Calculation

CONVERSION FACTORS 1 HP = 1 HP = 1 HP =

33,000 ft.lbs/min 746 Watts 42.4 BTU/Min.

=

550 ft lbs/sec

Amps -------------------------746 (---------------------------- ) Volts x decimal efficiency

Electric HP =

=

Amps -------- ( 120 V) @ 86 % efficiency 7.2

=

Amps -------- ( 240 V) @ 86 % efficiency 3.6

=

Amps -------- ( 550 V) @ 75 % efficiency 1.8

H2O = 62.4 lb/ft3

=

8.34 # / gal

1 Imp.Gal = 4.546 liters 1 lb -------3000 ft2

F C

=

grams 1.64 ---------Meter2

9 = ---- C + 32 5 5 = (F – 32) x ----9

1 SANTOSH SWAR

BASIS WEIGHT CONVERSIONS Offset (3300 ft2) x 1.48 Bond (1300 ft2) x 3.76 Liner (1000 ft2) x 4.89 News (3000 ft2) x 1.63

= = = =

GSM GSM GSM GSM

CONVERSION FACTORS FROM lb in in in ft/min (fpm) cfm cfm/in cfm/in2 cfm/in2/1000 fpm oz oz/ft2

TO OBTAIN kg mm cm m m/min (mpm) m3/hr m3/hr/cm m3/hr/cm2 3 m /hr/cm2/100 mpm g g/m2

MULTIPLY BY 0.454 25.4 2.54 0.0254 0.305 0.589 0.232 0.91 0.278 28.3 305.6

1) M/C PRODUCTION CALCULATION Let

M/c M/c

speed in Deckle in

mt/min mt

SPEED x DECKLE x GSM x 60 2 (mt/min) (mt) (Gm/mt ) (min/hr) M/C PROD = -----------------------------------------------------(Tons/Hr) 1000000 (Gm/Ton)

M/C PROD (Tons/hr)

=

Speed x Deckle x GSM x 60 ---------------------------------------106

EXAMPLE Let

M/c Speed Deckle GSM

= = =

200 mt/min 3.2 mt 50

2 SANTOSH SWAR

. . . M/C PROD =

=

200 x 3.2 x 50 x 60 ------------------------106 1920000 -----------106

= 1.92 Tons/hr

M/C Production in kg units Speed x Deckle x GSM x 60 M/C Prod. = ------------------------------------(kg/hr) 103 200 x 3.2 x 50 x 60 = --------------------------103 1920000 = ----------103 = 1920 kgs/hr

2)

REAM WEIGHT CALCULATION REAM WEIGHT = PAPER SIZE x GSM x NO.OF SHEETS (Gm) = Length x Width x GSM x 500 (Let) (Mt) (Mt) GM ( ----- ) mt2 Length x Width x GSM x 500 (cm) (cm) GM ( ----- ) m2 REAM WEIGHT = ---------------------------------------------(kg) 107 EXAMPLE : Let

Paper size GSM No.of sheets

= = =

40 cm x 60 cm 50 500

3 SANTOSH SWAR

Ream Weight

=

40 x 60 x 50 x 500 -------------------------107

60000000 = ---------------107 = 6 kg

3) CALCULATION OF GSM Any samples of any size can be expressed in terms of GSM.

GSM =

Where,

W l b

10000 x W ----------------l x b = = =

Weight of paper in gms. Length of paper in cms. Width of paper in cms.

EXAMPLE : Let a paper sample of dimensions 10 cm x 15 cm That means

l = 10 cm B = 15 cm Weight of this sample, W = 0.72 gm

GSM =

10000 x 0.72 --------------------10 x 15 7200 = ------150 = 48 GSM

Second Expression 10 cm x 15 cm size of paper has 0.72 gms 4 SANTOSH SWAR

i.e. 10 cm x 15 cm = 0.72 gms 150 cm2

=

0.72 gms

1 cm2

=

0.72 -----150

=

0.0048 gms

=

0.0048 gms

=

0.0048 gms

1 cm2

1 ---------- m2 10000

. . 1 m = 100 cm . 1 m2 = 10000 cm2 1 2 1 cm = ------ m2 10000

1 m2 = 48 gms TEMPLET EXPRESSION

25 cm 20 cms

Area of templet size sample = 2 x 25 cm x 20 cm = 1000 cm2 Let

weight of this sample is 5.8 gms 1000 cm2 = 5.8 gms 1 cm2

=

1 cm2 1 --- m2 104 1 m2 1 m2

=

5.8 -------- gms 1000 0.0058 gms

=

0.0058 gms

= =

0.0058 x 104 gms 58 gms

5 SANTOSH SWAR

4) THEORETICAL HEAD V ----100 -------------K

Theoretical Head =

Where, V K Head K

5)

Spouting velocity (fpm) Constant (see table)

In of H2O 1.932

Ft of H2O 23.184

In of Hg 26.345

Pressure PSIG 53.623

In of Hg 513.3

Pressure PSIG 732.3

SPOUTING VELOCITY V Where, V h k Head K

6)

= =

2

= k = = =

h Spouting velocity (fpm) Theoretical head Constant (see table)

In of H2O 139.2

Ft of H2O 481.5

HEAD CALCULATION V2

=

2 gh

V2 h = ------2g Where, h = Height of stock in head box V = Velocity (M/c speed in m/min) g = Acceleration due to gravity (9.81 m/sec2)

6 SANTOSH SWAR

EXAMPLE Let M/c speed V = 200 m/min V2 h = ------2g m2 200 x 200 -----min 2 ------------------------2 x 9.81 m -----Sec2 40000 m x sec 2 ------- -----------19.62 min2

=

=

=

m x sec2 2038.74 ------------3600 sec2

=

2038.74 ----------3600

= =

0.5663 m 56.63 cm

. . 1 min2 . = 1 min x 1 min = 60 sec x 60 sec 3600 sec2

m

Short Method 200 x 200 ------------2 x 9.81

m2 ------ x min2

=

200 x 200 ------------2 x 9.81

m x sec2 ------------min2

=

200 x 200 -------------2 x 9.81

100 cm x sec2 -----------------3600 sec2

h =

sec2 ------m

7 SANTOSH SWAR

=

200 x 200 x 100 ---------------------- cm 2 x 9.81 x 3600

=

200 x 200 ----------- cm 706.32

=

. .

100 . --------------------- = 706.32 2 x 9.81 x 3600

56.63 cms.

QUICK FORMULA Speed x Speed h = -------------------(in cms) 706.32

Where h in cms Speed = in m/min.

EXAMPLE M/c Speed 1) 180 M/min ,

180 x 180 h = -------------- = 45.87 cms 706.32

2) 200 M/min ,

h =

200 x 200 ------------ = 56.63 cms 706.32

3) 250 M/min ,

h =

250 x 250 ------------ = 88.49 cms 706.32

7) FORMING LENGTH GUIDELINES Dwell time in seconds between head box slice and first flat box or dandy roll:    

Wire speed < 12 fpm : 1.5 – 2.0 seconds multiply forming length in feet by 40 (1.5 sec) or 30 (2 sec) to determine M/c speed that can be supported conventional drainage table. Wire speed > 1200 fpm : 1.0 seconds. Multiply forming length by 60 to obtain M/c speed potential. 42 lb. Liner : 1.25 seconds. Multiply forming length by 48 to obtain M/c speed potential. Foodboard : 2 seconds. Multiply forming length by 30 to obtain M/c speed potential.

8 SANTOSH SWAR

8) JET VELOCITY CALCULATION (Stock speed at slice) Jet Velocity ______ V = Cv  2 gh Where, V Cv g h

= = = =

Stock speed at slice Coefficient of velocity discharge Acceleration due to gravity Head of stock

If h is measured close to the slice and V is measured at the vena-contracta of the Jet, Cv is approximately 1.0 for most slices. ______ V =  2 gh Friction losses will reduce Cv possibly to around 0.98.

EXAMPLE Let head of stock

h = 57 cms

Then Jet Velocity V

______________________ =  2 x 9.81 m x 57 cm -----Sec2 ______________________ =  2 x 9.81 x 57 m x cm -----Sec2

V

____________________________ =  57 x 2 x 9.81 m x 1/100 m -------------1/3600 min2

( . . 100 cm = 1 m

1 1 cm = ------- M ) 100

9 SANTOSH SWAR

Similarly 3600 sec2 = 1 min2 1 Sec2 =

1 ------ min2 3600

______________________ =  57 x 2 x 9.81 x 3600 M2 ------ ---100 Min2 ______________________ =  57 x 2 x 9.81 x 36 M2 ---Min2 __________________ =  57 x 706.32 M2 ---Min2 ______________ =  40260.24 M2 ---Min2 M = 200.6 -----Min.

QUICK FORMULA

Jet Velocity (In m/min)

____________ =  Head x 706.32 (in cms)

EXAMPLE 1) h = 46 cms,

2) h = 89 cms,

____________ V =  46 x 706.32 _________ =  32490.72 = 180.25 M/min ___________ V =  89 x 706.32 _________ =  62862.48 = 250.72 M/min

10 SANTOSH SWAR

9) CRITICAL SPEED OF CALENDAR ROLL (fpm) Ro ___________ C.S. = 4.12 x 106 x -----  Ro2 + Ri2 L2 Where, C.S. = Critical speed (fpm) Ro = Outside radius (inches) Ri = Inside radius (inches) L = Centerline to centerline bearing (inches) (assume L = face + 40 inches)

10) RETENTION 1)

Net consistency Retention % = -------------------------- x 100 Head box consistency

Head box consistency - Tray consistency 2) Retention % = ----------------------------------------------------- x 100 Head box consistency

3) Overall Retention =

Filler in sheet ----------------------------- x 100 % Filler added to furnish

Filler in Sheet 4) First – Pass Retention = ----------------------- x 100 % Filler in head box

11) FLOW RATE OF SLICE Q

=

= = =

Av V ,

Where Av = Area of cross section at Vena-contracta Av = Ca As Ca = Coefficient of contraction As = Areas of cross section at the Slice opening. _____ Ca As Cv  2 g h _____ Ca Cv As  2 g h _____ Cq As  2 g h where, Cq = Coefficient of volume discharge. ______ Q = Cq As  2 g h 11 SANTOSH SWAR

12) FLUID VELOCITY Velocity =

GPM x 0.321 -----------------A

Where, Velocity in fps A = Area in (inches 2) NOTE : This formula is for savealls and general paper flow, since there is no orifice Coefficient included.

13) ROLL SPEED RPM =

Where,

3.82 ( V ) ------------Do

RPM = V = Do =

revolutions per minute Speed (fpm) Roll outside diameter (inches)

14) FORMING BOARD SETTING 12 V COS A _______________ ----------------- (  V2 Sin2 A + Zyh - V Sin A) g _____________ = 0.37267 V COS A (V2 Sin2 A + 64.4 h - V Sin A)

X =

Where, X V A g h

= = = = =

Distance of slice to lead forming board blade Initial Jet Velocity Jet angle 32.2 ft/S2 Jet to wire height

12 SANTOSH SWAR

15) WATER REMOVAL BY A TABLE ROLL 1st Method DU K q = -------F2 Where, q = D = U = F =

K

=

Water removed by a table roll per unit time & width Diameter of roll Wire speed A drainage factor (proportional to basis weight) determined by the Sag of wire, air content, thickness and porosity of mat, stock freeness, head box consistency, degree of flocculation and evenness of formation. Exponent defining the effect of speed on drainage, characteristics of type and quality of pulp (varies between 0.3 – 1.2).

Second Method STOCK

h B

Wire

A

C R

Extracted water

Table Roll (mechanism of water extraction) According to Mr.Cowan, the quantity of water that is being removed from the wire from A to B is equal to 4 K2 R gh = ------------V Where

K

=

R g h V

= = = =

Drainage coefficient (dependent on the sheet weight & type of wire) Radious of the roll Acceleration due to gravity Head of stock suspension above the wire Velocity of the wire

13 SANTOSH SWAR

16)

VACUUM PUMP CAPACITY (CFM) PV

=

nRT

Where, P V n T R

= = = = =

Absolute pressure, lb.ft2 = (Psi gauge + 14.7) x 144 Total gas volume, ft3 Weight of gas, lbs Absolute temperature  R = F + 460 Gas constant, lbf x ft / (lbn x  R ) Ra (air) = 53.3 Rw (water)vapour = 85.8

P1V1

=

P2V2

V2

=

29.92 - P1 (  Hg) ----------------------- x V1 (CFM) 29.92 - P2 ( Hg)

or for temperature cooling effects : P1V1 ------T1

=

P2V2 ------T2

17) DRYER SURFACE REQUIRED CALCULATION For normal types of dryers, the following empirical formula can be used for obtaining a rough value of dryer surface required. (Exact value will depend on the quality of paper and constructional details etc.)

L

SWd = K ------(t – 100)

Where, L

=

Peripherial length in meters of dryers in contact with paper During drying. 14 SANTOSH SWAR

K S W

= = =

d t

= =

Constant value around 0.05 Speed of M/c in M/min Basis weight of paper in gm ---m2 Thickness of dryer shell in centimeters Temperature of ingoing steam in C

EXAMPLE Let

M/c speed S = 180 M/Min GSM W = 50 Dryer shell thickness d = 2.2 cm t = 150 C

L =

0.05 x 180 x 50 x 2.2 -----------------------------(150 – 100)

=

990 -----50

=

19.8 meters to be required for paper drying

18) HEAD BOX FLOW RATE (GPM/inch) GPM / inch = S.O. x V x 0.052 x C Where V S.O. C

= = =

Spouting velocity (fpm) Slice opening (inches) Orifice coefficient (see table for approximate values)

Type Nozzle A B

C 0.95 0.75 0.70 15 SANTOSH SWAR

C

Type

0.60

A) Low angle, Converflo

ß

Type B) High angle

ß Type C) Straight

ß

19) NO OF DRYERS REQUIRED CALCULATION GIVEN DATA (TPD, Sheet width, Dryer diameter) Production Rate = 40 TPD Sheet width (to dryer) in inches W = 200 inches

W

=

200 ----- = 16.67 foot 12 16 SANTOSH SWAR

Dryer diameter, d = 60 inches 60 = ---- 5 foot 12 Moisture to dryer = 37 % dry Moisture to reel = 94 % dry

Hourly production

=

400 -----24

Tons/hr

400 Convert it into lb/hr, ------- x 2000 = 33333 lb/hr 24 ( 1 US Ton = 2000 lbs)

Now water to be removed 94 33333 x ( ---37

__

1)

Dryer surface required @ 2.8 lb water/hr/ft2 Evaporation rate Required dryer surface

=

51350 -------2.8

= 18339 ft2

Now Area of single dryer surface,  dw 22 = --- x 5 x 16.67 = 261.95 ft2 7 = 262 ft2/dryer No.of dryer required 18339 -------262

= 69.99 = 70 dryers

20) PRESS IMPULSE 5 x PLI 17 SANTOSH SWAR

PI = --------Speed Where, PI = PLI = Speed =

Press impulse (Psi – Sec) Nip pressure (Pli) Nip speed (fpm)

21) TORQUE Tq = Force x Radius Where Tq = Force = Radius =

Torque (inch – pounds) in pounds in inches

22) WR2 OF A ROLL WR2 = (0.000682) (W) (L) (Do4 – Di4) Where, WR2 W L Do Di

= = = = =

in (lbs – ft2) Density (pounds/inches 3) Length (inches) Outside diameter (inches) Inside diameter (inches)

23) STOCK FLOW THROUGH THE PIPE CALCULATION Q

=

Vrc

Where, Vrc Q d

24)

= = =

1 ----4

=

 d2 x Vrc

4Q ---- d2

Velocity of flow Total quantity Pipe diameter

IDEAL DRAINAGE IN WIRE PART a) Theoritically after forming board the drainage should be 80 to 85 % of stock thickness ( i.e. Slice opening). 18 SANTOSH SWAR

b) At half of the forming zone, it should be 40 % of slice stock thickness c) Before dandy it should be 20 to 25 % of slice stock thickness

25) STOCK THICKNESS ON FORMING FABRIC Basis weight T = -----------------------------------J Consistency x R x ( ---- ) W Where, T

=

R

=

Thickness of stock on table in cm. Basis weight in g/cm2 % Consistency in --100 Retention from that point down the rest of the machine

J/W

=

Jet/Wire ratio = 1.0 except at slice

i.e., overall retention of a machine with slice opening of ½  making 50 gsm at 0.6 % slurry and Jet/Wire ratio of 0.95.

R =

0.0050 -------------------------------( 0.0060 ) x 1.6 x 0.95

= 73 %

26) CALCULATION OF WIRE LENGTH Ls = 2 l +

Where, Ls l D1 D2 K EXAMPLE L D1

 ---- (D1 + D2) + K 2

= = = = =

Length of wire (in mm) Distance between center of breast roll to couch roll (in mm) Diameter of breast roll (in mm) Diameter of couch roll (in mm) A constant, 130 mm for fourdrinier wire part.

= =

12,500 mm 450 mm 19 SANTOSH SWAR

D2

=

Then Ls

800 mm = = = = =

 2 x 12500 + ----- (450 + 800) + 130 2 25000 + 1.571 (1250) + 130 25000 + 1963.75 + 130 27093.75 mm 27.093 mtr.

27 a) DRAG LOAD

The term drag load resulted from the necessity of fabric manufacturers to monitor the power used to drive the fabric. It is a measure of the increase in tension ( T) of the fabric as a result of the suction forces pulling the fabric against the foil surfaces, the iovac surfaces & the hivac surfaces. T+ T

T

Suction Couch Drive Power = VOLT x AMP

DRAG LOAD =

0.8 ---------1000

VA ------UW

Kilonewtons ----------------Meter

Where V A U W

= = = =

Volt AMP Fabric speed (M/S) Fabric width (M)

lb { To convert in ------ , in

KN ------- x 5.71 M

=

lb -------- } inch

b) DRAG LOAD - CONVENTIONAL V x A x 0.8 DL = -----------------------0.0226 x U x W

DL in Pli

Wehre, V

=

Drive volts (V) 20 SANTOSH SWAR

A U W

c)

= = =

Drive AMPs (AMPS) Nominal fabric speed (fpm) Nominal fabric width (inches)

DRAG LOAD CALCULATION Safe drag load is 10 – 12 HP/Meter width of the wire/100 m/min wire speed If it is beyond 15 HP then it is alarming

VOLT x AMP x 49 (Constant) DRAG LOAD = -------------------------------------------(In kg/cm) WIRE WIDTH x WIRE SPEED (in mm) (in m/min) Volt x AMP

= WATT

746 WATT = 1 HP

28) DRAG LOAD – BETWEEN COMPONENTS IN THE FABRIC Vn DL = (---Vs

__

1)

(EM + Ts)

Where, DL Vn Vs EM

= = = =

Drag load (Pli) Fabric speed at point n in fabric run (fpm) Fabric speed on slack side of fabric run (fpm) Fabric elastic modulus (young) at temperature

EMr

=

T ~ EMr – KT Elastic modulus at reference temperature r (Pli)

K

=

Modulus ------------------------Temperature constant

Ts

=

Slack side tension (Pli)

Pli (----- ) ºF

21 SANTOSH SWAR

29) DANDY DIAMETER CALCULATION 1) Open type 15 – 20 % of wire width

2) Journal typs 10-15 % of wire width

Below 240 m/min dandy of diameter equal to 10 % wire width may be used. At higher speed the diameters should be more because with very high number of revolutions it throws water causing damage to the web. For Wove Dandy M/C Speed ------------------------------------- x Maximum number of revolutions

Dia of Dandy =

V D = -------n Where,

D V n

= = =

Diameter in mm M/c speed in m/min Maximum number of revolution for a wove Dandy & it should be taken as 150 rev/min.

V D = --------------3.142 x 150 V D = ------477

EXAMPLE Let M/c speed 400 D = ------477 Wire Speed (m/min) Dia of dandy (in mm)

V = 400 m/min

= 850 mm

80 300

150 400

250 500

300 600

400 800-1000

22 SANTOSH SWAR

30) DANDY ROLL REVOLUTION PER MINUTE Wire Speed (fpm) RPM = --------------------------------------- x Dandy roll diameter (ft) Where  = 3.142 RPM Target = 125 – 150 RPM

31) SIZE PRESS ROLL REVOLUTION PER MINUTE

RPM =

Web Speed (fpm) ---------------------------------------------3.142 x Size press Roll diameter (ft)

Maximum 250 rpm

32)

TONS PER DAY (T/D)

T/D =

Capacity (gpm) x % Bone dry consistency -----------------------------------------------------16.65

33) CENTRICLEANER DESIGN CALCULATION GIVEN DATA - FINISHING PROD = 30 TPD M/C PROD = 33 TPD Ton 33 --------- Convert it into kg/min Day 33 x 1000 = 33000 kg 24 hrs x 60 min = 1440 min

23 SANTOSH SWAR

33000 kg 33 TPD = ------------1440 min

= 22.9 kg/min

Bs factor = 1.45 (68.9 wire retention

100 ------68.9

= 1.45)

Bs factor x prod. 22.9 x 1.45 = 10 % reject = 5 % vent reject=

33.22 3.32 1.66 ----38.20

kg/min kg/min kg/min kg/min

kg/min x 100 38.20 x 100 Now LPM = ------------------ = ------------------ = 4775 lpm Consistency 0.8 Through put/leg

= 500 lpm

No of legs required

=

Primary legs Secondary legs Tertiary legs

4775 --------500 = = =

= 9.55 = 10 legs

10 Nos. 3 Nos. (30 % of primary legs) 1 No. (30 % of secondary legs)

Pressure drop = 1.4 kg/cm2

WEIR FLOW – CONTRACTIONS

34)

RECTANGULAR

WEIR

WITH

END

Q (ft2 H2O/Sec) = 3.33 (L - 0.2 H) H 1.5 Where,

35)

L

=

H a

= =

Length of weir opening in feet (should be 4 – 8 times H) Head on weir in feet ( ~ 6 ft back of weir opening) at least 3 H (end contraction)

WEIR FLOW – TRIANGULAR NOTCH WEIR WITH END CONTRACTIONS 24 SANTOSH SWAR

4 ______ Q = C ( -----) L H  2 gH 15 Where L H C a

= = = =

Width of notch in ft at H distance above apex Head of water above apex of notch in feet 0.57 Should be not less than ¼ L (end contraction)

For 90 notch the formula is : Q

=

2.438 H 5/2

For 60 notch the formula is : Q

=

1.4076 H 5/2

36) WASTED VOLUME OF THE COUCH ∆P Wv = DA x DW x U x t x ---------P Where, Wv DA DW U t ∆P P

= = = = = = =

Wasted volume (in ft3/min. or m3/sec) Drilled Area (in %) Drilled width (in inch or M) Machine speed (in ft/min. or M/minute) Shell Thickness (in inch or cm) Suction Pressure (couch vacuum) in inch Hg) Pressure (in inch Hg)

EXAMPLE

Wv

=

DA DW U t ∆P P

= 50 % = 286 inch = 300 fit/min = 2.5 inch = 24 inch Hg = 30 inch Hg

= or or or or or

=

DA x DW x U x t

50/100 = 0.5 7.26 M 914.6 M/min 6.35 cm. Or 0.0635 M 81.3 K Pa 101.6 K Pa x ∆P ----P

24 inch Hg 0.5 x 286 inch x 3000 fit/min. x 2.5 inch x ---------------25 SANTOSH SWAR

30 inch Hg

=

286 2.5 0.5 x ----------- ft x 3000 ft/min x -------- fit x 0.8 12 12 5948 ft3/min

=

81.3 KPa Wv = 0.5 x 7.26 M x 914.6 M/min x 6.35 cm x ---------101.6 KPa

914.6 M = 0.5 x 7.26 M x ------------ ------- x 60 Sec

6.35 ---------- M x 0.8 100

= 2.81 m3/S

37) COUCH VACUUM EXPANSION VOLUME CFM = (V) (b) (S) (E) (m) Where, V b S E

= = = =

m

=

P2 P1

= =

M/c speed, fpm Roll shell face width, feet Hole depth, feet % Open area of shell P2 Expansion factor, -----P1

0.9

-1

ambient pressure,  Hg absolute Suction box vacuum, Hg absolute 26 SANTOSH SWAR

38)

FORMATION – BLADE PULSE FREQUENCY F

=

V ------5x

Where, F V

= Formation – blade pulse frequency (in cycle/sec) = Wire speed (fpm) = blade spacing, tip to tip (inches)

Optimum frequency for formation improvement “ F > 60 cycle/sec

39) PAPER WEB DRAW Draw (%) =

Where, SF S1

SF – S1 -------------- x 100 S1

= =

Final Speed, fpm Initial Speed, fpm

40) EFFLEX RATIO CONCEPT ER

=

SLICE JET SPEED -----------------------WIRE SPEED

Efflex ratio should be 0.9 – 1.0 for better runnability of M/C. EXAMPLE : Let M/c speed Head in the head box . . .

Slice jet speed

= =

200 M/min 50 cms

=

 50 x 706.32

=

 35316 187.9 = 188 M/min.

27 SANTOSH SWAR

ER

=

Slice Jet Speed ------------------M/C Speed 188 --------200

=

=

0.94

JET VELOCITY VS WIRE SPEED IF IF IF

Jet velocity > Wire speed Jet velocity < Wire speed Jet velocity = Wire speed

Floading problem GSM drastically changed Real fiber orientation not occur

So Jet velocity is kept slightly less than M/c speed for real fiber orientation.

41) PAPER ON ROLL (feet)  (OD2 – ID2) Ft of paper = ---------------------48 x Caliper OD, ID and Caliper in inches.

42) PAPER CALIPER (inches) Basis weight Paper caliper = -----------------------Area x 144 x Density Where Caliper in inches Basis weight (lbs/Area), Example : 30 lb/3000 ft2 Area (ft2) Density (lbs/in3), see table below Average paper density Grade Coated & supered Coated only News print

Density lb/ in3 0.042 0.038 0.023 28 SANTOSH SWAR

Fine paper Liner board Board (coated)

0.029 0.024 0.028

43) MASS OF PAPER ON REEL CALCULATION  Mass of paper = ------ (D2-d2) x W x Apparent Density (in kg) 4  GSM 2 2 Mass of paper = ------ (D -d ) x W x ----------(in kg) 4 Thickness Where,

D = Parent roll dia (in m) d = Empty spool dia (in m) W = Reel width (deckle) (in m) Thickness in mm

UNIT CALCULATION  GSM 2 2 ------ (D -d ) x W x ----------4 Thickness

=

GM ----m2 -------------mm

m2

x

m

m3

x

0.001 kg --------------m2 x .001 m

Kg m x -------m3 3

x

. . 1 GM = .001 kg 1 mm = .1 cm = .001 m

= kg

EXAMPLE Let a parent roll of deckle 3 meter. GSM = 50 Thickness = 0.075 mm 29 SANTOSH SWAR

. . .

50 = ------ = 666.67 kg/m3 0.075

AD

(Apparent density) Parent roll circumference  D = 3.82 m 3.82 D = --------- = 1.2157 m 3.142 D2 = 1.478 m2 Empty spool circumference  d = 1.11 m 1.11 d = --------- = 0.3533 m 3.142 d2 = 0.1248 m2  GSM 2 2 Mass of the Roll = ------ (D -d ) x W x ----------4 Thickness 3.142 = -------- (1.478 – 0.1248) x 3 x 666.67 4 = 0.7857 x 1.3532 x 3 x 666.67 = 2126.429 kg.

44) HORSE POWER HP

Where,

=

T N

TN --------------63,000 = =

torque (inch – pounds) speed (rpm)

45) TENSION HP Tension HP

=

fpm x Pli x inches of width ----------------------------------------33,000

46) APPROXIMATION FOR VACUUM COMPONENT IN PLI WHEN TAKING NIP IMPRESSIONS (Pliv). 30 SANTOSH SWAR

Pliv

=

Vacuum box width x Vacuum --------------------------------------3

Vacuum box width (inches) Vacuum (inches of Hg)

47) KWH CALCULATION TYPE OF METHODS 1. By taking 80 % efficiency 2. By Amp reading method 1st Method Motor capacity = 10 KW 80 % efficiency, 10 x 80 % = 8 KWH 2nd Method KWH =

Where

 3 V I COS  ----------------------------1000

V = I = COS  =

Input voltage (in volt) Current (in Amp) Power factor

EXAMPLE If 10 KW motor taking load 12 Amp Input voltage V = 410 V COS  = 0.95 (power factor)  3 V I COS  1.732 x 410 x 12 x 0.95 KWH = --------------------- = ----------------------------1000 1000 =

8.09  8 KWH

48) HYDRAULIC PUMP HORSE POWER (HP) Hydraulic Pump HP = 583 x PSI x GPM x 10-6

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In centrifugal pumps or blowers A) Capacity varies directly with speed B) Head varies as the square of speed C) Horse power varies as the cube of speed.

49) STANDARD HEAD BOX FLOW RATE (GPM/inch) GPM/inch =

( B.D. Ton/24 hr/in) (16.76) (1.5 – Tray Consistency) ---- ------------------------------------------------------------1.5 x Net consistency

Where, Net consistency = Head box consistency - Tray consistency

50) TISSUE HEAD BOX FLOW RATE (GPM/inch) GPM/inch

T.O. V

= =

=

T.O x V ----------- = 19.25

T.O. x V x 0.0052

Throat opening (inches) Spouting velocity (fpm)

51) CALCULATION OF LENGTH OF BELT The percentage of power transmission through pulley and belt largely depends on the length of belts. If the belt is tight the pulley will also run tightly. Its bush bearings shall also worn out easily. On the other hand if length of a belt is in excess of the need, it will slip frequently and result in loss of power. In order to determine the right length of belt, the following formula are applied. Indications L= C= D= d=

Requisite length of the belt Distance from the center Diameter of the larger pulley Diameter of the smaller pulley

Length of Open Belt 32 SANTOSH SWAR

1) For pulley of equal diameter L =  D + 2C 2) For pulley of different diameter  _________________ L = ----- (D+d) + 2  C2 + D - d 2 2 -------2 Length of Cross Belt

1) For pulley of equal diameter ________ L =  D + 2  D2 + C2 2) For pulley of different diameter  _________________ L = ----- (D+d) + 2  C2 + D + d) 2 2 -------2

52) TANK SIZING AND CAPACITY # / ft3 x volume Tons = ---------------------2000

=

% B.D. x Volume ----------------------1.6 x 2000

Volume = 3200 x # tons/% B.D. US Gallons = Volume / 7.4805 Where # / ft3 Volume 1 US Gallon % B.D.

= = = =

Weight of dry stock at % consistency Volume of tank in cubic feet 2.31 Cu.inches Percent consistency of stock

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53) LOAD FACTOR OF WIRE Ph K = ------F Ph K = ------b.L

(.

. F = b.L. ) .

Where, Ph F b L

= = = =

Production of paper/hr (in kg/hr) Working surface of paper m/c wire Working width of paper web on the wire (in m) Distance from the axis of breast roll to the axis of couch roll (in m) (working length of Wire).

54) INTERPRETING THE NIP IMPRESSION (Ne2 – Nc2) (D1 + D2) C = -------------------------------2 D1D2 Where, C Ne Nc D1 D2 OR

= = = = =

Change in total crown of two rolls (inches) Nip width at the ends (inches) Nip width at the center (inches) Top roll diameter (inches) Bottom roll diameter (inches)

if rolls have equal diameters

C =

Ne2 - Nc2 -------------D

NOTE : If C is minus, then the nip is over crowned. EXAMPLE Let us assume that we have two 30 inch (762 mm) diameter rolls and we find that the nip widths are 0.9 inches (22.9 mm) on the ends and 0.7 inches (17.8 mm) at the center under the loading which we desire to run the rolls. 34 SANTOSH SWAR

Then Nc Ne D

= = =

0.7 inches (17.8 mm) 0.9 inches (22.9 mm) 30 inches (762 mm)

(0.9)2 - (0.7)2 C = ------------------30

0.81 - 0.49 -----------------30

=

0.32 = -----30

= 0.011 inch (2.8 mm)

55) HEAD LOSS IN STOCK PIPES Pressure drop in pipes conveying 2 % - 6 % consistency stock is given by K x 0.0915 x C 1.89 x Q 0.364 x L H = ------------------------------------------------D 2.06 Where, H = Head loss in feet of water / feet of pipe K = A constant depending of the type of stock (for bleached sulphite = 0.9 unbleached sulphite = 1.0 Cooked ground wood & kraft ground wood = 1.4 oven dry cy %. Q = Flow of stock L = Pipe length D = Pipe diameter (in inch) For pipes made of 2 or more section of different diameter and length, the pressure drop is given as

H = K x 0.0915 x C

1.89

x Q

0.364

x

L1 --------- + D1 2.06

L2 --------D2 2.06

56) VACUUM PLI K N/M (SUCTION ROLLS) Suction rolls present a problem in that part of the core bending or distortion load is the result of the application of vacuum. This can be addressed either by increasing the PLI KN/M to compensate for the vacuum or by sealing off the section box area with plastic and applying an amount of vacuum equal to that normally run in the roll. If the increased PLI KN/M method is used, the original equipment supplier should be contacted to obtain the correct amount to be used. If this information is not readily available, the incremental PLI KN/M addition can be approximated by using the following formula : 35 SANTOSH SWAR

Vacuum PLI KN/M = 0.4912 x W x V x F Where

W V F

= = =

The width, in inches (mm) of the vacuum box. The vacuum level, in inches (mm) of mercury. Box seal efficiency factor ( F = 0.9 for most suction rolls).

Only 70-75 % of the vacuum PLI KN/M is used as an addition to the applied loading.

57) BELT WIDTH IN FLAT PULLEY bo

=

Where, bo P C2

= = =

C3 = FUN = V

=

P x C2 x C3 x 1000 ---------------------------FUN x V Belt width Kilowatt of motor Over load factor (50 % of normal load of motor) For paper Industry C2 = 1.2 (constant factor) Ratio between both pulley Belt type i.e. 40 (40 means 1 cm of belt take 40 kg load) Belt speed d1 x n1 V = ----------- m/sec 19100 d1 n1

= =

pulley dia Motor rpm

(For cone pulley C3 is not required)

58)

% WEAROUT OF WIRE 1) For single layer synthetic wire Original caliper - Average used caliper % Wear = --------------------------------------------------- x 100 0.7 x Weft  2) For double layer synthetic wire Original caliper - Average used caliper % Wear = --------------------------------------------------- x 100 36 SANTOSH SWAR

0.85 x Weft  3)

For Metal wire Original caliper - Average used caliper % Wear = --------------------------------------------------- x 100 0.7 x Wrap 

NOTE :- Synthetic wire is weft runner so weft  is taken for calculation & metal wire is wrap runner so wrap  is taken for calculation. EXAMPLE Single layer synthetic wire Original caliper Average used caliper Weft 

= = =

0.565 mm 0.4 mm 0.28 mm

0.565 – 0.4 % Wear = --------------------- x 100 0.7 x 0.28 = 84 %

59) PRODUCTION RATE (Off Machine) Production (lbs/hr) = Factor x speed x Deckle x Basis weight Where,

Grade Liner board Bond Cover Index Bristol Offset Manuscript Wrapper News print

Production Factor Speed Deckle Basis weight

= = = = =

lbs/hour From table Feet/minute inches at reel lbs/ream

Ream size

No.of sheets

17 x 22 20 x 26 25 ½  x 30 ½ 22 ½ x 28 ½ 25 x 38 18 x 31 24 x 36 24 x 36

500 500 500 500 500 500 480 500

Ft2/Ream 1000 1300 1805 2700 2110 3300 1948 2880 3000

Factor 0.005 0.00385 0.00277 0.00185 0.00225 0.00152 0.00258 0.00174 0.00167

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60) M/C EFFICIENCY CALCULATION

M/C Efficiency = (%)

=

Total Actual Prod. (MT) ---------------------------------- x 100 Total theoretical prod. (MT) Actual Production (MT) x 106 -------------------------------------------------------- x 100 GSM x Speed x Deckle x Total running hours (M/Min) (M) (in min)

Actual Production (MT) x 108 M/C Efficiency = --------------------------------------------------------------GSM x Speed x Deckle x Total running hours (M/Min) (M) (in min)

EXAMPLE

Actual Production (T) 2T 11.3 T 5.4 T 12.7 T

GSM

Speed

Deckle

Total Running Efficiency Time (in min.) (%)

54 65 60 50

170 155 160 180

3.04 3.05 3.08 3.04

1.15’ = 75’ 6.10’ = 370’ 3.05’ = 185’ 9.00’ = 540’

95.55 99.39 98.72 85.56

61) CALCULATION OF NIP LOAD ON PRESS Area of Intensity Lever Edge x No.of  Roll weight Cylinder x pressure x sides (kg) (cm2) (kg/cm2) Nip Load on Press = --------------------------------------------------------------------------------(kg/cm) Face length (in cm) 38 SANTOSH SWAR

Area of Cylinder   2 2 This is D ----- or (D – d ) ----4 4 2

1)

Fulcrum Air for Loading  Here Area of cylinder is (D2 – d2) -----4

2) 

Air for

Here Area of

Loading

 cylinder is D --4 2

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3)

Air for Loading  Here Area of cylinder is (D2 – d2) = ----4

Intensity Pressure Intensity pressure can be found from pressure gauge reading. Lever Edge

(It is only ratio)

Y

X 1000

500

Y ( 1000) = X (500)

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X 1000 = ------ = ------Y 500

= 2

2)

Y

X 14

26 

(14  + 26 ) Y (14  + 26 )

=

X ---Y

X (26) 40 ----26

=

= 1.538

Roll Weight If the loading role is lower side then the role weight to be subtracted & if it is on upper side the role weight to be added. 1)

2)

Here loading roll is on lower side, So roll weight is -ve.

Here loading roll is on upper side, So roll weight is +ve.

EXAMPLE

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Y

X (14 + 26)

Given Data Roll weight = 2 T = 2000 kg Intensity pressure = 5 kg/cm2 Face length = 220 cms Cylinder bore (in cms) D = 25 cm Piston rod dia (in cms) d = 5 cm Distance from roll center to fulcrum = 26  Distance from loading edge to fulcrum = 40

Now

Area of cylinder is  2 2 (D – d ) ----4 3.142 = (252 – 52) ----------4 = 600 x 0.7855 = 471.3 cm2

Lever edge Y (14 + 26) X --Y

=

= X (26) 40 ---26

= 1.538

Unit load on roll 42 SANTOSH SWAR

=

=

Area of x Intensity x Lever x No.of  Roll weight Cylinder pressure edge sides -----------------------------------------------------------------------------Face length 471.3 cm2 x 5 kg/cm2 x 1.538 x 2 - 2000 kg --------------------------------------------------------------220 cm 7248 kg - 2000 kg ------------------------------220 cm

=

=

5248 -----220

=

23.8 kg/cm

kg/cm

2)

Y

| 1000

500

Given Data Roll weight = 2 T (2000 kg) Face length = 320 cms Intensity pressure = 6 kg/cm2 Cylinder bore = 10 inch = 25.4 cm Distance between loading edge to fulcrum = 1000 cms 43 SANTOSH SWAR

Distance between fulcrum to roll center = 500 cms  Are of cylinder D2 ----4 3.142 2 = (25.4 cm) x -----------4 = 645.16 x 0.7855 cm2 = 506.8 cm2 Lever edge Y (1000)

=

X (500)

X ---Y

1000 -----500

= 2

=

Nip load on press

=

Area of x Intensity x Lever x No.of  Roll weight Cylinder pressure edge sides -----------------------------------------------------------------------------Face length

=

506.8 x 6 x 2 x 2 + 2000 --------------------------------------------------------------320

=

12163.2 + 2000 --------------------------------------320

=

14163.2 ---------320

= 44.26 kg/cm

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62) MAXIMUM SPEED OF COUCH Required Data 1) 2) 3) 4) 5) 6) 7)

Dia of couch (d) (in metres) Couch gear box teeth details (ratio) Couch gear box pulley  ( Max. (in mm), Min. (in mm) Line shaft couch cone pulley  ( Max. (in mm), Min. (in mm) Main motor pulley  (in mm) Main motor RPM Line shaft pulley  ( in mm)

EXAMPLE Main motor pulley  = Main motor RPM = Line shaft pulley  =

360 mm 1500 840 mm

360 x 1500 . . Line shaft pulley RPM = ---------------840 = 642.8 Couch Gear Box pulley  Maximum = 760 mm Minimum = 710 mm Shaft Couch Cone Pulley  , Max. = 460 mm Min. = 410 mm 735 . . Ratio = -------- = 1.68 435

Mean = 735 mm

Mean = 435 mm

 1.7

642.8 Hence gear box pulley RPM = ---------------- = 378.1 1.7 62 Gear box teeth details 62 & 19 i.e. -------- = 3.263 19 378.1 Gear box out put RPM = ---------- = 116 RPM 3.263 Now Dia of couch = 0.66 metres . . Speed of couch =  d x RPM 45 SANTOSH SWAR

= 3. 142 x 0.66 x 116 = 240 M/min.

63) WATER EVAPORATED AT DRYER Water evaporated at Dryer =

Final Dryness -------------------- -1 Production Initial Dryness

EXAMPLE Let Final dryness = 95 % Paper web entering the dryer Section of dryness = 38 % Production = 1200 kg/hr 95 Water evaporated at = ( ---- - 1) x 1200 Dryer/hr 38 = 1.5 x 1200 = 1800 kg . . .

1800 kg water evaporated at dryer Section for produce 1200 kg paper in 1 hour.

64) FOURDRINIER SHAKE Shake Number =

Where,

Amplitude x (Frequency)2 -------------------------------Wire speed

Amplitude in inches Frequency in strokes/minute Wire speed in feet/minute

Optimum shake number is generally over 30 – 60

65) HEAD BOX APPROACH SYSTEM STOCK VELOCITIES V (fps) =

Stock flow (gpm) -------------------- x 0.0007092 Pipe Radius (ft) 2 Stock flow (gpm) 46 SANTOSH SWAR

=

--------------------- x 0.321 Area of pipe (in2)

Acceptable Range 7 – 14 fps

66) “ L ” FACTOR (lbs. Paper / ft2 dryer surface /hour) SW “ L” Factor = ----------------(“C” Value) N Where, L S W N

# paper /ft2 dryer surface/hr M/c speed (fpm) Basis weight (lbs./3000 ft2) Number of dryers

= = = =

4 ft 5 ft 6ft

“C” Value 628.3 785.4 942.5

67) EVAPORATION RATE (lbs H2O/ft2 dryer surface/hour) BD Out - BD In Evaporation Rate (Ev) = L ( ---------------------------) BD Out BD Out x BD In

Where, Evaporation = # H2O /ft2 dryer surface / hr BD = Percent bone dry

68) DEFLECTION OF A ROLL – OVER FACE (Normally used for crown calculations)

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| | | | | | | | | |

B

F |

|

| | | | | | | | | |

d WF3 (12 B – 7F) d = --------------------------384 E I

Where, d W F B E I Do Di

= = = = = = = = =

deflection (inches) overface Resultant unit load of shell (pounds/inch) Shell face (inches) Centerline to centerline bearings (inches) Modulus of elasticity (lb/in2) Moment of inertia (inches 4) 0.0491 ( Do4 - Di 4) Outside diameter (inches) Inside diameter (inches)

69) APPROXIMATE CRITICAL SPEED OF A ROLL

C.S.

=

55.37 Do (0.9) ------------------_____  d

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Where, C.S. Do d

70)

= = =

Critical speed (fpm) Out side diameter of roll (inches) Roll deflection (inches) over face due to roll weight only (not to include externally applied forces) (See previous formula)

RIMMING SPEED (5’ & 6’ dryers) 2160 Remming speed (fpm) = ( 5720 - -------- ) L 1/3 D 30000 Rimming speed (M/min) = (260 - -----------) L D Where D in mm L in mm Where, D L

= =

0.33

Inside diameter of roll (feet) Condensate film thickness (feet)

71) NATURAL FREQUENCY OF SINGLE DEGREE OF FREEDOM SYSTEM 3.127 F = ---------_____  d Where

F d

= =

Natural frequency (cycles/second) Static deflection due only to weight of body (No externally applied forces)

Wt.of dry material 72) Consistency = ------------------------------- x 100 % Wt. of Suspension

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