To Kill a Mockingbird on Commitment Essay Example For Students

To Kill a Mockingbird: on Commitment Essay To Kill a Mockingbird: Essay on Commitment BY Firegtr11411 The Play of To Kill a Mockingbird Paragraph Plan for Critical Essay Task: Choose a play in which a character shows commitment to principles, ideals and individuals despite difficulties he encounters. Show what difficulties the character encounters and explain why he remains committed. A play in which a character shows commitment is the play of To Kill a Mockingbird, based on the novel by Harper Lee and dramatised by Christopher Sergel. The play, set in the American Deep South of the 1930, is about a white lawyer, Atticus Finch, who encounters difficulties in the orm of prejudice and hatred when he decides to defend a black man called Tom Robinson, accused of sexually assaulting a white woman. This essay will show how Atticus remains committed to his principles, ideals and individuals in the play by exploring setting, characterisation and key incidents. The difficulties the main character encounters are rooted in the setting of the play. We will write a custom essay on To Kill a Mockingbird: on Commitment specifically for you for only $16.38 $13.9/page Order now The action takes place in a village called Maycomb, a tired old town dominated by poverty, in Alabama in 1935. Racism was entrenched in the Deep South at the time and is highlighted in the frequent use of the highly offensive term nigger by the townspeople. Atticus is aware that his defence of a black man will create conflict for him and his family and reminds his children We will be fighting our friends, preparing them for a potential backlash. Atticus is a character who is committed to the principles of tolerance and empathy. He values empathy highly. When his children harass their recluse neighbour Boo, Atticus does not punish them but reminds them to show more empathy: miou see, you never really understand a person until you consider things from his point of view. Until you climb into his skin and walk around in it, showing hat he acts according to his principles. He is not frightened to stand up for his beliefs and, possibly surprisingly, is very tolerant in a prejudiced and bigoted society. He points out that One thing doesnt abide by majority rule a persons conscience. He is very courageous in pursuing this principle and is even willing to risk his own safety in the case of Tom Robinson he defends him from a lynch mob. Mr Cunningham, one of the villagers, warns him: mfou know what we want. Stop aside from the door, Mr Finch. It is impressive to see how Atticus, sitting outside the Jail ith a chair and a light, reading a book, remains calm, non-confrontational and defensive to defuse this situation full of difficulties. Atticus is a character that acts as an example to the audience as he lives by his commendable beliefs. Atticus decision to defend Tom shows he is committed to the accused despite encountering hostility. This decision is based on his belief in the fundamental strength of the American justice system. He declares in our courts all men are created equal but is at the same time aware of the limitations of that principle due to the potentially racist views f the Jury. He skilfully defends Tom and openly acknowledges Mayella Ewells motive for lying as she has broken a taboo: She has committed no crime, but she broke a rigid code of our society, a code so severe that whoever breaks it is hounded out from our midst as unfit to live with. He has some sympathy for Mayellas lie but also reminds her that it has put Toms freedom, even his life, at risk. During Atticus defence, he exposes Bob Ewell, Mayellas father, as a brutal and controlling liar and on by Bob, he stays calm and remarks dryly: l wish Bob Ewell wouldnt chew obacco. The trial creates enormous pressure on Atticus because of emotive issues involved but he remains principled and composed and delivers an impressive defence. .udb0e4d2c5a3d51aa59899b4e58dcccc0 , .udb0e4d2c5a3d51aa59899b4e58dcccc0 .postImageUrl , .udb0e4d2c5a3d51aa59899b4e58dcccc0 .centered-text-area { min-height: 80px; position: relative; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 , .udb0e4d2c5a3d51aa59899b4e58dcccc0:hover , .udb0e4d2c5a3d51aa59899b4e58dcccc0:visited , .udb0e4d2c5a3d51aa59899b4e58dcccc0:active { border:0!important; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .clearfix:after { content: ""; display: table; clear: both; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .udb0e4d2c5a3d51aa59899b4e58dcccc0:active , .udb0e4d2c5a3d51aa59899b4e58dcccc0:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .centered-text-area { width: 100%; position: relative ; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .udb0e4d2c5a3d51aa59899b4e58dcccc0:hover .ctaButton { background-color: #34495E!important; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .udb0e4d2c5a3d51aa59899b4e58dcccc0 .udb0e4d2c5a3d51aa59899b4e58dcccc0-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .udb0e4d2c5a3d51aa59899b4e58dcccc0:after { content: ""; display: block; clear: both; } READ: Chevron EssayDespite the shocking guilty verdict, the audience sense that a little victory has been achieved by Atticus as Bob Ewells true nature has been highlighted. Later on in the play Bob Ewell attacks Jem and Scout, an event a key incident and the climax of the play that Atticus did not foresee because he always tries to see the best in people. He falsely believes that Jem has killed Bob and wants his son to face he consequences, stating that he wont hush up but it becomes apparent Boo, the neighbour, killed Ewell to protect the children. It is a compliment to Atticus principled parenting that Scout at the end of the play a more mature person agrees with sheriffs attempt to protect their neighbour Boo, remarks, l mean, itd be sort of like shooting a mockingbird, wouldnt it. She acknowledges Boo does not do any harm to anybody with his lifestyle like a mockingbird and shows the kind of empathy in that Atticus wants his children to have for other people. At the end of the lay some of the conflict created by Tom Robinsons trial is resolved but with the prevalence of racism it is never far from the surface. In conclusion, Atticus is an impressive and admirable character who remains true to his ideals and principles of empathy, tolerance and equality in difficult circumstances created by the racism, intolerance and hatred around him. Instead of being a token lawyer for Tom, he stays true to the principle of equality before the law and delivers a committed and impressive defence. Atticus principles and actions act as a beacon of hope for more tolerance in a hostile environment.

Ernest Hemingway and his significance of landscapes, from The Complete Short Stories.

Ernest Hemingway and his significance of landscapes, from The Complete Short Stories. This essay was written about the significance of landscape using his "Big Two Hearted River" and "Hills Like White Elephants."The Significance of LandscapeHemingway uses landscape to reflect his characters emotions and also as a secondary character in his stories. Two good examples of this are Hemingway's "Big Two Hearted River" and "Hills Like White Elephants."In "Big Two Hearted River," Nick was the only human character but the scenery definitely played a huge role. In the beginning of the story Hemingway describes an old town that had been burnt to ashes. "There was no town, nothing but the rails and the burned over country." (pg. 163) Leaving you with a sense of loneliness and despair, Hemingway uses this scenery to paint a picture of not only how the town looked but how the character must have felt. Throughout his hike toward the country, away from the burnt over town, the scenery had started to change and so did the mood.English: Hemingway posing for a dust jacket photo ..."The re was nothing but the pine plain ahead of him, until the far blue hills that marked the Lake Superior height of land." (pg. 164) Using the pine plain, and the blue hills to describe what's ahead, gives you a sense of relief, like there's a light at the end of the tunnel.Hemingway's use of old verses new allows you to feel a sense that when ever something dies, something is born, or when ever there is an end, there must be a new beginning. Hemingway describes the river and the woods as being the image of everything that is unblemished and whole. The closer he gets to the river, the closer to he gets to the serenity he had been searching for. "He sat on the...

Shakespeare's The Tempest Essay Example | Topics and Well Written Essays - 2000 words

Shakespeare's The Tempest - Essay Example However, in the story, Caliban appears to be knowledgeable but subjugated due to which, all his qualities are suppressed. The character of Caliban points towards closeness to nature, colonialism and slavery. The character of Caliban is described by Shakespeare as a ‘Savage and deformed slave’ that characterizes the upbringing of the creature (Graf 3). Caliban is nurtured by Prospero as a slave who makes use of the creature for his own benefits. He teaches Caliban his language with the help of which, Caliban can tell him about the marvels of the island. Caliban is native to the island and knows better than anyone else about the island. Prospero killed Caliban’s mother and take hold of young Caliban, about which, he says, Prospero taught Caliban how to use language and considering this as a service, he told Prospero about the island and its features that were hidden previously. After gaining knowledge of the details of the island, Prospero enslaves Caliban and considers him as an abhorred creature who knows nothing. For Prospero, Caliban is only a slave who is savage and deformed. He is far from civilization and knows nothing about it. Prospero calls Caliban with many abusive words and he regards him as a slave only created for his service such as he says, Prospero uses his power to keep control on the creature. Prospero makes Caliban to do what he likes and takes away his freedom and liberty. Caliban is allowed only to do what Prospero likes. Caliban is deprived of his right over the island and Prospero after making the island colonized regards it his property and the native of the island, Caliban becomes his slave. Prospero belongs to some other place and in reality, the island does not belong to Prospero but he takes hold of it because of his power. Caliban is the true dweller of the island and acknowledges this fact as he says, Caliban describes the fact that the island is usurped by Prsopero and he is

Relationships between primates and their environments Research Paper

Relationships between primates and their environments - Research Paper Example The sexual dimorphism between females and males is quite small in body size. In contrast to other species of apes, pair bonding is displayed by gibbons and they tend to live in small family groups comprising of one monogamous pair of adults along with their offspring. They are diurnal, however, in contrary to great apes, gibbons don’t form nests and instead they lean on tree branches during night. Gibbons feed on ripe fruits but they also eat insects, birds and leaves. The most prominent aspect of gibbons is their singing conduct used for mate attraction as well as territorial defense, and their style of arborous locomotion termed as â€Å"brachiation†. However, on ground they walk on two feet (Hohmann, Robbins and Boesch, 2006). Amongst the entire great apes genus, only Orangutans are found in Asia. Their average adult height is 1.5m while mean adult weight is 110 kg. However, the sexual dimorphism is quite significant in weight amongst males and females. The female adults have 1.25m average height and 45kg average weight. Orangutans are extensively solitary while strong social linkages exist exclusively amongst female orangutans and their offspring. Both female and male Orangutans either reside as individuals in a determined home territory or as transitory individuals, and within determined ranges there is only one dominant male that breeds primarily and safeguards the female population within his territorial dominion from coerced copulations. They are diurnal but spend most of their time on trees. They also exhibit â€Å"brachiation† and quadrupedal fist-walking. Every night a nest is being built on tree by Orangutans for sleeping. They mainly feed on leaves, bark and shoots and rarely on bird†™s eggs, insects or small vertebrates (Stanford, Allen and Antà ³n,

Kant, Mill, and Nietzsche Essay Example | Topics and Well Written Essays - 750 words

Kant, Mill, and Nietzsche - Essay Example However, Kant and Mill also differ in a sense that Kant states that a person’s moral is innate as in a priori reasoning and Mill suggests, however selfishly, that morals are essential to a person’s happiness as people abide to morals for them to be happy. Now, for Nietzsche being good – at least morally good, is a social construct as the upper class and the people who are in control are defining the word with their preferences and activities. It can change as to who controls the social power. With this notion, evil, or bad, is also arbitrary, as the social class also determines which are bad and those things are generally the ones that do not work for them. For example, if you are an upper class gentleman and your slaves are working well for you, would you consider slavery immoral? No, because it is good for you. You might even think that slavery is important. But if you are a slave, slavery for you is evil because you suffer. However, who holds the social power? For the longest time, it was the upper class, which is why slavery did not become illegal for a long time. Then when the revolution happened, many people were affected by it. And those â€Å"many people† helped slavery become a â€Å"bad† thing. ... Now these imperatives in turn, determine everyone’s duty. Because of this, morals are not arbitrary because it is universal and it is true to everyone. For Mill on the other hand, has a similar belief to Kant – everyone has innate morals; morals are not arbitrary. But unlike Kant, Mill states that these morals are used to maximize happiness and reduce suffering. Utilitarian in most ways, he basically says that the right things bring happiness and wrong things bring suffering. In his words, â€Å"actions are right in proportion as they tend to promote happiness; wrong as they tend to produce the reverse of happiness†. Therefore, if an action is morally right if it makes the doer really happy. Therefore, happiness is not a social construct but an innate faculty in humans, and because it is innate, it cannot be arbitrary. Now, given the scenario above, what would each of the philosophers do given the situation? For Nietzsche, the nurse would just be in a wrong place at a wrong time. If the society urges euthanasia, then she would not have been stigmatized by having outrages towards her action. She will not be treated as a criminal. She’s having these negative connotations mainly because in her society, the people in power have deemed that her action is â€Å"immoral†. If we are taking Kant’s point of view, where all morals are innate, then we can conclude that the nurse is indeed wrong, as there are more people who were outraged by her actions than the people who weren’t. For Kant, there are universal ethical standards that cannot be broken. Now these ethical standards are about killing. Obviously, killing is not very accepted as it has produced crime and outrage. If we are talking about Mill’s point of view, then we should ask the nurse if she was

A Major Application Area Of Thermodynamics Engineering Essay

A Major Application Area Of Thermodynamics Engineering Essay A major application area of thermodynamics is refrigeration, which is the transfer of heat from lower temperature region to a higher temperature one. The devices that produce refrigeration are called refrigerators, and the cycles on which they operate are called refrigeration cycles. The most frequently used refrigeration cycle is a vapour-compression refrigeration cycle in which the refrigerant is vaporized and compressed alternatively and is compressed in the vapour phase. There are number of refrigerants which can be used in here, but the most commonly used on a commercial scale is a R12 (used in this experiment as well). The thermodynamics of ideal vapour compression cycle can be analyzed on a temperature versus entropy diagram as depicted in Figure 1. At point 1 in the diagram, the circulating refrigerant en- ters the compressor as a saturated vapour. From point 1 to point 2, the vapour is isentropically compressed (i.e., compressed at constant entropy) and exits the compressor as a superheated va- pour. From point 2 to point 3, the superheated vapour travels through part of the condenser which removes the superheat by cooling the vapour. Between point 3 and point 4, the vapour travels through the remainder of the condenser and is condensed into a saturated liquid. The condensation process occurs at essentially constant pressure. Between points 4 and 5, the saturated liquid refrigerant passes through the expansion valve (throttling device) and undergoes an abrupt decrease of pressure. This process results in the adia- batic flash evaporation and auto-refrigeration of a portion of the liquid (typically, less than half of the liquid flashes). The adiabatic flash evaporation process is isenthalpic (i.e., occurs at con- stant enthalpy). Figure 12 Temperature Entropy diagram 1 www. wikipedia.org/wiki/Refrigeration 2 http://upload.wikimedia.org/wikipedia/commons/f/f7/RefrigerationTS.png UMAR DARAZ Page 3 of 22 Thermodynamics Lab 2 Between points 5 and 1, the cold and partially vaporized refrigerant travels through the coil or tubes in the evaporator where it is totally vaporized by the warm air (from the space being refrigerated) that a fan circulates across the coil or tubes in the evaporator. The evaporator operates at essentially constant pressure. The resulting saturated refrigerant vapour returns to the compressor inlet at point 1 to complete the thermodynamic cycle. The area under the process curve on T-s diagram represents the heat transfer for internally reversible processes. The area under the process curve 5-1 represents the heat absorption in the evaporator, the area under the process 2- 4 represents the heat rejection in the condenser. In the ideal vapour compression refrigeration cycle all the heat losses and disruptions are being ignored, but in actual refrigeration cycle, we need to take these losses into consideration as they have been mentioned in this report later. The Hilton refrigeration laboratory unit R714 is capable of following entities;  · Investigation of the variation in refrigerator duty or cooling ability for various condens- ing temperature and the heat delivered to the cooling water with variation in condensing temperature. We can also investigate the variation in refrigeration coefficient of per- formance for the various condensing temperature.  · Investigation of the variation in coefficient of performance based on electrical, shaft and indicated power, determination of the overall heat transfer coefficient for the condenser cooling coil and performance of the thermostatic expansion valve.  · Investigation of the heat delivered to the cooling water with variation in condensing tem- perature, coefficient of performance as a heat pump for various condensing temperature, as well as power input based on electrical, shaft and indicated power. The important aspect of this report is to demonstrate the two laws of thermodynamics i.e. first and second law of thermodynamics. The first law is simply an expression of the conservation of energy principle, and it asserts that energy is thermodynamic property. Qout = Wnet + Qin Equation (1) In this experiment the Qin is provided by input voltage, this input is used to do the net work done on the refrigerant by compressor and motor, and the result of this produces the heat which is being removed by the condenser i.e. Qout. The second law of thermodynamics asserts that energy has quality and quantity, and actual processes occur in the direction of decreasing quality of energy. UMAR DARAZ Page 4 of 22 Thermodynamics Lab 2 Aims and objectives: The Hilton refrigeration laboratory unit R712 has been designed to allow students to fully investigate the performance of a vapour compression cycle under various conditions of evaporator load and condenser pressure. The main objectives of this laboratory are listed below;  · The demonstration of application of the First and second law of thermodynamics.  · The introduction of to refrigeration plant and calculate the various coefficient of perform- ance.  · Investigation of system losses, this includes motor, compressor, evaporator and con- denser losses. These losses (friction, heat losses) occur only in practical/commercial refrigerator, there are no losses in ideal vapour compressor refrigerator. UMAR DARAZ Page 5 of 22 Thermodynamics Lab 2 Apparatus The figure shown below looks like a refrigeration laboratory unit R712 (not exactly it) and it consists of the following components; Figure 23 Refrigeration laboratory unit Panel: High quality glass reinforced plastic on which the following components are mounted. Refrigerant: R12 Digital Thermometer: A device that measures temperature. Wattmeter: Allows measurement of the power input to either evaporator or motor. Voltage Controller: To vary evaporator load. Variable Area Flow meters: Variable area types to indicator R12 and H2O flow rates. Pressure Gauges: To indicate R12 pressure in evaporator and condenser. Spring Balance and Tachometer: These two together allow measurement of power required to drive the compressor. Expansion Valve: Thermostatically controlled type i.e. throttling device. Evaporator: Electrically heated device i.e. heat exchanger Compressor: (Internally mounted) Twin cylinder belt driven unit, along with spring balance force system. Condenser: A device or unit used to condense vapor into liquid. It is also called heat exchanger. Motor: A machine that converts electricity into a mechanical motion. 3 www.p-a-hilton.co.uk/English/Products/ Refrigeration__2_/refrigeration__2_.html UMAR DARAZ Page 6 of 22 Thermodynamics Lab 2 Procedure4 In prior performing an experiment the most important things to do are, to measure the atmos- pheric pressure, which would be added to the gauge pressure to get an absolute pressure for both condenser and evaporator, and to balance the two tips of the spring balance force, being applied on the compressor. In failure to do these things would cause a sufficient amount of error in the final results. In this experiment the condenser pressure is being kept constant i.e. 900KPa. Step-1 Turn on the refrigeration plant using one of the control breakers, and setting the evaporator voltage i.e. 40 100 volts, at the same time balancing the two tips of compressor load and set the condenser pressure to 900KPa, using rota-meter. Step-2 Record the following values; Evaporator Amps (1-2.42A), from wattmeter, compressor speed using tachometer, water and refrigerant flow rate using flow meter. Step-3 Record the spring balance force, reading directly from the scale. The hot water is in the tubes is indicated by red and cold water is indicated by blue sign in the refrigeration plant. Step-4 The flow rate is controlled by a throttling device (valve), the small changes in opening and closing the valve, effect the condenser pressure. Step-5 The temperature values of the refrigerant at different stages in the whole cycle at constant pressure is given by temperature dialler. Now we had all the values we needed, now we changed evaporator Amps value, recorded rest of the values as mentioned earlier and repeated the whole experiment for three to four times. The Refrigeration Laboratory Unit has three controls. Firstly a combined miniature circuit breaker and switch turns on both the compressor motor and the supply to the electrically heated evaporator. A combined variable area water flow meter and valve allow control of the condenser pressure and a panel mounted voltage controller allows control of the evaporator load from zero to full power. Refrigerant R12 vapour is drawn into the compressor from the evaporator mounted on the front of the panel. Work is done on the gas in the compressor and its pressure and temperature are raised. This hot, high pressure gas discharges from the compressor and flows into the panel mounted water cooled condenser, where heat is removed from it. This liquid then flows through a thermostatic expansion valve. Here it passes through a controlled orifice, which allows its pressure to fall from that of the condenser to that of the evaporator. The refrigerant has a satu- rated vapour phase at this point. The voltage across the heater elements may be varied from zero to that of the mains supply voltage by adjustment of a voltage controller situated on the front panel. Measurement of the power is carried out by a panel mounted digital wattmeter. 4 http://www.p-a-hilton.co.uk/R714-Edition-2-GREY.pdf UMAR DARAZ Page 7 of 22 Thermodynamics Lab 2 Results The observation table below shows all the values of different components in the refrigeration plant along with input indices and output indices, enthalpy of the cycle and losses in the system. The calculations required to get those results (to complete the table) are also listed after this table below. 1 Condenser pressure (gauge) Pc KNm-2 900 900 900 2 Evaporator pressure (gauge) Pe KNm-2 -20 20 40 3 Condenser pressure (Abs) Pc KNm-2 1001.663 1001.663 1001.663 4 Evaporator pressure (Abs) Pe KNm-2 81.663 121.663 141.663 5 Compressor suction t1 0 C -23.5 -22.6 -5.2 6 Compressor delivery t2 0 C 59.9 68.5 69.4 7 Liquid leaving condenser t3 0 C 31.6 34.8 33.8 8 Evaporator inlet t4 0 C -32 -23.6 -19.1 9 Water inlet t5 0 C 23.8 21.6 21.4 10 Water outlet t6 0 C 41.2 38.6 39.5 11 Water flow rate Mw g s-1 1.5 5.0 6.0 12 R 12 Flow rate Mr g s-1 0.7 1.5 1.9 13 Evaporator Volts Ve V 40 70 100 14 Evaporator Amps Ie I 1 A 1.70 A 2.42 A 15 Motor Volts Vm V 235 232 232 16 Motor Amps Im A 3.6 3.6 3.6 17 Spring balance Force F N 5.5 7.5 8.2 18 Compressor speed nc rpm 477 474 473 UMAR DARAZ Page 8 of 22 Thermodynamics Lab 2 19 Motor Speed = 3.17 ÃÆ'- nc Nm rpm 1512.09 1502.58 1449.71 20 h1 KJ/Kg 340 345 360 21 h2 KJ/Kg 385 400 420 22 h3 KJ/Kg 225 240 250 23 h4 KJ/Kg 160 170 180 24 Qe,Elec = Ve ÃÆ'- Ie W 40 119 242 25 Qe, R 12 = Mr(h1 h4) W 126 262.50 342 26 Wc = 0.0172ÃÆ'-FÃÆ'-Nm W 143.043 193.832 204.467 27 Power factor at shaft (power Wc) pf 0.43 0.48 0.52 28 Wm = Vm. Im. pf W 363.78 400.89 434.31 29 Wc = Mr (h2 h1) W 31.5 82.50 114.0 30 Q cond = Mr (h2 h3) W 112 240 323 31 Qw = Mw ÃÆ'- 4.18 (t6 t5) W 109.09 376.20 428.87 32 CoPnet = Qe, Elec / Wm 0.109 0.296 0.557 33 CoP R12 = (h1 h4)/(h2 h1) 4.0 3.1818 3.00 34 t41 can be found by (t1 t4) 0 C 8.5 1.00 13.9 35 CoP (te-t2) = t41 / (t2-t41) 0.165 0.015 0.250 36 Motor loss = Wc Wm W -220.73 -207.06 -229.84 37 Compressor loss = Wc-Wc W -111.54 -110.33 -90.47 38 System loss = Qcond Qw W 2.91 -136.20 -105.87 39 System loss = Qe, R12 Qe,Elec W 86 143.50 100.0 UMAR DARAZ Page 9 of 22 Thermodynamics Lab 2 Figure 3 A graph represents the relationship between net CoP and evaporator temperature Figure 4 A comparison of different losses of the system in one graph against Evaporator temperature The fluctuation and randomness in the graphs is because of the poor calibration and less number of repeated results (less tests provide less information), and most of the recorded results are based on guessed values. Calculations To find absolute pressure, we need an atmospheric and gauge pressure of the component. Now for two individual components,  · Condenser As we know Patm = à Ã‚ gh = 13600 kg/m3 ÃÆ'- 9.81 m/s2 ÃÆ'- 762 ÃÆ'-10-3m = 101.663ÃÆ'-103 Kg / ms2 = 101.663 KN/m2 Hence Pgauge,cond = 900 KN/m2 Pabs,cond = Patm + Pgauge,cond = 101.663 + 900 = 1001.663 KN/m2  · Evaporator As Patm = à Ã‚ gh = 13600 kg/m3 ÃÆ'- 9.81 m/s2 ÃÆ'- 762 ÃÆ'-10-3m = 101.663ÃÆ'-103 Kg / ms2 = 101.663 KN/m2 i. Pgauge,Evap = -20 KN/m2 Pabs,Evap = Patm + Pgauge,Evap Therefore = 101.663 + (-20)= 81.663 KN/m2 ii. Pgauge,Evap = 20 KN/m2 Pabs,Evap = Patm + Pgauge,Evap = 101.663 + (20)= 121.663 KN/m2 iii. Pgauge,Evap = 40 KN/m2 Pabs,Evap = Patm + Pgauge,Evap Therefore = 101.663 + 40 = 141.663 KN/m2 To find Qw (Heat removal from condenser) As we repeated the experiment three times, so water flow rate have three different values, hence we need to find Qw at three points, Qw = Mw ÃÆ'- 4.18 (t6 t5) When Mw = 1.5 gs-1, t6 = 41.2 0C, t5 = 23.8 0C Qw = 1.5 ÃÆ'-4.18 (41.2 23.8) = 109.098 W UMAR DARAZ Page 11 of 22 Thermodynamics Lab 2 As Qw = Mw ÃÆ'- 4.18 (t6 t5) When Mw = 5.0 gs-1, t6 = 39.6 0C, t5 = 21.6 0C So Qw = 5.0 ÃÆ'-4.18 (39.6 21.6) = 376.2 W Qw = Mw ÃÆ'- 4.18 (t6 t5) When Mw = 6.0 gs-1, t6 = 38.5 0C, t5 = 21.4 0C Qw = 6.0 ÃÆ'-4.18 (38.5 21.4) = 428.87 W To find Wc (work done by the compressor or a shaft loss) The work done by the compressor depends on spring balance force and motor speed, hence to get more work done out of the compressor we need to increase any of the above mentioned parameters. Therefore Wc = 0.0172ÃÆ'-FÃÆ'-Nm i. Wc = 0.0172ÃÆ'-5.5ÃÆ'-1512.09 = 143.043 W ii. Wc = 0.0172ÃÆ'-7.5ÃÆ'-1502.58 = 193.832 W iii. Wc = 0.0172ÃÆ'-8.2ÃÆ'-1449.71 = 204.467 W To find Wm (work done by the motor on a shaft to rotate) The work done by the motor is a product of voltage provided, amount of current flowing the motor and power factor of the shaft, which gives us the following values; Wm = Vm ÃÆ'- Im ÃÆ'- pf i. Wm = 235 ÃÆ'- 3.6 ÃÆ'- 0.43 = 363.78 ii. Wm = 232 ÃÆ'- 3.6 ÃÆ'- 0.48 = 400.89 iii. Wm = 232 ÃÆ'- 3.6 ÃÆ'- 0.52 = 434.31 UMAR DARAZ Page 12 of 22 Thermodynamics Lab 2 To find CoPnet (Total coefficient of performance of refrigerant) CoPnet = Qe, Elec / Wm By substituting different values of electric input heat energy (artificial input energy) and the work done by the motor, we get net coefficient of performance of the cycle, i. CoPnet = 40 / 363.78 = 0.109 = 11% ii. CoPnet = 119 / 400.89 = 0.296 = 30% iii. CoPnet = 242 / 434.31 = 0.557 = 56% To find CoP (te-t2) This is the coefficient of performance of ratio of temperature values at point 1-4 and difference of it, to the temperature of the refrigerant after compression, so we get following CoP (te-t2) = t41 / (t2-t41) i. CoP (te-t2) = 13.9 / (69.4 13.9) = 0.250 = 25% ii. CoP (te-t2) = 8.5 / (59.9 8.5) = 0.165 = 16% iii. CoP (te-t2) = 1.0 / (68.5 1.0) = 0.015 = 1.5% To find Qe, R 12(Heat removal from Evaporator) The given equation is à ¢Ã¢â€š ¬Ã‚ ¦ Qe, R 12 = Mr (h1 h4) By substituting different values of enthalpy, which we recorded from a pressure enthalpy diagram, so we get i. Qe, R 12 = 0.7 (340 160) = 126.0 ii. Qe, R 12 = 1.5 (345 170) = 262.5 iii. Qe, R 12 = 1.9 (360 180) = 342.0 UMAR DARAZ Page 13 of 22 Thermodynamics Lab 2 To find Wc (Input work done or compressor work loss) The input work done by the compressor can be calculated by finding flow rate of the refrigerant R12 and the difference of enthalpy of refrigerant before and after the compression. Wc = Mr (h2 h1) Substituting all three values of the above parameters (variables), we get i. Wc = 0.7 (385 340) = 31.5 ii. Wc = 1.5 (400 345) = 82.5 iii. Wc = 1.9 (420 360) = 114 To find Q cond (Heat loss by the condenser) Similarly heat loss by the condenser is a product of refrigerant flow rate to the difference of enthalpy values of it, before entering and leaving the condenser, we get Q cond = Mr (h2 h3) Now, using above stated equationà ¢Ã¢â€š ¬Ã‚ ¦ i. Q cond = 0.7 (385 225) = 112 ii. Q cond = 1.5 (400 240) = 240 iii. Q cond = 1.9 (420 250) = 323 To find CoPR12 (Coefficient of performance of refrigerant) CoP R12 = (h1 h4)/(h2 h1) Coefficient of performance of refrigerant is a ratio of all the enthalpy values in the cycle, here note that for ideal vapour compression refrigeration cycle h3 = h4 Hence we get i. CoP R12 = (340 160) / (385 340) = 4.00 ii. CoP R12 = (345 170) / (400 345) = 3.1818 iii. CoP R12 = (360 180) / (420 360) = 3.00 UMAR DARAZ Page 14 of 22 Thermodynamics Lab 2 Systems losses Motor loss = Wc Wm = 143.043 363.78 = -220.75 = 193.832 400.89 = -207.06 = 204.467 434.31 = -229.84 Compressor loss = Wc-Wc = 31.5 143.043 = -111.54 = 82.5 -193.832 = -110.33 = 114 204.467 = -90.47 System loss = Qcond Qw = 112 109.09 = 2.91 = 240 376.20 = -136.20 = 323 428.87 = -105.87 System loss = Qe, R12 Qe,Elec = 126 40 = 86.00 = 262.5 119 = 143.50 = 342 242 = 100.00 UMAR DARAZ Page 15 of 22 Thermodynamics Lab 2 Discussion of Results The observation table of results has been listed on page 8 9, and it is followed by all the calculations required to complete the table or to get the results. The experiment has been repeated three times, so all the results (values have been listed three times. In the calculation section the system losses and heat energy are shown as negative val- ues, its because the work is done on the system and heat is being removed from that particu- lar system, in this case its condenser. The positive values of system loss and heat energy shows that heat is being add in the system and work is done by the system, and in this case its evaporator. The condenser pressure i.e.900 KPa, was not exactly 900 KPa. As we were set- ting the pressure manually, so in the whole experiment the pressure was 900 KPa  ± 10%, it was because of the fluctuation in the gauge needle, so we assumed the considered pressure. The compressor pressure applied by spring balance force, affected the work done of the com- pressor on the refrigerant R12, because to get an accurate compressor work done, the two tips of the spring balance should be in balance (level), but during an experiment we were getting random values (results), so then I realised that something is wrong, so I looked at all the components of the refrigeration plant, and I found that the two tips of the spring were not bal- ance. Hence to get right results we had to redo the experiment. The throttling device or valve has a huge impact on condenser pressure, because by opening or closing i.e. changing a flow rate make a considerable amount of difference on condenser pressure and evaporator tem- perature. Motor loss refers to the consumption of electrical energy not converted to useful mechanical energy output, but in this case energy loss means less input energy to the compressor, which means a refrigerant would be less compressed by a compressor, so less heat would be re- moved by the condenser, and even after passing through the valve the refrigerant would still have a high temperature and pressure, hence less refrigeration would occur in a vapour com- pression cycle. Therefore we need to take into account power losses in the electric motor. In order to study this process more closely, refrigeration engineers use this pressure en- thalpy diagram shown in Figure 5. This diagram is a way of describing the liquid and gas phase of a substance. Enthalpy can be thought of as the quantity of heat in a given quantity, or mass of substance. The curved line is called the saturation curve and it defines the boundary of pure liquid and pure gas, or vapour. In the region marked vapour, its pure va- pour. In the region its marked liquid, it is a pure liquid. If the pressure rises so that we are considering a region above the top of the curve, there is no distinction between liquid and va- pour. Above this pressure the gas cannot be liquefied. This is called the Critical Pressure. In the region underneath the curve, there is a mixture of liquid and vapour. UMAR DARAZ Page 16 of 22 Thermodynamics Lab 2 3 2 4 1 Figure 65 Pressure Enthalpy diagram Evaporator Pressure line Condenser pressure line stage (Not a straight line) Isobar Condensation stage sion valve R12 Evaporation process 5 http://www.mvsengineering.com/chapter18.pdf UMAR DARAZ Page 17 of 22 Isentropic Compression R12 passing through Expan- Thermodynamics Lab 2 At the inlet of the compressor the temperature (t1) is the same as temperature of refrigerant R12 at the outlet of the evaporator. So we go straight from that temperature of left side of the doom (saturated liquid) to the right side of the doom (saturated vapour line), and then following the temperature gradient line, we go down and record the enthalpy value at that temperature and pressure. Similarly for the stage 2, we find h2 on x-axis. When the refrigerant leaves the condenser, it obtains a saturated liquid phase (left side of the doom), so taking the reference of condenser pressure line (red line), we take a straight line parallel to the y-axis, and wherever it meets the x-axis gives a value of enthalpy (h3) at stage three. In actual refrigerant plant, enthalpy at stage 3 and stage 4 is not same, but for the sake of calculation we assume that its an ideal condition and enthalpy at these two points is same. Test 1 As Compressor suction = t1 = -23.5 0C and condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h1 = 340 KJ/Kg Compressor delivery = t2 = 59.9 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h2 = 385 KJ/Kg Here Liquid leaving condenser = t3 = 31.6 0C And Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h3 = 225 KJ/Kg As mentioned earlier that h3 = h4 (Ideal condition) Hence the enthalpy h4 = 225 KJ/Kg But using temperature at evaporator inlet, t4 = -32 0C, we get Actual enthalpy value, h4 = 160 KJ/Kg Test 2 As Compressor suction = t1 = -22.6 0C and condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h1 = 345 KJ/Kg (from above p-h diagram) Compressor delivery = t2 = 68.5 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence using Figure 4, we get enthalpy h2 = 400 KJ/Kg Here Liquid leaving condenser = t3 = 34.8 0C And Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h3 = 240 KJ/Kg UMAR DARAZ Page 18 of 22 Thermodynamics Lab 2 As mentioned earlier that h3 = h4 (Ideal condition) Hence the enthalpy h4 = 240 KJ/Kg But using temperature at evaporator inlet, t4 = -23.6 0C, we get Actual enthalpy value using figure 4, h4 = 170 KJ/Kg Test 3 As Compressor suction = t1 = -5.2 0C and condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h1 = 360 KJ/Kg Compressor delivery = t2 = 69.4 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h2 = 420 KJ/Kg Here Liquid leaving condenser = t3 = 33.8 0C And Condenser Pressure (Abs) = Pc = 1001.663 KNm-2, Evaporator Pressure = 40 KPa Hence the enthalpy h3 = 250 KJ/Kg As mentioned earlier that h3 = h4 (Ideal condition) Hence the enthalpy h4 = 250 KJ/Kg But using temperature at evaporator inlet, t4 = -19.1 0C, we get Actual enthalpy value at this stage, h4 = 180 KJ/Kg 6However the expansion of the high pressure liquid, process 5 1 above is non reversible. Notice that Expansion is a constant enthalpy process. It is drawn as a vertical line on the P-h diagram. No heat is absorbed or rejected during this expansion, the liquid just passes through a valve, like water coming out of a tap. The difference is that because the liquid is saturated at the start of expansion by the end of the process it is partly vapour. Point 1 is inside the curve and not on the curve as described in the Evaporation process. At point 4 it starts to condense and this continues until point 5 when all the vapour has turned into liquid. Point 5 is saturated liquid. If more heat is removed, the liquid cools. It is then called sub-cooled liquid. Hence h4 is on a saturated liquid line (left side of the doom), and does not appear in a vapour compression cycle, and this is the case in all three tests. 6 http://www.alephzero.co.uk/ref/vapcom.htm#ph UMAR DARAZ Page 19 of 22 Thermodynamics Lab 2 As there is no moving part in the whole refrigeration plant apart from motor shaft of a compressor, so work done by them is zero, i.e. w = 0 So using steady state energy equation, we get W Q = h2 h1 Equation (2) As W =0, so equation (1) becomes Q = h2 h1 Or Q = h1 h2 Equation (3) The coefficient of performance or COP (sometimes CP), of a heat pump (i.e. refrigerator) is the ratio of the change in heat at the output (the heat reservoir of interest) to the supplied work.To find Cop value of refrigeration plant as well as for the refrigerant is a good practice, because this will illustrate that how much efficient of these two are. 7It takes a lot of heat to evaporate liquid. In other words a small amount of liquid circulating in a refrigerator can perform a large amount of cooling. This is one reason why the vapour compression cycle is widely used. The refrigeration system can be small and compact. Also from a practical point of view heat exchange is much better when using change of state evaporation and condensation. However the expansion of the high pressure liquid, process 5 1 above is non reversible. And so the efficiency of this cycle can never even approach Carnot efficiency. 7 http://www.alephzero.co.uk/ref/practcop.htm UMAR DARAZ Page 20 of 22 Thermodynamics Lab 2 Conclusion 8The vapour-compression cycle is used in most household refrigerators as well as in many large commercial and industrial refrigeration systems but the efficiency of this cycle can never even approach Carnot efficiency, because of its low coefficient of performance. In the refrigeration plant the operating parameters can be varied by adjustment of condenser cooling water flow and electrically heated evaporator supply voltage. Components have a low thermal mass resulting in immediate response to control variations and rapid stabilisation. Instrumentation includes all relevant temperatures, condenser pressure, evaporator pressure, refrigerant and cooling water flow rates, evaporator and motor power, motor torque and com- pressor speed. The most of components of refrigeration plant used in this experiment (R712) are manually calibrated scales (not digital), and based on this poor calibration all the recorded results are being guessed on the base of individual judgment, which is wrong most of the time. Anyway a small amount of liquid circulating in a refrigerator can perform a large amount of cooling. This is one reason why the vapour compression cycle is widely used. The enthalpy values which are being recorded directly from enthalpy pressure diagram (Figure 4), and based on how unclear that diagram is, I would say it is not a great source of information, but still we use this to find enthalpy. The system (refrigeration plant) has some losses, which have described earlier in this report, this includes motor loss, condenser and evaporator loss. In conclusion, I would like to say that by doing this experiment I learnt a great amount of knowledge, about refrigeration plant, and how it works, what kind of cycle more often use for this, how much efficient is this and how to calculate the different losses in this system. I would say by understanding the operation of this small scale refrigeration plant, I think I would be able to operate on an industrial scale refrigeration plant, because the basic principle is same. 8 http://www.alephzero.co.uk/ref/vapcomcyc.htm UMAR DARAZ Page 21 of 22 Thermodynamics Lab 2

Essay --

What does famine mean? It is a situation in which people do not have enough food, according to Webster’s New World Dictionary. The great potato famine, however, was far more than just a lack of food. Its effects were felt around the world due to environmental conditions, agriculture practices and economic and political factors (Japikse, 1994). Famine still exists in today’s society, as does the potato blight from the largest crop manufacturers to the small family gardens in Wisconsin, but nothing comes close to the disaster experienced by the Irish. It was the injustices that caused the famine, Gorta Mor, The Great Hunger of 1845-52, as called by the Irish (Daly, 1996). The potato famine of 1846 was one of the biggest natural disasters in Irish history. The â€Å"explosive disease† said William Fry, Ph.D., caused by Phytophtera infestans, is a condition that prevents growth and destroys the plant by disease. P. infestans is a fungus-like pathogen that infects and destroys the leaves, stems, potato seed and the potato itself, turning it black and slimy (Craig, 1998). The first signs of blight are when leaves get brown specks on them that start to get white, hairy growth around them (Damsker, n.d.). In addition, leaves and stems withered very quickly. This if followed by a terrible smell. It spreads as a spore and grows rapidly in moist, warm and humid conditions. Today, the blight disease still affects potato crops. However, an application of the plant fungicide, metalaxyl, or copper sulfate mixtures and pesticide will prevent and eliminate or, at least, decrease the possibility of its appearance (Daly, 1996). In fact, accordi ng to Gibbon’s article in 2013, the pathogen and its host of dried leaves ar... ...Irish and British people had bitter feelings towards one another. Many Irish were angry the English government did next to nothing to prevent the famine. Then when it happened the government turned their heads. The anger caused a rebellion in 1848 by a group called the Young Ireland party, saying Ireland wanted its own government (Results of the Great Famine, 2014). In conclusion, famine is still a part of the world today, but Gorta Mor of this level of devastation does not occur anymore. The same issues of environment, agriculture practices, economic and political factors are still part of famine today along with the appearance of the potato blight, too. Truth be told, the difference between then and now, however, is that technology is the medium to deliver images and descriptions faster and quicker so as to not let such devastation be experienced.