Storage Tank Design
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Pamantasan ng Lungsod ng Maynila College of E of Engineering and Technology Chemical Engineering Department
WATER STORAGE TANK Design Description: Description: Bulk storage of liquids is generally handled by closed tanks to prevent escape escape of volati volatile le and contam contamina inatio tion. n. In some some ins instan tances ces,, such such as water water storage, storage, where contaminati contamination on and dilution dilution are not a factor, factor, large reservoirs reservoirs can be employed. employed. Natural terrain, terrain, concrete-w concrete-walled alled excavations, excavations, or concrete concrete tanks are the typical construction. Reinforced-wall design is required and the conc concre rete te must must be wate waterp rpro roof ofed ed with with a su suit itab able le pain paintt to prev preven entt any 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: Considerations: 1)
Capacity of the tank
2)
Type of mate aterial being hand andled
3)
Class ssiificati ation of of ttan ank k to to be be us used
4)
Material of construction
Data and Assumptions: Assumptions: 1)
The amount of water stored in the tank is 2952.77 kg H 2O.
2)
The The tan tank k is is ven ventted for for an an eas easy y flo flow w of of wat water er..
3)
Assum As sume e H= H= 4 4/3 /3 D, D, si sinc nce e this this is a comm common on rati ratio o used used for for tan tank k desi design gns. s.
4)
All Allowan owanc ce of of 20 20% as as saf safet ety y fac facto torr is is use used. d.
Design Requirements: Requirements: 1) 2)
Volume of the tank
3)
Height and diameter of the tank
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4)
Working pr pressure
5)
Shell thickness
6)
Head thickness
7)
Depth of the head
8)
Volume of the head
9)
Surface area of the head
10)
Bottom th thickness
Design Calculations: Calculations: The density and the mass of the water are as follows: ρH2O=1000 kg/m3 mH2O=2952.77 kg H2O
Therefore the volume of the tank is, VH2O=2952.77 kg H2O1000 kg/m3 VH2O=2.95 m3
Basis: per batch of operation
1)
Volume of the tank
Calculating for the volume of water in the tank and assuming 20% allowance as safety factor, Vtank=VH2O(1.20) Vtank=3.54 m3
Use 3.60 m3 water storage tank.
2)
Height an and di diameter
To compute for the standard ration of the water storage, assume H=4/3 D, Vtank=π4D2H
But, H=4/3 D Therefore, the diameter and height of the tank are: Vtank=π3D3 3.54=π3D3 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 for diameter and height, respectively.
3)
Working pressure
Ptotal=Poptimum+Hρ
Since the tank is vented, Poptimum= 14.7 psi, and H=2.00m (6.57 ft) Therefore, Ptotal=14.7lbsin2+6.57ftft2144 Ptotal=14.7lbsin2+6.57ftft2144 in262.4 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, ts=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 = 13 1300 000 0p psi si
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(for (for lowlow-car carbon bon nickel nickel steel, steel, PED, PED, p.69 p.69))
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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
a nd
Economics
for
Chemical
Engineering by Peters, p. 542) Substituting to the equation, ts=PD+C2Se-P ts=17.55 lbs/in26.57ft12 lbs/in26.57ft12 inft+1/16 in23,250lbsin20.80-17.55lb in23,250lbsin20.80-17.55lbsin2 sin2 ts=0.27 in ≈6.78 mm
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 th=Plw2Se
Refer to Eqn. 4-6, p.86, PED Since:
Di = 1.50m (59.04 in)
Do=Di+2ts Do=59.58 in
From PED, p.69 Crown radiusL=Di-6 (in in.) L=53.04 in kr=knuckle radius=0.06 Do kr=3.57 in
Calculating the ratio, R = kr/L S pinning
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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, th=PLW2Se th=0.32 in ≈8.18 mm
Use 9 mm as head thickness.
6)
Depth of the head (h):
From Eqn. 4-14, p. 92, PED, h=L-L2-D24 h=53.04-53.042-59.0424 h=8.97 in ≈0.23 m
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=1.05h23L-h (all values in inches)
Substituting the previously computed values to the above equation would give, V=12685.26 in3≈0.21 m3
Use 0.30 m3 as volume of the head.
8)
Surface area of the head (A):
From Eqn. 4-16, p.92, PED, A=6.28 hL A=2987.83 in2≈1.93 m2
Use 2.00 m2 for surface area of the tank.
9)
Bottom thickness:
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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).
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SODIUM HYDROXIDE TANK Design Description: 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 concrete must be water waterpro proofe ofed d with with a sui suitab table le paint paint to preven preventt any possibility of leaking.
Desin Selection: S pinning
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This tank is selected to store and supply the sodium hydroxide needed in the alkaline sorbing (degumming) process.
Design Considerations: Considerations: 1)
Capacity of the tank
2)
Type of mate aterial being hand andled
3)
Class ssiificati ation of of ttan ank k to to be be us used
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4)
Material of construction
5)
Quant uantiity of mat mate erial ial mo moved ved per per uni unit ti time
Data and Assumptions: Assumptions: 1)
The The tank tank is clos closed ed to avoi avoid d con conta tami mina nati tion on of of the the NaOH NaOH solu soluti tion on..
2)
Density of sodium hydroxide is 2100 kg/m3
3)
Assum As sume e H= H= 4 4/3 /3 D, D, si sinc nce e this this is a comm common on rati ratio o used used for for tan tank k desi design gns. s.
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4)
All Allowan owanc ce of of 20 20% as as saf safet ety y fac facto torr is is use used. d.
5)
Mass of NaOH is 93.60 kg (refer to the material balance)
Design Requirements: Requirements: 1) 2)
Volume of the tank
3)
Height and diameter of the tank
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4)
Working pr pressure
5)
Shell thickness
6)
Head thickness
7)
Depth of the head
8)
Volume of the head
9)
Surface area of the head
10)
Bottom th thickness
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Design Calculations: Calculations: The density and the mass of the sodium hydroxide are as follows: ρNaOH=(2100kgm3) mNaOH=93.60 kg
Therefore the volume of the tank is, VNaOH=93.60 kg 2100 kg/m3 S pinning
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VNaOH=0.04 m3
Basis: five days of operation (15 batches)
1)
Volume of the tank
Calculating for the volume of water in the tank and assuming 20% allowance as safety factor, Vtank=VNaOH1.2(15 Vtank=VNaOH1.2(1 5 batches ) Vtank=0.79 m3 S pinning
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Use 0.8 m3 water storage tank.
2)
Height an and di diameter
To compute for the standard ration of the water storage, assume H=4/3 D, Vtank=π4D2H
But, H=4/3 D Therefore, the diameter and height of the tank are: Vtank=π3D3 S pinning
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0.79=π3D3 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, respectively.
3)
Working pressure
Ptotal=Poptimum+Hρ
Since the tank is not vented, Poptimum= 14.7, and H=1.21 m ( 3.98 ft) S pinning
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Therefore, Ptotal=14.7+3.98ftft2144 Ptotal=14.7+3.98ftft2144 in2130.92 lbsft3 Ptotal=18.32 psi
Use 19 psi as working pressure.
Plate Design: 4.) Shell thickness
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For shell thickness, use Eqn. 4-3, Process Equipment Design by Hesse and Rushton, p. 85, ts=PD+C2Se-P
Where: S = ultimate tensile strength P = maximum allowable working pressure D = diameter S pinning
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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 S pinning
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Fa = radiograph factor Fr = stress relieving factor Fs = ultimate strength factor Fm = material factor Thus, Su = 9 900 000 0 psi psi
(for (for Stai Stainl nles ess s ste steel el type type 30 304, 4, Timm Timmer erhau haus) s)
Fs = 25%
(Table 4-2, PED, p.84)
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Fm = 1.00
(for high tensile strength carbon steel, PED, p.81)
Fr = 1.00 Fa = 1.00
(if st stress re relieving, ra radiographing iis s no not re 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, S pinning
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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, ts=PD+C2Se-P ts=18.32 psi(35.88 in)+1/16 in22250lbsin20.80-18.32 in22250lbsin20.80-18.32 psi S pinning
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ts=0.18 in ≈4.66 mm
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 th=Plw2Se
Refer to Eqn. 4-6, p.86, PED S pinning
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Since:
Di = 0.91 m (35.88 in)
Do=Di+2ts Do=36.24 in
From PED, p.69 Crown radiusL=Di-6 (in in.) L=29.88 in kr=knuckle radius=0.06 Do S pinning
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kr=2.17 in
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, S pinning
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th=PLW2Se th=0.27 in ≈6.95 mm
Use 7 mm as head thickness.
1)
Depth of the head (h):
From Eqn. 4-14, p. 92, PED, h=L-L2-D24 S pinning
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h=29.88-(29.88)2-35.8824 h=5.99 in ≈0.15m
Use 0.2 m as the head depth of the tank.
2)
Volume of the head (V):
From Eqn. 4-15, p.92, PED, V=1.05h23L-h (all values in inches) S pinning
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Substituting the previously computed values to the above equation would give, V=3151.44 in3= 0.05 m3
Use 0.1 m3 as volume of the head.
3)
Surface area of the head (A):
From Eqn. 4-16, p.92, PED, S pinning
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A=6.28 hL A=1124.00 in2=0.73m2
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Use 0.80 m2 as the surface area of the head.
4)
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).
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BELT CONVEYOR WITH SPRAY WASHER Design Description: Description: Belt Belt conv convey eyor or,, as the the name name su sugg gges ests ts,, cons consis ists ts of endl endles ess s belt belts, s, suitably suitably supported and driven, driven, which carry carry or transport transport solids solids from place to S pinning
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place. Belts are made of canvas, reinforced rubber or balata and strip steel. Stri Strip p stee steell is empl employ oyed ed for for conv convey eyin ing g mate materi rial als s thro throug ugh h furn furnac aces es.. Belt 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 the belt belt vari varies es from from 14 to 16 in, in, and and the the numb number er of idle idlers rs varie aries s corres correspon pondin dingly gly.. This This spacin spacing g ranges ranges from from about about 5 feet feet for narrow narrow belts belts
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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.
Design Considerations: Considerations: S pinning
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1)
Capa apacity of the belt conveyor
2)
Lengt ength h of trav trave el/l l/lengt ength h of the the belt belt
3)
Type of mate aterial being hand andled
4)
Speed of the conveyor
5)
Number of spray nozzles
Data and Assumptions: Assumptions: 1)
Capacity based from material balance is 968.80 kg
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2)
Norm Normal al spe speed ed rang range e of bel beltt conv convey eyor or is is betw betwee een n 200 200 to 400 400 ft/ ft/mi min n (from (from Uni Unitt Operations by Brown, p. 55) 3)
4)
Ratio of feed to wash water is 1:2
The belt width is 36 in (3 ft) (from (from Perry’s Chemical Chemical Engineering Engineering Handbook section 21-10 5)
The spray washer has 9 nozzles
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6)
The lump size of feed is 18 in for 36 in width belt conveyor (Unit Operations by Brown, p. 58)
7)
The The saf safet ety y fac facto torr for for belt belt conv convey eyor or is 15 % (Ti (Timm mmer erha haus us,, p. 36 36))
Design Requirements: Requirements: 1)
Capacity of spray nozzles
2)
Capacity of belt conveyor
3)
Power requirement
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Design Computations: Computations: From material balance: Inlet capacity is 968.80 kg/batch of water hyacinth stalks
1) Capacity Capacity of spray spray nozzl nozzle: e: V=MD
Where:
M= mass of water D = density of water
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Since water to feed ratio is 2:1, M=2(968.80 kg) M=1937.60 kg H2O D=1000 kg/m3 V= 1.94 m39 nozzles V= 0.22m3nozzle
Use 0.30 m3 of water per spray nozzle for washing. S pinning
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2) Capacity Capacity of the belt belt conveyor conveyor (T) T=968.80 kgbatch×1 batch8 hours×2.2lbskg×1 hours×2.2lbskg×1 ton2000 lbs T=0.13tonhr 266.42kghr
Giving an allowance factor of 15%, for future expansion, T=0.13tonhr(1.15) T=0.15tonhr 305.90kghr S pinning
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Use 310 kg/hr as belt conveyor capacity.
3) Power requiremen requirementt Using equations for power requirement (Unit Operations by Brown, p.58) for plain bearings, Hp=FL+LoT+0.03WS+T∆z990
Where:
Hp = horsepower required F=
S pinning
friction factor, 0.05 for plain bearings
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L=
length of conveyor between terminal pulleys, ft.
Lo = 100 for plain bearings S= T = ∆z
speed of belt, fpm capacity of belt conveyor, ton/hr
= increase in in elevation elevation of material, material, ft
W = mass mass of moving moving parts includi including ng belts belts and idl idlers ers per foot foot distance centers of terminal pulleys (both runs), lbs S pinning
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Pamantasan ng Lungsod ng Maynila College of E of Engineering and Technology Chemical Engineering Department
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=1lbin-ft36 in(2 runs) W=72lbsft
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Water ter H yacinth) Stalks and of Alkaline Sorbed Fibers from Eichhornia Crassipes ( W a E xtruded Polyethylene T erephthalate PET erephthalate ( PET ) Bottles for the Production of T extile extile Y arn arn
Pamantasan ng Lungsod ng Maynila College of E of Engineering and Technology Chemical Engineering Department
Assuming L = 30 ft, since the length of the conveyor should be greater than the length of the spray washer, Hp=0.0530ft+1000.15tonhr+0.037 Hp=0.0530ft+1000 .15tonhr+0.0372lbsft200 2lbsft200 ftmin +(0)0.15990 Hp=2.84 hp
Use 3 Hp since it is commercially available.
S pinning
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