a) The length of the warp
b) The total length of yarn in the warp
c) Amount the warper is to be paid, if the rate of warping is 7 paisa per 1000 yds of warp.
Solution
Weight of warp on the beam = 369-69 = 300 lbs
a) The length of the warp
=(Weight of warp×count×20) ÷ (No. of ends)
=(300×20×840) ÷ 420
=12000 yds.
b) The total length of yarn in the warp
= Length of warp in hanks×count×No. of ends) ÷ No. of ends
= (300×20×420) ÷ 420
= 6000 hanks.
c) Amount to be paid = 12000 × 0.07 = 840 taka.
A super speed beam warper with a warpingspeed of 800 yds/min is preparing a standard warp of 525 ends. If the count of the yarn is 30s & overall efficiency is 84%. Calculate the following.
I. Total length of warp produced per day of 8 hours.
II. Number of beams produced per day of 8 hours.
III. Total weight of yarn in lbs warped per day of 8 hours.
IV. Weight of yarn on a beam.
The length of warp on each beam is required to be 44352 yds. Ignore waste.
Solution
I. Actual length of warp in yds/hr= Calculated production × 60 × Efficiency
= 880 × 60 × (84÷100)
= 44352 yds
Therefore, actual production pre day of 8 hrs= 8 × 44352 = 354816 yds.
II. No. of beams produced per day of 8 hrs in yds
= (Total length of warp produced per day of 8 hours in yds) ÷ ( Length of warp on a beam in yds)
= (354816÷44352)
= 8 beams
III. Total weight of yarn in lbs warped pre day of 8 hrs
=(354816×525) ÷ (840×20)
= 7392 lbs.
IV. Weight of yarn on a beam in lbs
=(Total weight of yarns in lbs warped in 8hrs) ÷ ( No of beam produced)
=(7392÷8)
= 924 lbs
A modern high speed beam warping machine produces 8 beams each containing 222720 yards of warp per day of 8 hrs. If the calculated warping speed of the warper is 580 yds/min. calculate its efficiency.
Solution
Actual production in yds/hrs
=(222720÷8)
= 27840 yds
Calculated production in yds/hrs
Therefore, Efficiency
= (Actual production) ÷ (Calculated production) × 100
=(27840÷34800) × 100
=80%.
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