Calculation

SANTOSH SWAR 1

CONVERSION FACTORS

1 HP = 33,000 ft.lbs/min = 550 ft lbs/sec

1 HP = 746 Watts

1 HP = 42.4 BTU/Min.

Amps

Electric HP = --------------------------

746

(---------------------------- )

Volts x decimal efficiency

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 grams

-------- = 1.64 ----------

3000 ft2 Meter2

9

F = ---- C + 32

5

5

C = (F – 32) x -----

9

SANTOSH SWAR 2

BASIS WEIGHT CONVERSIONS

Offset (3300 ft2) x 1.48 = GSM

Bond (1300 ft2) x 3.76 = GSM

Liner (1000 ft2) x 4.89 = GSM

News (3000 ft2) x 1.63 = GSM

CONVERSION FACTORS

FROM TO OBTAIN MULTIPLY BY

lb kg 0.454

in mm 25.4

in cm 2.54

in m 0.0254

ft/min (fpm) m/min (mpm) 0.305

cfm m3/hr 0.589

cfm/in m3/hr/cm 0.232

cfm/in2 m3/hr/cm2 0.91

cfm/in2/1000 fpm m3/hr/cm2/100 mpm 0.278

oz g 28.3

oz/ft2 g/m2 305.6

1) M/C PRODUCTION CALCULATION

Let M/c speed in mt/min

M/c Deckle in mt

SPEED x DECKLE x GSM x 60

(mt/min) (mt) (Gm/mt2) (min/hr)

M/C PROD = ------------------------------------------------------

(Tons/Hr) 1000000

(Gm/Ton)

Speed x Deckle x GSM x 60

M/C PROD = ----------------------------------------

(Tons/hr) 106

EXAMPLE

Let M/c Speed = 200 mt/min

Deckle = 3.2 mt

GSM = 50

SANTOSH SWAR 3

. 200 x 3.2 x 50 x 60

. . M/C PROD = -------------------------

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

( ----- )

m 2

REAM WEIGHT = ----------------------------------------------

(kg) 107

EXAMPLE : Let Paper size = 40 cm x 60 cm

GSM = 50

No.of sheets = 500

SANTOSH SWAR 4

40 x 60 x 50 x 500

Ream Weight = --------------------------

107

60000000

= ----------------

107

= 6 kg

3) CALCULATION OF GSM

Any samples of any size can be expressed in terms of GSM.

10000 x W

GSM = -----------------

l x b

Where, W = Weight of paper in gms.

l = Length of paper in cms.

b = 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

10000 x 0.72

GSM = ---------------------

10 x 15

7200

= -------

150

= 48 GSM

Second Expression

10 cm x 15 cm size of paper has 0.72 gms

SANTOSH SWAR 5

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

150 cm2 = 0.72 gms

0.72

1 cm2 = ------

150

= 0.0048 gms

1 cm2 = 0.0048 gms

1

---------- m2 = 0.0048 gms . . 1 m = 100 cm

10000 . 1 m2 = 10000 cm2

1

1 cm2 = ------ 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

5.8

1 cm2 = -------- gms

1000

1 cm2 = 0.0058 gms

1

--- m2 = 0.0058 gms

104

1 m2 = 0.0058 x 104 gms

1 m2 = 58 gms

SANTOSH SWAR 6

4) THEORETICAL HEAD V 2

-----

100

Theoretical Head = --------------

K

Where, V = Spouting velocity (fpm)

K = Constant (see table)

Head In of H2O Ft of H2O In of Hg Pressure PSIG

K 1.932 23.184 26.345 53.623

5) SPOUTING VELOCITY

V = k h

Where, V = Spouting velocity (fpm)

h = Theoretical head

k = Constant (see table)

Head In of H2O Ft of H2O In of Hg Pressure PSIG

K 139.2 481.5 513.3 732.3

6) HEAD CALCULATION

V2 = 2 gh

V2

h = -------

2 g

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)

SANTOSH SWAR 7

EXAMPLE

Let M/c speed V = 200 m/min

V2

h = -------

2 g

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 ------------- . . 1 min2

3600 sec2 . = 1 min x 1 min

= 60 sec x 60 sec

3600 sec2

2038.74

= ----------- m

3600

= 0.5663 m

= 56.63 cm

Short Method

200 x 200 m2 sec2

h = ------------- ------ x -------

2 x 9.81 min2 m

200 x 200 m x sec2

= ------------- -------------

2 x 9.81 min2

200 x 200 100 cm x sec2

= -------------- ------------------

2 x 9.81 3600 sec2

SANTOSH SWAR 8

200 x 200 x 100

= ---------------------- cm

2 x 9.81 x 3600

200 x 200 100

= ----------- cm . . --------------------- = 706.32

706.32 . 2 x 9.81 x 3600

= 56.63 cms.

QUICK FORMULA

Speed x Speed Where h in cms

h = -------------------- Speed = in m/min.

(in cms) 706.32

EXAMPLE

M/c Speed

180 x 180

1) 180 M/min , h = -------------- = 45.87 cms

706.32

200 x 200

2) 200 M/min , h = ------------ = 56.63 cms

706.32

250 x 250

3) 250 M/min , h = ------------ = 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.

SANTOSH SWAR 9

8) JET VELOCITY CALCULATION (Stock speed at slice)

Jet Velocity

______

V = Cv 2 gh

Where, V = Stock speed at slice

Cv = Coefficient of velocity discharge


h = 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

1

( . . 100 cm = 1 m 1 cm = ------- M )

100

SANTOSH SWAR 10

Similarly 3600 sec2 = 1 min2

1

1 Sec2 = ------ 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 = Head x 706.32

(In m/min) (in cms)

EXAMPLE

____________

1) h = 46 cms, V = 46 x 706.32

_________

= 32490.72

= 180.25 M/min

___________

2) h = 89 cms, V = 89 x 706.32

_________

= 62862.48

= 250.72 M/min

SANTOSH SWAR 11

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 Net consistency

1) Retention % = -------------------------- x 100

Head box consistency

Head box consistency - Tray consistency

2) Retention % = ----------------------------------------------------- x 100

Head box consistency

Filler in sheet

3) Overall Retention = ----------------------------- 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

SANTOSH SWAR 12

12) FLUID VELOCITY

GPM x 0.321

Velocity = ------------------

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

3.82 ( V )

RPM = -------------

Do

Where, RPM = revolutions per minute

V = Speed (fpm)

Do = Roll outside diameter (inches)

14) FORMING BOARD SETTING

12 V COS A _______________

X = ----------------- ( V2 Sin2 A + Zyh - V Sin A)

g

_____________

= 0.37267 V COS A (V2 Sin2 A + 64.4 h - V Sin A)

Where,

X = Distance of slice to lead forming board blade

V = Initial Jet Velocity

A = Jet angle

g = 32.2 ft/S2

h = Jet to wire height

SANTOSH SWAR 13

15) WATER REMOVAL BY A TABLE ROLL

1st Method

DU K

q = --------

F2

Where,

q = Water removed by a table roll per unit time & width

D = Diameter of roll

U = Wire speed

F = 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.

K = 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 = Drainage coefficient (dependent on the sheet weight & type of

wire)

R = Radious of the roll


h = Head of stock suspension above the wire

V = Velocity of the wire

SANTOSH SWAR 14

16) VACUUM PUMP CAPACITY (CFM)

PV = nRT

Where,

P = Absolute pressure, lb.ft2 = (Psi gauge + 14.7) x 144

V = Total gas volume, ft3

n = Weight of gas, lbs

T = Absolute temperature R = F + 460

R = Gas constant, lbf x ft / (lbn x R )

Ra (air) = 53.3

Rw (water)vapour = 85.8

P1V1 = P2V2

29.92 - P1 ( Hg)

V2 = ----------------------- x V1 (CFM)

29.92 - P2 ( Hg)

or for temperature cooling effects :

P1V1 P2V2

------- = -------

T1 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.)

SWd

L = K -------

(t – 100)

Where,

L = Peripherial length in meters of dryers in contact with paper

During drying.

SANTOSH SWAR 15

K = Constant value around 0.05

S = Speed of M/c in M/min

W = Basis weight of paper in gm

----

m2

d = Thickness of dryer shell in centimeters

t = 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

0.05 x 180 x 50 x 2.2

L = ------------------------------

(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 = Spouting velocity (fpm)

S.O. = Slice opening (inches)

C = Orifice coefficient

(see table for approximate values)

Type C

Nozzle 0.95

A 0.75

B 0.70

SANTOSH SWAR 16

C 0.60

Type 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

200

W = ----- = 16.67 foot

12

SANTOSH SWAR 17

Dryer diameter, d = 60 inches

60

= ---- 5 foot

12

Moisture to dryer = 37 % dry

Moisture to reel = 94 % dry

400

Hourly production = ------ Tons/hr

24

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 ( ---- __ 1)

37

Dryer surface required @ 2.8 lb water/hr/ft2

Evaporation rate

Required dryer surface

51350

= -------- = 18339 ft2

2.8

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

-------- = 69.99 = 70 dryers

262

20) PRESS IMPULSE

5 x PLI

SANTOSH SWAR 18

PI = ---------

Speed

Where, PI = Press impulse (Psi – Sec)

PLI = Nip pressure (Pli)

Speed = Nip speed (fpm)

21) TORQUE

Tq = Force x Radius

Where Tq = Torque (inch – pounds)

Force = in pounds

Radius = in inches

22) WR2 OF A ROLL

WR2 = (0.000682) (W) (L) (Do4 – Di4)

Where, WR2 = in (lbs – ft2)

W = Density (pounds/inches 3)

L = Length (inches)

Do = Outside diameter (inches)

Di = Inside diameter (inches)

23) STOCK FLOW THROUGH THE PIPE CALCULATION

1

Q = ----- d2 x Vrc

4

4 Q

Vrc = -----

d2

Where, Vrc = Velocity of flow

Q = Total quantity

d = Pipe diameter

24) IDEAL DRAINAGE IN WIRE PART

a) Theoritically after forming board the drainage should be 80 to 85 % of stock

thickness ( i.e. Slice opening).

SANTOSH SWAR 19

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 = Thickness of stock on table in cm.

Basis weight in g/cm2

%

Consistency in ---

100

R = 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.

0.0050

R = -------------------------------- = 73 %

( 0.0060 ) x 1.6 x 0.95

26) CALCULATION OF WIRE LENGTH

Ls = 2 l + ---- (D1 + D2) + K

2

Where, Ls = Length of wire (in mm)

l = Distance between center of breast roll to couch roll (in mm)

D1 = Diameter of breast roll (in mm)

D2 = Diameter of couch roll (in mm)

K = A constant, 130 mm for fourdrinier wire part.

EXAMPLE

L = 12,500 mm

D1 = 450 mm

SANTOSH SWAR 20

D2 = 800 mm

Then Ls = 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

0.8 VA Kilonewtons

DRAG LOAD = ---------- ------- -----------------

1000 UW Meter

Where

V = Volt

A = AMP

U = Fabric speed (M/S)

W = Fabric width (M)

lb KN lb

{ To convert in ------ , ------- x 5.71 = -------- }

in M inch

b) DRAG LOAD - CONVENTIONAL

V x A x 0.8

DL = ------------------------ DL in Pli

0.0226 x U x W

Wehre,

V = Drive volts (V)

SANTOSH SWAR 21

A = Drive AMPs (AMPS)

U = Nominal fabric speed (fpm)

W = Nominal fabric width (inches)

c) 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 = (---- __ 1) (EM + Ts)

Vs

Where,

DL = Drag load (Pli)

Vn = Fabric speed at point n in fabric run (fpm)

Vs = Fabric speed on slack side of fabric run (fpm)

EM = Fabric elastic modulus (young) at temperature

T ~ EMr – KT

EMr = Elastic modulus at reference temperature r (Pli)

Modulus Pli

K = ------------------------- (----- )

Temperature constant º F

Ts = Slack side tension (Pli)

SANTOSH SWAR 22

29) DANDY DIAMETER CALCULATION

1) Open type 2) Journal typs

15 – 20 % of wire width 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

Dia of Dandy = --------------------------------------

x Maximum number of revolutions

V

D = --------

n

Where, D = Diameter in mm

V = M/c speed in m/min

n = 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 V = 400 m/min

400

D = ------- = 850 mm

477

Wire Speed (m/min) 80 150 250 300 400

Dia of dandy (in mm) 300 400 500 600 800-1000

SANTOSH SWAR 23

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

Web Speed (fpm)

RPM = ----------------------------------------------

3.142 x Size press Roll diameter (ft)

Maximum 250 rpm

32) TONS PER DAY (T/D)

Capacity (gpm) x % Bone dry consistency

T/D = ------------------------------------------------------

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

SANTOSH SWAR 24

33000 kg

33 TPD = ------------- = 22.9 kg/min

1440 min

100

Bs factor = 1.45 (68.9 wire retention ------- = 1.45)

68.9

Bs factor x prod.

22.9 x 1.45 = 33.22 kg/min

10 % reject = 3.32 kg/min

5 % vent reject= 1.66 kg/min

-----

38.20 kg/min

kg/min x 100 38.20 x 100

Now LPM = ------------------ = ------------------ = 4775 lpm

Consistency 0.8

Through put/leg = 500 lpm

4775

No of legs required = --------- = 9.55 = 10 legs

500

Primary legs = 10 Nos.

Secondary legs = 3 Nos. (30 % of primary legs)

Tertiary legs = 1 No. (30 % of secondary legs)

Pressure drop = 1.4 kg/cm2

34) WEIR FLOW – RECTANGULAR WEIR WITH END

CONTRACTIONS

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

Where,

L = Length of weir opening in feet

(should be 4 – 8 times H)

H = Head on weir in feet ( ~ 6 ft back of weir opening)

a = at least 3 H (end contraction)

35) WEIR FLOW – TRIANGULAR NOTCH WEIR WITH END

CONTRACTIONS

SANTOSH SWAR 25

4 ______

Q = C ( -----) L H 2 gH

15

Where L = Width of notch in ft at H distance above apex

H = Head of water above apex of notch in feet

C = 0.57

a = 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 = Wasted volume (in ft3/min. or m3/sec)

DA = Drilled Area (in %)

DW = Drilled width (in inch or M)

U = Machine speed (in ft/min. or M/minute)

t = Shell Thickness (in inch or cm)

∆ P = Suction Pressure (couch vacuum) in inch Hg)

P = Pressure (in inch Hg)

EXAMPLE

DA = 50 % = 50/100 = 0.5

DW = 286 inch or 7.26 M

U = 300 fit/min or 914.6 M/min

t = 2.5 inch or 6.35 cm. Or 0.0635 M

∆ P = 24 inch Hg or 81.3 K Pa

P = 30 inch Hg or 101.6 K Pa

Wv = DA x DW x U x t x ∆ P

-----

P

24 inch Hg

= 0.5 x 286 inch x 3000 fit/min. x 2.5 inch x ----------------

SANTOSH SWAR 26

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 6.35

= 0.5 x 7.26 M x ------------ ------- x ---------- M x 0.8

60 Sec 100

= 2.81 m3/S

37) COUCH VACUUM EXPANSION VOLUME

CFM = (V) (b) (S) (E) (m)

Where,

V = M/c speed, fpm

b = Roll shell face width, feet

S = Hole depth, feet

E = % Open area of shell

P2 0.9

m = Expansion factor, ------ - 1

P1

P2 = ambient pressure, Hg absolute

P1 = Suction box vacuum, Hg absolute

SANTOSH SWAR 27

38) FORMATION – BLADE PULSE FREQUENCY

V

F = -------

5 x

Where,

F = Formation – blade pulse frequency

(in cycle/sec)

V = Wire speed (fpm)

= blade spacing, tip to tip (inches)

Optimum frequency for formation improvement “

F > 60 cycle/sec

39) PAPER WEB DRAW

SF – S1

Draw (%) = -------------- x 100

S1

Where, SF = Final Speed, fpm

S1 = Initial Speed, fpm

40) EFFLEX RATIO CONCEPT

SLICE JET SPEED

ER = ------------------------

WIRE SPEED

Efflex ratio should be 0.9 – 1.0 for better runnability of M/C.

EXAMPLE :

Let M/c speed = 200 M/min

Head in the head box = 50 cms

.

. . Slice jet speed = 50 x 706.32

35316

= 187.9 = 188 M/min.

SANTOSH SWAR 28

Slice Jet Speed

ER = -------------------

M/C Speed

188

= ---------

200

= 0.94

JET VELOCITY VS WIRE SPEED

IF Jet velocity > Wire speed Floading problem

IF Jet velocity < Wire speed GSM drastically changed

IF Jet velocity = Wire speed 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 Density lb/ in3

Coated & supered 0.042

Coated only 0.038

News print 0.023

SANTOSH SWAR 29

Fine paper 0.029

Liner board 0.024

Board (coated) 0.028

43) MASS OF PAPER ON REEL CALCULATION

Mass of paper = ------ (D2-d2) x W x Apparent Density

(in kg) 4

GSM

Mass of paper = ------ (D2-d2) 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

------ (D2-d2) x W x -----------

4 Thickness

GM

-----

m2

= m2 x m x --------------

mm

0.001 kg . . 1 GM = .001 kg

m3 x --------------- 1 mm = .1 cm

m2 x .001 m = .001 m

Kg

m3 x -------- = kg

m3

EXAMPLE Let a parent roll of deckle 3 meter.

GSM = 50

Thickness = 0.075 mm

SANTOSH SWAR 30

. 50

. . AD = ------ = 666.67 kg/m3

0.075

(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

Mass of the Roll = ------ (D2-d2) 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

TN

HP = ---------------

63,000

Where, T = torque (inch – pounds)

N = speed (rpm)

45) TENSION HP fpm x Pli x inches of width

Tension HP = -----------------------------------------

33,000

46) APPROXIMATION FOR VACUUM COMPONENT IN PLI WHEN

TAKING NIP IMPRESSIONS (Pliv).

SANTOSH SWAR 31

Vacuum box width x Vacuum

Pliv = ---------------------------------------

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

3 V I COS

KWH = -----------------------------

1000

Where V = Input voltage (in volt)

I = Current (in Amp)

COS = 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

SANTOSH SWAR 32

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)

( B.D. Ton/24 hr/in) (16.76) (1.5 – Tray Consistency)

GPM/inch = ---- -------------------------------------------------------------

1.5 x Net consistency

Where,

Net consistency = Head box consistency - Tray consistency

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

T.O x V

GPM/inch = ----------- = T.O. x V x 0.0052

19.25

T.O. = Throat opening (inches)

V = 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 = Requisite length of the belt

C = Distance from the center

D = Diameter of the larger pulley

d= Diameter of the smaller pulley

Length of Open Belt

SANTOSH SWAR 33

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 = Weight of dry stock at % consistency

Volume = Volume of tank in cubic feet

1 US Gallon = 2.31 Cu.inches

% B.D. = Percent consistency of stock

SANTOSH SWAR 34

53) LOAD FACTOR OF WIRE

Ph

K = -------

F

Ph

K = ------- ( . . F = b.L. )

b.L .

Where,

Ph = Production of paper/hr (in kg/hr)

F = Working surface of paper m/c wire

b = Working width of paper web on the wire (in m)

L = 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 = Change in total crown of two rolls (inches)

Ne = Nip width at the ends (inches)

Nc = Nip width at the center (inches)

D1 = Top roll diameter (inches)

D2 = Bottom roll diameter (inches)

OR if rolls have equal diameters

Ne2 - Nc2

C = --------------

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.

SANTOSH SWAR 35

Then Nc = 0.7 inches (17.8 mm)

Ne = 0.9 inches (22.9 mm)

D = 30 inches (762 mm)

(0.9)2 - (0.7)2 0.81 - 0.49 0.32

C = ------------------- = ------------------ = ------

30 30 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

L1 L2

H = K x 0.0915 x C 1.89 x Q 0.364 x --------- + ---------

D1 2.06 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 :

SANTOSH SWAR 36

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

Where W = The width, in inches (mm) of the vacuum box.

V = The vacuum level, in inches (mm) of mercury.

F = 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

P x C2 x C3 x 1000

bo = ----------------------------

FUN x V

Where, bo = Belt width

P = Kilowatt of motor

C2 = Over load factor

(50 % of normal load of motor)

For paper Industry C2 = 1.2 (constant factor)

C3 = Ratio between both pulley

FUN = Belt type i.e. 40

(40 means 1 cm of belt take 40 kg load)

V = Belt speed

d1 x n1

V = ----------- m/sec

19100

d1 = pulley dia

n1 = 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


% Wear = --------------------------------------------------- x 100

SANTOSH SWAR 37

0.85 x Weft

3) For Metal wire


% 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 = 0.565 mm

Average used caliper = 0.4 mm

Weft = 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, Production = lbs/hour

Factor = From table

Speed = Feet/minute

Deckle = inches at reel

Basis weight = lbs/ream

Grade Ream size No.of sheets Ft2/Ream Factor

Liner board 1000 0.005

Bond 17 x 22 500 1300 0.00385

Cover 20 x 26 500 1805 0.00277

Index 25 ½ x 30 ½ 500 2700 0.00185

Bristol 22 ½ x 28 ½ 500 2110 0.00225

Offset 25 x 38 500 3300 0.00152

Manuscript 18 x 31 500 1948 0.00258

Wrapper 24 x 36 480 2880 0.00174

News print 24 x 36 500 3000 0.00167

SANTOSH SWAR 38

60) M/C EFFICIENCY CALCULATION

Total Actual Prod. (MT)

M/C Efficiency = ---------------------------------- 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)

GSM Speed Deckle Total Running

Time (in min.)

Efficiency

(%)

2 T 54 170 3.04 1.15’ = 75’ 95.55

11.3 T 65 155 3.05 6.10’ = 370’ 99.39

5.4 T 60 160 3.08 3.05’ = 185’ 98.72

12.7 T 50 180 3.04 9.00’ = 540’ 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)

SANTOSH SWAR 39

Area of Cylinder

This is D2 ----- or (D2 – d2) -----

4 4

1)

Fulcrum

Air for

Loading

Here Area of cylinder is (D2 – d2) ------

4

2)

Air for Here Area of

Loading cylinder is D2 ---

4

SANTOSH SWAR 40

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)

SANTOSH SWAR 41

X 1000

= ------ = ------- = 2

Y 500

2)

Y X

14

26

(14 + 26 )

Y (14 + 26 ) = X (26)

X 40

---- = ----- = 1.538

Y 26

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, Here loading roll is on upper side,

So roll weight is -ve. So roll weight is +ve.

EXAMPLE

SANTOSH SWAR 42

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

(D2 – d2) -----

4

3.142

= (252 – 52) -----------

4

= 600 x 0.7855

= 471.3 cm2

Lever edge

Y (14 + 26) = X (26)

X 40

--- = ---- = 1.538

Y 26

Unit load on roll

SANTOSH SWAR 43

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

= ------ kg/cm

220

= 23.8 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

SANTOSH SWAR 44

Distance between fulcrum to roll center = 500 cms

Are of cylinder D2 -----

4

3.142

= (25.4 cm) 2 x ------------

4

= 645.16 x 0.7855 cm2

= 506.8 cm2

Lever edge

Y (1000) = X (500)

X 1000

---- = ------ = 2

Y 500

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

= ---------- = 44.26 kg/cm

320

SANTOSH SWAR 45

62) MAXIMUM SPEED OF COUCH

Required Data

1) Dia of couch (d) (in metres)

2) Couch gear box teeth details (ratio)

3) Couch gear box pulley ( Max. (in mm), Min. (in mm)

4) Line shaft couch cone pulley ( Max. (in mm), Min. (in mm)

5) Main motor pulley (in mm)

6) Main motor RPM

7) Line shaft pulley ( in mm)

EXAMPLE

Main motor pulley = 360 mm

Main motor RPM = 1500

Line shaft pulley = 840 mm

360 x 1500

. . Line shaft pulley RPM = ----------------

840

= 642.8

Couch Gear Box pulley Maximum = 760 mm Mean = 735 mm

Minimum = 710 mm

Shaft Couch Cone Pulley , Max. = 460 mm Mean = 435 mm

Min. = 410 mm

735

. . Ratio = -------- = 1.68 1.7

435

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

SANTOSH SWAR 46

= 3. 142 x 0.66 x 116

= 240 M/min.

63) WATER EVAPORATED AT DRYER

Final Dryness

Water evaporated at Dryer = -------------------- -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

Amplitude x (Frequency)2

Shake Number = --------------------------------

Wire speed

Where, 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

Stock flow (gpm)

V (fps) = -------------------- x 0.0007092

Pipe Radius (ft) 2

Stock flow (gpm)

SANTOSH SWAR 47

= --------------------- 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 = # paper /ft2 dryer surface/hr

S = M/c speed (fpm)

W = Basis weight (lbs./3000 ft2)

N = Number of dryers

“C” Value

4 ft 628.3

5 ft 785.4

6ft 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)

SANTOSH SWAR 48

| B |

| |

| |

| |

| |

| |

| F |

| | | |

| |

| |

d

WF3 (12 B – 7F)

d = ---------------------------

384 E I

Where,

d = deflection (inches) overface

W = Resultant unit load of shell (pounds/inch)

F = Shell face (inches)

B = Centerline to centerline bearings (inches)

E = Modulus of elasticity (lb/in2)

I = Moment of inertia (inches 4)

= 0.0491 ( Do4 - Di 4)

Do = Outside diameter (inches)

Di = Inside diameter (inches)

69) APPROXIMATE CRITICAL SPEED OF A ROLL

55.37 Do (0.9)

C.S. = -------------------

_____

d

SANTOSH SWAR 49

Where,

C.S. = Critical speed (fpm)

Do = Out side diameter of roll (inches)

d = Roll deflection (inches) over face due to roll weight only

(not to include externally applied forces)

(See previous formula)

70) RIMMING SPEED (5’ & 6’ dryers)

2160

Remming speed (fpm) = ( 5720 - -------- ) L 1/3

D

30000

Rimming speed (M/min) = (260 - -----------) L 0.33

D

Where D in mm

L in mm

Where, D = Inside diameter of roll (feet)

L = Condensate film thickness (feet)

71) NATURAL FREQUENCY OF SINGLE DEGREE OF FREEDOM

SYSTEM

3.127

F = ----------

_____

d

Where F = Natural frequency (cycles/second)

d = Static deflection due only to weight of body

(No externally applied forces)

Wt.of dry material

72) Consistency = ------------------------------- x 100 %

Wt. of Suspension

SANTOSH SWAR 50

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