Parrot In the Oven essays

Parrot In the Oven essays Parrot in the Oven by Victor Martinez is a story told by a teenage Mexican American boy, Manny, who is attempting to find his place in a society full of disappointment. It is a story highly influenced by the race and culture of its protagonist. The crises and problems that Manny experiences like love, violence, sibling conflicts, gangs, family problems, and money shortage, are universal obstacles in teenage life. Set in the projects, Manny gives a very realistic account of what it is like to grow up as a minority in a poor, dysfunctional home. Receiving no real direction from his family, Manny battles with what type of man he should and will become. He is tempted by gang life, but at the same time, he seems to have a pure heart that prohibits him from falling too far. This is how life is spent everyday for a Latino boy like Manual Hernandez. Each and every day Manny learned something new about his family, and how to deal with the constant situations. He joined the gang thinking that theyre going to be his family, its good for him, and it can help him with his problems. Victor Martinez writes about a poor, destitute family, and how a little boy tries to live his own lifestyles. He was put in a position to do things he wouldn't normally do. When Eddie, one of the leaders of the gang, decides to steal an innocent lady's purse, Manual is suddenly confuses as to rather go with Eddie or do the right thing. Manual decides to do the right thing, and not go along with Eddie, and the reader is left with a relieving feeling. Victor's writing stays with you all throughout the book, and yet it seems to never leave. Victor Martinez puts hard , struggling emotions into his writing, and it takes his readers to another level. At the beginning of the story, Manny has the goal of buying a baseball glove and to achieve that goal he goes to pick chilies. After working for hours, he is able to pick enough chilies to buy himself a baseball gl...

Cute Guy Quotes

Cute Guy Quotes If you see a really cute guy and want to approach him, what would you do? Turn on your charm and hope that he notices you, or go right up to him and ask him out? Whatever you do, you need to know how guys think. Here are some cute guy quotes to get you started. Work your way into the mind of cute guys and you will never go wrong with your approach. Alphonse Karr: If men knew all that women think, they would be twenty times more daring.Mark Twain: Man is the only creature who has a nasty mind.William Shakespeare, Much Ado About Nothing: O, what men dare do! What men may do! What men daily do, not knowing what they do!Mason Cooley: Young men preen. Old men scheme.Robert Elliott Gonzales: Even the most staid and respectable husband likes for his wife to think he is a devil among the women.Francesca M. Cancian: Part of the reason that men seem so much less loving than women is that mens behavior is measured with a feminine ruler.George Bernard Shaw: Power does not corrupt men; fools, however, if they get into a position of power, corrupt power.Julius Charles Hare: The greatest truths are the simplest, and so are the greatest men.Karen Blixen: What is man, when you come to think upon him, but a minutely set, ingenious machine for turning, with infinite artfulness, the red wine of Shiraz into urine?Jean Giraudoux: When you see a woma n who can go nowhere without a staff of admirers, it is not so much because they think she is beautiful, it is because she has told them they are handsome. Buddha: A dog is not considered a good dog because he is a good barker. A man is not considered a good man because he is a good talker.Oscar Wilde: A man who marries his mistress leaves a vacancy in that position.Buddha: A wise man, recognizing that the world is but an illusion, does not act as if it is real, so he escapes the suffering.Will Cuppy: All modern men are descended from a worm-like creature, but it shows more on some people.Albert Einstein: Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.Richard J. Needham: Every woman needs one man in her life who is strong and responsible. Given this security, she can proceed to do what she really wants to do fall in love with men who are weak and irresponsible.Anais Nin: I, with a deeper instinct, choose a man who compels my strength, who makes enormous demands on me, who does not doubt my courage or my toughness, who does not believe me naive or innocent, who has the courage to treat me like a woman. Linda Ellerbee: If men can run the world, why cant they stop wearing neckties? How intelligent is it to start the day by tying a little noose around your neck?Rita Mae Brown: If the world were a logical place, men would ride side saddle.Henry David Thoreau: In the long run, men hit only what they aim at. Therefore, they had better aim at something high.Helen Rowland: It takes a woman twenty years to make a man of her son, and another woman twenty minutes to make a fool of him.Groucho Marx: Man does not control his own fate. The women in his life do that for him.Jim Backus: Many a man owes his success to his first wife and his second wife to his success.Laura Swenson: Men are like a deck of cards. Youll find the occasional king, but most are jacks.Kathleen Mifsud: Men are like a fine wine. They all start out like grapes, and its our job to stomp on them and keep them in the dark until they mature into something youd like to have dinner with.Carrie Latet: Men like a woman with a daring tongue. Thats a double-entendre, which reminds me they also like a menage a trois. Ninon de Lenclos: Men lose more conquests by their own awkwardness than by any virtue in the woman.Roger Woddis: Men play the game; women know the score.E. H. Chapin, Living Words: Physically, man is but an atom in space and a pulsation in time. Spiritually, the entire outward universe receives significance from him, and the scope of his existence stretches beyond the stars.Maureen Murphy: Some men are so macho theyll get you pregnant just to kill a rabbit.Gloria Steinem: Some of us are becoming the men we wanted to marry.Francis Bacon: The desire of excessive power caused the angels to fall; the desire of knowledge caused men to fall.Natalie Wood: The only time a woman really succeeds in changing a man is when he is a baby.

Criminal Justice Assignment Example | Topics and Well Written Essays - 250 words - 5

Criminal Justice - Assignment Example When she was trying to help her, people came following the screams of the injured girl. The boy managed to lay the blame of the attack on Getrude and she had no defense due to her shyness. Soon, everyone believed it was her who attacked the girl and was labeled a â€Å"vampire† while other labeled her a â€Å"monster†. At first, it was very hard on her as she sat all alone feeling very hated and ashamed of her reputation. However, after some time, her attitude changed when she saw that people treated her with fear and she started developing a feeling of authority and importance. She later accepted the labels she was given and did her best to exceed the expectations of all her peers by being angry for no reason, hedonistic, impulsive and lacking self-control. She made it clear that anyone who wronged her will suffer the outcome. The labels she got, combined with the lack of family support and peer support when she was very vulnerable, propelled her to become unruly and

Contemporary Issues in Accounting Essay Example | Topics and Well Written Essays - 3500 words

Contemporary Issues in Accounting - Essay Example Toyota Motor Corporation, Japan's largest and the world's #4 carmaker by 2003 sales (after General Motors, Ford, and Daimler Chrysler), had a wide range of products and strong brand names with high quality image. Toyota's growing reputation for quality and the very small numbers of technical problems in its vehicles generated interesting customer loyalty and a growing demand for its products. Toyota management was managing the company's inventory, costs and capacity very successfully and was applying cost reduction programs very well. Toyota had riving ambition to become greener. The company made a hybrid-powered (gas and electric) sedan- the prius- that had already been snapped up in U.S. and European markets. Toyota also made huge investments in developing fuel-cell technology for its vehicles. Its gas-powered cars, pick-ups, minivans, and SUVs included such models as the Camry, Celica, Corolla, 4Runner, Echo, Land Cruiser, Sienna, the luxury Lexus line, and a full-sized pick-up tr uck, the v-8 Tundra. With its wide distribution channels, strong channel efficiency and effectiveness, Toyota was successfully competing with the world's upper three auto makers and poised to replace GM in the top spot this decade. Toyota was known world-wide for its up-to-date vehicles, strong vehicle design, comfortableness, safety, strong resistance to wind and rollover, low fuel consumption, presence of electronic and other devices in the vehicles, and strong reputation for luxury. Surveys, however, rated the attractiveness and comfort of its passenger cars as mediocre. Also rated mediocre was the off-road excellence of its SUVs. Toyota was a leader in technological improvements, such as drive, production and vehicle construction technology and had a solid ability to design and innovate new products, to differentiate its products, to innovate new vehicle lines, and to extend existing vehicle lines. Six stakeholders of Company: In the case of Toyota the stakeholders or the users of the annual reports include "present and potential investors, employees, lenders, suppliers and other trade creditors, customers, governments and their agencies and the public. They use financial reports in order to satisfy some of their different needs for information" (Australian Accounting Standards Board, 2004). The improvement in public scrutiny and the controlled market discipline is largely dependent upon the meaningful and accurate disclosure of information. This not only helps the shareholders but also helps the organisation to conduct business in a safe and efficient manner by achieving their targets through improving their risk management processes. The researchers find many gaps in the appropriate disclosure of risk by the organisations. Many surveys have been

Enron collapse A look back Essay Example for Free

Enron collapse A look back Essay Enron was formed as a result of merging with another company and it became a successful corporate. The joy of the business owners is to see how it grows fast and to attract more investors. There are rules and regulation that governs the corporate financial report that is open for inspection by potential investors (Folger, 2011). The audit of these financial reports should disclose the accurate financial state of the company and this should be made known to the stakeholders of the company. The stakeholders of a company play an important role in progress of the business and the going concern of the company (Sterling, 2002). The company of Enron did not manage its debts and therefore looked for means of hiding the truth from its stakeholders so as to continue making profit. Â   Â   Â   Â   Â   The aim of a business is to make profit and be able to pay the debts of the creditors and also attract investors who are interested in the business. Most investor relay on the financial statement to determine whether to invest or not to invest (Folger, 2011). The Enron Company was a big company that was famous and successful before its fall. The corporate attracted many investors since they financial report showed how the business was growing at high rate (Bauer, 2009). However the corporate management did not disclose the true and fair view of the financial reports. The financial report of a company should not mislead the shareholders or its members. Â   Â   Â   Â   Â   Moreover, in the Enron scandal there were some cases in which it showed misconduct of its financial reporting since the corporate did not display true and fair financial accountings to its stakeholders. The corporate used financial fabrication and mark- market accounting to hide its actual debts and real financial situation (Folger, 2011). These reports made the investor believe that the corporate was making profit while it was making losses in real sense. It is also a form of fraud to stakeholders since it cannot meet all its debts and in case of winding up majority of the investors and shareholder would suffer greatly. The Enron scandal was deemed to be great since it had huge debts to settle and its assets could not settle these debts. Â   Â   Â   Â   Â   Indeed, financial misconduct affects a large group of stakeholder and leaves a great mark that cannot be erased (Sterling, 2002). For example the shareholders of Enron corporate were highly affected and suffered a loss of billions that were not recovered since the corporate went bankrupt and the assets of the business were also false in existence. The investors also suffered greatly from these financial misconduct, they lost their resources. In addition, the employees suffered greatly by losing billions of pension benefits due to the misconduct of financial reports which led to the bankruptcy of the Enron therefore could no longer pay them (Folger, 2011). Â   Â   Â   Â   Â   The financial statement of a company is very important to the investors, it gives an over view of the stability of the business and its ability to pay debts. The Enron corporate failed to disclose true and fair view financial statements by hiding its real financial reports and the investors were misled and also due to many investments made by the company lead to its bankruptcy (Sterling, 2002). The purpose of accurate financial reports is to help the investors and also the company to know to what extent they should contract or invest in other businesses. It also helps in managing of the company’s debt thus making profit but the Enron was only interested in making a lot of profits that led to their down fall. Â   Â   Â   Â   Â   Â  Ethics are rules that govern every business and its members on how to conduct their daily roles in the company (Brady, Dunn, 1995). In other words, the managers of this corporate owe their loyalty to its stakeholders and their interest ought to be the interest of the business (Bauer, 2009). In deontology of the Enron corporate, the management had a duty and obligations to display the true statement of finance and also to operate the business in the interest of the stakeholders and not their interest. As the leaders they ought to make sure that the going concern of the business is kept and the assets of the company are secured. In addition, the duty of the Enron managers was to disclose the true and fair view of financial reports (Bauer, 2009). Â   Â   Â   Â   Â   Â  On other hand, utilitarian is a form ethic that is used to show the positive side of the organization, for example by disclosing the false statement to stakeholders thus blinding them of the real situation (Folger, 2011). The Enron leaders used different methods to conceal the truth about its debts and faked the profits. They made the business look attractive and therefore more investors invested in the business. At the end, the ethics rules were violated and the leaders had a role to play .Enron had a role to disclose the truth which could have rescued the company (Bauer, 2009). If at the beginning, the company revealed the truth it would not have ended bankrupt and the employees would have secured their jobs. Finally, the company had a duty to disclose the true financial statements and also save the company from falling and the scandals would have been avoided. The companies should put into practice the ethics governing the corporates. Therefore, to pre vent any future happenings such as the past frauds in the company, there has been an enhanced regulation as well as oversight in the company (Folger, 2011). References Bauer, A. (2009). The Enron scandal and the Sarbanes-Oxley-Act. München: GRIN Verlag. Brady, F. N., Dunn, C. P. (1995). Business meta-ethics: An analysis of two theories. Business Ethics Quarterly, 385-398. Folger, J. (2011). The Enron collapse: A look back. Investopedia, December 1. Retrieved October 25, 2014, at http://www.investopedia.com/financial-edge/1211/the-enron-collapse-a-look-back.aspx Sterling, T. F. (2002). The Enron scandal. New York: Nova Science Publishers. Source document

Proposal for International Foods at the University Dining Services Essa

Proposal for International Foods at the University Dining Services Pedro Cabrera came to State University from Puerto Rico. He decided to be an exchange student at State because he was interested in experiencing life in the United States, and because he was excited about the excellent veterinarian program offered at State. While he loves attending college and living in the United States, there are some things that he would like to change, or rather introduce to State University. Pedro admitted that he feels homesick at times, a feeling not uncommon to new freshmen. He confessed that one of the most important things that he misses about his home is his mother's home cooking. Pedro confessed that the food that the State University dining service provides, does not offer a very wide selection of international food, especially none that are specific to Puerto Rico. Therefore, Pedro would like to propose to the State University dining service that a wider variety of international be incorporated into the daily dining menus. Comfort for international stud ents, variety and accessibility for all students, and a positive aspect of State University are some of the reasons why the introduction of international foods would be beneficial in the residence hall dining services. Pedro shyly admitted before that he feels homesick at times for foods that he eats in Puerto Rico. Like many other students, international or not, food can be a great comfort when students feel homesick. Many students who are not international students can at least relate to the kinds of foods that are found at the dinning services, even if they are not exactly the same as a home cooked meal. For international students, however, the food served at the dining ser... ...idence hall dining services. On an everyday basis, State University tries to help students adjust to college life in one way or another. The dining service would be taking a large step towards assisting international students in this matter. Not only this, the dining service would join in the campaign to educate students in something new, each and everyday, by exposing all students to new international foods. Plus, at the same time, they would be saving all students a great deal of time and money, and not to mention, keeping them excited about eating in the dining services on a daily basis! The proposed way of incorporating international foods to dining service is both simple and efficient. The introduction of new and different international foods to residence hall dining services would benefit everyone who has the pleasure of dining at Iowa State University.

Thermodynamic Analysis and Performance Characteristics

AEROSPACE ENGINEERING SCHOOL OF MECHANICAL ENGINEERING AND DESIGN THE THERMODYNAMIC ANALYSIS AND PERFORMANCE CHARACTRISTICS OF A TURBOFAN JET ENGINE By J. E, Ibok 2011 Supervisor: Dr Lionel Ganippa ABSTRACT This work focuses on the performance analysis of a twin spool mixed flow turbofan engine. The main objective was to investigate the effects of using hydrogen, kerosene and natural gas fuel on the performance characteristics such as net thrust, specific fuel consumption and propulsive efficiency of the turbofan.Another aim of this work was to introduce the concept of exergy and thermoeconomics analysis for twin spool mixed flow turbofan engine and show the components that contributes the most to the inefficiency of the engine. A generic simulation was carried out using Gas Turb 11 software to obtain reasonable analysis results that were verified with a real-time JT8D-15A turbofan engine. The parametric analysis was done for constant value of mass flow rate of fuel and constant turb ine inlet temperature for all three fuels.The result were rightfully obtained for these analysis cases and discussed accordingly. Brunel University Mechanical Engineering Academic Session: 2010/2011 Name of Student: Johnson Essien Ibok Supervisor:Dr Lionel Ganippa Title: The Performance Characteristics and Thermodynamics Exergy and Thermoeconomics analysis of a Twin Spool Mixed Flow Turbofan Engine Operating at 30,000ft at M0 0. using Kerosene, natural Gas and Hydrogen Fuel. Abstract: This work focuses on the performance analysis of a twin spool mixed flow turbofan engine. A generic simulation was carried out using Gas Turb 11 software to obtain reasonable analysis results that were verified with a real-time JT8D-15A turbofan engine. The parametric analysis was done for constant value of mass flow rate of fuel and constant turbine inlet temperature for all three fuels.The result were rightfully obtained for these analysis cases and discussed accordingly. Objectives: The main aim of this work is to conduct the parametric cycle simulation of a twin spool mixed flow turbofan engine and investigate the performance characteristics of it. Another aim of this work is to show the effects of using hydrogen, Kerosene and natural gas fuel on the overall performance of the twin spool mixed flow turbofan engine.Also, the purpose of this work is to introduce the use of the second law of thermodynamics analysis known as exergy and thermoeconomics in analysis the twin spool mixed flow turbofan engine Background/Applications: This work is applicable in so many ways when it comes to the overall performance optimization and feasibility analysis of a jet engine. This work relates to the aerospace and aviation industries since the turbofan engine is amongst the vast number of jet engine used in propulsion of aircrafts.There is increasing pressure in the aviation industry to reduce pollution and depletion of energy resources while at the same time maintaining reasonable investment cost and high overall performance. Hence, this research was conducted in hopes of coming up with a new solution to this problem. Conclusions: The main conclusion drawn from the performance analysis is that hydrogen fuel produced the highest thrust level and the lowest specific fuel consumption between the three fuels for a constant mass flow rate of fuel.Kerosene fuel generated thrust level can be increased if it is mixed with a small amount of hydrogen. The Exit jet velocity ratio remained constant despite the increasing bypass ratio for all three fuels at constant mass flow rate of fuel. Using the exergetic analysis showed that the combustion chamber and the mixer contributed the most to the inefficiency of the turbofan engine. The amount of exergy transferred into the turbofan engine by hydrogen was depleted in the smallest ratio compared to natural gas and kerosene for constant mass flow rate of fuel.The thermoeconomics analysis showed that it is preferable to use local based co st evaluation to quantity specific thermoeconomics cost of thrust than the global method since the value was lower. Results: The results obtained from the simulation using Gas Turb 11 produced an error range of 0. 25% – 8. 5% when verified with the actual test data of the JT8D-15A turbofan engine. The results obtained for the analysis defined a reference design point at which the parametric analysis was conducted on. The analysis was done in three cases as shown clearly in the test matrix in table 1 below.Analysis| Parameters being varied| Parameters Kept Constant| Performance Characteristics| case 1| * Bypass ratio * Turbine Inlet temperature| * HPC Pressure Ratio * LPC Pressure Ratio * Fan Pressure Ratio| * Velocity ratio * Fuel-Air-ratio * Turbine inlet temperature * Net thrust * Specific Fuel Consumption * Thermal efficiency * Propulsive efficiency| case 2| * Bypass Ratio * Three different fuelsmH2mCH4mC12H23| * Mass flow rate of fuel * HPC Pressure Ratio * LPC Pressure R atio * Fan Pressure Ratio| | Case 3| * Bypass Ratio * Three different fuelsmH2mCH4mC12H23| * Turbine inlet temperature * HPC Pressure Ratio * LPC Pressure Ratio * Fan Pressure Ratio| | Table 1 The Test matrix of the Parametric Analysis. The exergy analysis was done for the parametric analysis of case 2 and case 3 where the exergy destruction rates, exergetic efficiency, exergy improvement potential rate and fuel depletion ratio were calculated. The distribution of these results throughout each component of the turbofan engine was represented with bar charts and Grassmann diagram. The thermoeconomics analysis was conducted for analysis case 2 using kerosene fuel.The specific thermoeconomics cost of thrust was calculated using global and local based cost evaluation methods. ACKNOWLEDGEMENTS First of all, I would like to thank my parents for their financial support and encouragement because without them I would not be here and be able to do this work. I am deeply thankful to my supervi sor, Dr Lionel Ganippa for believing in me and giving me the opportunity to work with him in this field of study. I am also thankful to him for giving the necessary guidance and advice and his enthusiasm and innovative ideas inspired me. Finally, I would like to thank Mr Joachim Kurzke for providing me with the necessary software needed for my dissertation. Table of ContentsAcknowledgements i Contents ii List of Notations and Subscripts iv List of Tables vi List of Figures vi Chapter 1: Introduction1 1. 1. Aims and Objectives2 1. 2. Computational Modeling3 Chapter 2: Jet Engines4 2. 1. Performance characteristics4 2. 1. 1. Thrust4 2. 1. 2. Thermal Efficiency5 2. 1. 3. Propulsive efficiency5 2. 1. 4. Overall efficiency6 2. 1. 5. Specific Fuel Consumption6 2. 2. Fuel and Propellants For Jet Engines7 Chapter 3: Turbofan Jet Engines †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ †¦8 3. 1. Introduction 8 3. 2. Classification of Turbofan Engines9 3. 3. Major Components of a Turbofan Engine10 3. 3. 1. Diffuser10 3. 3. 2. Fan and Compressor11 3. 3. 3. Combustion Chamber12 3. 3. 4. Turbine13 3. 3. 5. Exhaust Nozzle14 3. 4.Thermodynamic Process and Cycle of a Twin Spool Mixed Flow Turbofan Engine15 Chapter 4: Mathematical and Gas turb 11 Modeling of the turbofan Engine18 4. 1. Station Numbering and Assumptions18 4. 2. Design Point Cycle Simulation of the Turbofan Engine18 4. 3. Off-design Point Cycle Simulation of the Turbofan Engine21 4. 3. 1. Module/Component Matching 22 4. 3. 2. Off-Design Point Component Modeling22 Chapter 5: Methodology, Results and Discussions26 5. 1. General Relationship equations of the Major Parameters27 5. 2. Results and Discussions of Parametric cycle Analysis of Case 129 5. 3. Results and Discussions of Parametric Cycle Analysis of Case 235 5. 4.Results and Discussions of Parametric Cycle Analysis of Case 343 Chap ter 6: Exergy and Thermoeconomics Analysis of the Turbofan Engine49 6. 1. Exergy Analysis49 6. 1. 1. Exergy Analysis Modeling 50 6. 1. 2. Exergy and Energy Balance Equations of the Components58 6. 1. 3. General Relationships in Exergetic Analysis of the Turbofan Engine60 6. 1. 4. Results and Discussions61 6. 1. 5. Grassmann Diagram72 6. 2. Thermoeconomics Analysis74 6. 2. 1. Thermoeconomics Analysis Modelling74 6. 2. 2. Global Based Cost Evaluation76 6. 2. 3. Local Based Cost Evaluation77 6. 2. 4. Results and Discussion of the Thermoeconomics Analysis78 Chapter 7 Conclusions and Future Work80 Reference Appendix A Exergy Analysis Results Appendix B Thermoeconomics Analysis resultsList of Notations and Units ?| Isentropic efficiency| ?| Total Pressure ratio| m| Mass Flow Rate (kg/s)| f| Fuel/Air Ratio| M| Mach Number| Pt| Total pressure (kPa)| Tt| Total Temperature (K)| NCV| Net Calorific Value (MJ/kg)| Ht| Total Enthalpy (kJ/kg)| V| Velocity (m/s)| ?| Bypass Ratio| T| Static Temperat ure (K)| P| Static Pressure (kPa)| N| Actual Spool Speed (RPM)| Nc| Corrected Spool Speed (RPM)| mc| Corrected Mass Flow Rate (kg/s)| R| Universal Gas Constant (kJ/kmolK)| ?0| Standard Chemical Exergy (kJ/kmol)| Ex| Exergy Rate (MW)| xi| Mole Fraction| cp| Specific Heat at Constant Pressure (kJ/kgK)| ?| Ratio of Chemical Exergy to NCV| ?| Exergetic Efficiency| | Fuel Depletion Ratio| W| Power Rate of Work done (MW)| List of Subscripts| | LPT| Low Pressure Turbine| HPT| High Pressure Turbine| CC| Combustion Chamber| HPC| High Pressure Compressor| LPC| Low Pressure Compressor| d| Diffuser| noz| Nozzle| mix| Mixer| dest| Destruction Rate| 0, ambFAR| Ambient conditionFuel-Air-Ratio| CH| Chemical| PH| Physical| KN| Kinetic| PN| Potential| IP| Exergy Improvement Potential Rate (MW)| CRF| Cost Recovery Factor| c| Specific Thermoeconomic Cost (MJ/kg)| STD| Standard Temperature and Pressure| TIT| Turbine Inlet Temperature| TSFC| Thrust Specific Fuel Consumption (g/kNs)| SFC| Specific Fuel Co nsumption| p| Propulsive| TH| Thermal|O| Overall| T| Thrust| equip| Equipment| PEC| Capital Cost of Equipment| List of Tables Table 1 input parameters for Design Point Cycle Simulation on Gas Turb 1119 Table 2 Comparison table for the Actual Test Data and Simulated Data using gas Turb 1121 Table 3 Comparison Table for Actual Test Data and Simulated Off-Design Point data Using gas Turb 11. 25 Table 4 Equivalence Ratio of the three Fuels Combustion Processes†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ 62 Table 5 Assumed Capital costs of Each Component of the Turbofan Engine. 75 Table 6 Flow of Specific Thermoeconomics Cost in all the Components 79 List of Figures Figure 1 Classification of Turbofan Engine9Figure 2 Layout of Forward Fan Twin Spool Mixed Flow Turbofan16 Figure 3 T-S Diagram for the Forward Fan Twin Spool Mixed Flow Turbofan17 Figure 4 Design Point Cycle Simulation Algorithm Using Gas Turb 1120 Figure 5 Example of a Compressor Performance Map/Cu rve24 Figure 6 Effects of Varying Bypass Ratio at Constant Values of TIT on Fuel-Air-Ratio30 Figure 7 Effects of Varying Bypass Ratio at Constant Values of TIT on Exit Velocity Ratio30 Figure 8 Effects of Varying Bypass Ratio at Constant Values of TIT on LPT Exit Pressure Ratio31 Figure 9 Effects of Varying Bypass Ratio at Constant Values of TIT on Net Thrust32 Figure 10 Effects of Varying Bypass Ratio at Constant Values of TIT on Specific Fuel Consumption33 Figure 11 Effects of Varying Bypass Ratio at Constant Values of TIT on Propulsive Efficiency34 Figure 12 Effects of Varying Bypass Ratio t Constant Values of TIT on Thermal Efficiency35 Figure 13 T-S diagram of using Hydrogen Fuel when the bypass Ratio is increased36 Figure 14 Variation of Fuel-Air-Ratio with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels37 Figure 15 Variation of TIT with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels37 Figure 16 Variation of Exit Velocity Ratio with Byp ass Ratio at Constant Fuel Flow Rate using three different Fuels38 Figure 17 Variation of LPT Exit Pressure Ratio with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels39 Figure 18 Variation of Net Thrust with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels40 Figure 19 Variation of Specific Fuel Consumption with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels41 Figure 20 Variation of Thermal Efficiency with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels42 Figure 21 Variation of Propulsive Efficiency with Bypass Ratio at Constant Fuel Flow Rate using three different Fuels43 Figure 22 Variation of Fuel-Air-Ratio with Bypass Ratio at Constant TIT using the three Different Fuels44 Figure 23 Variation of Exit Velocity Ratio with Bypass Ratio at Constant TIT using the three Different Fuels44 Figure 24 Variation of LPT Exit Pressure Ratio with Bypass Ratio at Constant TIT using the three Different Fuels45 Figure 25 Variation of Net Thrust with Bypass Ratio at Constant TIT using the three Different Fuels46 Figure 26 Variation of Specific Fuel Consumption with Bypass Ratio at Constant TIT using the three Different Fuels46 Figure 27 Variation of Propulsive Efficiency with Bypass Ratio at Constant TIT using the three Different Fuels47 Figure 28 Variation of Thermal Efficiency with Bypass Ratio at Constant TIT using the three Different Fuels48 Figure 29 Variation of Exergy Destruction Rate Using the three Fuels for Analysis Case 262 Figure 30 Variation of Exergy Destruction Rate Using the three Fuels for Analysis Case 364 Figure 31 Variation of Exergetic Efficiencies Using the three Fuels for Analysis Case 266 Figure 32 Variation of Exergetic Efficiencies Using the three Fuels for Analysis Case 367 Figure 33 Distribution of Exergy Improvement potential Rate Using the three Fuels for Analysis Case 268 Figure 34 Distribution of Exergy Improvement potential Rate Using the three Fuels for Analysis Case 369 Figure 35 variation of Fuel Depletion ratio using the Three Fuels for Analysis Case 270 Figure 36 variation of Fuel Depletion ratio using the Three Fuels for Analysis Case 371 Figure 37 Grassmann Diagram for the Exergetic analysis of Case 2 using kerosene Fuel for the Turbofan engine. 72 Chapter 1 Introduction Jet engines are complex thermodynamic systems that use a series of non-linear equation to define their thermodynamic processes and they operate under the principle of Brayton cycle.Brayton cycle is a cycle that comprises of the compressor, combustor and turbine working as a unit. Additionally, the major parameters that dictate the operational conditions of the engine at any point during the process are the relative altitude and Mach number. Mach number is the ratio of the velocity of the jet engine to the speed of sound. Basically, the main purpose of this type of thermodynamic system in aerospace industry is to accelerate a jet of air and as a result, generate enough thrust needed for flight. In addition, the design of jet engines is dependent of what purpose it will be used for in order to derive its maximum performance.For instance, in military application, jet engines are required to generate maximum thrust in minimum response time which consumes a lot of fuel whereas commercial jet engines are required to less noise generative, less fuel consuming and at the same time have high overall efficiency (El-sayed, 2008). There are certain factors that jet engine manufacturers take into consideration when designing jet engines which are the operating cost, engine noise, environmental emissions, fuel burn and overall efficiency. Accordingly, this has caused a global market competition for engine manufacturers like Rolls Royce, Pratt and Whitney, General Electric and CFM on who can produce the most efficient jet engines.In fact, Pratt and Whitney Company is working on a geared turbofan jet engine that they believe will reduce fuel burn, produce lesse r noise and emit less toxics while General Electric is coming up with simpler â€Å"ecore† jet engines that will be more fuel efficient than the current jet engines with as much as almost two fifths of current jet engines (Cassidy, 2008). Taking all that has been said into consideration, it can easily be asserted that by reducing the fuel consumption of the jet engine, the total temperature at the turbine blades will reduce thereby increasing the operating life and overall efficiency of the engine. Also, the total cost of the engine can be cut down. Indeed, Dr Pallan cited in (Ward, 2007) stated that reducing the fuel consumption by as little as 1% is highly longed after by engine manufacturers and this can result in very significant increase in the overall performance.In a general point of view, it can be said that the maximum point of achievement for jet engine manufacturers would be to design an engine that consumes the minimum amount of work in the compressor unit while g enerating the maximum amount of work in the turbine unit at minimum fuel supply. The main purpose of this work is to analyse the thermodynamic processes and performance of a jet engine using a simulation tool, exergy and thermoeconomics concept. 1. 1. Aims and Objectives The main objective of this work is to carry out the thermodynamic analysis and show the performance characteristics of a turbofan jet engine. In this work, the vivid explanation of the thermodynamics processes and cycle of each component of the turbofan engine starting from the diffuser to the nozzle will be covered. Also, the first and second law of thermodynamics with other laws will be applied extensively throughout this work.However, in the aspect of performance characteristics of the turbofan engine, a generic simulation will be carried out on a twin spool mixed flow turbofan engine. To relate this work to real life application, a JT8D-15A turbofan engine manufactured by Pratt and Whitney Company will be used a s the twin spool mixed flow turbofan for the simulation using the original design data. Indeed, the simulation tool that will be used is GasTurb 11 which was designed by Joachim Kurke and for more details on how it works can be found in (Kurke, 2007). This work will use the reference design point of the twin spool mixed flow turbofan at sea level with maximum take-off thrust to obtain the operating point of 30,000ft at M0 0. using the off-design performance simulation which will serve as the operating design point for the analysis in this work since the engine will spend most of its time in the cruise phase between 30000ft to 38000ft. The purpose of carrying this generic simulation of the turbofan engine is to investigate the effects of varying bypass ratio and turbine inlet temperature (thermal limit parameter) on the performance characteristics of the turbofan engine. In other words, the parametric cycle studies of the turbofan engine. This investigation will be done for three dif ferent cases which case 1 will be studying the effects of varying bypass ratio and turbine inlet temperature on the performance characteristics of the turbofan engine when some of the design choices are kept constant.The second case of study will be the comparison of the performance characteristics of the turbofan engine when three different fuels (kerosene, natural gas and Hydrogen) are used at the same mass flow rate using the same design point in case 1. Finally, the third case of study will be the comparison of the performance characteristics of the turbofan engine when the three fuels are undergoing the same combustion process that is constant turbine inlet temperature for the design point in case 1. This aspect of this analysis is very important owing to the growing problem of greenhouse effect and depletion of energy resources. In fact, statistics by the intergovernmental panel shows that aerospace industry is amongst one of the fast growing sources of greenhouse effect and t hat the emission of carbon dioxide will increase to five times what it is presently which is 3% (Symonds, 2005).Based on this, using alternative fuels like hydrogen and natural gas can tend to reduce pollution and consumption of energy resources risk and this work aims to show how that can be achieved while the overall efficiency of the engine is still high. Another approach of analysis in this work will be the use of the second law of thermodynamics analysis also known as exergy and thermoeconomics. This aspect of analysis of the turbofan engine will be done for the parametric analysis of case 2 and case 3 in efforts to also compare the three fuels that are being considered and show which fuel will cause the turbofan engine components to be most inefficient or have the most irreversibility.This analysis will be done by calculating the exergy relationships such as exergy transfer rates, exergy destruction rates, exergetic efficiencies, exergy improvement potential rates, and fuel de pletion ratios. Furthermore, the exergy analysis will be represented in a Grassmann diagram for parametric analysis case 2 of study. However, as for the thermoeconomics analysis of the turbofan engine, only parametric analysis case 3 studies will be done for only kerosene fuel and this work will aim to show how to use concept of local and global evaluation of thermoeconomic cost. 1. 2. Computational Modelling It will be very expensive and time wasting to design and develop new aircraft engine whenever an optimization or analysis wants to be done.In fact, Caoa Y, Jin, Meng and Fletcher (2005) stated that new ways should be developed to reduce aircraft engine design, maintenance and manufacturing cost in order to have effective worldwide market competition. Surprisingly, computer modelling is one approach of reducing manufacturing cost and time wasting. Computational modelling can simply be defined as the use of computer codes to replicate a typical system using some of its original d ata in order to analyse the system at varying conditions. The other side of the medallion shows simulation. There are many types of simulation tools normally used in simulating gas turbines such as Matlab/simulink, Modelica, Gas Turb 11, NPSS and many more. However, the simulation tool that will be adopted for the purpose of this dissertation is Gas Turb 11 designed by Joachim Kurzke.Gas Turb 11 is a language oriented program with a command prompt that calculates the output data without using block diagrams or graphical interface. It is user friendly in a sense that it is easy to find the tools library and to substitute data in for simulation. The Gas Turb 11 is specifically designed for simulation of all kinds of gas turbines starting from power generators to jet engines. Gas Turb 11 usually carries out two types of analysis which are the on design cycle point simulation and off-design cycle point simulation. Engine design point cycle simulation involves the study of comparing gas turbines of different geometry. This cycle design point must be defined before any other simulation can be done.On the other hand, off-design performance cycle point simulation involves the study of the behaviour of a gas turbine with known geometry. This cycle outlines the performance characteristics of each component such as performance maps, Overall efficiency. The type of simulation that will be done in this dissertation will involve the off-design and design point cycle. Chapter 2 Jet Engines 2. 1. Performance Parameter of Jet Engines 2. 2. 1. Thrust Thrust is the way of quantifying the ability of a jet engine to effectively utilise the energy added to it in order to propel or push itself forward in the opposite direction of the exiting jet in the exhaust nozzle.In other words, it is the reactive force to the force imparted by the exiting jet in the nozzle in accordance to Isaac Newton’s third law of motion. It is the most important parameter that has to be obtained for any jet engine and it depends heavily on the ingested mass of air, exiting velocity and pressure, the area of the nozzle, the flight velocity and ambient conditions. In fact, the mathematical expression for thrust which incorporates these factors is shown below as. Thrust=meVe-m0V0+Pe-P0Ae Where, e=the exit conditions at the exhaust nozzle, 0=ambient conditions at the inlet me=m0+mfuel Momentum Thrust=meVe; This is the thrust obtained from the reaction of the hot exhaust gases high velocity.Momentum Drag= m0V0 ; This the friction or drag force caused by the high velocity ingestion of air mass at the inlet. Pressure Thrust=Pe-P0Ae; This force is generated as a result of the higher exit static pressure compared to the ambient pressure which pushes back at the engine. Gross Thrust=meVe+Pe-P0Ae; It is the maximum obtainable positive thrust a jet engine can have when the drag forces are ignored. Special Cases of Thrust Take-off Thrust It is the thrust a jet engine can generate with its o wn power at static or low power setting which means the momentum drag component of thrust is ignored and the power of the engine at this point is equivalent to zero.This can be used to explain why the thrust of an engine at take-off condition is usually higher than at cruise condition since there is no momentum drag and effects of varying ambient condition. This only applies to turbojet, turbofan, and turboprop jet engines but when it comes to ramjet and scramjet, the air flow has to be accelerated by a booster system before it can start producing a positive take-off thrust. Pressure Thrust Component This is the thrust generated as a result of the static pressures of the exiting jet and ambient environment. In ideal cases where the nozzle has perfectly expanded the jet exit pressure to that of the ambient condition, the pressure thrust component will disappear which this case is not possible in reality.However, if the nozzle is choked which indicates that the ambient pressure is low er than the exit pressure of the jet, the pressure thrust component will have a positive effect on the net thrust. Also, if the nozzle tends to over expand the jet because of low energy addition to the jet and the exit pressure is lower than the ambient pressure, the pressure thrust component will have a negative effect on net thrust. 2. 2. 2. Thermal efficiency It is simply the measure at which energy in the engine system is converted. In other words, it is the measure at which total energy supplied to the engine system as heat transfer is converted to kinetic energy.In another way, it can easily be said to be the ratio of the power generated in the engine airflow to the rate at which energy is supplied in the fuel. ?TH=Power Generated in the Engine AirflowRate of Energy Supplied in the Fuel =12? meVe2-12? m0V02mfuel? NCV 2. 2. 3. Propulsive efficiency It is a measure at which kinetic energy possessed by air as it passes through the engine is converted into power of the propulsion of the engine. In mathematical terms, it is simply known as the ratio of thrust power to the power generated in the engine airflow. ?p=Thrust PowerPower Generated in the Engine Airflow = T? V012? meVe2-12? m0V02 2. 2. 4. Overall EfficiencyAs the name overall depicts, it is the resultant efficiency of a jet engine can have which is simply the product of the thermal and propulsive efficiencies. In mathematical terms, it is represented as shown below. ?O=? TH p =12? meVe2-12? m0V02mfuel? NCV? T? V012? meVe2-12? m0V02 =T? V0mfuel? NCV 2. 2. 5. Specific Fuel Consumption Specific fuel consumption as any other performance characteristics is a ratio and surprisingly it has a major effect on the economics of the aircraft as it is used to determine the aircrafts flight ticket costs. Specific fuel consumption has different expressions depending on what type of jet engine it is. For instance, in ramjet, turbojet and turbofan jet engines, it is the measure of the fuel mass flow rate to the thrus t force generated.Also, it is sometimes called the thrust specific fuel consumption (TSFC). TSFC=mfT However, in turbopropeller jet engines, it is the ratio of the fuel mass flow rate to the power generated in the engine shaft by the turbomachinery. It is sometimes referred to as the brake-specific fuel consumption (BSFC). TFSC=mfSP 2. 2. Fuel and Propellants for Jet Engines Fuels can implicitly be defined as substances used to add heat energy to a system through combustion or other processes. Fuels are mostly hydrocarbons like kerosene, diesel, petrol, alcohol, paraffin and butane and can also be in the form of individually free reactive molecular substances like hydrogen or chemical composites like natural gas, coal, wood.The gaseous state substances used as fuels such as hydrogen, and natural gas (94% methane and 6% ethane) are usually made into a cryogenic state as in liquefied at very low temperature because of their low boiling point. It can easily be asserted by anyone that t he only purpose that fuels have in jet engines is to add energy but little do they know that the purposes grows as the speed of the aircraft increases. For instance, Kerrebrock (2002) stated that supersonic aircrafts which attains very high stagnation temperature that can create destabilization to the airframe structure, engine component and organic substances like lubricants, uses its fuel as a coolant to this parts or components.The energy added by the fuel burned per unit mass of air flow is called the heating value of the fuel and it is a very crucial parameter to be defined before any combustion process analysis is done on a jet engine since it shows how complete the combustion process is through efficiency. The heating value can either be said to be higher or lower depending on if the water product of combustion is a vapour or a liquid. Since the combustion process in jet engine produces vaporised water, the lower heating value of the fuel is used. The most frequently used fue ls for jet engines are kerosene jet A1, A2, JP10 and many more but diesel can also be used. The disadvantages of these fuels are their inevitable emission of toxic substances that contribute to greenhouse effect and their risk of depletion.Accordingly, this has been the driving force for the use of alternative fuels such as cryogenic hydrogen and natural gas which is believed will reduce toxic emissions. Besides, hydrogen is a carbon-free energy carrier and possesses almost no risk of toxic emission since most of its combustion product will be water Chiesa and Laozza (2005). Chapter 3 Turbofan Jet Engine 3. 1. Introduction Between 1936 and the next decade when turbofan engines were invented, people showed little or no interest in them as they described them to be a complicated version of a turbojet engine. However, in 1956, the benefits of turbofan engines started to be noticed as major companies like Rolls-Royce and General Electric began manufacturing them.Since then, it is been o ne of the most used jet engine for commercial purposes because of its low fuel consumption and less noise production. In fact, it has been concluded to be the most reliable jet engine ever manufactured El –Sayed (2008). The turbofan jet engine gas generator unit comprises of a fan unit, compressor section, combustion chamber and turbine unit. Fundamentally, a turbofan jet engine operates as a result of the compressors pressuring air and supplying it afterwards for further processing. The majority of the pressurised air is bypassed around the core of the engine through a duct to be mixed or exhausted whereas the rest of it flows into the main engine core where it combusts with the fuel in the combustion chamber.The hot expanded gas products from the combustion process passes through the turbine thereby rotating the turbine as it leaves the engine. Consequently, the rotating turbine spins the engine spool which in turn rotates the other turbo machinery in the engine. This cause s the front fan to pressurise more and more air into the engine for the process to start all over again in continuous state. The turbofan engine is believed to be the perfect combination of the turboprop and turbojet engine and as a result, its advantages are usually compared to that of the turboprop and turbojet. In fact, Kerrebrock (1992) said that turbofan engine provides a better way of improving the propulsive efficiency of a basic turbojet.It is asserted that at low power setting, low altitude condition and low speed, the turbofan engine is more fuel efficient and has better performance than a turbojet engine. Unlike turboprop engine where vibration occurs in the propeller blades at relative low velocities, the fan in the turbofan engine can attain high relative velocities of Mach 0. 9 before vibration occurs. Also, since the fan in turbofan engines has many blades, it is more stable than the single propeller so even if the vibration velocity is reached, the vibration will not destabilize the airflow because the vibrations are almost negligible. Since the flow into the diffuser of the turbofan is usually subsonic, there very slim chances of shock waves being developed at the entrance. 3. 2. Classification of Turbofan EnginesThere are various types of turbofan engine ranging from high and low bypass ratio, afterburning and non-afterburning, mixed and unmixed flow with multi-spool, after fan and geared or ungeared. The classification of the various types of turbofan engines is shown below in figure 1. Nonetheless, the type of turbofan engine that would be used for the purpose of this dissertation is a forward fan two spool mixed flow turbofan engine. This type of turbofan engine was chosen because it is the compromise of a simple and complex turbofan engine. This is said because it comprises of almost all the classes of a turbofan which are low bypass ratio, forward fan with mixed flow, twin spool with ungeared fan.Moreover, because of the mixed flow intro duced, it produces additional thrust in the hot nozzle compared to the high bypass and it can also permit the addition of afterburner which produces a lot of thrust while consuming a lot of fuel which makes it suitable for military application which shows little worry on fuel consumption. In essence, carrying out a study on this type of turbofan engine will be of great relevance to the military air force sector especially if new research is discovered. TURBOFAN ENGINES Low Bypass Ratio Aft Fan Forward Fan Nonafterburning Afterburning High Bypass Ratio Geared Fan Single Spool Short Duct Ungeared Fan Two Spool Mixed Fan and Core Flow Unmixed Flow Long Duct Three SpoolFigure 1 Classification of Turbofan Jet Engines (El-sayed, 2008) 3. 3. Major Components of Turbofan Engine 3. 4. 1. Diffuser or Inlet Diffuser is the first component that air encounters as it flows into the engine. Basically, the purpose of a diffuser is to suck in air smoothly into the engine, reduce the velocity of the air, increase the static pressure of the air and finally, supply the air in a uniform flow to the compressor. Given the fact that overall performance of an engine is highly dependent on the pressure supplied to the burner, it is necessary to design a diffuser that incurs the minimum amount of pressure loss.To demonstrate this, Flack (2005) stated that if the diffuser incurs a large total pressure loss, the total pressure in the burner will be reduced by the compressor total pressure ratio time this loss. In other words, a small pressure drop in the diffuser can translate into a significant drop in the total pressure supplied to the burner. Another point taken into consideration when designing a diffuser is the angle because if the angle is too big, there will be tendency of eddy flow generation due to early separation. The major causes of pressure losses in the diffuser are as follows. First, losses due to generation of shock waves outside the diffuser and it majorly occur in super sonic diffusers.Secondly, the loss due to the unfavourable or adverse pressure gradient of the diffuser geometry which makes the flow separate a lot earlier and generates eddies. This separation causes a convergent area which makes the velocity not to be reduced by much. Due to the separation, the wall shear deteriorates the static pressure even further. Further analysis done by El-Sayed (2008), describes ways of accounting for this losses like using Fanno line flow and combined area and friction. Thermodynamic Process Equation In this analysis, the loss due to heat transfer is negligible so the process can be adiabatic. The initial kinetic energy is used to raise the static pressure p0 to the total pressure ? =pt2pt0 (inlet pressure recovery) efficiency ? d=IdealReal=ht2s-h0ht2-h0 assuming the gas is ideal and the specific heat at constant pressure is constant efficiency ? d=Tt2s-T0Tt2-T0 simplifying the equation given that ht0=ht2=ht2s and Tt2=Tt0and pt2s=pt2 TtT0=1+? -12M02 and T tT0=ptp0? -1? pt2p0=1+ ? d? -12M02 -1 3. 4. 2. Fan And Compressors Compressor is a very crucial component for the operation of an engine in the sense that it prepares the air for the combustion process in the burner. The main purpose of a compressor as the first rotating component is to use its rotating blades to add kinetic energy to the air and later translate it into total pressure increase.There are basically two types of compressors which are the centrifugal and the axial compressor. Firstly, centrifugal compressor as the name implies changes the direction of an axial airflow to a radial outflow of the air. It was the early compressors adapted in jet engines. It comprises of three main parts which are the impellers, the diffusers and the compressor manifold. The purpose of the impeller is to change the direction of the flow from axial to radial and at the same time increases its static pressure. The diffuser slows down the airflow and further increase the static pressure as it is supplied axially by the compressor manifold to the combustion chamber.The centrifugal compressor is advantageous because the cost of manufacturing it is low compared to axial compressor and as a result is suitable for small engines like turboshafts and turboprops. It is also advantageous because the pressure ratios at single stage are higher than that of the axial compressor. The centrifugal compressor has the tendency of attaining low flow rates and as a result is ideally suitable for helicopters and small aircrafts which require low flow rates. On the other hand, the centrifugal compressor cannot attain high pressure ratio and so it is not suitable when high peak efficiency is required. It incurs a lot of losses due to the change in direction. Secondly, an axial compressor is the most reliable type of compressor and is usually applied when higher pressure ratios of up to 40:1 are required.An axial compressor does not change the axial flow direction of the air but increases the total pressure. Indeed, an axial compressor comprises of three major components which are the rotor with blades, stator can and the inlet guide vane. A stage is a combination of a stator and a rotor. The assembly of the full rotor blade and stator can form the number of stages in a compressor and the greater the number of stages, the higher the total pressure ratio. In this arrangement, the air flows into the inlet guide vane and then into the rotor and stator assembly where compression starts. Also, the length of the rotor and stator reduces along the whole unit which signifies a reduction in volume which induces the increase in pressure.A fan or low pressure compressor is a type of axial compressor but the only differences are that the blades are longer, the total pressure ratio is lower than the typical compressor and the number of stages is usually 1 or 2. The main purpose of creating a fan is to compress more air and to create a bypass air which can be used to generate addition thrust or used for mixing process. Fan Equation Process Given that, isentropic efficiency ? fan= Ideal CycleActual cycle=ht3s-ht2ht3-ht2 Since the specific heat is constant, the equation deduces to ? fan=Tt3s-Tt2Tt3-Tt2 Simplifying the equation whenpt3s=pt3, Tt3sTt2=pt3pt2? -1? , ? fan=pt3pt2 and ? fan=Tt3Tt2 ? fan=? fan? -1? -1? fan-1 Bypass Ratio=msma where ms is the bypass flow rate and ma is the engine core flow rate.For the high pressure compressor, the equations remain the same as that of the fan except the changes in station numbering and the bypass ratio. 3. 4. 3. Combustion Chamber/ Burner The combustion chamber as the Brayton cycle implies is the only source of heat energy addition to the system. Accordingly, the combustion chamber causes very significant increase in the temperature of the air which results in the air gaining enormous internal energy. This energy gained is extracted to be used to power the turbine while the rest is used to create highly accelerated gases from the nozzle. There are three types of combustor namely; the can combustor, the annular combustor and the cannular combustor.The main considerations when designing a combustion chamber is to ensure that the combustion process is complete with no fuel waste, the combustor should have long life materials because any failure can lead to engine explosion. The other consideration is that the air must be heated enough above the ignition fuel temperature in order to ensure stoichiometric combustion. Equations of the Combustion Chamber In the real process of the combustor, total and static pressure drops and the temperature also drop. The major causes of pressure losses are the high level of irreversibility or non-isentropic process and viscous effects in the burner. The burner pressure ratio ? =pt5pt4Burner temperature ratio ? b=Tt5Tt4 Since no work is done only heat transfer, the efficiency of the burner is analysed using the heating value NCV of the fuel used. Thus, efficiency ? b=hea t addedHeating value of fuel=ma+mfht5-maht4NCVmf Given that f=mfma, ? b= 1+fht5- ht4NCVf Equivalence Ratio of combustion It is the ratio of the actual fuel to air ratio of the combustion process to the stoichiometric fuel to air ratio. This ratio produces a means of classifying the combustion process to show whether it is a lean, rich or stoichiometric combustion. The mathematical expression for this is as shown below ? =Actual FARStiochiometric FAR 1 Rich combustion process 3. 4. 4. Turbine Turbine can simply be said to be the antonym of a compressor. In response, a turbine extracts molecular kinetic energy from the air and uses it to drive the turbo machineries which results in the pressure and temperature of the air to drop. If truth be told, Flack (2005) asserted that the turbine uses 70% to 80% of the total energy gained by the air in the combustion chamber to drive the turbo machineries while the remaining 20% to 30% is used to generate thrust in the nozzle.Since the geometry of a turbine have favourable pressure gradient unlike the compressor which is adverse, the efficiency of the turbine is usually very high. Since the turbine is the opposite of the compressor, it has exactly the same configuration of rotor and stator but the volume increase across it which induces the pressure drop. One major problem faced when design a turbine is the deterioration of the blades due to high inlet temperature from the combustion chamber. Based on this, (Song et al. 2002) demonstrated that General Electric uses about 16. 8% of the compressor air to cool the turbine blades of GE 7f engine. Turbine Equation Analysis Given that, Turbine efficiency ? T=ActualIdeal=ht6-ht5ht6s-ht5 T=Tt6-Tt5Tt6s-Tt5 Simplifying the equation given that pt6s=pt6 Tt6sTt5=pt6pt5? -1? ?T=pt6pt5 ? T=Tt6Tt5 ?T=? T-1? T? -1? -1 3. 4. 5. Exhaust Nozzle The nozzle is the final component of the jet engine that the air passes through. The main purposes of the nozzle is to add extra acceleration to the h igh velocity exiting air, reduces its total pressure to that of ambient condition and finally generate sufficient thrust. There are two conditions that occur in the exit of the nozzle depending on the ambient pressure. The first condition is termed under-expansion which occurs when the ambient pressure is less than the exit pressure of the gases.The result of this is that the exit velocity will be lower than it normally is and this makes the momentum component of thrust to be lower than ideal. On the other hand, it will create a positive thrust component for the pressure terms. The second case termed as overexpansion which occurs when the ambient pressure is greater than the exit pressure of the gases. Consequentially, the opposite of what happens in the under-expansion condition occurs where the pressure term is lower and the momentum is higher. Nozzle efficiency ? n=ActualIdeal=ht8-h9ht8-h9s=Tt8-T9Tt8-T9s for constant specific heat Using the steady state energy equation and balanc ing it out, U9=2ht8-h9 . When specific heat is constant U9=2cpTt8-T9 p9pt8=T9sTt8? -1? T9Tt8=11+? -12M92 p9pt8=11+? -12M92-1+ ? n ? n 3. 4.Thermodynamic Process and Cycle of Twin Spool Mixed Flow Turbofan Engine Before any explanation is done from Figure 2, the blue arrows represent the incoming air into the diffuser and the red represent the air flow into the core of the engine while the black arrow represent the bypass air flow through the fan. Finally, the brown arrow represents the air flow after the bypass air and the core air flow have mixed. Based on the arrangement of the turbofan engine in figure 2, it can be seen that air at ambient condition is sucked into the diffuser where the air velocity is reduced and some of its kinetic energy is used to increase the static pressure to the total pressure. The air exiting the diffuser enters the fan or low pressure compressor where it is compressed. Indeed, the molecules of the air gains kinetic and internal energy by colliding rapid ly with one another and as a result increase the enthalpy and static pressure.Also, in the fan, some of the compressed air is bypassed through a duct to be used for the mixing process later while the rest of the air enters into the high pressure compressor of the engine core. In the high pressure compressor, the air is further compressed where the enthalpy and pressure increases as it is released into the combustion chamber. Also, in the high pressure compressor, some of the air mass flow rate is bled out to be used to cool the turbine blades and for air conditioning in the aircraft. In the combustion chamber, the incoming fuel reacts with the air in an oxidation process at constant pressure where the by-product gases gain molecular kinetic energy thereby increasing the enthalpy.This high temperature gases escapes into the high pressure turbine where it is expanded and the gases lose some of their kinetic molecular energy as it enthalpy and static pressure reduces. In other words, i t can be said that the molecular kinetic energy of the gases is being converted to mechanical work which is used to power the high pressure spool. Consequently, the gases enters into the low pressure turbine where it is further expanded to a lower pressure and enthalpy as their molecular kinetic energy is converted to mechanical work to power the low pressure spool. These gases escaping from the low pressure turbine enters the mixing zone or mixer after it has lost most of its total enthalpy and mixes with the bypassed cold air from the duct to further reduce its enthalpy as that of the cold air increases.In other words, the cold air absorbs some of the heat energy from the hot gases until they both attain equilibrium enthalpy. The mixture of the cold air and hot gases both escape at the same equilibrium enthalpy and pressure through the nozzle where their velocity is increased and the pressure is reduced considerably to that of the ambient condition. Furthermore, the exhausted high velocity gases is used to produced thrust for propulsion according to Newton’s third law of motion (In every action, there is equal and opposite reaction). 2 4. 5 6 4 13 0 HPC DIFFUSER FAN/LPC HPT LPT NOZZLE COMBUSTION CHAMBER 2. 5 3 5 8 16 BYPASS DUCT HP Spool LP Spool MIXING ZONEFigure 2 Layout of a Forward Fan Twin Spool Mixed Flow Turbofan Engine P0 P3 P4. 5 P5 P8 P6 P2. 5 P2 P13 P4 ENTROPY (S)(kJ/kg) TEMPERATURE (K) Figure 3 T-S Diagrams for the Forward Fan Twin Spool Mixed Flow Turbofan Engine Chapter 4 Mathematical and Gas Turb 11 Modelling of the Engine 4. 1. Station Numbering and Assumptions Station numbering is a very crucial step that has to be taken when analysis of any thermodynamic system involving many processes is to be done. Moreover, station numbering contributes immensely to showing how the properties of one process relate to another and how the interaction between these processes derives the functional relationship of the thermodynamic system.Returning to the work in hand, the station numbering system that has been adopted for this work on a JT8D-15A turbofan engine is in accordance with the Aerospace Recommended Practice (ARP) and it is shown in figure 2. Assumptions The following assumption were made based on Mattingly (2002) and Kurzke (2007) in order to perform the modelling as listed below * The air flow through the engine is assumed to be steady and one dimensional * The fan and the low pressure Compressor are driven by the low pressure turbine * The overall engine is assumed to have no bleeds in mass flow or power off-take in turbine. * The nozzle of the engine is choked which means the exit pressure will be greater than the ambient pressure. The air is assumed to act as a half ideal gas where the specific heat and ratio is dependent on temperature only. * The areas of each station of the engine is assumed to be constant 4. 2. Design Point Cycle Analysis of the Turbofan Engine The off-design or performance cycle analysis cann ot be done without the design point cycle being defined. The design point cycle in this analysis is obtained using exactly the same data used in the actual test analysis for a JT8D-15A turbofan engine operating at sea level with maximum take-off thrust as shown in (â€Å"JT8D Typical Temperature and Pressure†) and (â€Å"ICAO†). Some of the input parameters such as the isentropic efficiencies and pressure ratios from the actual test data had to be calculated.Since not all the input parameters were given from the actual test data, some of the parameters like inlet corrected mass flow rate, diffuser pressure ratio and efficiency; mechanical spool efficiency had to be guessed in order to complete the analysis and the data are represented below in Table 1. With all the Input Parameter being specified as shown in table 1, the design point cycle simulation of the JT8D-15A turbofan Engine using the Gas Turb 11 software can then be performed. All the steps taken to model the m ixed flow turbofan engine on Gas Turb 11 is clearly represented in the algorithm shown in figure 3 below. COMPONENT| INPUT PARAMETER| | DIFFUSER| Pressure Ratio (? d)| 1| | Inlet Corrected Mass Flow Rate (mc2)| 138. 618 kg/s| FAN| Pressure Ratio (? fan)| 2. 054| | Isentropic Efficiency (? fan)| 0. 78| | Bypass Ratio (? )| 1. 08| Low Pressure Compressor (LPC)| Pressure Ratio (? LPC)| 4. 7| | Isentropic Efficiency (? LPC)| 0. 88| | Nominal Low Pressure Shaft Speed (NLP)| 8160RPM| High Pressure Compressor (HPC)| Pressure Ratio (? HPC)| 3. 77| | Isentropic Efficiency (? HPC)| 0. 864| | Nominal Low Pressure Shaft Speed (NHP)| 11420RPM| Combustion Chamber (cc)| Pressure Ratio (? CC)| 0. 934| | Isentropic Efficiency (? CC)| 0. 99| | Burner Exit Temperature (TIT)| 1277. 15K| High Pressure Turbine (HPT)| Isentropic Efficiency (? HPT)| 0. 9| | HP Spool Mechanical efficiency (? m)| 1| Low Pressure Turbine (LPT)| Isentropic Efficiency (? LPT)| 0. 91| | LP Spool Mechanical efficiency (? m)| 1| T able 1 Input Parameters for the Design Point Cycle Simulation STARTSpecify all the input data gotten from the actual test data as shown in Table 1 Run the Gasturb 11 software and select mixed flow turbofan from the drag down Tab list. Set the scope to ‘More’, set the Calculation Mode as Design and click ‘Run’ Choose the Units to either Imperial or SI and Select the type of fuel from to drop down list to Kerosene, Natural Gas or Hydrogen Estimate the inlet Corrected mc2 Mass Flow rate to the FAN/LPC Choose ‘Single Cycle’ for ‘Select a Task ‘Option and click ‘Run’ Check if the Thrust, SFC, ? HPT, ? LPT and EPR are within (0-10) % of the actual test Experiment END YES NO Figure 4 Design Point Cycle Simulation Algorithm Using Gas Turb 11 Verification of the Design Point simulation ResultsSince not all the input parameters were specified in the actual test data and some of them had to be guessed, it is without any doubt that errors are bound to generate in the simulation results using the Gas Turb 11 software. In order to ensure that the errors accumulated in the simulation were within range, the major output parameters obtained such as net thrust, fuel flow rate, Engine exit pressure ratio, etc were compared to the actual test data as shown in Table 2 and the error range was calculated to be between 0. 25% to 8. 5% which is within an acceptable range. PARAMETERS| ACTUAL TEST DATA| SIMULATED DATA USING GASTURB 11| Net Thrust| 69307. 74| 69320| Engine Exit Pressure Ratio P8P0| 2. 09| 2. 167|Burner Fuel Flow| 1. 100843| 1. 09781| HPT pressure Ratio (? HPT)| 0. 415| 0. 449| LPT Pressure Ratio (? LPT)| 0. 3294| 0. 3514| HPT temperature Ratio (? HPT)| 0. 8097| 0. 8435| LPT temperature Ratio (? LPT)| 0. 7718| 0. 793| Table 2 Comparison Table for the Actual Test Data and Simulated Data Using GasTurb 11 4. 3. Off-Design Point Cycle Simulation of the Turbofan Engine The off-design or performance cycle simulatio n takes into account the concept of module matching of each component through performance maps. This cycle analysis enables the determination of different operating point of the engine at a given design point of the engine.Considering the work in hand, the design point have been defined and verified for the JT8D-15A turbofan engine operating at sea level with maximum take-off thrust which means that different operating points of the engine can be defined with the concept of off-design module matching of the engine. Indeed, the off-design operating point that was considered for the parametric analysis in this work was 30,000ft at M0 0. 8 for the turbofan engine. The off-design modelling of the JT8D-15A engine for the operating point of 30,000ft at M0 0. 8 based on the reference design point defined earlier is clearly demonstrated as follows. The off-design performance cycle simulation may contain some errors because of the component performance maps that were used for the simulation. 4. 3. 1. Module/Component Matching This process only applies to the off-design performance cycle point of the engine.It can simply be defined as the act of synchronising each component of a jet engine to coexist as a unit in order to derive the overall performance characteristics of the jet engine. Component matching involves the process closely studying the ramifications of the actual jet engine overall performance behaviour on the components major characteristics such as pressure ratio, temperature ratio, efficiency and spool speed. This process introduces the concept of empirically determined component performance maps that establishes the relationship between the thermodynamic properties and the geometry of the jet engine itself. 4. 3. 2. Off-Design Component Modelling Diffuser The diffuser was assumed to be adiabatic and the pressure ratio ? d=1 The Isentropic Efficiency was assumed to be 1 For Sea Level,Pamb=101325pa , Tamb=288. 15K For 30,000ft and M0 0. 8, Tamb=288. 15-0. 0 065? 9144 =288. 15-59. 436 =228. 71K Pamb=101325? Tamb288. 155. 2561 =30. 09kpa Tt1=228. 71? 1+? -12M02 =228. 71? 1+1. 4-12? 0. 82 =258K pt1p0=1+ ? d? -12M02 -1 pt1=30. 09? 1+ 1? 1. 4-120. 821. 41. 4-1 pt1=45. 8674kPa pt1=pt2 Tt1=Tt2 Fan and Low Pressure Compressor The inlet corrected mass flow rate is estimated as 138. 618kg/s , As for the off design simulation using the component performance maps for the altitude of 30000ft and Mach no. 0. 8, the actual spool speeds and inlet mass flow rate are calculated based on the estimated inlet corrected mass flow rate as shown below.Low and High pressure spool mechanical efficiency is assumed to be=1 HP spool Speed=11420RPM, LP spool Speed=8160RPM m2=Pt2PSTD? mc2Tt2TSTD =45. 878101. 325? 138. 618258288. 15 Actual Mass flow rate m2=66. 3323kg/s N=Tt2TSTD? NcLP=228. 71288. 15? 8160=7722 RPM The calculated actual mass flow rate and spool speed were used to evaluation the isentropic efficiency and the pressure ratio of the LPC for that operatin g condition from the compressor performance map. Figure 5 Example of a Compressor Performance Map/Curve The diagram above in figure 4 depicts a typical compressor performance map that was used for the off-design point analysis in this work.It can be seen that the x-axis represents the inlet corrected mass flow rate mc2 into the compressor, the y-axis represents the compressor pressure, the red contour lines represents the isentropic efficiencies and the black curved lines represent the relative corrected spool speed. To add to that, the red dash line that ends the speed lines and efficiency lines represent the surge margin which is also known as the stall line that must be avoided since the flow will become unstable in that region. In this work, the inlet corrected mass flow rate and spool speed were calculated which were interpolated on the performance map to obtain the pressure ratio and the isentropic efficiency.For instance, the yellow dot on the map represents a design point tr aced for a given pressure ratio, High Pressure Compressor The inlet corrected mass flow rate into the HPC mc2. 5=mc21+? mc2. 5=138. 6182. 08=66. 64kgs m2. 5=Pt2. 5PSTD? mc2. 5Tt2. 5TSTD N=Tt2. 5TSTD? NcHP The same equation used for the LPC is used to calculate the actual mass flow rate and spool speed which is used to evaluate the isentropic efficiency and pressure ratio when it is operating at an altitude of 30000ft at M0 =0. 8. Verification of the off-design modelling for 30000ft at Mo 0. 8 In order to verify the simulation result gotten for the operational design point of 30000ft at M0 0. , the actual test data results gotten from Mattingly, Heiser and Pratt (2002) for the same operating condition was compared. Due to the difficulties in obtaining a lot of output parameters for this operating point, the result will be verified with only the net thrust generated and the specific fuel consumption. Indeed, the error accumulated was 1. 71% for the net thrust and 0. 83% for the specif ic fuel consumption. PARAMETERS| ACTUAL TEST DATA| SIMULATED DATA USING GASTURB 11| Net Thrust (lb)| 4920| 4836| Specific Fuel Consumption(lb/lbh)| 0. 779| 0. 7855| Table 3 Comparison Table for the Actual Test Data and Simulated Off-design Data Using GasTurb 11 Chapter 5Methodology, Results and Discussions Given that the design point of the JT8D-15A turbofan engine at sea level has been obtained and verified with the actual test data, the operating point of 30000ft at M0 0. 8 was simulated and obtained which now served as the design point for the analysis in this work. Moreover, the procedure taken to define this design point of 30000ft at M0 0. 8 of the JT8D-15A turbofan engine has been clearly stated earlier which gives the permission to conduct the parametric cycle study of the turbofan engine. The parametric cycle studies were done for three different cases for the operational design point of 30000ft at M0 0. of the JT8D-15A turbofan engine as explained as follows. 1. The first parametric analysis case 1 aim to create an understanding of the effects of varying major design parameters on the performance parameters of the turbofan engine when some of the design choices are kept constant. In other words, the bypass ratio and thermal limit parameter (turbine inlet temperature) were varied when the design choices such as the compressor pressure ratio, fan pressure ratio and isentropic efficiencies were kept constant in order to investigate their effects on the performance parameters such as the net thrust, specific fuel consumption, propulsive efficiency, thermal efficiency, and fuel-air-ratio.Much interest is shown nowadays in using alternative fuels like hydrogen and Natural gas in efforts to reduce the cancer known as pollution and the risk of depletion of energy resources. Based on this, conducting a research that focuses of comparing different fuels consumption rate, their risk of pollution and their contribution to the performance of the engine will be re ally valuable. Based on this, a parametric analysis had to be done on the JT8D-15A turbofan engine using three different fuels which are the design point fuel kerosene, hydrogen and natural gas. Since the original design point of the JT8D-15A turbofan was obtained using kerosene fuel, the design points of using hydrogen and natural gas was obtained using the same design choices as that of kerosene.Now that the design points of the JT8D-15A turbofan engine had been defined when using the three different fuels, it had given a go ahead to perform whatever parametric cycle studies of the turbofan engine using the three fuels. In order to compare the performance characteristics of the turbofan engine when it is using the three different fuels, different approaches had to be devised to compare them effectively on a rational basis which defines the last two parametric analysis cases as follows. 2. The second case of parametric analysis was that the fuel flow rate would be kept constant for the three fuels that would be used as the bypass ratio is varied with design choices remaining the same. 3.The third case of study was to make the energy supply into the combustion chamber of the turbofan engine the sa

Project Management and Critical Path Essay

What is project management, and what are its main objectives? What is the relationship between tasks and events, or milestones? What is a work breakdown structure? How do you create one? What are task patterns, and how can you recognize them? Compare the advantages and disadvantages of Gantt and PERT/CPM charts. Define the following terms: best-case estimate, probable-case estimate, and worst- case estimate, and describe how project managers use these concepts. How does a project manager calculate start and finish times? What is a critical path, and why is it important to project managers? How do you identify the critical path? What are some project reporting and communication techniques? What is risk management, and why is it important? In Poor Richard’s Almanac, Benjamin Franklin penned the familiar lines: â€Å"For the want of a nail the shoe was lost, for the want of a shoe the horse was lost, for the want of a horse the rider was lost, for the want of a rider the battle was lost, for the want of a battle the kingdom was lost — and all for the want of a horseshoe nail.† Looking at the outcome in hindsight, could project management concepts have avoided the loss of the kingdom? Explain your answers. At Countywide Construction, you are trying to convince your boss that he should consider modern project management techniques to manage a complex project. Your boss says that he doesn’t need anything fancy, and that he can guess the total time by the seat of his pants. To prove your point, you decide to use a very simple example of a commercial con- struction project, with eight tasks. You create a hypothetical work breakdown struc- ture, as follows: Prepare the foundation (10 days). Then assemble the building (4 days). When the building is assembled, start two tasks at once: Finish the interior work (4 days) and set up an appointment for the final building inspection (30 days). When the interior work is done, start two more tasks at once: landscaping (5 days) and driveway paving (2 days). When the landscaping and driveway are done, do the painting (5 days). Finally, when the painting is done and the final inspection has occurred, arrange the sale (3 days). Now you ask your boss to estimate the total time and write his answer on a piece of paper. You look at the paper and see that his guess is wrong. 1. What is the correct answer? 2. What is the critical path? 3. Create a Gantt chart that shows the WBS. 4. Create a PERT/CPM chart. ————- What are the five questions typically used in fact-finding? What additional question can be asked during this process? What is a systems requirement, and how are systems requirements classified? What are JAD and RAD, and how do they differ from traditional fact-finding methods? What are their pros and cons? What is total cost of ownership (TCO), and why is it important? Provide examples of closed-ended, open-ended, and range-of-response questions. What are three types of sampling, and why would you use them? What is the Hawthorne Effect? Why is it significant? What is a functional decomposition diagram (FDD) and why would you use one? Explain how to create an FDD. What are agile methods, and what are some pros and cons of this approach? To what three different audiences might you have to give a presentation? How would the presentation differ for each? ——– A group meeting sometimes is suggested as a useful compromise between interviews and questionnaires. In such a group meeting, one systems analyst meets with and asks questions of a number of users at one time. Discuss the advantages and disad- vantages of such a group meeting. ——- Elmwood College Situation: The school is considering a new system that will speed up the registration process. As a systems analyst, you are asked to develop a plan for fact-finding. 1. List all the possible techniques that you might use. 2. Describe an advantage for each technique. 3. Suppose the development budget is tight. How might that affect the fact-finding process? 4. What are five important questions to use during fact-finding?

AP Chemistry FRQ How to Ace the Free Response Questions

AP Chemistry FRQ How to Ace the Free Response Questions SAT / ACT Prep Online Guides and Tips Practicing free-response questions is one of the best things you can do to improve your AP Chemistry score. Not only will you excel on the free-response section, but you'll also know the material so well that the multiple-choice questions will be a piece of cake. In this article I'll tell you all about the free-response section of the exam, give you some tips on how to solve AP Chemistry FRQs, and go through a couple of sample problems from recent exams so you can get a feel for what they're like! What’s the Format of the AP Chemistry Free-Response Section? The free-response section of the AP Chemistry exam seems intimidating because it’s longer than free-response sections on most other AP tests, and it includes lots of calculations and experiments that you have to interpret.The most important thing you can do is remain calm and stay focused and methodical in your approach to each question.It’s not as scary or difficult as it looks if you’ve prepared well for the test and use your common sense! Here's an overview of the format of the free-response section: 105 minutes (1 hour 45 minutes) Calculator use permitted Seven questions total Three long response worth 10 points each Four short response worth 4 points each The long response questions always come first! You’ll be tested on the following skills (which may be tied to any topic in the AP Chemistry curriculum): Experimental design Analyzing real lab data to identify patterns and explain phenomena Creating or analyzing diagrams of molecules and atoms to explain observations Translating between different representations of data Following logical steps to analyze and solve problems OK, that all makes sense, but how exactly do you solve these long, complicated questions? In the next section,I’ll go through a step-by-step guide for how to approach the AP Chemistry FRQs. How to Solve AP Chemistry Free-Response Questions It's important to have a game plan for the free-response section. My first piece of advice is not to feel obligated to do the questions in order!Take a couple of minutes (no more than 5-10) at the beginning of the section to look through everything and decide which question you want to tackle first.It’s best to start with your strengths so you'll have more time at the end for challenging questions. You should spend a maximum of 20 minutes on each long free-response question and 10 minutes on each short free-response question.Keep an eye on the time so it doesn’t get away from you! Here are some basic steps you should take to solve free-response questions: Step 1: Figure Out What You Know First, assess the information the question gives you.It can be confusing to extract the data that’s embedded in the introduction over and over again as you go through different parts of the question.Make things easier for yourself by writing down the values you’re given next to the question so that they’re easily accessible when you need them.You should also take time to understand (at least on a general level) the experiment being described so that you don’t feel confused and overwhelmed when you start reading the question. Step 2: Dive Into the Question For each part of the question, read the instructions and ask yourself the following: Do you need to do any calculations? Decide which equations you’ll need (if applicable), and write them down.Do the necessary calculations based on the numbers you extracted in the previous step and any numbers you were given in this part of the question.Make sure you show your work! Don’t erase your calculations, and double-check everything to make sure you have the correct units and your answer makes sense logically. Does the question ask you to justify or explain your answer? If so, DON’T ignore these instructions.In many cases, you’ll only get points for your answer if you can explain it adequately.Use concrete evidence to back up your response (we’re talking hard data).Even if something seems obvious to you, spell it out as clearly as possible to ensure that you earn those points! Do you have to draw a diagram? Even though neither of the free-response questions I'll go over in the next section requiresthis, some questions will ask you to draw diagrams.For example: If you have to answer a question like this, try to be as clear as possible.Draw out your answer on scrap paper first if you feel unsure so that the final product is neat and unambiguous.Here’s the answer, by the way: Step 3: Double Check Look back at the question to make sure you didn’t miss anything or leave out any explanations.Reread each part and connect it directly to its corollary in your response so you’re scooping up all the points you possibly can!You should also walk through how you found each answer to make sure you didn’t make any weird errors you missed the first time around. Finally, check again for appropriate units. Errors in unit conversion are common silly mistakes that are really, really frustrating if you knew how to do the problem correctly otherwise. Did you remember to convert milliliters to liters? AP Chemistry FRQExamples I’ll go through the solution process for a sample question of each type so you can get a better idea of what the test will be like.Notice that these questions look complex and overwhelming at first, but if you stay calm and break them down methodically, they don’t end up being that bad! Long Free-Response Sample Question Here’s a sample long free-response question from the 2014 exam: There’s a ton of information here, so let’s start from the beginning.What we know from the short intro is that the pH of a 0.20 M 50 mL sample of propanoic acid is 2.79 at 25 degrees Celsius.There's alsoan equation that shows how the acid reacts with water and which products are created by this reaction. Now we're ready to tackle part (a): Where’s the conjugate acid-base pair in the equation?There are two pairs that you could potentially list for this answer: CH3CH2COOH (acid) and CH3CH2COO- (base) OR H30+ (acid) and H20 (base) We know that the first compound is propanoic acid, and the loss of the hydrogen atom through the reaction creates the basic compound on the right side of the equation.Conversely, the water in the first half of the equation is a base that becomes an acid when it picks up the hydrogen atom from the propanoic acid. To get this point, you would need to label each compound indicating which is the acid and which is the base. Always read instructions carefully, or you may lose out on points - I can't say this enough! Let’s move onto part (b): What’s the value of Ka for propanoic acid at the temperature indicated in the question?We just need to plug some values into the equation for Ka,which is given to you on the formula sheet for the test: Hmm, looks like we can’t plug in the values yet because we don’t know the molarity of the H30+, which would have to go in the [H+] spot in the equation.We can find that value using this other equation from the formula sheet: This means: -pH = log[H+]10-pH = [H+]10-2.79 = [H30+]1.6 x 10-3 M = [H30+] This value for the molarity of H30+is equal to the molarity of CH3CH2COO-. The same amount of each must be created by the equation because the hydrogen atoms are removed and added in a 1:1 ratio.Armed with this new information, we can go back to the Ka equation: Let's plug in our values: Ka = [H30+][CH3CH2COO-] / [CH3CH2COOH]Ka = (1.6 x 10-3 M)(1.6 x 10-3 M)/ 0.2 MKa = (1.6 x 10-3 M)2/0.2 MKa = 1.3 x 10-5 For part (b), you could earn three points total: One for correctly solving for [H30+] One for plugging the right values into the Ka equation One for solving for Ka correctly OK, now for part (c)! Oh crap, this one has PARTS WITHIN THE PART.Don’t panic; you got this. It’s just true or false plus answer explanations!The explanations on these types of questions are very important. If you don’t explain your answer adequately, you won’t get any points even if the answer itself is correct.You can earn two points total on this question, one for each answer AND explanation. Part (i): In solution, the OH ions from the NaOH will react with the CH3CH2COOH to form water and CH3CH2COO- like so (hydrolysis reaction): The pH of the resulting solution will be GREATER than 7 because of the formation of the new basic compound at equivalence. That means it’s false! Part (ii): If two acid solutions have the same pH, but one is with hydrochloric acid, and the other is with propionic acid, would the first solution necessarily have a lower molar concentration of the HCl? HCl is a strong acid that ionizes completely in solution while propionic acid only partially ionizes.Fewer moles of HCl are needed to produce the same molar concentration of H30+ and reach an equivalent pH level to the propionic acid solution.This one is true! The next part of the question offers up a new scenario, so let’s take stock of what we've learned from the added description.So, the student titrates 25 mL of a ~mystery solution~ (mysterious squiggles added for dramatic flair) of propanoic acid with 0.173 M NaOH and reaches the endpoint of the titration after 20.52 mL of the NaOH has been added. Now onto part (d)! Based on this information, part (d) asks us to figure out the molarity of the propanoic acid. First, how many moles of NaOH were put into the solution?We can find this by multiplying the total volume of NaOH solution by its molarity: (0.02052 L NaOH) x (0.173 mol NaOH / 1 L NaOH) =3.55 x 10-3mol A total of 3.55 x 10-3 moles of NaOH were put into the solution.Since the titration reached the equivalence point at this time, that means that the number of moles of NaOH added would have to be the same as the number of moles of propanoic acid in the original solution.If we divide 3.55 x 10-3 mol propanoic acid by the number of liters of acid in the original solution, we will get the molarity: 3.55 x 10-3 mol propanoic acid / 0.025 L propanoic acid = 0.142 M For this part, you get one point for correctly calculating the number of moles of acid at the equivalence point and one point for providing the correct molarity. Part (e) is a critical thinking question about a new experiment. Would the student have to use a different indicator to figure out the concentration of a solution of an acid with pKa of 4.83?Based on ourKa calculations in part b, we can use one of the equations on the formula sheet to figure out pKa for the original propionic acid and compare the two values. pKa = -logKapKa = -log(1.3 x 10-5)pKa = 4.89 The twopKa values of4.83 and 4.89 are pretty close to one another, so you wouldn’t need to use a different indicator in the new titration.The correct response is to disagree with the student’s claim.You get one point here for disagreeing with the claim and explaining why, and you get a second point for directly comparing the two pKa values. Short Free-Response Sample Question Here’s a sample short free-response question, also from the 2014 exam: As you can see, the â€Å"short† questions aren’t really that short, but they’re not as involved as the long ones.There isn't as much information to digest, and each of the parts of the question is more direct.Each part of this question is worth one point (4 points total). Your response must include the correct answer and the correct justification/methodology to earn points! Starting with part (a): This is a PV = nRT question!Since we want the number of moles of CO2, we’re solving for n.P is 1.04 atm, V is 1.00 L, R is the gas constant (0.08206 L atm mol−1 K−1), and T is 00 K. n = PV/RTn = (1.04 atm)(1.00 L)/(0.08206 L atm mol−1 K−1)(00 K) n = 0.05 mol CO2 Moving onto part (b): In experiment 1, the original number of moles of CaCO3 would be equivalent to 50.0 g / (100.09 g/mol). The 100.09 g/mol number was calculated by adding up the atomic weights of the elements in the compound.This calculation gives us 0.500 mol CaCO3 total. If all of it had decomposed, the figure we calculated in part a for the number of mols of CO2 produced would also be 0.500 mol, but it was only 0.05 mol.This discrepancy means that the student’s claim has to be false! Now let’s tackle part (c): What would happen if more gas was added to the container and the pressure went up to 1.5 atm? Would it go back down to 1.04 atm afterward? Equilibrium was reached in both experiments, and it resulted in a final pressure of 1.04 atm.The reaction would just adjust to the added gas by shifting towards the reactant.The pressure would go back down to the equilibrium figure of 1.04 atm as the excess CO2 was consumed.The final pressure would still be equal to 1.04 atm. Finally, we’ll answer part (d): Can we find Kp with the information we’re given?Yes!The pressure of the CO2 in this experiment determines the equilibrium constant as well because it’s the pressure of the gas at equilibrium.There’s only one gas involved in the reaction, and we already know its equilibrium pressure, which means we also know the value of the constant. Kp = 1.04 How to Practice AP Chemistry Free-Response Questions You can find AP Chemistry FRQsfrom previous years (and their solutions) on the College Board site. The test changed starting in 2014 (seven free-response questions instead of six, and no questions asking you just to balance equations), so keep in mind that only the 2014 and 2015 questions will be completely accurate representations of what you can expect to see on your test. Here's a link to the most recent questions and answer explanations: 2014 and 2015 AP Chemistry Free-Response Questions Note that you have to login to your College Board account for access to the 2015 questions. Even if you don't have an account yet, it's easy enough to create one (and you'll need it eventually anyways!). You can also practice with free-response questions from earlier versions of the exam. While these aren't completely aligned with the current structure of the test, they're still good tools to use in practicing your skills: AP Chemistry Free-Response Questions: 1999-2013 There is no login required for access to these questions. You can also check out my article that lists all the AP Chemistry practice tests and quizzes that are available online (coming soon!). There are a few more unofficial practice tests that include free-response questions modeled after the questions on the real exam. And use a calculator when you practice (you get one for the free-response section on the real test)! Definitely don't use a quill, though. That part of this image is irrelevant and mystifying. Conclusion The free-response section is the most challenging part of the AP Chemistry exam for most students. To do well, you need to have a strong understanding of all the major concepts covered in the course and be able to apply them to a variety of experimental scenarios. Most of these questions look intimidating, but they're manageable if you take them one step at a time and break them down into smaller chunks. Here's a review of how to solve free-response questions: Step 1: Figureout what you know Write down any data that's included in the question Step 2: Diveinto the question Figure out which formulas you need Do the necessary calculations Justify your responses Draw diagrams if applicable Step 3: Double check Make sure your units of measurement are correct Verify that your answers make sense logically Practice your skills on free-response questions from past AP tests until you feel comfortable. Remember, questions from tests before 2014 will be slightly different from the current free-response questions; base your expectations for the real test off the material from 2014 and 2015! What's Next? If you're looking for more AP Chemistry practice, check out my list of the best review books for this year's test.Almost all of them include free-response questions modeled after the current format of the exam. For a holistic look at the most effective way to study for in-class assessments and the final exam, read my ultimate study guide for AP Chemistry. Are you debating whether to start studying now or put it off a little longer? Learn more about when you need to get serious about your study plans for AP tests. Want to improve your SAT score by 160 points or your ACT score by 4 points? We've written a guide for each test about the top 5 strategies you must be using to have a shot at improving your score. Download it for free now: