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).