Storage Tank Design

Pamantasan ng Lungsod ng MaynilaCollege of Engineering and Technology

Chemical Engineering Department

WATER STORAGE TANK

Design Description:

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

escape of volatile and contamination. In some instances, such as water

storage, where contamination and dilution are not a factor, large reservoirs

can be employed. Natural terrain, concrete-walled excavations, or concrete

tanks are the typical construction. Reinforced-wall design is required and the

concrete must be waterproofed with a suitable paint to prevent any

possibility of leaking.

Design Selection:

This tank is selected to supply the water needed by the spray washer,

reactor (degumming machine), the sink – and – float tank and, hot washing

tank

Design Considerations:

1.) Capacity of the tank

2.) Type of material being handled

3.) Classification of tank to be used

4.) Material of construction

Data and Assumptions:

1.) The amount of water stored in the tank is 2952.77 kg H2O.

2.) The tank is vented for an easy flow of water.

3.) Assume H= 4/3 D, since this is a common ratio used for tank designs.

4.) Allowance of 20% as safety factor is used.

Design Requirements:

1.) Volume of the tank

2.) Height and diameter of the

tank

3.) Working pressure

4.) Shell thickness

5.) Head thickness

Spinning of Alkaline Sorbed Fibers from Eichhornia Crassipes (Water Hyacinth) Stalks and Extruded Polyethylene Terephthalate (PET) Bottles for the Production of Textile Yarn



6.) Depth of the head

7.) Volume of the head

8.) Surface area of the head

9.) Bottom thickness

Design Calculations:

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

ρH 2O=1000kg /m3

mH2O=2952.77 kg H 2O

Therefore the volume of the tank is,

V H 2O=2952.77kg H 2O

1000kg/m3

V H 2O=2 .95m3

Basis: per batch of operation


Calculating for the volume of water in the tank and assuming 20% allowance

as safety factor,

V tank=V H 2O(1.20)

V tank=3 .54m3

Use 3.60 m3 water storage tank.

2.) Height and diameter

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

V tank=π4D2H

But, H=4/3 D

Therefore, the diameter and height of the tank are:

V tank=π3D3

3 .54= π3D3

D=1.50m (4 .92 ft ); H=2.00m(6.57 ft)

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





Ptotal=Poptimum+Hρ

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

Therefore,

Ptotal=14.7 lbs

¿2+ (6.57 ft )( ft 2

144 ¿2 )( 62.4 lbsft3 )Ptotal=17.55 psi

Use 18 psi as working pressure.

Plate Design:

4.) Shell thickness

For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and

Rushton, p. 85,

t s=( PD+C2Se−P )

Where:

S = ultimate tensile strength

P = maximum allowable working pressure

D = diameter

C = allowance for corrosion

e = efficiency

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

Equipment Design by Hesse and Rushton, p. 84,

S=Su×Fa×Fr× Fs×Fm

Where:

Su = ultimate tensile strength

Fa = radiograph factor

Fr = stress relieving factor

Fs = ultimate strength factor

Fm = material factor

Thus,




Su = 13000 psi (for low-carbon nickel steel, PED, p.69)

Fs = 25% (Table 4-2, PED, p.84)

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

Fr = 1.00

Fa = 1.00 (if stress relieving, radiographing is not required,

PED, p.88)

Substituting to the equation of maximum allowable tensile strength,

S = 13,000 x 1.00 x 1.00 x 1.00 x 0.25 = 3250 psi

For double-butt joint,

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

For corrosion allowance,

C = 1/16 in (Plant Design and Economics for Chemical

Engineering by Peters, p. 542)

Substituting to the equation,

t s=( PD+C2Se−P )t s=¿

t s=0.27∈≈6 .78mm

Use 7 mm shell thickness of the tank

5.) Head thickness

For thickness of head:

A standard dished head was chosen for simplicity and availability

t h=Plw2Se

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

Since: Di = 1.50m (59.04 in)

Do=Di+2t s

Do=59 .58∈¿

From PED, p.69

Crownradius (L )=Di−6 (¿∈.)




L=53 .04∈¿

k r=knuckle radius=0.06D o

k r=3.57∈¿

Calculating the ratio, R = kr/L

R=0.07

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

W=1.80

With S = 3250 psi, as calculated previously, the value of the head thickness,

using the equation 4-6 of PED,

t h=P LW2 Se

t h=0.32∈≈8.18mm

Use 9 mm as head thickness.

6.) Depth of the head (h):

From Eqn. 4-14, p. 92, PED,

h=L−√L2−D 2

4

h=53.04−√ (53 .04 )2−(59 .04 )2

4

h=8 .97∈≈0.23m

Use 0.30 m as the head depth of the tank.

7.) Volume of the head (V):

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

V=1.05h2 (3 L−h )(all values∈inc hes)

Substituting the previously computed values to the above equation would

give,

V=12685 .26¿3≈0.21m3

Use 0.30 m3 as volume of the head.




8.) Surface area of the head (A):


A=6.28h L

A=2987 .83¿2≈1 .93m2

Use 2.00 m2 for surface area of the tank.

9.) Bottom thickness:

The same values and calculation of thickness of head were done on

thickness of bottom, since thead = tbottom.

Hence, tbottom = 8.18 mm (9 mm).




SODIUM HYDROXIDE TANK

Design Description:

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

escape of volatile and contamination. Reinforced-wall design is required and

the concrete must be waterproofed with a suitable paint to prevent any

possibility of leaking.

Desin Selection:




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

in the alkaline sorbing (degumming) process.


1.) Capacity of the tank


3.) Classification of tank to be used




4.) Material of construction

5.) Quantity of material moved per unit time


1.) The tank is closed to avoid contamination of the NaOH solution.

2.) Density of sodium hydroxide is 2100 kg/m3

3.) Assume H= 4/3 D, since this is a common ratio used for tank designs.




4.) Allowance of 20% as safety factor is used.

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



2.) Height and diameter of the

tank


4.) Shell thickness

5.) Head thickness




6.) Depth of the head

7.) Volume of the head

8.) Surface area of the head

9.) Bottom thickness

Design Calculations:

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

ρNaOH=(2100 kgm3

)




mNaOH=93.60kg

Therefore the volume of the tank is,

V NaOH=93.60kg

2100kg/m3

V NaOH=0.04m3

Basis: five days of operation (15 batches)





Calculating for the volume of water in the tank and assuming 20% allowance

as safety factor,

V tank=V NaOH (1.2 )(15batches)

V tank=0.79m3

Use 0.8 m3 water storage tank.

2.) Height and diameter

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




V tank=π4D2H

But, H=4/3 D

Therefore, the diameter and height of the tank are:

V tank=π3D3

0 .79= π3D3




D=0 .91m (2.99 ft ); H=1.21m(3.98 ft )

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

respectively.


Ptotal=Poptimum+Hρ

Since the tank is not vented, Poptimum= 14.7, and H=1.21 m ( 3.98 ft)

Therefore,




Ptotal=14.7+ (3 .98 ft )( ft2

144¿2 )( 130.92 lbsft3 )Ptotal=18 .32 psi

Use 19 psi as working pressure.

Plate Design:

4.) Shell thickness




For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and

Rushton, p. 85,

t s=( PD+C2Se−P )

Where:

S = ultimate tensile strength

P = maximum allowable working pressure




D = diameter

C = allowance for corrosion

e = efficiency

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

Equipment Design by Hesse and Rushton, p. 84,

S=Su×Fa×Fr× Fs×Fm

Where:




Su = ultimate tensile strength

Fa = radiograph factor

Fr = stress relieving factor

Fs = ultimate strength factor

Fm = material factor

Thus,

Su = 9000 psi (for Stainless steel type 304, Timmerhaus)




Fs = 25% (Table 4-2, PED, p.84)

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

Fr = 1.00

Fa = 1.00 (if stress relieving, radiographing is not required,

PED, p.88)

Substituting to the equation of maximum allowable tensile strength,

S = 9000 x 1.00 x 1.00 x 1.00 x 0.25 = 2250 psi




For double-butt joint,

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

For corrosion allowance,

C = 1/16 in (Plant Design and Economics for Chemical

Engineering by Peters, p. 542)

Substituting to the equation,




t s=( PD+C2Se−P )t s=¿

t s=0.18∈≈4 .66mm

Use 5 mm shell thickness of the tank.

5.) Head thickness:

For thickness of head:




A standard dished head was chosen for simplicity and availability

t h=Plw2Se

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

Since: Di = 0.91 m (35.88 in)

Do=Di+2t s

Do=36 .24∈¿




From PED, p.69

Crownradius (L )=Di−6 (¿∈.)

L=29 .88∈¿

k r=knuckle radius=0.06D o

k r=2 .17∈¿

Calculating the ratio, R = kr/L

R=0.07




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

W=1.80

With S = 2250 psi, as calculated previously, the value of the head thickness,

using the equation 4-6 of PED,

t h=PLW2Se

t h=0.27∈≈6 .95mm




Use 7 mm as head thickness.

6.) Depth of the head (h):

From Eqn. 4-14, p. 92, PED,

h=L−√L2−D 2

4




h=29.88−√(29.88)2−(35.88 )2

4

h=5 .99∈≈0.15m

Use 0.2 m as the head depth of the tank.

7.) Volume of the head (V):





V=1.05h2 (3 L−h )(all values∈inc hes)

Substituting the previously computed values to the above equation would

give,

V=3151.44 ¿3=0.05m3

Use 0.1 m3 as volume of the head.

8.) Surface area of the head (A):





A=6.28h L

A=1124 .00¿2=0.73m2

Use 0.80 m2 as the surface area of the head.

9.) Bottom thickness:




The same values and calculation of thickness of head were done on

thickness of bottom, since thead = tbottom.

Hence, tbottom = 6.97 mm (7 mm).




BELT CONVEYOR WITH SPRAY WASHER

Design Description:

Belt conveyor, as the name suggests, consists of endless belts,

suitably supported and driven, which carry or transport solids from place to

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

Strip steel is employed for conveying materials through furnaces. Belt

conveyors are adapted to wide varieties and quantities of materials; require




relatively low power and can transport solids for a long distance. The width

of the belt varies from 14 to 16 in, and the number of idlers varies

correspondingly. This spacing ranges from about 5 feet for narrow belts

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

spray washer.

Design Selection:




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 stalks at the same time they are being conveyed.


1.) Capacity of the belt conveyor

2.) Length of travel/length of the belt





4.) Speed of the conveyor

5.) Number of spray nozzles


1.) Capacity based from material balance is 968.80 kg

2.) Normal speed range of belt conveyor is between 200 to 400 ft/min

(from Unit Operations by Brown, p. 55)

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




4.) The belt width is 36 in (3 ft) (from Perry’s Chemical Engineering

Handbook section 21-10

5.) The spray washer has 9 nozzles

6.) The lump size of feed is 18 in for 36 in width belt conveyor (Unit

Operations by Brown, p. 58)

7.) The safety factor for belt conveyor is 15 % (Timmerhaus, p. 36)





1.) Capacity of spray nozzles

2.) Capacity of belt conveyor

3.) Power requirement

Design Computations:

From material balance:

Inlet capacity is 968.80 kg/batch of water hyacinth stalks

1.) Capacity of spray nozzle:




V=MD

Where: M= mass of water

D = density of water

Since water to feed ratio is 2:1,

M=2(968 .80 kg)

M=1937 .60kg H 2O




D=1000 kg/m3

V= 1 .94m3

9nozzles

V=0.22m3

nozzle

Use 0.30 m3 of water per spray nozzle for washing.




2.) Capacity of the belt conveyor (T)

T=968 .80kgbatch

×1batch8hours

×2.2lbskg

×1ton2000 lbs

T=0.13 tonhr (266 .42 kghr )

Giving an allowance factor of 15%, for future expansion,

T=0.13 tonhr

(1.15)




T=0.15 tonhr (305 .90 kghr )

Use 310 kg/hr as belt conveyor capacity.

3.) Power requirement

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

plain bearings,




H p=F (L+Lo ) (T +0.03WS )+T ∆ z

990

Where: Hp = horsepower required

F = friction factor, 0.05 for plain bearings

L = length of conveyor between terminal pulleys, ft.

Lo = 100 for plain bearings

S = speed of belt, fpm




T = capacity of belt conveyor, ton/hr

∆z = increase in elevation of material, ft

W = mass of moving parts including belts and idlers per foot

distance centers of terminal pulleys (both runs), lbs

Computing for W,

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




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

foot

W=1 lb¿−ft

¿

W=72 lbsft




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

the length of the spray washer,

H p=

(0.05 ) (30 ft+100 ){0.15 tonhr +[ (0.03 )( 72lbsft )( 200 ftmin )]}+(0) (0.15 )

990

H p=2.84hp

Use 3 Hp since it is commercially available.





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Storage Tank Design