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## Warp Sizing Calculation in Weaving

Warp Sizing Calculation in Weaving
Abdullah Nur Uddin Rony
Textile Engineering College, Noakhali

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,
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,

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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