operation to attain maximum weaving efficiency, especially for blended & filament yarns. Duo to sizing, increases the elasticity of yarn, yarn strength, the weight of the yarn, smoothness, frictional resistance. In this article, we will know about warp sizing calculation in weaving.
Formula for Warp Sizing Calculation
1. Total wt. of size on warp = wt. of sized warp – wt. of unsized warp.
2.The wt. of size to be put on warp = wt. of unsized warp x % of size reqd to be put on the warp.
………………………………………….length of warp in yds
3. The wt. of unsized warp = ……………………………….. x no. of ends + wt. of size.
…………………………………………………840 x count
…………………………………………Wt. of size
4. Wt. of sized warp = ……………………………… x 100%
………………………………..Wt. of unsized warp
…………………………………………Wt. of size
5. % of size on warp = ……………………………….. x 100%
………………………………….Wt. of unsized warp
…………………………………………….length of warp in yds
6. Count of sized yarn = ………………………………………………… x no. of ends
……………………………………..840 x wt. of sized warp (in lbs)
………………………………………………………………………………100
7. Count of sized yarn = count of unsized yarn x …………………………
………………………………………………………………………100 + % of size
Warp Sizing Mathematical Problem & Solution
Problem-1:
Calculate the production of a slasher sizing m/c from the following particulars:
Circumference of drawing roller = 29.25” PPM of drawing roller = 36
Efficiency = 80%
Production/8hrs = ?
Solution:
Production,
….Circumference of drawing roller x rpm of drawing roller x 60 min x hr x efficiency
= ……………………………………………..…………………………………………..………………………….. yds
………………………………………………………36 x 100
….29.25 x 36 x 60 x 8 x 80
= …………………………..……… yds
………………36 x 100
Production/8hrs = 11232 yds (ANS)
Problem-2:
Calculate the production in-lb of a slasher sizing m/c from the following particulars;
Circumference of drawing roller = 29.25 inches.
PPM of drawing roller = 36
No. of warp ends = 2100
Yarn count = 32
Efficiency = 80%
Production/ 8 hrs = ?
Solution:
Production,
…π x Dia of drawing roller x rpm of drawing roller x 60 min x hr x eff. x no. of warp ends
= ……………………………………………………………………………………………………..…………………….. lb
……………………………………………36 x 840 x yarn count x100
= 877.5 lb (ANS)
…..877.5
= ………… kg
…..2.204
= 398.14 kg (ANS)
Problem-3:
A warp containing 2800 ends is required to be sized to 25%. The length of the sized warp on the beam is required to be 1080 yds. If the counts of the yarn 40s. Find-
i) The wt. of the size to be put on the warp of the given length.
ii) The wt. of sized warp
iii) The count (Ne) of sized warp.
Solution:
i) The wt. of the size to be put on the warp,
…..of unsized warp
= ………………………. x % of size
……..840 x count
…..2800 x 1080
= ………………….. x 25%
……..840 x 40
= 22.5 lbs (ANS)
ii) wt. of sized warp = wt. of unsized warp + wt. of size on it
= 90 + 22.5
= 112.5 lbs (Ans.)
iii) The count of the sized warp,
………..length of warp in yds
= …………………………………………. x no. of ends
…..840 x wt. of size warp in-lbs
…2800 x 1080
= …………………
….840 x 112.5
= 32 s (ANS)
OR, Count of the sized warp,
……………………………………….100
= Count of unsized x …………………………
………………………………….100 + %size
……………100
= 40 x ……………
………..100 + 25
= 32 s (ANS)
Problem-4:
The calculated production of a high-speed slasher is 100 yds per min. If the efficiency of the m/c is 75%, calculate the followings-
a. The actual production per day of 8 hrs.
b. Total length of yarn if the total ends are 3520.
c. The total wt. of sized warp, if it is sized to 10% & the count of unsized is the 40s.
Solution:
a. Calculated production per day of 8 hrs = 100 x 60 x 8 yds.
= 48000 yds.
………………………………………………………………….75
The actual prodn per day of 8hrs = 48000 x ……….. yds
…………………………………………………………………100
= 36000 yds (ANS)
b. The total length of yarn sized = Total length of warp x no. of ends.
= 36000 x 3250 yds.
= 117000000 yds. (Ans)
……………………………………….Total length of warps
c. Total wt. of sized warp = ………………………………… + 10%
………………………………………………840 x count
…..117000000
= …………………. + 10%
….840 x count
= 3482 + 10%
= 3830 lbs (ANS)
Problem-5:
A warp containing 2800 ends, is required to be sized to 25%. The length of the sized warp on the beam is required to be 1080 yds. If the count of the yarn is the 40s. Find out –
a. The wt. of the size to be put on the warp of the given length.
b. The wt. of sized warp.
c. The count of the sized warp.
Solution:
a. The wt. of the size to be put on the warp,= wt. of unsized warp x % of size required to be put.
…1080 x 2800
= ………………… x 25%
…….840 x 40
= 22.5 lbs (ANS)
b. Wt. of sized warp = wt. of unsized warp + wt. of size
…………………………………….length of warp in yds
Wt. of unsized warp = ……………………………………… x no. of ends.
……………………………………………….840 x 40
….1080 x 2800
= ………………….. lbs
……..840 x 40
= 90 lbs
Wt. of sized warp = 90 + 22.5
= 112.5 lbs. (ANS)
……………………………………………………………………100
c. Count of sized warp = count of unsized x ………………….
………………………………………………………………100 + size%
………………100
= 40 x ……………….
………….100 + 25
=32 s (ANS)
Problem-6:
A warp containing 2400 ends of 44s sized to 10%. If the sized warp wt. lbs. Calculate the length of the sized warp & total length of sized yarn.
Solution:
………………………………………………………………………100
Count of the sized warp = count of unsized x …………………
………………………………………………………………..100 + %size
…………….100
= 44 x ……………
…………100 + 10
= 40 s
Total length of sized warp = 120 x count x 120 x 40 x 4800 hanks. (Ans)
…………………………………………Total length of warp
Total length of yarn sized = …………………………….
……………………………………………….No. of ends
…….4800
= …………..
…….2400
= 2 hanks (ANS)
Problem-7:
The wt. of sized yarn on a beam was found to be 82.5 lbs. The beam contains 1050 yds of warp whose count before sizing was 50s. If the no. of ends in warp is 3000. Calculate –
a. The wt. of size on the yarn.
b. The % of size put on the yarn
c. Count of the sized yarn.
Solution:
a.The wt. of size on the yarn = wt. of sized warp – wt. of unsized warp
…………………………………….length of warp in yds
Wt. of unsized warp = …………………………………….. x no. of ends
……………………………………………840 x count
…1050 x 3000
= ………………….
…….840 x 50
= 75 lbs
Wt. of size = 82.5 – 75 = 7.5 lbs (Ans)
……………………………………………………………..of size x 100%
b) Percentage of size put on the yarn = ……………………………….
………………………………………………………….wt. of unsized warp
……7.5
= …….. X 100
……75
= 10% (ANS)
……………………………………………………………………100
c) Count of sized warp = count of unsized x ………………..
……………………………………………………………..100 + %size
…………….100
= 50 x ………………
………….100 + 10
= 45.45 s (ANS)
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Author of this Article: Abdullah Nur Uddin Rony Textile Engineering College, Noakhali Facebook: Abdullah Rony
Helpful post. Thanks for sharing.