Spinning calculations

(SPINNING CALCULATION-1)
MEASUREMENT OF MOISTURE
All textile materials are more or less hygroscopic. They contain a certain amount of moisture
depending upon the relative humidity of surrounding atmosphere.
Standard moisture regain of different materials is given below:
 Cotton 8.5%
 Wool 17%
 Viscose 11-13%
 Silk 11%
 Jute 14%
 Polyester 0.3-0.4%
 Acrylic 3-4%
 Nylon 4.2%
Absolute Humidity:
Weight of water in a unit volume of moist air. It is usually denoted by gm/m3
or grain/ft3
.
Relative Humidity:
It is the ratio of actual vapor pressure divided by the saturated vapour pressure multiply by 100.
R.H% = 98.6 –
Original weight:
This is the wt. Of material in its original condition containing any level of moisture. It is usually
denoted by O.W.
Dry weight:
The weight of material without any moisture. It is denoted by D.
MUHAMMAD GHUFRAN (05TE-16) Page 1

Oven dry weight:
The weight of material dried at oven state. The temperature in the oven is 1050
C.
Correct invoice weight:
It is the weight of the material at standard moisture regain. It is given as:
C.C.W = Dry wt. × 8.5%
C.C.W =
Regain:
Weight of moisture in a material expressed as %age of oven dry weight.
Moisture content:
Weight of moisture based on original weight of sample expressed in % age.
Relation between Regain & moisture Content:
Proof:
We know that
By dividing D in nominator & denominator

Similarly we can proof that,
Q. Oven dry wt. Of 500 grains of lint was found to be 400grains, calculate:
1. Wt. Of moisture
2. CCW of lint
3. R%
4. M.C%
Sol.
1. Weight of moisture = 500 – 400 = 100 grains
2. CCW = D × 1.085
CCW =400× 1.085 = 434 grain
3.

=25%
4.
Q. Moisture regain, if cotton was found to be 10.5. What will be the value of content % age?
Sol.
R = 10.5 %
We know that,
COTTON PURCHASE
Q. A spinning unit has purchased lint of 20,000kg at 8.5% moisture regain. Calculate the
reduction in wt. If the moisture regain is found to be 10.2%.

Sol.
Dry wt. = 20,000 × 10.2/100
= 17960kg
CCW = D × 1.085
= 17960 × 1.085
= 19486.60kg
Reduction in wt. = 20,000-19486.60
= 513.40kg
Q. A consignment of 400kg of lint cotton was dispatched to a station where oven dry wt.
Of 1000gm sample was found to be 850gm. Calculate CCW for which the supplier is to be paid
by the consigny. Also calculate the value of R and M and excessive moisture regain %age.
Sol.
1000gm = 1kg
850gm = .85kg
Dry wt. Of consignment = dry wt of sample × total wt of consignment/original wt of sample
= .85 × 400 /1 = 340kg
CCW. = D × 1.085
= 340 × 1.085 = 368.9kg
Wt of moisture = W = 400 – 340 = 60kg
= 60×100/340 = 17.65%
M = W×100/(W+D)

= 60 × 100/(400) = 15%
E.M% = moisture regain – std. regain
= 17.65 – 8.5 = 9.15%
Q. A representative sample of 9.5 oz. Drawn from a consignment of 1200 lbs of lint
cotton gave the oven dry wt of 8.5 oz. Determine the CCW of consignment and the amount of
money to be paid by the purchaser if the price per lbs is 45Rs.
Sol.
9.5oz = 0.59375 lbs
8.5oz = 0.53125 lbs
Dry wt of consignment = dry wt of sample × total wt of consignment/original wt of sample
= 0.53125 × 1200 /0.59375 = 1073.69 lbs
CCW. = D × 1.085
= 1073.69 × 1.085 = 1164.95 lbs
Price per lbs = 45Rs
Total price = 1164.95 × 45
= 52422.63 Rs
Q. A representative sample is taken from the consignment of 2 lots with bale wt of 170
kg. Sample wt was 13oz and dry wt was 11oz find out the conditioned wt of consignment and
conditioned price if price per lbs is 60 Rs. Also calculate M.R and M.C.
Sol.
Total wt of consignment = 2 × 100 ×170
= 34000kg
Dry wt. Of consignment = dry wt of sample × total wt of consignment/original wt of sample
= 11 × 34000 /13 = 28769.2kg
CCW. = D × 1.085

= 28769.2 × 1.085 = 31214.6 kg
Price for original wt = 34000 × 60 × 2.2046 = 4497384 Rs
Price for conditioned wt. = 31214.6 × 60 × 2.2046 = 4128942 Rs
Saving = 4497384 – 4128942 = 368442 kg
R% = W×100/D
= 2×100/11 = 18%
M% = W×100/ (W+D)
= 2 × 100/ (13) = 15%
MOISTURE & COUNT CALCULATIONS
Q. A sample of cotton yarn has a count of 40S
. A certain amount of moisture has been
added & count becomes 38S
. What %age of moisture has been added?
We know that,
Hence,
Wt. of 40S
= 1/40= 0.025 lbs
Wt. of 38S
= 1/38= 0.0263 lbs
Moisture wt. = 0.0263-0.025 = 0.0013157lbs
= 5.26%
Q. A spindle point yarn of 40s contains 5% moisture, what will be the count of yarn when
moisture added becomes 8.5% which is the standard moisture?

Wt. of yarn = 1/40 = 0.025 lbs
Excess moisture added = 8.5 - 5 = 3.5%
Amount of moisture added = 0.000875 lbs
CCW = 0.025 + 0.000875 = 0.025875 lbs
Hence,
Count = 1/0.025875 = 38.65S
Q. Calculate the conditioned count of 500m of cotton yarn in tex. If its oven dry weight is
10gm, what is count at dry wt?
Dry wt = 10gm
Length = 500m = 0.5km
= 10.85gm
Tex count = no of gms/1km
Conditioned count = 10.85/0.5 = 21.7tex
Dry count = 10/0.5 = 20tex
Q. Calculate conditioned count of 20km cotton yarn, if its oven dry weight is 800gm. Also
calculate denier count?
Dry wt = 800gm

= 868gm
Tex count = 868/20 = 43.4tex
We know,
Denier = Tex × 9
Denier count = 43.4 × 9 = 390.6denier
Q. Calculate the MR, MC% & CCW of 100gm of cotton yarn if its oven dry weight is 94gm.
Also calculate tex and denier counts at dry wt & CCW. Take 1 hank of yarn?
O.W = 100gm
D = 94gm
W = 100 – 94 = 6gm
= 6.38%
= 6%
= 102gm
 Count at dry wt:
Wt. = 94/453.6 = 0.207lbs

English count = 1/0.207 = 4.82s
Tex count = 590.5/4.82 = 122.4 tex
Denier count = 5315/4.82 = 1101 denier
 Count at conditioned wt:
Wt. = 102/453.6 = 0.225lbs
English count = 1/0.225 = 4.45s
Tex count = 590.5/4.45 = 132.8 Tex
Denier count = 5315/4.45 = 1195 denier
Q. A warper beam contains 500 ends of cotton yarn each end has a length of 3500 yds.
The yarn contains moisture up to extent of 5% above the standard moisture and wt is now
100lbs. calculate the count of yarn on dry wt, CCW, & present?
Sol.
No of ends = 500
Length of one end = 3500 yds
Total length = 500 × 3500 = 1750000 yds
We know that,
Length (yds) = wt (lbs) × 840 × count
So,
At 5% excess moisture, moisture present will be 13.5%.
Hence,
113.5 lbs material has dry wt = 100 lbs

100 lbs material has dry wt = 100 × 100/113.5 = 88.10 lbs

BLENDING & MIXING CALCULATIONS
Q. The price of a cotton A is 60 Rs/Kg with a weight of 1000Kg.while price of cotton B is
55 Rs/Kg with a weight of 2000 Kg. Calculate the average price of mixing of two cottons.
Soln:
Pa = 60 Rs Wa = 1000
Pb = 55Rs Wb = 2000
= 56.67 Rs/Kg
Q. The average price of mixing is Rs 40/Kg.The average price of component A with a
weight of 2500 Kg is Rs 45/Kg.What should be wt of component of B with a price of 35
Rs/Kg.
Solution:
(2500+Wb)×40 = 112500 + 35 Wb

100000 + 40 Wb = 112500 × 35 Wb
(40 - 35) Wb = 112500 - 100000
Wb = 12500/5 = 2500 Kg
Q. The Price of cotton A is 64Rs/Kg While Price of Cotton B is 70Rs/Kg .If total
quantity of two cottons is same , what will be average price of mixing.
Solution:
Suppose wt of samples = W
We know that,
Q. The average mixing price per kg in a mill is Rs 45.The price of components A is 38Rs/Kg
while price of 2nd
component 38Rs/Kg.
Calculate the followings.
 Percentage of individual component for getting required price
 Total no of bales of individual components, if total bales available are 50000.
Slon:
Given data
Pm = 45
Pa = 38
Pb = 48

% Age of components
A= 3/10×100=30%
B= 7/10×100=70%
Bales of component A= 3/10×50000
= 15000bales
Bales of component B= 7/10×50000
=35000bales
Q. The average mixing price for making PC yarn is Rs 52/Kg.The price of 1Kg cotton is
42Rs while price of 1Kg polyester is 64Rs calculate
 Percentage of two components
 Total no of bales of individual components, if total quantity is 100000.
Cotton bale = 170Kg
Polyester bale = 200Kg
Soln:
Given data
Pm = 52
Pc = 42
Pp = 64

Cotton =12/22×100=54.55
Polyester =10/22×100=45.45
Cotton bales =12/22×100000=54545Kg
=321bales
Polyester bales =10/22×100000=45454.5Kg
=228bales
Q. Total quantity is 25000kg.Calculate quantity for each component while
Pm = 42
Pa = 48
Pb = 43
Pc = 38
Soln:
Qa = 4/15×25000=6666.67kg
Qb = 4/15×25000=6666.67Kg
Qc = 7/15×25000=11666.67Kg

PIPING CALCULATIONS
Q. The main pipe in a blow room have to feed 3-branch pipes, their respective dias are
2.5”, 1.5”, and 1.25”. calculate dia of main pipe?
Soln:
Dm = (D1)2
+ (D2)2
+ (D3)2
Dm = (2.5)2
+ (1.5)2
+ (1.25)2
= 6.25 + 2.25 + 1.56
= 10.0625 = 3.17”
Q. In a ventilated system a rectangular suction duct of 10” × 12” is to be replaced a
circular duct 25% greater in capacity. What will the dia of the circular duct?
Soln:
Area of rectangular duct = 10 × 12 = 120 in2
25% greater capacity area
A = 120 + (120 × 25/100) = 150 in2
A =
D2
= 150 × 4/ = 190.98 in2
D = 13.82”
Q. A tower of blow room is 82
ft in section discharging into base by pipe with particulars:
 One pipe of 15” dia having linear air speed of 3000 ft/min.

 One pipe of 9” dia having linear air speed of 2000 ft/min.
 Six pipes of 12” dia having linear air speed of 1500 ft/min.
Calculate the velocity of air flow in ft/min in the tower?
Soln:
Area of tower = 8 × 8 = 64 ft2
Volume = area × velocity
V1 = π/4(15/12)2
× 3000 = 3681.6 ft3
V2 = π/4(9/12)2
× 2000 = 883.57 ft3
V3 = π/4(12/12)2
× 1500 = 7068 ft3
Air volume in tower = sum of air volume in branches
V = V1 + V2 + V3
= 3681.6 + 883.57 + 7068
= 11633.17 ft3
Velocity = volume/area
= 11633.17/64 = 181.77 ft/min
Q. A room measuring 40 × 30 × 15 yds is ventilated by 2 pipes of 18” each. If 10% of the
space in room is taken up by machinery and it is desired to change the air twice every hour.
Calculate the linear air speed in the pipes in ft/min?
Soln:
Volume of the room = 40 × 30 × 15 = 18000 yd3
Useful volume = 18000 × 0.9 = 16200 yd3
Area of pipes = 2 × π/4 (18/36)2
= 0.3927 yd2

Volume = area × velocity
Velocity = 16200/0.3927
= 41252.9 yds/30min
= 4125.3 ft/min
GROWTH RATE CHECKING
Growth rate checking is used to find out fiber demage due to the beaters. In growth rate
checking increase in percentage of short fibers is determined after beating action of beater, if
there is increase in percentage of short fibers after beating growth rate is not good and its
mean that fiber is demaging up by the beater.
COEFFICIENT OF VARIATION OF LAP
CV% of blow room ranges from 1-1.5. Normally it is of about 1.3%.
Mean:
It is the average value of two or more values e.g. X1, X2, X3, X4,……… are any values there mean is
calculated as:
X = X1 + X2 + X3 + X4+…./n
Standard diviation:
The standard deviation is the square root of the mean of the squares of the deviations of the
observations from their mean.
Standard deviation = = (∑(X – X)2
/n)

Coefficient of variation:
By expressing the standard deviation as a percentage of tegh mena we obtain the coefficient of
variation, C.V.% .
Coefficient of variation, CV% = (standard deviation × 100/mean)
WASTE CALCULATIONS IN BLOW ROOM
Waste removed from blow room is very important regarding cleaning efficiency of blow room.
Waste = lint + trash
Input = output * (100/100 – W%)
Output = input – waste
Waste% = waste * 100/input
Cleaning efficiency = waste extracted * 100/total waste
Q. 80 bales of cotton each of 167kg, are being fed daily in a 2-scutcher blow room line.
Actual production per 2-scutcher per day is 12000kg. calculate the total quantity of waste
throughout whole blow room line & waste %age.
Soln:
Input = 80 * 167 = 13360 kg
Output = 12000 kg
Waste = 13360 – 12000 = 1360 kg
Wase% = 1360 * 100/13360 = 10.18%
Q. The production per scutcher per hour in a 2-scutcher blow room line is 325 lbs. waste
from whole blow room line is 700 kg/day. Calculate the no of bales required per day for given
blow room line. (1 bale = 167 kg).

Soln:
Waste = 700 kg
Output = 325 * 2 * 24/2.2046 = 7076.4 kg/day
Input = 7076.4 + 700 = 7776.4 kg/day
No of bales/day fed = 7776.4/167 = 47 bales
Q. Production per hour from scutcher of 3-scutcher blow room line is 310 lbs. waste %age
of blow room is 6.5%. calculate the no. of bales to be fed in given blow room per month
provided that one bale is os 168 kg.
Soln:
Blow room output per day = 310 * 3 * 24/2.2046 = 10124.7 kg
(100 – 6.5) input = 1012474.5 kg
Input = 1012474.5/93.5 = 10828.6 kg/day
Hence no. of bales fed per month
= 10828.6 * 30/168
= 1934 bales
Q. 63 bales of cotton are being fed daily in a blow room line consisting of 2-scutchers.
The blow room is to be kept stopped for one hour after each shift for maintenance. If waste
%age is 6.5% , what will be the production per scutcher in kg/hour, while 1 bale = 166kg.
Soln:
Input = 63 * 166 = 10458 kg/day
Waste = 10458 * 6.5/100 = 679.77 kg
Output = 10458 – 679.77

= 9778.23 kg/day
Prod. Per scutcher per day = 9778.23/2*21
= 232.82 kg/hour
Q. 80 bales of cotton are being fed in a blow room line daily. If the pure weight of cotton
per bale is 168 kg and waste% of blow room line is 6%.
 What will be the wt. of cotton lap received per day from scutcher.
 Calculate the no. of laps prepared per day if the wt. of one lap is 22kg.
 If lap rod is 2kg then calculate no. of pure lint laps.
Soln:
Input = 80 * 168 = 13440kg
Waste = 6%
Waste = 13440 * 6/100
= 806.4 kg
Output = 13440 – 806.4
= 12633.6 kg/day
This is cotton lap received per day
No of laps/day = 12633.6/22
= 574.2 laps
Pure lint laps/day = 12633.6/20
= 631.68 laps
Q. The total material fed into a step cleaner was 800 lbs/hr. The cleaning effeciency of
machine was found to be 35%. The total material delivered per hour was found to be 2% less

than the fed material. Calculate total quantity of waste as well as trash per hour. If there is
80% trash of total waste.
Soln:
Input = 800 lbs/hr
Output = 800 (1 - .002) = 784 lbs/hr
Waste = 800 – 784 = 16 lbs
Waste% = 16 * 100/800 = 2%
Trash% = 80% of waste
Trash = 16 * 80/100
= 12.8 lbs
Q. A lap of 50 lbs with a length of 55yds is being prepared on scutcher in 4.5 min. if the
waste %age of blow room line 7%, what should be required no. of bales to be fed daily, if bale
wt = 170kg
Soln:
50 lbs are prepared in = 4.5 min
In one hour = 50 * 60/4.5
= 666.67 lbs
Production per day = 666.67 * 24
= 16000 lbs/day
Hence,
Input = 16000 (100/93) = 17204.3 lbs/day
No of bales = 17204.3/2.2046*170
= 46 bales

PRODUCTION CALCULATIONS
Q. Calculate Blow Room Production, if number of scutchers is 3,shell roll speed is 13rpm shell
roll dia is 9”, lap weight is 12oz/yd, and efficiency is 85%.
Soln.
Blow room Production = shell roll dia* shell roll rpm*3.14* oz/yd* 60*3*8* number of scutchers* efficiency
36 * 16
Now putting values in the formula
Blow room Production = 9 * 3.14 *13 * 12 *60 *3 * 8* 3* 0.85
36 * 16
Blow room Production = 9 *3.14 * 13 * 12 * 60* 3 * 8* 3 *0.85
36 * 16
= 16188232.3/576
Blow Room Production = 28100 lb/day
Q. If Blow Room Production is 30000 and number of scutchers are 3, shell roll dia is 9”, lap weight is
13oz/yd, and efficiency is 85% then calculate shell roll speed
Sol
36 * 16
Blow room Production = 9 * 3.14 * rpm * 13 *60 *3 * 8* 3* 0.85
36 * 16
30000 = 9 *3.14 * rpm * 13 * 60* 3 * 8* 3 *0.85

36 * 16
rpm = 30000 * 36 * 16
9 *3.14 *13 *60 *3 * 8 * 3 * 0.85
rpm = 17280000/1349019.36
Shell roll speed = 13rpm
Q. If Blow Room Production is 30000 and number of scutchers are 3, shell roll speed is 13rpm, shell
roll dia is 9”, and efficiency is 85% then calculate lap weight.
Soln.
36 * 16
Blow room Production = 9 * 3.14 * 13 * oz/yd *60 *3 * 8* 3* 0.85
36 * 16
30000 = 9 *3.14 * 13 * oz/yd * 60* 3 * 8* 3 *0.85
36 * 16
oz/yd = 30000 * 36 * 16
9 *3.14 *13 *60 *3 * 8 * 3 * 0.85
oz/yd = 17280000/1349019.36
lap weight = 13 oz/yd

Q. In a spinning mill there are 30 ring frames each have 480 spindles and producing 20s count with TM
4.1. Spindle speed is 1600rpm and efficiency is 90%. The yield age of mill is 83%. In blow room there
are 3 scutchers with shell roll speed is 12rpm wt of bale is 170kg. Vary the ozs/yard calculate:
a) Total no. of bales required?
b) No. of lap/scutcher(Lap length is 50 yards)?
c) Total no of laps?
d) Balance the production of blow room with reference to ring department?
Soln.
Ring production (ops) = (Spindle speed x 0.254)/TPI x count
= 16000 x 0.254/18.33 x 20
= 11.1 x 0.90 x 3
= 29.92ozs
Production = 1.8lbs
= 1.8 x 480 x 30 = 25920lbs
= 260bags
Total wt required = 25920/0.83 = 31228.90
= 31228.90/2.24 = 13194/170
= 82bales
Total no of Lbs = 31228.90
Deducting 6% waste = 31228.90 x 0.94
= 29355.10lbs
Production = (shell roll Dia x rpm x π x ozs/yard x 60 x 8 x 3) x 3
36 x 16
= (9 x 12 x 3.14 x 11.5 x 60 x 8 x 3) x 3
36 x 16

= 29349.5lbs
1 lap = 50yards
1 lap = 11.50ozs/yard
1 lap = 50 x 11.5 = 36lbs
16
Total no. of laps = 29349.5/36
= 816 laps
No. of laps/ Scutcher = 816/3
= 272 laps
Q. Prepare the production plan of a spinning mill with the given statically data While the yield
age is 82 %, bale wt. is 170 Kg., blow room has 2 Scutcher, efficiency 85 % and waste %age is
6%.
Count
Spindle
speed(rpm)
Tm Tpi
Efficiency
%
Frames
No. of
Spindle/frame
OPS
Production
(bags per
day)
20 16000 4.1 18.33 90 10 480 9.98 89.82
30 16500 4.2 23.0 94 8 480 5.71 41.11
40 17000 4.3 27.1 95 15 480 3.78 51.03
50 18000 4.5 31.8 96 5 480 2.76 12.42

Soln.
TPI
For 20 Nec
Tpi =™×√C
= 4.1×√20
=18.33
For 30 Nec
Tpi = ™×√C
= 4.2×√30
= 23.0
For 40 Nec
Tpi = ™×√C
= 4.3×√40
= 27.1

For 50 Nec
Tpi = ™×√C
= 4.5×√50
= 31.81
OPS
For 20 Nec
Sp Speed ×0.254×ή where 0.254= 60×8×16/ 36×840
OPS= TPI × Ct
16000× 0.254×0.90
= 18.33×20
= 9.98
For 30 Nec
Sp Speed ×0.254×ή where 0.254= 60×8×16/ 36×840
OPS= TPI × Ct
16500× 0.254×0.94
= 23.0×30
= 5.71

For 40 Nec
Sp Speed ×0.254×ή where 0.254= 60×8×16/ 36×840
OPS= TPI × Ct
17000× 0.254×0.95
= 27.1×40
= 3.78
For 50 Nec
Sp Speed ×0.254×ή where 0.254= 60×8×16/ 36×840
OPS= TPI × Ct
18000× 0.254×0.96
= 31.8×50
= 2.76
Production (Bag /day)
For 20 Nec
Production(bags/day)= OPS × 0.90 × No. of Frames Whereas 0.90 =3×480/16 ×100
= 9.98 × 0.90 × 10
= 89.82
For 30 Nec
Production(bags/day)= OPS × 0.90 × No. of Frames Whereas 0.90 =3×480/16 ×100
= 5.71 × 0.90 × 8
= 41.11

For 40 Nec
Production (bags/day)= OPS × 0.90 × No. of Frames Whereas 0.90 =3×480/16 ×100
= 3.78 × 0.90 × 15
= 51.03
For 50 Nec
Production (bags/day)= OPS × 0.90 × No. of Frames Whereas 0.90 =3×480/16 ×100
= 2.76 × 0.90 × 5
= 12.42
Total production of ring = 89.82 + 41.11 + 51.03 + 12.42
=194.38 bags /day
Bale required = ?
Production of ring × 45.36
Bale required = yieldage × bale wt.
194.38 × 45.36
= 0.82 × 170
= 63.25 ~ 64 bales per day

Blow room production = 64 / 1.06 = 60.37 bales per day
Suppose OZ / yd = 13
NπD × 60 × 24 × ή × oz/yd × no. of scucher
Blow room production = 36 × 16
60.37 × 170 × 2.2046 = N × 3.1416× 9 × 60 ×24 ×0.85 × 13 × 2
36 × 16
60.37 × 170 × 2.2046 × 36 × 16
N = 3.14 × 9 × 60 × 24 × .85 × 13 × 2
N = 14.49 ~ 15 (RPM)

L = 100*72*54*16*33*9*π/1*112*13*70*35*36
L.L = L.L.C * L.L.C.M
L =0. 628*L.L.C
Q. The value of LLC is .628 size of L.L.C.M is 80T. Time for lap formation is 4.25 min. calculate
the production of the blow room line consisting of 2-scutchers with a hank of 0.0017?
Soln.
L.L.C.W = .628
L.L.C.W=80T
Time =4.25 minutes
Hank of cap =0.0017
Scutcher = 2

Production per hour =?
L =0.628*80
L =50.24 yards
Count = no. of hanks/lb
No. of lbs for one Scutcher = 50.24/(840*0.0017) = 35.18lbs/4.25 min
= 35.18 * 60/4.25 = 496.69 lbs/hr
Production of blow room line = 496.69 * 2/2.2046
= 450.6 kg/hour
Q. Production per hour from a 3-scutcher blow room line is 800 kg. Lap wt is 15 oz/m and
time for making full lap is 4.5 min. Calculate value of L.L.C.W When L.L.C is 0.628.?
Soln.
W = 15 oz/m= 15/1.0936 = 13.72oz/yd
Production per Scutcher = 800 * 2.2046/3
= 587.87 lbs/hr
= 587.87*16/(60*13.72)
= 11.43 yds/min
L.L = L.L.C.W*L.L.C
Total lap length = 11.43*4.5
= 51.46 yds
L.L.C.W = 51.46/0.628 = 82T

CARD CALCULATIONS

Q. calculate following from sketch given below; when cylinder is rotating at 180 rpm. Pully on
cyliner shaft is 18” driving a pulley 7” on taker-in; hank sliver is 0.135s and production
efficiency is 84% .
 Production in kg/hr
 Prod constant
 Induvidual draft
 Total draft
 Drft constant
 Size of dcp for giving actual draft of 85 with a waste of 5%
 Condensation factor
 Size of barrow wheel for gettingh 5% higher prod
 Time required to fill a can with a size of 18” * 40”, if can capacity is 40lbs
 Density of packing can
Soln;
Production=
π*2/36[180*18.25*7.25*5.25*20*216*31*20*20/7.25*7.25*10.25*100*30*15*20*20]*60*.84/840*0.135
= 53.55 lbs/hr = 24.3 kg/hr
Taking barrow wheel as one tooth:
π*2/36[180*18.25*7.25*5.25*1*216*31*20*20/7.25*7.25*10.25*100*30*15*20*20]*60*1/840*1
= 0.0043 lbs/hr
=0.001953 kg/hr
D1 = 48*2.5/18*6 = 1.11
D2 = 120*40*27/18*40*2.5 = 72
D3 = 216*4/30*27 = 1.067
D4 = 31*2/15*4 = 1.033
M.D = 48*120*216*31*2/18*18*30*15*6=88.18
D.C= 48*120*216*31*2/18*1*30*15*6 = 1587.2
A.D = M.D * (100/100-W)
When A.D = 85
M.D = A.D * (100-W/100)
= 85 * (100-5/100) = 8.75
New DCP = 1587.2/80.75 = 19.66 = 20T
Surface speed of cylinder = π*50*180 = 28274.3 in/min
Surface speed of doffer = π*27*[180*18.25*5.25*20/7.25*10.25*100] = 39.37 in/min
Condensation factor = S.S of cylinder/S.S of doffer

= 28274.3/3937 = 7.18
New prod with 5% increase = 24.3 * 1.05 = 25.52 kg/hr
25.52 = 0.001953*Barrow wheel * 84/0.135
B.W = 25.52*0.135/0.001953*84 = 21T
Production = 53.55 lb/hr
53.55 lbs = 60 min
40 lbs = 60*40/53.55 = 45 min
Density = mass/volume
Volume = πd2
*L/4
= 3.14*(18)2
*40/4
= 10178.76 inch3
Density = 40/10178.76 = 0.0039 lbs/in2
Q. If the doffer speed of flat card is 40rpm and its diameter is 27”, grains/yards are 65,
Tension draft is 1.03, and the efficiency of card is 85% then calculate the production of card in
pounds per day if there are 8 cards in department.
Data:
Doffer speed= 40rm
Doffer dia= 27”
Tension draft = 1.03
Sliver weight = 65 grains/yds
Efficiency %= 85
No. of cards = 8
Production (lbs/day) =?
Solution:
Since we know the production formula
Card production (lbs/day)
doffer surface speed*grains/yds*60*8*3*efficiency %age*no.of cards
= 36*7000

40*27*3.14*65*60*8*3*.85*8
= 36*7000
= 8565lbs/day
Question:
If the production of card is 8565lbs/day, doffer diameter is 27”, grains/yards are 65, Tension draft is
1.03, and the efficiency of card is 85% then calculate the doffer speed if there are 8 cards in
department.
Data:
Production (lbs/day) =8565lbs/day
Doffer dia= 27”
Tension draft = 1.03
Sliver weight = 65 grains/yds
Efficiency %= 85
No. of cards = 8
Doffer speed= ?
Solution:
Since we know the production formula
Card production (lbs/day)
doffer surface speed*grains/yds*60*8*3*efficiency %age*no.of cards
=

36*7000
doffer speed*dia*3.14*grains/yds*60*8*3*efficiency %age*no.of cards
=
36*7000
So,
card production*36*7000
Doffer speed =
dia*3.14*grains/yds*60*8*3*efficiency %age*no.of cards
8565*36*7000
Doffer speed =
27*3.14*65*60*8*3*.85*8
Doffer speed = 39.9rpm
Approximately 40rpm
Q. 65 bales of cotton are feed in 2 scutcher daily; each bale contains 165 kg cotton. Waste
%age through out the b/r line is 6.5%. Doffer speed is 30 rpm. Tension draft is 1.07.
Production effy is 88%. Sliver wt, per yard is 65 grain. Calculate the total no. Of carding m/c
required per scutcher for proper balance of production. Waste %age of card is 6%.
Solution:
B/R INPUT = NO.OF BALES *WT, OF ONE BALE
=65*170
=11050 Kg
B/R WASTE = 11050*6.5/100

= 718.25 Kg
OUT PUT OF B/R = 11050-718.25 Kg
= 10331.75 Kg
PRODUCTION PER SCUTCHER = 10331.75/2
= 5165.875 Kg
NOW PRODUCTION OF CARD MACHINE = n*π*d*60*24*W*T.D*effy/36*7000
= 3.14*27*30*60*24*65*1.07*.88/36*7000
= 889.52 lb/day
INPUT REQUIRED FOR CARD = 889.52*100/94
= 946.29 lb
NO. OF CARDS REQUIRED FOR ONE SCTHURE = 5165.875*2.2046/ 946.29
= 12 CARDS
Q. IF SPEED OF DOFFER IS 40RPM WITH .125Nec COUNT AND EFFY IS 88% CALCULATE CARD
PRODUCTION AND ALSO CALCULATE NUMBER OF SCTHURES REQUIRED TO MEET CARD
PRODUCTION, IF WASTE %AGE OF B/R IS 6.5% AND OF CARD IS 6%? TENSION DRAFT IS 1.07.
IF SHELL ROLLER RPM IS 12. AND OUNCES PER YARD IS ALSO13.
SOLUTION:-
PRODUCTION OF CARD MACHINE = n*π*d*60*24*T.D*effy/36*840*ct
=3.14*27*40*60*24*1.07*.88/36*840*0.125
=1064.38 lb
card input = 1064.38*100/94
= 1132.32lb
PRODUCTION OF 15 CARDS = 1132.32*15
= 16984.8 lb
BLOW ROOM PRODUCTION = n*π*d*60*24*Wt,*effy*no. of scuthres/36*16
16984.8 = 3.14*9*12*60*24*13*.88*no. of scthures/36*16

REQUIRED NO. OF SCTHURES = 1.99 ~ 2

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