overall design

Design of a 100 MT per day

Sulfuric Acid (98.5 wt%) Plant

by

Hla Tun (0502004)

Mir Rezwan Rahman (0502030)

Muhammad Rejwan Billah (0502040)

Joyanta Goswami (0502050)

Supervised by

Dr. Syeda Sultana Razia

Associate Professor

Department of Chemical Engineering

Bangladesh University of Engineering & Technology (BUET)

Dhaka, Bangladesh

i

Acknowledgement

As the authors of this design project we are especially grateful to Dr. Syeda Sultana

Razia, Associate Professor, Department of Chemical Engineering, BUET, for her

prospective guidance and support throughout the course of the design project. We are

also grateful to B.M. Sirajeel Arifin and K.M. Tanvir Ahmmed, Lectureres, Department of

Chemical Engineering, for accompanying us during the industrial visit of sulfuric acid

plant and for their assistance regarding our project design. Moreover, we would like to

thank the engineers of Crescent chemicals for assisting us with industrial data of

sulfuric acid plant.

Table of Contents

Page

Acknowledment

i

Chapter 1 Definition of the Project 1

Chapter 2 Design Basis 3

Chapter 3 Process Description 6

Chapter 4 Process Block Diagram (PBD) 10

Chapter 5 Process Flow Diagram (PFD) 12

Chapter 6 Material Balance 14

Chapter 7 Energy Balance 31

Chapter 8 Water & Steam Balance 55

Chapter 9 Power Balance 59

Chapter 10 Stream Table 71

Chapter 11 Equipment List 75

Chapter 12 Equipment Sizing 78

Chapter 13 Detailed Engineering Design 141

13.1 Converter 142

13.2 Interpass Absorption Column 163

13.3 Waste Heat Boiler-2 182

13.4 Storage Tank 193

Chapter 14 Plot Plan & Equipment Layout 199

Chapter 15 Estimation of Total Capital Investment and

Production Cost

202

Chapter 16 Economic Analysis 213

References 217

Chapter One

Definition of the Project

2

A sulfuric acid plant is to be set-up at Pubile, Gazipur, having a production capacity of

100 Mt commercial grade sulfuric acid (98.5wt% H2SO4) on a summer day (temperature

37oC and relative humidity 100%), corresponding to 32000 Mt sulfuric acid (98.5%

H2SO4) per year (320 working days per year is assumed) by double absorption sulfur

burning contact process, including all offsite, auxiliaries, utilities and supporting facilities

using 95% pure imported granulated elemental sulfur from China.

Raw materials

i. 95% pure sulfur ( impurities include gypsum, carbon and ashes)

ii. Ambient air

iii. Demineralized water for dilution of sulfuric acid

Chapter Two

Design Basis

4

1. Meteorological data for the site

i) Temperatures

Maximum temperature - 34oC (April)

Average annual temperature -27oC

Lowest temperature -14oC (December)

ii) Rainfall

Annual rainfall – 1875 mm

Average monthly rainfall – 156 mm

Maximum rainfall – 337 mm (August wettest month)

Minimum rainfall – 5 mm ( December driest month)

iii) Humidity

Average annual relative humidity – 65.8%

Average monthly humidity (maximum) – 79% (June)

Average monthly humidity (minimum) – 45% (March)

iv) Wind

Average wind speed – 9.7 km/h

Design wind speed – 125 km/h

Design wind loading – 80 kg/m2

v) Design conditions

Dry bulb temperature – 37 oC

Wet bulb temperature - 37 oC

Relative humidity – 100%

2. Utility

i) Steam:

Low pressure steam – 150o C at 4.5 kgf /cm2

ii) Water – Ground water at 20oC treated to soft water for cooling

5

iii) Major electric power requirements:

Blower – 222.8 kW

Pump1 – 0.0074 kW

Pump2 – 0.0147 kW

Pump3 – 0.43 kW

Pump4 – 0.075 kW

Pump5 – 0.122 kW

Melter – 51.8 kW

3. Raw material:

95% pure granulated sulfur

4. Process air – 2 atm (absolute)

Chapter Three

Process Description

7

Selection of Process for Sulfuric Acid Production

The two major processes for production of sulfuric acid are: chamber process and

contact process. The chamber process, the older of the two processes, is formerly used

to produce much of the acid used to make fertilizers; Chamber process produces a

relatively dilute acid (62 wt%–78 wt% H2SO4). As chamber process produce dilute acid,

which is the most important disadvantage of this process, in all the new sulfuric acid

plants the contact process is used. In contact process it is possible to produce 98.5

wt% H2SO4 and oleum (fuming sulfuric acid, 100% sulfuric acid with sulfur trioxide

dissolved in it) can also be produced.

Process Description

The main stages of sulfuric acid production includes burning sulfur to produce sulfur

dioxide gas, conversion of sulfur dioxide to sulfur trioxide and absorption of sulfur

trioxide gas by concentrated sulfuric acid where the water portion of acid reacts with

sulfur trioxide to produce sulfuric acid. The chemical equations representing the main

stages are:

At the initial stage elemental sulfur is converted to liquid sulfur in a melting unit where

sulfur is heated to temperatures of 140ºC using electrical heating and supplement

steam heating. The molten sulfur is pumped in a steam jacketed pipeline to the furnace

where it is sprayed through nozzles. The sprayed sulfur comes in contact with moisture-

free air from drying tower and reacts vigorously with oxygen to produce sulfur dioxide.

8

The reaction of sulfur and oxygen is an auto-ignition reaction and is highly exothermic

resulting in the temperatures of outlet gases to raise about 936ºC. The air entering the

furnace is the supplying route of air for the entire process. Thus air is supplied in surfeit

at the entrance of the furnace. The required amount of excess air is determined based

on the concentration of sulfur dioxide gas in the outlet of the furnace which is

maintained around 9-12%.

The high temperature outlet gas from the furnace is cooled to temperatures of 407ºC

by passing through a waste heat boiler before being processed in the converter. The

steam produced in the waste heat boiler is used as heating fluid in other heat

exchanger or used to drive turbo-generator unit to produce electrical power. The

process gas from the waste heat boiler is further passed through a hot gas filter to

remove any remaining dust or other solid impurities.

In the four stage converter the process gas containing a mixture of SO2 and air

undergoes oxidation at each stage in presence of catalyst vanadium pentaoxide

arranged in fixed beds. Since the oxidation of SO2 to SO3 is exothermic and as this

equilibrium oxidation reaction favors lower temperature, intermediate cooling of product

gases from each stage are carried out by circulating gases through waste heat boilers,

economizers and superheaters.

The gas from the third bed of the reactor, after intermediate cooling, is delivered to the

intermediate absorption tower where most of the SO3 present is absorbed using 98.5%

sulfuric acid from circulation tank. Similarly, the gas from the last stage of reactor is

passed to the final absorber after intermediate cooling for final absorption of SO3 using

98.5 wt% sulfuric acid. The acid products from the both absorbers are 100% fuming

sulfuric acid and are delivered to the circulation tank where they are diluted to 98.5wt%

H2SO4. The implementation of double absorption technique achieves 99.7% conversion

of SO2 to SO3 with intermediate absorber absorbing 95 wt% of total SO3 produced.

9

The fuming sulfuric acid from both the absorbers enters the circulation tank where the

acid is diluted to 98.5 wt% H2SO4. The 98.5 wt% sulfuric acid is then pumped to the

absorbers and air drying unit as absorbent for SO3 and moisture respectively. A portion

of the 98.5 wt% sulfuric acid from circulation tank is passed to the product storage

tank.

Chapter Four

Process Block Diagram

11

Drying

Tower

Furnace

Melter

HE-1HE-2

Interpass

absorption

Tower

Final

Absorptio

n Tower

Circulation

Tank

Elemental

Sulfur

Converter

Air

Product Acid

98.5 wt %

Dilution

Water

To Stack

Figure 1: Process Block Diagram for Double Absorption Contact Sulfuric Acid Plant (Monsanto Enviro-Chem)

Chapter Five

Process Flow Diagram

1

4

5

6

9

12

13

14

15

16

18 19

20

21

22

23

24

3

2

29

30

31

32

34

39

Air

Solid

Sulfur

To

Stack

Dilution

Water

Dilution

Water

Q-120

R-130

Q-121

D-110

D-140

D-150

E-131

E-132

E-133E-142

E-143

E-151

E-152

E-141

E-162

F-154

F-144

F-160

L-122

E-153

L-145

E-111

G-112

7

8

10

2625

27

11

33

17

28

35

37

38

L-146

L-161

36

Figure 2: Process Flow Diagram for Double Absorption Contact Sulfuric Acid Plant (Monsanto Enviro-Chem)

Chapter Six

Material Balance

15

Basis: 100 MT H2SO4 (98.5 wt%) production per day

Product Stream

Amount of H2SO4 = wt%)H2SO4(98.51

H2SO4985.0

1

1000%)5.98(H2SO4100

kg

kg

MT

kg

day

wtMT

= 98500 kg H2SO4 / day

= 4104.17 kg H2SO4 / hr

Molar flow rate of H2SO4 in outlet stream = 4298

4214217.4104

SOHkg

SOHkmol

hr

SOHkg

= 41.88 kmol H2SO4 / hr

Amount of Water = wt%)H2SO4(98.51

H2O015.0

1

1000%)5.98(H2SO4100

kg

kg

MT

kg

day

wtMT

= 1500 kg H2O/ day = 62.5 kg H2O/ hr

Molar flow rate of H2O in outlet stream = OHkg

OHkmol

hr

OHkg

218

2125.62

= 3.47 kmol H2O / hr

Overall Sulfur(S) Balance

Required amount of pure sulfur = 421

14288.41

SOHkmol

Skmol

hr

SOHkmol

= 41.88 kmol S/ hr

Amount of raw material sulfur required = 41.88 x 0.95 = 44.08 kmol/h

16

Theoretical O2 required

S + O2 SO2

SO2 + 0.5 O2 SO3

Overall Reaction: S + 1.5 O2 SO3

Amount of theoretical O2 supplied = Skmol

Okmol

hr

Skmol

1

25.188.41

= 62.82 kmol O2 / hr

Theoretical air supplied = 221.0

1288.62

Okmol

airkmol

hr

Okmol

= 299.43 kmol air / hr

For 50% excess air supply,

Molar flow rate of dry air (DA) = 1.5299.43 kmol air /hr

= 449.145 kmol air/ hr

Molar flow rate of N2 = airkmol

Nkmol

hr

airkmol

1

279.0145.449

= 354.48 kmol N2/ hr

Molar flow rate of O2 = airkmol

Okmol

hr

airkmol

1

221.0145.449

= 94.32 kmol O2/ hr

17

DRYING TOWER (D-110)

Assumptions: a. Humid air at 37oC

b. Relative Humidity = 100% [assuming worst condition]

c. Acid circulation is half of that in the absorber.

= 257.578 kmol H2SO4 soln (98.5wt%) / hr

At 37oC and 100% relative humidity, moisture content = 0.0096 kg H2O/ kg Dry air

=

= 0.01538 kmol H2O / kmol DA

264.486 kmol H2SO4

soln (97.976 wt%)/hr /hr

257.578 kmol H2SO4

soln (98.5 wt%)/hr Dry Air

Moist Air

Drying

Tower

18

Amount of inlet moisture (H2O) =

= 6.908 kmol H2O/ hr

Compositions of 98.5wt% H2SO4 soln is 0.923 kmol H2SO4/kmol and 0.077 kmol

H2O/kmol

H2SO4 Balance: 257.578 0.923 = Px

Px = 237.744……………………(1)

H2O Balance: 257.578 0.077 + 6.908= P(1-x)

P(1-x) = 26.742

P – Px = 26.742

P – 237.744 = 42.502 [from equation (1)]

P = 264.486 kmol/hr

from equation (1)

x =

= 0.8989 mol%

Molar flow rate of outlet H2SO4 soln = (257.578 + 6.908) kmol H2SO4 soln

= 264.486 H2SO4 soln

1 kmol H2SO4 = 98 kg H2SO4

0.8989 kmol H2SO4 = 98 0.8989 = 88.0922 kg H2SO4

(1-0.8989) kmol H2O = 1.8198 kg H2O

wt% of H2SO4 solution =

= 97.976 wt%

19

FURNACE (Q-121)

S + O2 SO2

Amount of O2 that reacts = 41.88 kmol O2/hr

Molar flow rate of SO2 = 41.88 kmol SO2/hr

Molar flow rate of O2 at furnace outlet = (94.23 – 41.88) kmol O2/hr

= 52.35 kmol O2/hr

Molar flow rate of N2 at outlet stream = 354.48 kmol N2/hr

Outlet composition:

0.093 kmol SO2 / kmol

0.117 kmol O2 / kmol

0.790 kmol N2 / kmol

20

INTERPASS ABSORPTION TOWER (D-140)

Total amount of SO3 produced [including 4th pass converter] = (0.997 41.88) kmol

SO3/hr = 41.754 kmol SO3/hr

In intermediate absorption tower 90-95% of the total amount of SO3 produced is

absorbed.

Assuming 95% absorption,

Total amount of SO3 absorbed in the Interpass absorption tower = (0.95 41.754)

kmol SO3/hr = 39.666 kmol

SO3/hr

Interpass

Absorption

Tower

0.923 kmol H2SO4/kmol

0.077 kmol H2O/kmol

41.500 kmol SO3/hr

0.377 kmol SO2/hr

31.600 kmol O2/hr

354.480 kmol N2/hr

L1 (98.5 wt% H2SO4)

L2 (100 wt% H2SO4)

21

Amount of SO3 in the gas outlet from the absorber = 41.50 – 39.666 = 1.834 kmol

SO3/hr

In 1 kg 98.5 wt% H2SO4, amount of H2SO4 = 0.985 kg

Amount of H2SO4 in mole = 0.985/98 = 0.01 kmol H2SO4

Amount of H2O in mole = 0.015/18 = 0.00083 kmol H2O

Composition of H2SO4 in 98.5 wt% H2SO4 =

= 0.923 mol H2SO4/ kmol soln

Composition of H2O = 0.077 kmo H2O/kmol soln

SO3 + H2O H2SO4

Amount of water need to produce 98.5wt% H2SO4 = 39.667 kmol H2O/hr

L1 =

= 515.156 kmol H2SO4 soln (98.5wt %) / hr

L2 =

= 515.156 kmol H2SO4/hr

22

CONVERTER (R-130)

1st pass catalyst bed:

SO2 + 0.5 O2 SO3

Assuming, 56% conversion of SO2 [ref: McKetta]

Molar flow rate of SO3 at outlet stream = 0.56 41.88 = 23.45 kmol SO3/ hr

Molar flow rate of SO2 at outlet stream = 41.88 – 23.45 = 18.43 kmol SO2/hr

Molar flow rate of O2 at outlet stream = 52.35 – 23.45 0.5 = 40.63 kmol O2/ hr


1st Pass

41.88 kmol SO2 / hr

52.35 kmol O2/hr

354.48 kmol N2/hr

1st Pass 41.88 kmol SO2 / hr

52.35 kmol O2/hr

354.48 kmol N2/hr 18.43 kmol SO2 / hr

40.63 kmol O2/hr

23.45 kmol SO3/hr

354.48 kmol N2/hr

23

2nd pass catalyst bed:

Assuming, 87% conversion (Ref: McKetta)

Molar flow rate of SO2 at outlet stream = 41.88 (1 - 0.87) = 5.44 kmol SO2/hr

Molar flow rate of SO3 at outlet stream = 23.45+18.43–5.44=36.44 kmol SO3/ hr

Molar flow rate of O2 at outlet stream=40.63–(18.43–5.44) 0.5

=34.135 kmol O2/ hr


2nd Pass

40.63 kmol O2/hr

23.45 kmol SO3/hr

354.48 kmol N2/hr

18.43 kmol SO2/hr

2nd Pass

23.45 kmol SO3/hr

354.48 kmol N2/hr

18.43 kmol SO2/hr

40.63 kmol O2/hr

36.44 kmol SO3/hr

5.44 kmol SO2/hr

34.135 kmol O2/hr

354.48 kmol N2/hr

24

3rd pass catalyst bed:

Assuming 99.1% conversion (according to McKetta)

Molar flow rate of SO2 at outlet stream = 41.88 (1 - 0.991)

= 0.377 kmol SO2/hr

Molar flow rate of SO3 at outlet stream = 36.44 + (5.44 – 0.377)

= 41.50 kmol SO3/ hr

Molar flow rate of O2 at outlet stream = 34.135 – (5.44 – 0.377) 0.5

= 31.60 kmol O2/ hr


3rd Pass

36.44 kmol SO3/hr

5.44 kmol SO2/hr

34.135 kmol O2/hr

354.48 kmol N2/hr

3rd Pass

36.44 kmol SO3/hr

34.135 kmol O2/hr

354.48 kmol N2/hr

5.44 kmol SO2/hr

41.50 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

25

4th pass catalyst bed:

Assuming 99.7% conversion (according to McKetta)

Molar flow rate of SO2 at outlet stream = 41.88 (1 - 0.997)

= 0.126 kmol SO2/hr

Molar flow rate of SO3 at outlet stream = 1.834 + (0.377 – 0.126)

= 2.085 kmol SO3/ hr

Molar flow rate of O2 at outlet stream = 31.60 – (0.377 – 0.126) 0.5

= 31.475 kmol O2/ hr


4th Pass

1.834 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

4th Pass

1.834 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

2.085 kmol SO3/hr

0.126 kmol SO2/hr

31.475 kmol O2/hr

354.480 kmol N2/hr

26

FINAL ABSORPTION TOWER (D-150)

Assuming 99.9% SO3 is absorbed.

Required amount of water = 0.999 2.087 = 2.085 kmol SO3/hr

L1 =

kmol H2SO4 (98.5wt%)/hr = 27.077 kmol H2SO4(98.5wt%)/hr

L2 = 0.923 27.077 + 2.085 = 27.077 kmol H2SO4 (100wt%)/hr

Final

Absorption

Tower


0.077 kmol H2O/kmol

L1 (98.5wt% H2SO4)

L2 (100 wt% H2SO4)

2.085 kmol SO3/hr

0.126 kmol SO2/hr

31.475 kmol O2/hr

354.480 kmol N2/hr

27

Thus, Outlet gas stream is as follow:

Amount of SO3 at outlet stream = 2.087 – 2.085 = 0.002 kmol SO3/hr

Amount of SO2 at outlet stream = 0.126 kmol SO2/hr

Amount of O2 at outlet stream = 31.475 kmol O2/hr

Amount of N2 at outlet stream = 354.48 kmol N2/hr

28

CIRCULATION TANK (F-154)

Acid Balance:

27.077 = 27.077 0.923 + L2 0.923

L2 = 2.259 kmol H2SO4(98.5wt%)/hr

Water Balance:

L1 = 0.077 L2 + 0.077 27.077

L1 = 2.259 kmol H2O/hr


0.077 kmol H2O/kmol

L2 kmol H2SO4 (98.5wt%)/hr

L1 kmol H2O/hr


0.077 kmol H2O/kmol

27.077 kmol H2SO4 (98.5wt%)/hr

Final

Absorption

Tower

Circulation

Tank

27.077 kmol H2SO4(100%)/hr

29


L1 kmol H2O/hr


0.1011 kmol H2O/kmol

DT_out

Drying Tower in

IAT_ in

515.156 kmol H2SO4 (98.5 wt%)/hr


0.077 kmol H2O/kmol

IAT_out

515.156 kmol H2SO4 (100 wt%)/hr

FAT_circ_tank_out


0.077 kmol H2O/kmol

2.259 kmol H2SO4 (98.5wt%)/hr

Dilution Water

Iinterpass

absorber and

Drying

Tower

Circulation

Tank

Product, P


0.077 kmol H2O/kmol

257.578 kmol H2SO4

soln (98.5 wt%)/hr

264.486 kmol H2SO4 (97.976 wt%)/hr

30

Acid Balance:

515.156 + 264.486 0.8989 + 2.259 0.923 = 257.578 0.923 + P 0.923 +

515.156 0.923

P 0.923 = 41.754 kmol H2SO4 /hr

P = 45.273 kmol H2SO4(98.5wt%)/hr

Water Balance:

264.486 0.1011 + L1 + 2.259 0.077 = 257.578 0.077 + 45.237 0.077

+515.156 0.077

L1 = 36.07 kmol H2O/hr

Chapter Seven

Energy Balance

32


Reaction: H2O (g) + H2SO4(l) = H2SO4(l) + H2O(l) ; H = -80MJ per kmol H2O(g)

Slightly weakened acid

Heat evolved for 6.908 kmol H2O/hr = -80 6.908 = -552.64 MJ/hr

Now,

-Q = mCpT

= 42.43oC

Outlet air Temperature = 42.43 + 37 = 79.43oC

449.145 kmol DA/hr

365 K

Dry Air 350 K

Moist Air @ 37oC

264.486 kmol H2SO4

soln (97.976 wt%)//hr

257.578 kmol H2SO4

soln (98.5 wt%)/hr

Drying

Tower

6.908 kmol H2O/hr

33

MELTER (Q-121)

For Solid Sulfur,

Hin =

= 0.2802 kJ/mol

For Liquid Sulfur at 130oC,

Hout = 4734 kJ/kmol

Thus heat load, Q = 41.88×4734 – 41.88×103×0.2802 = 186525.14 kJ/hr

140oC 37

oC

41.88 kmol Solid S/hr 41.88 kmol S(l)/hr

34

FURNACE (Q-120)

Reaction: S(l) + O2(g) = SO2(g) ; = -298.3kJ/mol

Cp (sulfur) = 0.23 Btu/lb.oF = 962.32 J/kg.oC

= 30.79 kJ/kmol.K

= 354.82 103

= 562.609 103 kJ/hr

= 94.32 103

= 152.407 103 kJ/hr

= 41.88 103 0.0308 (140-25) = 148.311 103 kJ/hr

ToC

41.88 kmol SO2/hr

140oC

41.88 kmol S/hr

79.43oC

354.48 kmol N2/hr

94.23 kmol O2/hr

52.35 kmol O2/hr

354.48 kmol N2/hr

35

Now, for adiabatic system,

Δ = - =0

-41.88 103 298.3 + - (562.609 103 + 152.407 103 +

148.311 103) = 0

= 13.693 106

41.88 103

+ 52.35 103

+ 354.82 103

= 13.693 106

T = 936.439oC

36

WASTE HEAT BOILER-1 (E-131)

ΔH = 41.88 103

+ 52.35 103

+ 354.82 103

= -6.062 106 kJ/hr

407oC 937.44

oC

41.88 kmol SO2/hr

52.35 kmol O2/hr

354.48 kmol N2/hr

37


ΔH = 23.45 103

+ 18.43 103

+ 40.63 103

+ 354.82 103

= -2.526 105 -1.334 105 – 1.909 105 – 15.504 105

= -21.273 105 kJ/hr

23.45 kmol SO3/hr

417oC 557.5

oC

18.43 kmol SO2/hr

40.63 kmol O2/hr

354.48 kmol N2/hr

38

HEAT EXCHANGER-1 (E-133)

From temperature 500.14oC to 385oC,

Q = = 36.44 103

+ 5.44 103

+ 34.135 103

+ 354.82 103

= -3.155 105 - 0.318 105 - 1.303 105 - 12.602 105

= -17.378 105 kJ/hr

372oC

385oC

500.14oC

1.834 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

ToC

36.44 kmol SO3/hr

5.44 kmol SO2/hr

34.135 kmol O2/hr

354.48 kmol N2/hr

39

Now,

1.834 103

+ 0.377 103

+ 31.6 103

+ 354.82 103

= 17.378 105

T = 224.78oC

40


From temperature 26.5oC to 224.78oC

Q = = 1.834 103

+ 0.377 103

+ 31.6 103

+ 354.82 103

= 21.28 103 + 3233.08 + 190.75 103 + 2.067 106

= 2.282 106 kJ/hr

224.78oC

ToC

417oC

1.834 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

26.5oC

41.5 kmol SO3/hr

0.377 kmol SO2/hr

31.6 kmol O2/hr

354.48 kmol N2/hr

41

Now,

41.5 103

+ 0.377 103

+ 31.6 103

+ 354.82 103

= -3.058 106

T = 262oC

42

ECONOMIZER (E-143)

262°C 207°C

Q =

= 41.5 103

+ 0.377 103

+ 31.60 103

+ 354.48 103

= -7.87 105 kJ/hr

41.5 kmol SO3/hr

0.377 kmol SO2/hr

31.60 kmol O2/hr

354.48 kmol N2/hr

43


From Perry‟s Chemical Engineers‟ Handbook, 8th Edition, Table 2-174,

At 20oC, Cp for 91.81 mol% H2SO4 = 0.3787 kcal/kg.oC

Cp for 94.82 mol% H2SO4 = 0.3554 kcal/kg.oC

Thus, Cp for 98.5wt% or 92.3 mol% H2SO4 = 0.3749 kcal/kg.oC

= 144.06 kJ/kmol.oC

515.156 kmol (98.5 wt% H2SO4)/hr@77oC 388.291 kmol/hr@T

oC

0.08138 kmol O2/kmol

0.00097 kmol SO2/kmol


0.91292 kmol N2/kmol

0.077 kmol H2O/kmol


Interpass

Absorption

Tower

427.957 kmol/hr@207oC

515.156 kmol H2SO4/hr@107oC





44

For Acid,

= 515.156 kmol/hr 144.06 kJ/kmol.oC (77-25)oC = 3859095.42 kJ/hr

=

= 6310635.35 kJ/hr

Reaction: SO3(g) + H2O(l) = H2SO4 (aq) ; ∆H = -130MJ/kmol SO3(g)

Thus amount of heat released = (427.957×0.09697 – 388.291×0.00472)×130×103

= 5156609.7 kJ/hr

For gas stream,

10−12 3) +

0.377×103×25207(38.91×10−3+3.904×10−5 −3.104×10−8 2+8.606×10−12 3)

+

31.6×103×25207(29.10×10−3+1.158×10−5 −0.6076×10−8 2+1.311×10−12 3)

+

354.48×103×25207(29.00×10−3+0.2199×10−5 +0.5723×10−8 2−2.871×10−12 3)

= 2.507×106 kJ/hr

Now,

T = 26.5oC

45






= 144.06 kJ/kmol.oC

Final

Absorption

Tower


0.077 kmol H2O/kmol





27.077 kmol (98.5 wt% H2SO4)/hr

27.077 kmol H2SO4/hr

388.168 kmol/hr



0.0815 kmol O2/kmol

0.9181 kmol N2/kmol

386.083 kmol/hr

46

For Acid,

= 27.077 kmol/hr 144.06 kJ/kmol.oC (77-25)oC = 202837.06 kJ/hr

=

= 331692 kJ/hr

Reaction: SO3(g) + H2O(l) = H2SO4 (aq) ; ∆H = -130MJ/kmol SO3(g)

Thus amount of heat released = (388.168×0.00538 – 386.083×0.00000518)×130×103

= 271224.71 kJ/hr

For gas stream,

10−12 3) +

0.124×103×25207(38.91×10−3+3.904×10−5 −3.104×10−8 2+8.606×10−12 3)

+

31.477×103×25207(29.10×10−3+1.158×10−5 −0.6076×10−8 2+1.311×10−12 3)

+

354.48×103×25207(29.00×10−3+0.2199×10−5 +0.5723×10−8 2−2.871×10−12 3)

= 2.09×106 kJ/hr

Now,

T = 219.2oC

47

SUPERHEAT ECONOMIZER (E-151)

From temperature 374oC to 207oC

Q =

= 2.085 103

+ 0.126 103

+ 31.475 103

+ 354.48 103

= -1.97 106 kJ/hr

207oC 374

oC

2.085 kmol SO3/hr

0.126 kmol SO2/hr

31.475 kmol O2/hr

354.48 kmol N2/hr

48

COOLER-1 (E-152)





= 144.06 kJ/kmol.oC

Q = - =

= – 1.014 105 kJ

103oC 77

oC

27.077 kmol H2SO4 (98.5wt%)/hr


0.077 kmol H2O/kmol

49

COOLER-2 (E-141)





= 144.06 kJ/kmol.oC

Q = - =

= – 2.576 106 kJ

100.14oC 77

oC

772.7 kmol H2SO4 (98.5wt%)/hr


0.077 kmol H2O/kmol

50

COOLER-3 (E-162)





= 144.06 kJ/kmol.oC

Q = - =

= – 4.444 105 kJ

100.14oC 32

oC

45.273 kmol H2SO4 (98.5wt%)/hr


0.077 kmol H2O/kmol

51






= 144.06 kJ/kmol.oC

2.259 kmol H2O/hr

ToC

107oC 32

oC

27.077 kmol H2SO4(100%)/hr

2.259 kmol H2SO4 (98.5wt%)/hr


0.077 kmol H2O/kmol

27.077 kmol H2SO4 (98.5wt%)/hr


0.077 kmol H2O/kmol

Final

Absorption

Tower

Circulation

Tank

ToC

52

=

= 331.691 103 + 1192.3 = 332.884 103 kJ/hr

Assuming adiabatic mixing,

Q = 0 = - =

- 332.884 103

0 = 4226.14 (T - 25) – 332.884 103

T = 103oC

53


=

= 6.3106 106 + 19.038 103 + 25.384 103 + 2.499 106

= 8.854 106 kJ/hr

=

= 817.971 144.06 (T-25) = 117.837 103 (T-

25)

36.07 kmol H2O/hr



DT_out @ 365K

ToC

(515.156+45.237+257.578) =

817.971 kmol H2SO4 (98.5 wt%)/hr


0.077 kmol H2O/kmol

IAT_out @ 380K

515.156 kmol H2SO4 (100 wt%)/hr

FAT_circ_tank_out

@103oC


0.077 kmol H2O/kmol

2.259 kmol H2SO4 (98.5wt%)/hr

Dilution Water @ 32oC

Interpass

absorber and

Drying

Tower

Circulation

Tank

264.486 kmol H2SO4 (97.976 wt%)/hr

54

Assuming adiabatic mixing,

Q = 0 = - = 117.837 103 (T-25) - 8.854 106

0 = 117.837 103 (T-25) - 8.854 106

T = 100.14oC

Chapter Eight

Water & Steam Balance

56


Ground water temp = 20 0 C = 68 0 F

Temperature of steam = 100 0 C = 212 0 F (Let)

Q1 = Q2

6.062 109 = m × S × Δθ + m×

= m × (S × Δθ +

= m × (4200 × (100-20) +

= m × 2593000

m = 2337.83 kg/hr



Temperature of steam = 100 0 C = 212 0 F (Let)

Q1 = Q2

21.273 108 = m × S × Δθ + m×

= m × (S × Δθ +

= m × (4200 × (100-20) +

= m × 2593000

m = 820.4 kg/hr

57

ECONOMIZER (E-143)


Temperature of water = 80 0 C = 176 0 F (Let)

Q1 = Q2

7.87 108 = m × S × Δθ

= m × S × Δθ

= m × 4200 × (80-20)

= m × 252000

m = 3123 kg/hr


Inlet steam temp = 100 0 C

Temperature of outlet steam = 150 0 C (Let)

Q1 = Q2

1.97 106 = m × S × Δθ

m ×2100 × (150 – 100) = 1.97 106

m = 18.76 kg/hr.

58

COOLER-1 (E-152)

Ground water temp = 20 0 C

Temperature of water = 35 0 C

1.014 108= m Δθ

= m (35-20)

m = 1609.52 kg/hr

COOLER-2 (E-141)



2.576 109 = m Δθ

= m (70-20)

m = 12266.66 kg/hr

COOLER-3 (E-162)



4.444 108 = m Δθ

= m (50-20)

m = 3526.98 kg/hr

Chapter Nine

Power Balance

60

BLOWER (G-112)

Blower outlet pressure = 2 atm

Assuming isothermal compression

Work = nRT ln(P2/P1)

= 449.145×1000 mol air/hr × 8.314JK-1mol-1 × 310K × ln(2/1)

= 802.386×106 J/hr

= 222885 J/s

= 298.77 hP

61

PUMP-1 (L-122)

Density of Liquid Sulfur = 1.819 g/cm3 = 1819 kg/m3

Flow rate of liquid sulfur = 41.88 kmol/hr

= 1340.16 kg/hr

Viscosity of liquid sulfur at 140oC = 7.52 cP

Volumetric flow rate of liquid sulfur = 0.737 m3/hr

Assuming 2in nominal pipe size and 40 schedule number.

Internal dia for 4in nominal size pipe = 2.067 in =0.053 m

Total power head,

Wo=g∆Z+∆p/ρ+ΣF

Sum ΣF =

Reynolds no,

Re = ρ

=

= 121

Mass flow rate,

m= 0.3722 Kg/s

Volumetric flow rate,

Q=0.377/1819 = 0.0002 m3/s

Area, A=

=

=0.0022 m2

Velocity, V=Q/A

=

= 0.093 m/s

62

Friction factor,

f=

=0.1322

Equivalent length,

L=2 m

ΣF=

=0.030 J/kg

Here, ∆p/ρ=0

Elevation Head,

∆Z=1.5 m

Total power head,

Wo=9.8*1.5+0.030

=14.745 J/kg

Hence total H.P=14.745x0.3722/746 H.P

=0.01 hp

63

PUMP-2 (L-153)

Using Nomograph for the estimation of optimum economic pipe dia = 4 in

Thus we will select pipe with 4in nominal pipe size and 40 schedule number.

Internal dia for 4in nominal size pipe = 4.026 in =0.1023m

Total power head,


Sum ΣF =

Reynolds no,

Re = ρ

=

=318

Mass flow rate,

m=27.077 x 0.923 x1000x98x10-3/3600

=0.68 Kg/s


Q=0.68/1825 = 3.72 x 10-4 m3/s

Area, A=

=

=8.1x 10-3 m2

Velocity, V=Q/A

=

= 4.6 x 10-2 m/s

Friction factor,

f=

=0.051

64

Equivalent length,

L=2.5 m

ΣF=

=0.00531 J/kg

Here, ∆p/ρ=0

Elevation Head,

∆Z=2.2 m

Total power head,

Wo=9.8*2.2+0.00531

=21.57 J/kg

Hence total H.P=21.57x0.68 /746 H.P

=0.02 hp

65

PUMP-3 (L-145)

Using Nomograph for the estimation of optimum economic pipe dia = 5.5118 in

Thus we will select pipe with 6in nominal pipe size and 40 schedule number.

Internal dia for 4in nominal size pipe = 6.065 in =0.154 m

Total power head,


Sum ΣF =

Reynolds no,

Re = ρ

=

= 7300

Mass flow rate,

m=19.713 Kg/s


Q=19.713/1825 = 0.0108 m3/s

Area, A=

=

=0.0186 m2

Velocity, V=Q/A

=

= 0.58 m/s

Friction factor, from Moody‟s diagram, assuming smooth pipe,

f = 0.034

66

Equivalent length,

L=2.5 m

ΣF=

=0.371 J/kg

Here, ∆p/ρ=0

Elevation Head,

∆Z=2.2 m

Total power head,

Wo=9.8*2.2+0.371

=21.93 J/kg

Hence total H.P=21.93x19.713 /746 H.P

=0.58 hp

67

PUMP-4 (L-146)

Using Nomograph for the estimation of optimum economic pipe dia = 0.007m =0.276 in

Thus we will select pipe with

in nominal pipe size and 40 schedule number.

Internal dia for

in nominal size pipe = 0.364 in =0.0092 m

Total power head,


Sum ΣF =

Reynolds no,

Re = ρ

=

= 111

Mass flow rate,

m=0.58 Kg/s


Q=0.058/1825 = 0.0000316 m3/s

Area, A=

=

=0.000066 m2

Velocity, V=Q/A

= 0.475 m/s

Friction factor,

f=

=0.144

68

Equivalent length,

L=5 m

ΣF=

=35.35 J/kg

Here, ∆p/ρ=0

Elevation Head,

∆Z=4.75 m

Total power head,

Wo=9.8*4.75+35.35

=81.9 J/kg

Hence total H.P=81.9 x 0.58 /746 H.P

=0.1 hp

69

PUMP-5 (L-161)

Using Nomograph for the estimation of optimum economic pipe dia = 0.02m =0.7874 in

Thus we will select pipe with in nominal pipe size and 40 schedule number.

Internal dia for in nominal size pipe = 1.049 in =0.0266 m

Total power head,


Sum ΣF =

Reynolds no,

Re = ρ

=

= 2038 < 2100

Mass flow rate,

m=1.154 Kg/s


Q=1.154/1825 = 0.00063 m3/s

Area, A=

=

=0.00056 m2

Velocity, V=Q/A

= 1.134 m/s

Friction factor,

f=

=0.0079

70

Equivalent length,

L=11 m

ΣF=

=8.4 J/kg

Here, ∆p/ρ=0

Elevation Head,

∆Z=10 m

Total power head,

Wo=9.8*10 + 8.4

=106.4 J/kg

Hence total H.P=106.4 x 1.154 /746 H.P

=0.164 hp

Chapter Ten

Stream Table

72

Components Stream 1 Stream 2 Stream 3

T = 37oC T = 37oC T = 37oC

Flow Rate (kmol/hr)

Composition (kmol/kmol)

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3

SO2

O2 94.32 0.207 94.32 0.207 94.32 0.207

N2 354.48 0.778 354.48 0.778 354.48 0.778

H2O (g) 6.908 0.015 6.908 0.015 6.908 0.015

Total 455.708 1.000 455.708 1.000 455.708 1.000


T = 79.43oC T = 37oC T = 140oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


S(s) 41.88 1.00

S(l) 41.88 1.00

SO2

O2 94.32 0.21

N2 354.48 0.79

Total 448.80 1.00 41.88 1.00 41.88 1.00


T = 140oC 936.44oC 407 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


S(l) 41.88 1.00

SO3

SO2 41.88 0.093 41.88 0.093

O2 52.35 0.117 52.35 0.117

N2 354.48 0.790 354.48 0.790

Total 41.88 1.00 448.71 1.000 448.71 1.000


557.5 oC 417 oC 500.14 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3 23.48 0.054 23.48 0.054 36.44 0.085

SO2 18.43 0.042 18.43 0.042 5.44 0.013

O2 40.63 0.093 40.63 0.093 34.135 0.079

N2 354.48 0.811 354.48 0.811 354.48 0.823

H2SO4

Total 437.02 1.000 437.02 1.000 430.495 1.000

73


385 oC 417 oC 262 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3 36.44 0.085 41.50 0.0553 41.50 0.0553

SO2 5.44 0.013 0.377 0.0005 0.377 0.0005

O2 34.135 0.079 31.60 0.0421 31.60 0.0421

N2 354.48 0.823 354.48 0.4721 354.48 0.4721

H2SO4

Total 430.495 1.000 750.837 1.00 750.837 1.00


207 oC 26.5 oC 224.78 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3 41.50 0.0553 1.834 0.005 1.834 0.005

SO2 0.377 0.0005 0.377 0.001 0.377 0.001

O2 31.60 0.0421 31.60 0.081 31.60 0.081

N2 354.48 0.4721 354.48 0.913 354.48 0.913

H2SO4

Total 750.837 1.00 388.291 1.000 388.291 1.000


372 oC 374 oC 207 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3 1.834 0.005 2.805 0.0072 2.805 0.0072

SO2 0.377 0.001 0.126 0.0003 0.126 0.0003

O2 31.60 0.081 31.475 0.0809 31.475 0.0809

N2 354.48 0.913 354.48 0.9116 354.48 0.9116

H2SO4

Total 388.291 1.000 388.886 1.000 388.886 1.000


219.2 oC 107 oC 25 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


SO3 0.002 5.18×10-6

SO2 0.126 0.0003

O2 31.475 0.0815

N2 354.48 0.9181

H2SO4 27.077 1.00 24.992 0.923

H2O (l) 2.085 0.077

Total 386.083 1.000 27.077 1.00 27.077 1.00

74


25 oC 77 oC 32 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


H2SO4 24.992 0.923 24.992 0.923

H2O (l) 2.085 0.077 2.085 0.077 2.259 1.00

Total 27.077 1.00 27.077 1.00 2.259 1.00


103 oC 32 oC 100.14 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


H2SO4 2.085 0.923 713.233 0.923

H2O (l) 0.174 0.077 36.07 1.00 59.501 0.077

Total 2.259 1.000 36.07 1.00 772.734 1.000


100.14 oC 77 oC 77 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


H2SO4 713.233 0.923 237.744 0.923 475.489 0.923

H2O (l) 59.501 0.077 19.834 0.077 39.667 0.077

Total 772.734 1.000 257.578 1.000 515.156 1.000


365K 103 oC 107 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


H2SO4 237.746 0.8989 2.085 0.923 515.156 1.00

H2O (l) 26.740 0.1011 0.174 0.077

Total 264.486 1.00 2.259 1.000 515.156 1.00


100.14 oC 100.14 oC 32 oC

Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


Flow Rate (kmol/hr)


H2SO4 45.787 0.923 45.787 0.923 45.787 0.923

H2O (l) 3.486 0.077 3.486 0.077 3.486 0.077

Total 45.273 1.000 45.273 1.000 45.273 1.000

Chapter Eleven

Equipment List

76

Columns & Towers

Item No. Equipment Required No. Type

D-110 Drying Tower 1 Packed Bed

D-140 Interpass Absorption Tower 1 Packed Bed

D-150 Final Absorption Tower 1 Packed Bed

Reactors


R-130 Converter 1 Multiple Catalyst Bed

Furnaces & Process Heater


Q-120 Furnace 1 Gun spray

Q-121 Melter 1 Heater

Heat Exchanger


E-131 WHB-1 1 Shell-and-tube

E-132 WHB-2 1 Shell-and-tube

E-133 Heat Exhanger-1 1 Shell-and-tube

E-142 Heat Exchanger-2 1 Shell-and-tube

E-143 Economizer 1 Shell-and-tube

E-151 Superheat Economizer 1 Shell-and-tube

E-152 Cooler-1 1 Shell-and-tube



Storage Vessels & Circulation Tanks


F-154 FAT Circulation Tank 1 Vessel

F-144 IAT and DT Circulation Tank 1 Vessel

F-160 Storage Tank 1 Vessel

77

Pumps


L-122 Pump-1 1 Centrifugal





Other Equipments


H-111 Filter 1 Electrostatic precipitator

G-112 Blower 1 Centrifugal

Chapter Twelve

Equipment Sizing

79


Average wt per kmol of solution at inlet = 0.923 kmol H2SO4 + 0.077 kmol H2O

= 0.923 × 98 + 0.077 × 18

= 91.84 kg/kmol

Mass flow rate of liquid, L = 257.578 kmol/hr × 91.84 kg/kmol

= 23655.96 kg/hr

= 6.57 kg/sec

257.578 kmol (98.5 wt% H2SO4)/hr

0.077 kmol H2O/kmol


Drying

Tower

264.486 kmol (97.976 wt% H2SO4)/hr

94.32 kmol O2/hr

354.82 kmol N2/hr

6.908 kmol H2O/hr



94.32 kmol O2/hr

354.82 kmol N2/hr

80

Mass flow rate of gas, G = 6.908×18 + 94.32×32 + 354.82×28

= 13077.544 kg/hr

= 3.6327 kg/sec

Specific Gravity of a 98.5wt % H2SO4 = 1.825 [from „Sulfuric Acid Manufacturing‟ by

Devenport]

Density, ρL = 1.825 × 1000 kg/m3

= 1825 kg/m3

Weight of gas per kmol, M = 0.015×18 + 0.207×32 + 0.778×28

= 28.678 kg/kmol

ρg =

= 0.9986 kg/m3

μL = 5.2 cP [from Devenport]

Packing Material

1.5 inch Ceramic Intalox Saddles

From McCabe Smith,

Bulk density = 39 lb/ft3

Total area = 59 ft2/ft3

Porosity = 0.76

Packing factor, Fp = 52 ft2/ft3

81

Diameter Evaluation

=

= 0.0423

From fig 15-7 [Peters & Timmerhaus, 5th Edition]

= 0.2

G2 =

= 9.4155

G = 3.068 kg/s.m2

v =

= 3.07 m/s

According to the literature, packed column operates at a vapor velocity that is 70% to

90% of the flooding velocity.

Therefore, allowable vapor velocity, v = 0.9 3.07 = 2.763 m/s

Volumetric flow rate of gas =

= 3.638 m3/s

=

=1.3166 m2

=

= 1.295 m

82

Height Evaluation

NTU calculation:

y1 = 0.01515 kmol H2O/kmol

y2 = 0.00001 kmol H2O/kmol

NOG =

But reaction in the interface is so instantaneous that y* = 0

NOG =

= ln

= ln

= 7.323

HTU calculation:

According to the equation developed by Fuller et al. [reference: Coulson & Richardson,

volm 6, 3rd Edition]

Dv =

Ma = molecular weight of H2O = 18

Mb = molecular weight of H2SO4 = 98

T = average H2SO4 temperature = 360K

P = 101325 Pa

Va = Diffusion volume co-efficient for H2O = 1.98 2 + 5.48 = 9.44

Vb = Diffusion volume co-efficient for H2SO4 = 1.98 2 + 17 + 5.48 = 42.88

Dv =

= 0.621×10-5 m2/s

Gv = Gas phase mass velocity =

= 2.759 kg/m2-s

83

av = Packing surface area per unit volume of packing = 59 ft2/ft3 = 193.57 m2/m3

μv = 2.595 × 10-5 kg/m-s

According to the correlation of Onda et al. [Ref: Perry‟s Chemical Engineering

Handbook, pp 5-80, table 5-24],

=

Here,

C1 = 5.23

=

= 82.768

=

= 1.611

= = 0.0184

From the correlation,

=

= 5.23 82.768 1.611 0.0184 = 12.832

kG = 12.832 193.57

0.621 10-5

= 0.523 10-3

V =

= 0.107

84

HG =

=

= 1.06 m

HOG = HG +

HL = HG = 1.06 m

Thus, H = HOG NOG

= 1.06 7.323

= 7.74 m

= 25.4 ft

85

CONVERTER (R-130)

The partial pressures of the various species are numerically equal to their mole fractions

since total pressure is one atmosphere.

Catalyst Bed1

Gas Initial Moles Moles at

Conversion f

Mole fraction at fraction conversion

f

SO2 41.88 41.88(1-f)

41.88(1-f)/ (448.71 - 20.94f)

N2 354.48 354.48 354.48/(448.71 - 20.94f)

O2 52.35 52.35 - 20.94f 52.35 - 20.94f/(448.71 - 20.94f)

SO3 0.0 41.88f 41.88f/(448.71 - 20.94f)

448.71 - 20.94f 1.00

Inlet molar flow rates in lb-moles/s

SO2 = 0.026 lb-moles/s

O2 = 0.032 lb-moles/s

N2 = 0.217 lb-moles/s

86

Average molecular weight lb/lb-mol

Inlet mass velocity,

lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)

Molal mass velocity of reactant A (SO2) :

lb-moles/ft2

87

Catalyst Bed 1

function [T]= temp1(f); T=(680+(287.3*f))/(1+(0.023*f));

function [b]=beta1(T,f); x=((1.875*10.^11)*(52.35-20.94*f)*(1-f).^0.5)/(((448.71-

20.94*f).^0.5)*exp(15601.4/T)); y=(f*((52.35-20.94*f).^0.5)*(8.13*10.^15))/(((1-f).^0.5)*exp(26975.3/T)); b=(448.71-20.94*f)/(x-y);

function design1 t(1)=680; z(1)=0; for i=1:14 n=i-1; f=0:.04:0.56; if t(i)>860; break end t(i+1)=temp1(f(i+1)); b(i)=beta1(t(i),f(i)); b(i+1)=beta1(t(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); z(i+1)=z(i)+dz; end disp('data for 1st bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,

' f1', 't1', 'z1') fprintf('%s\n', ' ' ) A=[f(1:i+1)',t(1:i+1)',z(1:i+1)']; disp(A)

Data for 1st catalyst bed are given below:

f1 t1 z1

0.00 680.0000 0.0000

0.04 690.8564 0.3070

0.08 701.6929 0.5297

0.12 712.5059 0.6934

0.20 723.3062 0.9071

0.24 744.8405 0.9773

0.28 755.5781 1.0315

f1 t1 z1

0.32 766.2961 1.0740

0.36 776.9945 1.1077

0.40 787.6734 1.1348

0.44 798.3329 1.1571

0.48 808.9729 1.1756

0.52 819.5937 1.1914

0.56 830.1951 1.2054

Outlet temperature = 830K

Conversion = 56% of inlet

Catalyst bed height = 1.21 ft

88

Catalyst Bed 2

Gas Initial Moles Moles at Conversion f

Mole fraction at fraction conversion f

SO2 18.43 18.43(1-f)

18.43(1-f)/ (436.99 – 9.215f)

N2 354.48 354.48 354.48/(436.99 – 9.215f)

O2 40.63 40.63 – 9.215f (40.63 –9.215f)/(436.99 – 9.215f)

SO3 23.45 23.45 +18.43f (23.45+18.43f)/(436.99 – 9.215f)

436.99 – 9.215f 1.00






Average molecular weight


lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)

89


lb-moles/ft2

lb-mols/ft2-s

90

Catalyst Bed 2

function[T]=temp2(f) T=(690+(126.02*(f-0)))/(1+(0.0102*(f-0)));


9.215*f).^0.5)*exp(15601.4/T)); y=((23.45+18.43*f)*((40.63-9.215*f).^0.5)*(6.67*10.^14))/(((1-

f).^0.5)*exp(26975.3/T)); b=(436.99-9.215*f)/(x-y);

function design2 T(1)= 690; Z(1)= 0; for i=1:10 f=0:0.07:0.70; if T(i)>850; break end T(i+1)=temp2(f(i+1)); b(i)= beta2(T(i),f(i)); b(i+1)=beta2(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 2nd bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,

' f1', 't1', 'z1') fprintf('%s\n', ' ' ) A=[f(1:i+1)',T(1:i+1)',Z(1:i+1)']; disp(A)

Data for 2nd catalyst bed are given below:

f1 t1 z1

0.00 690.0000 0.0000

0.07 698.3228 0.3348

0.14 706.6337 0.6064

0.21 714.9328 0.8294

0.28 723.2201 1.0151

0.35 731.4956 1.1721

f1 t1 z1

0.42 739.7593 1.3074

0.49 748.0112 1.4268

0.56 756.2515 1.5357

0.63 764.4801 1.6404

0.70 772.6969 1.7509

Outlet temperature = 772 K



91

Catalyst Bed 3

Gas Initial Moles Moles at Conversion f Mole fraction at fraction conversion f

SO2 18.43 5.44(1-f)

5.44(1-f)/ (430.50-2.72f)

N2 354.48 354.48 354.48/(430.50-2.72f)

O2 40.63 34.135 – 2.72f (34.135– 2.72f)/( 430.50-2.72f)

SO3 23.45 36.44+5.44f (36.44+5.44f)/( 430.50-2.72f)

430.50-2.72f 1.00








lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)


92

lb-moles/ft2

lb-mols/ft2-s

93

Catalyst Bed 3

function [T]= temp3(f) T=(658+(36.92*(f-0)))/(1+(0.003*(f-0)));


2.72*f).^0.5)*exp(15601.4/T)); y=((36.44+5.44*f)*((34.135-2.72*f).^0.5)*(4.17*10.^15))/(((1-

f).^0.5)*exp(26975.3/T)); b=(430.5-2.72*f)/(x-y);

function design3 T(1)= 658; Z(1)= 0; for i=1:31 f=0:0.03:0.93; if T(i)>860; break end T(i+1)=temp3(f(i+1)); b(i)= beta3(T(i),f(i)); b(i+1)=beta3(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 3rd bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,


94

Data for 3rd catalyst bed are given below:

f1 t1 z1

0.00 658.0000 0.0000

0.03 659.0483 0.3047

0.06 660.0964 0.6040

0.09 661.1443 0.8983

0.12 662.1920 1.1879

0.15 663.3340 1.4731

0.18 664.2869 1.7543

0.21 665.3340 2.0319

0.24 666.3810 2.3061

0.27 667.4278 2.5774

0.30 668.4744 2.8462

0.33 669.5208 3.1129

0.36 670.5670 3.3779

0.39 671.6130 3.6418

0.42 672.6588 3.9051

0.45 673.7045 4.1683

f1 t1 z1

0.48 674.7500 4.4323

0.51 675.7952 4.6978

0.54 676.8403 4.9656

0.57 677.8852 5.2370

0.60 678.9299 5.5133

0.63 679.9744 5.7962

0.66 681.0188 6.0880

0.69 682.0629 6.3916

0.72 683.1069 6.7113

0.75 684.1507 7.0532

0.78 685.1942 7.4272

0.81 686.2376 7.8504

0.84 687.2809 8.3578

0.87 688.3239 9.4034

0.90 689.3667 10.4393

0.93 690.4094 10.5245




95

Catalyst Bed 4


SO2 0.377 0.377(1-f)

0.377(1-f)/ (388.29-0.189f)

N2 354.48 354.48 354.48/(388.29-0.189f)

O2 40.63 31.60 - 0.189f (31.60-0.189f)/(388.29-0.189f)

SO3 23.45 1.834+0.377f (1.834+0.377f)/(388.29-0.189f)

388.29-0.189f 1.00






Average molecular weight 0 0047 80 0 00097 64 0 0814 32 0 913 28

28 606

Inlet mass velocity, 0 238 28 606

144 4

0 0602 lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)

96


lb-moles/ft2

lb-mols/ft2-s

97

Catalyst Bed 4

function [T]=temp4(f) T=(640+(0.785*(f-0)))/(1+((0.63*10.^-6)*(f-0)));


0.189*f).^0.5)*exp(15601.4/T)); y=((1.834+0.377*f)*((31.6-0.189*f).^0.5)*(9.66*10.^17))/(((1-

f).^0.5)*exp(26975.3/T)); b=(388.29-0.189*f)/(x-y);

function design4 T(1)=640; Z(1)=0; for i=1:11 f=0:0.06:0.66; if T(i)>850; break end T(i+1)=temp4(f(i+1)); b(i)= beta4(T(i),f(i)); b(i+1)=beta4(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 4th bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,


Data for 4th catalyst bed are given below:

f1 t1 z1

0.00 640.0000 0.0000

0.06 640.0471 0.0689

0.12 640.0942 0.1400

0.18 640.1412 0.2136

0.24 640.1883 0.2899

0.30 640.2354 0.3694

f1 t1 z1

0.36 640.2825 0.4524

0.42 640.3295 0.5395

0.48 640.3766 0.6314

0.54 640.4237 0.7291

0.60 640.4708 0.8338

0.66 640.5178 0.9473

Outlet temperature = 640.5 K



98

Total height of catalyst packing = 14.422ft

Weight of catalyst

Daisy ring catalyst diameter = 20 mm = 0.78 in

Reynolds number, Re

Ergun equation:

Inlet gas density,

lb/ft3

From Ergun equation,

lbm/ft-s2=0.110 psi

99


SHELL SIDE :

Fluid: WATER

Tavg : 140 0F

Viscosity, μ = 0.5cP

Thermal Conductivity, K = 0.381 Btu/ hr-ft-oF

Specific Heat Capacity, c = 0.45 Btu/lb-oF

Inlet Temperature, t1= 68 oF

Outlet Temperature, t2 = 212 oF

Mass Flow rate : 5154 lb/hr.

TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 1241.996 0F

Viscosity μ = 0.039 cP

Thermal Conductivity , K =0.0265 Btu/ hr-ft-oF


Inlet Temperature T1 = 1719.392 oF

Outlet Temperature , T2 = 764.6 oF

Mass Flow rate : 27906.35538 lb/hr.

SHELL SIDE:

Q = 5750000 Btu/ hr

100

LMTD CALCULATION:

Cold fluid on the shell side.

Hot fluid on the tube side.

LMTD =

= 1050.35 0 F

R =

= 6.6305

S =

= 0.087199163

From graph 18(Kern):

FT = 0.98

Corrected LMTD = 1050.35 × 0.98 = 1029.34383 oF

Now, Table 8(Kern) used for assuming UD

For, Gas – Water overall UD range is (2-50)Btu/ hr ft2 oF

Let, UD = 45 Btu/ hr ft2 oF

Heat transfer area calculation:

Q = UD ×A×(corrected LMTD)

A

=

= 124.042 ft2

101

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in

Length of tube, L = 16ft

Surface area per unit length, a‟‟ = 0.1963 ft

1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 40

Nearest count:

¾ in OD, 1 in square pitch

No. of tubes, = 52 (2-pass) [From table 9, Kern]

Shell ID 10 in (from table 9 KERN, corresponding value of No. of tubes)

So, A= 52 × 0.1963 × 16 ft2

= 163.3216 ft2

Corrected coefficient

=

= 34.17719289 Btu/ hr ft2 oF

For this value the calculated value is greater than the literature value.

So the total tube heat transfer area is 163.3216 ft2.

102


SHELL SIDE :

Fluid: WATER

Tavg : 140 0F







TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 909.050F




Inlet Temperature T1 = 1035.5oF

Outlet Temperature , T2 = 782.6oF


SHELL SIDE:

Q = 2010000 Btu/ hr

103

LMTD CALCULATION:



LMTD =

= 767.760 F

R =

= 1.75625

S =

= 0.148837209


FT = 0.98





Heat transfer area calculation


A

=

= 59.5 ft2

104

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 19

Nearest count:




So, A= 26 × 0.1963 × 16 ft2

= 81.66 ft2


=

= 32.761 Btu/ hr ft2 oF


So the total tube heat transfer area is ft2.

105


SHELL SIDE :

Fluid: GAS MIXTUER

Tavg : 828.626 0F

Viscosity, μ = 0.032 cP



Inlet Temperature, T1= 932.252 oF

Outlet Temperature, T2 = 725 oF


TUBE SIDE:

Fluid: GAS-GAS MIXTURE

Tavg : 569.102 0F

Viscosity μ =0.028 cP



Inlet Temperature t1 = 436.604oF

Outlet Temperature , t2 =701.6oF


SHELL SIDE:

Q = 1650000 Btu/ hr

106

LMTD CALCULATION:

Hot fluid on the shell side.

Cold fluid on the tube side.

LMTD =

= 258.4497749 0 F

R =

= 0.782094824

S =

= 0.534645555


FT = 0.85



Over all UD range is (2-50)Btu/ hr ft2 oF




A

=

= 166.616 ft2

107

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 53.05

Nearest count:




So, A= 52 × 0.1963 × 16 ft2

= 163.3216 ft2

108


SHELL SIDE:

Fluid: WATER

Tavg : 643.10F




Inlet Temperature, T1= 782.6 oF

Outlet Temperature, T2 = 503.6oF


TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 258.152 0F


Thermal Conductivity , K = 0.0265 Btu/ hr-ft-oF


Inlet Temperature t1 = 79.7oF

Outlet Temperature , t2 = 436.604 oF


SHELL SIDE:

Q = 2160000 Btu/ hr

109

LMTD CALCULATION:

Hot fluid on the shell side.

Cold fluid on the tube side.

LMTD =

= 383.63 0 F

R =

= 0.781722816

S =

= 0.507759283


FT = 0.88

Corrected LMTD = 383.63 × 0.88 = 337.595o F


Overall UD range is (2-50)Btu/ hr ft2 oF




A

=

= 142.37 ft2

110

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 45.33060455

Nearest count:




So, A= 52 × 0.1963 × 16 ft2 = 163.3216 ft2


=

= 39.228 Btu/ hr ft2 oF



111


Average wt per kmol of solution = 0.923 kmol H2SO4 + 0.077 kmol H2O

= 0.923 × 98 + 0.077 × 18

= 91.84 kg/kmol


= 47311.927 kg/hr

= 13.142 kg/sec

Interpass

Absorption

Tower


0.077 kmol H2O/kmol





515.156 kmol (98.5 wt% H2SO4)/hr


427.957 kmol/hr





388.291 kmol/hr

112

Mass flow rate of gas, G = 427.957 kmol/hr × [0.00088×64 + 0.09697×80 +

0.07384×32 + 0.82831×28]

= 14280.703 kg/hr

= 3.967 kg/sec


Devenport]

Density, ρL = 1.825 × 1000 kg/m3

= 1825 kg/m3

Weight of gas per kmol, M = [0.00088×64 + 0.09697×80 + 0.07384×32 +

0.82831×28]

= 33.3695 kg/kmol

ρg =

= 0.8473 kg/m3


Packing Material:


From McCabe Smith,



Porosity = 0.76

113


Diameter Evaluation:

=

= 0.0714


= 0.17

G2 =

= 6.7911

G = 2.606 kg/s.m2

v =

= 3.076 m/s





= 4.682 m3/s

=

=1.69 m2

=

= 1.467 m

114

Height Evaluation:

NTU calculation:

y1 = 0.09697 kmol SO3/kmol


NOG =


NOG =

= ln

= ln

= 3.0226

HTU calculation:



Dv =

Ma = molecular weight of SO3 = 80

Mb = molecular weight of water = 18

T = average water temperature = 365K

P = 101325 Pa

Va = Diffusion volume co-efficient for SO3 = 17.0 + 5.48 3 = 33.44

Vb = Diffusion volume co-efficient for H2O = 1.98 2 + 5.48 = 9.44

Dv =

= 2.793×10-5 m2/s


= 2.36 kg/m2-s

115


μv = 2.595 × 10-5 kg/m-s



=

Here,

C1 = 5.23

=

= 74.19

=

= 1.036

= = 0.0184


=

= 5.23 74.19 1.036 0.0184 = 7.3966

kG = 7.3966 193.57

2.793 10-5

= 1.336 10-3

V =

= 0.0707

116

HG =

=

= 0.273 m

HOG = HG +

HL = HG = 0.273 m

Thus, H = HOG NOG

= 0.273 3.0226

= 0.8266 m

= 2.712 ft

Pressure drop calculation:

Dp =

= 0.0074 m

vsm = 2.77 m/s

Here, NRe =

= 669

For dry packing,

= 0.2 in water/ft

117



= 0.923 × 98 + 0.077 × 18

= 91.84 kg/kmol


= 2486.75 kg/hr

= 0.6908 kg/sec

Final

Absorption

Tower


0.077 kmol H2O/kmol





27.077 kmol (98.5 wt% H2SO4)/hr


388.168 kmol/hr



0.0815 kmol O2/kmol

0.9181 kmol N2/kmol

386.083 kmol/hr

118


0.08109×32 + 0.91321×28]

= 13508.728 kg/hr

= 3.7524 kg/sec


Devenport]

Density, ρL = 1.825 × 1000 kg/m3

= 1825 kg/m3


0.91321×28]

= 34.8012 kg/kmol

ρg =

= 0.8836 kg/m3


Packing Material:


From McCabe Smith,



Porosity = 0.76

119


Diameter Evaluation:

=

= 0.00405


= 0.19

G2 =

= 7.9151

G = 2.813 kg/s.m2

v =

= 3.184 m/s





= 4.2467 m3/s

=

=1.482 m2

=

= 1.3736 m

120

Height Evaluation:

NTU calculation:



NOG =


NOG =

= ln

= ln

= 6.946

HTU calculation:



Dv =




P = 101325 Pa



Dv =

= 2.793×10-5 m2/s


= 2.532 kg/m2-s

121


μv = 2.595 × 10-5 kg/m-s



=

Here,

C1 = 5.23

=

= 77.94

=

= 1.017

= = 0.0184


=

= 5.23 77.94 1.017 0.0184 = 7.6278

kG = 7.6278 193.57

2.793 10-5

= 1.378 10-3

V =

= 0.0728

122

HG =

=

= 0.273 m

HOG = HG +

HL = HG = 0.273 m

Thus, H = HOG NOG

= 0.273 6.946

= 1.896 m

= 6.22 ft

123

ECONOMIZER (E-143)

SHELL SIDE :

Fluid: WATER

Tavg : 122 0F






Mass Flow rate : 3123 lb/hr.

TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 454.10F

Viscosity μ = 0.026cP






SHELL SIDE:

Q = 746000 Btu/ hr

124

LMTD CALCULATION:



LMTD =

= 332.08 0 F

R =

= 0.917

S =

= 0.247933884


FT = 0.97

Corrected LMTD = 332.08 × 0.97 = 322.12 o F






A

=

= 51.46 ft2

125

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 16.38

Nearest count:




So, A= 26 × 0.1963 × 16 ft2 = 81.6608 ft2


=

= 28.358 Btu/ hr ft2 oF



126


SHELL SIDE :

Fluid: WATER

Tavg : 257 0F







TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 554.9 0F







SHELL SIDE:

Q = 1870000 Btu/ hr

127

LMTD CALCULATION:



LMTD =

= 285.05 0 F

R =

= 3.34

S =

= 0.182481752


FT = 0.94







A

=

= 174.21 ft2

128

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 55

Nearest count:




So, A= 52 × 0.1963 × 16 ft2 = 163.3216 ft2


=

= 42.67 Btu/ hr ft2 oF



129

FAT CIRCULATION TANK (F-154)

Assuming acid accumulated for 12 hour

Thus weight of acid at tank for 30min = (27.077+2.259)×12 =352 kmol

= 352 kmol×91.84 kg/kmol

= 32330.62 kg 98.5 wt% acid solution

Density of 98.5 wt% sulfuric acid = 1825 kg/m3

Volume of 98.5 wt% sulfuric acid = 32330.62/1825 = 17.72 m3

Let, the diameter of the tank = D

And height to diameter ratio = 2

or, H = 2D

Now,

D3 = 11.28

D = 2.24 m

Thus, Height = 4.48 m

130

IAT & DT CIRCULATION TOWER (F-144)

Assuming acid accumulated for 30min or half an hour

Thus weight of acid at tank for 30min = (515.156+45.273+257.578)×0.5 = 409 kmol

= 409 kmol×91.84 kg/kmol

= 37562.56 kg 98.5 wt% acid solution

Density of 98.5 wt% sulfuric acid = 1825 kg/m3

Volume of 98.5 wt% sulfuric acid = 37562.56/1825 = 20.58 m3

Let, the diameter of the tank = D

And height to diameter ratio = 2

or, H = 2D

Now,

D3 = 13.10

D = 2.36 m

Thus, Height = 4.72 m

131

COOLER-1 (E-152)

SHELL SIDE :

Fluid: WATER

Tavg : 81.5 0F



Specific Heat Capacity, c = 1 Btu/lb-oF



Mass Flow rate, W = 3548.384 lb/hr.

TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 194 0F







SHELL SIDE:

Q = 96100 Btu/ hr

132

LMTD CALCULATION:



LMTD =

= 112.21 0 F

R =

= 1.733

S =

= 0.181


FT = 0.94

Corrected LMTD = 112.21 × 0.94 =105.48 o F






A

=

= 20.25 ft2

133

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in

Length of tube, L = 8 ft


1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 13

Nearest count:




So, A= 26 × 0.1963 × 8 ft2 = 40.8304 ft2


=

= 22.32 Btu/ hr ft2 oF



134

COOLER-2 (E-141)

SHELL SIDE :

Fluid: WATER

Tavg : 90.5 0F







TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 191.426 0F







SHELL SIDE:

Q = 2440000 Btu/ hr

135

LMTD CALCULATION:



LMTD =

= 100.92 0 F

R =

= 0.9256

S =

= 0.312


FT = 0.96

Corrected LMTD = 100.92 × 0.96 = 96.88 o F






A

=

= 252.021 ft2

136

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 80

Nearest count:




So, A= 76 × 0.1963 × 16 ft2 = 238.701 ft2


=

= 105.58 Btu/ hr ft2 oF



137

COOLER-3 (E-162)

SHELL SIDE:

Fluid: WATER

Tavg : 95 0F







TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 150.926 0F







SHELL SIDE:

Q = 421000 Btu/ hr

138

LMTD CALCULATION:



LMTD =

= 48.011 0 F

R =

= 2.271

S =

= 0.374


FT = 0.87

Corrected LMTD = 48.011× 0.87 = 41.77 o F






A

=

= 224.09 ft2

139

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 71

Nearest count:




So, A= 76 × 0.1963 × 16 ft2 = 238.701 ft2


=

= 42.24562344 Btu/ hr ft2 oF



140

STORAGE TANK (F-160)

Amount of Sulfuric acid in 6 days =100x6x1000

=600000 Kg

Density of acid = 1825 Kg/m3

Volume of the acid = 328.767 m3

Capacity =328.77 m3

Type: Closed cylindrical pressure vessel

Assuming length-diameter ratio = 1.55

Length = 10 m

Diameter = 6.47 m

Chapter Thirteen

Detailed Engineering Design

Converter

Interpass Absorption Column

Waste Heat Boiler-2

Storage Tank

13.1

Converter Design

Equipment Sizing

Mechanical Design

Engineering Drawing

P & ID Diagram

143

For the one-dimensional plug flow model, assuming adiabatic operation, the energy

equation is given as:

………………………. ( 1)

The material balance equation is given as:

…………………………………… (2)

………………………………………….. (3)

…………………………………………… (4)

From the two above equations:

…………………………….. (5)

Combining equations 2 and 5 with vA=-1 gives:

Where rm is

144

The partial pressures of the various species are numerically equal to their mole fractions

since total pressure is one atmosphere.

Catalyst Bed1

Gas Initial Moles Moles at

Conversion f

Mole fraction at fraction conversion

f

SO2 41.88 41.88(1-f)

41.88(1-f)/ (448.71 - 20.94f)

N2 354.48 354.48 354.48/(448.71 - 20.94f)

O2 52.35 52.35 - 20.94f 52.35 - 20.94f/(448.71 - 20.94f)

SO3 0.0 41.88f 41.88f/(448.71 - 20.94f)

448.71 - 20.94f 1.00





Average molecular weight lb/lb-mol

145


lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)


lb-moles/ft2

146

Catalyst Bed 1

function [T]= temp1(f); T=(680+(287.3*f))/(1+(0.023*f));


20.94*f).^0.5)*exp(15601.4/T)); y=(f*((52.35-20.94*f).^0.5)*(8.13*10.^15))/(((1-f).^0.5)*exp(26975.3/T)); b=(448.71-20.94*f)/(x-y);

function design1 t(1)=680; z(1)=0; for i=1:14 n=i-1; f=0:.04:0.56; if t(i)>860; break end t(i+1)=temp1(f(i+1)); b(i)=beta1(t(i),f(i)); b(i+1)=beta1(t(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); z(i+1)=z(i)+dz; end disp('data for 1st bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,

' f1', 't1', 'z1') fprintf('%s\n', ' ' ) A=[f(1:i+1)',t(1:i+1)',z(1:i+1)']; disp(A)

Data for 1st catalyst bed are given below:

f1 t1 z1

0.00 680.0000 0.0000

0.04 690.8564 0.3070

0.08 701.6929 0.5297

0.12 712.5059 0.6934

0.20 723.3062 0.9071

0.24 744.8405 0.9773

0.28 755.5781 1.0315

f1 t1 z1

0.32 766.2961 1.0740

0.36 776.9945 1.1077

0.40 787.6734 1.1348

0.44 798.3329 1.1571

0.48 808.9729 1.1756

0.52 819.5937 1.1914

0.56 830.1951 1.2054

Outlet temperature = 830K



147

Catalyst Bed 2

Gas Initial Moles Moles at Conversion f

Mole fraction at fraction conversion f

SO2 18.43 18.43(1-f)

18.43(1-f)/ (436.99 – 9.215f)

N2 354.48 354.48 354.48/(436.99 – 9.215f)

O2 40.63 40.63 – 9.215f (40.63 –9.215f)/(436.99 – 9.215f)

SO3 23.45 23.45 +18.43f (23.45+18.43f)/(436.99 – 9.215f)

436.99 – 9.215f 1.00








lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)

148


lb-moles/ft2

lb-mols/ft2-s

149

Catalyst Bed 2

function[T]=temp2(f) T=(690+(126.02*(f-0)))/(1+(0.0102*(f-0)));


9.215*f).^0.5)*exp(15601.4/T)); y=((23.45+18.43*f)*((40.63-9.215*f).^0.5)*(6.67*10.^14))/(((1-

f).^0.5)*exp(26975.3/T)); b=(436.99-9.215*f)/(x-y);

function design2 T(1)= 690; Z(1)= 0; for i=1:10 f=0:0.07:0.70; if T(i)>850; break end T(i+1)=temp2(f(i+1)); b(i)= beta2(T(i),f(i)); b(i+1)=beta2(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 2nd bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,


Data for 2nd catalyst bed are given below:

f1 t1 z1

0.00 690.0000 0.0000

0.07 698.3228 0.3348

0.14 706.6337 0.6064

0.21 714.9328 0.8294

0.28 723.2201 1.0151

0.35 731.4956 1.1721

f1 t1 z1

0.42 739.7593 1.3074

0.49 748.0112 1.4268

0.56 756.2515 1.5357

0.63 764.4801 1.6404

0.70 772.6969 1.7509




150

Catalyst Bed 3


SO2 18.43 5.44(1-f)

5.44(1-f)/ (430.50-2.72f)

N2 354.48 354.48 354.48/(430.50-2.72f)

O2 40.63 34.135 – 2.72f (34.135– 2.72f)/( 430.50-2.72f)

SO3 23.45 36.44+5.44f (36.44+5.44f)/( 430.50-2.72f)

430.50-2.72f 1.00








lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)


151

lb-moles/ft2

lb-mols/ft2-s

152

Catalyst Bed 3

function [T]= temp3(f) T=(658+(36.92*(f-0)))/(1+(0.003*(f-0)));


2.72*f).^0.5)*exp(15601.4/T)); y=((36.44+5.44*f)*((34.135-2.72*f).^0.5)*(4.17*10.^15))/(((1-

f).^0.5)*exp(26975.3/T)); b=(430.5-2.72*f)/(x-y);

function design3 T(1)= 658; Z(1)= 0; for i=1:31 f=0:0.03:0.93; if T(i)>860; break end T(i+1)=temp3(f(i+1)); b(i)= beta3(T(i),f(i)); b(i+1)=beta3(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 3rd bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,


153

Data for 3rd catalyst bed are given below:

f1 t1 z1

0.00 658.0000 0.0000

0.03 659.0483 0.3047

0.06 660.0964 0.6040

0.09 661.1443 0.8983

0.12 662.1920 1.1879

0.15 663.3340 1.4731

0.18 664.2869 1.7543

0.21 665.3340 2.0319

0.24 666.3810 2.3061

0.27 667.4278 2.5774

0.30 668.4744 2.8462

0.33 669.5208 3.1129

0.36 670.5670 3.3779

0.39 671.6130 3.6418

0.42 672.6588 3.9051

0.45 673.7045 4.1683

f1 t1 z1

0.48 674.7500 4.4323

0.51 675.7952 4.6978

0.54 676.8403 4.9656

0.57 677.8852 5.2370

0.60 678.9299 5.5133

0.63 679.9744 5.7962

0.66 681.0188 6.0880

0.69 682.0629 6.3916

0.72 683.1069 6.7113

0.75 684.1507 7.0532

0.78 685.1942 7.4272

0.81 686.2376 7.8504

0.84 687.2809 8.3578

0.87 688.3239 9.4034

0.90 689.3667 10.4393

0.93 690.4094 10.5245




154

Catalyst Bed 4


SO2 0.377 0.377(1-f)

0.377(1-f)/ (388.29-0.189f)

N2 354.48 354.48 354.48/(388.29-0.189f)

O2 40.63 31.60 - 0.189f (31.60-0.189f)/(388.29-0.189f)

SO3 23.45 1.834+0.377f (1.834+0.377f)/(388.29-0.189f)

388.29-0.189f 1.00






Average molecular weight 0 0047 80 0 00097 64 0 0814 32 0 913 28

28 606

Inlet mass velocity, 0 238 28 606

144 4

0 0602 lb/ ft2-s

Cp = 0.250 cal/g°K (650K-850K)

155


lb-moles/ft2

lb-mols/ft2-s

156

Catalyst Bed 4

function [T]=temp4(f) T=(640+(0.785*(f-0)))/(1+((0.63*10.^-6)*(f-0)));


0.189*f).^0.5)*exp(15601.4/T)); y=((1.834+0.377*f)*((31.6-0.189*f).^0.5)*(9.66*10.^17))/(((1-

f).^0.5)*exp(26975.3/T)); b=(388.29-0.189*f)/(x-y);

function design4 T(1)=640; Z(1)=0; for i=1:11 f=0:0.06:0.66; if T(i)>850; break end T(i+1)=temp4(f(i+1)); b(i)= beta4(T(i),f(i)); b(i+1)=beta4(T(i+1),f(i+1)); db=0.5*(b(i)+b(i+1)); dz=db*(f(i+1)-f(i)); Z(i+1)= Z(i)+dz; end disp('data for 4th bed are given below:-') fprintf('%2s\t %2s\t %2s\t %2s\t %2s\t %2s' ,


Data for 4th catalyst bed are given below:

f1 t1 z1

0.00 640.0000 0.0000

0.06 640.0471 0.0689

0.12 640.0942 0.1400

0.18 640.1412 0.2136

0.24 640.1883 0.2899

0.30 640.2354 0.3694

f1 t1 z1

0.36 640.2825 0.4524

0.42 640.3295 0.5395

0.48 640.3766 0.6314

0.54 640.4237 0.7291

0.60 640.4708 0.8338

0.66 640.5178 0.9473

Outlet temperature = 640.5 K



157

Total height of catalyst packing = 14.422ft

Weight of catalyst

Daisy ring catalyst diameter = 20 mm = 0.78 in

Reynolds number, Re

Ergun equation:

Inlet gas density,

lb/ft3

From Ergun equation,

lbm/ft-s2=0.110 psi

158

Mechanical Design of Reactor

Calculation of thickness of reactor wall

Reactor internal diameter, Di = 12ft = 3.66 m

Material of construction – Stainless steel (SS 304)

Mechanical Properties of SS 304:

Tensile stress = 586 MPa (RTP)

0.2% Yield stress = 241 MPa (RTP)

Design stress (580oC), f = 93.6 MPa = 13576 psi

Internal working pressure, Pi = 2.0 atm = 29.38 psi

Joint efficiency, J = 0.80

Corrosion allowance, Ca = 2 mm = 0.079 in

Thickness of cylindrical reactor wall,

Calculation of thickness of reactor head (Ellipsoidal head)


Thickness,

159

Volume of material required for fabrication

Volume of material for cylindrical section,

ft3

Blank diameter, Bd

Volume of material for ellipsoidal head, Ve

in3 = 3.85 ft3

Weight of material, Wm

Skirt design

Dead weight of vessel (catalyst, attachments, vessel material),

Dw = 3457288.6 lb=1568200 kg

Outside diameter of skirt, Dok = vessel outer diameter = 3.66 m

Stress due to dead weight, fd

kg/m2

Skirt height

Wind load for lower part of the vessel, Plw

= 598.08(3.66)

Where,

k = coefficient depending on shape factor (0.7 for cylindrical surface)

P1= wind pressure for the lower part of the vessel (40 to 100 kg/m2)

Bending moment, Mw

Stress due to wind load, fw

kg/m2

Where, Z = modulus of section of skirt cross-section

160

Stress at base due seismic load, fb

kg/m2

W = total dead weight of vessel

C = seismic coefficient (0.08)

H = total vessel height

Axial compressive stress on skirt, Sa

Allowable compressible stress is determined from:

Young‟s Modulus for SS 304, E = 207 GPa = 2.11x1010 kg/m2

Yield strength of SS 304, y = 241 MPa = 24575160 kg/m2

Therefor,

Solving for tsk

tsk > 0.014 m or 0.551 inch

Again,

tsk> 0.0138 m or 0.54 inch

161

12 ft

3ft

1.97ft

R-130

1.205 ft

1.75 ft

10.52 ft

0.95 ft

0.274 in

0.54 in

68.42 ft

Catalyst

Bed

Manhole

Reactor bed

partition

Brick

insulation

Quartz

layer

Gas in

Gas in

Gas in

Gas in

Gas out

Gas out

Gas out

Gas out

Figure 3: Four Pass Reactor (R-130)

162

TI

TI

TI

TI

TC

TC

TI

TI

TI

PI

PI

PI

TI

TC

E 131

E 132

PI

Piping &

Instrumentation

Diagram for four pass

converter

(R 130)

R 130

E 133

TT

TT

TT

Figure 4: P&ID Diagram for Four Pass Converter (R-130)

13.2

Interpass Absorption Column Design

Equipment Sizing

Mechanical Design

Engineering Drawing

P & ID Diagram

164


= 0.923 × 98 + 0.077 × 18

= 91.84 kg/kmol


= 47311.927 kg/hr

= 13.142 kg/sec

Interpass

Absorption

Tower


0.077 kmol H2O/kmol





515.156 kmol (98.5 wt% H2SO4)/hr


427.957 kmol/hr





388.291 kmol/hr

165


0.07384×32 + 0.82831×28]

= 14280.703 kg/hr

= 3.967 kg/sec


Devenport]

Density, ρL = 1.825 × 1000 kg/m3

= 1825 kg/m3


0.82831×28]

= 33.3695 kg/kmol

ρg =

= 0.8473 kg/m3

μL = 5.2 cP [Ref: 3]

Packing Material:


From McCabe Smith,



Porosity = 0.76

166


Diameter Evaluation

=

= 0.0714

From fig 15-7 [Reference 16]

= 0.17

G2 =

= 6.7911

G = 2.606 kg/s.m2

v =

= 3.076 m/s





= 4.682 m3/s

=

=1.69 m2

=

= 1.467 m

167

Height Evaluation

NTU calculation:



NOG =


NOG =

= ln

= ln

= 3.0226

HTU calculation:

According to the equation developed by Fuller et al. [ref: 18]

Dv =




P = 101325 Pa



Dv =

= 2.793×10-5 m2/s


= 2.36 kg/m2-s

168


μv = 2.595 × 10-5 kg/m-s

According to the correlation of Onda et al. [Ref: 15],

=

Here,

C1 = 5.23

=

= 74.19

=

= 1.036

= = 0.0184


=

= 5.23 74.19 1.036 0.0184 = 7.3966

kG = 7.3966 193.57

2.793 10-5

= 1.336 10-3

V =

= 0.0707

169

HG =

=

= 0.273 m

HOG = HG +

HL = HG = 0.273 m

Thus, H = HOG NOG

= 0.273 3.0226

= 0.8266 m

= 2.712 ft

Pressure drop calculation

Dp =

= 0.0074 m

vsm = 2.77 m/s

Here, NRe =

= 669

For dry packing,

= 0.2 in water/ft

170

Mechanical Design of Absorber

Inner diameter of Vessel, Di = 1.467 m

Height of the packing required = 0.8266 m

Skirt Height = 2 ft

Density of material column = 7700 kg/m3

Wind Pressure = 130 kg/m2

Material

Carbon Steel

Permissible tensile stress, f = 950 kg/cm2

Thickness of shell

Thickness of Shell,

Here,

Working pressure = N/m2

Design pressure, p = = N/m2 = 0.1266 N/mm2

Inner diameter of Vessel, Di = 1.467 m

Permissible tensile stress, f = 950 kg/cm2 = 9320 N/cm2 = 93.2 N/mm2

Joint Efficiency (J) = 0.85

Corrosion Allowance, c = 3 mm

171

Hence,

= 1.03 mm + 3.00 mm

= 4.03 mm

We take thickness as 8 mm

So, outer dia of the shell, Do = = 1483 mm = 1.483 m

Calculation of thickness of absorber head

Assuming ellipsoidal head,

Internal working pressure, Pi = 1.25 atm = 18.375 psi


Corrosion allowance, Ca = 3 mm = 0.118 in

Permissible tensile stress, f = 950 kg/cm2 = 9320 N/cm2 = 93.2 N/mm2 = 13521 psi

Thickness,

172

Stress Analysis and Shell Thickness at Different Heights

Let X be the distance in “m” from the top of the shell, then

1. AXIAL STRESS DUE TO PRESSURE

Axial stress due to pressure,

=

= 7.8 N/mm2

=

= 79.52 kg/cm2

2. STRESS DUE TO DEAD LOAD

a) Compressive Stress due to weight of shell up to a distance X

Outer Diameter Of shell = D i + 2ts = 1.483 m

Density of Shell material, ρs = 7700 Kg /m3

Now,

=

= 0.77X kg/cm2

b) Compressive stress due to weight of insulation at height X

Material for Insulation = Asbestos

Thickness of insulation, tins = 100 mm

Density of insulation = 575 Kg/m3

Let,

Dins = Diameter of insulation,

Dm = Mean diameter of vessel

173

And, for large diameter column,

Dins = Dm

=

= 11500 kg/m2

= 1.15X kg/cm2

c) Compressive stress due to liquid in column up to height X

Density of liquid, ρl = 1825 Kg/m3

fdliq =

= 0.133 106 N/m2

= 1.36 kg/cm2

d) Compressive stress due to attachment

We have the following attachments in the absorber column

Piping weight

Head weight

Ladder

Head weight (approximately) = 200 kg

Weight of Ladder = 160X kg

Total compressive stress due to attachments fd is given by,

fd(attachments) =

=

= 1.09 + 0.698X kg/cm2

174

e) Stress due to Wind

fw =

=

=

To determine the value of X

Permissible Stress = 95 N/mm2

And,

ftmax = fwx + fap – fdx

Solving the above equation,

We get,

SUPPORT FOR ABSORBER

Skirt support is used to support the absorber column.

Material to be used = Structural steel (IS 800)

Inner Diameter of the vessel, Di = 1.467 m

Outer Diameter of the vessel, Do = 1.483 m

Height of the Packing = 0.8266 m

Density of carbon steel, ρs = 7700 kg/m3

175

Total weight = Weight of vessel + Weight of Attachments

=

= 3660 kg

Diameter of Skirt = 1.467 m

Considering the height of Skirt is 2 ft = 0.61 m

a. Stress due to Dead Weight

Thickness of the skirt support is tsk

Stress due to dead load

fd = Total Weight /π Dstsk

=

=

kg/m2

b. Due to wind load

The forces due to wind load acting on the lower and upper parts of the

vessels are determined as plw = kp1h1Do

puw = kp2h2Do

Where K is coefficient depending on the shape factor

k=0.7 for cylindrical surface

Plw = kP1h1D0 = 0.7×100×3.17×1.483

= 223.4 kg

Assumption, k=0.7

P1=Wind pressure for the lower part of the vessel (40 to 100 kg/m2)

176

Bending moment

Mw=0.5Plwh1+Puw(h1+0.5h2)

=0.5×223.4kg×3.17 m = 354.1 kg-m

Stress fwb =

=

=

=

kg/m2

c. Stress due to Seismic Load

Load F= CW

W is total Weight of vessel

C is Seismic Coefficient

C=0.15 for magnitude 8.25 earthquake

Where,

Rok is radius of skirt

Thus,

=

Maximum Stress at bottom of Skirt

ftmax = (fwb or fsb) – fdb

=

-

kg/m2

=

kg/m2

177

Permissible tensile Stress for structural steel = 140 N/mm2

tsk = 0.002mm

Maximum Compressive Stress

ftmax = (fwb or fsb) + fdb

=

+

kg/m2

=

kg/m2

Yield point = 200N/mm2

fc permissible<or = 1/3 Yield point

= 66.6 N/mm2

= kg/m2

tsk = 2.4 mm

so, we will assume a thickness of 4 mm

Maximum Compressive Stress between bearing plate and foundation

fc = Total Weight/A + Mw/Z

Let, bolt circle diameter = (skirt diameter +20) cm

= 1.467 + 0.2 = 1.667 m

Dski = 1.467 m

= 0.4923 m2

= 0.364 m3

178

Thus fc = 3660kg/0.4923m2 + 354.1 kg-m/0.364m3

= 8407.3 kg/m2

Permissible stress F in bending is 157.5 N/mm2 = 16.07×106 kg/m2

tB = 0.0084 m = 8.4 mm

179

Ceramic Intalox

Saddle Packing

2.7 ft

2.5 ft

4 inch Liquid In

Gas In10 inch

2.5 ft

10

inch

1.4 ft

4.8 ft

Liquid

Out

Gas Out

4 inch

2 ft

0.7 ft

Ceramic Intalox Saddle

Packing

Mist

Eliminator

Acid

Distributor

Packing

Support

Manhole

Figure 5: Interpass Absorption Column (D-140) with Packing

180

Mist Eliminator

(Mesh Pad Type)

Ladder Type Acid

Distributor

Packing Support

(Ceramic Beams)

Figure 6: Mist Eliminator, Liquid Distributor and Packing Support

181

DPT

LT

DPAH

LI

LLA

HLA

PIC FTFC

TT

TI

FI

TI

To Acid

Circulation

Tank

From Acid

Circulation

Tank

TC

Water In

TI

Water Out

Gas In

Gas Out

TA-102

HE-108

PA-103

Figure 7: Piping & Instrumentation Diagram for Interpass Absorption Column (D-140)

13.3

Waste Heat Boiler-2 Design

Equipment Sizing

Mechanical Design

Engineering Drawing

P & ID Diagram

183

SHELL SIDE :

Fluid: WATER

Tavg : 140 0F







TUBE SIDE:

Fluid: GAS MIXTURE

Tavg : 909.050F




Inlet Temperature T1 = 1035.5oF

Outlet Temperature , T2 = 782.6oF


SHELL SIDE:

Q = 2010000 Btu/ hr

184

LMTD CALCULATION:



LMTD =

= 767.760 F

R =

= 1.75625

S =

= 0.148837209


FT = 0.98





Heat transfer area calculation


A

=

= 59.5 ft2

185

TUBE SELECTION:

OD= ¾ in

BWG = 16

ID = 0.62 in



1 in square pitch.

No. of tubes,

Nt =

‟‟ =

= 19

Nearest count:




So, A= 26 × 0.1963 × 16 ft2

= 81.66 ft2


=

= 32.761 Btu/ hr ft2 oF

186

TUBE SIDE:

Flow area per tube, = 0.302 in2

(table 10 KERN)

=

=

= 0.0273 ft2

Gt=

=

= 1052956.595 lb/ hr ft2

Tube inner dia, D = 0.62/12 =0.0517 ft

Reynold‟s no ,

μ =

= 642299.3789

1000 [ref:11]

μ

[assume

μ

μ = 1]

= 1000

= 482.213 Btu/hr-ft2-F

= ×

= 482.213 ×

= 398.63 Btu/hr-ft2-F

187

SHELL SIDE

Clearance, C‟ = 0.25

Baffle- space, B =

=

in.

Pitch, Pt = 1 in square Pitch

Flow area, as =

=

= 0.0222 ft2

Mass velocity, Gs=

=

= 81390.26 lb/hr ft2

Equivalent Diameter, De =

π

π = 0.07895 ft.

Reynold‟s no , 5310.73

40 (figure 28 KERN)

μ

[assume

μ

μ = 1]

= 217.425 (Btu/hr*ft2*F)

Clean heat transfer coefficient, =

=

= 140.69

Dirt factor, =

=

= 0.0234

Theoretical = 0.0001

Here calculated value is greater than theoretical value.

188

Shell side pressure drop:

Re = 5310.73

f= 0.003 ft2/in2 (figure-29, Kern)

S = 0.985 (specific gravity of water at 1400 F)

No of crosses, N+1 =

=

= 120

=

= 0.667 ft

Δ =

=

( =1)

= 0.391635684 psi

Tube side pressure drop:

Re = 642299.3789

f= 0.00009 ft2/in2 [Ref:11]

S = 1 (specific gravity of air)

L =16 ft

n =2

=1

Pressure drop , Δ =

=

= 1.184 psi

189

=

From figure 27, Tube side return pressure loss curve [Ref:11]

=0.15 psi

Δ

= 1.2 psi

Δ = Δ + Δ

= (1.184 +1.2)

= 2.384 psi.

Mechanical Design

Shell thickness

Material of construction: Carbon- steel

No. of shell: 1

No. of passes: 2

Fluid:



Working pressure : 48 psia

190

Design pressure:

Adding 10 % to the working pressure:

=48 × 1.1

= 52.8

The components of the tubular heat exchanger are normally designed for one of the

following pressure:

75, 150, 300, 450, 600 (TEMA STANDARD)

Therefore, shell has to be designed for 75 psi

Shell inside diameter, D = 8 in

Maximum permissible stress for Carbon Steel, f = (791+14.7) psia = 805.7 psia

j = welding efficiency (welding joint factor) = 0.9

Shell thickness, t =

= 0.393 inch

The minimum thickness required is 3/8 or 0.375 in. (KERN – page 129)

So the shell thickness is 0.393 in

Support:

Horizontal cylindrical vessels are supported on saddles. Saddle supports for fixed tube

sheet exchanger: one of the support is fixed while the other is placed on the roller to

facilitate expansion of shell.

191

Figure 8: AEP Type Heat Exchanger with 25% baffle cut for E-132

192

E-132

PI

TIT

TCPI

PI

TI

TI

PI

Hot

fluid in

Hot

fluid out

Cold

fluid out

Cold

fluid in

Figure 9: Piping & Instrumentation Diagram for Waste Heat Boiler-2 (E-132)

13.4

Storage Tank Design

Equipment Sizing

Mechanical Design

Engineering Drawing

P & ID Diagram

194

Amount of Sulfuric acid in 10 days =100x6x1000

=600000 Kg

Density of acid = 1825 Kg/m3

Volume of the acid = 328.767 m3

Capacity =328.77 m3

Type: Closed cylindrical pressure vessel

Length = 10 m

Diameter = 6.47 m

Mechanical Design

Type: Standard fixed roof <Closed cylindrical pressure vessel>

Codes: API standard 650

Material of construction: Carbon steel <lined with acid resisting brick>

Storage time = 6 days

Mass of stored H2SO4 capacity = 328.77 m3

d=6.47 m

h=10 m

Tank design data

Design temperature = 400C

Design pressure= Pressure due to liquid head

Working pressure=29.4 psi

H2SO4 density=1825 Kg/cm2

Material of construction =Carbon steel S275

Design tensile stress = 275 MPa

195

Calculation of internal pressure

Internal pressure,

P= ρ (H-0.3) x 102

=1825(10-0.3) x102x 10-6

=1.77 Kg/cm2

Safety factor = 1.2

P=2.124 Kg/cm2

Design tensile stress for carbon steel is,

f=275 MPa

=2804.2 Kg/Cm2

Thickness t is

t =

=

=2.95 mm

Density =0.283 lb/in3

Modulus of elasticity, E =30x106 lb/in2

Weight of the roof per unit area=thickness x density

=

=0.033 lb/in2

Total pressure on the roof = superimposed load + weight of roof plate

=

=0.201 lb/in2

196

The slope of the roof,

Sinθ=

=

=0.55

θ=33.1 0

197

33.1 0

10 m

6.47 m

Brick

Lining

Inlet

Outlet

Roof

0.14m

0.14m

2.95 mm

vent

Figure 10: Acid Storage Tank (F-160)

198

LT

LI LLA HLA

FI

F-160

Figure 11: Piping & Instrumentation Diagram for Storage Tank (F-160)

Chapter Fourteen

Plot Plan & Equipment Layout

200

Housing

Colony

Garden

Officers

Club

Recreational

Facilities

Playground

Gu

ard

Car

Parking Canteen

Administration

Office

La

bo

rato

ry

an

d R

&D

Fire

Fighting

First

AidW

ork

sh

op

Raw Material

Storage

Waste

Disposal

Utility

Plant

Mo

sq

ue

Process Plant

Future

Expansion

N

S

Figure 12: Plot plan for sulfuric acid plant

201

Control

Room

Air Intake

NG

Burner

Melter

Waste

Heat

Boiler-1

Waste

Heat

Boiler-2

Drying

TowerInterpass

Absorption

Tower

Final

Absorption

Tower

HE-1 HE-2

4 pass

Converter

Furnace

FAT

Circulation

Tank

IAT & DT

Circulation

Tank

Storage

Tank

N

S

Figure 13: Equipment layout for sulfuric acid plant

Chapter Fifteen

Estimation of

Total Capital Investment

And Production Cost

203

Cost Estimation of Major Units:

Blower (G-112)

Capacity = 298.77 Hp

For 298.77 HP capacity, cost = $ 1.1 105 (From figure 12.28 ref: 16)

Drying Tower(D-110)

Tower height = 11 m

Tower inner diameter = 1.295 m

Thickness = 8 mm

Outer diameter = 1.303 m

Volume = 0.018 m3

Density of carbon steel = 7850 kg/ m3

Weight of carbon steel = V × ρ = 0.018 × 7850 = 1413 kg

Form figure 15.10 for 1413 kg cost = $ 2 104

Waste heat boiler 1 (E-131)

Heat transfer surface area = 163.32 ft2 = 15.17 m2

Form figure 14.19 (ref: 16)

For 15.17 m2 surface area purchased cost = $ 8 103

Waste heat boiler 2(E-132)



For 7.586 m2 surface area purchased cost = $ 3.2 103

204

Converter(R-130)

Converter height = 20.85 m

Converter inner diameter = 3.66 m

Converter Thickness = 7 mm

Converter Outer diameter = 3.667 m

Converter Volume = 0.84 m3


Weight of carbon steel = V × ρ

= 0.81 × 7889 = 6626 kg

Form figure 15.10 for 6626 kg cost = $ 1 105

Heat exchanger 1(E-133)




Heat exchanger 2(E-142)




205

Final Absorbsion Tower (D-150)

Tower height =3.633 m


Thickness = 8 mm


Volume = 0.063 m3



= 0.063 × 7850 = 494 kg

Form figure 15.10 for 494 kg cost = $ 1.1 104

Intermediate Absorption Tower(D-140)

Tower height =2.56 m


Thickness = 8 mm


Volume = 0.095 m3



= 0.095 × 7850 kg =745.75 kg

Form figure 15.10 for 745.75 kg cost = $ 1.4 104

206

Economizer (E-143)



For 7.586 m2 surface area purchased cost = $ 3.2 103

Super heater economizer(E-151)




Circulation tank (F-154)

Tank height = 4.48 m

Tank inner diameter = 2.24 m

Tank thickness = 8 mm

Tank outer diameter = 2.248 m

Tank volume = 0.126 m3



= 0.126 × 7850 = 992 kg


207

Circulation tank ( F-144):

Tank height = 4.72 m

Tank inner diameter = 2.36 m

Tank thickness = 8 mm

Tank outer diameter = 2.368 m

Tank volume = 0.14 m3



= 0.14 × 7850 = 1100 kg


Pumps

From figure 12.20 (ref: 16)

according to the capacity

Pump 1 (L-122) , cost = $ 2000

Pump 2 (L-153), cost = $ 2200

Pump 3 (L-145), cost = $ 3000

Pump 4 (L-146), cost = $ 2200

Pump 5 (L-161), cost = $ 2200

Cooler 1(E-152)




208

Cooler 2(E-141)




Cooler 3(E-162)




Storage tank(F-160)

Tower height = 10 m


Thickness = 8 mm


Volume = 0.3 m3



= 0.3 × 7850 = 2355 kg


209

Summary of the equipment costs

No. Name of the Equipment Cost (In January 2002)

01. Filter $ 1.5 104

02. Blower $ 1.1 105

03. Drying Tower $ 2 104

04. Furnace $ 3.5 104

05. Melter $ 2.26 104

06. Waste Heat Boiler – 1 $ 8 103

07. Waste Heat Boiler – 2 $ 3.2 103

08. Converter $ 1 105

09. Heat Exchanger – 1 $ 8 103

10. Heat Exchanger – 2 $ 8 103

11. Final Absorption Tower $ 1.1 104

12. Intermediate Absorption Tower $ 1.4 104

13. Economizer $ 3.2 103

14. Superheat Economizer $ 8 103

15. FAT Circulation Tank $ 1.5 104

16. IAT & DT Circulation Tank $ 1.6 104

17. Pump – 1 $ 2,000

18. Pump – 2 $ 2,200

19. Pump – 3 $ 3,000

20. Pump – 4 $ 2,200

21. Pump – 5 $ 2,200

22. Cooler – 1 $ 3 103

23. Cooler – 2 $ 5 103

25. Cooler – 3 $ 5 103

26. Storage Tank $ 2.6 104

Cost of Major Equipments $ 4,47,600

210

Cost Index for 2011 is about 1309.37

Thus cost of major equipments = ($ 4,47,600 × 1309.37)/1116.9 =$ 5,24,732

Table: Estimation of Capital Investment

Factor Costs ($)

Direct Costs

Purchased equipment delivered 1.00 5,24,732

Purchased equipment installation 0.47 2,46,624

Instrumentation and controls 0.36 1,88,903

Piping 0.68 3,56,817

Electrical systems 0.11 57,720

Buildings 0.18 94,451

Yard improvements 0.10 52,473

Service facilities 0.70 3,67,312

Total direct plant cost 3.60 18,89,035

Indirect Costs

Engineering and supervisions 0.33 1,73,161

Construction expenses 0.41 2,15,140

Legal expenses 0.04 20,989

Contractor‟s fee 0.22 1,15,441

Contingency 0.44 2,30,882

Total indirect costs 1.44 7,55,614

Fixed Capital Investment 5.04 26,44,653

Working Capital 0.89 4,67,012

Total Capital Investment 5.93 31,11,665

211

Let, total product cost = C

Table: Estimation of total product cost

1. Manufacturing cost = Direct production cost + Fixed charges + Plant overhead

costs

i. Direct production costs

No. Components Costs ($)

01. Raw materials (10-80% of total product cost) 0.23 C

02. Operating labor (10-20% of total product

cost)

0.15 C

03. Direct supervisory and clerical labor (10-20%

of total product cost)

0.10 C

04. Utilities (10-20% of total product cost) 0.15 C

05. Maintenance and repairs (2-10% of fixed

capital investment cost)

0.025 FCI

06. Operating supplies (0.5-1% of fixed capital

investment)

0.005 FCI

07. Laboratory charges (10-20% of operating labor)

0.024 C

08. Patents and royalties (0-6% of total product cost)

0.005 C

ii. Fixed charges (10-20% of total product cost) = 0.1 C

iii. Plant overhead cost (5-15% of total product cost) = 0.05 C

Thus, manufacturing cost = 0.809 C + 0.03 FCI

212

2. General expenses = administrative costs + distribution and selling costs + R&D

costs

i. Administrative costs (2-5 % of total product cost) = 0.02 C

ii. Distribution and selling costs (2-20 % of total product cost) = 0.06 C

iii. R&D costs (about 5% of total product cost) = 0.05 C

iv. Financing (0-10% of total capital investment) = 0.05 TCI

Thus, General expenses = 0.13 C + 0.05 TCI

Hence, Total production cost = Manufacturing cost + General expenses

Or, C = 0.809 C + 0.03 FCI + 0.13 C + 0.05 TCI

Or, C = 0.939 C+ 234923

Or, 0.061 C = 234923

Or, C = 3851194

Total production cost = $ 3851194

Chapter Sixteen

Economic Analysis

214

Calculation of total income from sales

Production of H2SO4 (98.5 wt%) = 4104.17 kg/hr = 3,59,52,500 kg/yr

Market price = 10 tk per kg

Annual sales = 359,52,500 kg/yr × 10 tk/kg × $ 1/70 tk

= $ 51,36,072/yr

Calculation of rate of return

VAT = 2.5 %

Total sales after VAT = $ 50,07,670/yr

Total annual income = $ 50,07,670 - $ 38,51,194= $ 11,56,476/yr

Tax on income = 20%

Net profit after tax = $ 9,25,181/yr

Percent rate of return = net profit after tax/total capital investment

= 9,25,181/31,11,665

= 29.73 %

Turnover ratio

Turnover ratio = gross annual sales / fixed-capital investment

= 51,36,072/26,44,653

= 1.94

215

Depreciation of FCI

Project life = 15 yrs

Salvage Value at the end of the project life = 10% of FCI

Depreciable FCI = 0.9 × FCI = $ 23,80,188

Now depreciation by sinking fund, with 15% interest rate

= $ 23,80,188 (A/P, 15%, 15)

= $ 4,07,053

Table: Calculation of Payback Period

End of year k Net Cash Flow Cumulative PW at i=0%/yr through Year k

Present Worth of Cash Flow at i=15%/yr

Cumulative PW at i=15%/yr through Year k

0 -31,11,665 -31,11,665 -31,11,665 -31,11,665

1 925181 -2186484 804505 -2307160

2 925181 -1261303 699570 -1607590

3 925181 -336122 608322 -999268

4 925181 +589059 528975 -470293

5 925181 459978 -10315

6 925181 399981 389666

216

Payback period

Payback period (without interest) = 4 yr

Payback period (with interest) = 6 yr

Calculation of IRR by PW method

Let, IRR = i%

By PW method

-TCI + Annual Income (P/A, i%, 15) + Salvage value (P/F, i%, 15) = 0

Or, - $ 31,11,665 + $ 9,25,181 (P/A, i%, 15) + $ 2,64,465 (P/F, i%, 15) = 0

Or, - $ 31,11,665 + $ 9,25,181

+ $ 2,64,465

= 0

Or, i = 0.292

Thus, IRR = 29.2 %

Since IRR>MARR, the project is acceptable.

NPV Calculation with MARR=15%

NPV (15%) = -TCI + Annual Income (P/A, 15%, 15) + Salvage value (P/F, 15%, 15)

= - $ 31,11,665 + $ 9,25,181 (P/A, 15%, 15) + $ 2,64,465 (P/F, 15%, 15)

= - $ 31,11,665 + $ 5409876 + $ 32,501

= $ 23,30,712 > 0

So the investment is economically acceptable.

References

218

1. Backharst, R. J., and Harker, H. J., "Process Plant Design", Heinemann

Educational Books, London, 1973.

2. Daugherty, R.L., Franzani, J.B., and Finnemore, E.J., "Fluid Mechanics with

Engineering Applications", 1989.

3. Devenport, W.G., King, M. J., “Sulfuric Acid Manufacture”, Elsevier Science,

2005

4. Duecker, W.W., West, J.R., “The Manufacture of Sulfuric Acid”, Reinhold

Publishing Corporation, New York, 1959.

5. Felder, M.R., and Rousseau, R.W., "Elementary Principles of Chemical

Engineering", John Willy & Sons, 2nd Edution.

6. Foust, S., Alan; Wenzel, A.L., Clump, W.; C.; Maus Louis, and Andersen B., L.,

"Principles of Unit Operation", John Willy & Sons, 1980.

7. George, T.A., "Shrieves Chemical Process Industries", McGraw-Hill Book Co., 5th

Edition, International Edition.

8. Halman, P.J., "Heat Transfer", McGraw-Hill Book Co., 7th Edition, SI Units.

9. Hill, Charles G., “An Introduction to Chemical Engineering Kinetics & Reactor

Design”, John Wiley & Sons, 1977

10. Joshi, M.V., "Process Equipment Design", The Macmillan Company, 1976.

219

11. Kern, D.Q., "Process Heat Transfer", McGraw-Hill International Book Co., Japan,

International Student Edition, 1983.

12. Ludwig, E.E., "Applied Process Design For Chemical And Petrochemical Plants",

Volume Two, Gulf Publishing Company, U.S.A. Second Edition.

13. Meketta, J., John, "Piping Design Handbook", Marcal Dekkar Inc., New York,

1992.

14. Meketta, J., John, and Connigham A., W., "Encyclopedia of Chemical Processing

and Design", Marcal Dekkar Inc., New York, 1984.

15. Perry, Robert, H.; Green D., W., and Maloney, James O., "Perry's Chemical

Engineer's Hand Book", McGraw-Hill, Inc., 7th International Edition, April 1997.

16. Peters, M., S., and Timmerhaus, K., D., "Plant Design and Economics for

Chemical Engineer", McGraw-Hill, Inc., 4th Edition.

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Publishing Corp, Washington

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Butterworth Heinemann, 3rd Edition, 2003

overall design

Documents

design project

project design

design basis

mt sulfuric acid

oc relative humidity

detailed engineering

day sulfuric acid

average annual temperature