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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.

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Oven dry weight:

The weight of material dried at oven state. The temperature in the oven is 1050C.

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

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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.

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=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%.

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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)

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= 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

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= 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?

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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

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= 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

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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

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

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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

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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

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% 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

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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

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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.

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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

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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)

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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).

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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

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= 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

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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

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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

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 * rpm * 13 *60 *3 * 8* 3* 0.85

36 * 16

30000 = 9 *3.14 * rpm * 13 * 60* 3 * 8* 3 *0.85

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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.

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 * 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

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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

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= 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%.

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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

(SPINNING CALCULATION-1)

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

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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

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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

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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

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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)

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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

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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

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CARD CALCULATIONS

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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 = 1.11D2 = 120*40*27/18*40*2.5 = 72D3 = 216*4/30*27 = 1.067D4 = 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 = 85M.D = A.D * (100-W/100) = 85 * (100-5/100) = 8.75New 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

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= 28274.3/3937 = 7.18New prod with 5% increase = 24.3 * 1.05 = 25.52 kg/hr25.52 = 0.001953*Barrow wheel * 84/0.135B.W = 25.52*0.135/0.001953*84 = 21T

Production = 53.55 lb/hr53.55 lbs = 60 min40 lbs = 60*40/53.55 = 45 min

Density = mass/volumeVolume = 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

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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

=

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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

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= 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

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REQUIRED NO. OF SCTHURES = 1.99 ~ 2

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