Mathematical Problems and Solutions in Warping

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
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

For more clear concept you can read this article:
Formula and Examples of Warping Calculation

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).