Subscribe For Free Updates!

We'll not spam mate! We promise.

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).

If You Like Then Please Share It
SOCIALIZE IT →
FOLLOW US →
SHARE IT →

0 comments:

Post a Comment