Leadership principles Essay Example | Topics and Well Written Essays - 750 words

Leadership principles - Essay Example This paper will discuss various aspects of a good way to lead in work-related situations. Leadership entails establishing a clear and candid vision. Palmisano (2008) argues that leaders should create and share vision with others within an organization so that they can follow it towards attaining predetermined goals. A leader should provide the methods, knowledge, and information to all members in order to realize the established vision. More importantly, leading an organization in a good way involves balancing and coordinating the conflicting interests of stakeholders and members. Various studies have shown that conflicts in an organization are inevitable and therefore a leader should have the ability of dealing with the conflict in a proper manner (Bonnici, 2011). A leader should step up in times of conflicts and be able to act creatively in addressing conflict in an amicable way. The process of resolving a conflict should include identifying the root cause of the problem and findin g ways of addressing those root causes in order to ensure that conflict of that particular nature does not recur in the future (Palmisano, 2008). ... At the same time, a leader is required to be polite and have empathy in his communication in order to ensure that he connects well with workers’ emotions and feelings. While a leader should communicate from the top to his sub-ordinates, he should ensure that bottom-up communication strategy is employed in the organization. This is important for making sure that employees have avenue to share their thoughts and ideas regarding how organizational goals and objectives can be achieved (Palmisano, 2008). It is also important in helping to understand some of the challenges that employees are experiencing and therefore assist in formulating better ways of addressing them. The most likeable thing about leadership is that one is able to take a lead in making a positive impact in an organization. Considering that one may fail to achieve the desired organizational goals, it is important that a leader should be competent and be willing to learn continuously how to become the best leader ( Bonnici, 2011). Leadership also involves frustrating or difficult situations ; a leader may experience a difficult or frustrating situation in leadership such as resistance to change. Change is an inevitable leadership process, which is aimed at transforming an organization to have better performance and achieve its goals much effectively and efficiently (Palmisano, 2008). However, in many cases change process face resistance from employees and even from some members of the management. This is usually the case because those opposed to change want to maintain status quo for their interests rather than for the interest of the organization. Since resistance to change can be a big obstacle in achievement of short and long-term goals, it is important that a leader should overcome this

Strategic management Essay Example | Topics and Well Written Essays - 2000 words - 7

Strategic management - Essay Example Strategic planning helps in formulating the business approach to a context. As shown in figures 1 and 2, it is seen that the strategy stems out of the context which influences the strategy and the actions that lead to a solution on the strategy and finally, provides the results which in turn leads to altering the context. The strategy should get altered based on the change in the context thus altered. Strategies are worked out using various tools. One of the oldest methods involved using the SWOT analysis for bringing out the strengths, weaknesses, opportunities and the Threats perceived by the company and the business2. The strategy would primarily enhance and capitalise on the strengths of the company for tapping the opportunities and to counter the threats perceived while at the same time would offset the weaknesses perceived. This would help the company to realise its objectives of enhancing the profitability of the company. Porter’s Five Forces helped the strategists to model the environment and the business is positioned in it to understand the effect of the environment. This was perceived to have been made up of the five forces that Porter projected. He further qualifies some of the approaches to strategy as generic. There are three types of generic strategies that are used by companies, employing the common economic forces that are in play in the market. These are: Cost Leadership by providing the best cost for a product or a service, product differentiation and thereby commanding higher prices and finally, identifying its own niche products for a specific product-segment thereby monopolising the entire business for that product or at least dominating it. While these strategies help the company to move forward, without a basic strategy the company stutters. As can be seen from the figure 2, the markets are created by companies which fall under any one of the strategies depicted in it, either consciously or otherwise. The perfect competition would

The Latino population in the US Essay Example for Free

The Latino population in the US Essay The Latino population that reside in the US have several cultural beliefs and values which are very important for the US Healthcare delivery system to understand and take into account. The policy makers should be aware of the various cultural characteristics that the Latino population develops. Family relationships form a very important part of the Hispanic culture, and frequently it is found that the Latin population would be staying together in the US as a very strong family. Hence, the entire family would be taking strong care of the pregnant women and her child. The family members would be advising the mother to sleep well, consume a nutritious diet and take a relaxing walk. The family members would be preventing the pregnant woman from smoking, doing heavy exercises, consuming alcohol and using medications. Pregnant women are given a lot of respect in the Latino culture and during this period a lot of care is given to them. The Hispanics consider pregnancy to be a very normal phenomenon. They would be frequently seeking prenatal management and the healthcare units. However, they also feel that pregnancy does not require any kind of extra-ordinary treatment. During the pregnancy the mother is given want she craves to eat, as they feel that any refusal would result in the baby developing birthmarks. The mother is given several traditional teas to reduce the pain that develops during labor. The mother is not allowed to move around during an eclipse as the baby is at a high-risk of developing cleft lip and cleft palate. The pregnant women would be considering putting a red band around her waist to prevent the baby from developing cleft lip or cleft palate. Usually, the pregnant woman’s mother or the mother-in-law would be available to the pregnant woman during her pregnancy as a supportive measure. The pregnant woman is also advised to walk during the period as the baby in the womb is bound to stick to the wall of the uterus. Medical interventions such as medications administration are not advisable during their pregnancy. The mother would approach the healthcare unit for nursing care late during the labor stage. During the postpartum period, the mother is given certain Home remedies known as ‘purgantes’, which effectively prevent the development of postpartum depression. The woman is advised to take adequate rest during the nursing period and is also expected to take care of the child (Hawaii Community College, 2005). In the US, about 54 % of the Hispanic elders belong to the Mexican groups; about 14 % belong to the Cuban, and the remaining to the other Spanish-speaking nations. About 5 %of all the Hispanic Americans belong to the elder’s age group, as on in the year 1990. However, with an increase in the life span, and more number of migrations to the US, there is an increase in the Hispanic elder group in the US. About one-fifth of the elders belonging to the Hispanic group are living below the poverty line. They are facing a lot of discrimination due to the social and economic status, poor education levels, high unemployment rates, lack of security, etc. They are deprived of proper healthcare facilities and a majority of them do not have insurance coverage. About 28 % of the Hispanic elders have a very poor health status. About 85 % had a long-standing health problem and about half of them were not able to function properly due to disease. Studies have shown that the health problems begin earlier in the Hispanic groups compared to the White population. The life span of the Hispanic groups is between 55 to 60 years. They visit the physician only if the health problem becomes severe or life threatening. They do not believe in preventive medicine. Although the elders require institutional care, about 10 % are institutionalized. This is about 23 % in the White population. One reason for low institutionalized rates is because the children provide care for the parents in their old age as per the customs and traditions. Many of the Hispanic elders receive home care even in the dying stages of life. The family ties with elders are quite high compared to the White population. The children would be offering economic and financial help to the elders. Besides, socially, the Hispanic communities respect the elders, and many positive interactions are generated between the community members and the elders. Even widows and widowers are given a lot of support by their family members, and this would help them (Socrates, 2007). The Hispanic populations are experiencing several problems in the US arising from disability and lack of provision of education. Poor research has been done by the healthcare authorities in the US, to understand the health needs of the disabled Hispanic population and providing them with adequate rehabilitative services. NGOs and governmental organizations have not collaborated in an effort to provide an efficient network to help the disabled Hispanic population. The problems that the Hispanic disabled population are facing are much similar to that of the other ethnic minority groups. Their socio-economic situation may be very poor and their living conditions may be very bad indeed. Educational levels and the lifestyle may be detrimental to their health needs. Many of the disabled children may leave school, as they are unable to manage their problems. In the US, about 20 % of the Hispanic population are disabled. The Hispanic disabled groups are not provided with proper rehabilitative, educational, vocational, or health services. A lot of research needs to be conducted by the health authorities to determine their health needs, and accordingly develop a system that could aid them. The outcome of rehabilitative services for the disabled should be improved. Factors that result in a poor outcome should be identified and addressed appropriately. Language and cultural barriers also need to be addressed. Healthcare professionals need to become more culturally competent. Professionals belonging to the Hispanic populations should be recruited in the healthcare system to help disabled individuals. A social network should be formed which could help the disabled populations. Besides, educational and vocational training programs should also be organized (Wong-Hernandez, 1997). Hence, it can be found that at present the Hispanic population is not utilizing the US healthcare system extensively to solve their health problems. Their health status at the moment is very poor and their health needs are high. The health department should develop a strong health policy that could help the Hispanic population with special needs. Healthcare personnel who are culturally competent should be employed. They should be able to interact in the local language and also understand their problems, beliefs, values and attitudes. This would give the Hispanic population greater confidence in the healthcare system and utilize it more frequently. This would also help to improve their health status and subsequently lead to increased use of preventive and family medicine. References: Arons, B. Chavez, N. (2001, January), Cultural Competence Standards in Managed Care Mental Health Services: Four Underserved/Underrepresented Racial/Ethnic Groups, Retrieved on July, 24, 2007, from SAMHSA Web site: http://mentalhealth. samhsa. gov/publications/allpubs/SMA00-3457/intro. asp Griggs, Shirley, and Dunn, R. (1995). Hispanic-American Students and Learning Style. Emergency Librarian 23 (2, Nov-Dec): 11-16. http://library. adoption. com/education/hispanic-american-students-and-learning-style/article/4281/1. html Hawaii Community College (2005), Hispanic, Retrieved on July 24, 2007, from Hawaii Community College Web site: http://www. hawcc. hawaii. edu/nursing/RNHispanic_04. html National Alliance for Hispanic Health, Duran, D. G. , Reyes, C. , Villarruel, A. , Brana-Lopez, A. R. , Gomez, P. , Mora, J. , Paz, J. (2001). Quality Health Services for Hispanics: The Cultural Competency Component. DHHS, no. 99-21. ftp://ftp. hrsa. gov/hrsa/QualityHealthServicesforHispanics. pdf Purves, H. (2003), Cultural Factors the Health of North Carolina Latinos. North Carolina Institute of Medicine Web site: http://www. nciom. org/projects/latino/latinopub/C3. pdf Socrates (2007), Hispanic American Elderly, Retrieved on July 24, 2007, from Socrates Web site: http://socrates. berkeley. edu/~aging/ModuleMinority2. html Wong-Hernandez, L. (1997), Building Networks in the Latino Community: A Mechanism for Empowerment, Retrieved on July 24, 2007, from San Jose State University Foundation Web site: http://www. dinf. ne. jp/doc/english/Us_Eu/ada_e/pres_com/pres-dd/lucywong. htm

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

Causes For The Fall of the Roman Empire :: The Fall of the Roman Empire

What major events led to the eventual decline and fall of the Roman Empire? Categorized between internal and external factors with broad reasoning, doesn’t lend itself to just a few events as the cause for the actual fall. From the internal factors: socio-economic problems and political corruption with the emperors and senate with their selfish, indulgence lifestyles with gladiator games being a major expense from the coffers, moral decline impacted the richest Romans with immorality, various outlandish sexual behaviors, gambling on most any activities and public lewd/sexual acts in the Colosseum. Education became only for the rich and usually only males. The basic standards of ethics and values were lost with total disregard of human and animal life, cheap slave labor lead to major unemployment for the working class plebeians that stressed the continued divide from the rich patricians. From the external factors: Constant wars and heavy military spending, the great Roman army was excessively expensive, over-stretched trying to keep the experienced warriors, recruiting more and more soldiers then turning to hiring mercenaries and barbarians. With the adding of more soldiers from conquered lands, those barbarians became more like Romans after a short time and less able to fight other barbarians from other countries as a result. With the army stretched thin at all the borders, when an issue grew it could become overwhelming quickly as with the Visogoths, once they were allowed to settle on the south side of the Danube, poverty living conditions and starvation led them to attack, then move to sack the city of Rome. Then there were the natural disasters such as famine, earthquakes and plagues. With influx of barbarians and many newcomers from over-run countries, the manifestation of serious sickness, transferring of plagues and lack of consistent medical care perpetu ated all natural disasters many times over.

Did you say Library Anxiety? - Part One :: Essays Papers

Did you say Library Anxiety? - Part One Most people are familiar with the terms test anxiety, math anxiety, performance anxiety, computer anxiety, or even social anxiety. But mention "library anxiety" and you'll likely get a response similar to, "Library what?" Library anxiety is not a well-known phenomenon, even among librarians. The bulk of research on library anxiety has concentrated on the problem as it applies to university students, but it’s not hard to imagine that it manifests itself in library patrons across the board. Where did this idea come from, how can librarians identify it, what steps can be taken to reduce it and what can the library community learn from it? Although it has been cited in the literature as far back as 1972 , the term library anxiety was first identified in 1986 by Constance A. Mellon. Virtually every article or study on the subject since then has referenced Mellon’s work in this area. Her studies showed that most students felt that other students knew more about library searching than they did and that to ask for help would be to reveal their stupidity. She also found that contact with reference librarians was more effective in alleviating library anxiety than the bibliographic instruction sessions conducted by their teachers. There are other names in the field such as Carol C. Kuhlthau, who found that students’ ability to process information from the aspects of mental, creative and physical locating operations is hampered by their feelings, thoughts, and actions. In 1992, Sharon L. Bostick devised a valid and reliable instrument to measure Mellon’s theory of library anxiety. The basis of her doctoral dissertation, she developed a 43 item, 5 point Likert-format test instrument that defines levels of library anxiety. Her instrument showed that it is possible to identify library anxiety and to measure it quantitatively. She identified five factors that contribute to library anxiety: 1) Affective Barriers; 2) Mechanical Barriers; 3) Comfort with the Library; 4) Knowledge of the Library; and 5) Barriers with staff. "Affective barriers" measures the feelings of adequacy when using the library. As we will see, affective barriers come in to play with all of the other factors, each of which will be described in greater detail. Mechanical Barriers: The ability to locate and use library equipment is hampered by the physical barriers libraries present. Students search for copy machines and upon locating them they learn that they need specific change to use them, or must purchase a copy card.

Geographic Distribution of Natural Resources for Energy Production Essa

Energy Distribution It is commonly known that natural resources are not distributed evenly around the world. As a result, energy is not distributed evenly around the world. In the present climate of the global economy, the distribution of energy presents conflicts between growing industry and already established industry. As a result, countries such as China, which are beginning to consume more resources, are turning to former methods of polluting natural resources such as coal because the resources they used to use are not as widely available. This is also taking place in other parts of the world. As the resources of crude oil deplete, oil companies are turning to new fuels such as ethanol without considering the side effects, similar to the way that China is blindly turning to coal without considering the pollution that is the result of coal. As a result of the rapid development and the sudden need for more energy in China, the production of energy in China is also different from that of the United States. Since China is a developing country, it has more of a chance to shift to alternate power sources more easily than the United States could; however, China needs rapid energy, and so it turned to coal rather than resources such as solar power or wind power. China is essentially at the stage that the United States was during the Industrial Revolution, whereas the United States has moved on to cleaner sources of energy such as petrol and oil. Another advantage that the United States has over developing nations is the use of alternate methods of power such as Nuclear Power Plants. Developing nations such as China do not have as easy an access to nuclear power due to the uncertainty as to the communist nature of the cou... ...op=793c8805q2fjrfq2fjgdloq5eddcmjmeeq22jeq3bjmbjq2aycfq5eyzcq2adyznjzoq2azjmbcq51q5eq2fq2ayfq7bsc0n>. â€Å"Coal power in China†. Wikipedia. 28 July 2008 . "Nuclear power in France." Wikipedia. 28 July 2008 . "Buying a Car: Gas vs. Diesel." Gas Vs. Diesel. 28 July 2008 . Pimentel. "Ethanol Fuel from Corn Faulted as ‘Unsustainable Subsidized Food Burning’." Ethanol. 28 July 2008 . Wallace, Ed. "Ethanol: A Tragedy in 3 Acts." Business Week. 28 July 2008 . Cohen, Matthew. "The Problem With Ethanol." Tampa Bay Online. 29 July 2008 .