Fourth Year - Chemical Process Eng. Branch Petroleum Refinery Eng. Lectures By Dr. Adnan A. Abdul Razak

Transcription

1 Fourth Year - Chemical Process Eng. Branch Petroleum Refinery Eng. Lectures By Dr. Adnan A. Abdul Razak

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3 Petroleum Processing Overview

4 Petroleum Processing Overview Actually, crude oil straight from the ground has some value, but not a lot. Table 1 shows the history of petroleum before Before 1859, oil that was mined or that simply seeped up out of the ground was used to water-proof ships, as an adhesive in construction, for flaming projectiles, and in a wide variety of ointments. After 1859, petroleum became more and more important to the world s economy, so important that today, without a steady flow of oil. It provides fuels and lubricants for our trucks, trains, airplanes, and automobiles. Ships are powered by fuel oil derived from petroleum. Bottom-of-the-barrel petroleum derivatives pave our roads and provide coke for the steel industry.

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6 Petroleum Processing Overview At the end of 2003, the world was consuming 78 million barrels of oil per day. In August 2005, that volume of petroleum was worth $4.6 billion per day, or $1.7 trillion per year. The availability of kerosene got a sudden boost on August 27, 1859, when Edwin L. Drake struck oil with the well he was drilling near Titusville, Pennsylvania.

7 History of Petroleum Processing

8 Modern Petroleum Processing All refineries are different. Oil refining separates everything into useful substances. The petroleum refining industry converts crude oil into more than 2500 refined products, including liquefied petroleum gas, gasoline, kerosene, aviation fuel, diesel fuel, fuel oil, lubricating oil and petrochemical industry feedstock. Most refineries perform the seven basic operations named

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10 Petroleum Refinery Feed-Stocks and Products

11 What is Petroleum or Crude Oil? Petroleum (Latin Petroleum derived from Greek πέτρα (Latin petra) - rock + έλαιον (Latin oleum) - oil) or crude oil is a naturally occurring liquid found in formations in the earth consisting of a complex mixture of hydrocarbons (mostly alkanes) of various lengths The approximate length range is C5H12 to C18H38. Any shorter hydrocarbons are considered natural gas or natural gas liquids, while long-chain hydrocarbons are more viscous, and the longest chains are paraffin wax. In its naturally occurring form, it may contain other nonmetallic elements such as sulfur, oxygen, and nitrogen. Crude oil vary from light coloured volatile liquids to thick, dark oils-so viscous that difficult to pump. To use the different parts of the mixture, they must be separated refining. What is natural gas? A mixtures of hydrocarbons with small molecules. These molecules are made of atoms of C and H i.e. CH4.

12 Why are oil and gas so useful? Oil is a liquid. Meaning that oil may be transported and delivered through pipes. Compare oil to coal-coal is a solid, which comes in lumps. To get it, miners have to work underground. HC with small molecules makes good fuels. Methane (smallest molecules, gas) used for cooking, heating and generating electricity. Gasoline, diesel, jet fuel and fuel oil are all liquid fuels. HC molecules can be split up into small ones, built up into bigger ones, altered in shape or modified by adding other atoms. Even the thick black tarry residue left after distillation is useful bitumen (for road surfacing and roofing). Due to its high energy density easy transportability relative abundance It has become the world's most important source of energy since the mid-1950s. Petroleum is also the raw material for many chemical products, including pharmaceuticals, solvents, fertilizers, pesticides, and plastics; the 16% not used for energy production is converted into these other materials.

13 Origin of Petroleum Theories broadly classified into non-biogenic & biogenic : Non-biogenic : from inorganic sources Metal carbides + H2O..> Hydrocarbons CaCO3 --->CaC2--->Acetylene--->Petroleum hydrocarbons Reaction of CO2, in presence of alkali and alkaline earth metals, with water is also postulated to form hydrocarbons Theory did not receive much recognition Biogenic : from organics, by bacterial transformation : Organic matter (carbohydrates/proteins//lipid/ ligninboth from plant & animal origin >Decay in presence and/or absence of air into HC rich sediments which in presence of micro organism undergoes biological/physical and chemical alterations to form Kerogen (geopolymer) which may be coaly or sapropelic Sapropelic Kerogen under high pressure and temperature further gets converted into Oil & Gas. Widely accepted theory

14 Formation of Oil and Gas Where have crude oil and natural gas come from?

15 Accumulation of Oil and Gas The oil, gas, and salt water occupied the pore spaces between the grains of the sandstones. Whenever these rocks were sealed by a layer of impermeable rock, the cap rock, the petroleum accumulating within the pore spaces of the source rock was trapped and formed the petroleum reservoir.

16 Fourth Year - Chemical Process Eng. Branch Petroleum Refinery Eng. Lectures By Dr. Adnan A. Abdul Razak

17 Refinery Feedstocks and Products

18 Chemical Composition of Petroleum Petroleum is a mixture of hundreds of hydrocarbons of all type with water, salts, sulfur and nitrogen containing compounds and some metal complexes. The elementary composition of crude oil usually falls within the following Table 2.1 Elemental composition of crude oils ranges. The proportion of hydrocarbons in the mixture is highly variable and ranges from as much as 97% by weight in the lighter oils to as little as 50% in the heavier oils and bitumens. Four different types of hydrocarbon molecules appear in crude oil. The relative percentage of each varies from oil to oil, determining the

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20 Paraffins The paraffin series of hydrocarbons is characterized by the rule that the carbon atoms are connected by a single bond and the other bonds are saturated with hydrogen atoms. The general formula for paraffins is C n H 2n+2. The simplest paraffin is methane, CH 4, followed by the homologous series of ethane, propane, normal and isobutane, normal, iso-, and neopentane, etc. (Fig. 1). Figure( 1 ) Paraffins in crude oil.

21 Saturated alkanes: (n-alkane and i-alkane) General formula C n H 2n+2 Boiling point and density increase with increasing no. of C atoms. Branched alkanes (iso-alkanes) is very small in quantity Boiling point of straight chains > iso-alkanes with the same # of C

22 Naphthenes or cycloparaffins Cycloparaffin hydrocarbons in which all of the available bonds of the carbon atoms are saturated with hydrogen are called naphthenes. Typical examples of these are cyclopentane, cyclohexane, etc. (Figure 2). General formula C n H 2n for one ring compounds Figure 2: Naphthene compounds

23 Aromatics The aromatic series of hydrocarbons is chemically and physically very different from the paraffins and cycloparaffins (naphthenes). Aromatic hydrocarbons contain a benzene ring which is unsaturated but very stable and frequently behaves as a saturated compound. Some typical aromatic compounds are shown in (Figure 3). Figure 3: Aromatic compounds

24 Olefins Olefins do not naturally occur in crude oils. However, they are formed during its processing. They are very similar to paraffins, but they exhibit double bonds, usually one per molecule (Figure 4), although some di-olefins (two double bonds in the same molecule, ( Figure 5) can be found. Figure 4: Olefin compounds Figure 5: Di-olefin compounds

25 Sulfur Compound (might be present in inorganic and organic forms) (hydrogen sulfide, sulfides, disulfides, elemental sulfur). Each crude oil has different amounts and types of sulfur compounds (sulfur concentration can range from 0.1 to more than 8 weight percent ), but as a rule the proportion, stability, and complexity of the compounds are greater in heavier crude-oil fractions. Hydrogen sulfide is a primary contributor to corrosion in refinery processing units. Other corrosive substances are elemental sulfur and mercaptans. Moreover, the corrosive sulfur compounds have an obnoxious odor. The combustion of petroleum products containing sulfur compounds produces undesirables such as sulfuric acid and sulfur dioxide. Catalytic hydrotreating processes such as hydrodesulfurization remove sulfur compounds from refinery product streams. Sweetening processes either remove the obnoxious sulfur compounds or convert them to odorless disulfides, as in the case of mercaptans.

26 Oxygen Compounds less than 1% (found in organic compounds such as carbon dioxide, phenols, ketones, carboxylic acids) Carboxylic (OH-C=O bonded to a benzene ring) Phenolic (OH bonded to a benzene ring) Oxygen compounds are responsible for petroleum acidity

27 Nitrogen Compounds less than 1% (basic compounds with amine groups) carbazole (2 benzene rings separated by 1 N atom) neutral Quinoline (2 benzene rings with 1 N atom on 1 ring) basic Nitrogen oxides can form in process furnaces. The decomposition of nitrogen compounds in catalytic cracking and hydrocracking processes forms ammonia and cyanides that can cause corrosion.

28 Metals less than 1% (nickel, iron, vanadium, copper, arsenic) Often found in crude oils in small quantities and are removed during the refining process. Burning heavy fuel oils in refinery furnaces and boilers can leave deposits of vanadium oxide and nickel oxide in furnace boxes, ducts, and tubes. It is also desirable to remove trace amounts of arsenic, vanadium, and nickel prior to processing as they can poison certain catalysts.

29 Salts less than 1% (sodium chloride, magnesium chloride, calcium chloride. Crude oils often contain inorganic salts such as sodium chloride, magnesium chloride, and calcium chloride in suspension or dissolved in entrained water (brine). These salts must be removed or neutralized before processing to prevent catalyst poisoning, equipment corrosion, and fouling. Salt corrosion is caused by the hydrolysis of some metal chlorides to hydrogen chloride (HCl) and the subsequent formation of hydrochloric acid when crude is heated. Hydrogen chloride may also combine with ammonia to form ammonium chloride (NH4Cl), which causes fouling and corrosion.

30 Naphthenic Acids Some crude oils contain naphthenic (organic) acids, which may become corrosive at temperatures above 450 F when the acid value of the crude is above a certain level.

31 Basic products: Petroleum refineries are large, capital-intensive manufacturing facilities with extremely complex processing schemes. They convert crude oils and other input streams into dozens of refined (co-)products, including: The basic products from fractional distillation are:

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33 Liquid petroleum gas (LPG) : Used for heating, cooking, making plastics Small alkanes (1 to 4 carbon atoms) Commonly known by the names methane, ethane, propane, butane Boiling range < 90 degrees Fahrenheit / < 27 degrees Celsius Often liquified under pressure to create LPG (liquified petroleum gas)

34 Gasoline - motor fuel: Its main use is as fuel for internal combustion engines. Liquid Mix of alkanes and cycloalkanes (5 to 7 carbon atoms) Boiling range = degrees Fahrenheit / degrees Celsius

35 Naphtha : Intermediate that will be further processed to make gasoline Mix of 6 to 10 carbon atom alkanes Boiling range = degrees Fahrenheit / degrees Celsius

36 Kerosene : Used for heating, fuel for jet engines and tractors; starting material for making other products Liquid Mix of alkanes (10 to 15 carbons) and aromatics Boiling range = degrees Fahrenheit / degrees Celsius

37 Gas oil or Diesel distillate: Used for diesel fuel and heating oil; starting material for making other products Liquid Alkanes containing carbon atoms Boiling range = degrees Fahrenheit / degrees Celsius

38 Lubricating oil : Used for motor oil, grease, other lubricants Liquid Long chain (20 to 50 carbon atoms) alkanes, cycloalkanes, Aromatics Boiling range = 572 to 700 degrees Fahrenheit / 300 to 370 degrees Celsius

39 Heavy gas or Fuel oil : Used for industrial fuel; starting material for making other products Liquid Long chain (16 to 40 carbon atoms) alkanes, cycloalkanes, Aromatics Boiling range = degrees Fahrenheit / degrees Celsius

40 Residuals : Coke, asphalt, tar, waxes; starting material for making other products Solid Multiple-ringed compounds with 40 or more carbon atoms Boiling range = greater than 800 degrees Fahrenheit / 565 degrees Celsius

41 Fourth Year - Chemical Process Eng. Branch Petroleum Refinery Eng. Lectures By Dr. Adnan A. Abdul Razak

42 Thermo-physical Properties of Petroleum Fractions and Crude Oils

43 Classification of Petroleum Objective of Crude Oil Characterisation / Crude Assay Marketing / Pricing of Crude Oil Transportation of Crude Oil Design of Grass Root Refinery Processing of a new crude oil in an operating Refinery Expansion / Modification of an operating Refinery Primary / Secondary operations Optimization of the product yields Value Addition

44 The petroleum industry generally classifies crude oil By the geographic location it is produced in (e.g. West Texas, Brent, or Oman), Its API gravity (an oil industry measure of density), and By its sulfur content.

45 According to Location West Texas Intermediate (WTI), a very high-quality, sweet, light oil delivered at Cushing, Oklahoma for North American oil Brent Blend, comprising 15 oils from fields in the Brent and Ninian systems in the East Shetland Basin of the North Sea. The oil is landed at Sullom Voe terminal in the Shetlands. Oil production from Europe, Africa and Middle Eastern oil flowing West tends to be priced off the price of this oil, which forms a benchmark Dubai-Oman, used as benchmark for Middle East sour crude oil flowing to the Asia-Pacific region Tapis (from Malaysia, used as a reference for light Far East oil) Minas (from Indonesia, used as a reference for heavy Far East oil) The OPEC Reference Basket, a weighted average of oil blends

46 According API Gravity Specific gravity and API (American Petroleum Institute) gravity are expressions of the density or weight of a unit volume of material. The specific gravity is the ratio of the weight of a unit volume of oil to the weight of the same volume of water at a standard; both specific gravity and API gravity refer to these constants at 60 ⁰F(16 ⁰ C) API = Sp. gr Sp. gr = API Light Crude Oil >31 Mixed Based Heavy crude<22 API is a major factor for Crude pricing

47 According to Sulfur Content Crude oil naturally contains sulfur compounds. Crudes are classed as sweet or sour depending on their sulfur content. If a crude has less than 0.5% sulfur in it, it is considered to be "sweet". If has greater than 2.5% sulfur, it is "sour". A crude with a sulfur content between these two endpoints is called "intermediate".

48 Light crude oil is more desirable than heavy oil since it produces a higher yield of gasoline, While sweet oil commands a higher price than sour oil because it has fewer environmental problems and requires less refining to meet sulfur standards imposed on fuels in consuming countries.

49 According to types of hydrocarbons ( parafffins, naphthenes, and aromatics). This rating is important to the refinery since the value of the crude oil decreases from classification 1 to 6 Crude Classifications (in order of decreasing value): 1) Paraffinic Crudes paraffins + naphthenes > 50% paraffins > naphthenes paraffins > 40% 2) Naphthenic Crudes paraffins + naphthenes >50% naphthenes > paraffins naphthenes >40% 3) Paraffinic Naphthinic Crudes aromatics < 50% paraffins < 40% naphthenes < 40% 4) Aromatic Naphthenic Crudes aromatics > 50% 5) Aromatic - Intermediate Crudes aromatics > 50% paraffins >10% 6) Aromatic Asphaltic Crudes naphthenes > 25% paraffins < 10%

50 Quantitative Basis Approach based on determining the base of a crude oil by one of the following methods : i) US Bureau of Mines Classification based on degree API: Key Fraction Boiling point Pressure API Note No atm. > 40 (Paraffinic Base) 33<API >40 (Intermediate Base) API < 33 (Napthene Base) No (40-mm)Hg 1 atm > 30 (Paraffinic Base) API < 22 (Napthene Base) 22<API<30 (Intermediate Base) The presence of wax is noted by cloud point (if below 5) it indicates little wax (Wax-free)

51 Characterization Factor: (C.F), (K) The most widely used index is characterization factor (Watson, Nelson and Murphy). It was originally defined as: K = 3 T MABP sp. gr In which: TMABP is the molal average boiling point ( R) Sp. gr. : is the specific gravity at 60 0 F It has since related to viscosity, aniline, temperature, molecular weight, critical temperature, percentage of hydrocarbon etc. K (Paraffinic Base) K< 11.5 (Napthene Base) K between (Intermediate Base)

52 ii) Bureau of Mines Correlation Index (BMCI): Like (C.F) related to boiling point and gravity C. I = sp. gr TMABP The correlation index is useful in evaluating individual fractions from crude oils. The CI scale is based upon straight-chain paraffins having a CI value of 0 and benzene having a CI value of 100. The CI values are not quantitative, but the lower the CI value, the greater the concentrations of paraffin hydrocarbons in the fraction; and the higher the CI value, the greater the concentrations of naphthenes and aromatics.

53 Analysis of Crude Petroleum When a refining company evaluate its own crude oils to determine the most desirable processing sequence to obtain the required products, Its own laboratories will provide data concerning the distillation and processing of the oil and its fractions. The first step in refinery is distillation in which the crude oil separated into fractions according to its boiling point. There are at least three types of distillation curves or ways of relating vapor temperature and percentage vaporized. 1. True-boiling-point (T.B.P) 2. Equilibrium or Flash Vaporization (EFV) 3. ASTM or no fractionating distillation

54 1. True-boiling-point (T.B.P) Fractional, run only on crude oil, batch. Fig (4.7) This test enlightens the refiners with 1. All possible information regarding the percentage quantum of fractions, 2. Base of crude oil and 3. The possible difficulties beset during treatment operation etc. Distillation characteristics of a crude are assessed performing a preliminary distillation called "True Boiling Point" analysis (TBP). True boiling point (TBP) and gravity-mid percent curves can be developed: The first is the portion of the distillation at atmospheric pressure and up to F (275 0 C) end point, the second at 40 mm Hg total pressure to F (300 0 C) end point. Table (1) The distillation temperatures reported in the analysis be corrected to 760 mm Hg pressure, by use of chart Fig. (1) The gravity mid- percent curve is plotted on the same chart with TBP. Fig.(3)

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56 Table (1) U.S Bureau of Mines crude petroleum analysis (From Gary and Handwerk, 2001)

57 Figure (1) Boling point at 760 mmhg versus boiling point at 40mmHg (From Gary and Handwerk, 2001)

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59 2. Equilibrium or Flash Vaporization (EFV). The feed material is heated as it flows continuously through a heating coil. As vapor is formed it kept cohesively with liquid at some temperature and a sudden release of pressure quickly flashes or separates the vapor from the mixture without any rectification. By successive flash evaporation like this the stock can be progressively distilled at different increasing temperatures. a curve of percentage vaporized vs. temperature may be plotted. Travels along in the tube with remaining liquid until separation is permitted in a vapor separator or vaporizer. By conducting the operation at a series of outlet temperature, a curve of percentage vaporized vs. temperature may be plotted.

60 3. ASTM or no fractionating distillation: (no fractional, run on fractions). It is supposed to be like EFV, a non fractionating distillation system. It is a simple distillation carried out with standard ASTM flasks 100,200,500 ml flasks. The data obtained is similar to TBP data Fractionation column like (IP24 or D86).

61 TBP, EFV and ASTM curves for a 39.7 degree API distillate

62 Calculation of (K) (Characterization Factor) for The Whole Crude 1. Calculate the TVABP using 20, 50, and 80 volume % TBP temperature. 2. Calculate the 10 to 70% slope of the whole curve. 3. Using a proper correction factor, convert TVABP to TMABP. (or some time given ): (TMABP=TVABP-ΔT) 4. Constract a spg mid percent curve and evaluate the spg for the whole crude. 5. K is found as a function of TMABP and sp. gr.

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64 Conversion between ASTM and TBP Distillation The following equation suggested by Riazi and Daubert where a and b are constants varying with percent of liquid sample distilled as given in Table 3 TBP is true boiling point temperatures at 0, 10, 30, 50, 70, 90, and 95 volume percent distilled, in degrees Rankin. ASTM D86 isthe observed ASTM D86 temperatures at corresponding volume percent distilled, in degrees Rankin. Table 3

65 Exampl e Solution Application of the API method is straightforward using equation (3) and the constants in Table 3

66 As can be seen from Figure E3.1, the TBP distillation curve is below the ASTM curve at volume distilled below 50% and above it for volume distilled above 50%.

67 Distillation Fractions: All the components that boil between the two specified temperature which called the cut point Cut Point Each fraction has an IBP and EP on ASTM curve because of un efficient fractionation the IBP of heavier fraction is interrelated with the EP of lighter fraction.

68 Figure 4: Cut points and end points.

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70 H.W (1): 4000 BPD of (35 0 API) crude oil having the given TBP data is available. TBP ⁰F % vol. Distilled 0 API % Sulfur ۸٥ ۱ ۱۱۰-۱۸۰ ۱۳ ٦۳-۳۸٥ ۳۰ ٤۹ 0.1 ٥۱۰ ٥۰ ۳۸ 0.3 ٦۲۰ ٦۳ ۲۳ 0.5 ۷٥۰ ۷۳ ۲۰ 0.8 ۱۰۰۰ ۸٤ ۱۷ 1.5 ۱۰۰۰ + ۱۰۰ ۱۱ 2.3 a) Draw an assay curve. b) Evaluate the given crude ; TMABP=TVABP- 120 ( 0 F ) c) Select TBP cut temperature for the products to be obtained from distilling this crude and estimate their yields.

71 H.W (2) : For the given crude oil ( API) ; a) Evaluate the given oil. b) Select TBP cut points for the products to be obtained from processing this crude in an atmospheric distillation unit and estimate the %yield for each cut. % vol. Distilled TBP (⁰F ) API % S

72 H.W (3) : For the given crude oil ( API ),sulfur percent 0.15 ; a) Evaluate the given oil. b) Select TBP cut points for the products to be obtained from processing this crude in an atmospheric distillation unit and estimate the %yield for each cut. Stage 1- Distillation at atmospheric pressure 751 mm Hg Fraction No Cut Temp. ⁰F Percent Distilled Sum. Percent Sp. gr. 60/60 ⁰F API 60 ⁰F

73 Stage 2- Distillation at atmospheric pressure 40 mm Hg Fraction No. 11 Cut Temp. ⁰F 392 Percent Distilled 7.3 Sum. Percent 53.1 Sp. gr. 60/60 ⁰F API 60 ⁰F

74 Lect./4a Petroleum refinery

75 Degassing and Dehydration In a producing oilfield the fluid obtained at the wellhead is submitted to : 1- Dehydration operation 2- Degassing operation

76 Degassing and Dehydration 1- Degassing: - At the high pressure existing at the bottom of the producing well, crude oil contains great quantities of dissolved gases. When crude oil is brought to the surface, it is at a much lower pressure. Consequently, the gases that were dissolved in it at the higher pressure tend to come out from the liquid.

77 Degassing and Dehydration Two-Phase Gas Oil Separation High-pressure crude oils containing large amount of free and dissolved gas flow from the wellhead into the flow line, which routes the mixture to the GOSP. (This is usually done by admitting the well fluid into a gas oil separator plant (GOSP) through which the pressure of the gas oil mixture is successively reduced to atmospheric pressure in a few stages). In the separator, crude oil separates out, settles, and collects in the lower part of the vessel.

78 Degassing and Dehydration Gas goes out the top of the separators to a gas collection system, a vapor recovery unit (VRU), or a gas flow line. Crude oil, on the other hand, goes out the bottom and is routed to other stages of separation, if necessary, and then to the stock tank (Fig. 1).

79 Degassing and Dehydration

80 Degassing and Dehydration Figure 2 : Pressure-drop profile for a typical GOSP in the Middle East. - Volatile components either sent to consumer centers as natural gas or used to repressurize the oilfield. - The stripped liquid components are marketed as stabilized natural gasoline.

81 CRUDE OIL (DESALTING) *The fluid produced at the wellhead consists usually gas, oil, free water, and emulsified water (water oil emulsion). * The gas remove by degassing process * First step in the refining process is to remove salt and solids to reduce corrosion, plugging, and fouling of equipment and to prevent poisoning of the catalysts in processing units. *contaminants must be removed by desalting (dehydration). *If the salt content of the crude is greater than 10 lb/1000 bbl (expressed as NaCl), the crude requires desalting to minimize fouling and corrosion caused by salt deposition on heat transfer surfaces and acids formed by decomposition of chloride salts.(<1000 lb/bbl).

82 Typical Configuration Atmospheric Column w/o Preflash

83 Degassing and Dehydration The basic principles for the treating process are as follows: 1. Removal of free water. 2. Resolution of Emulsified Oil.

84 Degassing and Dehydration Free water is simply defined as that water produced with crude oil and will settle out of the oil phase if given little time. There are several good reasons for separating the free water first: 3. Minimization of corrosion because free water comes into direct contact with the metal surface, whereas emulsified water does not. 2. Reduction of heat input when heating the emulsion (water takes about twice as much heat as oil). 1. Reduction of the size of flow pipes and treating equipment.

85 Degassing and Dehydration

86 Degassing and Dehydration

87 Degassing and Dehydration Oil emulsions are mixtures of oil and water. In general, an emulsion can be defined as a mixture of two immiscible liquids, one of which is dispersed as droplets in the other (the continuous phase), and is stabilized by an emulsifying agent.

88 Degassing and Dehydration In the oil field, crude oil and water are encountered as the two immiscible phases together. They normally form water-in-oil emulsion (W/O emulsion), in which water is dispersed as fine droplets in the bulk of oil. This is identified as type C in Figure 2.

89 Degassing and Dehydration For two liquids to form a stable emulsion, three conditions must exist: 1. The two liquids must be immiscible. 2. There must be sufficient energy of agitation to disperse one phase into the other. 3. There must be the presence of an emulsifying agent.

90 Degassing and Dehydration Some of the common emulsifiers are as follows: 1. Asphaltic materials 2. Resinous substances 3. Oil-soluble organic acids 4. Finely dispersed solid materials such as sand, carbon, calcium, silica, iron, zinc, aluminum sulfate, iron sulfide, and so on.

91 Crude possessing emulsifying To increase the coalescence of water drops. (Soda ash, sodium hydroxide, salt of fatty acids petroleum sulfonates which assist coalescence of water droplets).

92 Resolution of Emulsified Oil consists of three consecutive steps: A. Breaking the emulsion: Using chemicals followed by settling can break some emulsions. Other emulsions require heating and allowing the water to settle out of the bulk of oil. More difficult (tight) emulsions require, however, both chemicals and heat, followed by coalescence and gravitational settling. Basically, a dehydration process that utilizes any or a combination of two or more of the treatment aids mentioned earlier (heating, adding chemicals) is used to resolve water-oil emulsions.

93 B. Coalescence: This involves the combination of water particles that became free after breaking the emulsion, forming larger drops. C. Gravitational settling and separation of water drops: The larger water droplets resulting from the coalescence step will settle out of the oil by gravity and be collected and removed.

94 Lect./4b Petroleum refinery

95 1.Introduction: Furnaces are used throughout the industry to provide the heat, using the combustion of fuels. These fuels are solid, liquid or gaseous. Furnaces consist essentially of an insulated, refractory lined chamber containing tubes. Tubes carry the process fluid to be heated, and sizes are device for burning the fuel in air to generate hot gases. A great variety of geometries and sizes are used. However, all furnaces have in common the general feature of heat transfer from hot gas source to a cold sink.

96 *2.Types of furnaces used in process plant *An industrial furnace or direct fired heater, Fig.(1), is an equipment used to provide heat for a process or can serve as reactor which provides heats of reaction. *Furnace designs vary as to its function, heating duty, type of fuel and method of introducing combustion air. However, most process furnaces have some common features.

97 *Fuel flows into the burner and is burnt with air provided from an air blower. *There can be more than one burner in a particular furnace which can be arranged in cells which heat a particular set of tubes. *The flames heat up the tubes, which in turn heat the fluid inside in the first part of the furnace known as the radiant section or firebox. *In this chamber where combustion takes place, the heat is transferred mainly by radiation to tubes around the fire in the chamber. *The heating fluid passes through the tubes and is thus heated to the desired temperature. *The gases from the combustion are known as flue gas. *After the flue gas leaves the firebox, most furnace designs include a convection section where more heat is recovered before venting to the atmosphere through the flue gas stack.

98 Middle of radiant section Convection section Furnace burner

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101 Heating of Crude Oil Pipe Still Heater Pipe heaters can be categorized into three types: 1) Box/ Rectangular. 2) Cylindrical. 3) Radiant Wall All these furnaces have got separate radiation section and convection section. The most universal classification is based on direction of tubes as well as shape of furnace and mode of application of heat.

102 Heating of Crude Oil In most of the furnaces, the direction of tubes is horizontal as in all box type heaters and vertical in cylindrical stills. Radiant walls also use horizontal tubes; however tubes can be placed vertically also. The radiant section design is based on Stefan's law of radiation: Q r = bat 4 = ba( T G 4 T 4 B ) A: area of radiating surface, ft 2 b : 1.72x10 9 Btu/ o F ft 2 hr. at black bod conditions. T: absolute temperature of the surface, o F

103 Heating of Crude Oil For a satisfactory design, the following schedule of heat distribution may be employed: Type of heat Percent Convection heat transfer 30-50% Radiant heat transfer 45-60% Losses (Furnace) 5% Stack losses 12% Design of a furnace radiation section is based on Hottle, Wilson method as:- and radiant heat absorption is given

104 R = G 1+ 1 Q R= % heat absorbed in radiant section. / S α A cp x100 Q = Total heat developed by flame G= Air /fuel ratio (wt. basis). α = Factor to convert actual exposed surface to cold surface: for two rows at spacing 2 OD for one rows at spacing 2 OD. If Q in Kj / hr S=14200 Area in m 2 Q in Btu/hr S=4200 Area in ft 2 Q in Kcal/hr S=6930 Area in m 2

105 Heating of Crude Oil A cp = LN C 12 A cp = area of furnace wall that has tube mounted on it (ft 2 ). α A cp = equivalent cold plane surface (ft 2 ) L= length. C= Center to center spacing. N= Number of tube per row. A = LnN D 12 A= Projected area D= Tube diameter ( in )

106 Heating of Crude Oil n= no. of rows A = A cp = na A n cp D C C D RQ = Aq q= rate of heat absorption per square foot of projected tube area Q = Aq R = na ( D / C cp ) R q

107 x S C D n R q G R α + = 2 2 ) ( ) (1 ) ( G S R R n C D q = α For a most commercial case D/C= 0.5, n=2 2 2 ' ) ( ) ( G S R R xq = ' q n a x D C x q = Heating of Crude Oil

108 Heating of Crude Oil Example:- A petroleum stock at a rate of 1200 bbl/hr. of sp. gr is passed through a train of heat exchangers and is allowed to enter directly the radiant section of box type heater at 220 o C. The heater is designed to burn 3500 kgs per hour of refinery off gases as fuel. The net heating value of fuel is 47.46x10 3 Kj per kg. The radiant section contains 150 sq. meters of projected area of one row of tubes (10.5 cm, 12 m long and spaced at 2 OD). >> Find the outlet temperature of the petroleum stock? Where, α=0.88 Air fuel ratio= 25 Average Specific heat of stock=2.268 Kj/Kg o C.

109 Heating of Crude Oil Solution Total heat liberated (Q) = m fuel * NHV = 47.46*10 3 * 3500 = 1.66*10 8 Kj/hr. Projected area of one tube (L * D) = 12 * No. of tubes = 150 / ( 12 * ) = 120 tubes α A cp = 0.88 * 120 * * 2 * 12 = 266 Sq. m. = Heat absorption %(R) G 1 Q / S α A cp x

110 Heating of Crude Oil = 1+ 25* * = 44% Outlet temperature of the stock:- Q = m Cp Δt 0.44*1.66*10 8 = 1200 * 200 * * * Δt Δt =157 o C So the outlet temperature is equal to = 377 o C

111 Heating of Crude Oil Example A pipe still uses 7110 lb/hr of a cracked gas (Net Heating Value (NHV) Btu/lb). The radiant section contains 1500 sq ft of projected area, and the tube (5 in. outside diameter) are spaced at a center-to-center distance of 10 in. there is only one row of radiant tubes, and they are 40 ft long. The ratio of air to fuel is (21-30 percent excess air). a) What percentage of the heat liberation is absorbed in the radiant section? b) How many Btu are absorbed per hour through each square foot of projected area?

112 Heating of Crude Oil Solution Total heat liberated(q)= m fuel * NHV A = LnN A = 1500 A cp = LN D 12 N = number of tubes C 12 = 7110 * = Btu/hr = *5/12 = 90 = 40 * 90 *10 / 12 = 3000 α A cp = 0.88 * 3000 = 2640 sq ft

113 Heating of Crude Oil Heat absorption %(R) = = G * 21* 1 Q / αa S 1 cp 1.46x x100 6 = 45.8% Heat absorption in radiant section = *146 *10 6 = Btu/hr Heat absorbed / sq ft projected area = q = /1500 = Btu/hr.ft 2

114 Heating of Crude Oil H.W (1) A furnace is to be designed for a heat duty of 30 x 10 6 Btu/hr and efficiency of 75%. The furnace is fired with gaseous fuel at a rate of 17 lb air/lb fuel (NHV = Btu/lb). The tube are arranged in two rows and are of 5 in OD., 40 ft length and 2x OD. Spacing, heat rate of Btu/hr of projected area is recommended. calculate: 1) % heat absorbed in radiation section (R %). 2) Heat absorbed in the convection section. (State any assumptions used). 3) The number of tubes in the radiation section.

115 Solution a) D n (1 R) q( ) = C α R 2 S ( ) G ( 10 2 ) (1 R) = R ( ) = (1-R) 2 R= 47.55% Heat absorbed in radiation zone = R x Q= X 40x10 6 = Btu b) q x A = R x Q x A= A= ft 2 A= L n N D/12 N= x12/( 40x2x5) N= 16 C) % Heat absorbed in the convection section= 100%- (Heat absorbed in radiation zone+ Losses (Furnace)+ Stack losses) % Heat absorbed in the convection section= 100%- ( )% %Heat absorbed in the convection section =32.45% Heat absorbed in the convection section= X 40x10 6 = 13.1x10 6 Btu

116 Heating of Crude Oil H.W (2) 7000 lb/hr of cracked gas of Btu/lb NHV is used as a fuel in a furnace. The radiant section absorbed Btu/hr ft 2 of projected area. The tubes are 5 in. OD., 10 in. spacing, and 20 ft long. They arranged in two rows. The air to fuel ratio is Calculate : 1) the number of tubes in the radiation section. 2) the amount of heat absorbed in this section.

117 Lect./4c Petroleum refinery

118

119 Atmospheric Topping Unit After desalting, the crude oil is pumped through a series of heat exchanger and its temperature raised to about 550 o F (288 o C ) by heat exchange with product and reflux streams. It is then further heated to about 750 o F (399 o C ) in a furnace (i.e. direct fired heater or "pipe still" ) and charged to flash zone of atmospheric fractionators. The furnace discharge temperature is sufficiently high to cause vaporization of all products with drawn above the flash zone + about 20% of the bottom product. The 20% "over flash" allows some fractionation to occur on the trays just above the flashing zone by providing internal reflux in excess of side stream withdrawals.

120 Atmospheric Topping Unit In many petroleum distillations, steam is admitted to the space in which vaporization occurs, the steam reduce the partial pressure in the vapor by Dalton's law, the boiling point of a material may be reduced in only two ways: 1) the pressure may be reduced. 2) or some inert gas such as steam may be introduced. The distillation causes the fractions to separate in increasing order of boiling point. The top product being highly volatile has to be condensed in a reflux condenser. Some portion of the condensed fraction goes back as reflux. All other fractions form the side draw products of distillation column. There fractions are usually classified as heavy naphtha, kerosene, gas oil.

121 Atmospheric Topping Unit Bottom product of atmospheric column is now again routed through a furnace to reach a temperature of 350 to 400 o C and is allowed to flash in a vacuum column, vacuum gas oil, heavy diesel, lubrications oil cuts / pressure distillates shall be the side cuts.

122 Atmospheric Topping Unit

123 Atmospheric Topping Unit Heat and Material Balances 1) The vapor liquid feed enters the tower at a high temperature, and the product are withdrawn at lower temperature, hence heat must be removed, and it is referred as " reflux heat ". 2) The most satisfactory temperature datum is the vaporizer temperature because this temperature can be accurately estimated and is the temperature about which the entire design of tower, and pipe still hinges. 3) By using this datum plane, the heat balance consists simply of the sensible heat required to: a) Cool each product from vaporizer temperature to its withdrawal temperature. b) Condense the products that are withdrawal as liquid.

124 Atmospheric Topping Unit Example (1) : Heat Balance of a Fractionating Tower A heat balance of the simple tower system shown in Fig. below will be computed to determine the amount of heat that must be removed to keep the tower in thermal balance. The capacity is 1200bbl per day (2100 gal per hour of a 12.1 to 12.2 Characterization Factor crude oil. At 576 o F the gasoline, naphtha, kerosene, and gas oil are vapor and the reduced crude oil is a liquid. A sufficient quantity of heat must removed from the vapors to cool them as vapors to the temperature at which they are withdrawn from the tower and to condense the naphtha, kerosene, and gas oil at their withdrawal temperature.

125 Atmospheric Topping Unit Volume % API Lb/gal Gal/hr Lb/hr 50 % bp Mol. wt Latent heat Gasoline Naphtha Kerosene Gas oil Reduced crude Loss Crude

126 Atmospheric Topping Unit

127 Atmospheric Topping Unit Solution : Energy Balance Basis =1hr Sensible heat Btu Gasoline (vapor) 3415 * ( ) * 0.56 = Naphtha (vapor) 754 * ( ) * 0.55 = Kerosene(vapor) 2765 * ( ) * 0.57 = Gas oil (vapor) 1530 * ( ) * 0.59 = Reduced crude (liquid) 5610 * ( ) * 0.72 = Steam (vapor) 567 * ( ) * 0.5 =

128 Atmospheric Topping Unit Latent heat Gasoline ( withdraw as vapor ) Btu Naphtha 754x113 = Kerosene 2765x100= Gas oil 1530x90= Total heat to be removed

129 Atmospheric Topping Unit 1) Cold Reflux Kinds of Reflux : 2) Hot Reflux 2) Circulating Reflux 1) Cold Reflux : Is defined as reflux that is supplied at some temperature below the temperature at the top of the tower. Each pound of this reflux removes a quantity of heat equal to the its latent heat and the sensible heat required to raise it temperature from the storage tank temperature to the temperature at the top of the tower Q = mλ + mc pl T

130 Atmospheric Topping Unit 2) Hot Reflux:- Admitted to the tower at the same temperature. Reflux or over flow from plate to plate in the tower is essentially hot reflux because it is always substantially at its boiling point. Hot reflux capable of removing only the latent heat because no difference in temperature is involved Q = mλ 3) Circulating Reflux : It is not vaporized. It is only able to remove the sensible heat that is represented by its change in temperature as it circulate. This reflux is withdrawn from the tower as a liquid at a high temperature as a liquid and is returned to the tower after having been cooled. Q = mc pl T

131 Atmospheric Topping Unit

132 Atmospheric Topping Unit Example (2) : (Quantity of Reflux) A tower fractionating system is such that Btu/hr of reflux heat must be removed. Example (1) illustrates the method of determining the reflux heat. How many pounds of (1) hot (2) cold, and (3) circulating reflux are required? Solution Hot reflux: λ ( Gasoline ) =123 Btu/lb lb of hot reflux = /120 = lb/hr mole hot reflux = 15500/110 = 141 mole gasoline = 3415/110 = 31 moles vapor = = 172 moles steam = 567/18 = 31.5 total moles at the top of the tower = = 203.5

133 Atmospheric Topping Unit total pressure at the top of the tower = 780 mm Hg the partial pressure in the gas phase is (172/203.5) * 780 = 660 mmhg ( according to Daltons law P i = y i P T ) The dew point of 100% gasoline on EFV curve = 296 o F (at 760 mm Hg) >> At 660 mm Hg the temperature is calculated according to Claussius Clapeyron Eq. ln p p R= Btu/lbmole. o R o 1 = λ ( R T o 1 T ) ln = 120x ( T )

134 Atmospheric Topping Unit T=284 o F The actual top temperature when using hot reflux = 286 o F Cold reflux : Assume storage tank at 80 o F Lb cold reflux = = (286 80) *0.58 Moles cold reflux = 7950/110 = 72.3 Mole gasoline 31 Moles vapor Moles steam 31.5 Total moles p = * 780 = 600 mm Hg The equilibrium temperature of 296 o F corrected to 600 mmhg= 275 o F

135 Atmospheric Topping Unit Circulating reflux : Assuming the reflux is cooled from 286 to 200 o F. Mole gasoline 31 Mole steam 31.5 Total moles 62.5 Partial pressure = *780 = 387 Correction 296 to 387 mm pressure gives 253 o F ( )*0.605 mmhg Lb circulating reflux = = Moles reflux = 10403/110 = 94.6

136 Atmospheric Topping Unit Side-Draw Temperature: 1) The method of calculating side-draw temperature is much the same as the calculation of the top temperature except that complications arise because of the presence of the low boiling materials that pass the draw plate. 2) Making heat balance upon the drawn plate. 3) In practice, steam and vapor of lighter products are usually present, and hence the effect of these vapors on the final condensation temperature must be estimated. The lighter vapors extended from materials boiling at almost the same temperature as the side-draw product to materials that are substantially fixed gases.

137 Atmospheric Topping Unit 4) Those vapor materials which are far above their boiling point behave as fixed gases and lower condensation point by Dalton's law of partial pressures, just as steam does, but those vapor materials which are at or near their boiling point are not effective in reducing the partial pressure. 5) Arbitrarily, the vapors materials that will be condensed at the second or higher draw plate above the plate under consideration may be considered to act as fixed gases. 6) Also, the vapor constituting the materials that is withdrawn from the draw plate above the one under consideration are assumed to have no effect at all on the partial pressure. 7) thus in a tower producing, gasoline, naphtha, kerosene, and gas oil, at the kerosene draw plate the gasoline vapor would be considered as a fixed gas, whereas naphtha vapor would assumed to have no effect on the condensation point.

138 Atmospheric Topping Unit Example (3) : (Calculation of Side Temperature) This example is a continuation of examples (1) and (2). The temperature of the kerosene plate will be computed.(actual temperature = 420 o F)

139 Atmospheric Topping Unit Solution : Heat balance on kerosene plate, quantity of reflux and reflux (or vapor reflux) must be determined. Cool gasoline (vapor) = 3415 * ( ) * 0.58 = Cool naphtha (vapor) = 754 * ( ) * 0.57 = Cool kerosene(vapor) = 2765 * ( ) * 0.57 = Cool gas oil (vapor) = 1530 * ( ) * 0.58 = Reduced crude (liquid) = 5610 * ( ) * 0.72 = Cool steam = 567 * ( ) * 0.5 = Condense kerosene = 2765 * 100 = Condense gas oil = 1530 * 90 = Reflux heat at kerosene plate

140 Atmospheric Topping Unit QT Moles internal reflux = = = ( λxm ) 185x100 Moles fixed gases Steam 31.5 Gasoline 31 Naphtha no effect Total moles vapor = = Assume tower pressure at kerosene plate = 950 mmhg P i = P T * y I Partial pressure 78.6 = *950 = 530 mmhg ln p p o 1 = λ ( R T o 1 T ) W K

141 Atmospheric Topping Unit ln = 100* ( ) T T= 414 o F T actual = 420 o F H.W: Repeat the example above, recalculate the temperature of the naphtha plate, assume a tower pressure of 810 mmhg?

142 Atmospheric Topping Unit H.W : A) Calculate the amount of Hot, Cold and circulating reflux if the storage temperature is 100 o F and Cp v = 0.6 and Cp l = 0.7 Btu/lb. o F B) Check the top tower temperature if hot reflux is used. The dew point of gasoline is 296 o F and the pressure at the top plate is 780 mm Hg. Fraction Lb\hr M.w. λ Temperature o F Gasoline Kerosene Gas oil R.C Steam C.O

143 Atmospheric Topping Unit Calculation of The Diameter of Distillation Column. Example (4): See examples 1, 2 and 3 the quantities and conditions will be taken from these examples. Solution: Density of vapor at top of column (the reflux in the column is always hot reflux) Mole gasoline 31 Mole hot reflux 141 Mole steam Total moles P=780 mm Hg T=286 o F nrt 203.5*379*( ) x760 3 Volume of vapors = = = ft P 780*520

144 Atmospheric Topping Unit Mass of vapor = mass of gasoline + mass of hot reflux + mass of steam = = lb/hr ρ V =19482 / = lb/ft 3 Assume density of liquid ρ l = 42.7 lb/ft 3 w a = K ρ v ( ρl v ρ ) Assume K= 735, K is constant dependent primarily on the tray spacing. w = a lb 2040 hrxft 2 mass A = = = massvelocity ft 2

145 Atmospheric Topping Unit 4A 4x9.55 D = = = π π 3.5 ft 3 To check the vapor velocity at top (3.5 ft/sec) v u = = = A 9.55* ft / sec, u is ok

146 Atmospheric Topping Unit H.W: Lb/hr of 34 o API crude oil at 650 o F is fed to an atmospheric distillation unit. Steam at a rate of 600 lb/hr and 850 o F is used. The fraction obtained were lb/hr gasoline (MW=110, λ=120 ) at 310 o F; lb/hr kerosene ( MW=185, λ=108 ) at 420 o F; lb/hr gas oil ( MW=270, λ=95 ) at 510 o F. The residue is withdrawn at 510 o F. Assume C PL = 0.7,C PV = 0.6 Btu/lb. o F. a) Check the top tower temperature if the dew point of gasoline is 296 o F and the pressure at the top plate is 780 mm Hg. b) Calculate the diameter of the tower if K=735 and ρ l = 42.7 lb/ft 3.

147 H.W 1000 BPD of 32 o API crude oil at 650 o F is fed to an atmospheric distillation unit. Steam at a rate of 600 lb/hr and 530 o F is used. The fractions obtained were 3500 lb/hr gasoline (MW=110, λ=120) at 310 o F; 3000 lb/hr kerosene (MW=185, λ=100) at 420 o F; 1500 lb/hr gas oil (MW=240, λ=90) at 510 o F. The residue is withdrawn at 510 o F. Assume CPL=0.7, CPV=0.6 Btu/lb o F. and the vaporizion temperature=576 0 F. Given: 1 BPD = 42gal, ρ water = lb/gal Check the gas oil plate temperature if the dew point and bubble point of gas oil are 600 o F and 580 o F respectively and the pressure at the withdrawn plate is 990 mm Hg.

148 Lect./5 Conversion Processes

149

150 Thermal cracking: Is defined as the thermal decomposition, under pressure, of large HC molecules to form smaller molecules. Lighter, more valuable HC may be obtained from such relatively low value stocks as heavy gas oils (boiling up to 540 o C (1005 o F)) and residues. These processes are considered as upgrading processes for vacuum residue.

151 Requirements of Thermal cracking: Residual fractions (bottom of barrel) are the least valuable streams of a refinery. Nearly % of the typical C.O in Iraq contains C fractions. Worldwide limited reserve of sweet crude oil. Disposal problems due to stringent environmental regulations. Decreasing demand of fuel oil. Gradually increasing demand of middle distillates.

152 Advantages of Thermal cracking: It is simple and cost effective process. However rapid uncontrolled thermal cracking produces undesirable products like gas and coke. Limitation of catalytic cracking for resid processing : 1. In resid feed stock presence of high molecular weight compounds like resins, asphaltenes and metals it unsuitable for catalytic cracking, FCC can process resid feed only to certain limit. 2. Metal and sulfur compounds in resid feedstock as catalyst poison.

153 Coking: Coking are severe thermal cracking operations, most commonly used carbon rejection process that upgrades residues to a wide range of lighter H.C gases and distillates through thermal cracking. The byproduct of coking process is petroleum coke. The goal of coker operation is to maximize the yield of clean distillates and minimize the yield of coke.

154 Feed stock of coking process: Wide variety of feed stocks ( can have considerable metals (nickel &vanadium), sulfur, resin and asphaltens. Most contaminants exit with coke. Typical feed is vacuum resid.

155 The main produts: Off-gas from which LPG is recovered. Naphtha may be used as gasoline blending agent although its O.N = RON. Gasoil may be catalytic cracking Coke the main uses of petroleum coke: 1. Fuel 2. Manufacture of anode for electrolytic cell reduction of alumina 3. Direct use as chemical carbon source for manufacture of elemental phosphorus, calcium carbide, and silicon carbide. 4. Manufacture of electrodes for use in electric furnace production of elemental phosphorus, titanium dioxide, calacium carbide, and silicon carbide. 5. Manufacture of graphite.

156 The major coking processes: 1) Delayed Coking Semi-continuous process. Feed :atm. Residue, cracked tar, and heavy cycle oil Feed heated and transferred to large soaking (or coking drum). Long residence time need to allow the cracking reactions to complete. 1. Moderate heating : o C ( ) o F 2. Soak drum : o C ( ) o F 3. Residence time : until they are fill of coke. 4. Coke is removed hydraulically 5. Coke yield = 30 %

157 Fig 1: Delayed coking unit

158 2) Fluid Coking or flexicoking A continuous process which uses the fluidized- solids technique to convert residues to more valuable products. The residue is coked by being sprayed into a fluidized bed of hot, fine coke particles. The use of a fluid bed permits the coking reaction to be conducted at higher temperature and shorter contact times than those in delayed coking: Fluidized bed with steam. Severe heating o C ( ) o F at 10 psig. Higher yields of light ends. Less coke yields ( 20 % for fluid coking and 2 % for flexicoking)

159 Coke & liquid yields may be estimated by simple equations:

160 Example ( 1 ) : Develop preliminary estimate of product yields on the processing of Rc of BPD capacity. Conrad son carbon = 19%, 2.3% S, API = Solution : (lb/bbl) Feed BPD API lb/ hr lb/ hr wt% S lb/hr (348.56) RC Products wt% lb/ hr lb/ hr S Coke wt % Gas (C4 - ) wt% Gasoline wt% Gas oil wt%

161 Visbreaking: Visbreaking is a relatively mild thermal cracking operation mainly used to reduce the viscosities and pour points of vacuum tower bottoms to meet the requirements of fuel oil reduce the amount of cutting stock required to dilute the residue to meet the specifications. It is also used to increase catalyst cracker feed stocks and gasoline yields.

162 Visbreaking: The principal reactions which occur during the visbreaking operation are : 1) Cracking of the side- chains attached to cyclo-paraffin and aromatic rings. 2) Cracking of resins to light HC (primarily olefins) and compounds which convert to asphaltenes. 3) At temperature above 900 o F some cracking of naphthene rings.

163 Visbreaking: There are two types of visbreaker operation 1) Coil or furnace cracker Uses high furnace outlet temperature ( o F), and reaction time from 1-3 minutes. The feed is heated in a furnace or coil and quenched as it exits the furnace with gas oil or tower bottoms to slop the cracking reaction. 2) Soaker The feed leaves the furnace at ( ) o F and pass through a soaking drum which provides an additional reaction time, before it is quenched.

164

165

166 Catalytic cracking is the most important and widely used refinery process for converting heavy oils into more valuable gasoline and lighter products. Catalytic cracking breaks complex hydrocarbons into simpler molecules in order to increase the quality and quantity of lighter, more desirable products and decrease the amount of residuals. This process rearranges the molecular structure of hydrocarbon compounds to convert heavy hydrocarbon feedstock into lighter fractions such as kerosene, gasoline, LPG, heating oil, and petrochemical feedstock. Catalytic cracking is similar to thermal cracking except that catalysts facilitate the conversion of the heavier molecules into lighter products. Use of a catalyst (a material that assists a chemical reaction but does not take part in it) in the cracking reaction increases the yield of improved-quality products under much less severe operating conditions than in thermal cracking.

167 There are three basic functions in the catalytic cracking process: 1. Reaction: Feedstock reacts with catalyst and cracks into different hydrocarbons; 2. Regeneration: Catalyst is reactivated by burning off coke; and 3. Fractionation: Cracked hydrocarbon stream is separated into various products.

168 Catalytic Cracking

169 Fluid Catalytic Cracking (FCC) The most common process is FCC, in which the oil is cracked in the presence of a finely divided catalyst which is maintained in an aerated or fluidized state by the oil vapors. The fluid cracker consists of a catalyst section and a fractionating section that operate together as an integrated processing unit. The catalyst section contains the reactor and regenerator, which, with the standpipe and riser, forms the catalyst circulation unit. The fluid catalyst is continuously circulated between the reactor and the regenerator using air, oil vapors, and steam as the conveying media. A typical FCC process involves mixing a preheated hydrocarbon charge with hot, regenerated catalyst as it enters the riser leading to the reactor.

170 The charge is combined with a recycle stream within the riser, vaporized, and raised to reactor temperature (900-1,000 F) by the hot catalyst. As the mixture travels up the riser, the charge is cracked at psi. In the more modern FCC units, all cracking takes place in the riser. The "reactor" no longer functions as a reactor; it merely serves as a holding vessel for the cyclones. This cracking continues until the oil vapors are separated from the catalyst in the reactor cyclones. The resultant product stream (cracked product) is then charged to a fractionating column where it is separated into fractions, and some of the heavy oil is recycled to the riser. Spent catalyst is regenerated to get rid of coke that collects on the catalyst during the process. Spent catalyst flows through the catalyst stripper to the regenerator, where most of the coke deposits burn off at the bottom where preheated air and spent catalyst are mixed. Fresh catalyst is added and worn-out catalyst removed to optimize the cracking process.

171 Fluid Catalytic Cracking

172

173 Moving-Bed Catalytic Cracking The moving-bed catalytic cracking process is similar to the FCC process. The catalyst is in the form of pellets that are moved continuously to the top of the unit by conveyor or pneumatic lift tubes to a storage hopper, then flow downward by gravity through the reactor, and finally to a regenerator. The regenerator and hopper are isolated from the reactor by steam seals. The cracked product is separated into recycle gas, oil, clarified oil, distillate, naphtha, and wet gas.

174 Catalyst Both systems use basically similar catalysts but produced in a different form, in the shape of beds for moving bed and fine powder for fluidized bed. 1. Acid treated clays ground to a powder 2. Synthetic silica- alumina catalysts of higher activity (amorphous) 3. Crystalline synthetic silica alumina catalyst called zealots or molecular sieves. The advantages of zealots over the natural and synthesis amorphous catalyst are 1) Higher activity. 2) Higher gasoline yields at a given conversion. 3) Production of gasoline containing a larger % of paraffinic and aromatic HC. 4) Lower coke yield. 5) Increased iso-butane production. 6) Ability to go for higher conversion per pass without over cracking.

175 PROCESS VARIABLES

176 Catalytic cracking is differences with respect Thermal Cracking uses a catalyst lower temperature lower pressure more flexible different reaction mechanisms : ionic vs. free radical High thermal efficiency Good integration of cracking and regeneration High yields of gasoline and other distillates Low gas yields High product selectivity Low n-alkane yields High octane number Chain-branching and high yield of C4 olefins High yields of aromatics

177 Lect./6 Upgrading Naphtha

178

179 Catalytic Reforming Catalytic reforming is the process of transforming C7 C10 hydrocarbons with low octane numbers to aromatics and iso-paraffins which have high octane numbers. It is a highly endothermic process requiring large amounts of energy. Depending on the properties of the naphtha feedstock (as measured by the paraffin, olefin, naphthene, and aromatic content) and catalysts used, reformates can be produced with very high concentrations of toluene, benzene, xylene, and other aromatics useful in gasoline blending and petrochemical processing. Hydrogen, a significant by-product, is separated from the reformate for recycling and use in other processes. A catalytic reformer comprises a reactor section and a product recovery section.

180

181 Role of Reformer in the Refinery and Feed Preparation The catalytic reformer is one of the major units for gasoline production in refineries. It an produce 37 wt% of the total gasoline pool. Other units such as the fluid catalytic cracker (FCC), the methyl ter-butyl ether (MTBE) production unit, alkylation unit and isomerization unit, also contribute to this pool. The straight run naphtha from the crude distillation unit is hydrotreated to remove sulphur, nitrogen and oxygen which can all deactivate the reforming catalyst. The hydrotreated naphtha (HTN) is fractionated into light naphtha (LN), which is mainly C5 C6, and heavy naphtha (HN) which is mainly C7 C10 hydrocarbons. Light naphtha (LN) is isomerized in the isomerization unit (I). Light naphtha can be cracked if introduced to the reformer. The role of the heavy naphtha (HN) reformer in the refinery is shown in Figure 5.2. Hydrogen, produced in the reformer can be recycled to the naphtha hydrotreater, and the rest is sent to other units demanding hydrogen.

182

183 Reforming Reactions (PONA) {Paraffin, Olefin, Naphthene, Aromatic} 1) P isomerizes to some extent converted to N, and N subsequently converted to aromatics. 2) O saturated to form P which then react as in (1) {hydro-cracking}. 3) N converted to aromatics. {dehydrogenation}. 4) A unchanged.

184 Process Technology There are several commercial processes available for reforming. 1. Semi-regenerative Fixed Bed Process :The name semi-regenerative comes from regeneration of the catalyst in the fixed bed reactors after shut down by burning off the carbon formed on the catalyst surface. (Low capital cost). 2. Continuous Regenerative (moving bed) CCR Platforming :Catalyst can be regenerated continuously and maintained at a high activity. (Higher capital cost). 3. Cyclic :compromise between the two extremes having a swing reactor for regeneration.

185 All of the reforming catalyst contains platinum supported on Platinum serve as a catalytic site for hydrogenation and Catalyst a silica alumina base. In most cases rhenium is combined with platinum to form a more stable catalyst which permits operation at lower pressure. dehydrogenation reactions. Chlorinated alumina provides an acids site for isomerization and hydro- cracking reactions and cyclization.

186

187 Example (2)

188 Example (3) On processing 1200 ton / day of 27 API catalyst crackers feed stock at a temperature of 450 o C, pressure =1050 mm Hg the following products were obtained: Products wt % API Mw Gases C 5+ gasoline TCGO Coke Given that: WHSV = 0.7 hr -1, Linear velocity of vapor (U) = 0.3 m/s, ρ catalyst = 420 Kg /m 3 Calculate: a) diameter of the cracker, b) weight of catalyst needed, c) conversion, and d) efficiency. Solution : 1200 ton/day * 1000 Kg/ton m feed = = Kg /hr 24 hr/day m gases = 0.15 * / 3600 =2.08 Kg/s m + C5 gasoline = 0.55 * 50000/ 3600 =7.64 Kg/s m TCGO = 0.26 * 50000/ 3600 = 3.61 Kg/s Total moles of vapor = (2.08/ 32) + (7.64/ 110) + (3.61/ 260 ) n = Kg mole / s

189 22.4 (m 3 / kg mole) * 760 mm Hg R = = Kg mole * 273 n R T * *( ) V= = = 6.35 m 3 /s P 1050 П V 6.35 m 3 /s A =----- D 2 = = = m 2 4 U 0.3 m/s m catalyst = 50000/0.7 = Kg V catalyst = m/ ρ = 170 m 3 4* 170 H= = 8 m 3.14 * (5.19) 2

190

191 Alkylation combines low-molecular-weight olefins (primarily a mixture of propylene and butylene) with isobutane in the presence of a catalyst, either sulfuric acid or hydrofluoric acid. The product is called alkylate and is composed of a mixture of high octane, branched-chain paraffinic hydrocarbons. Alkylate is a premium blending stock because it has exceptional antiknock properties and is clean burning. The octane number of the alkylate depends mainly upon the kind of olefins used and upon operating conditions.

192

193 Isomerization is the process in which light straight chain paraffins of low RON (C6, C5 and C4) are transformed with proper catalyst into branched chains with the same carbon number and high octane numbers. The hydrotreated naphtha (HTN) is fractionated into heavy naphtha between C ( F) which is used as a feed to the reforming unit. Light naphtha C5-80 C (C5-180 F) is used as a feed to the isomerization unit. Isomerization Reactions: Isomerization is a reversible and slightly exothermic reaction: n-paraffin i-paraffin The isomerization reactions are slightly exothermic and the reactor works in the equilibrium mode. There is no change in the number of moles and thus the reaction is not affected by pressure change. Better conversions are achieved at lower temperature The conversion to iso-paraffin is not complete since the reaction is equilibrium conversion limited. It does not depend on pressure, but it can be increased by lowering the temperature. However operating at low temperatures will decrease the reaction rate. For this reason a very active catalyst must be used.

194 Isomerization Catalysts There are two types of isomerization catalysts: 1. The standard Pt/chlorinated alumina with high chlorine content, which is considered quite active, 2. and the Pt/zeolite catalyst.

195 Lect./7 Product Blending

196 Product Blending Blending purpose is to allocate the available blending components in such a way to meet product demands and specification at the least cost. We now review how the properties of mixtures are estimated based on the properties of the components. Octane Blending: True octane numbers do not blend linearly. It is therefore necessary to use especial blending octane numbers to obtain linear expressions. The blending is performed on a volumetric average basis. The formula used for calculation is: B t ON t = n B i ON i ) i = 1 (.. 1 where B t = total amount of blended gasoline, bbl ON t = desired octane of blend B i = bbl of component i ON i = blending octane number of component i.

197 contributing to product t and blending octane numbers, respectively. The practice has been to use the following expression for the blending octane number: ON = MON + RON.. 2 i ( i i ) / 2 where MON and RON are the motor and research octane numbers, respectively. Note that the true octane number is the one obtained using a CFR test engine. For example, consider a 30% isomerate and 70% reformate blend. Isomerate has the following octane values: MON=81.1, RON=83.0, whereas reformat has the following octane numbers: MON=86.9 and RON=98.5. When blended in the proportion given above, the blended pool has ON= Pool Octane : is the average octane of the total gasoline production of the refinery, if the regular, mid- premium, and super- premium gasolines are blended together. Posted Octane numbers (PON): are the arithmetic average of the motor octane number (MON) and research octane number (RON).

198 Reid vapor pressure: The desired RVP of a gasoline is obtained by blending n-butane with C 5 (380 o F) with C 5 (193 o F) naphtha. The amount of n-butane required to give the needed RVP is calculated by: M = n t ( RVP) t i = 1 M i ( RVP) i.. 3 Where: M t = total moles blended product (RVP) t = specification RVP for product, psi M i = moles of component i (RVP) i = RVP of component i, psi or kpa The desired RVP for a blended gasoline is obtained by adding n-butane to reach the desired value.

199 Example (1): calculate the amount of n-butane to be added to following base stock to achieve an RVP of 10 psi( n- butane: MW=58, RVP=52). Base stock LSR gasoline BPD 4000 Lb/hr MW 86 RVP (psi) 11.1 Reformate Alkylate FCC gasoline Solution: Total Base stock LSR gasoline BPD 4000 Lb/hr MW 86 mol/hr 457 mol% 21 RVP 11.1 PVP 2.32 Reformate Alkylate FCC gasoline Total

200 Butane requirement : (use Eq. (3) ) (2179) (5.38) + M (52) = (2179+ M) (10) M= M M= 240 moles n-butane required. The above method requires obtaining the molecular weight of each of the streams involved, which could be a problem sometimes, although there are good ways of estimating such molecular weights. To makes matter simpler, one can use the method developed at Chevron. In this method Vapor blending indices (VPBI), which work well. The RVP of a blend is then calculated using the following volume averaging formula: ( RVP ) = V i ( RVP) blend i.. 4 In the case where the volume of the butane to be blended for a given RVP is desired : A(VPBI) a + B(VPBI) b + C(VPBI) c W(VPBI) w = (Y+W) (VPBI) m... 5 where A= bbl of component a, etc W= bbl of n- butane (w) Y= A+B+C ( all component except n- butane ) (VPBI) m = VPBI corresponding to the desired RVP of the mixture w= subscript indicating n- butane.

201 Table 11-1 and 11-2 show the blending component values for different blending streams and the blending indexes as a function of RVP values. Table 11-1: Blending Component values (Gary and Handwerk, 2001)

202 Table 11-2: Blending Component values (Gary and Handwerk, 2001)

203 Example (2): Repeat Example (1) use vapor blending indices (VPBI) method. Component BPCD RVP VPBI Vol x VPBI n-butane W W LSR gasoline Reformat Alkylate FCC gasoline Total W W Given : VPBI of n- butane =138 For 10 psi RVP, (VPBI) m = 17.8 Solution 17.8 ( W) = W ( ) W = W= 1660 bbl n-butane required Total 10 psi RVP gasoline = = BPCD Although this differs slightly from the result in Example (1)

204 Example (3): Consider the following gasoline blending streams are available from the various units. It is desired to produce a split of premium and regular gasoline having 91 and 87 posted octane numbers respectively, with both having an RVP= to 10.2 psi. calculate the quantity of n- butane required to give the desired vapor pressure. Component Volume MON RON VPBI Isomerate Reformat FCC gasoline Light hydrocrackate Alkylate Polymer Total Given : VPBI of n- butane =138 For 10.2 psi RVP, (VPBI) m = 18.2 Solution Starting with the given flow for all of the above streams and calculating the amount of n-butane to add to fix the RVP.

205 Component Vol. RVP VPBI Vol (VPBI) n- butane W W Isomerate Reformate FCC gasoline Light hydrocrackate Alkylate Polymer Total W W 18.2 ( W )= W W = 3937 bbl n- butane The total volume of 10.2 psi RVP premium gasoline = = BPCD

206 Octane calculations for pool Gasoline Component Volume Vol. fract. MON MON RON RON n- butane Isomerate Reformate FCC gasoline Light hydrocrackate Alkylate Polymer Total Pool octane [ ( MON + RON ) / 2 )] = PON This is not acceptable, as the octane requirement for pool gasoline is 89 PON. There are several ways of correcting this. Among the possibilities are : Increase severity of reforming to produce a 98.8 to 100 RON clear reformate. Use an octane blending agent, such as MTBE ( methyl tertiary butyl ether ) and ETBE ( ethyl tertiary butyl ether ).

207 Recalculating pool gasoline RVP and PON after adding sufficient MTBE to increase the PON to 89.0 gives the following. Component Vol. RVP VPBI Vol (VPBI) n- butane W W Isomerate Reformate FCC gasoline Light hydrocrackate Alkylate Polymer MTBE Total W W W=18.2 ( W) 119.8W= = W=3984 bbl Total pool 10.2 RVP, 89.0 PON gasoline= BPCD

208 Other properties: Several other properties of blend pools (viscosities, aniline point, pour points, flash points) can be estimated using a technique similar to that of the Chevron Method for RVP, that is: P t = n i = 1 ( v P ).. 6 i i where v i is the volume fraction of blending stream i as above, and P t as well as P i are the blending properties of the product and the blending streams, respectively. The blending properties are, of course, compiled in tables much in the same way as in the case of RVP. These additional properties are important for Diesel blending. Finally, properties like sulfur or nitrogen content are monitored and blended linearly with percentages.

209 Home work (1) : Using the value from the following table, calculate the number of barrels of n- butane that have to be added to a mixture of 1250 barrels of HSR gasoline, 750 barrels of LSR gasoline, and 620 barrels of C 5 FCC gasoline to produce a 9.0 psi Reid vapor pressure. What are the research and motor octane number of the blend? Component Volume MON RON RVP VPBI HSR gasoline LSR gasoline C 5 FCC gasoline Given : VPBI of n- butane =138, MON=92.0, RON=93 For 8 psi RVP, (VPBI) m = 13.4

210 Home work (2) : Calculate the octane number of the final blend and amount of n- butane needed for producing a 9.5 psi RVP gasoline from 5100 BPSD of LSR gasoline, 3000 BPSD light hydrocrackate,4250 BPSD alkylate, BPSD heavy hydrocrackate, BPSD C 5 FCC gasoline,14200 BPSD of 96 RON reformat, and 2500 BPSD of polymer gasoline. Component Volume MON RON RVP VPBI LSR gasoline light hydrocrackate alkylate heavy hydrocrackate C 5 FCC gasoline Reformate RON Polymer Given : VPBI of n- butane =138 For 9.5 psi RVP, (VPBI) m = 17.6

211 Home work (3) : A regular gasoline is to be blended using the components listed in the table below. The target (ON) specification is 90. Using the information provided, determine the volume of Reformate required meeting the blending specification. Component Hydrocracate Alkylate Reformate Straight run naphtha Volume (barrels) RON MON

212 Lect./8 Supporting Processes

213 Supporting Processes Not directly involved in the processing of petroleum based fuels Processes» Hydrogen production & purification» Acid gas treating» Sulfur recovery» Gas processing units» Water treatment

214 Hydrogen Production & Purification Many refineries produce sufficient quantities of hydrogen for hydro treating from their naphtha-fed platinum catalyst reforming operations. however, require more hydrogen than is produced by their catalytic reforming units This supplemental hydrogen requirement can be Provided by one of two processes: 1.Steam reforming of light ends such as methane (natural gas), ethane, or propane 2.Partial oxidation of heavy hydrocarbons such as fuel oil

215 Steam reforming Steam reforming for hydrogen production is accomplished in four steps: 1. Reforming. This involves the catalytic reaction of methane with steam at temperatures in the range of 1400 to 1500 F ( C), according to the following equation: CH4 + H2O CO + 3H2 (1) This reaction is endothermic and is carried out by passing the gas through catalyst-filled tubes in a furnace. The catalyst usually is in the form of hollow cylindrical rings ranging up to 3/4 inch in diameter. It consists of 25 to 40% nickel oxide deposited on a low-silica refractory base. 2. Shift conversion. More steam is added to convert the CO from step 1 to an equivalent amount of hydrogen by the following reaction: CO + H2O CO 2 + H2 (2) This is an exothermic reaction and is conducted in a fixed-bed catalytic reactor at about 650 F (343 C). Multiple catalyst beds in one reactor with external cooling between beds are commonly employed to prevent the temperature from getting too high. The catalyst used is a mixture of chromium and iron oxide.

216 3. Gas purification. The third step is removal of carbon dioxide by absorption in a circulating amine or hot potassium carbonate solution. Several other treating solutions are in use. The treating solution contacts the hydrogen and carbon dioxide gas in an absorber containing about 24 trays, or the equivalent amount of packing. Carbon dioxide is absorbed in the solution, which is then sent to a still for regeneration. 4. Methanation. In this step, the remaining small quantities of carbon monoxide and carbon dioxide are converted to methane by the following reactions: CO + 3H 2 CH4 + H2O CO 2 + 4H 2 CH4 + 2H 2O (3) This step is also conducted in a fixed-bed catalytic reactor at temperatures of about 700 to 800 F (427 C). Both reactions are exothermic and, if the feed concentration of CO and CO 2 is more than 3%, it is necessary to recycle some of the cooled exit gas to dissipate the heat of reaction. The catalyst contains 10 to 20% nickel on a refractory base.

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218 Partial oxidation of fuel oils Partial oxidation of fuel oils is accomplished by burning the fuel at high pressures (800 to 1300 psig) with an amount of pure oxygen which is limited to that required to convert the fuel oil to carbon monoxide and hydrogen. Enough water (steam) is added to shift the carbon monoxide to hydrogen in a catalytic shift conversion step. The resulting carbon dioxide is removed by absorption in hot potassium carbonate or other solvents. Ideally the partial oxidation reactions are as follows: 2CnHm + no 2 2nCO + mh2 2nCO + 2nH 2O 2nCO 2 + 2nH2

219 ACID GAS REMOVAL Gases from various operations in a refinery processing sour crudes contain hydrogen sulfide and occasionally carbonyl sulfide. Some hydrogen sulfide in refinery gases is formed as a result of conversion of sulfur compounds in processes such as hydrotreating, cracking, and coking. Recent air pollution regulations, however, require that most of the hydrogen sulfide be removed from refinery fuel gas and converted to elemental sulfur. In addition to hydrogen sulfide many crudes contain some dissolved carbon dioxide which through distillation finds its way into the refinery fuel gas. These components hydrogen sulfide and carbon dioxide are generally termed acid gases. They are removed simultaneously from the fuel gas by a number of different processes, some of which are:

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221 Hydrogen sulphide and carbon dioxide readily combine with aqueous solutions of certain alkanol amines at temperatures usually close to ambient, and may be driven off from the fat solutions by heating to about 100 o C. The reaction with hydrogen sulphide is : 2 RNH 2 + H 2 S Amine Treating Unit (RNH 3 ) 2 S Amine (mono- di and tri ethanol amines and methyl di ethanol amine)

222 The conventional equipment, comprising a bubble- cup tower together with a bubble cup tower for regeneration. The treating temperature is 5 to 10 o C above the dew point of the gas to ensure that no hydrocarbons liquid condenses out of the plant. This process is the most widely used method for the regenerative removal of H 2 S from both gases and liquids. Its use is not only in refineries but also for oilfield treatment of natural gases and LPG. The choice of the proper amine and solution depends on the composition of the gas to be treated and the final purity desired.

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224 Sulphur Recovery Acid gas streams from hydrodesulphurization containing H2S are sent to sulphur recovery unit (Claus unit). Furthermore, sulphur removal is carried out by tail gas clean up schemes. The purpose of removing the sulphur is to reduce the sulphur dioxide (SO2) emissions in order to meet environmental guidelines.

225 Claus Process The Claus process is the most significant elemental sulphur recovery process from gaseous hydrogen sulphide. Gases with an H2S content of over 25% are suitable for the recovery of sulphur in the Claus process. Hydrogen sulphide produced, for example, in the hydrodesulphurization of refinery products is converted to sulphur in Claus plants. The main reaction is 2H2S + O2 2S + 2H2O The Claus technology can be divided into two process stages : thermal and catalytic. In the thermal stage, hydrogen sulphide partially oxidized at temperatures above 850 0C (1562 F) in the combustion chamber. This causes elemental sulphur to precipitate in the downstream process gas cooler. If more oxygen is added, the following reaction occurs: 2H2S + 3O2 2SO2 + 2H2O This gas contains 20 30% of the sulphur content in the feed stream. Activated alumina or titanium dioxide is used. The H2S reacts with the SO2 and results in gaseous, elemental sulphur. This is called the Claus reaction: 2H2S + SO2 3S + 2H2O

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227 GAS PROCESSING UNIT

228 Gaseous Fuels The Importance of Gaseous Fuel - Generally very clean burning. Little soot. - Easy to burn - No grinding or atomisation. Excellent mixing - No problems with erosion or corrosion - No ash - The gas is easy to clean. E.g. if sulphur is present, it may be easily removed prior to combustion. - Simplest combustion plant of all - Burners - Problems with distribution and storage - Explosion risk - very volatile. - Relatively costly. Offset by cheaper and more efficient plant.

229 Classification of Gaseous Fuels (A) Fuels naturally found in nature: Natural gas Methane from coal mines (B) Fuel gases made from solid fuel Gases derived from Coal Gases derived from waste and Biomass From other industrial processes (Blast furnace gas) (C) Gases made from petroleum Liquefied Petroleum gas (LPG) gases Gases from oil gasification (D) Gases from some fermentation process Refinery

230 Natural Gas Natural gas is a gaseous fossil fuel consisting primarily of methane but including significant quantities of ethane, butane, propane, carbon dioxide, nitrogen, helium and hydrogen sulfide. It is found in oil fields and natural gas fields, and in coal beds. When methane-rich gases are produced by the anaerobic decay of non-fossil organic material, these are referred to as biogas. Natural gas is often informally referred to as simply gas, especially when compared to other energy sources such as electricity. Before natural gas can be used as a fuel, it must undergo extensive processing to remove almost all materials other than methane. The by-products of that processing include ethane, propane, butanes, pentanes and higher molecular weight hydrocarbons, elemental sulfur, and sometimes helium and nitrogen.

231 Chemical composition The primary component of natural gas is methane (CH4), the shortest and lightest hydrocarbon molecule. It also contains heavier gaseous hydrocarbons such as ethane (C2H6), propane (C3H8) and butane (C4H10). as well as other sulphur containing gases, in varying amounts. Nitrogen, helium, carbon dioxide and trace amounts of hydrogen sulfide, water and odorants can also be present. Mercury is also present in small amounts in natural gas extracted from some fields. The exact composition of natural gas varies between gas fields. Organosulfur compounds and hydrogen sulfide are common contaminants which must be removed prior to most uses. Gas with a significant amount of sulfur impurities, such as hydrogen sulfide, is termed sour gas; gas with sulfur or carbon dioxide impurities is acid gas.

232 Natural gas processing plant There are a great many ways in which to configure the various unit processes used in the processing of raw natural gas. The block flow diagram, Fig. (3), is a generalized, typical configuration for the processing of raw natural gas from non-associated gas wells. Figure 3: Natural gas processing plant