Analogous Structures in Evolution

Analogous Structures in Evolution There are many types of evidence supporting evolution, including studies in the molecular biology field, such as DNA, and in the developmental biology field. However, the most commonly used types of evidence for evolution are anatomical comparisons between species. While homologous structures show how similar species have changed from their ancient ancestors, analogous structures show how different species have evolved to become more similar. Speciation Speciation is the change over time of one species into a new species. Why would different species become more similar? Usually, the cause of convergent evolution is similar selection pressures in the environment. In other words, the environments in which the two different species live are similar and those species need to fill the same niche in different areas around the world. Since natural selection works the same way in these environments, the same types of adaptations are favorable, and individuals with favorable adaptations survive long enough to pass down their genes to their offspring. This continues until only individuals with favorable adaptations are left in the population. Sometimes, these types of adaptations can change the structure of the individual. Body parts can be gained, lost, or rearranged depending on whether their function is the same as the original function of that part. This can lead to analogous structures in different species that occupy the same type of niche and environment in different locations. Taxonomy When Carolus Linnaeus first began classifying and naming species with taxonomy, the science of classification, he often grouped similar-looking species into similar groups. This led to incorrect groupings compared to evolutionary origins of the species. Just because species look or behave the same doesnt mean they are closely related. Analogous structures dont have to share the same evolutionary path. One analogous structure might have come into existence long ago, while the analogous match on another species may be relatively new. They may go through different developmental and functional stages before they are fully alike. Analogous structures are not necessarily evidence that two species came from a common ancestor. It is more likely they came from two separate branches of the phylogenetic tree and may not be closely related at all. Examples The human eye is very similar in structure to the eye of the octopus. In fact, the octopus eye is superior to the humans in that it doesnt have a blind spot. Structurally, that is the only difference between the eyes. However, the octopus and the human are not closely related and reside far from each other on the phylogenetic tree of life. Wings are a popular adaptation for many animals. Bats, birds, insects, and pterosaurs all had wings. But a bat is more closely related to a human than to a bird or an insect based on homologous structures. Even though all these species have wings and can fly, they are very different in other ways. They just happen to fill the flying niche in their locations. Sharks and dolphins look very similar due to color, placement of their fins, and overall body shape. However, sharks are fish and dolphins are mammals. This means that dolphins are more closely related to rats than they are sharks on the evolutionary scale. Other types of evolutionary evidence, such as DNA similarities, have proved this. It takes more than appearance to determine which species are closely related and which have evolved from different ancestors to become more similar through their analogous structures. However, analogous structures themselves are evidence for the theory of natural selection and the accumulation of adaptations over time.

Walt WHitman essays

Walt WHitman essays Walt Whitman was looked upon as the forerunner of 20th Century poetry, praising democracy, and becoming a proclaimed poet of American democracy. He was known as the Son of Long Island, and he loved his country and everything about it. (Current, Williams, Freidel- page 292-293). Whitman lived during the time of the Civil War; a fact that increased his patriotism. Whitman was considered one of the most important American Poets of the 19th Century. (Encyclopedia of World Biography- page 249). He influenced the direction of 20th Century poets such as Erza Pound, William Carlos Williams, Carlos Sandberg, and Allen Ginsberg. Whitman praised democracy and spoke of the flesh as well as the spirit. (Encyclopedia of Biography- page 249). He rejected the normal rhyme and meter of poetry and wrote in free verse, relying on Native American language. In general, Whitman's poetry is idealistic and romantic. Whitman identified strongly with the outcasts of society. He said to one outcast, Not till the sun excludes you do I exclude you. (Lowen, Nancy- page 11) People hailed him as the most authentic voice of the United States of America. Edgar Allen Poe had said, The vitality and variety of his life was the mere reflection of the vitality and variety of the United States of America. Walter Whitman was born into a family of nine children and he had a rough childhood. The Whitman family first settled in the Huntington area by the middle of the seventeenth century. This helped him to write two of the worlds greatest literary works, There was a Child Went Forth and Song to Myself. (Lowen, Nancy- page 6). There was a Child Went Forth was about his siblings and his childhood. Out of nine children, only four survived to live to old age. He spoke of how his siblings died and how it affected his family. Whitman had one sibling who was insane, one who was se...

Law for business Essay Example | Topics and Well Written Essays - 1000 words

Law for business - Essay Example If the representations are received late, the director may instead ask that the said representations be read during the meeting. He may also be allowed to speak in the meeting on his protest against the intention to remove him as director. These are all provided for in Sections 168 (entitled Resolution to remove director) and 169 (entitled Director’s right to protest against removal) of the Companies Act 2006 (Companies Act 2006). Going to Cretins Ltd. (or Cretins for short), the plan of Joan and Mike to remove Phillip out as a director of the company will depend on its membership structure. Whatever it is, the process will have to be in accordance with the afore-cited Sections 168 and 169 of the Companies Act 2006. If the only three directors, Phillip, Joan and Mike, are also the only members or shareholders of Cretins and they own the same number of shares, the process will be very simple. Since it will be a concerted effort of Joan and Mike, all they have to do is first requisition for a general meeting of the members where they will submit a resolution to remove Phillip. Pursuant to the procedural due process mechanism set out in Section 169, Phillip has to be served with a notice of the intended resolution to remove him. Phillip is entitled to submit written representations to protest against his removal. During the said meeting, Phillip is likewise entitled to speak out. At any rate, all these remedies available to Phillip will become moot and academic because Joan and Mike will ultimately vote for his removal as is their plan. The requirement of the law is an ordinary resolution which means that the vote of the members needed for the subject removal is a simple majority (THE FREE LIBRARY BY FARLEX). As stated, the circumstances in the preceding paragraph are on the assumption that Joan, Mike and Phillip have the same number of

Australian Hospital Pharmacy -career options and challenges Assignment

Australian Hospital Pharmacy -career options and challenges - Assignment Example These collaborative efforts aim at coming up with the most appropriate method for dealing with a patient’s problems or health. Numerous hospitals focus on particular fields like oncology or surgery. From 1996 to 1999, the general population of graduated pharmacists in Australia increased from 13,834 to 14,747 (The Pharmacist, 2011, p. 731). Shortly after this rise, the Australian Institute of Health and Welfare projected a deficiency of pharmacy after 2010. On the contrary, health employers reported a surplus of pharmacists in Australia in 2011. There were 27,226 listed pharmacists by March 2013 in Australia. More than half of these pharmacists are over the age of 35 (The Pharmacist, 2011 p. 732). Hospital pharmacists early in their career in Australia can offer themselves an advantage in their lines of work market. These advantages are achievable by pursuing options in modern roles in Australian hospital pharmacy. The following is a list of the advantages. All professions make crucial contributions to the Australian community (Low, Hattingh, & Forrester, 2009, p. 231). Similarly, hospital pharmacists play a vital part in keeping patients healthy and industrious. Less than a decade ago, hospital pharmacy was a less rewarding profession than other types of pharmacy. Today, the odds have shifted. The part played by hospital pharmacists grew tremendously during the past five years. Today, hospital pharmacists operate in wards along with other medical professionals, playing an important part in the delivery of healthcare. Even though yearly income is still low for overall pharmacists in Australia, the working conditions for hospital pharmacists have improved significantly. The latest income survey indicated that the degrees of permanent job opportunities in hospital pharmacy are much more than for community pharmacists. In addition, their

Internet and Political Interactions Globally Essay

Internet and Political Interactions Globally - Essay Example This is achieved by organizing online donating stages where supporters and donors provide monetary assistance to a candidate. The funds play a crucial role in ensuring efficient running of campaigns. The availability of the internet creates connectivity among different persons, which encourages decentralization of power and resources. Therefore, the political front can efficiently distribute task different professionals to review policies with public participation. The internet provides the political scene with a wide audience who, in turn, become willing participants in current affairs. Citizens can effectively engage in public debates and forums, which seek to express their views regarding political policies as they are and how they affect them. Participation in political debates can serve as a way to petition policymakers into establishing better policies to the public as they wish. This encourages public participation in democracy as their right is exercised as guaranteed by the constitution. In addition, the internet provides a platform for constructive criticism, which may present itself as an anonymous comment thus protecting against victimization. The internet increases public access to information regarding events that shape the day’s political scene. This is facilitated by the availability of a wide audience over the internet through social sites and political blogs. This ensures an open flow of information and transparency in the political arena. With an adequate flow of information, the public becomes a political watchdog over brewing scandals and malpractices. This has played a significant role in ensuring accountability within politics to avoid public and negative publicity. Political blogs over the internet can serve as whistleblowers concerning ills among the political elite.  

Fluid Catalytic Cracking Process Engineering Essay

Fluid Catalytic Cracking Process Engineering Essay INTRODUCTION Fluid catalytic cracking process, which is now more than 60 years old, is the cornerstone of most of the petroleum refineries. It has proven to be the most-efficient process available for the conversion of gas oils and residue into more valuable lighter hydrocarbons. Many refiners consider the catalytic cracking process to be the highest profit generating unit in the entire refinery. In earlier times, Fluid Catalytic Cracking Unit (FCCU) was operated broadly in two modes, they are; Maximum gasoline mode Maximum distillate mode But with the advent of Reformulated gasoline (RFG), these are now operated in maximum olefin mode. FCCU is a very sophisticated unit with many factors affecting each other and the overall process. In some processes investigation of factors impact is done by changing one factor at a time while keeping other factors constant. In case of FCCU it is almost practically impossible to obtain a clear indication; as, change in one single factor leads to change(s) in one or more other factors. This whole phenomenon is a natural consequence of the heat balance of FCCU. If the unit is to operate at steady state, then the unit has to be in heat balance condition. At this stage the heat requirement in the reactor is satisfied by burning coke in the regenerator and transferring the energy to the reactor through circulating hot catalyst. Heat balance around the reactor-regenerator can be used to predict the effects of process changes although the exact degree of the changes may be difficult to estab lish. It is one step at a time thought process and rather difficult to pin down exact numbers without a careful study of yields and coke laydown rates as affected by changing variables. In this work a plant data is taken as reference and based on that, calculations have been done to find out the net heat of endothermic reactions occurring in the riser reactor, assuming that the unit is operating at steady state and that the riser is an isothermal one. Then as per the products slate, a 7-lumped model is considered from various literatures and based on the kinetics of reactions, rate equations are formed and with the knowledge of available kinetic parameters the differential temperature drops along the height of the riser are calculated. PROCESS DESCRIPTION More than a dozen types of FCCU are operating worldwide. But the basic designs of all these type remain the same. FCCU comprises of two parts; Riser reactor, in which catalytic cracking reactions occur Regenerator, in which burning of coke (deposited during cracking) from the catalytic sites is done Figure 1 shows a schematic diagram of a typical FCCU. The feed is preheated in a furnace and (Figure: 1- Schematic Diagram of a typical FCCU) injected at the bottom of the riser along with a small amount of steam. This steam helps in dispersion of feed, good atomization and reduces coke formation by decreasing the partial pressure of hydrocarbon vapours. The feed is subsequently vaporized when it comes in contact with the hot catalyst from regenerator. The hydrocarbon vapours so formed undergo endothermic cracking reactions on their way up through the riser. The expansion of product vapours occurs through the length of the riser and the gas velocity increases with decreasing gas density. Hot catalyst particles provide the sensible heat and latent heat requirements for vaporizing the liquid feed and also endothermic heat of reaction for the cracking reactions. After a certain distance from the entry zone of the riser, the liquid feed is completely vaporized. Cracking reactions continue with the vapours moving up in the riser and the temperature is dropped along the length of the riser due to endothermic nature of cracking. The catalytic cracking is started and also completed in a very short period of time inside the riser reactor in which the catalyst is pushed upward by incorporating steam at various locations along the length of the riser and hydrocarbon vapours. Mixture of catalyst and hydrocarbon vapour travels up in the riser into the reactors. Steams injected at different locations in the riser are as follows, Fluffing steam at the bottom of the riser Dispersion steam along with fresh feed injectors Riser dilution steam above the fresh feed injectors Dispersion steam along with recycle stream injectors Aeration steam into the riser J bend to fluidize the catalyst Along with this some other locations are there where steam is injected. They are as follows; Spent catalyst standpipe aeration steam Regenerated catalyst standpipe aeration steam Reactor quench steam Reactor dome steam Post riser quench steam Stripping steam into strippers Mixture of catalyst and hydrocarbon vapour is discharged from the riser to the riser cyclone assembly. The bulk of the spent catalyst is separated from product vapours in the cyclone assembly. If necessary the vapours leaving the riser cyclones are routed into secondary cyclone assembly located inside the reactor vessel. Separated catalysts flow through each cyclone dip leg into the stripper. Product vapours leave the reactor cyclones and flow into the main fractionator through the reactor overhead vapour line. Quench steam is injected inside the reactor vessel to reduce the temperature, so as to minimize post riser thermal cracking reactions and coke formation. Reactor dome steam is provided to sweep hydrocarbons and avoid dead areas on top of the reactor vessel that may lead to thermal cracking and coking in that area. The separated catalyst from the riser and reactor cyclone assemblies enters the catalyst stripper. As the catalyst flows down the stripper, it gets stripped off the entrained hydrocarbon vapours by the up flowing steam. Stripping enhances the product recovery and reduces the carryover of hydrocarbon to the regenerator along with the spent catalyst thereof. Fluffing steam ensures the fluidization of the circulating catalyst. Stripped catalyst from the stripper flows into the regenerator dense bed through the spent catalyst standpipe (SCSP). Catalyst level in the stripper is maintained by spent catalyst slide valve (SCSV). Aeration steam is provided in the SCSP to ensure proper flow and fluidization of spent catalyst. Coke adsorbed on the spent catalyst during cracking reaction is been removed in the regenerator by burning off the coke with air. Air is supplied from the air blower to the regenerator through multiple distributors. Air is also introduced at different locations of the regenerator, they are as follows; T-grid air Regenerated catalyst standpipe (RCSP) hopper aeration air RCSP aeration air Regenerator fluffing air at the bottom near the J bend The regenerator can be operated in two modes; Partial combustion mode Complete combustion mode For partial combustion mode, a CO boiler is needed to convert CO to CO2. The current discussion is for complete combustion mode regenerator. Flue gas from the regenerated dense bed flows to the two stage regenerator cyclone assembly. Here the entrained catalyst is separated from the flue gas. The separated catalyst flows back to the dense bed through cyclone dip legs. Flue gas from the cyclone flows out from top of the regenerator through a flue gas line. Total air flow to the regenerator is regulated based on the desired level of oxygen in flue gas. Too low O2 concentration will cause coke build up on regenerated catalyst and CO release from regenerator. Too high O2 concentration will lead to regenerator cooling. So, regenerator flue gas is regularly examined for O2, CO, CO2, NO2, SO2 analysis. FEED CHARACTERIZATION The only constant in FCC operation is the frequent change in feedstock quality. Thats why two feeds with similar boiling point ranges can exhibit huge differences in cracking performance and product yields. Feed characterization is one of the most important activities in monitoring the FCC process. Feed characterization is the process of determining physical and chemical properties of the feed. Understanding feed properties and also knowing their impact on units performance is an essential thing. Trouble shooting, catalyst selection, unit optimization and subsequent process evaluation, all depend on feedstock. Feed characterization relates product yields and qualities to feed quality. Analytical techniques like mass spectrometry are sophisticated and not practical for determining complete composition of FCC feedstock. Simpler empirical correlations are often used. They are as follows; oAPI gravity and UOP K Boiling range Average boiling point Carbon residue Metals Sulphur, Nitrogen and Oxygen oAPI gravity and UOP K It is a specific gravity relating the density of oil to the density of water. The empirical formula for this is; oAPI 131.5 (3.1) Feed to an FCC can range from 15o to 45o API. If the API gravity increases the charge stock will crack more readily and for the same reaction temperature there will be greater conversion. Secondly at a constant conversion level, there will be greater gasoline yield with slightly lower octane. A rough indication of the quantities of paraffin present is a characterization factor which relates boiling point to specific gravity, is called the UOP K factor. This is given by; (3.2) Where: CABP = cubic average boiling point, oR SG = specific gravity at 60 oF Higher the UOP K value more is the paraffinic nature of the feedstock. Boiling Range The boiling range of FCC feed varies from an initial point of 500oF to an endpoint of about 1000oF. There are two boiling point ranges which are used to describe the lighter material in the feed. They are; Per cent over 430oF Per cent over 650oF The first quantifies the amount of gasoline in the feed. The second one quantifies the light fuel oil in the charge. Average boiling point Average boiling point of the FCC feed depends on the average molecular weight. An increase in average boiling point and molecular weight will typically cause the following; The charge will crack more readily, so at constant reactor temperature conversion will increase At constant conversion, yield of C4 and lighter will decrease Olefinic content of the product will decrease Regenerator temperature will tend to rise At constant conversion, the gasoline yield will increase about 1% for an increase in the molecular weight of 20. Carbon residue The carbon residue of a feedstock is an indirect measure of its coke producing nature. Values may be determined by either Conradson or Ramsbottom methods. The carbon residue can be a useful number for determining possible contamination in storage. Entrainment in vacuum tower is a common cause of increased carbon residue. Colour may be used to approximately evaluate the carbon content of the feedstock. Darker stocks tend to have higher carbon residues. Metals Organometallic compounds in the FCC feed can cause serious overcracking if the metals deposit on the catalyst. The cleanliness of a chargestock is given by a metals factor: Fm = Fe + V + 10 (Ni + Cu) (3.3) Where: Fm = Metals Factor Fe = Iron concentration V = Vanadium concentration Ni = Nickel concentration Cu = Copper concentration All metal concentrations are ppm by weight in the feed. A factor of 1.0 is considered safe, over 3.0 indicate a danger of poisoning of catalyst. Sulfur, Nitrogen, Oxygen Sulfur is as undesirable in FCC feed as it is in the feed to most of the refining units, causing corrosion of the equipment and increased difficulty in treating products. At 50% conversion about 35% sulfur charged is converted to H2S, and at 70% conversion the figure will rise to 50%. Nitrogen produces NH3 and CN- in the reactors, and NOx and trace quantities of NH3 in the regenerator. These NH3 and CN- cause plugging and corrosion, while the NOx and NH3 in the flue gas cause environmental problems. Gas oil will absorb oxygen in storage unless the tanks are gas blanketed. This oxygen will combine with the compounds in the oil at about 450oF to form gum, which fouls heat exchangers. FCC REACTION CHEMISTRY Cracking reactions are predominantly catalytic, but some non-selective thermal cracking reactions do take place. The two processes proceed via different chemistry. The occurrence of both the reactions is confirmed by distribution of products. Catalytic cracking proceeds mainly via carbenium ion intermediates. There are three dominant reactions in cracking are catalytic cracking, isomerization, hydrogen transfer. The idealized reaction classes are tabled below with specific reactions to support them. (Table: 1 idealized reactions of importance in FCCU) Reaction classes Specific reactions Cracking n-C10H22 n-C7H16 + C3H6 ; 1-C8H16 2C4H8 Hydrogen transfer 4C6H12 3C6H14 + C6H6 ; cyclo-C6H12 + 3 1-C5H10 3n-C5H12 + C6H6 Isomerization 1-C4H8 trans-2-C4H8 ; n-C6H10 iso-C4H10 ; o-C6H4(CH3)2 m- C6H4(CH3)2 Transalkylation C6H6 + m- C6H4(CH3)2 2C6H5CH3 Cyclization 1-C7H14 CH3-cyclo-C6H11 Dealkylation Iso-C3H7-C6H5 C6H6 + C3H6 Dehydrogenation n-C6H14 1-C6H12 + H2 Polymerization 3C2H4 1-C6H12 Paraffin alkylation 1-C4H8 + iso-C4H10 iso-C8H18 Some of the reactions are endothermic in nature and some are exothermic in nature. Each reaction has a heat of reaction associated with it. The overall heat of reaction is the combination of both the types of heat of reactions. Though there are a number of exothermic reactions, then also the net reaction is endothermic. It is apparent that the type and magnitude of reactions have an impact on the heat balance of the unit. If the catalyst is with less hydrogen transfer characteristics, it will cause the net heat of reaction to be more endothermic. This in turn results in higher catalyst circulation and possibly a higher coke yield to maintain the heat balance. FCC UNIT MATERIAL BALANCE For this, a complete set of commercial plant data is used. The data is given in subsequent tables below; FEEDSTOCK (Table: 2 Properties of feed components) Feed Unit Hydrotreated VGO Un-hydrotreated VGO Light Coker Naphtha Quantity,TMTPA 3200 800 170 % of total feed wt% 76.74 19.18 4.08 Density @ 15oC gm/cc 0.894 0.932 0.6762 CCR wt% 0.1 1.2 Sulfur wt% 0.1 3.32 0.434 Hydrogen content wt% 13 Ni + V wppm 1 6.38 Nitrogen wppm 500 1594 30 ASTM Distillation, vol.% D-1160, oC D-1160, oC D-86, oC IBP 366 349 36 5 374 379 10 385 394 43 30 420 435 49 50 443 468 57 70 485 508 65 90 545 556 75 95 576 573 FBP 620 609 86 Bromine no. 107.86 Paraffins vol.% 46.7 Olefins vol.% 43.38 Naphthenes vol.% 7.25 Aromatics vol.% 2.68 RON, clear 79.4 Diene value 5.31 WATSON K 12.436 MW 82.001 PRODUCT YIELDS (Table: 3- product yields, Ex-reactor and Perfect fractionator basis) Products wt % Weight (lbs. /hr.) H2S 0.39 4309 Hydrogen 0.041 606 Methane 1.06 11710 Ethane 1.54 17010 ethylene 1.76 19442 Dry gas 4.401 48768 Propane 2.86 31592 Propylene 9.66 106708 n-butane 1.69 18668 i-butane 5.52 60976 butenes 7.47 82516 LPG 27.2 300460 LCN 14.50 160174 MCN 23.40 257978 HCN 3.90 43082 LCO 16.45 181713 CLO 4.75 153347 COKE 5.01 - OPERATING CONDITIONS (Table: 4- Operating conditions for the Unit) Riser-Reactor Unit Value Fresh heavy feed rate (VGO) m3/hr. 533.4 Fresh light feed rate (Coker naphtha) m3/hr. 30.2 CLO recycle m3/hr. 46 Riser top temperature oC 540 Riser top pressure Kg/cm2 1.5 Feed preheat temperature oC 350 Regenerator Air to regenerator (dry basis) Nm3/hr. 310717 Regenerator pressure Kg/cm2 1.9 Dense bed temperature oC 640 Dilute bed temperature oC 654 Flue gas temperature oC 657 Blower discharge temperature oC 226 Stripper Stripping steam rate Kg/hr. 5000 Stripping steam temperature oC 290 Stripping steam pressure Kg/cm2 10.5 Base temperature oC 0 Ambient temperature oC 35 Flue gas composition MW= 30.6 O2 vol. % 2.49 CO vol. % 0.005 CO2 vol. % 15.58 N2 vol. % 81.83 SO2 vol. % 0.085 SO3 vol. % 0.01 Now using the above data, amount of oxygen that was consumed by burning the hydrogen in coke is estimated. All the gas calculations are based upon 100 moles of flue gas. The oxygen consumed for H2O is given by the expression; O2 consumed = * (vol. % of N2 in flue gas) 2 * (vol. % of O2 in flue gas) 2 * (vol. % of CO2 in flue gas) (vol. % of CO in flue gas) (5.1) So, O2 consumed = * (81.83) 2 * (2.49) 2 * (15.58) (0.005) = 7.36 The weight of the hydrogen and carbon in the coke are calculated; Weight = 2.016 * (7.36) + 12.01 * (15.58+0.005) = 202.01 The temperature differentials are calculated; (oF basis) ΆTRR = (Regenerator dense bed temperature Riser outlet temperature) (5.2) = 1184 1004 ΆTRR = 180 ΆTRB = (Regenerator fluegas temperature Blower discharge temperature) (5.3) = 1215 439 ΆTRB = 776 ΆTRS = (Riser outlet temperature Stripping steam temperature) (5.4) = 1004 554 ΆTRS = 450 The weight combined feed ratio is calculated as; (Flow rate)CLO * (Density)CLO * 2.204 CFR = (5.5) (Flow rate)Fresh feed * (density)fresh feed * 2.204 = CFR = 0.074 The stripping steam and inert gases carried to the reactor by the regenerated catalyst are calculated on a weight per pound fresh feed basis; Steam = (5.6) Steam = 0.01 Inert gases = (5.7) Inert gases = 0.007 The amount of hydrogen in the coke is calculated as; Hydrogen in Coke, wt % = [2.016 * 7.36 / 202.01] * 100 % = 7.35 wt. % The air to coke ratio is; Air to coke, wt/wt = (2897/202.01) * (81.83/79) Air to coke, wt/wt = 14.85 lbs air / lb coke Where; 2897 is the molecular weight of air multiplied by 100 (basis of 100 moles of flue gas) The weight of coke per hour may be calculated as; Weight of coke, lbs/hr. = (4591) * 193.23 / 14.85 = 59738.6 lbs/hr. Where; (310717 Nm3/hr. = 5178.62 Nm3/min. = 193.23 MSCFM 4591 = air rate conversion factor from MSCFM to lbs/hr.) So, weight % of coke is then; wt. % coke = * 100% = (59738.6 / 1104941.7) * 100 % wt. % coke = 5.41 In the product yield table, the coke wt. % is indicted as 5.01 wt%. But it is calculated as 5.41 wt. %. Now the overall weight balance is as follows; OVERALL WEIGHT BALANCE INPUT:- = Fresh feed + Coker naphtha + CLO recycle = {(533.4 * 0.8 * 894 * 2.204) + (533.4 * 0.2 * 932 * 2.204)} + (30.2 * 676.2 * 2.204) + (46 * 808 * 2.204) = 1186860.1 lbs. / hr. OUTPUT:- = Total product yields + coke = 1149831 + 59738.6 = 1209569.6 lbs. / hr. So, error in weight balance is calculated as; = INPUT OUTPUT = (1186860.1 1209596.6) lbs. / hr. = 22736.5 lbs. / hr. = 1.88 wt. % = 98.12 % closure Now combustion heat of coke is determined as follows; (at hottest temperature = flue gas temperature = 1215oF) ΆHcomb = [(X) (vol. % of CO in flue gas) + (Y) (vol. % of CO2 in flue gas) + (Z) (vol. % of O2 consumed)] / (weight if hydrogen and carbon in coke) (5.8) = [(48000) * (0.005) + (169743) * (15.58) + (106472) * (7.36)] / 202.01 ΆHcomb = 16971.8 Btu / lb coke Where; X = heat of combustion of CO at 1215oF Y = heat of combustion of CO2 at 1215oF Z = heat of combustion of H2O at 1215oF There is correction factor for the hydrogen in coke, this is given as; Correction factor, C = 1133 (134.6) (wt. % hydrogen) (5.9) = 1133 (134.6) (7.35) = 143.7 The net heat of combustion after using the correction factor is; -ΆHC = 16971.8 + 143.7 Btu / lb coke -ΆHC = 17115.5 Btu / lb coke At this point the reactor and regenerator heat balances are calculated. The catalyst supplies the heat to the reactor. The regenerator heat balance is calculated first using a basis of one pound of coke at the hottest regenerator temperature. The reactor heat balance is based on one pound of fresh feed. HEAT BALANCE REGENERATOR HEAT (Figure: 2- Regenerator heat In Out scheme) HEATREG = ΆHCOMB. ΆHCOKE ΆHAIR ΆHRADIATION LOSS (6.1) Now, ΆHCOKE = heat required to raise coke to combustion temperature = (0.4) * (ΆTRR) (6.2) ΆHAIR = heat required to raise air to combustion temperature = (lb air / lb coke) * (0.26) * (ΆTRB) (6.3) ΆHRADIATION LOSS = 250 Btu / lb coke So, HEATREG = 17115.5 {(0.4) * (180)} {(14.85) * (0.26) * (776)} 250 HEATREGHEATREG = 13797.4 Btu / lb coke -ΆHCSo, regenerator efficiency = *100% (6.4) = 80.6 REACTOR HEAT (Figure: 3- Reactor heat In Out scheme) HEATRX = ΆHFRESH FEED + ΆHRECYCLE + ΆHSTRIPPING STEAM + ΆHREACTION + ΆHRADIATION LOSS + ΆHINERTS (6.5) ΆHFRESH FEED, ΆHRECYCLE = heat required to raise fresh feed recycle to reactor temperature ΆHSTRIPPING STEAM = heat required to raise steam to reactor temperature = ΆTRS * (0.485) * (lb steam / lb fresh feed) (6.6) ΆHRADIATION LOSS = 2 Btu / lb fresh feed ΆHINERTS = heat of inert gases carried from regenerator to reactor by regenerated catalyst = ΆTRR * (-0.275) * (lb inerts / lb fresh feed) (6.7) HEATRX = (enthalpy of fresh feed at riser outlet temperature enthalpy of fresh feed at preheat temperature) + CFR (enthalpy of recycle feed at riser outlet temperature enthalpy of recycle feed) + ΆTRS * (0.485) * (lb steam / lb fresh feed) + 2 Btu / lb fresh feed + ΆTRR * (-0.275) * (lb inerts / lb fresh feed) + ΆHREACTION = (745 460) + 0.074 * (745 460) + 450 * (0.485) * 0.01 + 2 + 180 * (-0.275) * 0.007 + ΆHREACTION HEATRX = 310 + ΆHREACTION Note:- Enthalpies for the fresh feed and the recycle feed were calculated by taking respective UOP K values, oAPIs and the temperatures from the API technical data book. Regenerator heat is calculated on a one lb of coke basis. This can be converted to one lb of fresh feed by use of weight % of coke term. So, HEATRX = HEATREG () (6.8) ΆHREACTION + HEATRX = HEATREG () + ΆHREACTION (6.9) ΆHREACTION = HEATREG () + ΆHREACTION + HEATRX (6.10) But HEATRX = + ΆHREACTION Putting this relation in equation (6.10), the equation changes to ΆHREACTION = HEATREG () ΆHREACTION = 13797.4 * 310 ΆHREACTION = 436.44 Btu / lb fresh feed So, HEATRX = 310 + 436.44 HEATRX = 746.44 Btu / lb fresh feed (0.275) (ΆTRR)Cat / Oil (wt. / wt.) = HEATRX (6.11) Cat / Oil (wt. / wt.) = 15 lb Catalyst / lb Oil Catalyst circulation rate = (Cat / Oil) * (lb fresh feed / hr.) (6.12) = 15 * 1104941.8 CCR = 16574127 lbs. / hr. = 7524 MT/ hr. Overall heat flow scheme for the whole FCCU can be shown as below; (Figure: 4- Typical FCCU heat balance scheme) Now, the net total endothermic heat of reaction is calculated through empirical formulae. But we took the assumption as the riser is an isothermal one. Practically it is not isothermal. The temperature at the base of the riser is higher than what is at the top of the riser or at the riser outlet. This is because the cracking reactions occurring along the length of the riser is endothermic in nature. So heat is being absorbed during the reaction and causes the temperature at that particular location to decrease. Gradually the temperature decreases and at the riser outlet the temperature is dropped significantly. In this context we can estimate the riser base temperature using empirical relations and therefore can estimate the drop in temperature at the next differential element up in the riser [DNS]. But before this a multi-lumped model is to be considered along with possible reaction schemes and there kinetic parameters. SEVEN LUMP KINETIC MODEL For this purpose a seven lump kinetic model proposed by Mehran Heydari et al. (2010) is used. They divided the model into seven lumps namely; VGO/Coker Naphtha, Clarified Oil, Light Cycle Oil, gasoline (LCN, MCN, and HCN), LPG, Dry gas and Coke. The schematic flow diagram is as follows; (Figure: 5- Seven lump kinetic model in FCCU) In order to develop a mathematical model for this particular system, certain assumptions has to be taken, they are as follows; The riser is an one dimensional ideal plug flow reactor with no radial and axial dispersion Reactor is an adiabatic riser Feed viscosity and heat capacities of all components are constant Fluid flow is not affected by the coke deposition on the catalyst Feed is vaporized instantaneously in the riser entrance All cracking reactions are taking place in the riser The model considers seven lumps and eighteen reactions and eighteen kinetic constants. Molecular weights of different lumps and boiling ranges are given [DNS] in the table below; (Table: 5- molecular weights and boiling ranges of lumps) j Lump Molecular weight (Kg/ Kmol) Boiling range (oC) 1 VGO 418.7 349 620 2 CLO 291 232 -567 3 LCO 226 170 392 4 GASOLINE 114 30 228 5 LPG 65 6 DRY GAS 30 7 COKE 12 Values of kinetic constants and activation energies along with heat of reactions for each reaction are given in the table below ([DNS], [Mehran Heydari], [Praveen ch. shishir sinha]); (Table: 6- reaction schemes with kinetic parameters) Reactions Rate constants (m3/ kg cat. hr.) Activation energy (KJ/Kmol) Heat of reaction (KJ/Kg) VGO CLO 14.93 50.73 45.821 VGO LCO 5.78 50.73 79.213 VGO GASOLINE 11.69 50.73 92.335 VGO LPG 3.59 16.15 159.315 VGO DRYGAS 0.35 16.15 159.315 VGO COKE 11.55 16.15 159.315 CLO LCO 5.78 50.73 56.314 CLO GASOLINE 0.94 46.24 128.571 CLO LPG 0.135 59.75 455.185 CLO DRYGAS 0.0135 59.75 455.185 CLO COKE 0.3272 59.75 455.185 LCO GASOLINE 0.5742 46.24 93.030 LCO LPG 0.0086 59.75 704.93 LCO DRYGAS 0.0009 59.75 704.93 LCO COKE 0.0596 59.75 704.93 GASOLINE LPG 0.0003 78.49 372.10 GASO DRYGAS 0.0001 78.49 372.10 LPG DRYGAS 0.0033 59.75 32.30 The riser model is assumed to be a two phase model

Henrik Isbens A Dolls House Essays -- A Dolls House Essays

1 A Dolls House   Ã‚  Ã‚  Ã‚  Ã‚  A Dolls House represents a women’s marital life from many years ago. The central theme of this play is Nora’’s rebellion against society and everything that was expected of her. Nora shows this by breaking away from all the standards and expectations her husband and society had set up for her. Women were not considered of importance to their husbands and that made women feel like in a â€Å"dolls house†, such as with Nora and her husband Helmer. In her time women were not supposed to be independent. They were to support their husbands, take care of the children, cook, clean, and make everything perfect around the house. Nora had two main rebellions; her taking out the loan, and when she left her family, and everything was all planned out by her. Nora’’s first rebellion was when she took out a loan so that she could pay for her husband, Torvald’s medical treatment. It was against the law for women to take out a loan without their husbands consent. When she did this she proved that she was not as submissive and helpless as Torvald thought she was. He called her â€Å"helpless†. A perfect example of Torvald’s control and Nora’s submissiveness was when she got him to re-teach her the tarantella. She already knew the dance but she acted as if she needed him to re-teach her the whole thing. When he says to her â€Å"Watching you swing and dance the tarantella makes my blood rush†. This shows that he is more interested in her physically than em...

Critical Views of Beowulf

Beowulf Critical views †¢One of the oldest and most important remains of the Anglo-Saxon literature is the epic poem of Beowulf. Its age is unknown; but it comes from somewhere between the 7th and the 10th centuries. It is like a piece of ancient armour; rusty and battered, and yet strong. The style of the epic poem is likewise simple- perhaps one should say, austere. Beowulf is indeed the most successful Old English poem because in it the elements, language, metre, theme, structure, are all most nearly in harmony.The author seems mainly bent upon telling us how his Sea-Goth slew Grendel and the Fire-drake. †¢The poem opens with an account of forefathers of Hrothgar the Scylding, king of Danes. He is the builder of Heorot, the hall where Beowulf contends with Grendel. The poem begins with the burial of Scyld, from whom the dynasty of Scyldings take its name. In ancient days, so ran the legend, scyld when he was child, was drifted in an open boat to the shores of Danes.When coming thus out of the secret of the Sea the bark touched the land, the folk found the naked child lying asleep in the midst of arms and gems and golden treasure, took him up and hailed him king. As he came alone and mysteriously out of the sea, so he passes away alone and mysteriously into the sea, and the introduction of the poem describes his burial. With as many treasures he brought, with so many they send him away when he died. And as the poem begins with this burial, so it ends with the burial of Beowulf.His burial is nothing mythic, nothing mystic surrounding it. Beowulf, dead after his fight with the dragon, and his gray hair lying around his hair, is borne to the top of the great cliff that overlooks the sea. The cliff has its own name, Whale’s Ness. †¢The epic is divided into three chief episodes. Yet these three episodes are well wrought and well diversified. They are not repetitions, exactly; there is a change of wrestling with Grendel in the night at Heorot and the descent underwater to encounter Grendel’s mother; while the sentiment of the Dragon is different again.But the great beauty, the real value, of Beowulf is in its dignity of style. †¢The word Grendel, as Lawrence points out, can be associated with the Old English grund, i. e. ground, bottom, or watery depths that we find the lurking-place of Grendel and his mother. †¢SIMILARITY WITH OTHER WORKS: Frederick Panzer in 1910 published the results of a careful study of over 200 folk-tales which have elements of resemblance to the Grendel story. These tales with all their variations of outline have enough in common. One of these is the tale of ‘The Bear’s Son’.From the varying versions of ‘The Bear’s Son, something like a central frame, or outline, can be reconstructed. An aged king builds a hall or house which is nightly haunted by a demon. The elder sons of the king are unable to overcome the invader, but the youngest son, formerly held in little esteem, wrestles with the monster and wounds him. The fight of the demon is marked by a trail of blood. An episode follows in which the hero fights in an underground lair of monsters often against a male and a female.His victory over them, sometimes by a use of a magic sword, frees captive maidens who return to the upper world. But the hero is abandoned by faithless companions, and must without aid contrive means of escape from the monster’s home. The tale often ends with the punishment of the traitors, and the marriage of the hero with one of the rescued maidens. Similarities in this outline to the Grendel episodes of the Beowulf are, of course, general rather than precise. But it seems clear that Panzer is correct in claiming that a relationship exists.

National Plan Essay

National Plan Essay National Plan Essay A National Plan On April 4, 1808, Albert Gallatin, Secretary of the Treasury in the administration of Thomas Jefferson, sent to the Senate his report on roads and canals. It begins with the assertion that " the general utility of artificial roads and canals . . . is universally admitted." The question is, who should build them. In some countries, "these improvements may often, in ordinary cases, be left to individual exertion, without any direct aid from Government." In the American case, two major circumstances," whilst they render the facility of communications throughout the United States an object of primary importance, naturally check the application of private capital and enterprise to improvements on a large scale." The first of these is the relative scarcity of capital. It is much more difficult than in Europe to attract investment by "prospects of remote and moderate profit." The second is " the extent of the territory compared to the population." With a sparse population, local traffic can not be counted on to make profitable a local improvement. In general, a canal will be unproductive unless it opens " a communication with a natural extensive navigation which will flow through that new channel." For this reason, " some works already executed are unprofitable; many more remain unattempt, because their ultimate productive- ness depends on other improvements, too extensive or too distant to be embraced by the same individuals." " The General Government ", declares the Report, "can alone remove these obstacles." Its resources are " amply sufficient for the completion of every practicable improvement." " With these resources, and embracing the whole Union, it will complete on any given line all the improvements, however distant, which may be necessary to render the whole productive, and eminently beneficial." he argument continues: The early and efficient aid of the Federal Government is recommended by still more important considerations. The inconveniences, complaints, and perhaps dangers, which may result from a vast extent of territory, can no otherwise be radically removed or prevented than by opening speedy and easy communications through all its parts. Good roads and canals will shorten distances, facilitate commercial and personal intercourse, and unite, by a still more intimate community of interests, the most remote quarters of the United States. No other single operation, within the power of Government, can more effectually tend to strengthen and perpetuate that Union which secures external in- dependence, domestic peace, and internal liberty. What, then, are the specific objects that on this argument require and justify action by the national government? Gallatin derives his answers from a broad view of the geography of the country. The main problems are to improve communications between the northern and southern states and to bring the settlers beyond the mountains into easy communication with the East. With respect to the former, he points out that the United States possesses" a tide water inland navigation . . . Which, from Massachusetts to the southern extremity of Georgia, is principally, if not solely, Interruptedly four necks of land." The four are Cape Cod, New Jersey between the Raritan and the Delaware, the peninsula between the Del- aware and the Chesapeake, and the "marshy tract, which divides the Chesapeake from Albemarle Sound." These should be cut by canals, which would total less than one hundred miles and would be useful " in peace or war" as protection against " storms and enemies". To this should be add ed " a great turnpike extending from Maine to Georgia . . . passing through all the principal seaports." The problem of communication with the West presents a greater difficulty From New York to southern Georgia; the two great ranges of the Appalachians block the way. " In the present state of science," it is useless to think of crossing them by canals. There are, however, places at

Free Essays on American Dreams

American Dream Compare â€Å"Hester Street† and â€Å"Eat a Bowl of Tea† Many years ago, people have immigrated to a new world where they can hope for a more beautiful existence, for the wealth, for the freedoms, for the better opportunities and most importantly, for the American Dreams. As each new era of foreigners migrate to America, they face the obstacle of conforming to mainstream America. As â€Å"Hester Street† and â€Å"Eat a Bowl of Tea† portrayed, immigrants come to this land of opportunities with the hopes and dreams of a better life for themselves or their families. In â€Å"Hester Street†, Jake, a Russian Jewish immigrant who lived in New York's Lower East Side for five years, leaving his wife behind, and taking up with a new woman and earning enough money to support his dance hall ways. On the other hand, in â€Å"Eat a bowel of tea†, Wah Gay is a traditional Chinese immigrant who owns a club in Chinatown, and sends all his money to his wife back in China, who he has not seen in 20 years because of the inequitable immigration laws that had prevented Chinamen bringing their women into the country. These immigrants fall within a lower social class, as a result they strive to conform to a more facilitating and suitable lifestyle. As they begin to build a new life in America, they face the process of assimilation. America holds an idea of a mainstream society; consequently those individuals not fitting this image are left with feelings of abandonment and insecurity. As a result, they feel pressured in achieving the American dream. Let’s look at the examples the movies give us. In â€Å"Hester Street†, Jake, a self-made Yankee, has abandoned the traditions of his culture by cutting off his beard and earlocks, and he has adopted the mannerisms of his new country, including a new girlfriend who runs a dance hall. When his wife Gitl and son Yossele join him from the Old World, Jake was embarrassed. He ... Free Essays on American Dreams Free Essays on American Dreams American Dream Compare â€Å"Hester Street† and â€Å"Eat a Bowl of Tea† Many years ago, people have immigrated to a new world where they can hope for a more beautiful existence, for the wealth, for the freedoms, for the better opportunities and most importantly, for the American Dreams. As each new era of foreigners migrate to America, they face the obstacle of conforming to mainstream America. As â€Å"Hester Street† and â€Å"Eat a Bowl of Tea† portrayed, immigrants come to this land of opportunities with the hopes and dreams of a better life for themselves or their families. In â€Å"Hester Street†, Jake, a Russian Jewish immigrant who lived in New York's Lower East Side for five years, leaving his wife behind, and taking up with a new woman and earning enough money to support his dance hall ways. On the other hand, in â€Å"Eat a bowel of tea†, Wah Gay is a traditional Chinese immigrant who owns a club in Chinatown, and sends all his money to his wife back in China, who he has not seen in 20 years because of the inequitable immigration laws that had prevented Chinamen bringing their women into the country. These immigrants fall within a lower social class, as a result they strive to conform to a more facilitating and suitable lifestyle. As they begin to build a new life in America, they face the process of assimilation. America holds an idea of a mainstream society; consequently those individuals not fitting this image are left with feelings of abandonment and insecurity. As a result, they feel pressured in achieving the American dream. Let’s look at the examples the movies give us. In â€Å"Hester Street†, Jake, a self-made Yankee, has abandoned the traditions of his culture by cutting off his beard and earlocks, and he has adopted the mannerisms of his new country, including a new girlfriend who runs a dance hall. When his wife Gitl and son Yossele join him from the Old World, Jake was embarrassed. He ...

The US Government Framework Assignment Example | Topics and Well Written Essays - 3000 words

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Fashion Photography As A Career Term Paper Example | Topics and Well Written Essays - 1250 words

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