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Technological Institute of the Philippines

SS-101 AMMONIA GAS SPHERICAL STORAGE TANK

Data and Assumptions:

1. Mass flow rate = 2,256.57 kg/hr

2. 20% Volumetric Allowance

EQUIPMENT SPECIFICATION
Ammonia Storage Tank Data EQUIPMENT TAG: SS-101
Sheet DESCRIPTION (FUNCTION)
To store Ammonia Gas.
Sheet No. 1
Operating Data 1
No. Required 5 Material of Carbon Steel 2
Construction (S42; Graded A)
Operating 30 ºC Operating 3.5 atm 3
temperature pressure
Contents Ammonia 4
Tank Capacity 8373.78 m3 5
Vessel data 6
Inside Diameter 25.760 m 7
Outside Diameter 25.785 m 8
Shell Thickness 12.689 mm 9
Circumference of the Sphere 81.006 m 11
Actual Dimension of the tank 12
Corrosion allowance 2 mm 13
Actual tank circumference 81.006 m 14
Actual Tank Diameter 25.785 m 15
Actual Inside Diameter 25.760 m 16
Actual No. of Layers 14 layers 17
Total no. of plates 57 plates 18
Inlet Nozzle Diameter 3 -inch, Schedule 5S 19
Outlet Nozzle Diameter 2 ½-inch, Schedule 80S 20
 Technological Institute of the Philippines

Design Requirements:

a. Material of Construction

b. Diameter

c. Shell Thickness

Design Calculations:

From Table 2-143, Perry’s Chemical Engineers’ Handbook, 8th Ed.:

T= 30˚C

kg
ρ = 9.107
m3

Capacity of the Tank (for the basis of 7 days, or 168 hrs)

kg
2256.57 x 168 hrs
V= hr = 𝟒𝟏𝟔𝟐𝟕. 𝟕𝟑 𝐦𝟑
kg
9.107 3
m

Solving for the diameter using spherical tank (Silla, 2003)

Calculating the number of tanks:

41627.73 m3
N= = 4.97 = 𝟓 𝐭𝐚𝐧𝐤𝐬
m3
8373.78
tank

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Tank Diameter

πD3
VTheo =
6

Solving for inside diameter of the vessel (using tank capacity)

πD3
8,373.78 m3 =
6

𝐃 = 𝟐𝟓. 𝟏𝟗𝟓 𝐦

Solving for the Surface Area of Sphere

SA = πD2

SA = 25.1952 π = 1994.245 m2

𝐒𝐀 = 𝟔𝟖𝟑𝟐. 𝟖𝟎 𝐟𝐭 𝟐

Standard size for plates is 6 x 20 ft.

Solving for number of plates

6832.80 ft 2
Nplates = = 56.940 = 𝟓𝟕 𝐩𝐥𝐚𝐭𝐞𝐬
ft 2
(6 x 20)
plate

Using N = 57 plates

120 ft 2
SA = 57 plates x = 𝟔𝟖𝟒𝟎 𝐟𝐭 𝟐
plate

𝐒𝐀 = 𝟐𝟎𝟖𝟒. 𝟖𝟑𝟐 𝐦𝟐

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2084.832 = πD2

𝐃 = 𝟐𝟓. 𝟕𝟔𝟎 𝐦

Solving for inside diameter of the vessel (using liquid level)

41627.73 m3 πD3
=
5 6

𝐃 = 𝟐𝟓. 𝟏𝟒𝟔 𝐦

25.760 m 3
Volumetric Allowance = [( ) − 1] x 100 % = 𝟖. 𝟐𝟐%
25.146 m

Design Pressure

From Ammonia Spherical Tank,

Pd = 3.5 atm = 354637.5 Pa

𝐏𝐝 = 𝟑𝟓𝟒𝟔𝟑𝟕. 𝟓 𝐏𝐚

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Solving Shell Thickness

From Table 13.2 Typical Maximum Allowable Stresses for Plates under ASME
BPV Code Sec. VIII (R K Sinnott)

N
Carbon Steel, S = 360 mm2

From Equation 13.39, (Chemical Engineering Design and Principles by

Coulson and Richardson)

Pd R i
t=
2SE − 0.2Pd

Where:

t = shell thickness

Pd = design pressure

Ri =internal diameter

S = design stress

E = joint efficiency

N
(0.3546375 ) (25760 mm)
t= mm2
N N
(2)(1)(360 2 ) − 0.2(0.3546375 )
mm mm2

𝐭 = 𝟏𝟐. 𝟔𝟖𝟗 𝐦𝐦

Outside Diameter

Do = Di + 2t

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Do = 25.760 m + 0.025378 m

𝐃𝐨 = 𝟐𝟓. 𝟕𝟖𝟓 𝐦

Nozzle Sizing

Basis:

As per Engineering Standard for Process Design (Standard Code IPS-PR-880)

1. For fluid inlet (Gas or Liquid): ρv2 < 1000

2. For gas outlet: ρv2 < 3600
3. For liquid outlet: v= 1 m/s

Feed Inlet Nozzle

Density (kg/m3) 9.107
ρv 2 (kg/ms2) 1000

1000 kg/ms2
v=√ = 10.479 m/s
9.107 kg/m3

Feed Flow Rate= 0.626825 kg/s

m 0.626825 kg/s
A= = = 6.568 x 10−3 m2
ρv kg m
(9.107 3 )(10.479 s )
m

4A 4(6.568 x 10−3 m2 )
Dn = √ = √ = 0.0914 m ≈ 91.447 mm
π π

Dn = 3.600 in, therefore use 3-inch Nominal Pipe Size, Schedule 5S.

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Gas Outlet Nozzle

Density (kg/m3) 9.107

ρv 2 (kg/ms2) 3600

Outlet Flow Rate= 0.338 kg/s

3600 kg/ms2
v=√ = 19.882 m/s
9.107 kg/m3

m 0.626825 kg/s
A= = = 3.462 x 10−3 m2
ρv kg m
(9.107 3 )(19.882 s )
m

4A 4(3.462 x 10−3 m2 )
Dn = √ = √ = 0.06639 m ≈ 66.39 mm
π π

Dn = 2.614 in, therefore use 2 1/2-inch Nominal Pipe Size, Schedule 80S.

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 Technological Institute of the Philippines

SS-102 CARBON DIOXIDE SPHERICAL STORAGE TANK

Data and Assumptions:

3. Mass flow rate = 4,994 kg/hr

4. 20% Volumetric Allowance

EQUIPMENT SPECIFICATION
Ammonia Storage Tank Data EQUIPMENT TAG: SS-102
Sheet DESCRIPTION (FUNCTION)
To store Carbon Dioxide Gas.
Sheet No. 1
Operating Data 1
No. Required 32 Material of Carbon Steel 2
Construction (S42; Graded A)
Operating 30 ºC Operating 3.5 atm 3
temperature pressure
Contents Ammonia 4
Tank Capacity 14950.186 m3 5
Vessel data 6
Inside Diameter 30.564 m 7
Outside Diameter 30.594 m 8
Shell Thickness 15.06 mm 9
Circumference of the Sphere 96.019 m 11
Actual Dimension of the tank 12
Corrosion allowance 2 mm 13
Actual tank circumference 96.019 m 14
Actual Tank Diameter 30.594 m 15
Actual Inside Diameter 30.564 m 16
Actual No. of Layers 14 layers 17
Total no. of plates 80 plates 18
Inlet Nozzle Diameter 5 -inch, Schedule 5S 19
Outlet Nozzle Diameter 3 1/3 inch, Schedule 80S 20

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Design Requirements:

a. Material of Construction

b. Diameter

c. Shell Thickness

Design Calculations:

From Table 2-143, Perry’s Chemical Engineers’ Handbook, 8th Ed.:

T= 30˚C

kg
ρ = 1.777
m3

Capacity of the Tank (for the basis of 7 days, or 168 hrs)

kg
4,994 x 168 hrs
V= hr = 𝟒𝟕𝟐𝟏𝟑𝟗. 𝟓𝟔 𝐦𝟑
kg
1.777 3
m

Solving for the diameter using spherical tank (Silla, 2003)

Calculating the number of tanks:

472139.56 m3
N= = 13.58 = 𝟑𝟐 𝐭𝐚𝐧𝐤𝐬
m3
14950.186
tank

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Tank Diameter

πD3
VTheo =
6

Solving for inside diameter of the vessel (using tank capacity)

πD3
14950.186 m3 =
6

𝐃 = 𝟑𝟎. 𝟓𝟔𝟒 𝐦

Solving for the Surface Area of Sphere

SA = πD2

SA = 30.5642 π = 2934.74 m2

𝐒𝐀 = 𝟗𝟔𝟐𝟖. 𝟒𝟑 𝐟𝐭 𝟐

Standard size for plates is 6 x 20 ft.

Solving for number of plates

9628.43 ft 2
Nplates = = 80.24 = 𝟖𝟎 𝐩𝐥𝐚𝐭𝐞𝐬
ft 2
(6 x 20)
plate

Using N = 80 plates

120 ft 2
SA = 80 plates x = 𝟗𝟔𝟐𝟖. 𝟒𝟑𝐟𝐭 𝟐
plate

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𝐒𝐀 = 𝟐𝟗𝟑𝟒. 𝟕𝟒𝟒 𝐦𝟐

2934.744 = πD2

𝐃 = 𝟑𝟎. 𝟓𝟔𝟒 𝐦

Solving for inside diameter of the vessel (using liquid level)

472139.56 m3 πD3
=
32 6

𝐃 = 𝟑𝟎. 𝟒𝟑𝟎 𝐦

30.564 m 3
Volumetric Allowance = [( ) − 1] x 100 % = 𝟏. 𝟑𝟑%
30.430 m

Design Pressure

From Ammonia Spherical Tank,

Pd = 3.5 atm = 354637.5 Pa

𝐏𝐝 = 𝟑𝟓𝟒𝟔𝟑𝟕. 𝟓 𝐏𝐚

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Solving Shell Thickness

From Table 13.2 Typical Maximum Allowable Stresses for Plates under ASME
BPV Code Sec. VIII (R K Sinnott)

N
Carbon Steel, S = 360
mm2

From Equation 13.39, (Chemical Engineering Design and Principles by

Coulson and Richardson)

Pd R i
t=
2SE − 0.2Pd

Where:

t = shell thickness

Pd = design pressure

Ri =internal diameter

S = design stress

E = joint efficiency

N
(0.3546375 ) (30564mm)
t= mm2
N N
(2)(1)(360 2 ) − 0.2(0.3546375 )
mm mm2

𝐭 = 𝟏𝟓. 𝟎𝟔 𝐦𝐦

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Outside Diameter

Do = Di + 2t

Do = 30.564 m + 0.03012 m

𝐃𝐨 = 𝟑𝟎. 𝟓𝟗𝟒 𝐦

Nozzle Sizing

Basis:

As per Engineering Standard for Process Design (Standard Code IPS-PR-880)

4. For fluid inlet (Gas or Liquid):2 < 1000

5. For gas outlet: ρv2 < 3600
6. For liquid outlet: v= 1 m/s

Feed Inlet Nozzle

Density (kg/m3) 9.107
ρv 2 (kg/ms2) 1000

1000 kg/ms2
v=√ = 10.479 m/s
9.107 kg/m3

Feed Flow Rate= 1.248 kg/s

m 1.248 kg/s
A= = = 0.013 m2
ρv kg m
(9.107 3 )(10.479 s )
m

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4A 4(0.013 m2 )
Dn = √ = √ = 0.129 m ≈ 128.65 mm
π π

Dn = 5.065 in, therefore use 5-inch Nominal Pipe Size, Schedule 5S.

Gas Outlet Nozzle

Density (kg/m3) 9.107

ρv 2 (kg/ms2) 3600

Outlet Flow Rate= 1.38 kg/s

3600 kg/ms2
v=√ = 19.882 m/s
9.107 kg/m3

m 1.387 kg/s
A= = = 7.660 x 10−3 m2
ρv kg m
(9.107 3 )(19.882 s )
m

4A 4(7.660 x 10−3 m2 )
Dn = √ =√ = 0.099m ≈ 98.76 mm
π π

Dn = 3.89 in, therefore use 3 1/3-inch Nominal Pipe Size, Schedule 80S.

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 Technological Institute of the Philippines

SODIUM
CHLORIDE
STORAGE
TANK
S-101

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 Technological Institute of the Philippines

EQUIPMENT SPECIFICATION

Sodium Chloride Storage Tank EQUIPMENT No. (TAG) ST-003

Data DESCRIPTION (FUNCTION)
Sheet To store Sodium Chloridesolution.
Sheet No. 1
Operating Data 1
No. Required 1 Material of Stainless steel 2
Construction 304
Operating 30 ºC Operating 3.5 atm 3
temperature pressure
Contents Sodium Chloride 4
3
Tank Capacity 1088.61 m 5
Vessel data 6
Height of cylinder 16.76 m 7
Shell thickness 18 mm 8
Head Thickness 8 mm 9
Bottom Thickness 16 mm 10
Circumference of the cylinder 12.193 m 11
Actual Dimension of the tank 12
Corrosion allowance 2 mm 13
Actual tank circumference 12.234m 14
No. of plates 4 plates 15
Actual Tank Diameter 7.789 m 16
Actual Inside Diameter 7.762 m 17
Actual Height of Shell 8.229 m 18
Actual No. of Layers 4 layers 19
Actual Tank Height 9.756 m 20
2
Actual Surface Area of 94.638 m 21
Hemispherical Head
Actual Surface Area Flat 47.709 m2 22
Bottom
No. of plates for Head and 22 plates 23
Bottom
Total no. of plates 26 plates 24
Inlet Nozzle 1 ½-inch, Outlet Nozzle 1 ½ -inch, 25
Diameter Sch 10S Diameter Schedule 160

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 Technological Institute of the Philippines

Data and Assumptions:

1. Mass flow rate = 6,435 kg/hr

2. 20% Volumetric Allowance

Design Requirements:

a. Material of Construction

b. Height

c. Diameter

d. Head Height

e. Shell Thickness

f. Head Thickness

Design Calculations:

From Table 2-103, Perry’s Chemical Engineers’ Handbook, 8th Ed.:

T= 30˚C

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 Technological Institute of the Philippines

kg
ρ = 1191.7
m3

Capacity of the Tank (for the basis of 7 days, or 168 hrs)

kg
6,435 x 168 hr
Vt = hr = 𝟏𝟎𝟖𝟖. 𝟔𝟏 𝐦𝟑
kg
1191.7 3
m

Tank Diameter

Solving for the diameter using hemispherical head (Silla, 2003)

πD3
VHemispherical =
12

VTheo = VCylinder + VHemispherical + Vcone

π 2 πD3 πD2 Hcone

VTheo = D H+ +
4 12 12

Where:

D = diameter

H = 2D

Hcone = 0.5 m

Solving for inside diameter of the vessel

π 2 πD3 πD2 (0.5)

VTheo = D (2D) + +
4 12 12

πD3 πD3 πD2 (0.5)

3
1088.61 m = + +
2 12 12

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𝐃 = 𝟖. 𝟑𝟖𝟐 𝐦

Solving for height of the vessel

H = 2D [Chemical Process Equipment by Couper]

H = 2 (8.382)

𝐇 = 𝟏𝟔. 𝟕𝟔 𝐦

Solving for number of plates

Circumference (C) = πD

C = 8.382 π = 26.163 m

𝐂 = 𝟖𝟓. 𝟖𝟑𝟕 𝐟𝐭.

Standard length for plates is 20 ft.

Solving for number of plates

85.837 ft
Nplates = = 4.29 = 𝟒 𝐩𝐥𝐚𝐭𝐞𝐬
ft
20
plate

Using N = 4 plates

ft
C = 4 plates x 20 = 80 ft
plate

1m
C = 80 ft x = 24.384 m
3.2808 ft

24.834 m
D= = 𝟕. 𝟕𝟔𝟐 𝐦
π

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πD3 πD3 πD2 (0.5)

Vt = + +
2 12 12

π(7.762)3 π(7.762)3 π(7.762)2 (0.5)

Vt = + +
2 12 12

Vt = 224.955 m3

224.955 m3
Volumetric Allowance = ( − 1) x 100 % = 𝟐𝟎. 𝟔𝟔 %
1088.61 m

Solving for number of levels

Standard height for plates is 6 ft.

3.2808 ft
16.76 m x
Nlevels = m = 9.16 = 𝟗 𝐥𝐞𝐯𝐞𝐥𝐬
ft
6
level

Using N = 9 levels

Surface Area of Cylinder:

6 ft 20 ft
SAcylinder = 9 levels x x 2 plates x = 2160 ft 2
level plate

2
2
1m
SAcylinder = 2160 ft x ( ) = 𝟐𝟎𝟎. 𝟔𝟕𝟓 𝐦𝟐
3.2808 ft

200.67 m2 = πDHnew

Solving for the new height:

Hnew = 𝟖. 𝟐𝟐𝟗 𝐦

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Surface Area of Hemispherical Head

πD2
SAhemispherical =
2

π(7.762)2
SAhemispherical = = 𝟗𝟒. 𝟔𝟑𝟖 𝐦𝟐
2

Surface Area of Cone

πDl
SAcone =
2

D 2
l = √( ) + Hcone
2
2

7.762 2
√
l= ( ) + (0.5)2
2

l = 𝟑. 𝟗𝟏𝟑 𝐦

π(7.762)(3.913)
SAcone = = 𝟒𝟕. 𝟕𝟎𝟗 𝐦𝟐
2

SAcone + SAcylinder = 248.384 m2

SAcone + SAcylinder = 2673.52 ft 2

Calculating the number of plates to be used:

1 plate
Nplates = 2673.52 ft 2 x = 22.279 = 𝟐𝟐 𝐩𝐥𝐚𝐭𝐞𝐬
6 ft x 20 ft

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𝐏𝐝 = 𝟑𝟓𝟒𝟔𝟑𝟕. 𝟓

Solving Shell Thickness

From Table 13.2 Typical Maximum Allowable Stresses for Plates under ASME
BPV Code Sec. VIII (R K Sinnott)

N
For Stainless Steel, S = 88.9 mm2

From Equation 13.40, (Chemical Engineering Design and Principles by

Coulson and Richardson)

Pi Di
t= +C
2Jf − Pi

Where:

t = shell thickness

Pi = internal pressure

Di =internal diameter

f = design stress

J = joint factor

C = corrosion allowance = 2mm

N
(0.3546375 ) (7762 mm)
t= mm2 + 2 mm
N N
(2)(1) (88.9 ) − (0.3546375 )
mm2 mm2

t = 17.513 ≈ 𝟏𝟖 𝐦𝐦

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Outside Diameter

𝟐𝐭 = 𝐃𝐨 − 𝐃𝐢

Do = Di + 2t

Do = 7.762 m + 0.036 m

𝐃𝐨 = 𝟕. 𝟕𝟗𝟖 𝐦

Solving Hemispherical Head Thickness

Head Design

From Equation 13.44, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd R
t=
2SE − 0.2Pd

Where:

Pd = Design Pressure

R = Inside radius

S = Maximum Allowable Stress

E = Joint Efficiency

N 7762
(0.3546375 ) ( 2 mm)
t= mm2
N N
2(1) (88.9 2 ) − 0.2 (0.3546375 )
mm mm2

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 Technological Institute of the Philippines

t = 7.744 mm ≈ 𝟖 𝐦𝐦

Bottom Design

From Equation 13.45, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd Dc 1
t= .
2fJ − Pd cos α

Where:

Pd = Design Pressure

Dc = Column Diameter

f = Design Stress

J = Joint Efficiency

α = half of the apex angle

N
(0.3546375 2 ) (7762 mm) 1
t= mm x
N N 7.762
2(1) (88.9 2 ) − (0.3546375 2 ) 1 m
mm mm cos [2 tan− ( 2 )]
0.5 m

t = 15.513 mm ≈ 𝟏𝟔 𝐦𝐦

Nozzle Sizing

Basis: As per Engineering Standard for Process Design (Standard Code IPS-PR-880)

1. For fluid inlet (Gas or Liquid): ρv2 < 1000

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2. For gas outlet: ρv2 < 3600

3. For liquid outlet: v= 1 m/s

Feed Inlet Nozzle

Density (kg/m3) 1191.7
𝜌𝑣 2 (kg/ms2) 1000

1000 kg/ms 2
v=√ = 0.92 m/s
1191.7 kg/m3

Feed Flow Rate= 1.7875 kg/s

m 1.7875 kg/s
A= = = 1.630 x 10−3 m2
ρv kg m
(1191.7 3 )(0.92 s )
m
4A 4(1.630 x 10−3 m2 )
Dn = √ = √ = 0.04556 m ≈ 45.56 mm
π π
Dn =1.794 in, therefore use 1 ½-inch Nominal Pipe Size, Schedule 10S
Liquid Outlet Nozzle

Density (kg/m3) 1191.7

Velocity (m/s) 1

Outlet Flow Rate= 1.7875 kg/s

m 1.7875 kg/s
A= = = 1.500 x 10−3 m2
ρv kg m
(1191.7 3 )(1 s )
m

4A 4(1.500 x 10−3 m2 )
Dn = √ = √ = 0.0437 m ≈ 43.70 mm
π π

Dn = 1.720 in, therefore use 1 ½ -inch Nominal Pipe Size, Schedule 160

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 Technological Institute of the Philippines

AMMONIUM
CHLORIDE
STORAGE
TANK
S-102

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Data and Assumptions:

3. Mass flow rate =1,580.61 kg/hr

4. 20% Volumetric Allowance

Design Requirements:

a. Material of Construction

b. Height

c. Diameter

d. Head Height

e. Shell Thickness

f. Head Thickness

Design Calculations:

From Table 2-103, Perry’s Chemical Engineers’ Handbook, 8th Ed.:

T= 30˚C

kg
ρ = 1064.1
m3

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Capacity of the Tank (for the basis of 7 days, or 168 hrs)

kg
1,580.61 x 168 hr x 1.2
Vt = hr = 𝟐𝟗𝟗. 𝟒𝟓𝟔 𝐦𝟑
kg
1064.1 3
m

Tank Diameter

Solving for the diameter using hemispherical head (Silla, 2003)

πD3
VHemispherical =
12

VTheo = VCylinder + VHemispherical + Vcone

π 2 πD3 πD2 Hcone

VTheo = D H+ +
4 12 12

Where:

D = diameter

H = 2D

Hcone = 0.5 m

Solving for inside diameter of the vessel

π 2 πD3 πD2 (0.5)

VTheo = D (2D) + +
4 12 12

πD3 πD3 πD2 (0.5)

299.456 m3 = + +
2 12 12

𝐃 = 𝟏𝟎. 𝟐𝟓𝟏 𝐦

Solving for height of the vessel

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H = 2D [Chemical Process Equipment by Couper]

H = 2 (10.251)

𝐇 = 𝟐𝟎. 𝟓𝟎𝟐 𝐦

Solving for number of plates

Circumference (C) = πD

C = 10.251 π = 32.204 m

𝐂 = 𝟏𝟎𝟓. 𝟔𝟓𝟖 𝐟𝐭.

Standard length for plates is 20 ft.

Solving for number of plates

105. 66 ft
Nplates = = 5.282 = 𝟓 𝐩𝐥𝐚𝐭𝐞𝐬
ft
20
plate

Using N = 4 plates

ft
C = 5 plates x 20 = 100 ft
plate

1m
C = 100 ft x = 30.480 m
3.2808 ft

30.480 m
D= = 𝟗. 𝟕𝟎𝟐 𝐦
π

πD3 πD3 πD2 (0.5)

Vt = + +
2 12 12

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π(9.702)3 π(9.702)3 π(9.702)2 (0.5)

Vt = + +
2 12 12

Vt = 1685.92 m3

1665.92 m3
Volumetric Allowance = ( − 1) x 100 % = 𝟒. 𝟔𝟑 %
299.456m

Solving for number of levels

Standard height for plates is 6 ft.

3.2808 ft
20.502 m x
Nlevels = m = 11.21 = 𝟏𝟏 𝐥𝐞𝐯𝐞𝐥𝐬
ft
6
level

Using N = 9 levels

Surface Area of Cylinder:

6 ft 20 ft
SAcylinder = 11 levels x x 2 plates x = 2690.52 ft 2
level plate

2
2
1m
SAcylinder = 2160 ft x ( ) = 𝟐𝟒𝟗. 𝟗𝟔𝟑 𝐦𝟐
3.2808 ft

249.963 m2 = πDHnew

Solving for the new height:

Hnew = 𝟖. 𝟐𝟎𝟎 𝐦

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Surface Area of Hemispherical Head

πD2
SAhemispherical =
2

π(9.702)2
SAhemispherical = = 𝟏𝟒𝟔. 𝟖𝟔 𝐦𝟐
2

Surface Area of Cone

πDl
SAcone =
2

D 2
l = √( ) + Hcone
2
2

147.86 2
√
l= ( ) + (0.5)2
2

l = 𝟕𝟑. 𝟗𝟑 𝐦

π(9.702)(73.93)
SAcone = = 𝟏𝟏𝟐𝟔. 𝟔𝟖 𝐦𝟐
2

SAcone + SAcylinder = 3817.20 m2

SAcone + SAcylinder = 12523.632 ft 2

Calculating the number of plates to be used:

1 plate
Nplates = 12523.632 ft 2 x = 104.3636 = 𝟏𝟎𝟒 𝐩𝐥𝐚𝐭𝐞𝐬
6 ft x 20 ft

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𝐏𝐝 = 𝟑𝟓𝟒𝟔𝟑𝟕. 𝟓

Solving Shell Thickness

From Table 13.2 Typical Maximum Allowable Stresses for Plates under ASME
BPV Code Sec. VIII (R K Sinnott)

N
For Stainless Steel, S = 88.9 mm2

From Equation 13.40, (Chemical Engineering Design and Principles by

Coulson and Richardson)

Pi Di
t= +C
2Jf − Pi

Where:

t = shell thickness

Pi = internal pressure

Di =internal diameter

f = design stress

J = joint factor

C = corrosion allowance = 2mm

N
(0.3546375 ) (9702mm)
t= mm2 + 2 mm
N N
(2)(1) (88.9 ) − (0.3546375 )
mm2 mm2

t = 21.390 ≈ 𝟐𝟏 𝐦𝐦

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Outside Diameter

𝟐𝐭 = 𝐃𝐨 − 𝐃𝐢

Do = Di + 2t

Do = 9.702 m + 0.043 m

𝐃𝐨 = 𝟗. 𝟕𝟒𝟓 𝐦

Solving Hemispherical Head Thickness

Head Design

From Equation 13.44, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd R
t=
2SE − 0.2Pd

Where:

Pd = Design Pressure

R = Inside radius

S = Maximum Allowable Stress

E = Joint Efficiency

N 9.745
(0.3546375 ) ( 2 mm)
t= mm2
N N
2(1) (88.9 2 ) − 0.2 (0.3546375 )
mm mm2

t = 9.723 mm ≈ 𝟏𝟎 𝐦𝐦

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Bottom Design

From Equation 13.45, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd Dc 1
t= .
2fJ − Pd cos α

Where:

Pd = Design Pressure

Dc = Column Diameter

f = Design Stress

J = Joint Efficiency

α = half of the apex angle

N
(0.3546375 2 ) (9702 mm) 1
t= mm x
N N 7.762
2(1) (88.9 ) − (0.3546375 ) 1 m
mm2 mm2 cos [2 tan− ( 2 )]
0.5 m

t = 19.392 mm ≈ 𝟏𝟗 𝐦𝐦

Nozzle Sizing

Basis:

As per Engineering Standard for Process Design (Standard Code IPS-PR-880)

4. For fluid inlet (Gas or Liquid): ρv2 < 1000

5. For gas outlet: ρv2 < 3600

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6. For liquid outlet: v= 1 m/s

Feed Inlet Nozzle

Density (kg/m3) 1064.1
𝜌𝑣 2 (kg/ms2) 1000

1000 kg/ms 2
v=√ = 0.97 m/s
1064.1 kg/m3

Feed Flow Rate= 0.4390 kg/s

m 0.4390 kg/s
A= = = 4.253 x 10−4 m2
ρv kg m
(1064.1 3 )(0.97 s )
m
4A 4(4.253 x 10−4 m2 )
Dn = √ = √ = 0.0233 m ≈ 23.27mm
π π
Dn =0.92 in, therefore use 1 inch Nominal Pipe Size, Schedule 10S

Liquid Outlet Nozzle

Density (kg/m3) 1064.1

Velocity (m/s) 1

Outlet Flow Rate= 0.4390 kg/s

m 0.4390 kg/s
A= = = 4.1256 x 10−4 m2
ρv kg m
(1064.1 3 )(1 s )
m

4A 4(1.500 x 10−3 m2 )
Dn = √ = √ = 0.0229 m ≈ 22.91 mm
π π

Dn = 0.902 in, therefore use 1 -inch Nominal Pipe Size, Schedule 160

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SODIUM
CARBONATE
SILO
S-103

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Data and Assumptions:

5. Mass flow rate =5,237.11 kg/hr

6. 20% Volumetric Allowance

Design Requirements:

a. Material of Construction

b. Height

c. Diameter

d. Head Height

e. Shell Thickness

f. Head Thickness

Design Calculations:

From Table 2-103, Perry’s Chemical Engineers’ Handbook, 8th Ed.:

T= 30˚C

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 Technological Institute of the Philippines

kg
ρ = 1327.4
m3

Capacity of the Tank (for the basis of 7 days, or 168 hrs)

kg
5,237.11 x 168 hr x 1.2
Vt = hr = 𝟕𝟗𝟓. 𝟑𝟗𝟏 𝐦𝟑
kg
1327.4 3
m

Tank Diameter

Solving for the diameter using hemispherical head (Silla, 2003)

πD3
VHemispherical =
12

VTheo = VCylinder + VHemispherical + Vcone

π 2 πD3 πD2 Hcone

VTheo = D H+ +
4 12 12

Where:

D = diameter

H = 2D

Hcone = 0.5 m

Solving for inside diameter of the vessel

π 2 πD3 πD2 (0.5)

VTheo = D (2D) + +
4 12 12

πD3 πD3 πD2 (0.5)

3
795.391 m = + +
2 12 12

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 Technological Institute of the Philippines

𝐃 = 𝟕. 𝟓𝟒𝟖 𝐦

Solving for height of the vessel

H = 2D [Chemical Process Equipment by Couper]

H = 2 (7.548)

𝐇 = 𝟏𝟓. 𝟎𝟗𝟓 𝐦

Solving for number of plates

Circumference (C) = πD

C = 15.095 π = 47.423m

𝐂 = 𝟏𝟓𝟓. 𝟓𝟖𝟔 𝐟𝐭.

Standard length for plates is 20 ft.

Solving for number of plates

155. 586 ft
Nplates = = 7.779 = 𝟖 𝐩𝐥𝐚𝐭𝐞𝐬
ft
20
plate

Using N = 8 plates

ft
C = 8 plates x 20 = 160 ft
plate

1m
C = 160 ft x = 48.768 m
3.2808 ft

48.768 m
D= = 𝟏𝟓. 𝟓𝟐𝟑 𝐦
π

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 Technological Institute of the Philippines

πD3 πD3 πD2 (0.5)

Vt = + +
2 12 12

π(15.524)3 π(15.524)3 π(15.524)2 (0.5)

Vt = + +
2 12 12

Vt = 6887.65 m3

6887.65 m3
Volumetric Allowance = ( − 1) x 100 % = 𝟖. 𝟔𝟔𝟎 %
795.391m

Solving for number of levels

Standard height for plates is 6 ft.

3.2808 ft
15.095 m x
Nlevels = m = 8.25 = 𝟖 𝐥𝐞𝐯𝐞𝐥𝐬
ft
6
level

Using N = 9 levels

Surface Area of Cylinder:

6 ft 20 ft
SAcylinder = 8 levels x x 2 plates x = 1920 ft 2
level plate

2
2
1m
SAcylinder = 1920 ft x ( ) = 𝟏𝟕𝟖. 𝟑𝟖 𝐦𝟐
3.2808 ft

178.38 m2 = πDHnew

Solving for the new height:

Hnew = 𝟕. 𝟓𝟐𝟐 𝐦

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 Technological Institute of the Philippines

Surface Area of Hemispherical Head

πD2
SAhemispherical =
2

π(15.523)2
SAhemispherical = = 3𝟕𝟖. 𝟓𝟎 𝐦𝟐
2

Surface Area of Cone

πDl
SAcone =
2

D 2
l = √( ) + Hcone
2
2

15.523 2
√
l= ( ) + (0.5)2
2

l = 𝟕. 𝟕𝟖 𝐦

π(15.523)(7.78)
SAcone = = 𝟏𝟖𝟗. 𝟕𝟎 𝐦𝟐
2

SAcone + SAcylinder = 2109.70 m2

SAcone + SAcylinder = 22708.07 ft 2

Calculating the number of plates to be used:

1 plate
Nplates = 22708.07 ft 2 x = 189.23 = 𝟏𝟖𝟗 𝐩𝐥𝐚𝐭𝐞𝐬
6 ft x 20 ft

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 Technological Institute of the Philippines

𝐏𝐝 = 𝟑𝟓𝟒𝟔𝟑𝟕. 𝟓

Solving Shell Thickness

From Table 13.2 Typical Maximum Allowable Stresses for Plates under ASME
BPV Code Sec. VIII (R K Sinnott)

N
For Stainless Steel, S = 88.9 mm2

From Equation 13.40, (Chemical Engineering Design and Principles by

Coulson and Richardson)

Pi Di
t= +C
2Jf − Pi

Where:

t = shell thickness

Pi = internal pressure

Di =internal diameter

f = design stress

J = joint factor

C = corrosion allowance = 2mm

N
(0.3546375 ) (1552.3mm)
t= mm2 + 2 mm
N N
(2)(1) (88.9 ) − (0.3546375 )
mm2 mm2

t = 5.10 ≈ 𝟓 𝐦𝐦

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 Technological Institute of the Philippines

Outside Diameter

𝟐𝐭 = 𝐃𝐨 − 𝐃𝐢

Do = Di + 2t

Do = 15.523 m + 0.01 m

𝐃𝐨 = 𝟏𝟓. 𝟓𝟑𝟑 𝐦

Solving Hemispherical Head Thickness

Head Design

From Equation 13.44, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd R
t=
2SE − 0.2Pd

Where:

Pd = Design Pressure

R = Inside radius

S = Maximum Allowable Stress

E = Joint Efficiency

N 1552.3
(0.3546375 2 )( 2 mm)
t= mm
N N
2(1) (88.9 ) − 0.2 (0.3546375 )
mm2 mm2

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 Technological Institute of the Philippines

t = 1.54 mm ≈ 𝟐 𝐦𝐦

Bottom Design

From Equation 13.45, (Chemical Eng’g Design and Principles by Coulson and
Richardson)

Pd Dc 1
t= .
2fJ − Pd cos α

Where:

Pd = Design Pressure

Dc = Column Diameter

f = Design Stress

J = Joint Efficiency

α = half of the apex angle

N
(0.3546375 2 ) (1552.3 mm) 1
t= mm x
N N 15.523
2(1) (88.9 2 ) − (0.3546375 2 ) 1 m
mm mm cos [2 tan− ( 2 )]
0.5 m

t = 3.10 mm ≈ 𝟑 𝐦𝐦

Nozzle Sizing

Basis:

As per Engineering Standard for Process Design (Standard Code IPS-PR-880)

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 Technological Institute of the Philippines

7. For fluid inlet (Gas or Liquid): ρv2 < 1000

8. For gas outlet: ρv2 < 3600
9. For liquid outlet: v= 1 m/s

Feed Inlet Nozzle

Density (kg/m3) 1327.4
𝜌𝑣 2 (kg/ms2) 1000

1000 kg/ms 2
v=√ = 0.87 m/s
1327.4 kg/m3

Feed Flow Rate= 1.454 kg/s

m 1.454 kg/s
A= = = 1.259 x 10−3 m2
ρv kg m
(1327.4 3 )(0.87 s )
m
4A 4(1.259 x 10−3 m2 )
Dn = √ = √ = 0.0400 m ≈ 40.04mm
π π
Dn =1.57 in, therefore use 1 1/2 inch Nominal Pipe Size, Schedule 10S

Liquid Outlet Nozzle

Density (kg/m3) 1327.4

Velocity (m/s) 1

Outlet Flow Rate= 1.455 kg/s

m 1.455 kg/s
A= = = 1.096x 10−3 m2
ρv kg m
(1327.4 3 )(1 s )
m

4A 4(1.096 x 10−3 m2 )
Dn = √ = √ = 0.0373 m ≈ 37.36 mm
π π

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Dn = 1.47 in, therefore use 1 1/2 -inch Nominal Pipe Size, Schedule 160

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