Jute Spinning Calculation

Mathematical Problem (Emulsion)

𝑜𝑖𝑙 𝑖𝑛 𝑒𝑚𝑢𝑙𝑠𝑖𝑜𝑛×𝑎𝑝𝑝𝑙𝑦
1) Oil in jute%=
100
Dollop.wt
2) Feed sliver wt/100 Yds= 100
Clock.lenght
Dollop.wt 100
3) Delivery sliver wt/100 Yds = 
Clock.lenght Draft

𝐷𝑜𝑙𝑙𝑜𝑝 𝑤𝑒𝑖𝑔ℎ𝑡 1
4) Production= Delivery roller speed (lb/min) × 𝑐𝑙𝑜𝑐𝑘 𝑙𝑒𝑛𝑔𝑡ℎ ×𝐷𝑟𝑎𝑓𝑡×emulsion%×
efficiency%×time

Problem-(01):
If Raw jute receives an application of 20% emulsion of which 15% is absorbed. Find
out the weight of a batch of 12 bales Raw jute (4000lbs = 1bale):
Solution:
Wt of 12 bales = 4000  12 = 48000 lbs.
100 lbs raw jute after batching wt = 115 lbs
115
1 lbs raw jute after batching wt =
100
115 48000
48000 lbs raw jute after batching wt =
100
= 55200 lbs (Ans)
Problem-(02):
If an emulsion is prepared as follows: Oil = 24%, Water = 74.5%, Emulsifier = 1.5%.
If raw jute receives an application of 20% emulsion of which 15% absorbed. Find out the
absorbed amount oil, water and emulsifier of raw jute (A batch containing 12bales,
4000lbs/bales)
Solution:
Wt of 12 bales = 4000  12 = 48000 lbs.

Oil:
100 lbs emulsion contains = 24 lbs oil
24
1 lb emulsion contains =
100
24  15
15 lbs emulsion contains =  3.6 lbs oil
100

100 lbs raw jute absorbs = 3.6 lbs

3 .6
1 lb raw jute absorbs =
100
3.6  48000
48000 lbs raw jute absorbs =
100
= 172.8 lbs.

Water:
100 lbs emulsion contains = 74.5 lbs water
74.5
1 lb emulsion contain =
100
74.5  15
15 lbs emulsion contain =
100
= 11.175 lbs
100 lbs jute absorbs = 11.175 lbs.
11.175
1 lb jute absorbs =
100
11.175  48000
48000 lbs jute absorbs =
100
= 536.4 lbs
Emulsifier:
100 lbs emulsion contains = 1.5 lb emulsifier
1 .5
1 lb emulsion contain =
100
1.5  15
15 lbs emulsion contain =
100
= .225 lb emulsifier

100 lbs raw jute absorbs = 3.225 lb emulsifier

3.225
1 raw jute absorbs =
100
.225 4800
48000 lbs raw jute absorbs =
100
= 10.8 lb emulsifier.

Problem-(03):
A jute spreader machine is running with following data:
Dollop wt = 1000 lbs, Clock gearing const = 10, Clock change pinion = 20T. Draft const =
350, DCP = 35T, Emulsification applied = 25%. Feed wt in lbs/100yds? Delivery wt/100
yds ?
Solution:
Clock length = clock gearing const  clock change pinion
= 10  20 = 200 Yds.
wt of feed jute in lbs/100Yds
1000
=  100  500 lbs/100Yds.
200
Draft.Const 350
Draft =   10
DCP 35
25
Emulsion applied =  500  125lbs/100Yds.
100
After application of emulsion wt of Total feed material = (500+125) = 625 lbs/100Yds
Feed sliver weight
Delivered Sliver Weight = lbs /100Yds
Draft
625
= lbs /100Yds
10
 = 62.5 lbs /100Yds

Problem-(04):
Find the production in lbs/hr of S. M/c delivered speed = 60 yds/min, Draft = 12,
Efficiency = 80%, emulsion applied = 20%. Wt of feed material = 6 lb/yds.
Solution:
6 100  20
Production in lbs/hour =  60  60   0.8
12 100
= 1728
Problem-(05):
Production of spreader M/C have dollop wt = 36lbs, Clock length = 6yds, delivery
speed = 60 yds/min, M/C draft = 12, emulsion applied = 12%
Solution:
delivery speed  dollop wt
Production of spreader machine in lbs/hour =  emulsion%
draft  clock lenght
36 100  12
= 60  60  
12  6 100
= 2016 Yds/hr.
 Jute carding

Mathematical Problem
Dollop.wt
1) Feed sliver wt / 100 Yds= 100
Clock.lenght
Dollop.wt 100
2) Breaker card machine delivery sliver wt/100 Yds = 
Clock.lenght Draft
3) Finisher card m/c delivery sliver wt/100 Yds =
Dollop.wt 100 Finisherdoubling
   wastage%
Clock.lenght Draft Draft

Surface speed of cylinder

4) Carding ratio=
Surface speed of worker

Surface speed of cylinder−Surface speed of worker

5) Carding efficiency = ×100
Surface speed of cylinder
6) Production= Surface speed of delivery roller ×36
60
×sliver weight/yds ×efficiency%
×wastage%

Problem (01):
Calculate the wt in lb/100Yds sliver from the data: Dollop wt = 30 lbs, Clock length
= 12.9Yds, Draft = 11.23. Evaporation loss = 4%.
Solution:
Dollop.wt 1 100  4
Delivery sliver wt per 100 Yds =   100 lbs
Clock .lenght Draft 100
30 1 96
=    100 lbs
12.9 11.23 100

= 19.88 lbs.
Problem (02):
Calculate production per hrs from the data: delivery Roller Speed = 18 ft/min,
Dollop wt = 30 lbs, Clock length = 19.5Yds, draft=12, Waste = 4%. Efficiency = 85%

Solution:

18 30 100  4
Production per hr =  60  0.85   lbs/hr
3 19.5 12 100
= 451.93 lbs/hr

Problem (03):
A breaker card is set to work with a clock length of 12.9 Yds and a draft of 12 in the
machine. If the dollop wt is 30 lbs. What will be the wt of delivered sliver in
lbs/100Yds.
Solution:

Dollop.wt 1
Sliver delivered per 100 Yds =   100 lbs
Clock.length Draft
30 1
=   100
12.9 12
= 19.38 lbs

Problem (04):
Calculate the wt of sliver in lbs/100Yds delivered form the finisher card from the data:
B/C delivered sliver = 24 lbs/100Yds
F/C draft = 15
F/C doubling = 12
Waste loss = 4%
Solution:
Wt of F/C delivers sliver/ Sliver delivered per 100 Yds from F/C
Doubling 100  waste%
= B / C.delivered.Sliver 
Draft 100
12 100  4
= 24  
15 100
= 148.432 lbs.
Problem (05):
Calculate wt of sliver in lb/100Yds delivered from the
F/C B/C dollop wt = 32 lb, B/C Clock length = 12.5 lb, B/C draft = 10, F/C
doubling = 12, F/C draft = 15.

32 1
Sliver delivered per 100 Yds from B/C =   100
12.5 10
= 25.6 lbs/100
Sliver deliver from F/C = lbs

Problem (06):
How many ends at the feed of the finisher card are required to obtain 14.22 lb/100
Yds sliver from finisher card with F/C draft = 15, B/C sliver production = 20 lb/100
Yds. . M.R. = 3%
Solution:
Sliver delivery from F/C in lbs/100Yds
F / C.doubling 100  loses
= delivery from B/C in lbs/100Yds  
F / C.draft 100
F / C.doubling 100  3
 14.22  20  
15 100
14.22  15  100
 F/C doubling = = 10.44  11
20  97
=11 ends will be required.

Jute Draw Frame

Problem 01: 3 finisher cards supply sliver to 2 1st D/F; Calculate the speed of the back rollers of
the drawing frames required to keep the cards and drawing frames in balanced. Each finisher
card produces 400 lb/hr of 16 lbs /100 yds sliver. Drawing frame has 2 heads, 2 delivery/head,
doubling 2:1.
Solution:
Production of 1 finisher card = 400 lb/hr.
Production of 3 finisher card = 400X3 =1200 lb/hr.
1200 lbs/hr is fed at 2 Drawing frame.
 1200
So, Amount of feed of each drawing frame= lbs / hr.
2
=600 lbs/hr.
600  100
Length of feed in each D/F = yds / hr.
16
600 100  3
 ft / min  187.5 ft / min .
16  60
As the finisher cards & the drawing frames are in balanced so, feed = Delivery.
187.5
The speed of back roller = ft / min .
2X 2X 2
=23.437 ft/min. Ans.
Problem 02: A high speed push bar D/F with 2 heads, 2 delivery/head, doubling 2:1, produces
600 lbs/hr of 8 lbs/100 yds sliver. Draft = 4, pitch of the faller bar = ½”, faller bar lead =10%;
Calculate the faller bar drops/min.
Solution:
600×100×3
Length of Production = ft/min
8×60
600 x100 x3x12

8 x60
 4500 inches/min.
4500
Sliver length production per delivery roller = inches / min .
2 x2
=1125 inches/min.
10
Surface speed of faller sheet = 281.25+ 281.25 X
100
=300.375
we know, Faller bar S.S.=Faller drops/min X Pitch.
Faller bar S.S.
Or faller drops/min =
Pitch.
309.375
=  618.75 Ans.
1
2
Problem 03: Calculate lbs/100 yds sliver from the D/F with particulars: F/C delivered
sliver……………..lbs/100 yds.
D/F draft ………………….4
D/F doubling ………….2:1.
Solution:
 D/F doubling
lbs/100 yds from D/F = lbs/100 yds from F/C X
D/F drafting
2
 16  8 Ans.
4
Problem 04: A 1st D/F, 2 heads, 2 del/head, doubling 2:1 Produces 600 lbs per hr. The m/c is
fed with rolls of sliver from the F/C. Each roll wt is 30 lbs of 14 lbs/100 yds sliver. Surafce
Speed of back roller is 30 ft/min. Find out the eff% of the m/c (d/f).
Solution:
Actual production=600 lbs/hr.
Each roll wt= 30 lbs.
30  100
Length of each roll=  214.28 yds  642.857 ft.
14
Surface speed of back roller= 30 ft/min.
642.857
So, time required to finish a fed roll =  21.428 min .
30
There are 8 rolls of sliver ( i.e. 2X2X2)
So, total wt = 8X30=240 lbs.
240
So, wt feed per min = lbs.
21.428
240 X 60
And wt feed per hr=  672lbs
21.428
600
So, efficiency =  100  89.28%
672

Jute Spinning

1. Time = 8 hr.
Flyer no. =100
Flyer R.P.M =3301
T.P.I = 3.35
Sliver from finisher D/F = .5 lbs/100 yds
Spinning frames draft = 9
Spinning frame efficiency = 80%
 Production=?
Solution:
Sliver from finisher D/F = 0.5 lbs/100 yds.
0.5X144
So, Spinning count=  8 lbs/spyndle
9
3301 X 60 X 8 X 80 X 100 X 8
 lbs/8hr.
Production 3.25 X 100 X 14400 X 36
=583.92 lbs/8hr. Ans.

2. If a jute spinning frame produces 8 lbs/spyndle Hessian warp yarn from 72

lbs/spyndle sliver then what will be the draft constant and twist constant? Consider
the following particulars;
D.C.P=27
T.C.P=32
T.P.I=4
Solution:
72
Draft= 9
8
Draft constant= Draft X D.C.P = 9 X 27 = 243.
Twist constant= T.P.I. X T.C.P = 4 X 32 = 128.

3. What will be the production for 8 hrs working of a spg. Frame of 100 spindles with
particulars as below:
R.P.M. of flyer =3300; TPI=3.35,
Sliver from finisher drawing frame 0.5 lbs/100 yds.
Spg. Frame efficiency = 80%

Solution:
14400
Yarn count = 0.5   8 lbs/spyndle.
9  100
3300  60  8 100  8  0.8
 prod n / shifts 
3.35  36 14400
= 583.747 lbs
4. Find the prodn /shift of 8 hrs of a jute mill having 28 frames of 100 spindles each running
with the following particulars
Flyer speed=3800 r.p.m
T.P.I=3.2
Count=8.6 lbs/spindle
Efficiency =84%
Comment on the quality of the yarn produced if 10% root cutting are used in the mixing
along with good fibre,
Solution:
3800  60  8  8.6  .84 100  28
Prodn /Shift=  22240.566lbsAns
3.2  36 14400