Yarn Sizing 6.5 Numerical Problems 1) 100 kg bone dry warp yarns were sized to the add on of 8% and dried to an overall (yarn and dry size) moisture content of 10%. Calculate the final weight of the sized yarns. Solution: Bone dry warp = 100 kg Pick up = 8% So,
Mass of dry size
100 = 8 Mass of bone dry warp
or,
Mass of dry size
100 = 8 100 kg
or, Mass of dry size = 8 kg
Total mass of bone dry (size +yarn) = 108 kg Moisture content = 10% So, 10 =
W 100 , where W is mass of water W 108
or, W = 12 kg So, final weight of sized yarn is = (108+12) kg = 120 kg
2) A 20 kg cotton warp, having moisture regain of 8.5%, is sized with a
paste of 15% concentration. If a 10% add on the bone dry weight of the yarn is aimed then what should be the wet pick up? How much water has to be evaporated so as to leave 8% moisture content in warp and size film? Solution: Moisture regain (MR) of cotton is 8.5 % or,
W 100 8.5 , where W= mass of water and D= oven dry mass of yarn D
or, W = 0.085 D . Total mass of yarn = oven dry mass of yarn + mass of water or, D + 0.085 D = 20 kg or, D = 18.43 kg Dry size added = D
add on % 100
18.43
10 1.843 kg 100
Now, mass of water (W) = (20 -18.43) kg = 1.57 kg
Wet pick up (WPU) =
Add on % = 10/15 = 0.667 Concentration %
Total mass of size paste picked up by yarn = 0.667 18.43 kg = 12.29 kg
Mass of water within size paste picked up by yarn =
100-Concentration% 12.29
100
100-15 12.29 = 10.45 kg 100 Total mass of water in the yarn after pick-up = Mass of water in the size paste + Mass of water originally present in yarn = (10.45 + 1.57) kg = 12.02 kg Water to be retained (for 8% moisture content in yarn and size film) =
8 1.843 8 18.43 92 92
= 1.76kg Therefore, the mass of water to be evaporated = (12.02 1.76) kg = 10.26 kg.
3) A 40 tex cotton yarn has add on of 8%. If the moisture regain of the yarn is 10% then determine the oven-dry mass of the size added per kg of the unsized yarn. Solution: Size add on = 8%, Moisture regain = 10% So,
W 100 10 , where W= mass of water and D= oven dry mass of yarn D
or, W =0.1D Now, total mass of unsized yarn = 1 kg So, D+W
= 1 kg
or, D+0.1D = 1 kg or, D = 0.909 kg Now, add on is 8%. So,
Mass of dry size
100 8 Mass of oven dry yarn (D)
or, Dry size = D x 0.08
= 0.073kg = 73 gm. 4) The stretch in a sizing machine is 3%, 5% and 2% in the creel zone, sizing zone and drying zone respectively. If the warp crimp in the woven fabric is 10% then determine the length of fabric that could be produced from 1000 m length warp sheet in each of the warpers beam. Solution: Total stretch in sizing can be calculated using the following multiplicative expression.
Total stretch = 1
3 5 2 1 1
100 100 100
= 1.103 So, length of sized yarn = (Ly) =1.1031000 = 1103 m Crimp % =10% So, 10
or,
Lf
Length of yarn(Ly )-Length of fabric(L f )
Length of fabric(L f )
Ly
1103 m =1002.7 m 1.1 1.1
Length of fabric is 1002.7 m
100
5) A sizing machine is running at 150 m/min with 6000 ends. The add on requirement is 12% and concentration of the size paste is 18%. If yarn count is 20 tex (without moisture) and residual moisture content in the sized yarn and film after drying is 10%, then calculate the number of drying cylinder required, if one drying cylinder can evaporate 4 kg water per min. Solution: Oven dry mass of yarn=
WPU
150 6000 20 kg = 18 kg 1000 1000
Add on % 12 = 0.667 Concentration % 18
Total mass of size paste = WPU Oven dry mass of yarn
=
12 18 kg = 12 kg 18
Mass of dry size = Mass of size paste concentration
Total mass of yarn and dry size = (18 + 120.18+) kg =
20.16 kg
So, for moisture content of 10%
Mass of water to be retained in yarn and size film =
1 20.16 kg 9
= 2.24 kg
100-Concentration% = 9.84 kg 100
Mass of water in size paste of 12 kg= 12
So, water to be evaporated = (9.84 - 2.24) kg/min = 7.6 kg/min
Drying capacity of one cylinder = 4 kg/min. So, 2 cylinders will be needed.
6) Determine the failure rate of sized yarns if the six yarns tested for abrasion breaks broke after 1025, 1550, 2232, 3785, 5608 and 7918 cycles. What is the reliability of the yarns at 2000 cycles if the reliability curve is exponential? Failure rate = number of failure/cycle = =
Total number of failure
r = Total number of cycles t
6 1025 1550 2232 3785 5608 7918
2.712 10-4 per cycle
Average life ( ) =
Reliability
=e =e
= 3686
(according to exponential failure where pdf of failure is =
2000 3686
= 0.581
7) Draw the life history curve for the sized yarns from the following information. Number of cycles 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90
Number of failures 35 7 6 5 5 6 8 9 19
Number of survivors 65 58 52 47 42 36 28 19 0
Life history curve is obtained by plotting failure rate (Y axis) against number of cycles (X axis). From the table it is observed that 35 failures have occurred between the 1st and 10th cycle. As the exact cycle number is not known for these 35 failures, it is assumed that all the failures have taken place at the mid-point of duration i.e. 5th cycle. Failure rate (R) =
Number of failures Mid point of observation number of failures +End point of observation number of survivers 35 0.042 35 5 65 10 6 R3(25.5) = 0.0035 6 25 52 30 5 R5(45.5) = 0.0022 5 45 42 50 8 R7(65.5) = 0.0032 8 65 28 70
