# Warping Mathematical Problems and Solutions

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.

#### 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

You may like: Formula and Examples of Warping Calculation

### Mathematical Problems and Solutions in Warping

1. In a super-speed warper machine, 200kg yarn is wrapped. If warp length is 20,000 m and the 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 machine 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 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).

```Author of this Article:
Md. Abu Sayed
Founder of Textile Apex
Pabna Textile engineering college, Pabna, Bangladesh
Email: [email protected]
Cell : +8801745214773```

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