Carding Section:
The first condition of spinning is to open the fiber at single-stage and which is possible only by carding. The quality of yarn especially depends on the carding action, so carding is called the heart of spinning. Here calculation plays a vital role. So we need to know about the cotton carding section problems and solutions.
Cotton Carding Section Problems and Solutions
Problems and Solutions of Cotton Carding:
1. Draft constant of a carding m/c is 2194 and if the DCP is 22T, then what will be the draft of the m/c?
Solution:
Here,
Draft constant of a carding m/c = 2194
DCP of carding = 22T
Draft of carding m/c =?
We know that,
…………..Draft constant
Draft = ……………………… = 2194/22 = 99.72 (ANS)
………………….DCP
2. If the draft of a carding m/c is 110 and DCP is 20T, then what will be the draft constant?
Solution:
Here,
Carding m/c draft = 110
Carding m/c DCP = 20T
Draft constant of carding m/c =?
We know that,
Draft constant = Draft X DCP
= 110 X 20
= 2200 (ANS).
3. If a carding m/c produces 64 grain/gauge sliver by 19T DCP, then what is the DCP needed to produce 70 grain/gauge wt. sliver?
Solution:
Here given,
Present DCP = 19T
Wt. of present sliver = 64 grain/gauge
Wt. of new sliver = 70 grain/gauge
New DCP =?
We know that,
……………………Present DCP X Wt. of new sliver
New DCP = ……………………………………………………
………………………….Wt. of present sliver
= (19 X 70)/ 64
= 20.78
= 21T (ANS)
4. Calculate the new sliver hank from the following data –
Present hank = 0.16
Present DCP = 18
New DCP = 24
Solution:
Here given,
Present hank = 0.16
Present DCP = 18
New DCP = 24
New sliver hank =?
We know that,
……………………………………………………………Present DCP
New sliver hank = Present sliver hank X ………………………
………………………………………………………………New DCP
= 0.16 X 18/24
= 0.12 (ANS).
5. A carding m/c produces 0.17 hank slivers from 13 oz (ounce)/gauge wt. lap by wasting 6%. What will be the mechanical draft of the machine?
Solution:
Here given,
Waste = 6%
Feeded lap wt. = 13 oz /gauge
Carded sliver hank = 0.17
Mechanical draft =?
………………….Lap length (gauge)………1 lb
Lap hank = ………………………….. X ………………….
………………………….840………………..Lap wt. (lb)
……..1…………….1
= ……….. X ……………
…….840………13/16
……….16
= ………………
….840 X 13
= 0.00146
Again,
Actual draft of carding m/c = Carded sliver hank / Lap hank
= 0.17/0.00146
= 116.44
Mechanical draft of carding m/c = Actual draft X (100 –Waste) / 100
= 116.44 X (100 – 6) / 100
= 109.45 (ANS).
6. If a carding m/c produces 70 grain/gauge sliver from 14 oz/gauge wt. lap. Then what will be the actual draft of the carding m/c?
Solution:
Here given,
Feed lap wt. = 14 oz/gauge
Wt. of delivered sliver = 70 grain/gauge
Actual draft =?
……………………………Lap wt. in oz
Lap wt. in grain = ……………………. X 7000
…………………………………16
= 14/16 X 7000 (1 lb = 7000 grain)
= 6125
Actual draft = Feeded lap wt. / Delivered sliver wt.
= 6125/70
= 87.5 (ANS)
7. In a carding m/c, rotation of doffer 15 rpm, doffer circumference is 85 inch, Wt. of produced sliver 65 grain/gauge, and efficiency is 85%. What is the production per day of the carding m/c (in metric ton)?
Solution:
Here given,
Rotation of doffer = 15 rpm
Doffer circumference = 85 inch
Wt. of produced sliver = 65 grain/gauge
Eff. Of carding m/c = 85%
Production per day =?
We know,
Production = Doffer circumference X Speed X Eff.%
= 85 X 15 X 85/100 inch/ min
= 85 X 15 X (60 X 24)/36 X 65 X 85/100 grain/ day
(As, 7000 grain = 1 lb and 2204 lb = 1 metric ton)
= 85 X 15 X (60 X 24)/36 X 65 X 1/ (7000 X 2204) X 85/100 Metric ton
= 0.183 Metric ton/day (ANS).
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8. In a carding m/c rotation speed of calendar roller 80/min, Dia of calender roller is 4”, Produced sliver hank is 0.14, Eff. Of carding m/c is 85%, If the tension draft of the m/c is 1.05 then calculate the production per shift (8 hr) of the carding m/c?
Solution:
From the given data,
Rotation speed of calendar roller = 80/min
Dia of calendar roller = 4 inch
Produced sliver hank = 0.14
Eff. Of carding m/c = 85%
Tension draft of carding m/c = 1.05
Production/hr =?
We know that,
Production / shift
= Circumference of calendar roller X (60 X 8) / (36 X 840 X Hank) X Tension draft X Eff.%
= 3.14 X 4 X 80 X (60 X 8)/(36 X 840 X 0.14) X 1.05 X 85/100
= 101.68 lb (ANS).
Author of this Article: Md. Abu Sayed Founder of Textile Apex Pabna Textile engineering college, Pabna, Bangladesh Email: [email protected] Cell : +8801745214773