DESIGN OF A PLANT FOR THE PRODUCTION OF 750,000 STANDARD M³/DAY OF HYDROGEN BY PARTIAL OXIDATION OF HEAVY OIL FEEDSTOCK

ABSTRACT

The design of a plant producing hydrogen by partial oxidation of oil feedstock has been carried out. Purity of hydrogen obtained from the process was set at 98%, with minimal levels of impurities. The main process involved include the partial oxidation of the heavy fuel oil feedstock in a flame lined reactor called a gasifier to produce hydrogen. The successive units in the plants are purification units to ensure that the product specification of 98% purity of hydrogen is met. A water gas shift converter was also present to improve hydrogen yield. Waste heat energy generated from the gasification process is used in the waste heat boiler to generate steam for the plant. These purification units include: Quencher, Initial H₂S scrubber, final H₂S removal, CO₂absorption.

The chemical and mechanical design of the main equipments in the plants was carried out and the results presented. Costing and HAZOP analysis of each item of equipment is also provided. Detailed economic and profitability analysis was done and general safety and environmental issues were addressed.

TABLE OF CONTENTS

CONTENTS                 PAGE

CERTIFICATION

DECLARATION

DEDICATION

LETTER OF TRANSMITAL

ACKNOWLEDGEMENT v

ABSTRACT vi

LIST OF TABLES

LIST OF FIGURES xix

LIST OF SYMBOLS xxi

CHAPTER ONE

1 INTRODUCTION

1.1 HYDROGEN GAS: BRIEF HISTORY AND MARKET VALUE

1.2 THE HYDROGEN ECONOMY

1.3 LIMITATIONS OF HYDROGEN PRODUCTION

CHAPTER TWO

2 LITERATURE REVIEW

2.1 THE GASIFIER

2.1.1 FLUIDISED BED GASIFIERS

2.1.2 MOVING BED GASIFIERS

2.1.3 ENTRAINED FLOW GASIFIERS

2.2 THE WASTE HEAT BOILER

2.2.1 TYPES OF WASTE HEAT BOILERS

2.2.2 WASTE HEAT BOILER DESIGN CONSIDERATIONS

2.3 QUENCHING

2.3.1 HORIZONTAL DESIGN: CFQ QUENCHER

2.3.2 VERTICAL DESIGN: CCQ-QUENCHER

2.3.3 DESCRIPTION OF DESIGN

2.4 SCRUBBING SYSTEMS (INITIAL H2S REMOVAL)

2.4.1 TYPES OF SCRUBBERS

2.4.2 SCRUBBING MECHANISM

2.4.3 SOLVENTS FOR SCRUBBING

2.4.4 ABSORBER DESIGN

2.5 CO CONVERSION STAGE

2.5.1 DESIGN AND OPERATION OF WATER GAS SHIFT REACTORS

2.6 SATURATOR/DESATURATOR. 30

2.6.1 THE SATURATOR 30

2.6.2 THE DESATURATOR 31

2.7 FINAL H2S REMOVAL (CHEMOSORPTION)

2.7.1 PRESSURE SWING ABSORBER DESIGN

2.8 CARBON DIOXIDE REMOVAL

2.8.1 THE UNIT OPERATION

2.8.2 EQUIPMENT

2.8.3 HOT POTASSIUM CARBONATE (BENFIELD)  FOR CO2 REMOVAL

2.9 ANCILLIARY EQUIPMENT 39

2.9.1 RAM PUMPS 39

2.9.2 HEAT EXCHANGERS  42

2.9.3 AIR-COOLED EXCHANGERS                       43

2.9.4 FINNED TUBES 44

2.9.5 PLATE HEAT EXCHANGERS 45

2.9.6 SHELL AND TUBE HEAT EXCHANGERS 47

2.9.7 MEAN TEMPERATURE DIFFERENCE ( DRIVING FORCE) 48

CHAPTER THREE

3 MATERIAL AND ENERGY BALANCE

3.1 MATERIAL BALANCE

3.2 ENERGY BALANCE AROUND CO CONVERSION STAGE

3.3 ENERGY BALANCE AROUND THE GASIFIER 58

CHAPTER FOUR                                                                                                                   61

4 CHEMICAL ENGINEERING DESIGN 61

4.1 TANK ONE (FEED STORAGE TANK FARMS) 61

4.2 TANK 2 (FRESH FEED SURGE DRUM) 62

4.3 TANK 3 (STORAGE TANK FARM) 62

4.4 THE HEAVY FUEL OIL GASIFIER 64

4.5 THE WASTE HEAT BOILER 67

4.6 THE QUENCHER 72

4.6.1 DESIGN PROCEDURES 73

4.6.2 TYPES OF PACKING 73

4.6.3 HEAT CAPACITY 74

4.6.4 COLUMN DIAMETER 76

4.6.5 HEIGHT OF THE COLUMN: Z 76

4.7 INITIAL  H2S REMOVAL 77

4.7.1 ASSUMPTIONS 77

4.8 CARBON MONOXIDE CONVERSION UNIT 79

4.8.1 REACTOR TYPE: FIXED BED REACTOR 80

4.8.2 CATALYST: CHROMIUM – PROMOTED IRON OXIDE 81

4.8.3 CATALYTIC CONVERTER SIZING 84

4.9 FINAL H2S REMOVAL STAGE 85

4.9.1 NORMAL OPERATION 86

4.9.2 REGENERATION OPERATION 86

4.9.3 THE EQUIPMENT 87

4.9.4 ASSUMPTION 89

4.9.5 REGENERATION 91

4.10 CARBON DIOXIDE REMOVAL 92

4.10.1PACKING MATERIAL: PALL RINGS (PLASTIC) 94

4.11 HEAT EXCHANGER DESIGN 96

4.11.1  THE PROPOSED DESIGN 96

4.11.2  OPTIMISATION 97

4.11.3  VISCOSITY CORRECTION FACTOR 97

4.11.4 OXYGEN STREAM PRE-HEATER. 98

4.11.5 AIR-COOLED EXCHANGERS 102

4.12    PUMP SIZING 108

4.12.1APPROXIMATE PUMP-SIZING CALCULATION PROCEDURE 109

4.12.2THE METERING RAM PUMP 111

CHAPTER FIVE 115

5 MECHANICAL DESIGN 115

5.1 THE GASIFIER 115

5.1.1 DESIGN PRESSURE 115

5.1.2 DESIGN TEMPERATURE 115

5.1.3 MATERIAL OF CONSTRUCTION 115

5.1.4 THICKNESS OF THE REFRACTORY BRICK 116

5.1.5 MINIMUM WALL THICKNESS 116

5.1.6 MINIMUM HEAD THICKNESS 117

5.1.7 VESSEL SUPPORT 117

5.2 THE WASTE HEAT BOILER 117

5.3 THE QUENCHER 118

5.3.1 MATERIALS FOR CONSTRUCTION 118

5.3.2 PACKING 118

5.3.3 DESIGN PRESSURE 118

5.3.4 DESIGN TEMPERATURE 119

5.3.5 TENSILE STRENGTH 119

5.3.6 DESIGN STRESS 120

5.3.7 CHOICE OF CLOSURE 120

5.4 THE INITIAL H2S ABSORBER 121

5.4.1 SHELL THICKNESS 121

5.4.2 SUPPORT FOR ABSORBER 124

5.5 THE CO CONVERTER 125

5.5.1 OPERATING AND DESIGN TEMPERATURES AND PRESSURES 125

5.5.2 VESSEL DIMENSIONS AND ORIENTATION 125

5.5.3 CONSTRUCTION MATERIALS 126

5.5.4 TYPE OF VESSEL HEAD 126

5.5.5 OPENINGS REQUIRED 127

5.5.6 HEATING AND COOLING JACKETS OR COILS 127

5.5.7 INTERNAL FITTINGS SPECIFICATION 127

5.6 THE FINAL H2S ABSORBER 128

5.6.1 DESIGN PRESSURE 128

5.6.2 WALL THICKNESS 129

5.6.3 SUPPORT 129

5.7 THE CO2 ABSORBER 129

5.7.1 SHELL THICKNESS 129

5.7.2 AXIAL STRESS DUE TO PRESSURE 130

5.7.3 STRESS DUE TO DEAD LOAD 130

5.7.4 EFFECT OF ATTACHMENTS 132

5.7.5 SUPPORT FOR ABSORBER 133

5.8 THE HEAT EXCHANGERS 133

5.8.1 FIRST EXCHANGER 134

CHAPTER SIX 138

6 PLANT HAZARD AND OPERABILITY STUDY 138

6.1 GASIFIER 138

6.2 WASTE HEAT BOILER 139

6.3 HEAT EXCANGERS 140

6.4 INITIAL H2S REMOVAL 141

6.5 SATURATOR/DESATURATOR 145

6.6 CO2 REMOVAL 147

6.7 STORAGE TANK 149

6.8 SAFETY 150

6.8.1 STORAGE TANK 150

6.8.2 THE HEAT EXCHANGERS 150

6.8.3 THE CO CONVERTER 151

6.8.4 SATURATOR – DESATURATOR 152

6.8.5 THE FINAL H2S ABSORBER 153

CHAPTER SEVEN 158

7 COSTING 158

7.1 ECONOMIC AND PROFITABILITY ANALYSIS: GASIFIER 158

7.2 ECONOMIC AND PROFITABILITY ANALYSIS: QUENCHER 161

7.3 COSTING (CO CONVERSION) 163

7.4 COST AND FEASIBILITY ANALYSIS (FINAL H2S REMOVAL) 164

7.5 COSTING AND PROFITABILITY (CO2 REMOVAL) 166

7.6 ESTIMATION OF COST (1ST HEAT EXCHANGER) 168

7.7 ESTIMATION OF COST (2ND HEAT EXCHANGER) 169

CHAPTER EIGHT                                                                                                                 170

8     CONCLUSION AND RECOMMENDATION                                                              170

8.1  CONCLUSION                                                                                                               170

8.2  RECOMMENDATION                                                                                                  171

REFERENCES          172

A.1 SIMULATION          175

A.2 PLANT AND SITE LAYOUT        176

A.3 PROCESS INFORMATION        177

A.4 CALCULATIONS                                                     178

LIST OF TABLES

TABLE	TITLE	PAGE
2.1	Classification of Gasifiers	10
2.2	Comparison of Calculated and Experimental Solubility Data for Scrubbing Solvents	23
3.1	Composition before H2S Removal	51
3.2	Composition after H2S Removal	51
3.3	Composition Out of WGS Reactor	53
3.4	Final H2S Remover Outlet	54
3.5	Percentage Composition of Constituents from Final H2S Removal Outlet	54
3.6	Composition of the Inlet into the CO Conversion Stage	57
3.7	Composition of the Outlet from the CO Conversion Stage	57
3.8	Energy Balance of Outlet Stream from Gasifier	59
3.9	Energy Balance of Inlet Stream to Gasifier	60
4.1	Tank 1 Dimensions	61
4.2	Tank 2 Dimensions	62
4.3	Tankage Dimensions	63
4.4	Rate Constant Values At Specific Temperatures	66
4.5	Equilibrium Constants at Specific Temperatures	66
4.6	Composition of Crude Gas Exiting The Gasifier (Wet Basis)	69
4.7	Composition of Inlet Stream Into Quencher	74
4.8	Composition of Outlet Stream from Quencher	75
4.9	Equilibrium Data for CO at Different Temperatures	80
4.10	Catalyst Properties of Chromium Promoted Iron Catalyst	82
4.11	Reactor Dimensions of Catalytic Converter	84
4.12	Composition of Inlet and Outlet Streams for CO2 Removal	93
4.13	Equilibrium Data for CO2 Removal	94
4.14	Summary of CO2 Absorber Dimensions	95
4.15	Summary of Proposed Heat Exchanger Design	96
4.16	Heat Transfer Coefficients with Corresponding Area And Face Velocity Values	105
4.17	Corresponding Face-Area and Face-Velocity Values	106
4.18	Approximate Frictional Pressure Drop Across Process	111
5.1	Tensile Strength Values At Different Temperatures	119
5.2	Design Stress Values at Different Temperatures	120
6.1	Hazop Analysis for Gasifier	138
6.2	Hazop Analysis or Waste Heat Boiler	139
6.3	Hazop Analysis for Heat Exchangers	140
6.4	Hazop Analysis for Initial H2s Removal	141
6.5A	Hazop Analysis for Saturator	145
6.5B	Hazop Analysis for Desaturator	146
6.5C	Hazop Analysis for Lines 1 And 2	146
6.6A	Hazop Analysis for Co2 Removal Inlet	147
6.6B	Hazop Analysis for Co2 Removal Outlet	148
6.7	Hazop Analysis for Storage Tanks	149
6.8	Flammability Limits of Different Compounds	155
7.1	Inflation Rates By Month And Year (1999-2011), US Inflation Calculator	159
7.2	Refractory Brick Density Calculation	160
7.3	Inflation Rates By Month And Year (2005-2011) (For Bare Vessel)	161
7.4	Inflation Rates By Month And Year (2005-2011) (For Packing Material)	162
7.5	Inflation Rates By Year (2005-2011) ( For H2s Removal Vessel)	165

LIST OF FIGURES

FIGURE	TITLE	PAGE
2.1	Shell Oil Gasifier	12
2.2	The Quenching Unit As Conditioned By Design Specifications	18
2.3	Schematic Diagram of a Packed Column	19
2.4	Structure of Sterically Hindered Amines	23
2.5	Adiabatic Stages in Water Gas Shift Reaction	26
2.6	Schematic Diagram Of Final H2s Removal Stage	32
2.7	Benfield Process For Co2 Removal	38
2.8	Ram Pump Schematics	39
2.9	Ram Pump Schematics	40
2.10	Ram Pump	41
2.11	Air-Cooled Exchangers	44
2.12	Finned Tubes	45
2.13	A Plate Heat Exchanger	46
2.14	Assemblage Of Baffles And Tubes Inside A Shell And Tube Heat Exchanger	48
2.15	Counter-Current Flow In A Heat Exchanger And The Temperature Profile Within The Exchanger	48
2.16	Co-current Flow Through A Heat Exchanger With The Temperature Profile Through The Equipment	49
4.1	The Heavy Fuel Oil Gasifier	64
4.2	The Quencher	72
4.3	The Initial H2S Scrubber	77
4.4	The CO Catalytic Converter	79
4.5	The Final H2S Remover	85
4.6	CO2 Removal	92
4.7	Air-Cooled Exchangers	103
4.8	Finned Tubes	104
5.1	Skirt Support	128
5.2	Double Plate With Gusset	128
5.3	Physical Layout Of A 2-Tube Pass Shell And Tube Exchanger	137
7.1	Cost Estimation Data For Shell And Tube Heat Exchangers	168

LIST OF SYMBOLS

2S left in output stream

B = amount of H₂S removed from aP₃

We are told that the H₂S is scrubbed to 15ppm

aP₃  = 0.0201K_mol/hr

so that

0.5 = 0.02011 + b

B = 0.4799 Kmol/hr

TABLE 3.1: COMPOSITION BEFORE H₂S REMOVAL

components	Kmol	Kmol %
H2	47.6	671.636
Co	42.1	594.031
Co2	8.3	117.113
N2	1.40	19.754
CH4	0.1	1.411
H2S	0.5	7.055
		1411

 

TABLE 3.2: COMPOSITION AFTER H₂S REMOVAL

components	Kmol	Kmol %
H2	671.636	47.839
Co	594.031	42.311
Co2	117.113	8.342
N2	19.754	1.401
CH4	1.411	00.1005011
H2S	0.02011	0.001432
	1403.9651	100.0000

From here, the scrubbed gas moves on to the water gas shift reactor, where CO is used up and H₂, CO₂ are produced.

Now, onto the CO conversion stage. The reaction stoichiometry ofWater gas shift is:

CO + H₂O = CO₂ + H₂

Assuming 99.8% conversion was obtained from the simulation we can obtain the composition of the output from the 2^nd stage CO conversion:

Overall conversion =

0.998 =

Where aP₅ is the molar flow of  CO out of the WGS reactor.

aP₅ = 1.881Kmol

Amount of CO reacted = 594.031 – 1.881

= 592.8429 Kmol

Also, using the reaction stoichiometry,

Amount of CO produced = 592.8429 Kmol

Amount of H₂ produced = 592.8429 Kmol

Amount of H₂O reacted = 592.8429 Kmol**

** is the total amount of water reacted. From the problem statement, we now that this is made up of water from the gasification process from the saturation and from the water fed into the WGS Reactor.

An oxygen atom balance around the Reactorshould tell exactly how much of this water came from the gasified and this will be done later.

TABLE 3.3: COMPOSITION OUT OF THE WGS REACTOR

Components	Kmol	Kmol %
H2	1264.4789	63.3335
CO	1.1881	0.05951
CO2	709.9559	35.5593
N2	19.754	0.9894
CH4	1.411	0.0707
H2S	0.02011	0.001007
	1996.541	100.001

From here, the gas mixture moves to the final H₂S removal stage, where H₂S is scrubbed to less this 1ppm. Assuming scrubbing of H₂S to 0.8ppm,

 

 

 

TABLE 3.4: FINAL H₂S REMOVAL OUTLET

Components	Kmol
H2	1264.4789
CO	1.1889
CO2	709.9559
N2	19.754
CH4	1.411
H2S	1.06583 x 10-3Kmol

We are required to produce 750,000 stdm³/day of feed i.e. 31250 stdm³/hr. Using the ideal gas relation, this is equivalent to 1277.92 Kmol /hr

Using this, we can calculate the % composition of all constituent except C_o2, since there are molar flows are already fixed

Table 3.5: PERCENTAGE COMPOSITION OF CONSTITUENTS FROM FINAL H2S REMOVAL OUTLET

H2	98.00%
Co	1.1889%
N2	1.55%
CH4	0.1104%
H2S	0.0000834%

CO₂ = 100 – (98.00 + 0.093 + 1.55 + 0.1104 + 0.0000834)

= 0.247%

Molar flow of CO₂ = x 1277.92

= 3.1503 Kmol /hr

To obtain feed flow rate:

Moles of C in quenched gas

= 591.636 () + 117.113 () + 1.411 ()

=287.842 Kmol of carbon

= 3454.104 kg

The mass of C calculated above amounts to 98.5% of the total carbon in the feed. Using this information, we can calculate the total weight of the Carbon in feed

0.985 C = 3454.104

C = 3506.705Kg

Now, we can calculate the total weight of feed:

85 wt. % of feed is carbon

0.85 (F) = 3506.705kg

F = 4125.535kg/hr feed

We can also calculate the flow rate of steam and air:

Mass of steam = 0.75 (1125.535)

= 3094.1511kg/hr

Now,

4125.535kg = + +

=  751.1912 kmol/hr

1.16kg of O₂ = 0.03625 mole of O₂

1.16kg of O₂ = 0.03625 mol of O₂/0.1821 mol feed

So that:

Total mole of (impure) feed =

= 149.551 mol of impure O₂

Actual mole of O₂ fed = 0.95 (149.551)

= 142.073 moles of O₂

= 4546.34kg of O₂

3.2 ENERGY BALANCE AROUND CO CONVERSION STAGE

TABLE 3.6 COMPOSITION OF THE INLET INTO THE CO CONVERSION STAGE

COMPONENT	COMPOSITION (Kmol)	Tref (K)	Href (J/gmol)	H380	∆H	MiHi
H	671.636	298	718	10977.2	10259.2	6890448.051
CO	594.031	298	728	11243.5	10515.5	6246532.981
C02	117.113	298	912	16200.5	15288.5	1790482.101
N2	19.254	298	728	11099.5	10371.5	199692.861
CH4	1.411	298	879	16800.5	15921.5	22465.237
H2S	0.02011	298	845	13522.3	12677.3	254.941
H2O	553438	298	837	13072.2	12235.2	653824617.6
TOTAL						668.97 ˣ 106

TABLE 3.7 COMPOSITION OF THE OUTLET FROM THE CO CONVERSION STAGE

COMPONENT	COMPOSITION (Kmol)	Tref (K)	Href (J/gmol)	H380	∆H	MiHi
H	1264.4789	298	718	10988.5	10278.5	12986830.54
CO	1.1888	298	728	11206.5	10478.5	12449.505
C02	702.9559	298	912	16231.5	15319.5	10768.93291
N2	19.754	298	728	11139.5	10411.5	205668.771
CH4	1.411	298	879	16819.5	15940.5	22492.0455
H2S	0.02011	298	845	13729	12884	259.097
H2O	52854	298	837	13171	12334	651901.236
TOTAL						675652768.9

 

To obtain the heat of reaction with respect to H₂ produced,

= (1264.4789 – 671.636) × (- 4.2 × 10⁴ KJ/hr Kgmol)

= 5928429Kgmol × 4.2 × 10⁴ KJ/ Kgmol

= – 24.899 × 10⁶ KJ

Q = ₀H₀ – ₁H₁ + ∆H_rn

= (675.652 × 10⁶ – 668.97 × 10⁶) + (- 24.899 × 10⁶)

= – 18.217 × 10⁶ KJ/hr

A total of 18.217 × 10⁶ KJ/hr of heat is given off in the CO conversion unit. This large amount of heat was utilised in the preheating of the unconverted gas in heat exchanger

2.       ENERGY BALANCE AROUND THE GASIFIER

The Geeral energy balance equation reduces to

ASSUMPTIONS

The enthalpy content of the crude gas leaving the gasifier is majorly due to the gasification reaction, which is:

But,

Ignoring all side reactions, the calorific value for the gasification reaction is equal to 4.29 x 10⁷KJ/Kg

The reference temperature of 50% is chosen for convenience, the general energy balance equation becomes;

TABLE 3.8: ENERGY BALANCE OF OUTLET STREAM FROM GASIFIER

Composition	Kgmol	Tref	Href, J/gmol	H1573 J/gmol		moHo
H2	671.636	323	1485.25	39064.25	37579	25.24 x 106
CO	594.031	323	1514.25	41796.75	40282.5	23.93 x 106
CO2	117.113	323	1965.25	66368.25	64403	7.54 x 106
N2	19.754	323	1513.25	41343.75	39830.5	786.81 x 103
CH4	1.411	323	1897.5	81440.75	79543.25	112.24x 103
H2S	7.055	323	1749.75	54078	52303.25	369.0 x 103
H2O	94.031	323	1749.75	51823.75	50074	29.74 x 106
C		323	–	–	–	87.72 x 106

 

 

TABLE 3.4 ENERGY BALANCE OF INLET STREAM TO GASIFIER

Htrh	Tref	Href, J/gmol	H1573 J/gmol	J/gmol	miHi	moHo
O2	323	1530.5	68.11	5280.5	1.95 x 106	369.07
N2	323	1513.25	6644	5130.75	99.66 x 103	19.425
H2O	323	1749.75	7752	6002.25	1.03 x 106	171.90
					3.080 x 106

is considered here to be the calorific value of the fuel oil

= 87.72

CHAPTER FOUR

4           CHEMICAL ENGINEERING DESIGN

1.       TANK ONE (FEED STORAGE TANK FARMS)

The feed storage tank farm will be designed in such a way that each tank will be able to hold about 5 days of feed in storage. There will be a total of ten feed storage tanks in the tank farm

• The tank is a vertical, cylindrical shaped storage tank with a fixed and floating roof

• Floating roof rises and falls with the liquid level inside the tank, thereby decreasing the vapor space above the liquid level. They are considered a safety requirement as well as a pollution prevention measure.
• A vertical tank is preferred over a horizontal tank since it takes up less space, is easily supported on concrete slabs and allows maximum mixing.
• The storage tanks are supported with a skirt support as they do not impose concentrated loads on the vessel shell
• A drain line is installed so that water at the bottom of the tank (caused by the difference between the internal and external temperature) can be regularly removed.

TABLE 4.1: TANK 1 DIMENSIONS

Top of shell height	100%	23.21m
Design fill level	95%	22.05m
Normal fill level	85%	19.73m
Minimum fill level	35%	8.12m
Bottom of shell height	0%	0

2.       TANK 2 (FRESH FEED SURGE DRUM)

This tank is similar to tank in function, configuration, material of construction and safety devices and structural supports used. The only difference is that it is a smaller tank used to aid the Start-up and shut down procedures. This tank is required to store enough heavy fuel oil for a 3 hour production run.

TABLE 4.2: TANK 2 DIMENSIONS

Top of shell height	100%	6.75m
Design fill level	95%	6.41m
Normal fill level	85%	5.74m
Minimum fill level	35%	2.36m
Bottom of shell height	0%	0m

3.       TANK 3 (STORAGE TANK FARM)

This tank is used for the storage of the final product gas, hydrogen before it is supplied to the immediate consumers. They are designed as spherical tanks since this shape is more ideal for storage of the gaseous product.

TANK DIMENSIONS

For each of the 50 tanks, assume that H = 3D

Similar to tanks 1 and 2, there are safety allowances in the height of the tank.

So that D = 44m

Installed storage capacity is now

TABLE 4.3: TANK STORAGE CAPACITIES

TANK	STORAGE CAPACITY
TANK 1 (FEED STORAGE TANK)	531.25m3
TANK 2 (FRESH FEED SURGE DRUM)	13.281m3
TANK 3 (PRODUCT STORAGE TANK)	75000m3

4.       THE HEAVY FUEL OIL GASIFIER

FIGURE 4.1 THE HEAVY FUEL OIL GASIFIER

The following reactions take place in the gasifier:

    (4.4.1)

  (4.4.2)

        (4.4.3)

        (4.4.4)

       (4.4.5)

The slowest reaction is reaction 4.4.4 (Carbon-steam reaction) and is therefore taken as the rate determining step for the reactor.

C + H₂O → H₂ + CO (rate-determining step)

The rate expression for this reaction is given as:

r = k_v

Where:

r = rate of reaction (in this case, the rate of disappearance of H₂O)

k_{v =} rate constant

K = equilibrium constant

C_H2O, C_H2 and C_CO = molar concentrations of C_H2O, C_H2 and C_CO respectively (kmol/m³)

Given the table below for rate constant, k_v (from reference source)

TABLE 4.4 RATE CONSTANT VALUES AT SPECIFIC TEMPERATURES

T (K)	952.38	1000
kv (cm3/gmol.s)	1.62	8.6

TABLE 4.5 COMPOSITION OF CRUDE GAS EXITING THE GASIFIER (WET BASIS)

COMPONENTS	AMOUNT (kmol/hr)	COMPOSITION
H2	671.636	0.4232
CO	594.031	0.3743
CO2	117.113	0.0738
CH4	1.411	0.0008
H2O	171.01	0.1077
H2S	7.055	0.0044
N2	19.754	0.0124
C	5.16	0.0033
Total	1587.17	1.0000

Height of the gasifier = 9.64 m

Diameter of the gasifier = 1.99 m

Total molar flow rate of the exit crude gas = 1587.17 kmol/hr

Volumetric flow rate of the exit crude gas =        =3347.88 m³/hr

Residence time, τ = Volume/Volumetric flow rate = = 0.0089 hr = 32.13s

Space velocity, υ = 1/ τ = 1/32.13s = 0.031 s^-1

 

5.       THE WASTE HEAT BOILER

The waste heat boiler will be designed as a special type of shell and tube heat exchanger called a water tube waste heat boiler, initially mentioned in the previous semester’s work. The water tube design is a very difficult one to obtain but has been selected because of its ability to withstand very large pressure generation like the one specified in this design. Additional energy recovery and fuel conservation can be realized with the use of the water tube boiler, by adding a boiler feed water economizer to the boiler design.

Before we start designing the exchanger, we need to determine the heat duty of the demineralised feed water i.e how much heat it will be required to absorb from the hot flue gas stream.

The calculations leading to the determination of in the above equation are shown below:

C_p-values are dependent on temperature, so we have to determine the C_p of all the individual components that make up the inlet stream (i.e. at 1300°C)

The heat capacity equations are:

C:        C_p = 11.18 + 1.095 × 10^-2 – 4.891 × 10^{5 -2}

H₂:       C_p =

CO:      C_p =

CO₂:     C_p =

N₂:       C_p =

CH₄:     C_p =

H₂S:     C_p =

Using these C_p values, we can calculate a bulk C_p for the stream as shown:

TABLE 4.6: BULK C_P VALUES FOR STREAM COMPONENTS

COMPONENT	Cp (J/KMOLOC)	MOLE FRACTION	MOLAR FLOW	IN-STREAM Cp (J/KMOLOC)
C	28.207	0.00362	5.157	0.2089
H2	32.585	0.47191	671.636	13.2328
CO2	58.747	0.08229	117.113	4.1593
CO	35.412	0.41739	594.031	12.7199
N2	35.222	0.01388	19.754	0.4191
CH4	135.761	0.00099	1.411	0.1222
H2S	51.4787	0.00496	6.702	0.0360

From the addition of all the values in the in-stream C_p column, we obtain:

C_p  = 35.92522 J/mol. °C

Substitute the value of C_p  above into Equation (2.1):

From the material balance calculations, we can see that

Inside heat transfer coefficient:

h_{i =}  1309379.395 W/m²k

Outside heat transfer coefficient

h_o = 2000Btu/ h_o ft^2oF

= 11333.33W/ m^2oC

Overall heat transfer coefficient:

Heat transfer area:

Mean temperature difference:

MTD =

Number of tubes:

Tube lunght, L:

Tube pitch:  P_i  =
shell diameter: =  1.6646m

Pressure drop (shell side):

Pressure drop (tube side):

Shell wall thickness determination:

P_i= 3000 kPa

D_i = 1.078 m

At the shell side mean temperature of 189.21^oC,

f = 134 × 10⁶N/m²

Therefore,

e = 3000000( 1.078)/ [2(134 × 10⁶) – 3000000]

= 12.2 mm

Other design specifications include:

• Design Pressure = 110% of operating Pressure = 3300 kPa
• Design temperature = Maximum shell temperature = 220^oC
• Corrosion Allowance = 4 mm
  6.       THE QUENCHER

FIGURE 4.2 THE QUENCHER

The quencher is designed as a packed column, so as to increase the contact area between the gas and the water and thus allow for efficient removal of residual carbon from the gas stream.

4.6.1        Design Procedures

The design of a packed column will involve the following steps:

1. Select the type and size of packing.

2. Determine the column height required for the specified separation.

3. Determine the column diameter (capacity), to handle the liquid and vapour flow rates.

4. Select and design the column internal features: packing support, liquid distributor, redistributors.

4.6.2        Types of Packing

The principal requirements of a packing are that it should:

• Provide a large surface area: a high interfacial area between the gas and liquid.
• Have an open structure: low resistance to gas flow.
• Promote uniform liquid distribution on the packing surface.
• Promote uniform vapour gas flow across the column cross-section.

Random packing and structured packing elements are more commonly used in the process industries.

 

DATA GIVEN

Exit Gas stream composition

Moles of exit and inlet gas stream

From material balance, input and exit gas stream composition is calculated.

DATA REQUIRED

Heat Capacity

Amount of water required to achieve the desired temperature change

Tower Diameter

Tower Height

TO OBTAIN MASS OF COOLING WATER REQUIRED:

Heat lost by crude gas= heat gained by cooling water

Entering Crude gas temperature=250^oC

Exiting Crude gas temperature=50^oC

Entering Temperature of cooling water= 25^oC

Exiting temperature of the cooling gas= this is assumed to be 90^oC

Inlet Stream Into Quencher

TABLE 4.7: COMPOSITION OF INLET STREAM INTO QUENCHER

COMPONENT	MOLE (Kmol)	MOLE FRACTION
C	5.16	0.00364
H₂	671.636	0.4743
CO	594.031	0.4195
CO2	117.113	0.0827
N₂	19.754	0.01395
CH₄	1.411	0.000996
H₂S	7.055	0.00498
	1416.16

 

4.6.3        HEAT CAPACITY

The heat capacity data can be obtained from heat capacity tables in available from literature

Taking a reference temperature of 15^oC.

The heat capacity equations are:

C:  Cp = 11.18 + 1.095 × 10^-2T – 4.891 × 10⁵T^-2

H₂: Cp =

CO: Cp =

CO₂: Cp =

N₂: Cp =

CH₄: Cp =

H₂S: Cp =

TABLE 4.8: COMPOSITION OF OUTLET STREAM FROM QUENCHER

COMPONENT	MOLE (Kmol)	MOLE FRACTION
H₂	671.636	0.476
CO	594.031	0.421
CO₂	117.113	0.083
N₂	19.754	0.014
CH₄	1.411	0.001
H₂S	7.055	0.005
	1411

 

Heat lost by the crude gas, ΔH is given by

_net dT

= 1039.525 KJ/Kmol

ΔH_net = (6984.618 × 1416.16) – (1039.525 × 1411) KJ/hr

= 2,340.157 KJ/s

Heat gained by the water = m

= 96,189.91 KJ/Kmol

Thus, Heat gained by water = (m/M x 96189.91KJ/Kmol)

Molecular weight of water = 18 Kg/Kmol

Heat gained by water = 96189.91/18 m = 5343.88 m

Since, heat lost by the crude gas = heat gained by water

Therefore,

5343.88m = 2340.157

m = 0.438 Kg/s

Allowing a minimum of 10 minutes hold-up of water,

Therefore, m = 0.438 Kg/s x 10mins x 60s = 262.748 Kg

4.6.4        COLUMN DIAMETER

D = (4S/π)^1/2

= (4 × 0.2185/3.1429) ^½

= 0.5275 m

 

4.6.5        HEIGHT OF THE COLUMN: Z

This is obtained, from Coulson and Richardson, using the following formula:

Z= H_OGN_OGS.F

= 3.0893 m

7.       INITIAL  H₂S REMOVAL

FIGURE 4.3 THE INITIAL H₂S SCRUBBER

4.7.1        ASSUMPTIONS

• No  NMP  in  Output  gas  stream
• Unit  operates  at  50%  of  the  flooding  gas  mass   velocity
• Flow rate  of  NMP  is  40%  more  than  the  minimum

Given that

Henry’s law constant for NMP = 0.97

Density of gas stream = 31.49kg/m³

Density of NMP = 1020kg/m³

H₂S concentration in the inlet stream = 0.5mol%

NMP has already been selected from the first semester preliminary design as our scrubbing liquid.

An energy balance is not require for this unit because there is very little temperature change between the unit’s inlet and exit streams and hence the unit can be assumed to be operating isothermally.

A packed column using 1 in. Raschig ring packing with a packing factor of 160 have been selected for use for this design. Raschig rings have been selected because they give a larger degree of contact between the gas phase and the liquid phase. It also gives a higher mass transfer area to work with.

Column height = 14m (approx.)

Column diameter = 2.44m (approx.)

 

8.       CARBON MONOXIDE CONVERSION UNIT

FIGURE 4.4 THE CO CATALYTIC CONVERTER

This unit converts carbon monoxide to carbon dioxide, a reaction which leads to the production of more hydrogen. It is required to carry out this reaction over chromium-promoted iron oxide catalyst. The catalyst vessel is a single shell with a dividing plate separating the two catalyst beds which constitute the two stages of conversion. The following reaction takes place in the CO catalytic converter:

CO + H₂O CO₂ + H₂    ΔH= – 41170.51J

Also provided are the basic data for the CO conversion section of the plant:

1. Space velocity

The space velocity through each catalyst stage be assumed to be 3500 volumes of gas plus steam measured at NTP per volume of catalyst per hour. It should further be assumed that use of this space velocity will allow a 10°C approach to equilibrium to be attained throughout the possible range of catalyst operating temperatures listed below.

2. Equilibrium data for the CO conversion reaction

TABLE 4.9 EQUILIBRIUM DATA FOR CO AT DIFFERENT TEMPERATURES

TEMP/K	600		700		800
KP	3.69 x 10-2		1.11 x 10-4		2.48 x 10-4

In the design of a catalytic converter, the following are the key issues to be considered: reactor type; catalyst type, catalyst volume and mass; the catalytic converter sizing; operating conditions (temperature and pressure) within the reactor; phase type; feed conditions (temperature and concentration).

4.8.1        Reactor Type: Fixed Bed Reactor

The fixed bed reactor is a tubular which is packed with solid granular catalyst on a fixed support. It is termed a fixed bed reactor because the reactor feed, whether liquid or gaseous feed passes through the catalyst bed to contact with the catalyst for chemical reactions, with the catalyst bed not being moved as the fluid flows feed through it. Based on the flow directions of the feedstock some of the hydro – processing reactors can be divided into “axial” reactors and “radial” reactors. In the axial reactor the feed flows along the axial direction for the reactions to take place. Generally the feed flows downwards. In the radial reactor the feed flows in the radial direction for the reactions to take place. The axial reactor is used when the pressure drop across a catalyst bed is small. Or otherwise the radial reactor should be used. Based on the temperature of the cylindrical shell some reactors can be divided into cold wall reactors and hot wall reactors. They are all fixed bed reactors. “Hot wall” reactor is not lined with any heat insulating linings. Because the temperature of the reactor shell is very high, the reactors must be made of alloy steel which can sustain high temperature and is highly resistant to hydrogen corrosion. The advantage is that more volume of such a reactor can be effectively used. A “cold wall” reactor is lined with heat-insulating material. The temperature of the reactor shell is quite low, ordinary low alloy steel can be used for making such reactors. Corrosion caused by feedstock, especially by hydrogen can be avoided.

4.8.2        Catalyst: Chromium – Promoted Iron Oxide

The overall reaction of a heterogeneous gas-solid reaction on a supported catalyst, such as the one involved in this design, is made up of a series of physical steps as well as the chemical reaction. First, there is mass transfer of reactant from the bulk gas phase to the external solid surface, after which there is diffusion from the solid surface to the internal active sites, activation of the adsorbed reactants; next is the most important step of chemical reaction, then desorption of products, internal diffusion of products to the external solid surface, and mass transfer to the bulk gas phase.

TABLE 4.10: CATALYST PROPERTIES OF CHROMIUM PROMOTED IRON CATALYST (Source: Chemical Reactor Design for Process Plants by Howard F. Rase, 1999)

PROPERTY	CHROMIUM – PROMOTED IRON OXIDE
Maximum operating Temperature (K)	890 0F
Tablet size
Bulk Density	70 lb / cu ft
Particle Density	126 lb / cu ft
Cost	20 $ / cu ft
Catalyst poisons	Inorganic salts, boron, oils or phosphorus compounds, liquid H20 is a temporary poison. Sulfur compounds in an amount greater than 50 ppm
Catalyst life	3 years and over depending on care in startup and during operation

The purpose of the chromium promotion is to minimize catalyst sintering by textural promotion.  During the course of the reaction, there are no significant side reactions.

The works of Xue et al confirm that Carbon is formed if the water-gas shift reaction occurs at the reaction stoichiometry and temperature. In order to achieve higher conversions with the use of a catalyst and also avoid coking, the H₂O / CO feed ratio must be higher than that required by the reaction stoichiometry. With this low H₂O / CO ratio, all possible side reactions would likely produce C or CH₄. Formulation of C blocks the catalyst sites causing catalyst deactivation and an increase in the pressure drop across the bed caused by plugging or fouling of the reactor. Formation of CH₄ would consume the hydrogen and alter the product composition which may give rise to potential difficulties in subsequent processes.

Activated Fe₃O₄ is pyrophoric, that is, it ignites spontaneously upon exposure to air, and as a result the catalyst must be re-reduced and stabilized by surface oxidation (using an inert gas with a low concentration of O₂) before being re-used.

4.8.2.1       Catalyst Volume (from space velocity):

The catalyst volume can be obtained from space velocity data given in the problem statement coupled with the simulation figures for the flowrate of gases into the catalytic converter.

FOR 1^ST STAGE:

4.8.2.2       Catalyst Mass (in 1^st stage of converter):

FOR 2^ND STAGE:

4.8.2.3       Catalyst Mass (in 2^nd stage of converter):

4.8.3        CATALYTIC CONVERTER SIZING

In carrying out the sizing of Fixed Bed Reactors, most designs approximate to plug flow. The simplest of such arrangements is adiabatic (Chemical Process Design and Integration by Robin Smith).

A summary of the reactor dimensions is provided in the table that follows.

TABLE 4.11: REACTOR DIMENSIONS OF CATALYTIC CONVERTER

CATALYST BED	LENGTH(M)	DIAMETER(M)	VOLUME(M3)
1st Stage	7.03	3.518	68.34
2nd Stage	12.22	3.518	118.83
REACTOR
1st Stage	8.795	3.518	85.50
2nd Stage	14.61	3.518	142.06
Total	23.405	3.518	227.56

 

 

 

 

9.       FINAL H₂S REMOVAL STAGE

FIGURE 4.5 FINAL H₂S SCRUBBER

The process:

The final H₂S removal stage could be divided into two operations namely:

1. Normal operation
2. Intermittent regeneration operation

4.9.1        NORMAL OPERATION

The converted gas leaving an air-cooled heat exchanger is fed to the unit together with oxygen. Iron oxide in the Lautamasse reacts with hydrogen sulphide in the converted gas to form iron sulphide. The iron sulphide is an intermediate product and it reacts with the oxygen as quickly as it is formed to produce iron oxide and sulphur deposits. This process occurs continuously and is represented by the equation.

2Fe₂O_3(s)  +  6H₂S    2Fe₂S₃   +     6H₂O

2Fe₂S₃      +  3O₂     2Fe2O3 +   6S

4.9.2        REGENERATION OPERATION

When the quantity of sulphur deposit overtime attain a value between 40 55% of the dry weight of the Lautmasse, the normal process will be stopped. Hot oxygen only is then continuously fed to the unit in a parallel in order to oxidize the sulphur deposit to sulphur(iv) oxide gas which can be converted to tetraoxosulphate(vi)acid in order to generate money.

S_(s) +  O_2(g)    SO_2(g)

The Lautamasse becomes ineffective when the regeneration process has been carried out for about 10times.

 

4.9.3        THE EQUIPMENT

The final H₂S removal is to take place in

1. Four vertical vessels
2. A separate outlet from each of the vessel converging SO₂ from regeneration
3. A separate inlet header to each of the column, conveying heated oxygen.
4. A minimum point where the oxygen and the converted gas is mixed.
5. Each column contains five trays each containing Lautamasse
6. The columns are to be identical.

The following was specified

• Height of column      = 18.5 m
• Diameter of column = 2.5 m
• Number of trays       = 5
• Total pressure drop = 35 KN/m²
• Number of columns = 4
• Locking lid type  = Autoclave

Unspecified Parameter are:

1. Plate type and its configuration
2. Mass of iron oxide per tray
3. Regeneration period
4. Regeneration duration
5.   Tray thickness
6. Tray spacing

Choice of tray – Bubble cap

Reasons:

1. It provides maximum contact between the gas and solid unlike the sieve and valve trays.
2. It allows for maximum occupation of the tray space volume by the Lautamasse powder.

1.   The riser hole size  –  5 mm (Coulson and Richardson )
2. Choice of material : Low alloy steel (Ni, Cr, Mo, V)

Properties

Tensile strength   –  550 N/mm²

Design stress       –   240 N/mm²

3.   Determination of number of holes

N = 25001 holes

(e) Determination of the mass of lautamasse and sulphur.

= 10402.6 Kg

When the quantity of sulphur deposit on the tray falls between 40-55% of the weight of the lautamsee the operation has to be stopped and then regeneration takes place.

Assuming the mass of sulphur and the lautamase is the ratio 1:0.4 that is assuming that sulphur is 50% of the total mass mixture.

Mass of sulphur

=3467.5kg

Mass of fe₂0_{3 =}

=6935kg

To calculate the time for regeneration:

4.9.4        ASSUMPTION

The absorption of H₂S in the four columns are 50%, 30%, 20%, 10%. The assumption is based on the fact that absorption decreases along the column. But we want to model the column base on the absorption of the first column.

Time after which regeneration will take place is

=15.8months

16months

After 16 months regeneration process has to be carried out.

Volume and height of sulphur and lautamasse determination.

Mass of sulphur = 3467.6kg

Density of sulphur = 2000kg/m³

Volume of sulphur =

=

Height =

=

= 0.3532m

Lautamasse;

Density of lautamasse = 675kg/m³

Mass of lautamasse = 6935kg

Volume =

= 10.274m³

Height =

= 2.0930m

This is a good value since the higher the height of the lautamasse, the lesser the effective tray spacing.

Total height of lautamasse and sulphur

= 2.0930 + 0.3532

= 2.45m

4.9.5        REGENERATION

Sulphur deposit after 16months per column = 3467.5kg 5

= 17337.5kg

Amount in moles=

= 541.797 moles

S + O  SO₂

541.8 kmol of O₂is required per column.

Since each column is designed assuming 50% of the total H₂S absorbed

(4 541.797kmol)SO₂ = 150% of the actual SO₂ produced

Also, (4 541.797kmol)SO₂ = 150% of the oxygen required

The actual SO₂ produced = 1444.8kmol

The actual O₂ required = 1444.8kmol

If regeneration is assumed to take place within 6hrs,

Flow rate of SO₂ regenerated = 240.8kmol/hr

10.   CARBON DIOXIDE REMOVAL

FIGURE 4.6 CO₂ REMOVAL

Mechanism of Removal: Absorption

Absorption Mechanism: Chemosorption(Chemical Absorption)

Equation of Reaction

K₂CO_3(aq) + H₂O_(l) + CO_2(g)  2KHCO_3(aq)

Basis : 4125.535 kg of Feed / hr

TABLE 4.12: COMPOSITION OF INLET AND OUTLET STREAMS FOR CO₂ REMOVAL

COMPONENT	yin	yout
H2	0.63326	0.9800
CO	0.00060	0.00093
C02	0.35555	0.00247
N2	0.00989	0.0155
CH4	0.00071	0.001104
H2S	5.33E-07	8.34E-07
TOTAL	1.000000	1.000000

ASSUMPTION: Only Carbon dioxide (CO₂)is absorbed and all the other gases act as inert in the 40% K₂CO₃ solution.

Mole Ratio Of Co₂ In The Gas Entering The Column

Mole Ratio Of Co₂  In The Gas Leaving The Column

Plotting the equilibrium curve for CO₂

                     TABLE 4.13: EQUILIBRIUM DATA FOR CO₂ REMOVAL

	X	Y	X	Y
0.025	0.1	0.0017	0.111111	0.001703
0.18	0.2	0.0122	0.25	0.012351
0.6	0.35	0.0408	0.538462	0.042535
1.7	0.5	0.1156	1	0.13071
4.0	0.68	0.2722	2.125	0.374004
7	0.75	0.4762	3	0.909126
10	0.8	0.6803	4	2.127932
11	0.9	0.748	9	2.968254

4.10.1    PACKING MATERIAL: Pall Rings (Plastic)

Pall rings are chosen as the F_p (Packing Factor), small column diameter, high quality with low price, low pressure drop.

The packing characteristics

D_p = 90mm

% = 92%

Summary of design

TABLE 4.14: SUMMARY OF CO₂ ABSORBER DIMENSIONS

COLUMN DIAMETER	1.3m
COLUMN HEIGHT	24.96m
PACKING TYPE	Plastic pall rings
DIAMETER OF PACKING	90mm
VOIDAGE	92%
SPECIFIC AREA	85m2/m3
PACKING FACTOR	17.07 ft-1
QTY OF SOLVENT REQUIRED	852.537 kmol/hr

11.   HEAT EXCHANGER DESIGN

4.11.1    THE PROPOSED DESIGN

TABLE 4.15: SUMMARY OF PROPOSED HEAT EXCHANGER DESIGN

EXCHANGER	FIRST	SECOND
Tube Layout	Split ring, floating head, 1 shell pass, and 1 tube pass of4.83 m length, 30 mm O.D., 26 mm I.D., and triangular pitch. Pitch-to-diameter ratio of 1.25	Split ring, floating head, 1 shell pass, and 4 tube passes of6.00 m length, 20 mm O.D., 16 mm I.D., and triangular pitch. Pitch-to-diameter ratio of 1.25
Total number of Tubes/Number of Tubes per pass	96/96	1308/327
Heat transfer area (based on outside diameter)m2	43	493
Shell Internal diameter (mm)	484.5	1064.4
Baffle spacing (mm)	97	212.88
Baffle cut	25%
Tube-side coefficient (W/m2 °C), clean.	252.97	102.42
Overall coefficient, estimated (W/m2 °C).	120.00	60.00
Overall coefficient required (W/m2°C).	149.85	67.81
Dirt/Fouling factors Tube-side/Shell-side(m2 °C /W)	0.0002 /0.0003	0.0002 /0.0003
Pressure drops, including drop over nozzles (estimated) Tube-side/Shell-side (kPa)	253/452	73/969

 

4.11.2    OPTIMISATION

There is scope for optimizing the design by reducing the number of tubes, as the pressure drops are well within specification and the overall coefficient is well above that needed. However, the method used for estimating the coefficient and pressure drop on the shell side(Kern’s method) is not so accurate, so keeping to this design will give some margin of safety.

4.11.3    VISCOSITY CORRECTION FACTOR

The viscosity correction factor was neglected when calculating the heat transfer coefficients and pressure drops. This is reasonable for gases as it has a relatively low viscosity.

Usually,

For such a small factor, the decision to neglect it was justified. Applying the correction would decrease the estimated heat transfer coefficient, which should not be a problem as the area required for heat exchange is well taken care of. It would also give a slight increase in the estimate.

4.11.4    OXYGEN STREAM PRE-HEATER.

4.11.4.1   FLUID ALLOCATION

Placing the higher temperature stream on the tube side will reduce the overall cost of design. Therefore, the saturated stream is placed in the tubes.

4.11.4.2   DUTY OF PRE HEATER

Let us assume that the amount of saturated stream used for pre-heating of the O₂ stream gives off ‘ONLY’ its Latent Heat to the O₂ – stream

Q= 877.04 ×10³ KJ/hr

= 243.622 ×10³ W

 

MASS OF SATURATED STREAM REQUIRED.

m_S =  0.1414 Kg/sec

 

 

MEAN TEMPERATURE DIFFERENCE (ΔTm)

ΔT_m= 0.986(114.823 0c)

= 133.215 ⁰c

4.11.4.3   ESTIMATION OF U (Overall Coefficient)

An estimation of 200W/m²C for the overall Heat Transfer Coefficient is chosen as suggested by literature.

4.11.4.4   PROVISIONAL AREA.

A  = 10.76m²

4.11.4.5   TUBE DIMENSION SELECTION

• Outer Diameter – 20mm
• Inner Diameter – 16mm
• Length of Tube – 6.10m
• Tube Effective Length- 6.0m

Small diameter tubes are chosen because they give more compact & cheaper exchangers. Use of longer tubes will reduce shell diameter thus lowering cost exchanger.

Heat Transfer Area of 1 tube = πd_oL

= π(20/1000)(6)

= 0.37699m²

4.11.4.6   NUMBER OF TUBES REQUIRED

Number of tubes = 28.542

=28 tubes

4.11.4.7   TUBE ARRANGEMENT SELECTION.

A Triangular Tube Arrangement pattern is chosen as it gives higher heat transfer rates as compared to others. The recommended tube pitch is 1.25 times the Tube Outside Diameter(do). Tube Bundle Diameter is therefore

= 170mm

4.11.4.8   SHELL DIAMETER DETERMINATION

Using a Split Ring Floating Head unit, the Diameter clearance between the Shell and Tube Bundle is 50mm

Therefore,

Shell Diameter = 170 +50 = 220mm

4.11.4.9   TUBE SIDE COEFFICIENT

A Horizontal Condenser with condensation in the tubes is the usual arrangement for pre-heaters &vaporizers using condensing steam as the heating medium. It is customary to design purposes. For air- free steam, a coefficient of 8000W/m² is used. Therefore,

h_I= 8000W/m²

4.11.4.10                   MASS VELOCITY IN SHELL (G_S)

G_S = 134.917Kg/m²s

4.11.4.11                   LINEAR VELOCITY (Us)

U_S = 3.20m/sec

 

4.11.4.12                   SHELL SIDE EQUIVALENT DIAMETER

For a Triangular pitch Arrangement

d_e = 1.10/d_o[pt²– 0.917d_o²]

= 14.2mm

 

4.11.4.13                   AVERAGE PHYSICAL PROPERTIES OF SHELL FLUID

• Viscosity (µ) = 2.61 ×10^-5 Pas
• Thermal Conductivity (K) = 3.448 × 10^-2 W/mK
• Heat Capacity (Cp) = 0.9818 ×10³ J/Kg ⁰C

4.11.4.14                   REYNOLD’s NUMBER ON SHELL SIDE

Re = 73402.99987

4.11.4.15                   PRANDTL’s NUMBER ON SHELL SIDE

Pr = 0.74318

4.11.4.16                   HEAT TRANSFER COEFFICIENT (SHELL SIDE)

h_s = 317.324W/m^{2 0}c

4.11.4.17                   OVERALL COEFFICIENT

U= 322.691 W/m^{2 0}c

4.11.4.18                   PRESSURE DROP CALCULATIONS

On the Shell Side

Re= 73402.99987

J_F= 3.6 × 10^-2

ΔP_s = 4(3.6×10^-2)(220/14.2)(6000/220)(42.07×3.20²)

= 26.2112 KN/m² which is very acceptable

On the Tube Side:

The pressure drop on the condensing side or tube side is difficult to predict as 2 phases are present & the vapour mass velocity is changing throughout the tube side.

4.11.5    AIR-COOLED EXCHANGERS

Air-cooled exchangers should be considered when cooling water is in short supply or expensive. They can also be competitive with water-cooled units even when water is plentiful. Frank (1978) suggests that in moderate climates air-cooling will usually be the best choice for minimum process temperatures above 65°C, and water-cooling for minimum process temperatures below 50°C. Between these temperatures a detailed economic analysis would be necessary to decide the best coolant. Air-cooled exchangers are used for cooling and condensing. Air-cooled exchangers consist of banks of finned tubes over which air is blown or drawn by fans mounted below or above the tubes (forced or induced draft). Typical units are shown in Figure 2 below. Air-cooled exchangers are packaged units, and would normally be selected and specified in consultation with the manufacturers. The equation for finned tubes given below can be used to make an approximate estimate of the area required for a given duty (Sinnott, 2005).

FIGURE 4.7 AIR-COOLED EXCHANGERS

4.11.5.1   FINNED TUBES

Fins are used to increase the effective surface area of heat-exchanger tubing. Many different types of fin have been developed, but the plain transverse fin shown in Figure 2 is the most commonly used type for process heat exchangers.

FIGURE 4.8: FINNED TUBE

Finned tubes are used when the heat-transfer coefficient on the outside of the tube is appreciably lower than that on the inside; as in heat transfer from a liquid to a gas, such as in air-cooled heat exchangers.

The fin surface area will not be as effective as the bare tube surface, as the heat has to be conducted along the fin. This is allowed for in design by the use of a fin effectiveness, or fin efficiency, factor.

(Sinnott, 2005)

     (4.11.1)

LMTD = 47.28°C

     (4.11.2)

Prandtl Number

     (4.11.3)

Thus, the Prandtl number for air at 216.4 °C = 0.69

Heat Duty Required

     (4.11.4)

Hence, the heat duty Q of the exchanger = 1.013 KJ/kg °C

Overall Heat Transfer Coefficient for the Design

     (4.11.5)

Trials with various values of U

TABLE 4.16: HEAT TRANSFER COEFFICIENTS WITH CORRESPONDING AREA AND FACE VELOCITY VALUES

S/N	U (KJ/hr m2°C)	A (m2)	vair (m/s)
1	6.127 × 104	965.56	3.00
2	8.169 × 104	724.167	4.00
3	1.225 × 105	482	6.00

In addition,

     (4.11.7)

A_face = face area (m²)

TABLE 4.17: CORRESPONDING FACE AREA AND FACE VELOCITY VALUES

vface (m/s)	Aface (m2)
3	1135.96
4	851.97
6	567.91

Substituting W_air in (4.11.7) into (4.11.4),

     (4.11.8)

ρ_air = density of air, 1.1762 kg/m³

So far, for heat exchanger parameters,

U = 1.2254 × 10⁵ KJ/hr m² °C

A = 482.78 m²

v_face = 6 m/s

A_face = 567.91 m²

4.11.5.2   FAN COVERAGE

= 293.96 m²

For the design also, we choose to use two fans rather than one, as this takes care of cooling over a bit of a longer distance of fan stages, and provides insurance in case of failure of one fan or the other.

4.11.5.3   FAN DIAMETER

With the two identical fans to be used the diameter is simply computed with the diameter, thus;

Diameter of fan = 13.45 m

General Properties of the Fans to be used as per industrial standards

Fan speeds v_fan = 70 m/s

V-belt drive of about 30 horsepower

Forced draft fan system will be used as it uses less power in blowing cool air volume

4.11.5.4   FIN AREA AND TUBE AREA ANALYSES

        (4.11.10)

     (4.11.11)

      (4.11.12)

D_fin = Diameter of fin = 12.7 mm

D_e = External diameter of the tube = 41.2 mm

D_i = Internal diameter of the tube = 37.2 mm

W = Width of tube = 4.00 m

L_fin = Length of fin = 68.2 mm

L = Length of tube = 15 m

A_HT = Total heat transfer area per length of tube = 0.01644 m²/m

Number of tubes per row of each metre in a tube bank = n^× =

With the chosen pitch, P = 47.625 mm; n^× = 20.997 ≈ 21 tubes per row

Number of rows n, chosen = 4; and given that W = 4.00 m,

Total number of tubes, N = n^× × W × n = 336 tubes.

12.   PUMP SIZING

Because liquids are incompressible, Equation 1 may be used to calculate the work required to pump a liquid. The kinetic energy term is small compared to the other terms and is neglected. Therefore, Equation 4.11.1 reduces to

g∆z + ∆p/ρ+ W+ E = 0,  defining points 1 and 2,we have,

g (z₂ – z₁,) + (p₂ – p₁,)/ρ + W + E = 0       (4.12.1)

The friction pressure loss term is split into two parts, one for the suction side of the pump, Es, and, the other for the discharge side of the pump, E_D. Thus, Equation 1 becomes, after solving for W.

W = g (z₁ – z₂) + (p₁ – p₂)/ρ – (Es + E_D)       (4.12.2)

Frequently, we must make a preliminary estimate of the pump work. Manufacturers do not stock all pumps and other expensive machinery because of the cost of carrying an inventory. The machinery is manufactured on receipt of an order from a customer. Manufacturing some process machinery, may take six months or longer. To save time in implementing a project will require ordering equipment having long delivery times before completing a detailed design. Also, the management of a firm will require an estimate of the cost of a project to prepare a budget or a proposal for a customer. Once the work is estimated, the pump shaft power,

P_p = mW/ ɳ_P          (4.12.3)                                                                

where,  ɳ_p – the pump efficiency, includes both the hydraulic and mechanical frictional losses. Below outlines a calculation procedure for calculating an approximate pump size.

4.12.1    Approximate Pump-Sizing Calculation Procedure

1. Define the flow system, i.e., locate points 1 and 2. The pressures p₁ and p₂ will be known at these points.

2. Locate the process equipment according to the rules-of-thumb which is zero above ground level for pumps.

3. Estimate z ₁ and z₂.

4. Estimate the frictional pressure losses Es and E_D according to the table 2 given below.

5. Calculate the pump work from Equation (4.12.2)

6. Calculate the pump shaft horsepower using Equation (4.12.3) and the pump efficiencies

given in the  excel spreadsheet.

7. Calculate the electric-motor horsepower using the motor efficiency given in

the spreadsheet.

8. Select a standard electric-motor horsepower using the spread sheet to obtain approximately

a 10% safety factor.

 

 

TABLE 4.18: APPROXIMATE FRICTIONAL PRESSURE DROP ACCROSS PROCESS EQUIPMENT

4.12.2    THE METERING RAM PUMP

The pumped fluid contacts only the inside surface of the tubing. There are no other valves, O-rings, seals or packing to worry about in a peristaltic pump. Therefore, the only compatibility to worry about in a peristaltic pump is the tubes for the fluid being pumped. Of all the design parameters, the compatibility is the first one that need to be considered.

The tube is made elastomeric in order to maintain the circular cross section of cycles of squeezing in the pump. This requirement eliminates a variety of non-elastomeric polymers that have compatibility with a wide range of chemicals, such as PTFE, polyolefins, PVDF etc. from consideration as material for pump tubing. Elastomers, both natural and synthetic, have common characteristics including elastic recovery after stress, flexibility, extrusion resistance, and relative impermeability to gases and liquids. Each class of elastomer has its own unique properties and performance that can be modified by the inclusion of other ingredients.

Reasons why tube is made elastomeric and not plastic are because elastomer provide :

1. Very good dimensional stability over a broad temperature range

2. High compression set resistance

3. Excellent extrusion resistance at high temperatures and pressures

4. Retention of sealing force during pressure and temperature cycling (a major problem with many plastics)

5. Elastomers are long-chain polymers connected by cross-links that impart strength, resilience and elasticity, and these cross-links are very stable to heat and high pressure. Plastics are also long-chain polymers, but they are not connected by chemical cross-links. Plastics acquire their strength when the chains orient with one another and become crystalline in regions. These regions may deform or melt under high pressure and temperature.

There are over 20 classes of elastomers. Of all these classes, the most ideal elastomer for this work is EPDM (ethylene propylene diene Monomer) + polypropylene (as in Santoprene).

Reasons why EPDM (ethylene propylene diene Monomer) + polypropylene (as in Santoprene) is the most ideal :

1. It has a wide service temperature range

2. EPM elastomers have excellent resistance to ozone, water and steam, alkalis and acids, salt solutions and oxygenated solvents

3. Very good balance of compression set, flex resistance, physical strength, low temperature flexibility, weather resistance properties, and resistance to acid condensates

4. Outstanding resistance to high heat; excellent resistance to oil, gasoline, hydraulic fluids and hydrocarbon solvents; very good impermeability to gases and vapor; very good resistance to weather, oxygen, ozone, and sunlight; good flame retardant.

5. EPDM + polypropylene have the best fatigue resistance compare to other elastomers

4.12.2.1   OCCLUSION

The minimum gap between the roller and the housing determines the maximum squeeze applied on the tubing. The amount of squeeze applied to the tubing affects pumping performance and the tube life – more squeezing decreases the tubing life dramatically, while less squeezing decreases the pumping efficiency, especially in high pressure pumping. Therefore, this amount of squeeze becomes an important design parameter.

The term “occlusion” is used to measure the amount of squeeze. It is either expressed as a percentage of twice the wall thickness, or as an absolute amount of the wall that is squeezed.

Mathematically;

Where;

For this design work , the wall thickness of the tube of the peristaltic pump(t) is 4mm and the minimum gap between the roller and the housing(g) is 1.6mm so the percentage occlusion is calculated using the above formular

Occlusion is one of the key peristaltic pump design parameter because it affects not only the efficiency of the fluid movement in the pump but the life span of the tube. The higher the occlusion, the more efficiently the pump moves fluid and the lower the occlusion, the less efficiently the pump moves fluid. The occlusion not only alters the efficiency of the pump but also the lifespan of the tubing. By increasing the occlusion, you are squeezing the tubing with greater force, which will wear out the tubing material faster. Prepare replacement tubing are provided because of the high occlusion , and because a replacement tubing may be needed is one of the reasons why EPM elastomer tubing is  the choice of tubing because it is inexpensive.

CHAPTER FIVE

5           MECHANICAL DESIGN

1.       THE GASIFIER

5.1.1        DESIGN PRESSURE

5.1.2        DESIGN TEMPERATURE

5.1.3        MATERIAL OF CONSTRUCTION

The wall of the gasifier is made of Austenitic stainless steel (Type 316) as in the storage tanks.

Lining Material (Refractory block):Refractory liners are used on the working face of entrained-flow slagging gasifiers that react coal, petroleum coke, heavy fuel oil or other feedstock with oxygen and water. The refractory liners protect the gasifier shell from elevated temperatures, corrosive slags, and thermal cycling during gasification. Refractory failure is primarily by two means which are corrosive dissolution and spalling. High chrome gasifier materials have evolved as the material of choice to line the hot face of gasifiers, yet the performance of these materials does need to improve by coating it with substances like magnesium oxide, aluminium oxide and/or chromium oxide (Bennet, 2004).

Therefore,

The refractory liner for the heavy fuel oil gasifier is: Chromium oxide coated with zirconium oxide and aluminium oxide (Cr₂O₃: 90%, Al₂O₃: 3%, ZrO₂: 7% in weight %)

5.1.4        THICKNESS OF THE REFRACTORY BRICK

Assuming that the ratio of the thickness of the vessel to the lining material is taken as 8:1, the thickness of the refractory lining is therefore = 90.6/8 mm = 11.325 mm

5.1.5        MINIMUM WALL THICKNESS

The minimum thickness required to resist the internal temperature of the gasifier is calculated thus,

5.1.6        MINIMUM HEAD THICKNESS

An Ellipsoidal head is chosen as the closure for this vessel because it is relatively cheaper than the torispherical head and it can withstand the design pressure of the vessel (Coulson and Richardson, 1999). The minimum thickness required is obtained thus,

Thickness of the head = 92.2 mm

5.1.7        VESSEL SUPPORT

Skirt Support is chosen because it is suitable for tall and vertical columns as it does not impose concentrated load on the vessel.

2.       THE WASTE HEAT BOILER

With the heat exchanger geometry defined, the mechanical design calculations must be done to ensure that the heat exchanger design is valid for the design pressure and conditions. The typical calculations are:

• Calculation of shell wall thickness.
• Calculation of nozzle wall thickness.
• Calculation of tube plate thickness.

SHELL WALL THICKNESS DETERMINATION

P_i= 3000 kPa

D_i = 1.078 m

At the shell side mean temperature of 192^oC,

f = 134 × 10⁶N/m²

Therefore,

e = 3000000( 1.078)/ [2(134 × 10⁶) – 3000000]

= 12.2 mm

Other design specifications include:

• Design Pressure = 110% of operating Pressure = 3300 kPa
• Design temperature = Maximum shell temperature = 220^oC
• Corrosion Allowance = 4 mm
  3.       THE QUENCHER

5.3.1        MATERIALS FOR CONSTRUCTION:

Carbon Steel is advised because, of its operability at not so high-pressure, because of its hardness and malleability and because it is most popular and therefore easy to obtain.

5.3.2        PACKING:

Interlox saddle packing was chosen because they are more efficient due to greater surface and improved hydrodynamics. They are made with a variety of holes and protrusions to enlarge the specific surface area. Because of their open structure and large specific surface area, mass transfer efficiency is high when proper distribution of liquid and gas over the cross section can be maintained. It also important to note that when ceramic construction of packing is suitable, saddles are the preferred packing.

5.3.3        DESIGN PRESSURE

Assuming 8% above operating pressure

= 1.08 x 4241.325 KN/m²

= 4580 .631 KN/m²

(4580.631-101.325) gauge

= 4479.306 kN/m² gauge

5.3.4        DESIGN TEMPERATURE

This is obtained as,

= Highest Temperature + allowance for uncertainty

Assuming an uncertainty of 80^oC

Therefore, design temperature= (250+ 80) ^oC

=330^oC

5.3.5        TENSILE STRENGTH

At 25^oC, the tensile strength of Carbon =210N/mm²

At 500^oC, the tensile strength of Carbon is 450 N/mm²

At 330^oC, the tensile strength of Carbon will be,

TABLE 5.1: TENSILE STRENGTH VALUES AT DIFFERENT TEMPERATURES

TEMPERATURE, OC	TENSILE STRENGTH, N/MM2
25	210
330	X
500	450

Interpolating we have,

(330 – 25) / (500 – 25) = (x – 200) / (450 – 210)

x=154+200

=  354N/mm²

5.3.6        DESIGN STRESS:

At 300^oC, the design strength of Carbon = 85N/mm²

At 350^oC, the design strength of Carbon is 80 N/mm²

At 330^oC, the design strength of Carbon will be,

(COULSON AND RICHARDSON, VOL 6)

TABLE 5.2: DESIGN STRESS VALUES AT DIFFERENT TEMPERATURES

TEMPERATURE, OC	DESIGN STRESS, N/MM2
25	85
330	X
500	80

Interpolating we have,

(330 – 300) / (350 – 300) = (x – 85) / (80 – 85)

X = τ = 82 N/mm²

Nominal design stress = 1.5 × = 1.5 x 82

= 123 N/mm²

5.3.7        CHOICE OF CLOSURE

Minimum wall thickness required for a cylindrical shell

e= Pi Di / 2f – Pi

Given that,

P_i = 4241.325 KN/m²

D_i = Internal diameter = 0.5275 m

therefore,

E = (4241.325 × 0.5275)

2 × 82000 – 4241.325

= 14 mm

Corrosion allowance = 2 mm

Minimum thickness of wall required = 14 mm + 2 mm = 16 mm

Choice of closure = Hemispherical head

Optimum thickness ratio for a hemispherical head to the thickness of the vessel = 0.6 (Coulson and Richardson, 2005)

Thickness of the hemispherical head = 16 x 0.6 mm = 9.6 mm

Type of Vessel Support = Skirt Support

 

4.       THE INITIAL H₂S ABSORBER

5.4.1        Shell Thickness

Thickness of shell,

where,

Inner diameter of vessel (D) = 1.3 m

Working pressure = 1.013 ×10⁶ N/m²

Design pressure (P) = 1.1× 1.013 x 10⁶ N/m²

Permissible Stress (f) = 115 × 10⁶ N/m²

Joint Efficiency (J) = 0.85

Corrosion allowance (c) = 2mm

hence,

So outer diameter of shell

1.                Axial Stress Due to Pressure

Axial stress due to pressure,

2.                Stress due to Dead Load

1.                 Compressive Stress due to weight of shell up to a distance X

 

2.                Compressive stress due to weight of insulation at height X

Insulator used is asbestos

Thickness of insulation, = 100mm

Diameter of insulation,

Density of insulation = 575 kg/m³

Mean diameter of vessel =

For large diameter column,

3.                 Compressive stress due to liquid in column up to height X

4.                Effect of attachments

1. Packing weight
2. Head weight
3. Ladder

Density of packing (Plastic pall rings) = 53 kg/m³

Packing weight =

= 690.11 X N

Head weight (approximately) = 35000 N

Weight of ladder = 1600 X N

5.4.2        SUPPORT FOR ABSORBER

Material used is structural steel (IS 800)

Skirt support is used

Inner diameter of the vessel = D_i = 4.69m

Outer diameter of the vessel = D_o = 4.70m

Height of the vessel = H = 10.054m

Density of carbon steel = = 7700kg/m³

Density of packing = = 53 kg/m³

5.       THE CO CONVERTER

The catalytic converter primarily functions to provide surface area required for the exothermic reaction involved in the water-gas shift reaction and also to house the catalyst required for the reaction. The materials that make entry into the converter include steam and flue gases which exit the initial H₂S removal stage and basically include hydrogen, carbon monoxide. Hydrogen sulphide, methane and nitrogen. The major aspects that will be considered in the mechanical design include:

• Operating and design temperatures and pressures
• Vessel dimensions and orientation
• Construction materials
• Type of vessel heads

5.5.1        Operating and Design temperatures and pressures

The operating temperature of the catalytic converter: 380 ⁰C

The operating pressure of the catalytic converter: 1200 kPa

The design temperature of the catalytic converter: 480 ⁰C (catalyst)

The design pressure of the catalytic converter: 1.1 1200 kPa = 1320 kPa

The maximum operating with respect to catalyst degeneration:  476.67 ⁰C

5.5.2        Vessel Dimensions and Orientation

The major dimensions have earlier been calculated in the chemical engineering design section of this unit; the diameter being 3.518m and total height of 23.41m. Minimum wall thickness – the thickness required to ensure that any vessel is sufficiently rigid to withstand its own weight, and any incidental loads; and for a diameter of about 3.5 m, this thickness is about 16 mm. Also, a 3mm corrosion allowance is added. The “corrosion allowance” is the additional thickness of metal added to allow for material lost by corrosion and erosion, or scaling.

5.5.3        Construction Materials

Construction materials for pressure vessels are usually plain carbon steels; low and high alloy steels, other alloys, clad plate, and reinforced plastics. Because the temperature of the reactor shell is very high, the reactors must be made of alloy steel which can sustain high temperature and is highly resistant to hydrogen corrosion. A “cold wall” reactor is lined with heat-insulating material. The temperature of the reactor shell is quite low, ordinary low alloy steel can be used for making such reactors.

Alloy steels contain one or more alloying agents to improve mechanical and corrosion-resistant properties over those of carbon steel.

Nickel increases toughness and improves low-temperature properties and corrosion resistance. Chromium and Silicon improve hardness, abrasion resistance, corrosion resistance, and resistance to oxidation.

Molybdenum provides strength at elevated temperatures.

 

 

5.5.4        TYPE OF VESSEL HEAD

The vessel head indicates the closure. Ellipsoidal heads will be made use of. The ellipsoidal head thickness can be calculated using the following formula:

where,  P_i – is the internal pressure; = 1200 kPa =1.2 Nmm^-2

D_i – the internal diameter; = 3.518 m

 J – the joint factor; which is equal to 1 as a result of the absence of joints

 f– the design stress at the specified temperature, equal to 180 N/mm²

Others include:

5.5.5        Openings required

Two openings are required:

An entry opening at the vessel top

An exit opening at the vessel bottom

Removable baffles for catalyst loading and discharge

5.5.6        Heating and Cooling Jackets or Coils.

Since its been established that a cold wall reactor will be used for the wall of the catalytic converter, implying an adiabatic arrangement to be ensured by incorporating on the reactor body an insulating material.

5.5.7        Internal fittings specification

The reactor will also consist of a reactor cylindrical shell, inlet diffuser, vapor-liquid distributor, scale removing basket, catalyst supporting plate, catalyst pipe, quench hydrogen box and a re-distributor, outlet accumulator, catalyst removal port and quench hydrogen pipe.

Also needed are Skirt supports of about 5 mm thickness;

FIGURE 5.1 SKIRT-SUPPORT

Base rings and anchor bolts for the supports.

FIGURE 5.2 DOUBLE PLATE WITH GUSSET

6.       THE FINAL H₂S ABSORBER

Design stress                  240N/mm₂

Construction material ;  low alloy steel

Tray thickness;              12mm

Design temperature;      30 – 45^oC

5.6.1        Design pressure:

This should be 10% greater than operating pressure

=

5.6.2        Wall thickness

e =

e = wall thickness

Pi= internal design pressure

Di= internal chamber

= design stress

e =

= 12.9 mm

Using corrosion allowance of 2mm

e= 14.9mm

5.6.3        Support

Skirttype is recommended

7.       THE CO₂ ABSORBER

Material for the absorber shell is Carbon steel

5.7.1        Shell Thickness

Thickness of shell,

where,

Inner diameter of vessel (D) = 1.3 m

Working pressure = 1.013 ×10⁶ N/m²

Design pressure (P) = 1.1× 1.013 x 10⁶ N/m²

Permissible Stress (f) = 115 × 10⁶ N/m²

Joint Efficiency (J) = 0.85

Corrosion allowance (c) = 2mm

Hence,

So outer diameter of shell

5.7.2        Axial Stress Due to Pressure

Axial stress due to pressure ,

5.7.3        Stress due to Dead Load

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

    

5.7.3.2       Compressive stress due to weight of insulation at height X

Insulator used is asbestos

Thickness of insulation, = 100mm

Diameter of insulation,

Density of insulation = 575 kg/m³

Mean diameter of vessel =

For large diameter column,

5.7.3.3       Compressive stress due to liquid in column up to height X

5.7.4        Effect of attachments

4. Packing weight
5. Head weight
6. Ladder

Density of packing (Plastic pall rings) = 53 kg/m³

Packing weight =

= 690.11 X N

Head weight (approximately) = 35000 N

Weight of ladder = 1600 X N

5.7.5        SUPPORT FOR ABSORBER

Material used is structural steel (IS 800)

Skirt support is used

Inner diameter of the vessel = D_i = 4.69m

Outer diameter of the vessel = D_o = 4.70m

Height of the vessel = H = 10.054m

Density of carbon steel = = 7700kg/m³

Density of packing = = 53 kg/m³

8.       THE HEAT EXCHANGERS

With the heat exchanger geometry defined, the mechanical design calculations must be done to ensure that the heat exchanger design is valid for the design pressure and conditions. The typical calculations are:

• Calculation of shell wall thickness.
• Calculation of nozzle wall thickness.
• Calculation of inner tube wall thickness.
• Calculation of expansion joint dimensions (to compensate for shell and tube side differential expansion due to temperatures differences.
• Calculation of tube plate thickness.

5.8.1        FIRST EXCHANGER

5.8.1.1       Shell Wall Thickness  (Corrosion Allowance = 2 mm):

 (Sinott, 2005)

Where f is the design stress, and P_i is the internal pressure.

Design pressure; take as 10 per cent above operating pressure,

= 31.9 bar

= 3.19 N/mm²

Typical design stress f = 540 N/mm²

e = = 1.70; with J = 0.85 for shell and tube exchangers

Adding corrosion allowance, 1.70 + 2 = 3.70, say 4.00 mm

5.8.1.2       Nozzle wall thickness:

Tube has been designed to provide entrance for nozzle = (0.2 shell diameter)

Thus nozzle diameter =

Nominal thickness at this condition is 0.5 in (=12.7mm) (Shilling et al., 1997)

5.8.1.3       Inner tube wall thickness:

With the tube dimensions selected according to standard,

Thickness =

 

 

 

5.8.1.4       Calculation of tube plate thickness:

The minimum plate thickness enough to withstand the bending and shearing force is gotten from literature as:

Minimum plate thickness = 0.75

5.8.1.5       OTHER SPECIFICATIONS

Number of baffles    Hence 

Baffle clearances

The edge distance between the outer-tube limit (OTL) and the baffle diameter has to be sufficient to prevent tube breakthrough due to vibration.

Baffle clearance = 11mm (for shell ID m) (Shilling et al., 1997)

Distance from tube O.D to centre of partition

Distance = 9.5mm (for ID)

Design temperature

Built to withstand 10% above highest stream temperature = (373.4 °C) × 1.1 ≈ 411 °C

Design pressure

Take as 10% above operating pressure = (30-1) × 1.1 = 39.1 bars

 

SECOND EXCHANGER

Shell wall thickness e (Corrosion Allowance = 2 mm):

(Sinott, 2005)

Design pressure; take as 10 per cent above operating pressure,

= 26.4 bar

= 2.64 N/mm²

Typical design stress f = 540 N/mm²

e = = 3.07; recall J = 0.85

Adding corrosion allowance, 3.07 + 2 = 5.07, say 6.00 mm

Nozzle wall thickness:

Tube has been designed to provide entrance for nozzle = (0.2 shell diameter)

Thus nozzle diameter=

Nominal thickness at this condition is 0.432in (=10.97mm)

Inner tube wall thickness:

Thickness=

Calculation of tube plate thickness:

Minimum plate thickness=0.75

OTHER SPECIFICATIONS

Number of baffles    Hence 

Baffle clearances

Baffle clearance=13mm (for shell IDm) (Shilling et al., 1997)

Distance from tube O.D to centre of partition

Distance = 9.5mm (for IDm)

 

Design temperature

Built to withstand 10% above highest stream temperature = (380.9 °C) × 1.1 ≈ 420 °C

Design pressure

Take as 10% above operating pressure = (25-1) × 1.1 = 26.4 bars

FIGURE 5.3: PHYSICAL LAYOUT OF A 2-TUBE PASS SHELL AND TUBE EXCHANGER

CHAPTER SIX

6           PLANT HAZARD AND OPERABILITY STUDY

1.       GASIFIER

GASIFIER INTENTION: to cause partial oxidation of the heavy oil feedstock.

TABLE 6.1:HAZOP ANALYSIS FOR GASIFIER

GUIDE WORD	DEVIATION	CAUSES	CONSEQUENCES	ACTION
NO	Flow (steam & gas)	Control valve stuck closed	Delays the reaction	Fit low temperature alarm
MORE	Flow (Steam/Oxygen)	Failure of ratio flow or failure of control valve	High reactor temperature Disaster	1)Alarms 2) Automatic shutdown
AS WELL AS	Composition	Refractory particles from reactor	Possible blockage of boiler tubes	Install filters up-stream of boiler

 

 

2.       WASTE HEAT BOILER

TABLE 6.2: HAZOP ANALYSIS FOR WASTE HEAT BOILER

GUIDE WORD	DEVIATION	CAUSES	CONSEQUENCES	ACTION
NONE	No tube side fluid flow	Failure of inlet tube side  valve to open	Process fluid temperature is not heated accordingly	Install Temperature indicator before and after the process fluid line Install TAH
MORE	More tube side fluid flow	Failure of inlet tube side valve to close	Output of Process fluid temperature too high	Install Temperature indicator before and after process fluid line Install TAL
LESS	Less tube side fluid	Pipe leakage	Process fluid temperature too low	Installation of flow meter
REVERSE	Reverse process fluid flow	Failure of process fluid inlet valve	Product off set	Install check valve (whether it is crucial have to check?)
CONTAMINATION	Process fluid contamination	Contamination in cooling water	Outlet temperature too low	Proper maintenance and operator alert

3.       HEAT EXCANGERS

TABLE 6.3: HAZOP ANALYSIS FOR HEAT EXCANGERS

GUIDE WORD	DEVIATION	CAUSES	CONSEQUENCES	ACTION
LESS	Less flow of shell side fluid	Pipe blockage	Temperature of process fluid increase	High Temperature Alarm
MORE	More shell side fluid flow	Failure of shell side valve	Temperature of process fluid decrease	Low Temperature Alarm
Corrosion	Corrosion of tube	Corrosive tube side fluid	Excess Heat transfer and crack of tube	Proper maintenance and operator alert
MORE OF	More pressure on tube side	Failure of service fluid valve	Bursting of tube	Install high pressure alarm
CONTAMINATION	Contamination of process fluid line	Leakage of tube and tube side fluid goes in	Contamination of process fluid	Proper maintenance and operator alert

4.       INITIAL H₂S REMOVAL

TABLE 6.4: HAZOP ANALYSIS FOR INITIAL H₂S REMOVAL

GUIDE WORD	DEVIATION	CAUSE	CONSEQUENCE	SAFETY ACTION
No	No supply of the gas into the absorber.	A stuck closed valve or faulty closed valve	Accumulation of the gas at the quencher or within the streamline	Valve CV2 closes
less	Less/ little supply of the crude gas into the absorber	Incomplete opening of the valve i.e. faulty valve Blockage within the stream line	Accumulation within the pipe Excess heat loss from the gas	CV2 closes to reduce the flow rate of entering NMP
more	More crude gas supply	A stuck open valve or faulty open valve	Insufficient H2S absorption Insufficient gas cooling	Open CV2 and CV3 Vent through relief valve
no	No supply of NMP into the absorber	A stuck closed valve or faulty closed valve	No H2S absorption occurs The crude gas flows to the saturator where it corrodes the equipment due to the H2S present	CV1 closes
less	Less supply of NMP into the absorber	Incomplete opening of the valve i.e. faulty valve Blockage within the stream line	Insufficient   H2S absorption occurs	Close CV1 to reduce gas flow rate
more	More NMP supply into the absorber	A stuck open valve or faulty open valve	Flooding of the absorber with NMP	Open CV4 and CV1.
no	No discharge of the scrubbed gas out of the tower	A stuck closed valve or faulty closed valve	Accumulation of the gas within the absorber Failure of equipment	Close CV1 AND CV2
less	little discharge of the scrubbed gas out of the tower	Incomplete opening of the valve i.e. faulty valve Blockage within the stream line	Accumulation within the pipe and outlet line Faulty valve equipment	Close valve CV1 and CV2 to reduce their flowrates
more	More scrubbed gas is discharged, i.e. if there is a pump sucking out the gaseous stream.	A stuck open valve or faulty open valve	Pressure drop within the tower	Open CV1, CV2 AND CV4
no	No discharge of NMP out of the tower	A stuck closed valve or faulty closed valve	Accumulation of NMP within the quencher Failure of equipment	Close CV2 and CV1
less	little discharge of NMP out of the tower	Incomplete opening of the valve i.e. faulty valve Blockage within stream	Accumulation within the pipe and outlet line Faulty valve equipment	Close CV1, CV2 and CV3 to reduce their flow rates

 

5.      
   SATURATOR/DESATURATOR

INTENTION – CONTACTING SYNGAS WITH HOT WATER

TABLE 6.5A: HAZOP ANALYSIS FOR SATURATOR

GUIDE WORD	DEVIATION	CAUSE	CONSEQUENCES & ACTION
Less of	Flow	Pump failure, blockage of pipe, valve failure	Syngas not contacted with water, high pressures, low level alarm, high pressure alarm
More of	Flow	Pump malfunction, valve stuck open	Flooding, channelling, high level alarm
As well as	Composition	Distributor malfunction, pump malfunction	Impure syngas.
No	Flow	Pump fail closed	Syngas not contacted with water, high pressures, low level alarm, high pressure alarm

 

 

DESATURATOR INTENTION – DRYING THE PRODUCT GAS FROM CO SHIFT CONVERTER

TABLE 6.5B: HAZOP ANALYSIS FOR DESATURATOR

GUIDE WORD	DEVIATION	CAUSE	CONSEQUENCES & ACTION
Less	Flow	Pump malfunction, pump fail closed	syngas not dried, high pressure alarm

 

LINE 1 & 2  INTENTION – supplies syngas in and out of the saturator or desaturator

TABLE 6.5C: HAZOP ANALYSIS FOR LINES 1 AND 2

GUIDE WORD	DEVIATION	CAUSE	CONSEQUENCES & ACTION
No	Flow	Valve failure, pump fail closed	Saturator mal-function
More	Flow	valve fail- open, pump malfunction	Less contact of syngas with water, high level alarm
Reverse	Flow	Pipe blockage	Pump malfunction, high pressure

 

6.      
   CO₂ REMOVAL

CO₂ REMOVAL INLET INTENTION – to purge syngas stream of CO₂

TABLE 6.6A: HAZOP ANALYSIS FOR CO₂ REMOVAL INLET

GUIDE WORD	DEVIATION	CAUSES	CONSEQUENCES
None	No Flow	Flow stopped upstream Line blockage Line fracture	No absorption in column. Entire process stops As for 1. Pressure buildup in pipe and secondary cooler As for 1. Gases escape to the surrounding
More of	More Flow	Increased feed	Possible reduction in absorption efficiency. May cause flooding
	Flooding	Flooding	Unit subject to high pressure, bursting discs may rupture, tail gas release
	More Temperature	Higher feed gas or liquid temperature	Decreased absorption, higher pollution
Less of	Less flow	Leaking inlet range	Gas escapes into surrounding
	Less temperature	Overcooling	Increased dissolved gases in acid

 

CO₂ REMOVAL OUTLET INTENTION – to send syngas stream of CO₂ to storage

TABLE 6.6B: HAZOP ANALYSIS FOR CO₂ REMOVAL OUTLET

KEYWORD	DEVIATION	POSSIBLE CAUSES	CONSEQUENCES
None	No Flow	No inlet gas flow Flooding in column Line fracture	No tail gas for expansion Pressure build up in column No absorption in column
More of	More Flow	Increased feed Decreased CO2 absorption	Transfer line subject to higher pressures. Tail gas emission levels up
	More Temperature	Higher feed gas or liquid temperature	Decrease absorption, higher pollution
Less of	Less flow	Leaking inlet range	Gas escapes into surrounding
Other	Maintenance	Equipment failure, flange leak, catalyst changeover in reactor, etc.	Process stops

 

7.       STORAGE TANK

STORAGE TANK INTENTION: The storage tanks 1 and 2 are designed to store heavy fuel oil enough for production of hydrogen for 5days and supply heavy fuel oil to the gasifier.

TABLE 6.7: HAZOP ANALYSIS FOR STORAGE TANKS

EQUIPMENT	GUIDE WORD	DEVIATION		CAUSE	CONSEQUENCES	ACTION
Line 1 Supplies HFO to the metering ram pump	NO	No flow	Faulty pump		1) Higher level in the tank 2) Explosion of tank	Fit a high-high alarm
Supplies only HFO to the pump	AS WELL AS	Composition	Corrosion of tank internals		1) Impurities in the stream that can lead to blockage of the pump 2) Over pressure of tank	Proper installation and maintenance

8.       SAFETY

6.8.1        STORAGE TANK

SAFETY CONSIDERATIONS FOR THE STORAGE TANKS

• Flame arresters are installed based on the temperature, pressure, or composition of the gases entering the arrester, to stop open fire from spreading out. They are also periodically inspected to make sure they are free of dirt, or corrosion.
• Like any other high pressure vessel, a pressure relief valve with a pressure higher than both the design and operating pressure, is installed on the storage tank.
• The tank content fuel air mixture is stored well above the flammability limits such that it cannot burn even if an ignition source were present.
• Scaffolds are installed to aid workers during maintenance activities
• The storage tanks are shielded from weather conditions by suitable protective structures.
• The tanks for the storage of the final product are kept within a secondary container called a bund with enough space to take a volume at least 110% of the total volume capacity of the tanks (4,125,000m³)

6.8.2        THE HEAT EXCHANGERS

Overpressure is the only common safety issue affecting shell and tube heat exchangers. They are pressure vessels and as such are subject to the same codes and practices as other pressure vessels. That means the ASME Boiler and Pressure Vessel Code, Section VIII, Pressure Vessels, Parts UG-125 to 136 dealing with pressure relief devices. This specification gives very clear guidelines concerning all aspects of pressure relief requirements and application.

Pressure relief must be provided for both the shell and tube sides. If the source of overpressure is from upstream, the relief valve for that stream is best placed on the inlet. Otherwise it does not matter much whether it is on the inlet or outlet so long as they are inside any control or isolation valves. It is not sufficient to put minimum stops on the valves as these are easily altered. Even if the stops are welded in place, the valve may be replaced at some future date and the modification forgotten. If careful analysis shows that there are no process, fire, or failure conditions that could possibly require relief valves, it is still strongly advised to install thermal reliefs on both sides of any exchanger that is capable of being blocked in. It may be argued that the fluid is gas or that the process is not capable of adding heat to the blocked in exchanger. This argument overlooks the various unanticipated conditions that may arise during testing and maintenance. A worst case scenario: A cooler was taken out of service and steam cleaned. No one had drained the cooling water which expanded in the tubes and ruptured the joints. True, good maintenance practice would have prevented this incident. But then an NPS ¾ relief valve would have provided a permanent solution and would have cost a lot less than the damage caused by its absence.

6.8.3        THE CO CONVERTER

Safe operation of a fixed bed catalytic reactor is paramount especially as our reactor is exothermic. Elements such as controllability, stability, risk and economic aspects need to be considered.  For the catalytic reactors, nominal conditions which are in the vicinity of safety limits are set so as to limit the hot-spot in the reactor and to excessive thermal stability to variation in the process parameters.

The following should be put in place for safe operation of the catalytic reactor.

1. Use of baffles in the reactor to prevent hot-spot: baffle plates are placed at the reactor entrance to deflect the jet of the incoming gases.
2. The height of the bed is limited to at least 1/2D to promote uniform flow distribution to avoid crushing the catalyst.
3. The reactor would be designed for a thickness that will give 100% safety factor.
4. Avoid steam condensation and minimize the re-oxidation of the catalyst upon shut down.
5. Approximately 1.0m should be added to the length in order to account for inert material between tangents at top and bottom and for free space above bed for distributor and work area.
6. For a stable operation, all heat generated by the exothermic reaction should be properly harnessed by the heat-exchanger so as to maintain the 10^oC approach to equilibrium in other to prevent  run-away reaction and also to prevent the formation of hydrocarbons (methane ) in large amount instead of the require hydrogen gas.

Thus, engineering safety covers the provision in the design of control systems, alarms, trips, pressure-relief devices, automatic shut-down system and firefighting equipment.

6.8.4        SATURATOR – DESATURATOR

SAFETY

For the saturator and desaturator syngas contains hydrogen gas which is flammable (flammability range 4.9 – 74.2). Hence the columns are prone to explosion and fire. This is so because hydrogen has a high burning rate, high buoyant velocity and high rate of vapour generation. Also inhalation of hydrogen can cause sleepiness and high-pitched voice in humans.

The vessels are operating at high pressures hence, the vessels are fitted with pressure relief valves for proper pressure control.

Preventive measures taken include:

• Adequate and secure water for fire fighting.
• Correct design of vessel.
• Use of pressure relief valves for pressure control.
• Use of corrosion resistant materials such as packing
• Compliance with standard codes such as the ASME code for the testing and design of the vessels
• Use of fails-safe instruments.

6.8.5        THE FINAL H₂S ABSORBER

Corrosion consideration

So₂ produced during regeneration combines with water to produce an acidic mixture that can badly corrode the SO₂ header . Hence, corrosion monitoring is required after regeneration and chemical injection corrosion inhibitors are required during regeneraton.

Operability study

1.                         The vessel: H₂S dry sorption vessel
2.                         Its lines;

• Inlet of flue gas plus O₂ line
• Outlet of treated flue gas
• Inlet of reheneration oxygen
• Outlet of SO₂

The Hazards

Toxicity:- H₂S is toxic; effects are acute and chronic

Lower toxic limit: 10 ppm

Inlet gas H₂S concentration: 15ppm

Lethal dose: 0.05% in air kills within 30 minutes to hour exposure

1.2 – 2.8 mg H₂S per liter of air kills instantly. Long term exposure to very small amounts of H₂S leads to chronic poisoning.

Symptoms: irritation of the mucous membrane, sensitivity to light, bronchitis, headaches, weariness, circulatory disturbances and loss of weight.

Causes of exposure in the unit

•             Loosely bound flanges
•             Pipe leakages
•             Depressurizing vessels (vents)
•             Depressurizing instruments such as pressure gauges

Safety precautions

Use appropriate personal protective equipment (PPE) such as nose masks during maintenance of measuring instruments, depressurization of vessels e.t.c

Ensure flanges are tightly bound with appropriately fitted gaskets. Welded joints preferred.

First Aid

Fresh air, artificial respiration, warmth, rest, transportation in an inclined position. Danger of suffocation if patient is unconscious.

Flammability

Pressurized hydrogen and CH₄ are the flammable substance in the unit. The flammability units are shown in the table below.

TABLE 6.8: FLAMMABILITY LIMITS OF DIFFERENT COMPOUNDS

MATERIAL	LOWER LIMIT	UPPER LIMIT
H2	4.1	74.2
H2S	4.3	45.0
CO	12.5	74.2
CH4	5.3	14.0

 

Safety precaution

The plant will conform to the recommendations of the British Standard, BS 5908 to prevent fire outbreaks

Explosion

The system is prone to unconfined cloud explosion. Here, a considerable quantity of the flue gas in the atmosphere would explode when ignited.

Safety precautions

•                         No smoking around the facility
•                         No naked flames
•                         Use of intrinsically safe cameras and devices
•                         No mobile phones allowed in the facility
•                         Flame traps in vent and flare lines

Pressure

This unit operates at about 2400kpa. Hence, any blockage in the process lines can result in over pressure, i.e exceeding the design pressure of 2458.5kPa.

Safety precaution

Indirectly actuated valves(electrical and pneumatic) which fail-open should be used as the pressure relief valves. Auxiliary bursting discs may also be installed for a faster rate of depressurization during major overpressure

Temperature deviation

Design temperature is set at 300^oc. The situation of excessively high temperature is highly unlikely. However it is possible.

Causes

If the outlet for SO₂ during regeneration is blocked, it could result to

•                        Overpressure due to temperature builup
•                        Overtemperature due to inhibited heat flow

Control

•                         Pressure relief valves
•                         Input heated O₂ flow controller
•                         Avoid restriction in the outlet SO₂ line

Noise

No rotating and moving parts exist in the system. Hence, noise levels are minimal.

Environmental consideration

16 month SO₂ emission:- scrubber should be installed, which uses Ca(OH)₂ to absorbed the SO₂ produced to form calcium sulphate, which can be sold to plaster of paris(POP) producing industries.

Flue gas vents should be routed to a central flaring system. A self-supported elevated flare system is recommended.

 

CHAPTER SEVEN

7           COSTING

1.       ECONOMIC AND PROFITABILITY ANALYSIS: GASIFIER

Bare cost of a vertical vessel of diameter 1.99m and a height of 9.64m is 24,000 dollar as at mid 2004 (Coulson and Richardson, 2005)

Purchase cost = Bare cost Material factor Pressure factor

Material factor = 2

Pressure factor = 2.28

Purchase cost = (24,000 2.28 2) dollar = 109,440 dollar

1dollar = N 151

109,440 dollar = N 16,525,440

Purchase cost of the vessel in 2004 = N16,525,440

The projected cost of the vessel in 2011 based on inflation rate is obtained as below:

TABLE 7.1: INFLATION RATES BY YEAR (1999-2011), US INFLATION CALCULATOR

YEAR	INFLATION RATE	PROJECTED COST (NAIRA)
2005	2.5%	16,938,596.00
2006	4.3%	17,666,934.77
2007	2.7%	18,143,942.01
2008	5%	19,051,139.11
2009	-1.4%	17,069,820.64
2010	1.1%	17,257,588.67
2011	4%	17,947,892.21

Therefore, the present purchase cost of the gasifier vessel = N17,.947,892.21

The estimation of the purchase cost of the lining material is carried out as follows:

Volume of the refractory brick = (29.88 –) m³

= (29.88 – 29.30) m³ = 0.576 m³

The table below shows calculation of the density of the refractory brick

TABLE 7.2: REFRACTORY BRICK DENSITY CALCULATION

COMPOUND	COMPOSITION (%)	DENSITY (Kg/m3)	% DENSITY
Cr2O3	90	4885	4396.5
Al2O3	3	248.67	7.4601
ZrO2	7	4825.14	337.7598
Total	100		4741.7199

Therefore,

The mass of the refractory brick () = 4741.7199 Kg/m³ x 0.576 m³ = 2730.18 Kg

From literature, the cost of the lining material = 4825 dollar/metric tonne

1016 Kg = 1 tonne

2730.18 Kg = = 2.687 tonnes

Therefore, the cost of the refractory brick = 4825 x 2.687 dollar = 12,965.69 dollar

1 dollar = 151 Naira

12,965.69dollar = 12,965.69 x 151 Naira = N1, 957,819.27

Therefore,

The total purchase cost of the gasifier = purchase cost of the vessel + purchase cost of the refractory block =  N17,.947,892.21 + N1, 957,819.27 = N19,905,711.48

2.       ECONOMIC AND PROFITABILITY ANALYSIS: QUENCHER

Bare cost of a vertical vessel of diameter 0.5275 m and a height of 3.0893 m is 9,715.32 dollar as at mid 2004 (Coulson and Richardson, 2005)

Purchase cost = Bare cost × Material factor × Pressure factor

Material factor = 1

Pressure factor = 1.8

Purchase cost = (9,715.32 × 1.8 × 1) dollar = 17,487.58 dollar

1dollar = N 151

17,487.58 dollar = N 2,640,623.98

Purchase cost of the vessel in mid 2004 = N 2,640,623.98

The projected cost of the vessel in 2011 based on inflation rate is obtained as below:

TABLE 7.3: INFLATION RATES BY MONTH AND YEAR (2005-2011) (FOR BARE VESSEL)

YEAR	INFLATION RATE (JUNE)	PROJECTED COST (NAIRA)
2005	2.5%	2,706,639.58
2006	4.3%	2,823,025.08
2007	2.7%	2,899,246.76
2008	5%	3,034,209.10
2009	-1.4%	2,991,730.173
2010	1.1%	3,024,639.21
2011	4%	3,145,624.78

Therefore, the present purchase cost of the gasifier vessel = N3,145,624.78

The estimation of the cost of packing material for the quencher is done as below

The cost of an intalox saddle packing as at mid 2004 = 829 dollar/m³ (Coulson and Richardson, 2005)

Height of packing = 1.2023 m/1.5 = 0.8015 m

Diameter of the vessel = 0.5275 m

Therefore, the volume of the packing = (π × 0.5275²/4) × 0.8015= 0.17516 m³

Thus,

The cost of the interlox saddle = 829 dollar/m³ x 0.17516 m³ = 145.21 dollar

145.21 dollar = N 21,926.53

The projected cost in the present year is estimated thus:

TABLE 7.4: INFLATION RATES BY MONTH AND YEAR (2005-2011) (FOR PACKING MATERIAL)

YEAR	INFLATION RATE (JUNE)	PROJECTED COST (NAIRA)
2005	2.5%	22,474.69
2006	4.3%	23,441.10
2007	2.7%	24,074.00
2008	5%	25,277.70
2009	-1.4%	24,923.81
2010	1.1%	25,197.97
2011	4%	26,205.89

Therefore the cost of the packing material = N 26,205.89

The total cost of the vessel along with the packing material = N 3,145,624.78 + N 26,205.89 = N 3,171,830.67

 

3.       COSTING (CO CONVERSION)

Cost of Catalyst

Volume of catalyst: 187.17 m³ or cu ft

Cost of catalyst = $20 per cu ft

Cost of procuring catalyst:

Cost of Equipment Materials

Reactor material of construction: Low alloy steel

Volume of reactor material: 227.56 m³

Density of low alloy steel: 8.44 g/cm³

Mass of reactor: g or 1920 tonnes

Low alloy steels (Cr-Mo) cost about 400 – 700 pounds per ton

 

Cost of securing reactor materials:

4.      
   COST AND FEASIBILITY ANALYSIS (FINAL H₂S REMOVAL)

Cost of equipment

Cost of low alloy steel = USD700  & 1 tone

Density = 7850kg/m³

External diameter = Internal diameter + 2(thickness)

External volume of column =

=92.56362m³

The 12mm was obtained by adding 2mm to the 10mm because of corrosion allowance

Internal volume of column =

=90.81166m³

Volume of steel = External – Internal

= 92.56362 – 90.81166

= 1.7520m³

Tray thickness = 12

Volume of steel tray =

= 0.05890m³

Number of Steel column = 4

Total number of trays = 5

Total volume of Steel = 4(1.7520) + 20(0.05890)

= 8.18610m³

Total Mass =

= 7850kg/m³

= 64261kg

=64.260 tonnes

Total cost of vessels + tray =700

= USD44,982 as at 2004

TABLE 7.5: INFLATION RATES BY YEAR (2005-2011) ( FOR H₂S REMOVAL VESSEL)

YEAR	INFLATION RATE	AMOUNT IN USD
2005	2.5%	46106.55
2006	4.3%	48089.13
2007	2.7%	49387.54
2008	5%	51856.92
2009	-1.4%	51130.92
2010	1.1%	51693.36
2011	4%	53761.09

The cost as at 2011 = USD53761.09

But cost of lautamasse = USD600/tone

Mass of lautamasse = 6935kg

= 6935kg

=6.935tonne

Cost of Lautamasse = USD(6.935

Cost of Lautamasse = USD4161

Total cost of Steel and Lautamasse = USD(53761.09 + 4161)

= USD57922.09         = NGN8.7 million

5.       COSTING AND PROFITABILITY (CO₂ REMOVAL)

Absorber parameters:

Where

Substituting 2028.91lb for W in the above equation,

For plastic pall rings of 3.54in diameter,

Total cost is given as:

0

6.       ESTIMATION OF COST (1^ST HEAT EXCHANGER)

The cost data given in Figure 8 can be used to make preliminary estimates. The base date is mid-1998, and the prices are thought to be accurate to within ±25%.

Purchased cost = (bare cost from figure) Type factor Pressure factor

(As at mid 1998)

Inflation-corrected value of Purchase cost from 1998 to 2011 gives $103971.17, which approximates to $104000.

FIGURE 7.1: COST ESTIMATION DATA FOR SHELL AND TUBE HEAT EXCHANGERS

 

7.      
   ESTIMATION OF COST (2^ND HEAT EXCHANGER)

The cost data given in Figure 8 will be used again for the cost calculation. The base date is mid-1998, and the prices are thought to be accurate to within ±25%.

Purchased cost = (bare cost from figure) Type factor Pressure factor

(As at mid 1998)

Inflation-corrected cost from 1998 to 2011 estimates to $582000.

 

 

CHAPTER 8

CONCLUSION AND RECOMMENDATION

8.1 CONCLUSION

The aim of this design project which is to design a plant producing 750,000 m³ of hydrogen of 95% purity by partial oxidation of oil feedstock has been achieved. The methodologies of HYSYS simulation plant were useful for carrying out the material and energy balance calculations required for the plant. From our material balance calculations, it is concluded that a 95% purity of hydrogen is obtained. Furthermore, it can be concluded from our energy balance calculations that the gasifier produces the largest amount of energy and the waste heat boiler consumes the largest amount of energy. The economic analysis shows the total cost and start-up capital required for the plant which tells us that it is profitable to set up the plant

8.2 RECOMMENDATION

It is hereby recommended that project design is carried out at lower levels, so as to give the students a well grounded approach to chemical engineering plant design.

Software development should be incorporated into the undergraduate prospectus because of the ease of computation that can be possible as a result of proficiency in the use of these applications.

Courses involving process design from other engineering departments should be taken by the students as plant design involves just more than chemical engineering.

Finally, supervisors should strive to take a more active interest in the design work of their respective students.

REFERENCES

Andrea Sella (2008),“Raschig’s Rings”. Chemistry World. page 83. Retrieved from wikipedia, the free encyclopedia date accessed: 05/07/2011.

Anon. (2011), Heat exchangers, Retrieved March 15, 2011 from www.wikipedia.com

Anon. (2011), Types of Heat Exchangers, Retrieved March 15, 2011 from   www.wikianswers.com

Don W.G. and Robert H.P. (2008).  Perry’s Chemical Engineering handbook, 8^th edition.

ISBN 0-07-142294-3.

Frank, O. (1978), Simplified design procedure for tubular exchangers, in Practical Heat Transfer, Chem. Eng. Prog. Tech. Manual (American Institute of Chem. Eng.)

John Wiley & Sons. ISBN 0-471-46480-5. Retrieved from wikipedia, the free encyclopedia date accessed: 05/07/2011.

Jones et al, July 1981, Gas quench cooling and solids separation process

Oak Ridge Associated Universities Raschig Rings for Criticality Control (1980s). Retrieved from www.orau.org/collection/micellaneous/raschigs.com. Date accessed: 05/07/2011.

Pitts D. R and Sissom L. E (1998), Schaum’s Outline of Theory and Problems of Heat      Transfer, Second Edition

Shilling R. L.,Kenneth J. B.,Patrick M. B.,Thomas M. F.,  Victor M. G.,Predrag S. H., Standiford F. C.,Klaus D. T. (1997), Perry’s Chemical Engineers’ Handbook, Seventh                                                       Edition,Chapter 11, Heat Transfer Equipment

Sinnott, R. K. (2005), Chemical Engineering, Volume 6, Fourth edition, Chemical   Engineering              Design

Seader, J.D. and Henley, Ernest J. (2006). Separation Process Principles (2nd Edition ed.). Sinnot, R.K. (2005a), Coulson and Richardson’s Chemical engineering series volume 6, 4^th   Edition, page 815.

Sinnot, R.K. (2005b), Coulson and Richardson’s Chemical engineering series volume 6, 4^th Edition, page 591.

Sinnot, R.K. (2005c), Coulson and Richardson’s Chemical engineering series volume 6, 4^th Edition, page 460.

Sinnot, R.K. (2005d), Coulson and Richardson’s Chemical engineering series volume 6, 4^th Edition, pages 609-616.

Staudinger et al, (1977), Process for quenching product gas of slagging coal gasifier www.freepatentsonline.com/5571295

Von Klock et al, (1986), Synthesis gas generation with prevention of deposit formation in                                                                                                                                                                                 exit lines

 

 

 

 

 

 

BLANK PAGE

 

 

 

 

 

 

APPENDIX ONE: SIMULATION

 

 

 

 

 

 

 

APPENDIX TWO: PLANT AND SITE LAYOUT

 

 

 

 

 

 

 

 

 

 

 

 

 

APPENDIX THREE: PROCESS INFORMATION

 

 

 

APPENDIX FOUR: CALCULATIONS

 

 

 

 

 

 

 

 

 

TANK STORAGE CAPACITY

The mass of feed required to produce 750,000m3 of Hydrogen per day is given by:

The tank is to store enough feed for the production of Hydrogen for 5days. Thus, the mass of feed to be stored is:

Thus, the required capacity of the tank is obtained from the equation below

Given the specific gravity of the feed at 50^oC = 0.9435,

Therefore, the required capacity of tank 1= Volume of feed

 

TANK DIMENSIONS

Assume that:

For safety reasons, an overfill level protection level and a minimum operating level is put into consideration.

H = 70% of the total height

TANK 2

This tank is similar to tank in function, configuration, material of construction and safety devices and structural supports used. The only difference is that it is a smaller tank used to aid the Start-up and shut down procedures. This tank is required to store enough heavy fuel oil for a 3 hour production run.

 

STORAGE CAPACITY

From the material balance,

Therefore the required volume capacity of the tank is given by:

 

TANK DIMENSIONS

Also, assuming that: 

Similar to tank 1, there is an overfill level and minimum fill level for safety purposes

H = 70% of the total height (H₁)

TANK 3

This tank is used for the storage of the final product gas, hydrogen before it is supplied to the immediate consumers. It is similar in configuration to tanks 1 and 2, but the material of construction is different.

STORAGE CAPACITY

According to the design problem statement, the volume of Hydrogen gas to be produced per day = 750,000m^3.

Assuming that the tanks are offloaded every 5 days such that the required capacity of the tank is

Due to the large capacity requirements of the tank, the total volume of the product is stored in 50 equal size tanks.

Thus, the capacity of each tank is

TANK DIMENSIONS

For each of the 50 tanks, assume that H = 3D

Similar to tanks 1 and 2, there are safety allowances in the height of the tank.

So that D = 44m

Installed storage capacity is now

7.7.1        THE HEAVY FUEL OIL GASIFIER

CHEMICAL ENGINEERING DESIGN

The following reactions take place in the gasifier:

     (1)

   (2)

         (3)

         (4)

        (5)

The slowest reaction is reaction 4 (Carbon-steam reaction) and is therefore taken as the rate determining step for the reactor.

C + H₂O → H₂ + CO (rate-determining step)

The rate expression for this reaction is given as:

r = k_v

Where:

r = rate of reaction (in this case, the rate of disappearance of H₂O)

k_{v =} rate constant

K = equilibrium constant

C_H2O, C_H2 and C_CO = molar concentrations of C_H2O, C_H2 and C_CO respectively (kmol/m³)

Given the table below for rate constant, k_v (from reference source)

T (K)	952.38	1000
kv (cm3/gmol.s)	1.62	8.6

RATE CONSTANT VALUES AT SPECIFIC TEMPERATURES

Temperature of the exit crude gas is taken as the reaction temperature in the gasifier

Therefore, reaction temperature = 1300^oC = (1300 + 273) K = 1573K

Extrapolating k_v values in the table above in order to get the value of k_v at 1573K, yields:

620.62 (6.98) = 47.62 (k_v – 1.62)

k_v = 92.59 cm³/gmol.s

k_v = 92.59 = 0.0926 m³/kmol.s

The table below is used to extrapolate for the value of the equilibrium constant, K at 1573K

T (K)	1300	1400
Log K	2.06	2.44

EQUILIBRIUM CONSTANT VALUES AT SPECIFIED TEMPERATURES

100 log K – 206 = 103.74

log K = 3.0974

K = 1251.411092

COMPOSITION OF CRUDE GAS LEAVING THE GASIFIER (FROM MATERIAL BALANCE CALCULATION: ON WET BASIS)

COMPONENTS	AMOUNT (kmol/hr)	COMPOSITION
H2	671.636	0.4232
CO	594.031	0.3743
CO2	117.113	0.0738
CH4	1.411	0.0008
H2O	171.01	0.1077
H2S	7.055	0.0044
N2	19.754	0.0124
C	5.16	0.0033
Total	1587.17	1.0000

The operating pressure in the gasifier is assumed to be 62 bar (6200KN/m²)

From equation of state,

C (Kmol) =, where P = pressure (= KN/m²), R = gas constant = 8.314 KJ/Kmol.K and T = temperature (= K)

Therefore,

C_H2O = = = 0.0511 Kmol/m³

C_{H2 =} = 0.2006 Kmol/m³

C_CO = = = 0.1774 Kmol/m³

And then,

r = 0.0926 = -0.0297

For a CSTR, the performance equation is given as:

V =

From material balance calculation,

Mass of steam entering the gasifier on a basis of 100kg/day feed of heavy fuel oil = 75kg

Therefore, actual mole of steam entering the gasifier

=   = 171.897 kmol/hr

Amount of steam leaving the gasifier = 171.01 kmol/hr

F_AO = F_H2Oo = 171.897 kmol/hr

X_A = X_H2O = = = 0.0052

Therefore,

V = = 29.88 m³

From literature, a standard 4.5 m-high gasifier can gasify up to 1.9 tonnes/hr of feed. This standard is then used to estimate the height of the heavy fuel oil gasifier.

1 tonne = 1016 kg

1.9 tonnes = (1016 x 1.9) kg = 1930.4 kg

Mass of heavy fuel oil feed = 4135.535 kg/day

1930.4 ——————————- 4.5m

4135.535 ————————— = 9.64m

and,

V = π

Therefore,

D = = = 1.99 m

So, in summary:

Height of the gasifier = 9.64 m

Diameter of the gasifier = 1.99 m

Total molar flow rate of the exit crude gas = 1587.17 kmol/hr

Volumetric flow rate of the exit crude gas =        =3347.88 m³/hr

Residence time, τ = Volume/Volumetric flow rate = = 0.0089 hr = 32.13s

Space velocity, υ = 1/ τ = 1/32.13s = 0.031 s^-1

 

MECHANICAL ENGINEERING DESIGN

 

DESIGN PRESSURE

DESIGN TEMPERATURE

MINIMUM WALL THICKNESS

The minimum thickness required to resist the internal temperature of the gasifier is calculated thus,

MINIMUM HEAD THICKNESS

An Ellipsoidal head is chosen as the closure for this vessel because it is relatively cheaper than the torispherical head and it can withstand the design pressure of the vessel (Coulson and Richardson, 1999). The minimum thickness required is obtained thus,

Thickness of the head = 92.2 mm

 

THE WASTE HEAT BOILER

The calculations leading to the determination of in the above equation are shown below:

C_p-values are dependent on temperature, so we have to determine the C_p of all the individual components that make up the inlet stream (i.e. at 1300°C)

The heat capacity equations are:

C:        C_p = 11.18 + 1.095 × 10^-2 – 4.891 × 10^{5 -2}

H₂:       C_p =

CO:      C_p =

CO₂:     C_p =

N₂:       C_p =

CH₄:     C_p =

H₂S:     C_p =

Using these C_p values, we can calculate a bulk C_p for the stream as shown:

BULK C_P VALUES FOR STREAM COMPONENTS

COMPONENT	Cp (J/KMOLOC)	MOLE FRACTION	MOLAR FLOW	IN-STREAM Cp (J/KMOLOC)
C	28.207	0.00362	5.157	0.2089
H2	32.585	0.47191	671.636	13.2328
CO2	58.747	0.08229	117.113	4.1593
CO	35.412	0.41739	594.031	12.7199
N2	35.222	0.01388	19.754	0.4191
CH4	135.761	0.00099	1.411	0.1222
H2S	51.4787	0.00496	6.702	0.0360

From the addition of all the values in the in-stream C_p column, we obtain:

C_p  = 35.92522 J/mol. °C

Substitute the value of C_p  above into Equation (2.1):

From the material balance calculations, we can see that

The next thing to do is try and obtain an overall heat transfer coefficient for the heat exchange process.

As mentioned previously, for a water tube waste heat boiler, the demineralised water supplied as utility goes in on the tube side while the hot process gases go in on the shell side. The configuration of the boiler and the fluids flowing in it determine the form that the overall heat transfer coefficient will take. For the water tube configuration:

Where:

f_s= steam fouling

f_g = gas fouling

d_o= external diameter

d_i  = internal diameter

k_{m =} thermal conductivity of tube (carbon steel)

h_i = inside heat transfer coefficient

h_o = outside heat transfer coefficient

From literature, heat exchanger tube diameters are specified between 16mm and 50mm. Waste heat boilers use tube internal diameters that are much closer to the upper limit than conventional heat exchangers. For this design:

Tube outer diameter,

Tube inner diameter,

h_i=

where

                                       =

=

=  90.855

                                h_i=

=  1309379.395 W/m²k

For gas streams, fouling factors range from 0.0002 to 0.0005. A factor of 0.0005 is selected as the design gas fouling factor so as to prepare for the highest possible degree of fouling.

For steam streams, fouling factors range from 0.0002 to 0.00067. A factor of 0.00067 is selected as the design gas fouling factor so as to prepare for the highest possible degree of fouling.

Steam outside heat transfer coefficient is:

h_o = 2000Btu/ h_o ft^2oF

= 11333.33W/ m^2oC

= 4.088e ⁴ KJ/hr. m^2oC

The thermal conductivity of carbon steel at the highest temperature quoted in literature is 47 W/m^{2 º}C

Substituting the aforementioned values into Equation 2.3, we obtain:

To obtain the heat transfer area, we need to first obtain the mean temperature difference between the inlet and the exit gas streams. Since there is a change of phase in the waste heat boiler, the standard log mean temperature difference cannot be applied, so a weighted mean temperature difference will be applied:

Recall that,

To obtain the flow rate of water required for the heat transfer, we need to first obtain the latent heat of vaporization of water at the design pressure specified.

From steam tables, we can obtain the enthalpy of saturated steam at the design pressure of 4140KN/m² using interpolation between values from the said steam table.

Enthalpy of saturated steam, at 254^oC:

ENTHALPY OF SATURATED STEAM

2801.5	250
	254.9
2785	275

Enthalpy of saturated liquid (water) at 254.9 ^oC:

ENTHALPY OF SATURATED LIQUID

1085.36	250
	254.9
1210.07	275

Latent Heat of vapourization,

=

= 1710.686 KJ/Kg

To obtain the water flow rate:

Heat loss by gas,

Q = m

Using the value of  obtained above, we can substitute into Equation 2.5 and obtain the mass flow rate of water, shown below:

= 31382.732 Kg/hr

By rule of thumb, design tubeside flow rates, for waste heat boilers usually range between 45kg/hr and 90kg/hr. Assuming    for this design:

Where

N_t = number of tubes

W = gas flow rate

w = tubeside flow rate

For the design, we select the number of tubes as 490 tubes

To obtain the tube length required,

Tube Side Area =

Using Equation 2.7, we can obtain the tube length required for the heat transfer.

To obtain the Tube bundle arrangement:

A triangular tube arrangement gives higher heat transfer rates.

Tube Pitch,

P_i = 125 d_o

= 1.25 (0.0508)

= 0.0635m

Tube bundle diameter is given by,

Where N_t  is the number of tubes

k_i and x_i, are constants dependent on tube arrangement

D_b = bundle diameter,

d_o = outer diameter of tube

=  1.58m

To obtain the clearance and hence the shell diameter, we refer to graph of clearance correlations to obtain correlated values of clearance and area. From the graph, we can see that our bundle diameter is out of the range for which a clearance value can be obtained, so we interpolate for its corresponding clearance value as shown overleaf:

CORRESPONDING AREA-CLEARANCE VALUES

AREA	CLEARANCE
1.2	77
0.85	70
1.58	x

To obtain the linear velocity of flow through the tubes:

For a two- pass flow regime in the boiler, the flow area will be about 480.36m²

= 5.58m/hr

To obtain the pressure drop (shell side):

(0.0936bar)

To obtain the pressure drop (tube side):

  (0.177bar)

 

 

 

QUENCHER

INLET STREAM INTO QUENCHER

COMPONENT	MOLE (Kmol)	MOLE FRACTION
C	5.16	0.00364
H₂	671.636	0.4743
CO	594.031	0.4195
CO2	117.113	0.0827
N₂	19.754	0.01395
CH₄	1.411	0.000996
H₂S	7.055	0.00498
	1416.16

HEAT CAPACITY

The heat capacity data can be obtained from heat capacity tables in available from literature

Taking a reference temperature of 15^oC.

The heat capacity equations are:

C:  Cp = 11.18 + 1.095 × 10^-2T – 4.891 × 10⁵T^-2

H₂: Cp =

CO: Cp =

CO₂: Cp =

N₂: Cp =

CH₄: Cp =

H₂S: Cp =

Multiplying these equations by the respective mole fraction of each component yield we have:

C: Cp × mass in terms of moles

= Cp × 0.00364

= 0.0407 + 3.986 × 10^-5T– 1780.324 T^-2

H₂: Cp × 0.4743

= 13.679 + 3.628 × 10^-5 T + 1.559 × 10^-6 T² – 4.125 × 10^-10 T³

CO: Cp × 0.4195

= 12.145 + 1.724 × 10^-3 T + 1.488 × 10^-6 T² – 9.23 × 10^-10 T³

CO₂: Cp × 0.0827

= 2.986 + 3.5 × 10^-3 T – 2.388 × 10^-6 T² + 6.173 × 10^-10 T³

N₂: Cp × 0.01395

= 0.405 + 3.068 × 10^-5 T + 7.984 × 10^-8 T² – 4.01 × 10^-11 T³

CH₄: Cp

=

H₂S:  Cp

=

Summing the heat capacity values for each component, we have

Cp_net = 29.457 + 5.462 × 10^-3T + 7.575 × 10^-7 T² – 7.856 × 10^-10T³ – 1780.324 T^-2

Heat lost by the crude gas, ΔH is given by

_net dT

ΔH = + 5.462 × 10^-3T + 7.575 × 10^-7 T² – 7.856 × 10^-10T³ – 1780.324 T^-2) dT

= 29.457 (250 – 15) + 0.002731 (250² – 15²) + 2.525 × 10^-7 (250³ – 15³) –1.964 ×10^-10                   (250⁴ – 15⁴) + 1780.324 (250^-1 – 15^-1)

= 6,984.618 KJ/Kmol

 

 

COMPOSITION OF OUTLET STREAM FROM THE QUENCHER

COMPONENT	MOLE (Kmol)	MOLE FRACTION
H₂	671.636	0.476
CO	594.031	0.421
CO₂	117.113	0.083
N₂	19.754	0.014
CH₄	1.411	0.001
H₂S	7.055	0.005
	1411

 

The heat capacity equations are:

H₂: Cp × 0.476

= 13.728 + 3.64 × 10^-5 T + 1.57 × 10^-6 T² – 4.14 × 10^-10 T³

CO: Cp × 0.421

= 12.19 + 1.73 × 10^-3 T + 1.494 × 10^-6 T² – 9.26 × 10^-10 T³

CO₂: Cp × 0.083

= 2.997 + 3.513 × 10^-3 T – 2.396 × 10^-6 T² + 6.195 × 10^-10 T³

N₂: Cp × 0.014

= 0.406 + 3.08 × 10^-5 T + 8.012 × 10^-8 T² – 4.019 × 10^-11 T³

CH₄: Cp 0.001

=

H₂S:  Cp 0.005

=

Summing the heat capacity values for each component, we have

Cp_net = 29.523 + 5.442 × 10^-3T + 7.668 × 10^-7 T² – 7.881 × 10^-10T³

Heat lost by the crude gas, ΔH is given by

_net dT

ΔH = + 5.442 × 10^-3T + 7.668 × 10^-7 T² – 7.881 × 10^-10T³) dT

= 29.523 (50 – 15) + 0.002721 (50² – 15²) + 2.556 × 10^-7 (50³ – 15³) – 1.97 × 10^-10      (50⁴ – 15⁴)

= 1039.525 KJ/Kmol

ΔH_net = (6984.618 × 1416.16) – (1039.525 × 1411) KJ/hr

= 2,340.157 KJ/s

Heat gained by the water = m

C_p for liquid water = 18.2964 + 47.212 x 10^-2 T – 133.88 x 10^-5 T² + 1314.2 x 10^-9 T³

Therefore,

Heat gained by water =

= 18.2964 (363 – 298) + 0.23606 (363² – 298²) – 4.463 × 10^-4 (363³ – 298³) + 3.286 ×10^-7            (363⁴ – 298⁴)

= 1189.266 + 101423.318 – 9536.786 + 3114.116

= 96,189.91 KJ/Kmol

Thus, Heat gained by water = (m/M x 96189.91KJ/Kmol)

Molecular weight of water = 18 Kg/Kmol

Heat gained by water = 96189.91/18 m = 5343.88 m

Since, heat lost by the crude gas = heat gained by water

Therefore,

5343.88m = 2340.157

m = 0.438 Kg/s

Allowing a minimum of 10 minutes hold-up of water,

Therefore, m = 0.438 Kg/s x 10mins x 60s = 262.748 Kg

COLUMN DIAMETER

The equation for the Column Diameter, from Handbook of Chemical Engineering Calculation is given as,

D = (4S/π)^1/2

Where,

S is the actual cross sectional area, which in terms of volumetric gas flow rate is

S = q/V

Where q = volumetric gas flow rate of inlet stream into the quencher

V = assumed superficial velocity (between 1 and 2 m/s)

To obtain the volumetric gas flow rate we have,

PV = nRT

Therefore, n = 1416.16 Kmol/hr

=  Kmol/s

q = (16.23 × 8.314 × 523) / (4140 + 101.325)

= 0.3934 m³/s

V = assumed to be 1.8 m/s

S = q/V

= 0.3934/1.8 = 0.2185 m²

Therefore,

D = (4S/π)^1/2

= (4 × 0.2185/3.1429) ^½

= 0.5275 m

 

HEIGHT OF THE COLUMN: Z

This is obtained, from Coulson and Richardson, using the following formula:

Z= H_OGN_OGS.F

Where,

H_OG assuming 3.0in of ceramic packing material, H_OG= 4.5ft

4.5 ft

=1.3716m

Assuming a Safety Factor of 1.5,

S.F= 1.5

N_OG is given by,

N_OG=ln (y₁/y₂)

Assuming 60% reduction in concentration of H₂S

Where,

y1= concentration of H₂S in entering gas stream= 0.00498Kmol

y2=concentration H₂S in exit gas stream=0.00199kmol

N_OG = ln (0.00498/0.00199)

= 0.9173

Using a 3 inch (76.2 mm) ceramic diameter packing, the corresponding height of transfer unit, H_OG = 4.5 ft

The height, Z = (H_OG) x (N_OG) x (SF)

Using a safety factor, SF of 1.5 for the quencher

Therefore,

Z = (4.5 x 0.3048) x (0.9173) x (1.5) = 1.887 m

Converting mass of water Kg to m³

V = (mass of water in Kg) ⁄density of water

V = 262.748/1000

= 0.2627 m³/s

Height in packing H = V/S

= 0.2627/ 0.2185= 1.2023 m

Height of entire column = 1.887 + 1.2023 m

= 3.0893 m

MECHANICAL ENGINEERING DESIGN

DESIGN PRESSURE

Assuming 8% above operating pressure

= 1.08 x 4241.325 KN/m²

= 4580 .631 KN/m²

(4580.631-101.325) gauge

= 4479.306 kN/m² gauge

DESIGN TEMPERATURE

This is obtained as,

= Highest Temperature + allowance for uncertainty

Assuming an uncertainty of 80^oC

Therefore, design temperature= (250+ 80) ^oC

=330^oC

 

TENSILE STRENGTH

At 25^oC, the tensile strength of Carbon =210N/mm²

At 500^oC, the tensile strength of Carbon is 450 N/mm²

At 330^oC, the tensile strength of Carbon will be,

TENSILE STRENGTH VALUES AT DIFFERENT TEMPERATURES

TEMPERATURE, OC	TENSILE STRENGTH, N/MM2
25	210
330	X
500	450

Interpolating we have,

(330 – 25) / (500 – 25) = (x – 200) / (450 – 210)

x=154+200

=  354N/mm²

DESIGN STRESS:

At 300 ^oC, the design strength of Carbon =85N/mm²

At 350 ^oC, the design strength of Carbon is 80 N/mm²

At 330 ^oC, the design strength of Carbon will be,

(COULSON AND RICHARDSON, VOL 6)

DESIGN STRESS VALUES AT DIFFERENT TEMPERATURES

TEMPERATURE, OC	DESIGN STRESS, N/MM2
25	85
330	X
500	80

Interpolating we have,

(330 – 300) / (350 – 300) = (x – 85) / (80 – 85)

X = τ = 82 N/mm²

Nominal design stress = 1.5 × = 1.5 x 82

= 123 N/mm²

CHOICE OF CLOSURE

Minimum wall thickness required for a cylindrical shell

e= PiDi / 2f – Pi

Given that,

P_i = 4241.325 KN/m²

D_i = Internal diameter = 0.5275 m

Therefore,

E = (4241.325 × 0.5275)

2 × 82000 – 4241.325

= 14 mm

Corrosion allowance = 2 mm

Minimum thickness of wall required = 14 mm + 2 mm = 16 mm

Choice of closure = Hemispherical head

Optimum thickness ratio for a hemispherical head to the thickness of the vessel = 0.6 (Coulson and Richardson, 2005)

Thickness of the hemispherical head = 16 x 0.6 mm = 9.6 mm

Type of Vessel Support = Skirt Support

 

INITIAL  H₂S REMOVAL

ASSUMPTIONS

• No  NMP  in  Output  gas  stream
• Unit  operates  at  50%  of  the  flooding  gas  mass   velocity
• Flow rate  of  NMP  is  40%  more  than  the  minimum

Given that

Henry’s law constant for NMP = 0.97

Density of gas stream = 31.49kg/m³

Density of NMP = 1020kg/m³

H₂S concentration in the inlet stream = 0.5mol%

NMP has already been selected from the first semester preliminary design as our scrubbing liquid.

An energy balance is not require for this unit because there is very little temperature change between the unit’s inlet and exit streams and hence the unit can be assumed to be operating isothermally.

A packed column using 1 in. Raschig ring packing with a packing factor of 160 have been selected for use for this design. Raschig rings have been selected because they give a larger degree of contact between the gas phase and the liquid phase. It also gives a higher mass transfer area to work with.

To  calculate  the   height  and  diameter of  the  column  to  be  used  for  this  absorption  process, we  need  to  determine  the  overall  gas  transfer  units, N_OG, as  well  as  H_OG.

To obtain Equilibrium outlet concentration,  x₁  at  y₁ = 0.5,

Where  M=Henry’s  law  constant  for  NMP

Concentration  Y₂  for   99.9797%  removal  of  H₂S from  the  inlet  stream:

,  Substitute = 0.5,

=

= 0.00020296

The ratio of the flow rate of NMP to the inlet gas flow rate, is estimated as

= 1.4

Substituting the values previously known and obtained into Equation 3.3, we obtain

= 1.4

= 1.3574

the expression for calculating is shown below:

Where A = absorption factor

But

Where m is the slope of the equilibrium curve, assuming it approaches zero.

= 1.4

Substituting this value into Equation 3.4 along with all equilibrium concentrations, we obtain:

= 22.593

From the value calculated above, we can go further to calculate the height of the column, h

Therefore, the column height that satisfies this design is approximately 46ft (14m).

Given that the physical characteristics of NMP are similar to those of water, some correlations of water systems can hence be used in calculations of NMP systems:

To obtain the flooding gas mass velocity,:

  (=)

Where

= viscosity of NMP                                  = raschig ring packing factor

= density of NMP

= density of gas stream

= 6.274

To obtain the actual gas velocity (assuming operation at 50% flooding):

(11293)

Hence, we can obtain the column diameter as

= 2.44m

In summary:

Column height = 14m (approx.)

Column diameter = 2.44m (approx.)

MECHANICAL DESIGN

Shell Thickness

Thickness of shell,

where,

Inner diameter of vessel (D) = 1.3 m

Working pressure = 1.013 ×10⁶ N/m²

Design pressure (P) = 1.1× 1.013 x 10⁶ N/m²

Permissible Stress (f) = 115 × 10⁶ N/m²

Joint Efficiency (J) = 0.85

Corrosion allowance (c) = 2mm

Hence,

So outer diameter of shell

3.                Axial Stress Due to Pressure

Axial stress due to pressure ,

4.                Stress due to Dead Load

5.                 Compressive Stress due to weight of shell up to a distance X

     

 

6.                 Compressive stress due to weight of insulation at height X

Insulator used is asbestos

Thickness of insulation, = 100mm

Diameter of insulation,

Density of insulation = 575 kg/m³

Mean diameter of vessel =

For large diameter column,

7.                Compressive stress due to liquid in column up to height X

8.                Effect of attachments

7. Packing weight
8. Head weight
9. Ladder

Density of packing (Plastic pall rings) = 53 kg/m³

Packing weight =

= 690.11 X N

Head weight (approximately) = 35000 N

Weight of ladder = 1600 X N

5.                SUPPORT FOR ABSORBER

Material used is structural steel (IS 800)

Skirt support is used

Inner diameter of the vessel = D_i = 4.69m

Outer diameter of the vessel = D_o = 4.70m

Height of the vessel = H = 10.054m

Density of carbon steel = = 7700kg/m³

Density of packing = = 53 kg/m³

 

 

7.7.2        CARBON MONOXIDE CONVERSION UNIT

Catalyst Volume (from space velocity):

The catalyst volume can be obtained from space velocity data given in the problem statement coupled with the simulation figures for the flowrate of gases into the catalytic converter.

FOR 1^ST STAGE:

 Catalyst Mass (in 1^st stage of converter):

Thus,

FOR 2^ND STAGE:

Catalyst Mass (in 2^nd stage of converter):

CATALYTIC CONVERTER SIZING

In carrying out the sizing of Fixed Bed Reactors, most designs approximate to plug flow. The simplest of such arrangements is adiabatic (Chemical Process Design and Integration by Robin Smith).

With the assumption of an elementary reaction, the rate of reaction can be proposed to be:

Inhibiting the undesired reverse reaction, leading to the elimination of the second term of the rate expression can be done by running the reaction at an appropriate temperature, removing product from the reaction vessel shortly after formation and operating at certain pressures.

Thus, this results into:

In terms of mole fraction of reactants entering into the reactor,

This provides grounds for assuming that is so large, it would relatively, be constant so that:

where,

= Specific reaction rate (constant) for component CO [=] s^-1 or h^-1

=

Reaction rate [=] kmol.m^-3.s^-1 or kmol.m^-3.h^-1

For a tubular flow reactor such as the one employed in the catalytic converter, the following performance equation is developed:

As a result,

substituting, we have:

since is constant,

Recalling that,

Thus,

Taking C_COo which is a constant into consideration, we have:

Thus, integrating the above expression, we have

where, c is an integration constant

Recall,

substituting V_o into the expression for V,

In the above expression, is the volumetric flowrate of the entering stream into the FIRST stage, and it is equal to .

Working with a basis of about 80% conversion in the first stage of the catalytic converter, X = 0.8

Explanations ensued on the appropriate to be used make way for literature data. The Howard Rase’s data from Girder which states that for the Iron – Oxide catalyst, Fe₃O₄

With the temperature being equal the equilibrium temperature in ⁰R of the catalytic converter stage in question, the temperature from the simulation is equal to 380.8which is equal to 1174⁰R

For the second stage, the expression translates into:

In the above expression, is the volumetric flowrate of the entering stream into the SECOND stage, and it is equal to .

Working with a basis of about 95% conversion in the second stage of the catalytic converter,  X₂ = 0.95 and X₁ = 0.80 in the first

Back to the rate constant dependent on temperature,

With the temperature being equal the equilibrium temperature in ⁰R of the catalytic converter stage in question, the temperature from the simulation is equal to 370.0which is equal to 1158⁰R

Thus, the following are the calculated reactor volumes:

1^st Stage of reactor  85.50

2^nd Stage of reactor 142.06

Standard reactor sizing reflected from Perry’s Handbook of Chemical Engineering reveal that for a standard plug flow tubular reactor,

will be assumed for this reactor

For a cylinder, volume,

For the first stage,    

This makes the height equal to m

Working with the reasoning that both stages of the reactor have to have the same diameter,

For the second stage, 3.518 m

From     ,

For the Height of catalyst bed:

1^st stage,

,

2^nd stage,

Type of vessel head

The vessel head indicates the closure. Ellipsoidal heads will be made use of. The ellipsoidal head thickness can be calculated using the following formula:

where,  P_i – is the internal pressure; = 1200 kPa =1.2 Nmm^-2

D_i – the internal diameter; = 3.518 m

 J – the joint factor; which is equal to 1 as a result of the absence of joints

 f– the design stress at the specified temperature, equal to 180 N/mm²

CHEMICAL ENGINEERING DESIGN (FINAL H2S REMOVAL)

The following was specified

Determination of number of holes

Using the correlation

A_p = perforated area = tray area

d_h = hole diameter

L_p = hole pitch = 3 for a triangular pitch

A_h= total hole area = n

A_P=,        d_{h =}5 x 10^-3

=4.9087m²

A_h =n

=0.000019634nm²

= 0.9

=0.9

=0.1

=

n=

=25001 holes

(e) Determination of the mass of lautamasse and sulphur.

The general equation for the thickness of a flat plate required to resist a given pressure load can be written in the form.

t = CD

=the maximum allowable stress ( the design stress)

D= the effective diameter

C= a constant, which depends on the edge support

T=0.43

Assuming a thickness of 10mm

10=0.43

=

P=0.00007684/m

P=20768N/

Using p=

Force=P

A=

=

=4.9087m²

F = 20768N/m ² x 4.9087 m²

=101945N

Mass =

= Kg

= 10402.6 Kg

When the quantity of sulphur deposit on the tray falls between 40-55% of the weight of the lautamsee the operation has to be stopped and then regeneration takes place.

Assuming the mass of sulphur and the lautamase is the ratio 1:0.4 that is assuming that sulphur is 50% of the total mass mixture.

Mass of sulphur= x 10402.5 kg

=3467.5kg

Mass of fe₂0_{3 =}

=6935kg

To calculate the time for regeneration:

Equation of reaction

2Fe₂0_3(s) + 6H₂S_(g)                2Fe₂S₃ + 6H₂0  (1)

2Fe₂0₃ + 30_2(g)   2Fe₂0₃ + 6S_(g)                   (2)

From the reaction

3 moles of H₂S(g) produced  6 moles of sulphur.

Number of moles of the fed =

=

=108.359kmol

Number of moles of the H₂S required to produced 2108.359 of H₂S is 108.369kmol from material balance

Molar flowrate of H₂S intel =0.0211kmol/hr

Molar flowrate of the outlet =1.06583 xkmol/hr

Amount of H₂S absorbed in flow rate

= (0.02011-1.0658x kmol/hr

=0.01904417kmol/hr

Assumption

The absorption of H₂S in the four columns are 50%, 30%, 20%, 10%. The assumption is based on the fact that absorption decreases along the column. But we want to model the column base on the absorption of the first column.

Time after which regeneration will take place is t

==

=11379.67hr

The molar flow rate (0.009522085kmol/hr) was calculated by multiplying 0.01904417kmol/hr x 50%

But 24hr = 1day

11379.76hr = x Days

x days=

=474.156 days

30 days=1mol

474.156 days =p mols

P =

=15.8months

16months

After 16 months regeneration process has to be carried out.

volume and height of sulphur and lautamasse determination.

Mass of sulphur = 3467.6kg

Density of sulphur = 2000kg/m³

Volume of sulphur =

=

Height =

=

= 0.3532m

Lautamasse;

Density of lautamasse = 675kg/m³

Mass of lautamasse = 6935kg

Volume =

= 10.274m³

Height =

= 2.0930m

This is a good value since the higher the height of the lautamasse, the lesser the effective tray spacing.

Total height of lautamasse and sulphur

= 2.0930 + 0.3532

= 2.45m

 

REGENERATION

Sulphur deposit after 16months per column = 3467.5kg 5

= 17337.5kg

Amount in moles=

= 541.797 moles

S + O  SO₂

541.8 kmol of O₂is required per column.

Since each column is designed assuming 50% of the total H₂S absorbed

(4 541.797kmol)SO₂ = 150% of the actual SO₂ produced

Also, (4 541.797kmol)SO₂ = 150% of the oxygen required

The actual SO₂ produced = 1444.8kmol

The actual O₂ required = 1444.8kmol

If regeneration is assumed to take place within 6hrs,

Flow rate of SO₂ regenerated = 240.8kmol/hr

7.7.3        CARBON DIOXIDE REMOVAL

Mechanism of Removal: Absorption

Absorption Mechanism: Chemosorption(Chemical Absorption)

Equation of Reaction

K₂CO_3(aq) + H₂O_(l) + CO_2(g)  2KHCO_3(aq)

CHEMICAL ENGINEERING DESIGN

Basis : 4125.535 kg of Feed / hr

COMPOSITION OF INLET AND OUTLET STREAMS FOR CO₂ REMOVAL

COMPONENT	yin	yout
H2	0.63326	0.9800
CO	0.00060	0.00093
C02	0.35555	0.00247
N2	0.00989	0.0155
CH4	0.00071	0.001104
H2S	5.33E-07	8.34E-07
TOTAL	1.000000	1.000000

ASSUMPTION: Only Carbon dioxide (CO₂)is absorbed and all the other gases act as inert in the 40% K₂CO₃ solution.

 

MOLE RATIO OF CO₂ IN THE GAS ENTERING THE COLUMN

 

MOLE RATIO OF CO₂  IN THE GAS LEAVING THE COLUMN

FLOW RATES OF GAS ENTERING = 1996.789 kmol/hr

FLOW RATES OF LIQUID LEAVING = 1277.92 kmol/hr

Typical absorber showing inlet and outlet gas and liquid streams

MAKING AN OVERALL MATERIAL BALANCE

Equation 1

Equation 2

Rearranging,

Equation 3

= flow rate of inert in gaseous stream

= flow rate of liquid stream

Solving Equation 2,

  as the entering  K₂CO₃ contains no CO₂

Plotting the equilibrium curve for CO₂

EQUILIBRIUM DATA FOR CO₂

	X	Y	X	Y
0.025	0.1	0.0017	0.111111	0.001703
0.18	0.2	0.0122	0.25	0.012351
0.6	0.35	0.0408	0.538462	0.042535
1.7	0.5	0.1156	1	0.13071
4.0	0.68	0.2722	2.125	0.374004
7	0.75	0.4762	3	0.909126
10	0.8	0.6803	4	2.127932
11	0.9	0.748	9	2.968254

STEPS INVOLVED IN OBTAINING

1. Plot a graph of X against Y
2. Trace the value of Y₁ across the graph and note the point of intersection with the equilibrium line.
3. Draw a line from Y₂to the point of intersection
4. Then find the slope of the line which is  .

From the graph,the value of = 0.2208

The calculated above is the amount of K₂CO₃ only, the amount of the whole solution

To find the ,

 

COLUMN DIAMETER CALCULATION

Gas Mass Flow rate at the bottom of column ( =

Liquid Mass Flow rate at the bottom of column ( =

The density of the gaseous feed,

(density of liquid )of 40% K₂CO₃ at temperature of 388.46K = 1359kg/m³ (Perry)

Also from data  viscosity of 40wt% K₂CO₃ at 388.46K = 0.9cP (Kohl)

 

 

PACKING MATERIAL: Pall Rings (Plastic)

Pall rings are chosen as the F_p (Packing Factor), small column diameter , high quality with low price, low pressure drop.

The packing characteristics

D_p = 90mm

% = 92%

F_p = 56m^-1 (17.0688ft^-1)

  Packing Height

Flow Parameter,

Using Figure 14.48 (Perry) for flow parameter of 0.0975 and of   Packing Height

Where

The Column Diameter is 1.3m

Height calculations

H_G (Height of the gas film transfer unit)

H_L (Height of the liquid film transfer unit)

Determine the values of the parameters S_cv, S_cl, L_w^×, f₁, f₂, f₃, k_3, y_h, f_h

Where,

S_cv = vapor phase Schmidt number

S_cl = liquid phase Schmidt number

L_w^×=liquid mass flow rate per unit column cross section

f₁ = liquid viscosity correction factor

f₂ = liquid density correction factor

f₃ = surface tension correction factor

Surface tension of water = 72.8 dynes/cm

Surface tension of liquid () = 91.4 dynes/cm

At assumed percentage flooding,

= = 11.8 kg/ m.s

At   ;

Also at assumed percentage flooding,

Take diameter correction term ^1.11 = 2.3

Estimate Z as 26m

Solving for

Solving for H_L

= 0.457

Colburn has suggested optimum values of to lie between 0.7 & 0.8

Therefore,

= 0.96

Now,   Column height= 0.96 x assumed Z

= 0.96 x 26

= 24.96

AUXILIARY EQUIPMENT: HEAT EXCHANGERS

BASIC DESIGN PROCEDURE

FIRST HEAT EXCHANGER DESIGN

STEP 1: SPECIFICATION

The process stream flows into the first heat exchanger, and it is required to raise the temperature of the unconverted gas stream from the saturator by heating against converted gas from the second stage of CO shift conversion reactor. The physical properties of the fluids are as presented in step 2 below.

Duty = 

= 901.1 kW

The same amount of heat duty transferred to the shell side

 

 

STEP 2: PHYSICAL PROPERTIES

Gas from second stage of catalytic converter

Inlet   Mean   Outlet

Flow           kg/hr             984100  984100  984100

Pressure  kPa   700   500   300

Temperature                ^oC             373.4   371.05     368.7

Specific Heat               KJ/kg ^oC 2.090              2.080   2.070

Thermal Conductivity mW/m ^oC    54.82              66.156   53.95

Density  Kg/m³            14.52              14.40   14.28

Viscosity  μNsm^-2  16.80              17.51   18.22

Gas from unconverted gas stream

Inlet   Mean   Outlet

Flow           kg/hr             29290   29290   29290

Pressure  kPa   3000       2750   2500

Temperature                ^oC             163.5   181.75     220.0

Specific Heat               KJ/kg ^oC 1.965              2.1472              1.965

Thermal Conductivity mW/m ^oC    75.34              79.851   86.28

Density  Kg/m³            13.83              13.60   13.37

Viscosity  μNsm^-2  17.56              17.55   17.54

 

STEP 3: OVERALL COEFFICIENT

For a gas-gas exchanger of this type, the overall coefficient, as suggested by literature, should range between 50 and 300W/m^{2 o}C (light oils).

 

STEP 4: EXCHANGER TYPE AND DIMENSIONS

An even number of tube passes is usually the preferred arrangement as this places the inlet and outlet nozzles at the same end of the exchanger, which simplifies the pipe work. Nonetheless, we use one shell pass and one tube pass.

                                                                                                                                F_t (correction factor) = 0.97

Assume U = 120W/m² °C

 

STEP 5: HEAT TRANSFER AREA

 

STEP 6: LAYOUT AND TUBE SIZE

A split-ring floating head exchanger is to be used for efficiency and ease of cleaning. Both shell and tube fluids are corrosive and the operating pressure is high, so a stainless steel can be used for the shell and tubes. Since the unconverted gas stream contains more CO than H₂, and the converted gas contains more H₂ than CO, thus the unconverted gas stream is more corrosive  due to the more corrosive nature of CO, and as such, the unconverted gas stream is passed through the tubes, and the converted, through the shell.

The following standard steel tube dimensions are to be used: 30mm outside diameter, 26mm inside diameter, 4.83m long tubes. A popular triangular pitch is to be used in the design, with pitch-to-diameter ratio = 1.25.

STEP 7: NUMBER OF TUBES

Area for heat transfer of one tube (External curved-surface area of tube sheet)

= 

Number of tubes = , Say 96

So, for 1 passes, tube per pass = 96

Tube internal cross-sectional area =

So area per pass =

Volumetric flow = 0.6300 m³/s

Tube side velocity, u_t =

This velocity is sufficient for tube operation (reasonable gas velocity should be from 15 to 30 m/s)

 

STEP 8: BUNDLE AND SHELL DIAMETERS

Selection Parameters for the Design of Tube Bundle Diameter (Sinott, 2005)

Where N_t= number of tubes

            D_b= bundle diameter, mm,

            d_o = tube outside diameter, mm .

From table above, k₁=0.319, n₁=2.142

D_b =

SHELL-BUNDLE CLEARANCE (SINOTT, 2005)

For split ring, floating head exchanger, the shell clearance from above figure is 51mm

= 426.5 + 58 = 484.5mm

 

 

 

 

STEP 9: TUBE-SIDE HEAT TRANSFER COEFFICIENT

  =

=

TUBE-SIDE HEAT-TRANSFER COEFFICIENTS

From FIGUREbelow j_h =

Neglecting viscosity correction factor,

STEP 10: SHELL-SIDE HEAT TRANSFER COEFFICIENT

Kern’s method will be used

The shell diameter as calculated is;

D_S=484.5 mm

As a first trial, take the baffle spacing =

=96.8mm

Calculating the area of cross flow A_sfor the hypothetical row of tubes at the shell equator, given by:

    

Also shell equivalent diameter (d_e) (hydraulic diameter), for triangular pitch is given by

 

Where P_t= tube pitch,

            d_o = tube outside diameter,

            D_s= shell inside diameter, m,

           l_B= baffle spacing, m.

=0.0180m²

=21.30mm (to be used to calculate the Reynolds number Re)

Shell velocity, u_s

=

= 34.41m/s

In using segmented baffles with a 25% cut, this should give reasonable heat transfer coefficient without too large a pressure drop.

SHELL-SIDE HEAT-TRANSFER COEFFICIENTS

From the above figure, j_h =

= 451.91

 

STEP 11: OVERALL HEAT TRANSFER COEFFICIENT

149.85

The calculated overall coefficient deviates from the assumed 120 value within 25%; as such, the equipment so designed has adequate area to handle the required duty.

STEP 12: PRESSURE DROPS CALCULATIONS

TUBE SIDE

Where  = tube-side pressure drop, N/m² (Pa),

               N_p = number of tube-side passes,

                u_t = tube-side velocity, m/s,

              L = length of one tube.

SHELL SIDE

where L = tube length,

          l_B= baffle spacing.

 

STEP 14: OPTIMISATION

There is scope for optimizing the design by reducing the number of tubes, as the pressure drops are well within specification and the overall coefficient is well above that needed. However, the method used for estimating the coefficient and pressure drop on the shell side(Kern’s method) is not so accurate, so keeping to this design will give some margin of safety.

VISCOSITY CORRECTION FACTOR

The viscosity correction factor was neglected when calculating the heat transfer coefficients and pressure drops. This is reasonable for gases as it has a relatively low viscosity.

Usually,

For such a small factor, the decision to neglect it was justified. Applying the correction would decrease the estimated heat transfer coefficient, which should not be a problem as the area required for heat exchange is well taken care of. It would also give a slight increase in the estimate.

SECOND HEAT EXCHANGER DESIGN

Applying the Above Steps for Heat Exchanger 2,

Fluid Properties:

Gas from first stage of catalytic converter

Inlet   Mean   Outlet

Flow            kg/hr             31220              31220   31220

Pressure    kPa   1200       950   700

Temperature                ^oC             380.9   377.15     373.4

Specific Heat               KJ/kg ^oC 2.1164              2.1033               2.0902

Thermal Conductivity mW/m ^oC    56.076              55.448   54.820

Density   Kg/m³            12.480              14.400   8.7527

Viscosity   μNsm^-2 17.904              17.7529  17.6018

Gas from unconverted stream

Inlet   Mean   Outlet

Flow           kg/hr             29290   29290   29290

Pressure   kPa   2500   1850   1200

Temperature                ^oC             220.0   285.0                350.0

Specific Heat               KJ/kg ^oC 1.9563              1.9665    1.9767

Thermal Conductivity mW/m ^oC    82.68      90.92    99.16

Density  Kg/m³          8.3043              14.400    1.0337

Viscosity  μNsm^-2  17.862              18.876    21.004

STEP 3: OVERALL COEFFICIENT

For a gas-gas exchanger of this type, the overall coefficient will be in the range of 20 to 300W/m^{2 o}C (Gases).

 

STEP 4: EXCHANGER TYPE AND DIMENSIONS

An even number of tube passes is usually the preferred arrangement as this positions the inlet and outlet nozzles at the same end of the exchanger, which simplifies the pipe work. So, we hereby use one shell pass and 4 tube passes.

C

                                      

By reading correction factor from literature with the values of R and S,

Assume U = 60 W/m^{2 o}C

 

STEP 5: HEAT TRANSFER AREA

STEP 6: LAYOUT AND TUBE SIZE

The ratio of H₂O and CO between both streams is still similar as with the first exchanger, and as such, the gas from the combustion chamber is passed through the shell, and the unconverted gas is passed through the tubes, as in the case of the former heat exchanger. Use 20mm outside diameter, 16 mm inside diameter, 6 m long tubes.

A triangular pitch will be used also. Pitch/diameter = 1.25.

 

STEP 7: NUMBER OF TUBES

Area of one tube (Neglecting thickness of tube sheet)

= 

Number of tubes = tubes

So, for 4 passes, tube per pass =327

Tube cross-sectional area   =

So area per pass=

Volumetric flow =

Tube side velocity, u_t =

This velocity is high enough to corrode the tube internal at pressures above atmospheric (reasonable gas velocity should be from 5 to 10 m/s)

 

STEP 8: BUNDLE AND SHELL DIAMETER

k₁ = 0.175, n₁ = 2.285; from Figure 4

D_b=

For split ring, floating head exchanger, the shell clearance from Figure 5 above is 73mm

= 991.39 + 73 = 1064.39mm

 

STEP 9: TUBE-SIDE HEAT TRANSFER COEFFICIENT

  =

=

From Figure 6, j_h = 2.5

=

Neglecting viscosity correction factor,

 

STEP 10: SHELL-SIDE HEAT TRANSFER COEFFICIENT

Kern’s method will be used

The shell diameter as calculated is;

D_S=1064.39 mm

As a first trial, take the baffle spacing =

l_B= 212.88 mm

Calculating the area of cross flow A_S for the hypothetical row of tubes at the shell equator, given by:

    

Also shell equivalent diameter (d_e) (hydraulic diameter), for triangular pitch is given by

,where the parameters are as listed above.

0.0564m²

14.22mm

Shell side velocity at operating condition== 11.15

G_s =

= 0.67

Using segmented baffles with a 25% cut.

From the above Figure 7, j_h =

= 124.93

STEP 11: OVERALL HEAT TRANSFER COEFFICIENT

67.81

The calculated overall coefficient deviates from the assumed 60 value with 13%.

STEP 12: PRESSURE DROP

TUBE SIDE

With  = tube-side pressure drop, N/m² (Pa),

N_p = number of tube-side passes,

u_t= tube-side velocity, m/s,

L =length of one tube.

 

SHELL SIDE

OXYGEN STREAM  PRE-HEATER

Fluid Allocation

Placing the higher temperature stream on the tube side will reduce the overall cost of design. Therefore, the saturated stream is placed in the tubes.

DUTY OF PRE HEATER

Let us assume that the amount of saturated stream used for pre-heating of the O2 stream gives off ‘ONLY’ its Latent Heat to the O2 – stream

Q= (mCpΔT)_O2 = (mL)_stream

At a mass temperature of 115 0c, the Cp of the O2 stream Cp = 0.9818KJ/Kg 0c

Q = 4701.58 kg/hr(0.9818J/Kg 0c)(210 – 20)

= 877.04 ×10³ KJ/hr

=243.622 ×10³ W

MASS OF SATURATED STREAM REQUIRED.

Q= m_sL_S

Where

L_S = Latent Heat of Vaporization of stream at specified conditions (1723KJ/Kg)

243.622×10³ = m_s(1723)

m_S = 877.04 ×10³/1723(KJ/hr)/(KJ/Kg)

= 509.02Kg/hr

= 0.1414 Kg/sec

Log MEAN TEMPERATURE DIFFERENCE (ΔTm)

ΔT_m = ΔT1 – ΔT2/ln(ΔT1/ΔT2)

=(254.9 -210)- (254.9- 20)/ ln[(254.9 – 210)/(254.9 – 20)]

ΔT_m = 114.823 ^oc

From Literature, F= 0.986

Corrected Tm = F(ΔT_m)

ΔT_m= 0.986(114.823 0c)

= 133.215 ⁰c

ESTIMATION OF U (Overall Coefficient)

An estimation of 200W/m²C for the overall Heat Transfer Coefficient is chosen as suggested by literature.

CALCULATION OF PROVISIONAL AREA.

A = 243.622 × 10³W/200(133.215 0C)

= 10.76m²

TUBE DIMENSION SELECTION

• Outer Diameter – 20mm
• Inner Diameter – 16mm
• Length of Tube – 6.10m
• Tube Effective Length- 6.0m

Small diameter tubes are chosen because they give more compact & cheaper exchangers. Use of longer tubes will reduce shell diameter thus lowering cost exchanger.

Heat Transfer Area of 1 tube = πd_oL

= π(20/1000)(6)

= 0.37699m²

NUMBER OF TUBES REQUIRED

Number of tubes required = Provisional area/Area of 1 tube

=10.75m²/0.37699m²

Number of tubes = 28.542

=28 tubes

TUBE ARRANGEMENT SELECTION.

A Triangular Tube Arrangement pattern is chosen as it gives higher heat transfer rates as compared to others. The recommended tube pitch is 1.25 times the Tube Outside Diameter(do). Tube Bundle Diameter is therefore

Nt = K(Db/do)ⁿ

28 = 0.249(Db/20)^2.207

Db= 169.95mm

=170mm

SHELL DIAMETER DETERMINATION

Using a Split Ring Floating Head unit, the Diameter clearance between the Shell and Tube Bundle is 50mm

Therefore,

Shell Diameter = 170 +50 = 220mm

TUBE SIDE COEFFICIENT

A Horizontal Condenser with condensation in the tubes is the usual arrangement for pre-heaters &vaporizers using condensing steam as the heating medium. It is customary to design purposes. For air- free steam, a coefficient of 8000W/m² is used. Therefore,

h_I= 8000W/m²

SHELL SIDE COEFFICIENT.

A_s = [(p_t – d_o)/p_s]D_SL_B

P_t = 1.25(d_o) = 25mm

L_B = D_S          (for Pre-heaters)

A_S = 25-20/25(220×10^-3)²

= 9.68×10^-3m²

 

MASS VELOCITY IN SHELL (G_S)

G_S= W_S/A_S

=4701.58/3600×1/9.68×10^-3

G_S = 134.917Kg/m²s

 

LINEAR VELOCITY (Us)

U_S = G_S/ρ

=(134.917Kg/m²s)/(42.07Kg/m³)

U_S = 3.20m/sec

 

SHELL SIDE EQUIVALENT DIAMETER

For a Triangular pitch Arrangement

d_e = 1.10/d_o[pt²– 0.917d_o²]

= 14.2mm

AVERAGE PHYSICAL PROPERTIES OF SHELL FLUID

• Viscosity (µ) = 2.61 ×10^-5 Pas
• Thermal Conductivity (K) = 3.448 × 10^-2 W/mK
• Heat Capacity (Cp) = 0.9818 ×10³ J/Kg 0C

REYNOLD’s NUMBER ON SHELL SIDE

Re = 42.07 (3.2)(14.2 × 10^-3)/(2.61 ×20^-5)

=73402.99987

PRANDTL’s NUMBER ON SHELL SIDE

Pr = (0.9818 ×10³)(2.61 ×10^-5)/3.448×10^-2

=0.74318

From literature, j_h = 2.3 × 10^-3

h_sd_e/k_f = j_hRePr ^1/3

h_s(14.2 × 10^-3)/3.448×10^-2 = 2.3 ×10^-3(73402.99987)(0.74318)^1/3

h_s = 317.324W/m^{2 0}c

OVERALL COEFFICIENT

Thermal conductivity of cupronickel alloys =50W/mK

Tube side scale Resistance = 0.052 × 10^-3m²K/W

Shell side scale resistance =0.14 ×10^-3 m²K/W

U^-1= (371.324)^-1 + 0.00014+20×10^-3ln(20/16)/2(50)+20(0.000052)/16+20/16(1/8000)

U^-1= 3.0989 ×10^-3

U = 322.691 W/m^{2 0}c

PRESSURE DROP CALCULATIONS

• On the Shell Side

Re = 73402.99987

J_F = 3.6 × 10^-2

ΔP_s = 4(3.6×10^-2)(220/14.2)(6000/220)(42.07×3.20²)

= 26.2112 KN/m² which is very acceptable

• On the Tube Side

The pressure drop on the condensing side or tube side is difficult to predict as 2 phases are present & the vapour mass velocity is changing throughout the tube side.

AIR-COOLED EXCHANGERS

     (1)

Where

T₁= the process side inlet temperature = 325°C

T₂= the process side outlet temperature = 40°C

t₁= the air-side inlet temperature = 25°C

t₂= the air-side outlet temperature = 216.4°C

LMTD = 47.28°C

     (2)

With

μ_o = the viscosity of air at 0°C = 1.78 × 10^-5 kg/m s

μ_t = the viscosity of air at t = 1.8829 × 10^-5 kg/m s

T_t = the set temperature = t = 216.4°C

T_o = the temperature of air at 0°C

Prandtl Number

     (3)

C_p = Specific heat capacity of air = 1.0130 KJ/kg °C

μ_air =Viscosity of air at 216.4 °C = 1.8829 × 10^-5 kg/m s

K = Thermal conductivity of air = 2.7143 × 10^-5 KJ/m °C

Thus, the Prandtl number for air at 216.4 °C = 0.69

Heat Duty Required

     (4)

W_air = mass flow rate of air = 1.443 × 10⁷ kg/hr

t_o =  outlet temperature of the air = 216.4 °C

t_i = inlet temperature of the air = 25 °C

Hence, the heat duty Q of the exchanger = 1.013 KJ/kg °C

Overall Heat Transfer Coefficient for the Design

     (5)

Where

U = Overall heat transfer Coefficient in KJ/hr m² °C

A = Effective heat transfer area in m²

Δt_m = Log mean temperature difference LMTD between the process stream and cooling air

The LMTD and Q values have been calculated as above

Also, A relates to other parameters as

      (6)

With V = volumetric flow of air = 1.2265 × 10⁷ m³/hr, v_air = v_face = velocity of air m/hr

Trials with various values of U

HEAT TRANSFER COEFFICIENTS WITH CORRESPONDING AREA AND AIR VELOCITY VALUES

S/N	U (KJ/hr m2°C)	A (m2)	vair (m/s)
1	6.127 × 104	965.56	3.00
2	8.169 × 104	724.167	4.00
3	1.225 × 105	482	6.00

HEAT TRANSFER COEFFICIENTS WITH CORRESPONDING AREA AND AIR VELOCITY VALUES

In addition,

     (7)

A_face = face area (m²)

CORRESPONDING FACE VELOCITY AND FACE AREA VALUES

vface (m/s)	Aface (m2)
3	1135.96
4	851.97
6	567.91

Substituting W_air in (7) into (4),

     (8)

ρ_air = density of air, 1.1762 kg/m³

So far, for heat exchanger parameters,

U = 1.2254 × 10⁵ KJ/hr m² °C

A = 482.78 m²

v_face = 6 m/s

A_face = 567.91 m²

FAN COVERAGE

   (9)

By standard, it is preferable to fan coverage ratio above 0.40, as a high ratio improves air distribution across the face of the tube bundle, but an excessively high value would imply excessive power input.

Here, we take fan coverage ratio for the design as 0.50

A_face = 567.91 m²

Thus,

Projected Area of Fan = 0.5 × A_face

= 0.5 × 567.91

= 293.96 m²

For the design also, we choose to use two fans rather than one, as this takes care of cooling over a bit of a longer distance of fan stages, and provides insurance in case of failure of one fan or the other.

Fan Diameter

With the two identical fans to be used the diameter is simply computed with the diameter, thus;

Diameter of fan = 13.45 m

General Properties of the Fans to be used as per industrial standards

Fan speeds v_fan = 70 m/s

V-belt drive of about 30 horsepower

Forced draft fan system will be used as it uses less power in blowing cool air volume

Fin Area and Tube Area Analyses

     (10)

     (11)

      (12)

D_fin = Diameter of fin = 12.7 mm

D_e = External diameter of the tube = 41.2 mm

D_i = Internal diameter of the tube = 37.2 mm

W = Width of tube = 4.00 m

L_fin = Length of fin = 68.2 mm

L = Length of tube = 15 m

A_HT = Total heat transfer area per length of tube = 0.01644 m²/m

Number of tubes per row of each metre in a tube bank = n^× =

With the chosen pitch, P = 47.625 mm; n^× = 20.997 ≈ 21 tubes per row

Number of rows n, chosen = 4; and given that W = 4.00 m,

Total number of tubes, N = n^× × W × n = 336 tubes.