**Warp Sizing Calculation in Weaving**

**Abdullah Nur Uddin Rony**

Textile Engineering College, Noakhali

Facebook: Abdullah Rony

**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)**