Warp Sizing Calculation in Weaving
Sizing: The process of applying a proactive adhesive coating upon the yarns surface is called sizing. This is the most important operation to attain maximum weaving efficiency especially for blended & filament yarns. Duo to sizing, increases elasticity of yarn, yarn strength, weight of the yarn, smoothness, frictional resistance.
Formula for Mathematical Problem
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 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
Problem-1:
Calculat 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 = 29.25inch.
PPM of drawing roller = 36
No. of warp ends = 2100
Yarn count = 32
Efficiency = 80%
Production/8hrs = ?
Solution:
Production,
Ï€ 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,
wt. 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 is 3520.
c. The total wt. of sized warp, if it is sized to 10% & the count of unsized are 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)
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 counts of the yarn is 40s. Find out –
a. The wt. of the size to be put on the warp of he 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)
wt. 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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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 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
Problem-1:
Calculat 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 = 29.25inch.
PPM of drawing roller = 36
No. of warp ends = 2100
Yarn count = 32
Efficiency = 80%
Production/8hrs = ?
Solution:
Production,
Ï€ 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,
wt. 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 is 3520.
c. The total wt. of sized warp, if it is sized to 10% & the count of unsized are 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)
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 counts of the yarn is 40s. Find out –
a. The wt. of the size to be put on the warp of he 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)
wt. 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)
