Assignment : 1
1. The count of the carded sliver is 0.12s Ne.
PROCESS-A (Lap doubling): 24 carded slivers are combined and a draft of 2.2 is applied
to get the sliver lap. 6 such laps are combined in the ribbon lap machine and draft of 6.8
is applied to obtain a final lap for comber.
PROCESS-B (Sliver doubling): 8 carded slivers are combined in a draw frame process and
a draft of 8.1 is applied. 36 such slivers are combined in a super lap former and a draft of
3.56 is applied to obtain a final lap for comber.
Calculate the total doubling, total draft and the weight per unit length (g/m) of the lap
produced in each process.
2. In the above case, the carded slivers have a mass variation of 12% CV. Additional CV%
introduced due to drafting in different machines are:
Sliver lap machine
Ribbon lap machine
:
:
1.5%
4.0%
Draw frame
Super lap former
:
:
7.5%
2.8%
Calculate the CV% of laps produced in Processes A and B.
3. The linear density of the feed lap is 60 g/m. The net forward movement of the detaching
roller per cycle is 30 mm. The feed length per cycle is 6 mm. Slivers from 4 combing
heads are combined and a draft of 10 is given. The resultant combed sliver count (N e) is
0.16. What is the noil extraction percentage?
Assignment 1
Solution
1. Process A
Total number of doublings
Total draft
Tex of the card sliver
Tex of the ribbon lap
So, g/m of the ribbon lap is
Process B
Total number of doublings
Total draft
Tex of the sliver
Tex of the comber lap
So, g/m of the ribbon lap is
= 24 x 6
= 2.2 x 6.8
= 144
= 14.96
= 590.5/0.12 = 4920.8
= 4920.8 x (24/2.2) x (6/6.8)
= 4736.6
= 4736.6/1000
= 47.4
= 8 x 36
= 8.1 x 3.56
= 288
= 28.84
= 4920.8 x (8/8.1)
= 4920.8 x (8/8.1)x (36/3.56)
= 4914.6/1000
= 49.1
= 4914.6
2. Process A
CV% of the comber lap
Process A
CV% of the comber lap
= 12.74%
= 14.43%
3. Count of the combed sliver
= 0.16s Ne
Therefore, the count of the fibre strand emerging out of single combing head
= 0.16 x(4/10)
= 0.064 s Ne
Tex of the fibre strand emerging out of single combing head
= 590.5/0.064
= 9226.6 Tex
= 9226.6/1000
= 9.23 g/m
Weight of fibres fed per cycle per head
Weight of fibres delivered per cycle per head
Therefore, Noil percentage
= 22.2 %
= 60 x (1/1000) x 6
= (30/1000) x 9.23
= 0.36 g
= 0.28 g