# Mathematical Problems of Sizing in Weaving

What is Sizing?
Sizing is the process of applying protective adhesive coating on the yarn surface. This is the most important segment of the weaving preparatory process. Better the quality of sizing higher the weaving efficiency & vice versa. It helps to increase the tensile or breaking strength of cellulose yarn. Sizing of yarn, increases the elasticity of yarn, weight of yarn, frictional resistance, etc. Mathematical problems of sizing in weaving will be discussed below:

Formulas of Sizing in Weaving:

A. Total length of yarn sized = Total length of warp X No of ends

B. Total wt of size on warp = Wt of sized warp – Wt of unsized warp

…………………………………………….Length of warp in yds
C. Count of sized yarn = ………………………………………………  X No of ends
……………………………………..840 X Wt of sized warp in-lbs

…………………………………………Wt of size
D. % of size on warp = ……………………………….X 100%
…………………………………..Wt of unsized warp

### Mathematical Problems of Sizing in Weaving

A. The calculated production of a high-speed slasher is 100 yds per min. If the efficiency of the machines is 75%. Calculate the following.

1. The actual production per day of 8 hrs
2. The total length of yarn if the total ends are 3250
3. The total wt of sized warp if it is sized to 10% and the count of unsized are 40’s

1. Calculated production per day of 8 hrs = 100 X 60 X 8 yds

= 48000 yds

…………………………………………………………………………..75
The actual production per day of 8 hrs = 48000 X ……….. yds
………………………………………………………………………….100

= 36000 yds (ANS)

2. The total length of yarn sized = Total length of warp X No of ends

= 36000 X 3250 yds

= 117000000 yds (ANS)

………………………………………………………Total length of warp in yds
3. The total wt of sized warp in lbs = ………………………………………….. + 10%
………………………………………………………………….840 X Count

….117000000
= ………………… + 10%
……840 X 40

= 3482 + 10%

= 3830 lbs (ANS)

B. The actual production of a high-speed slasher sizing machine is 32,400 yds per day of 8 hrs. If the Calculated speed is 90 yds per min, find out the efficiency.

Solution:

We know,
……………………………………………………32400
Actual production in yds per hour = ………….yds
………………………………………………………..8

= 4050 yds

Calculated production in yds per hr = 90 X 60 yds

= 5400 yds

……………………….Actual production
Efficiency = …………………………………… X 100%
…………………..Calculated production

…..4050
= ………… X 100%
…..5400

= 75% (ANS)

C. A Beam of 250 kg contains sized yarn of 15% takes up if the sized count 40.87 ‘s. Calculate unsized count.

Solution:

We know,

……………………………………………………………………………..100
Count of sized yarn = Count of unsized yarn X ……………………………
………………………………………………………………………100 + % of size

………………………………………………………..100
Or, 40.87 = Count of unsized yarn X …………………
……………………………………………………….100 + 15

Or, Count of unsized yarn = 47 ‘s (ANS)

```Author of this Article:
S. M. Hossen Uzzal
Production Officer
Monno Fabrics Ltd., Manikgonj
Email: [email protected]```

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