Mathematical Problems and Solutions of Sizing

Sizing:
Sizing is an important term in weaving technology. Sizing is done during beam preparation after winding and warping. Various types of sizing materials are applied on the yarn during sizing. Sizing is generally done on the warp yarns to protect the yarns. It is called the heart of weaving. In this article, we will solve the mathematical problems and solutions of sizing

Mathematical Problems and Solutions of Sizing (1)
Fig: Sizing Machine

Important formulas for sizing

………………………………………………..{Warp length (without sizing) X No. of warp}
Yarn count (Without sizing) = …………………………………………………………………………….
…………………………………………..{840 X Warp weight (without sizing)} (English count)

Yarn count (Sized)
= {Sized warp length (yds) X No. of warp} / {840 X Sized warp yarn weight (lbs)}

Required time to size,

……{Yarn length – Wastage} X {(100 + Increase %) / 100}
= ……………………………………………………………………………………
……………….Production of sizing machine per hour

Required no. of loom for a slasher sizing machine,

………………………………………….Production of Slasher m/c per hr (yds)
= ………………………………………………………………………………………………………………………………….
..{(100 + Wastage%) / 100} X {Production of a loom per hr (yds)} X {(100 + regain%) / 100}

Mathematical Problems and Solutions of Sizing

1. 25% starch is applied on warp having 2800 no. of yarns. Warp yarn length is 1080 yds after applied starch. If yarn count is in the 40s, then calculate:

a. Amount of starch in the warp
b. Warp yarn weight after sizing or applying starch
c. Warp yarn count after applying starch

Solution:

Given,

  • No. of warp yarn = 2800
  • Applied starch = 25%
  • Warp yarn length = 1080 yds
  • Yarn count before applying starch = 40s

We know that,

Warp weight before applying starch
= (warp length X No of warp) / (840 X Count)
= (1080 X 2800) / (840 X 40)
= 90 lbs

Starch weight = Warp weight without starch X Starch%
= 90 X (25/100)
= 22.5 lbs

Warp yarn weight after applying starch = Warp yarn weight without starch + Starch weight
= 90 + 22.5
= 112.5 lbs

………………………………………………………………….(Warp length X No. of warp)
Warp yarn count after applying starch = …………………………………………………………
………………………………………………………..(840 X Warp weight after applying starch)
= (1080 X 2800) / (840 X 112.5)
= 32s (ANS).

2. Wastage 50 yds, increase 1%, Actual production of sizing machine 1400 yds/hr. If yarn length is 14050 yds then calculate the required time to size one set beam.

Solution:

Here,

  • Wastage = 50 yds
  • Increased yarn length = 1%
  • Production of machine = 1400 yds/hr
  • Yarn length = 14050 yds

Required time to size one set beam =?

We get,

Actual yarn length of the beam,
= {Yarn length – Wastage} X {(100 + Increase %) / 100}
= {14050 – 50} X {100 + 1)/100}
= 14140 yds

Required time to size one set beam
= Actual length of yarn in beam / (Production of sizing machine/hr)
= 14140/1400
= 10.1 hr (ANS).

3. Production of a slasher sizing machine 1334 yds/hr, production of a loom 4 yds/hr, regain 6%, Wastage 5%. Calculate the required no. of the loom to keep going on the slasher sizing machine.

Solution:

Here given,

  • Production of a slasher sizing machine = 1334 yds/hr
  • Production of a loom = 4 yds/hr
  • Regain = 6%
  • Wastage = 0.5%

Required no. of the loom to keep going on slasher sizing machine =?

We get,

Yarn length of loom of a slasher sizing machine,
= {(100 + Wastage%) / 100} X {Production of a loom per hr (yds)} X {(100 + regain%) / 100}
= {(100 + .5)/100} X 4 X {(100 + 6)/100}
= 4.26 yds

Required no. of the loom to keep going on slasher sizing machine,
= Production of sizing machine per hr / Amount of yarn of loom per hr
= 1334/4.26
= 313

So, the Required no. of the loom is 313 (ANS).

Author of this Article:
Md. Abu Sayed 
Founder of Textile Apex 
Pabna Textile engineering college, Pabna, Bangladesh 
Email: sayed1952@facebook.com 
Cell : +8801745214773

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