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 efficiencyAmps = --- ( 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
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 = GSMCONVERSION 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
. 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
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
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
4) THEORETICAL HEAD
V 2 --- 100 Theoretical Head = --- KWhere, 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 gWhere, h = Height of stock in head box V = Velocity (M/c speed in m/min) g = Acceleration due to gravity
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
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.
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 g = Acceleration due to gravity 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
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
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.
______
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
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
g = Acceleration due to gravity
h = Head of stock suspension above the wire V = Velocity of the wire
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.
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
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
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
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) WR
2OF 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).
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
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,
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)
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
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
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
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
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
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
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
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
. 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 = --- 63,000Where, 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).
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)
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
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
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.
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 --- + ---
D12.06 D22.06
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 :
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
Original caliper - Average used caliper % Wear = --- x 100
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 = 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
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 = ---
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 ---
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)
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.
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
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 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
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
= 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
= --- x 0.321 Area of pipe (in2)
Acceptable Range 7 – 14 fps
66) “ L ” FACTOR (lbs. Paper / ft
2dryer 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 H
2O/ft
2dryer surface/hour)
BD Out - BD InEvaporation 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
| 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. = ---
_____
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
