**Mathematical Problems and Solutions in Warping**

**Md. Abu Sayed**

Founder of Textile Apex

Pabna Textile engineering college, Pabna, Bangladesh

**Email:**sayed1952@facebook.com

**Cell :**+8801745214773

**Warping:**

The warping is done to arrange packages of warp yarns of related length so that they can be collected on a single warpers beam, as a continuous sheet of yarns can be used for sizing or next processes.

Warping |

**Formulas for Warping:**

Warp yarn weight (kg) X 1000 X 1000

**Beam count of yarn (Tex System)**=

**……………………………………………**

No. of warp yarn X Warp yarn length (m)

Warp yarn length (yd) X No. of warp yarn

**Beam count of yarn**

**(English System)**=

**……………………………………………**

840 X Warp yarn weight (lb)

**Warp yarn length (yd)**= Warp length per beam X No. of beam per set X No. of set

**Production per shift of warping machine,**

60 X 8

= Surface speed of warping machine (yd/min) X

**…………………**X Eff% X Waste

840 X

**Yarn count**

**Or,**Surface speeds of warping machine (yd/min) X 60 X 8 X Eff% X Waste

**Beam yarn weight**= Beam weight with yarn – Empty beam weight

**Mathematical problems**

**1. In a super speed warper machine, 200kg yarn is wrapped. If warp length is 20,000 m and Total no. of warp yarn is 400 then determine the beam count of warp yarn in Tex.**

**Solution:**

Given that,

- Amount of yarn in beam = 200 kg = 2,00,000 gm
- Warp length = 20,000 m
- No. of warp yarn = 400
- Beam count of warp yarn =?

**We know that,**

Warp yarn weight (kg) X 1000 X 1000

**Beam count of yarn (Tex System)**=

**……………………………………………**

No. of warp yarn X Warp yarn length (m)

= (200 X 1000 X 1000) / (400 X 20,000)

= 25 Tex

**(ANS).**

**2. In a modern beam warping machine, warping speed per minute is 610 yd and Efficiency is 75%. Determine the length of yarn per 8 hr with the machine.**

**Solution:**

Here given,

- Speed of warping machine = 610 yd/min
- Warping machine efficiency = 75%
- Warping machines yarn length per 8 hr =?

**We know,**

**Warping machines production per 8 hr,**

= Surface speed of warping machine (yd/min) X 60 X 8 X Eff%

= 610 X 60 X 8 X 75/100

= 2,19,600 yd

**(ANS).**

**3. Calculate the time required to prepare of 8 warpers beam on 2 improves high speed warpers with warping speed of 560 yds (calculated) per min. The length of warp on each beam is required to be 36,000 yds. Efficiency 80%. Also find the efficiency.**

**Solution:**

Given that,

- Total length of warp in yds = 36,000 yds
- No. of beam = 8
- Calculated production in yds per m/c’s per hour = 36,000yds
- Efficiency = 80% = 80/100
- Required time =?

**We know,**

Total length of warp in yds X No. of beams

**Time required**=

**…………………………………………………….....…**………….(i)

Actual production in yds per m/c’s per hr X No. of m/cs

**Again we know,**

**Actual production per hour per machine in yds**= Calculated production in yds per m/cs per hr X Efficiency X 60

= 36,000 X 80/100 X 60 yds

= 26,880 yds

**(ANS).**

**Now from equation (i) we get,**

**Time required**= (36,000 X 8) / (26,880 X 2)

= 5.36 hr

**(ANS).**

**Again we know,**

**Efficiency**= (Actual production)/(Calculated production) X 100%

= 26,880/36,000 X 100%

= 74.67%

**(ANS).**

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