Storage Tank Design

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Pamantasan ngamantasan ng LLungsod ngungsod ng MMaynilaaynila C

College of ollege of EEngineering andngineering and TTechnologyechnology C

Chemicalhemical EEngineeringngineering DDepartmentepartment

WATER STORAGE TANK WATER STORAGE TANK Design

Design DescriptiDescription:on:

Bulk storage of liquids is generally handled by closed tanks to prevent Bulk storage of liquids is generally handled by closed tanks to prevent esc

escape ape of of volvolatiatile le and and concontamtaminainatiotion. n. In In somsome e insinstantancesces, , sucsuch h as as watwaterer stora

storage, where ge, where contamcontaminatiination on and and dilutdilution ion are not are not a a factorfactor, , large reservlarge reservoirsoirs can

can be be emplemployed. Natural oyed. Natural terrterrain, ain, concrconcrete-wete-walled excavatioalled excavations, ns, or or concrconcreteete tanks are the typical construction. Reinforced-wall design is required and the tanks are the typical construction. Reinforced-wall design is required and the co

concncrerete te mumust st be be wawateterprproroofofed ed wiwith th a a susuititabable le papainint t to to prprevevenent t ananyy possibility of leaking.

possibility of leaking. Design Selection: Design Selection:

This tank is selected to supply the water needed by the spray washer, This tank is selected to supply the water needed by the spray washer, reactor (degumming machine), the sink – and – float tank and, hot washing reactor (degumming machine), the sink – and – float tank and, hot washing tank

tank Design

Design ConsideratiConsiderations:ons: 1

1)) CCaappaacciitty y oof f tthhe e ttaannkk 2

2)) TTyyppe e oof f mmatateerriiaal l bbeeiinng g hhananddlleedd 3

3)) CCllaassssiiffiiccatatiioon on of tf tanank tk to bo be ue usseedd 4

4)) MMaatteerriiaal l oof f ccoonnssttrruuccttiioonn Data and

Data and AssumptionsAssumptions:: 1)

1) The amount of water stored in the tank is 2952.77 kg H The amount of water stored in the tank is 2952.77 kg H22O.O.

2

2)) ThThe te tanank ik is vs venentted ed fofor ar an en easasy fy flolow ow of wf wataterer.. 3)

3) AsAssumsume He H= 4= 4/3 D/3 D, si, sincnce the this is is a is a cocommmmon ron ratiatio uso used fed for tor tanank dek desisigngns.s. 4

4)) AlAlllowowanancce oe of 2f 200% a% as ss safafetety fy facactotor ir is us usesed.d. Design

Design RequirementRequirements:s: 1)

1) 2

2)) VVoolluumme e oof f tthhe e ttaannkk 3

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4)) WWoorrkkiinng pg prreessssuurree 5 5)) SShheelll l tthhiicckknneessss 6 6)) HHeeaad d tthhiicckknneessss 7 7)) DDeepptth h oof f tthhe e hheeaadd 8 8)) VVoolluumme e oof f tthhe e hheeaadd 9 9)) SSuurrffaacce e aarreea a oof f tthhe e hheeaadd 1 100)) BBoottttoom tm thhiicckknneessss Design

Design CalculatiCalculations:ons:

The density and the mass of the water are as follows: The density and the mass of the water are as follows:

ρH2O=1000 kg/m3 ρH2O=1000 kg/m3 mH2O=2952.77 kg H2O mH2O=2952.77 kg H2O

Therefore the volume of the tank is, Therefore the volume of the tank is,

VH2O=2952.77 kg H2O1000 kg/m3 VH2O=2952.77 kg H2O1000 kg/m3 VH2O=2.95 m3

VH2O=2.95 m3

Basis: per batch of operation Basis: per batch of operation 1

1)) VVoolluumme e oof f tthhe e ttaannk k

Calculating for the volume of water in the tank and assuming 20% allowance Calculating for the volume of water in the tank and assuming 20% allowance as safety factor, as safety factor, Vtank=VH2O(1.20) Vtank=VH2O(1.20) Vtank=3.54 m3 Vtank=3.54 m3 Use 3.60 m

Use 3.60 m33 _{water storage tank.}_{water storage tank.}

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2)) HHeeiigghht at annd dd diiaammeetteerr

To compute for the standard ration of the water storage, assume H=4/3 D, To compute for the standard ration of the water storage, assume H=4/3 D,

Vtank=π4D2H Vtank=π4D2H But, H=4/3 D But, H=4/3 D

Therefore, the diameter and height of the tank are: Therefore, the diameter and height of the tank are:

Vtank=π3D3 Vtank=π3D3 3.54=π3D3 3.54=π3D3 D=1.50 m 4.92 ft;H=2.00 m (6.57 ft) D=1.50 m 4.92 ft;H=2.00 m (6.57 ft)

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Use 1.60 m and 2.00 m

Use 1.60 m and 2.00 m for diameter and height, respectively.for diameter and height, respectively. 3

3)) WWoorrkkiinng g pprreessssuurree Ptotal=Poptimum+Hρ

Ptotal=Poptimum+Hρ

Since the tank is vented, P

Since the tank is vented, Poptimumoptimum= 14.7 psi, and = 14.7 psi, and H=2.00m (6.57 ft)H=2.00m (6.57 ft)

Therefore, Therefore,

Ptotal=14.7lbs

Ptotal=14.7lbsin2+6.57ftft2144 in262.4 in2+6.57ftft2144 in262.4 lbsft3lbsft3 Ptotal=17.55 psi

Ptotal=17.55 psi Use 18 psi as

Use 18 psi as working pressure.working pressure. Plate Design:

Plate Design: 4

4)) SShheelll l tthhiicckknneessss

For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and Rushton, p. 85, Rushton, p. 85, ts=PD+C2Se-P ts=PD+C2Se-P Where: Where:

S = ultimate tensile strength S = ultimate tensile strength

P = maximum allowable working pressure P = maximum allowable working pressure D = diameter

D = diameter

C = allowance for corrosion C = allowance for corrosion e = efficiency

e = efficiency

To find the maximum allowable tensile strength, use Eqn. 4-1 of Process To find the maximum allowable tensile strength, use Eqn. 4-1 of Process

Equipment Design by Hesse and Rushton, p. 84, Equipment Design by Hesse and Rushton, p. 84, S=Su×Fa×Fr×Fs×Fm

S=Su×Fa×Fr×Fs×Fm Where:

Where:

Su = ultimate tensile strength Su = ultimate tensile strength Fa = radiograph factor

Fa = radiograph factor Fr = stress relieving factor Fr = stress relieving factor Fs = ultimate strength factor Fs = ultimate strength factor Fm = material factor

Fm = material factor Thus,

Thus,

Su =

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F

Fs s = = 2255%% ((TTaabblle e 44--22, , PPEEDD, , pp..8844)) Fm

Fm = = 1.00 1.00 (for (for high high tensile tensile strength strength carbon carbon steel, steel, PED, PED, p.81)p.81) Fr = 1.00

Fr = 1.00 F

Fa a = = 11..0000 ((iif f ssttrreesss s rreelliieevviinngg, , rraaddiiooggrraapphhiinng g iis s nnoot t rreeqquuiirreedd,, PED, p.88)

PED, p.88)

Substituting to the equation

Substituting to the equation of maximum allowable tensile strength,of maximum allowable tensile strength, S = 13,000 x 1.00 x 1.00 x 1.00 x 0.25 = 3250 psi

S = 13,000 x 1.00 x 1.00 x 1.00 x 0.25 = 3250 psi For double-butt joint,

For double-butt joint, e

e = = 0.80 0.80 (based (based on on material material factor, factor, PED, PED, p.89)p.89) For corrosion allowance,

For corrosion allowance, C

C = = 11//116 6 iinn ((PPllaannt t DDeessiiggn n aannd d EEccoonnoommiiccs s ffoor r CChheemmiiccaall Engineering by Peters, p. 542)

Engineering by Peters, p. 542) Substituting to the equation, Substituting to the equation, ts=PD+C2Se-P

ts=PD+C2Se-P ts=17.55

ts=17.55 lbs/in26.57ft1lbs/in26.57ft12 2 inft+1/16 inft+1/16 in23,250lbsin23,250lbsin20.80-17.55lbin20.80-17.55lbsin2sin2 ts=0.27 in ≈6.78 mm

ts=0.27 in ≈6.78 mm

Use 7 mm shell thickness of the tank Use 7 mm shell thickness of the tank 5

5)) HHeeaad d tthhiicckknneessss For thickness of head: For thickness of head:

A standard dished head was chosen for simplicity and availability A standard dished head was chosen for simplicity and availability th=Plw2Se

th=Plw2Se

Refer to Eqn. 4-6, p.86, PED Refer to Eqn. 4-6, p.86, PED Since:

Since: Di Di = = 1.50m 1.50m (59.04 (59.04 in)in) Do=Di+2ts Do=Di+2ts Do=59.58 in Do=59.58 in From PED, p.69 From PED, p.69

Crown radiusL=Di-6 (in in.) Crown radiusL=Di-6 (in in.) L=53.04 in L=53.04 in kr=knuckle radius=0.06 Do kr=knuckle radius=0.06 Do kr=3.57 in kr=3.57 in

Calculating the ratio, R = k Calculating the ratio, R = krr/L/L

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R=0.07 R=0.07

For the value of W, from table 4-3, p.87, PED For the value of W, from table 4-3, p.87, PED W=1.80

W=1.80

With S = 3250 psi, as calculated previously, the value of the head thickness, With S = 3250 psi, as calculated previously, the value of the head thickness, using the equation 4-6 of PED,

using the equation 4-6 of PED, th=PLW2Se

th=PLW2Se

th=0.32 in ≈8.18 mm th=0.32 in ≈8.18 mm

Use 9 mm as head thickness. Use 9 mm as head thickness. 6

6)) DDeepptth h oof f tthhe e hheeaad d ((hh)):: From Eqn. 4-14, p. 92, PED, From Eqn. 4-14, p. 92, PED, h=L-L2-D24 h=L-L2-D24 h=53.04-53.042-59.0424 h=53.04-53.042-59.0424 h=8.97 in ≈0.23 m h=8.97 in ≈0.23 m

Use 0.30 m as the head depth of the tank. Use 0.30 m as the head depth of the tank. 7

7)) VVoolluumme e oof f tthhe e hheeaad d ((VV)):: From Eqn. 4-15, p.92, PED,

From Eqn. 4-15, p.92, PED,

V=1.05h23L-h (all values in inches) V=1.05h23L-h (all values in inches)

Substituting the previously computed values to the above equation would Substituting the previously computed values to the above equation would give,

give,

V=12685.26 in3≈0.21 m3 V=12685.26 in3≈0.21 m3 Use 0.30 m

Use 0.30 m33 _{as volume of the head.}_{as volume of the head.}

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8)) SSuurrffaacce e aarreea a oof f tthhe e hheeaad d ((AA)):: From Eqn. 4-16, p.92, PED,

From Eqn. 4-16, p.92, PED, A=6.28 hL

A=6.28 hL

A=2987.83 in2≈1.93 m2 A=2987.83 in2≈1.93 m2 Use 2.00 m

Use 2.00 m22 _{for surface area of the tank.}_{for surface area of the tank.}

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The same values and calculation of thickness of head were done on The same values and calculation of thickness of head were done on thickness of bottom, since t

thickness of bottom, since theadhead = t= tbottombottom..

Hence, t

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SODIUM HYDROXIDE TANK SODIUM HYDROXIDE TANK Design

Design DescriptiDescription:on:

Bulk storage of liquids is generally handled by closed tanks to prevent Bulk storage of liquids is generally handled by closed tanks to prevent escape of volatile and contamination. Reinforced-wall design is required and escape of volatile and contamination. Reinforced-wall design is required and the concr

the concrete must ete must be be watwatererproproofeofed d witwith h a a suisuitabtable le paipaint nt to to preprevenvent t anyany possibility of leaking.

possibility of leaking. Desin Selection: Desin Selection:

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This tank is selected to store and supply the sodium hydroxide needed This tank is selected to store and supply the sodium hydroxide needed in the alkaline sorbing

in the alkaline sorbing (degumming) process.(degumming) process. Design

Design ConsideratiConsiderations:ons: 1

1)) CCaappaacciitty y oof f tthhe e ttaannkk 2

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4)) MMaatteerriiaal l oof f ccoonnssttrruuccttiioonn 5

5)) QQuauantntiity ty oof mf matateerriaial ml mooveved pd per er ununiit tt tiimeme Data and

1) ThThe te tank ank is is clclososed ed to to avavoioid cd conontatamiminanatition oon of tf the he NaNaOH OH sosolulutitionon.. 2)

2) Density of sodium hydroxide is Density of sodium hydroxide is 2100 kg/m2100 kg/m33

3)

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4)) AlAlllowowanancce oe of 2f 200% a% as ss safafetety fy facactotor ir is us usesed.d. 5)

5) Mass of NaOH is 93.60 kg (refer to the material balance)Mass of NaOH is 93.60 kg (refer to the material balance) Design

Design RequirementRequirements:s: 1)

1) 2

2)) VVoolluumme e oof f tthhe e ttaannkk 3

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4)) WWoorrkkiinng pg prreessssuurree 5 5)) SShheelll l tthhiicckknneessss 6 6)) HHeeaad d tthhiicckknneessss 7 7)) DDeepptth h oof f tthhe e hheeaadd 8 8)) VVoolluumme e oof f tthhe e hheeaadd 9 9)) SSuurrffaacce e aarreea a oof f tthhe e hheeaadd 1 100)) BBoottttoom tm thhiicckknneessss

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Design

Design CalculatiCalculations:ons:

The density and the mass of the sodium hydroxide are as follows: The density and the mass of the sodium hydroxide are as follows:

ρNaOH=(2100kgm3) ρNaOH=(2100kgm3) mNaOH=93.60 kg mNaOH=93.60 kg

Therefore the volume of the tank is, Therefore the volume of the tank is,

VNaOH=93.60 kg 2100 kg/m3 VNaOH=93.60 kg 2100 kg/m3

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VNaOH=0.04 m3 VNaOH=0.04 m3

Basis: five days of operation (15 batches) Basis: five days of operation (15 batches) 1

1)) VVoolluumme e oof f tthhe e ttaannk k

Calculating for the volume of water in the tank and assuming 20% allowance Calculating for the volume of water in the tank and assuming 20% allowance as safety factor,

as safety factor, Vtank=VNaOH1.2(1

Vtank=VNaOH1.2(15 batches 5 batches )) Vtank=0.79 m3

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Use 0.8 m

Use 0.8 m33 _{water storage tank.}_{water storage tank.}

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2)) HHeeiigghht at annd dd diiaammeetteerr

To compute for the standard ration of the water storage, assume H=4/3 D, To compute for the standard ration of the water storage, assume H=4/3 D,

Vtank=π4D2H Vtank=π4D2H But, H=4/3 D But, H=4/3 D

Therefore, the diameter and height of the tank are: Therefore, the diameter and height of the tank are:

Vtank=π3D3 Vtank=π3D3

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0.79=π3D3 0.79=π3D3

D=0.91 m 2.99 ft;H=1.21 m (3.98 ft) D=0.91 m 2.99 ft;H=1.21 m (3.98 ft)

Use 1.0 m and 1.30 m for the diameter and height of the storage tank, Use 1.0 m and 1.30 m for the diameter and height of the storage tank, respectively.

respectively. 3

3)) WWoorrkkiinng g pprreessssuurree Ptotal=Poptimum+Hρ

Ptotal=Poptimum+Hρ

Since the tank is not vented, P

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Therefore, Therefore,

Ptotal=14.7+3.9

Ptotal=14.7+3.98ftft2144 8ftft2144 in2130.92 lbsft3in2130.92 lbsft3 Ptotal=18.32 psi

Ptotal=18.32 psi

Use 19 psi as working pressure. Use 19 psi as working pressure. Plate Design:

Plate Design:

4.) Shell thickness 4.) Shell thickness

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For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and Rushton, p. 85, Rushton, p. 85, ts=PD+C2Se-P ts=PD+C2Se-P Where: Where:

S = ultimate tensile strength S = ultimate tensile strength

P = maximum allowable working pressure P = maximum allowable working pressure D = diameter

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C = allowance for corrosion C = allowance for corrosion e = efficiency

e = efficiency

To find the maximum allowable tensile strength,

To find the maximum allowable tensile strength, use Eqn. 4-1 of use Eqn. 4-1 of ProcessProcess Equipment Design by Hesse and Rushton, p. 84,

Equipment Design by Hesse and Rushton, p. 84, S=Su×Fa×Fr×Fs×Fm

S=Su×Fa×Fr×Fs×Fm Where:

Where:

Su = ultimate tensile strength Su = ultimate tensile strength

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Fa = radiograph factor Fa = radiograph factor Fr = stress relieving factor Fr = stress relieving factor Fs = ultimate strength factor Fs = ultimate strength factor Fm = material factor

Fm = material factor Thus,

Thus, Su

Su = 9= 900000 p0 psisi (f(for or StStaiainlnlesess ss steteel el tytype pe 30304, 4, TiTimmmmererhauhaus)s) F

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Fm

Fm = = 1.00 1.00 (for (for high high tensile tensile strength strength carbon carbon steel, steel, PED, PED, p.81)p.81) Fr = 1.00

Fr = 1.00 F

Fa = a = 11..0000 ((iif sf sttrreesss rs reelliieevviinngg, r, raaddiiooggrraapphhiinng ig is ns noot rt reeqquuiirreedd,, PED, p.88)

PED, p.88)

Substituting to the equation

Substituting to the equation of maximum allowable tensile strength,of maximum allowable tensile strength, S = 9000 x 1.00 x 1.00 x 1.00 x 0.25 = 2250 psi

S = 9000 x 1.00 x 1.00 x 1.00 x 0.25 = 2250 psi For double-butt joint,

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e

e = = 0.80 0.80 (based (based on on material material factor, factor, PED, PED, p.89)p.89) For corrosion allowance,

For corrosion allowance, C

C = = 1/16 1/16 in in (Plant (Plant Design Design and and Economics Economics for for ChemicalChemical Engineering by Peters, p. 542)

Engineering by Peters, p. 542) Substituting to the equation, Substituting to the equation, ts=PD+C2Se-P

ts=PD+C2Se-P

ts=18.32 psi(35.88 in)+1/16

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ts=0.18 in ≈4.66 mm ts=0.18 in ≈4.66 mm

Use 5 mm shell thickness of the tank. Use 5 mm shell thickness of the tank. 5.) Head thickness:

5.) Head thickness: For thickness of head: For thickness of head:

A standard dished head was chosen for simplicity and availability A standard dished head was chosen for simplicity and availability th=Plw2Se

th=Plw2Se

Refer to Eqn. 4-6, p.86, PED Refer to Eqn. 4-6, p.86, PED

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Since:

Since: Di Di = = 0.91 0.91 m m (35.88 (35.88 in)in) Do=Di+2ts Do=Di+2ts Do=36.24 in Do=36.24 in From PED, p.69 From PED, p.69

Crown radiusL=Di-6 (in in.) Crown radiusL=Di-6 (in in.) L=29.88 in

L=29.88 in

kr=knuckle radius=0.06 Do kr=knuckle radius=0.06 Do

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kr=2.17 in kr=2.17 in

Calculating the ratio, R = k Calculating the ratio, R = krr/L/L

R=0.07 R=0.07

For the value of W, from table 4-3, p.87, PED For the value of W, from table 4-3, p.87, PED W=1.80

W=1.80

With S = 2250 psi, as calculated previously, the value of the head thickness, With S = 2250 psi, as calculated previously, the value of the head thickness, using the equation 4-6 of PED,

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th=PLW2Se th=PLW2Se

th=0.27 in ≈6.95 mm th=0.27 in ≈6.95 mm

Use 7 mm as head thickness. Use 7 mm as head thickness. 1

1)) DDeepptth h oof f tthhe e hheeaad d ((hh)):: From Eqn. 4-14, p. 92, PED, From Eqn. 4-14, p. 92, PED, h=L-L2-D24

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h=29.88-(29.88)2-35.8824 h=29.88-(29.88)2-35.8824 h=5.99 in ≈0.15m

h=5.99 in ≈0.15m

Use 0.2 m as the head depth of the tank. Use 0.2 m as the head depth of the tank. 2

2)) VVoolluumme e oof f tthhe e hheeaad d ((VV)):: From Eqn. 4-15, p.92, PED,

From Eqn. 4-15, p.92, PED,

V=1.05h23L-h (all values in inches) V=1.05h23L-h (all values in inches)

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Substituting the previously computed values to the

Substituting the previously computed values to the above equation wouldabove equation would give,

give,

V=3151.44 in3= 0.05 m3 V=3151.44 in3= 0.05 m3 Use 0.1 m

Use 0.1 m33 _{as volume of the head.}_{as volume of the head.}

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3)) SSuurrffaacce e aarreea a oof f tthhe e hheeaad d ((AA)):: From Eqn. 4-16, p.92, PED,

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A=6.28 hL A=6.28 hL

A=1124.00 in2=0.73m2 A=1124.00 in2=0.73m2

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Use 0.80 m

Use 0.80 m22 _{as the surface area of the head.}_{as the surface area of the head.}

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4)) BBoottttoom m tthhiicckknneessss::

The same values and calculation of thickness of head were done on The same values and calculation of thickness of head were done on thickness of bottom, since t

thickness of bottom, since theadhead = t= tbottombottom..

Hence, t

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BELT CONVEYOR WITH SPRAY WASHER BELT CONVEYOR WITH SPRAY WASHER Design

Design DescriptiDescription:on: Be

Belt lt coconvnveyeyoror, , as as ththe e naname me susuggggeseststs, , coconsnsisists ts of of enendldlesess s bebeltlts,s, suitab

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place. Belts are made of canvas, reinforced rubber or balata and strip steel. place. Belts are made of canvas, reinforced rubber or balata and strip steel. St

Stririp p ststeeeel l is is ememplployoyed ed fofor r coconvnveyeyining g mamateteririalals s ththrorougugh h fufurnrnacaceses. . BeBeltlt conveyors are adapted to wide varieties and quantities of materials; require conveyors are adapted to wide varieties and quantities of materials; require relatively low power and can transport solids for a long distance. The width relatively low power and can transport solids for a long distance. The width of

of ththe e bebelt lt vavariries es frfrom om 14 14 to to 16 16 inin, , anand d ththe e nunumbmber er of of ididlelers rs vvararieiess cor

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down to 3 feet for the widest belts. This conveyor is designed with a built-in down to 3 feet for the widest belts. This conveyor is designed with a built-in spray washer.

spray washer.

Design Selection: Design Selection:

The belt conveyor is selected to transport the water hyacinth stalks to The belt conveyor is selected to transport the water hyacinth stalks to the pressing equipment. It is designed with built – in spray washer to wash the pressing equipment. It is designed with built – in spray washer to wash the stalks at

the stalks at the same time they are being conveyed.the same time they are being conveyed. Design

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1)) CCapapaacciitty y oof f tthhe e bbeellt t ccoonnvveeyyoorr 2

2)) LLenengtgth h of of trtravaveel/l/llenengtgth h of of ththe e bebeltlt 3

4)) SSppeeeed d oof f tthhe e ccoonnvveeyyoorr 5

5)) NNuummbbeer r oof f sspprraay y nnoozzzzlleess Data and

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2)

2) NoNormrmal sal spepeed red rangange of be of belelt cot convnveyeyor ior is bes betwtweeeen 20n 200 to 40 to 400 f00 ft/t/mimin (frn (from Uom Uninitt Operations by Brown, p. 55)

Operations by Brown, p. 55) 3)

3) Ratio of feed to wash water is 1:2Ratio of feed to wash water is 1:2 4)

4) The belt width is The belt width is 36 in 36 in (3 ft) (3 ft) (fro(from Perry’s Chemicm Perry’s Chemical Engineeral Engineering Handbooking Handbook section 21-10

section 21-10 5)

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6)

6) The lump size of feed is 18 in for 36 in width belt conveyor (Unit Operations The lump size of feed is 18 in for 36 in width belt conveyor (Unit Operations by Brown, p. 58)

by Brown, p. 58) 7)

7) ThThe se safafetety fy facactotor fr for or bebelt lt coconvnveyeyor or is is 15 15 % (% (TiTimmmmererhahausus, p. , p. 3636)) Design

Design RequirementRequirements:s: 1

1)) CCaappaacciitty y oof f sspprraay y nnoozzzzlleess 2

2)) CCaappaacciitty y oof f bbeellt t ccoonnvveeyyoorr 3

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Design

Design ComputatioComputations:ns: From material balance: From material balance:

Inlet capacity is 968.80 kg/batch of

Inlet capacity is 968.80 kg/batch of water hyacinth stalkswater hyacinth stalks 1)

1) CapacCapacity of sity of spray npray nozzlozzle:e: V=MD

V=MD W

Whheerree:: MM= = mmaasss s oof f wwaatteerr D = density of water D = density of water

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Since water to feed ratio

Since water to feed ratio is 2:1,is 2:1, M=2(968.80 kg) M=2(968.80 kg) M=1937.60 kg H2O M=1937.60 kg H2O D=1000 kg/m3 D=1000 kg/m3 V= 1.94 m39 nozzles V= 1.94 m39 nozzles V= 0.22m3nozzle V= 0.22m3nozzle Use 0.30 m

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2)

2) CapacCapacity of the belity of the belt convet conveyor (T)yor (T) T=968.80 kgbatch×1 batch8

T=968.80 kgbatch×1 batch8 hours×2.2hours×2.2lbskg×1 ton2000 lbslbskg×1 ton2000 lbs T=0.13tonhr 266.42kghr

T=0.13tonhr 266.42kghr

Giving an allowance factor of 15%, for future expansion, Giving an allowance factor of 15%, for future expansion, T=0.13tonhr(1.15)

T=0.13tonhr(1.15)

T=0.15tonhr 305.90kghr T=0.15tonhr 305.90kghr

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Use 310 kg/hr as belt

Use 310 kg/hr as belt conveyor capacity.conveyor capacity. 3)

3) Power Power requirequiremenrementt

Using equations for power requirement (Unit Operations by Brown, p.58) for Using equations for power requirement (Unit Operations by Brown, p.58) for plain bearings,

plain bearings,

Hp=FL+LoT+0.03WS+T∆z990 Hp=FL+LoT+0.03WS+T∆z990 W

Whheerree:: HHpp = = horsepower horsepower requiredrequired

F

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L

L = = length length of of conveyor conveyor between between terminal terminal pulleys, pulleys, ft.ft. L

Loo = = 100 100 for for plain plain bearingsbearings

S

S = = speed speed of of belt, belt, fpmfpm T

T = = capacity capacity of of belt belt conveyor, conveyor, ton/hrton/hr

∆

∆z = z = increase iincrease in elevation elevation of mn of material, ftaterial, ft W

W = = masmass s of of movmoving parts incling parts includiuding ng belbelts and ts and idlidlers per ers per foofoott distance centers of terminal pulleys (both runs),

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Computing for W, Computing for W,

From table 16, Unit Operations by

From table 16, Unit Operations by Brown, p.58Brown, p.58

Approximate weight of belt conveyors = 1.0 lb/in of width per running Approximate weight of belt conveyors = 1.0 lb/in of width per running foot

foot

W=1lbin-ft36 in(2 runs) W=1lbin-ft36 in(2 runs) W=72lbsft

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Assuming L = 30 ft, since the length of the conveyor should be greater than Assuming L = 30 ft, since the length of the conveyor should be greater than the length of the spray washer,

the length of the spray washer, Hp=0.0530ft+1000

Hp=0.0530ft+1000.15tonhr+0.037.15tonhr+0.0372lbsft200 2lbsft200 ftmin ftmin +(0)0.15990+(0)0.15990 Hp=2.84 hp

Hp=2.84 hp Use 3 Hp since