Counting of reed count

• Reed determines the spacing of warp threads. normal reed used for weaving cotton materials is created of undersealed steel wires. they’re sure at every endways balks by pith bands, whose thickness determines the spacing of dents. generally brass wires square measure used rather than undersealed steel. reckoning of reed count and heald count in weaving can discuss below:

Required Count,         Dents Per in. within the famous system = …………………………………………………………. × count in a famous system Dents Per in. within the needed system

• REED COUNT choice IN WEAVINGBefore choosing the reed count for a given cloth piece of material, it’s necessary to understand regarding role of reed in weaving loom.
• The reed of a loom performs beating motion within the loom.
• The reed additionally decides the ends per in. within the cloth to be woven .

It additionally helps to keep up regular gap between 2 adjacent ends. “The range of dents per 2 inches is termed reed count available port system”. this method is extensively employed in weaving business. Now we have a tendency to come back on the strategy of reed choice in weaving. Before choosing the count of reed to be accustomed weave a given cloth, following cloth parameters square measure analyzed precisely:

• Dent ,arks and denting order showing within the fabric:
• A gap between ends left by dent wire of the reed within the cloth throughout weaving is termed dent mark. it’s clearly with the assistance of reckoning shut in the gray cloth. It gets very little tough to envision the dent mark within the cloth once process. AN skilled technician will see these dent mark in gray and processed cloth each with the assistance of reckoning glass. These dent marks decide the denting order of warp ends within the reed.
• Weft Regain This vital parameter of the material that plays a decisive role in creating choice of correct reed count in weaving. Technician ought to be terribly careful throughout thread regain analysis. “The thread regain is that the quantitative relation of distinction between straighten thread length and piece of material length to piece of material length expressed within the term of percentage”. The following 2 tables are used throughout reckoning Reed & Heald’s count in weaving. A system supported the amount of dents during a given house.

Name of System	Basis Of numbering
Stockport	Number Of dents per 2 inches
Radcliff	Number Of dents per 1 inch.
Huddersfield	Number Of dents per 1 inch.
Metric	Number Of dents per 1 decm.

The system based on the number of groups or beers in a given space.

Name of System	Basis Of numbering
Bolton	20 dents per 24.5 inches.
Bradford	20 dents per 36 inches.
Blackburn	20 dents per 45 inches.
Irish	100 dents per 40 inches.
Leeds	19 dents per 9 inches.
Macclesfield	100 dents per 36 inches.

All the above systems are convertible into dents per inch basis. Measuring the “Linear density” of yarn Yarn number ,density, decitex, cotton count and other terms are ways of expressing the linear density or weight  per unit length, of the yarn. For different types of yarn and for different fibers, there  different are different ways of expressing this property. Cotton Yarn Numbers  In general used to describe spun yarns, the english yarn number or english cotton yarn number indicates the number of 840 yard lengths in a pound of yarn. The lower the cotton yarn number, the heavier the yarn. following is the conversion from english yarn numbers, commonly provided on invoices, to metric yarn number. English yarn number*1.6933= Metric yarn number =The number of 1000 meter lengths in a kilogram of  yarn   the yarn number is usually shown in conjunction with the number of plies, for plies yarn, we  consider the yarn number of the individual signals yarns that make up a plied yarn, and convert it to metric.  Denier denier indicates the weight in gram of 9000 meters of yarn. Like the cotton yarn number ,it is an expressed of liner density for a yarn. But unlike the cotton yarn number the higher the denier is the heavier is the yarn. following is the conversion from denier, commonly given on invoices, to decitex Decitex=1.1111* denier  Tenacity Tenacity is the amount of force needed to break a yarn,devided by tbe denier,decitex,or some other measure of weight per unit length.  some common conversion related to tenacity Grams/denier*8.827=centinewtons per tex conversion factrors weight 1 kilogram=2.2046 1 ounce=28.35 Force 1 pound force=4.448 Newtons 1 kilogram force=9.807 newtons 1 kilogram  force=2.2046 pound force  Distance 1 inch= 2.54 centimeters 1 yard=0.9144 meter 1 kilometer=0.6214 miles Density NM=10000/Decitex NM=9000/Decitex Decitex=10*Tex Decitex=10000/NM Decitex= 5900/NE Decitex=1.111111*denier Decitex=14880000/Number of feet per pound Tenacity 1 gram/denier=8.827 centinewtons/tex 1 millinewton/decitex=1 centinewton/tex millinewtons/decitex=Centinewtons per tex LENGTH AND WEIGHT CONVERTION 1 yard  = 3.0 feet               = 36.0 inches 1 meter = 100 centimeter                = 3.281 ft                = 39.37 inches                = 1.0936 yard 1 foot      = 0.3333 yard                 = 12.0 inches                 =  30.48  centimeters                 =  0.3048 mt centimeters =10.0 millimeters                          =0.01 meter                          =0.3987 inches                           =0.328 foot 1 inch= 0.0278 yds             =0.0833 foot              =25.4 millimeter              =2.54 centimeter               =0.0254 mt 1 kilometer=1000 mt                       =909.363 yards 1 mile=1760 yard               = 1609.3 mts               =1.6093 km 1 pound=7000 grains                 =453.59 gms                 = 0.45 kg 1 kg=2.2 pounds         = 35.274 ounces           =15432.0 grains             =10o0 grams 1gm=15.43 grains          =0.03527 ounces            =1000 milli grams 1 quintal=100 kg                   =2204.6 pound 1 ton=  1016 kilograms 1 candy= 355.62 kg               Example No. 01: Find the count of 10s Irish reed into the Hudders  field System. Solution: 10s Irish, ….10 × 100 = ……………. ………40 = 25 dents per inch. 1 Huddersfield = 1 dent per inch. Therefore, 10s Irish, ……25 = ………. ……..1 = 25s Huddersfield. So, the count in Huddersfield System is 25s Example No. 02: Find the count of a 40s Radcliff in the Stockport System. Solution: Count, …..40 × 2 = ………….. ………1 = 80s Stockport. So, the count in Stockport System is the 80s Example No. 03: Find the number of ends per inch in a reed of 3/72s Bradford. Solution: 72s Bradford, ….72 × 20 = ………….. ends per inch ……….36 = 40 ends per inch. Therefore, the number of ends per inch = 3 × 4 = 120 ends per inch. Counting of heald count:  The number of heald eyes per inch across the healds in a set expresses the count of the healds. When a set contains 4 shafts, it is called a plain set. Example No. 04: Find the count of the healds that will be required for weaving a 6 shaft satin fabric using 72s Stockport reed, drawn 3 ends per dent. Solution: Number of ends per inch in the reed, ….3 × 72 = …………. ……..2 = 108 ends. Therefore, the rate of knitting, …..108 = ……… …….6 = 18 healds per inch on the ribband. Thus we require 6 healds of 72s plain set. Example No. 05: If 3 meter cloth is costing 25 Rs,how much will be the cost of 15 meter cloth. Solution: 3M     15M         25 RS  Cost of 15 meter cloth= 15* 25                                           ————–                                                   3                                           =  125 RS Example No. 06: I brought 15 textile books costing Rs 30, during the travel 3 are spoiled. Then how much the cost of 1 product? Solution: 12 products  are brought,3 aer spoiled ,then 15-3= 12 are in good condition. Now 9 products are costing me at Rs 30,how much is the cost of 1 product. 12 p                          1p                     30 Rs  Cost of 1 Product           1*30                                   =     ———                                                 12                                              =      2.5 Rs Example No. 07:  I brought 8 hooks from 1 shop at Rs 32,from other shop I brought two more which cost me only Rs 4.Then how much is the average cost of 3 hooks/? Solution:  I got total 10 hooks (8+2) Costing Rs 36(32+4) then the cost of 3 hooks  10 Hooks                             3 hook            36 Rs     cost of 3 hooks                       3* 36                                    =     ———————–                                                        10                                     =  10.8 Rs 

Calculation of Loom Production

 What is Loom? The loom is an apparatus/device for making fabric by weaving yarn or thread. It produces fabric by interlacing a series of length-wise, parallel yarns width a series of width-wise parallel yarns. To get the optimum output of the loom we must know about the loom production calculation. In this article, I have given many examples of the calculation of loom production.

Calculation of Loom Production

Problem-1: Production of the loom: PPM = 200, PPI = 50 Production per loom per minute =200/50 =4” Production per loom per month = 4 x 60 x 24 x 30 inch Production per loom per year at 70% efficiency = 4 x 60 x 24 x 30 x 12 x 0.7 inch = (4 x 60 x 24 x 30 x 12 x 0.7) / 36 yds = (4 x 60 x 24 x 30 x 12 x 0.7) / 39.37 m = 36868.68 m (ANS) Problem-2: PPM = 350 PPI = 60 No. of loom = 45 Time = 7 days Efficiency = 90% Find out production in m? Solution: …………………..350………………………………..90………..1 Production = …….. x (60 x 24 x 7) x 45 x …….. x ……….. m ……………………60………………………………..100……39.37 ….14288400000 = …………………… m ………236220 = 60487.68 m (ANS) Problem-3: PPM = 650 PPI = 58 No. of loom = 10 Time = 35 days Production = 110000 m Efficiency = ? Solution: ……………………..650 x 60 x 24 x 35 x 10 x η Production = …………………………………………… …………………………….58 x 100 x 39.37 ……………110000 x 58 x 100 x 39.37 => η = ………………………………………… ………………650 x 60 x 24 x 35 x 10 => η = 76.77% (ANS) Problem-4: Find out Wt. of yarn in fabric (warp & weft), from the following specification. …Warp count x Weft count .……………………………………….x Fabric width ……………..EPI x PPI or, ..20 x 22 .………….. x 56” ..60 x 58 Solution: In 5000 m fabric, wt of weft warp, ………60 x 56 x 5000 x 1.05 = ………………………………………. kg …..0.914 x 840 x 20 x 2.2046 = 521.09 kg (ANS) In 5000 m fabric, wt of weft, …..58 x 39.37 x 5000 x 56 x 1.05 = …………………………………………… Kg ……….36 x 840 x 22 x 2.2046 = 457.73 kg (ANS) Problem-5: Find out the wt. of required warp & weft to produce 5000m fabric from the following specification. …18 x 16 .…………… x 58” …66 x 62 Solution: ……………………………66 x 58 x 5000 x 1.05 x 1.1 Required warp = ………………………..……………………  =  725.28 kg (ANS) …………………………….0.9144 x 840 x 18 x 2.0246 …………………………..62 x 39.37 x 5000 x 58 x 1.05 x 1.1 Required weft = ……………………………………….……………….. = 766.49 kg (ANS) ……………………………………36 x 840 x 16 x 2.2046 Problem-6: Calculate the yarn consumption of a loom per hour running at 180 PPM for producing from the following specification. ….40 x 60 ………………. x 56” fabric ….72 x 48 Solution: …………………………………….180…………..1 Production/hour/loom = ……. X 60 x ……. ..……………………………………48…………..36 = 6.25 yd ………………………………….72 x 56 x 6.25 x 1.05 x 1.1 Req. amount of warp = …………………………………….. …………………………………………840 x 40 x 2.004 = 0.393 Kg ………………………………….48 x 36 x 6.25 x 56 x 1.05 x 1.1 Req. amount of weft = ……………………………………………… ……………………………………….36 x 840 x 60 x 2.024 = 0.175 Kg (ANS) Problem-7: Calculate the time required to complete a weavers beam having 1000yds of warp on it. The woven cloth is required to have 60 PPI. Loom speed is 200 and efficiency 80%. Assume any missing data. Solution: …………………………………….210….60 Production/loom/hour = ……. x ….. x 0.8 ……………………………………..60…..36 = 4.67 yds …………………….Length of cloth from the beam Time req. = …………………………………………………… …………………………..Actual production/hr Length of cloth from the beam, ……………………………………100 – crimp = (Length in yds of warp x ……………….) – Waste in yards …………………………………………..100 ………………100 – 6 = (1000 x …………….) – 6 …………………100 = 934 yds ……………………………..934 So, Time required = ………. ……………………………..4.67 = 200 hr (ANS) Problem-8: Calculate the actual yarn consumption per month of weaving mil having 100 looms of 250 ppm for the following fabric construction. Considering each point. …80 x 60 ……………..X 56” ..110 x 54 Production per month of 100 looms at 100% efficiency, ….250………………………….100 = …….. x (60 x 24 x 30) x ……. …..54…………………………….36 = 555556 yds ……………………………………….110 x 56 x 555556 x 1.05 x 1.1 Required amount of warp = ………………………………………… ……………………………………………..840 x 80 x 2.204 = 26687 Kg ………………………………………….54 x 36 x 555556 x 56 x 1.05 x 1.1 Required amount of weft = ………………..………………………………….. …………………………………………………36 x 840 x 60 x 2.204 = 17468 Kg (ANS) Problem-9: Find out thrust in Newton to drive a loom by using plate clutch system and conical clutch system. If F=425N, μ=0.45 and θ=20° Solution: ………………………………F………425 For plate clutch, P = …….. = ………. = 944.44 Newton ………………………………µ………0.45 …………………………………………..F sin θ……425 sin 200 For conical clutch system, P = …………. = ………………… = 323.02 Newton (ANS) ………………………………………………..µ…………….0.45 Problem-10: Calculate the production per shift of 10 hrs at 90% efficiency a weaving mill in average 22 picks per cm from the following information: No. of looms ………..picks/cm ………PPM …120 ……………………20 ……………….220 …16 ……………………..25 ………………..210 …110 ………………….. 28 ………………. 200 ………………………………………….220…………………….90………………1 Production at 22 picks/cm = ……… x (60 x 10) x …….. x 120 x ……. M …………………………………………..20…………………….100…………….100 = 6480 m ………………………………………….220…………………..90…………..1 Production at 22 picks/cm = ……… x (60 x 10) x ….. x 16 x ……. ……………………………………………20…………………..100…………100 = 824.73 m …………………………………………220…………………….90…………….1 Production at 22 picks/cm = ………x (60 x 10) x ……. x 110 x ……. …………………………………………..20……………………100………….100 = 5400 m Average production = (6480 + 824.734+5400) m = 12704.73 m (ANS) Problem-11: Calculate the “speed space” required to produce a elastic fabric having 30% shrinkage (width of finished fabric 50″) ……………………………………………..130 Required reed space = 11 x 50 x …….. ……………………………………………..100 = 65 inch (ANS) MCQ Technical details MCQ visit link face book whats up Instragram