How War Is Necessary Essay Example for Free

How War Is Necessary Essay War has been with mankind for many millenniums. The reasons for wars in mankind’s history have all been various, however one big reason for war is that countries want to grow bigger, by taking over one another. Throughout history, rulers would not be satisfied with the amount of land that they had or wanted to spread their beliefs around the world. So, they did it by trying to take over others. However, some didn’t want to give up their land or change, so they fought. Fighting these wars are, however, necessary no matter what people believe. This is because fighting them and winning will stop them from controlling others. Some wars that stopped countries from controlling others are the second world war, the war in Vietnam, and the Revolutionary war. However, there are claims that the Vietnam war could’ve been avoided, but there would’ve been consequences for going to war. War is necessary because it’s a way to try to stop countries from dominating over others and controlling them. There are many ways that war has stopped countries from controlling one another. Firstly, the Vietnamese war was necessary because, the war would’ve stopped the spreading of communism to neighboring asian countries, also known as the â€Å"domino theory†. The idea of communism started with the election of Ho Chi Minh in North Vietnam. Kennedy wasn’t concerned with Vietnam until Lyndon B Johnson called Ngo Dinh Diem the â€Å"Winston Churchill of Asia† because, Diem was the only one trying to resist the communists and Johnson promised to help him. Noam Chomsky, a US philosopher, stated that if a country better its economy due to communism, neighboring countries would try to improve their economy using communism, as seen when China influenced North Vietnam. Had communism been successfully spread to South Vietnam, then Thailand, Malaysia, Indonesia, etc. Also, other countries would’ve lost faith in the US for not protecting S.Vietnam in their time of need. This shows that the US needed to fight this war to stop the communistic ideals from spreading to countries and to keep the confidence of other countries around the world, so that they could help the US when the US needs them. The second world war as necessary becau se, it showed how the world responds to a fascist ruler controlling a major country, trying to claim the world. It also showed how the US would react when attacked by another major country. Hitler want to create a â€Å"master race†, which was blonde-haired, blue-eyed people. So when he came into power, he had convinced the german people that jewish people were less than humans and should be treated as such. Thus he started taking over various countries in Europe and started killing jewish people to start his plan. While that was happening, the US was maintaining neutrality and supplying Britain, Russia, and China with weapons until the japanese attacked pearl harbor on December 1941. The US then declared war on Japan and fought in the Pacific theater until August 15th, 1945. Had the axis powers won World War II, North America and South America would be surrounded by dictators ready to destroy democracy and take over the world. Also, Hitler’s plan would’ve proba bly been completed and the world would’ve been comprised of blonde, blue-eyed people. However, the war told us that when we are concerned with individualism, we become selfish and corrupted or under the influence of corrupted selfish nations. This war was needed because, when one tries to dominate and take over another country, we become blinded and end falling. We need teamwork to stay strong in the world. Finally, the Revolutionary War was necessary because, it allowed rights normally granted by the king, to be for every single individual. It also sparked other revolutions, along with stopping Britian from controlling the colonies. When the French Indian War concluded, King George III made the Proclamation of 1763, which stated that no colonists may settle west of the Appalachian and anyone other than Indians there had to move from the area. The Boston Massacre also brought them closer to the revolution, because British soldiers killed 5 civilians for calling them names. The Stamp Act of 1765, Townshend Act of 1767, and the Tea Act of 1773, were all attempt of Britian to try and control the colonies in North America. However after the war, the Declaration of Independence was signed and Britian let the colonies rule themselves. With this newfound freedom, the colonists were allowed to trade with anyone they wanted, colonize past the Appalachian Mountains, set up a new government and th e British moved out of the colonies. Had the colonists lost the war, the US would probably still be under British rule to this day. Even with all this evidence that war is necessary, people still see war as a terrible thing. Going back to the Vietnam War, the Vietnamese War could’ve been avoided because the US could’ve stayed neutral and let the problem in Vietnam blow over, instead of sacrificing 50,000 men. The war would’ve been avoided and Vietnam would’ve became a communist country along with possibly, part of Asia. Robert McNamara, defense chief under John F Kennedy and Lyndon B Johnson, says that he doubted Vietnam would let China or Russia use them as a base, but that’s the US had feared at the time. The US had feared that China or Russia would use them as a base they could use to control more parts of Asia. The Vietnamese could’ve probably fought the war themselves and probably resisted the communistic ways of China and/or Russia. The belief about if war is needed in the world or not cuts both ways. Either that it helps bring everlasting peace in the world closer or it’s useless and all it does is kill off innocent people. War is a big thing that stops countries from trying to gain total global domination or keeping them at bay until they give up. Until that happens, there will be allies who will attack at a moments notice when a country is trying to bite off more than it can chew.There are so many more examples of how war is necessary in the world that this would be a lot longer than it is already. The concept of war may seem bad, but in the end, it’s all worth it.

Reflective Essay on Empowering Healthcare Professionals

Reflective Essay on Empowering Healthcare Professionals This reflective account is going to explore a lecture that has changed my perception since the start of the nursing programme. I have chosen Gibbs (1998) reflective model to explore what I have learned, what my thoughts were before the lecture and changes. Therefore, to demonstrate an awareness of my learning outcome. I would also be covering how this has helped me to develop an understanding on what to do during my practical placement. The NMC code 2015 outlined standards and core values nurses must follow, therefore, our lectures focused around these to empower healthcare professionals to adapt to and deliver a quality service. The lecture I would be exploring emphasised on dignity, importance of dignity and how to encourage it. On a personal account, I thought I had a broad understanding of how to promote patients dignity. Nevertheless, after three hours of lesson on the subject, I have realized that there is a lot more to it. Dignity is a broad topic when properly studied. Dignity is a powerful tool that can determine a persons life and relationship with others suggested by Hinks, D. 2013. The poem what do you see? RCN, 2017, demonstrates the power of dignity. The poem represents a woman who is beseeching to be seen for who she is, not an old woman but someone who had lived a full life with feelings and emotions reinforcing the value of providing person centred care (RCN,2017). My feelings throughout the lesson changed considerably, leaving me to desire more of the lesson. Matti and Baillie (2011) proposed dignity as our innate value, merit and worth as human beings. To this end, the lecturer emphasized on respect making it clear that respecting people and making them feel worth is not just about caring for them, rather, it takes into consideration the whole process of how we approach and visually demonstrating dignity to the patients. I found that the NMC code (2015), stressed that as nurses, we have a duty of care to reach out for the physical, emotional, psychological, social and spiritual needs of our patients. So, give a holistic care by taking into consideration the individual as a whole. Hence, we need to prioritize patients care and dedicate time to them by preserving and promoting dignity through effective communication and supportive relationships. Emphasizing on these, it incorporated within me a sense of awkwardness and to realise how significan t it is for me as a person to be valued. Following the Poem What do you see? (RCN, 2017) I understood how patients are treated inhumanely. I felt like it is more dangerous to ignore patients emotion as this may have a huge impact on their personality It made me apprehended that, as a student nurse my responsibility is to work as part of a safeguarding team to raise concerns when I come across things that endanger patients worth and value according to the Local Government Association (2012). I found social care Act 2012 and CQC (2016) reinforcing on the link between dignity, quality and safeguarding indicating that the nurses, however, were not promoting dignity and therefore do not portray a good quality service so safeguarding is likely to decrease. The lecturer, also, underpinned the importance of personal appearance which connects to the concept of dignity because of the way uniform makes nurses feel and behave and how their appearance has an impact on patients supported by Chochinovs (2007) ABCD Framework (attitude, behaviour, compassion and dialogue) which reinforces essential things nurses need to be aware of when delivering care. Subsequently, this lesson appeared to pose more challenges. I felt highly challenged to question and evaluate on the type of staff I might be. I was challenged to think differently following Chochinovs (2007) self-awareness tool which got me thinking about if I would be able to provide a quality service to my patients while trying to promote dignity and what challenges I might face when dealing with safeguarding issues. However, after reflecting on myself and considering my attitude, behaviour, how I show empathy and compassion through communication, I felt confident to go out there and to demonstrate what I have learned with patients. I now feel like I am competent enough to fully and independently care for a person thus building a supportive relationship with them (Chochinovs 2007). Furthermore, this lecture helped me to comprehend the impact we as nurses can have on a patient as a person. In conclusion, dignity is how people feel, think and behave in relation to their worth or value. To treat someone with dignity is to show them as being worth and valued in a way that is exalt their diversity. Dignity may be endorsed or reduced by the physical setting, structural principles, approaches and conduct of others. When dignity is present, people feel in control, relaxed and able to make decisions for themselves, whereas when it is absent people feel devalued and lack control. Therefore, in my placement, I would do all the necessary things to ensure the value and worth of my patients, thus considering an approachable method of communication and rapport by making them feel at ease. Also, using person-centred care and empathy (RCN 2016). References: CARE QUALITY COMMISSION (CQC). 2010. Essential standards of quality and safety. London: HMSO. Chochinov (2007). Preserving patients dignity lends value to end of life: AHC Media: https://www.ahcmedia.com/articles/106003-preserving-patients-dignity-lends-value-to-end-of-life (accessed 24/02/17) Gibbs (1988) Reflective cycle. Available at: https://hhs.hud.ac.uk/lqsu/Sessionsforall/supp/Gibbs%201988%20reflective%20cycle.pdf (Accessed 24/02/17) Hick, D. (2013) What Is the Real Meaning of Dignity? Psychology today. Available at https://www.psychologytoday.com/blog/dignity/201304/what-is-the-real-meaning-dignity-0 (Accessed 27/02/17) LGA (2012) Dignity, quality and safeguarding adult: Establishing Local Health Watch Health. Available at http://www.local.gov.uk/c/document_library/get_file?uuid=d0235875-2da8-4a5c-a655-2f3600663f5dgroupId=10180 (Accessed 24/02/17) MATITI, M.R. and BAILLIE, L., eds., 2011. Dignity in Healthcare: a practical approach for nurses and midwives. London: Radcliffe. NURSING MIDWIFERY COUNCIL (NMC). 2015. The code: Standards of conduct, performance and ethics for nurses and midwives. London: NMC RCN (2017) Dignity and me: Available at: https://www2.rcn.org.uk/development/practice/cpd_online_learning/dignity_in_health_care/dignity_and_me (Accessed 24/04/15)

Premature Failure Of Road Network

Premature Failure Of Road Network Bahria town ltd started its development works in 1996 as a joint partner with Bahria foundation initially British Columbia were the consultants on the project. The extraordinary progress rate and high quality consultancy work of British Columbia (pvt) ltd was thought to be a big hurdle in the progress rate, eventually the agreement with the consultants terminated and Bahria Town (pvt) Ltd formed its own consultant wing. Unfortunately the consultancy wing failed to develop because of incompetent individuals who can really invest their heart and souls to address core issues .Site management and technical / top supervision issues were ignored .Today Bahria Town is facing problem of premature failure in its road network. Most of the road network has not been under projected traffic for which it has been designed; even then road failures are prominent .Most common failures depictive are settlement of road, flexural cracking, weathering of the road network. The mechanism of road failure is quite complex and it is tedious to identify the root cause of failure. The approach adopted was to analyze road network truly depictive of premature pavement failures, the representative sections were selected from the road network under study .Various field and laboratory test were performed on each section to determine the cause of premature pavement failures. The investigation revealed that mix produced from asphalt plant fails to meet specifications. The compaction of HMA and subsequent road layers is not adequate. The source gradation for aggregate base is improper .The Plasticity of fines is not in tolerance range. Pavement structural design depths were not executed on site besides poor workmanship and improper patching procedures. Keywords: Premature Failure, Flexural Cracking, Weathering, Source gradation. Undertaking I certify that research work titled To investigate the causes of Premature Failure of Road Network of Bahria Town to propose its Remedial Measuresis my own work. The work has not been presented elsewhere for assessment. Where material has been used from other sources it has been properly acknowledged/ referred. Tehseen Ellahi 2k9-MSc-Trans-05 Acknowledgements This research work is obviously a result of the initial encouragement and support in admission to the MS Transportation (Taxila) by Ehsan ul Haq, the Director General Planning and Design, Bahria Town (pvt) Ltd. Extraordinary help and support form Rana Zulfikar Ahmed Khan , Site Manager Bahria Town (pvt) Ltd.Continuous encouragement, and valuable input from Dr. M. A. Kamal, Director Taxila Institute of Transportation Engineering (TITE) and Dean, Faculty of Civil and Environmental Engineering, University of Engineering and Technology (UET) Taxila. There guidance, comments and suggestions from time to time, are gratefully acknowledged. 1.2.2 Rigid Pavements In rigid pavements the stress is transmitted to the sub-grade through beam/slab effect. Rigid pavements contain sufficient beam strength to be able to bridge over the localized sub-grade failures and areas of inadequate support. Factors effecting Pavement Performance There are numerous factors influencing the performance of a pavement, the following five are considered the most influential (Transportation research board, England; April 1985) 1.3.1 Traffic Traffic is the most important factor affecting pavement performance. The performance of pavements is mostly affected by the loading scale, arrangement and the number of load repetitions. The damage caused per pass to a pavement by an axle is defined relative to the damage per pass of a standard axle load, which is defined as a 80 kN single axle load (E80). Thus a pavement is designed to withstand a certain number of standard axle load repetitions that will result in a certain terminal condition of deterioration.(Kamal M.A. et al., 2009) 1.3.2 Moisture Moisture significantly reduces the supporting ability of gravel materials, especially the sub grade. Moisture enters the pavement structure through capillary action. The resulting action is the wet surface of particles, excessive movement of particles and dislodgment which ultimately results in pavement failures. (Terrel 1990) 1.3.3 Sub grade The sub grade is the lower layer of soil that supports the wheel loads. If the sub grade is not strong enough the pavement will show flexibility and finally the pavement will fail. Pavement will fail to perform ideally if the variation in particles behavior is not catered for in the design. 1.3.4 Construction quality Pavement performance is affected by poor quality construction, inaccurate pavement thicknesses, and adverse moisture conditions. These conditions stress the need for skilled staff and the importance of good inspection and quality control procedures during construction. Pavement performance is dependent on where, when and how maintenance is applied. No matter how good the pavement is built, it will deteriorate with time based upon the mentioned factors. The timing of maintenance is very important, if a pavement is allowed to deteriorate to a very poor condition, as illustrated by point B, then the added life compared with point A, is typically about 2 to 3 years. This added life is about 10 percent of the total life. The cost of repairing the road at B is four times of the cost required at A. The delay of maintenance hold implications, in that for the cost of repairing one poorly weathered road (Point B), four roads at point A would have to be postponed, which would mean that in a few years the rehabilitation cost could be 16 times as much. Thus, differing maintenance because of budget constraints will result in a significant financial penalty within a few years.(www.nra.co.za/live/content.php) History Bahria town is a modern township planned on an inspiration drawn from the home of American Society of civil engineers i.e the city of Reston, Virginia. The designing of its town ship is based on the most modern and strict criterion. It is located between the GT road and Islamabad Bahria town borders Safari Park on the northern side and is bounded to the south and west by Soan river and the Korang respectively. Town planning for Bahria Town has been done taking full advantage of the layout of the natural ground. Roads have been designed according to the traffic intensity rush hours. They have been standardized as 30, 40, 50, 60, 80 and Main Boulevards with the configuration of Pavement sidewalks and green areas.(www.bahriatown.com/index.php) Problem Statement Bahria Town (pvt) Ltd development represents a unique mark of distinction for Pakistan. Over a period of decade Bahria Town has emerged as Asias largest private property developers. The dilemma is that the capital involved is huge and to meet market demands common Engineering practices quality assurances are being ignored on account of time savings. One of the major problem , that Bahria Town that face today is related to premature failure of road network. Objectives The main objective of the study was to highlight the causes of pavement failures and to propose the remedial measures. Methodology Reconnaissance survey in study area to identify problems. Selection of test sections based upon road classification and distresses for deep testing analysis. Extraction of samples from the test sections for various laboratory testing. Comparison of various parameters between damaged and undamaged portions of test sections. Recommendations on the basis of investigation. Chapter 3 Chapter 3 Introduction to Study Area Pavement Evaluation Process Pavement evaluation is the first step in the development of pavement rehabilitation alternatives for the project .It is the process of learning the existing pavement system to understand the extent and the cause of problems prior to developing a rehabilitation plan.(www.pavementinteractive.org) Evaluation at Road Network Level Monitoring of the network is carried out at on a network level to define the status of an entire pavement network as part of the pavement management system .To achieve the said objective the road network of Bahria Town was divided in different zones .Preliminary reconnaissance survey was carried out to access the pavement condition of road network .The object was to confine the study and help prioritize and select the evaluation at the project levels. The road network under study is Safari valley. (Design report on Bahria Town, ESS.I.AAR Consultants) The justification behind selecting this study zone is that this zone is fully developed and in the possession of the residents therefore it is more realistic to study the road network performance of this zone. History The idea of Safari Valley was conceptualized in 2000 confirming to planning parameters of the cities of Ruston, Virginia, USA. Ruston being the American society of civil engineers has been planned in the most beautiful manner, the next year Safari Valley lunched another housing project. One of the aims of lunching this scheme was to provide a modern housing scheme with all the amenities for the general public at an affordable cost. Any middle class person desirous of buying a plot in Islamabad/Rawalpindi could not do so as the price in Islamabad/Rawalpindi is beyond the reach of the common man.. This scheme is planned and designed for the low-income people of the country to provide them affordable housing.(www.bahriatown.com/index.php) Location The Safari Valley is located in Southern part of Rawalpindi City, adjacent to Takht Pari forest on Japan Road. This site falls in Mauza Gali. The main access to this scheme is through Japan Road from G.T. Road, 3 Km from High Court towards Lahore.120 wide newly constructed Bahria Expressway along River Soan is another access to this project which is completed a year ago.(www.bahriatown.com/index.php) Untitled Figure 3: Study area Location in road network Road Network The proposed colony has been planned according to the contemporary principles of planning and design criteria of Tehsil Municipal Administration (TMA) for private housing schemes. Following three types of roads has been proposed. Primary Roads 120 Feet wide 80 Feet wide Collector Roads 65 Feet wide Streets 40 Feet wide The safari valley has 37.880 km of 40 wide roads, 6.083km of 65 wide roads, 1.23 km of 80 wide roads and 4.0 km of 120 wide roads. General Considerations It is desired by the consultant to provide the sub grade strength of the area in general for the construction of the internal roads. The purpose of the structural design is to limit the stresses induced in the sub grade by the traffic to a safe level at which the sub grade deformation is insignificant whilst at the same time ensuring that the road pavement layers themselves do not fail in any way within a specified period of time .In most design methods it is assumed that the routine and periodic maintenance is carried out during the design period of the road and that at the end of the design period , relatively low level of detoriation has occurred.(Structural design of Pavement at Safari valley, SS Soil explore Consultants) For the design of the flexible pavement the following factors should be kept in mind for guidance Economic Considerations Effect of climate Variability in the material Properties Construction Control Uncertainty in traffic forecasting Variability in material properties and construction control is generally much greater than desired by the engineer and must be taken into account explicitly in the design process. In practice only it is actually the variability of the sub grade strength that is considered and all other factors are controlled by setting out minimum acceptable values for the key properties by means of the specifications. Nevertheless, it is the task of the designer to estimate the likely variations in layer thickness and material strengths so that the realistic target values and tolerances can be set in the specifications to ensure the satisfactory road performances can be guaranteed as far as possible. Design basis The purpose of the structural design is to limit the stresses induced in the sub grade by the traffic. Estimating the amount of traffic and the cumulative number of equivalent standard axles that will use the road over the selected design life assessing the strength of the sub grade soil over which the road is built by selecting the most economical combination of the pavement materials and layer thickness that will provide the satisfactory service over the design life of the pavement when appropriate maintenance is carried out. In following paragraphs the component layers of a flexible pavement are referred in these terms: Surfacing This is the upper most layer of the pavement and will normally consist of bituminous surface dressing or a premixed bituminous material .When premixed materials are laid in two layers these are known as wearing course and base course (or binder course) Road Base This is the main load spreading layer of the pavement .It will normally consist of crushed stone or gravel ,or a gravelly soil ,decomposed rocks, sand and sandy clays stabilized with cement ,lime or bitumen. Sub Base This is the secondary load spreading layer underlying the road base .It will normally consist of material of lower quality than that used in the road base such as un processed natural gravels ,gravel-sand or gravel-sand-clay. This layer also serves as a spreading layer preventing contamination of the road base by the sub grade material Capping Layer Where very weak soils are encountered capping layer is sometimes necessary. This may consist of better quality sub grade material imported from elsewhere or existing sub grade material improved by mechanical and lime stabilization. Sub Grade This is the upper layer of the natural soil, which may be undisturbed local materials or may be soil excavated elsewhere and placed as fill. In either case it is compacted during the construction to give it adequate stability. Traffic In the present case no definite traffic pattern can be estimated as the construction period extends to a longer span .During the construction stage, the maximum traffic even over loaded trucks carrying mostly the construction material would apply. After the construction phase .the internal roads of the proposed project are subjected only to the light car traffic, which have very little destructive effect. The Sub grade Condition Following is the recommendations for the structural design of the bituminous surfaced roads for the proposed project. The existing sub grade at the site comprises of A 4 soil with PI range of 5 to 8.Determining the sub grade strength is necessary for the road construction and required by the design engineer for the internal light traffic roads, which are required to carry up to (assumed traffic) 0.5 million cumulative equivalent standard axles in one direction for the design life of 10 Years. Field investigation and Sampling All the field tests necessary for the design of the flexible pavements have been carried out .Test pit locations were selected so that overall picture of the sub surface can be examined .To do this samples from the different locations collected for the classification and California bearing ratio (CBR) .Following field and laboratory tests have been carried out in the detail Field density and moisture content Gradation analysis Sieve analysis Hydrometric analysis Hydrometric analysis AASHTO Soil Classification Laboratory compaction test Laboratory CBR on soaked conditions Appreciation of the sub grade condition The strength of the sub grade is commonly assessed in the form of California bearing ratio of the sub grade soil and is dependent on the type of the soil, its density and its moisture content The likely in situ strength of the sub grade is difficult to assess directly but its value can be obtained from the relationship between CBR, density and moisture content which must be measured in the laboratory for the soil in question, and form the knowledge of in situ density and equilibrium moisture content of the soil under the road. The density of the sub grade soil can be controlled under the road within limits by compaction at suitable moisture content at the time of the construction. The equilibrium moisture content of the sub grade soil is governed by the local climate and the depth of the water table below the road surfaces. For designing the thickness of the road pavement, the strength of the sub grade should be taken as that of the sub grade soil at the moisture content equal to the wettest moisture condition likely to occur in the sub grade after the road is opened to traffic. In the present case, field as well as the laboratory testing of different locations was carried out for gradation index and strength parameters and soaked CBR etc. The result of these tests are attached at the end of the report The CBR test shows the value of 3.5% having the representative design value of 90% and 95% modified by the AASHTO density .CBR value is considered to be unsatisfactory for the design of the flexible pavement .Therefore it is strongly recommended to provide capping layer over the existing sub grade soil to provide structural support and improve drainage conditions at the site. The thickness comes out to be 8 inches. Design Recommendations The pavement design of the internal roads of safari valley was calculated using the AASHTO Interim guide .Details of which are below: Method # 01 Out of the different methods available for calculating the road design, which cater for the repetition of the standard axle loads during the design life of the various traffic configurations expected on the road .This method caters for the site conditions and type of traffic likely to use the roads after the construction. The pavement design has been worked out as per Overseas Road Note No 31 (Transport and road research Laboratory, TRRL; Road note 31) The Local soil is moderately plastic for which average soaked CBR value was calculated to be 3.5 % and the same has been incorporated in the design calculations .In case of borrow /selected fill material is to be used for the making of the roads .The Laboratory CBR for that soil should not be less than 3.5 % against 96 hrs soaking. Method # 02 The pavement design of the internal roads of different categories is calculated using the simplified method as given in civil engineering handbook by Leonard Church Urquhart of which is given below. Design Procedure Using the graph (annexure A) against the clayey silt conforming to A-4 Soil and CBR of 3.5 % the total thickness of pavement above sub base is 15 inches. Keeping a minimum thickness of 8 inches for the base course and the wearing surface the sub base is required to be 7 inches. Since method 1 gave higher values method 1 was adopted. Flexible Pavement Distresses Roads have become important in our lives as a sole mean of communication. Modern roads are smooth, so people can travel easily from one place to another. Maintenance of road network is very important to ensure its continued efficiency and reliability. Normally roads are damaged due to environment affects, vehicular loadings and moisture.(Asphalt Institute , MS 16) Asphalt pavement distresses can generally be classified as one of the following type: Cracking Distortion Disintegration Skid hazard Surface treatment distresses Distresses caused can be related to: Wheel loads Environment Poor drainage Material deficiencies Construction related deficiencies External causes(Utilities) Cracking Cracking takes many forms .To make proper repairs, it is first necessary to determine the cause of cracking .Maintenance procedures generally depend upon the cause of distress, the crack width and the amount of cracking in the affected area. Reflective cracks These are cracks in asphalt overlays that reflect the crack pattern in the pavement structure underneath. The pattern may be longitudinal, transverse, diagonal or block. Reflective cracks are caused by vertical and horizontal movements in the pavement beneath the overlay, induced by expansion and contraction with temperature or moisture changes. They can also be caused by traffic or earth movement or by loss of moisture in sub grade by high clay content. Edge cracks These are longitudinal cracks 30 cm or so .They are caused due to lack of lateral support, settlement or yielding of the material beneath the cracked area .This may be the result of poor drainage ,frost heave or shrinkage from drying of the adjoining earth. They may be accelerated by concentration of heavy traffic near the edge of the pavement as well as heavy vegetation near the pavement edge. Block Cracking They are series of interconnected cracks forming the series of large blocks, 1 to 3 m. Frequently they are caused by volume change of the fine aggregate asphalt mix that have a high content of low penetration asphalt and adsorptive aggregate ,daily temperature cycles and aged asphalt. Block cracking is not load related. Alligator Cracking They are cracks that constitute to form series of blocks .They can be caused by various reasons such as excess deflection, sub surface moisture conditions, thin asphalt surface, excessive overloading, in adequate pavement design. If the asphalt surface is thin alligator cracking can quickly develop into potholing. Slippage Crack They are crescent shaped cracks resulting from the horizontal forces induced by the traffic. They result from the lack of bond between the surface layer and the courses beneath. The lack of bond may be due to dust, oil, rubber, dirt water or other non adhesive materials between the two courses. The Slippage cracks may result from the mixtures having a high sand content, as well as due to improper compaction. Linear Cracking This category includes categories such as joint cracks, construction joints, shoulder joint cracks and diagonal cracks. Transverse and diagonal cracks can result from low temperature contraction of the pavement or from the shrinkage of the cement bound base or sub grade soils .Longitudinal cracks in the wheel path may be fatigue related and eventually progress into alligator and a random occurring Longitudinal crack can be indicative of the sideways yielding sub grade or fill area. The cause of joint cracks (thermal and longitudinal) can be related to the thermal stresses or insufficient compaction. They can also be caused by a weak bond in the joint. Distortion Pavement distortion is the result of asphalt layer instability or granular base and sub base weakness. Distortion takes a number of forms: rutting, shoving, corrugation, depression and up heave. Rutting Ruts are channelized depressions in the wheel tracks of the pavement surface. Rutting results from consolidation, lateral movement of the sub grade, aggregate base or asphalt layers under traffic load. Rutting may occur in the sub grade and sub base due to insufficient design thickness, lack of compaction or weakness caused by moisture infiltration, down ward and lateral movement of the weak asphalt mixture under heavy wheel loads. Corrugations and shoving Corrugations and shoving are form of plastic movement typified by ripples across the asphalt pavement surface. They occur in the asphalt mixes that lack mix stability. It may also be caused due to excessive moisture in the granular base, contamination due to oil spillage or lack of aeration when placing mixes using emulsified and cut back asphalts. Settlement or grade depression Depressions are low areas of limited size that may be accompanied by cracking. They may be caused by traffic over loading or by consolidation, settlement or failure of the lower pavement layers. Up heave or swell Up heave is the localized upward displacement of the pavement due to the swelling of the sub grade. Up heave is most commonly caused by the expansion of ice in the lower courses of the pavement or sub grade. It may also be caused by the swelling effect of the moisture on the expansive soil. Utility cut or patch failure This is the failure of the utility installation or of a repaired area in the existing pavement. They usually are caused by lack of adequate compaction of the back fill, base or asphalt patch materials. Patch failures may also result from poor installation techniques, inferior materials or failure of the surrounding materials or under lying pavement. Disintegration Disintegration is the breaking up of the pavement into small, loose fragments. If the problem is not addressed the pavement disintegrates further until rehabilitation is required. Raveling/Weathering This is the progressive separation of the aggregate particles from the pavement surface downwards and from the surface inwards. Raveling usually occurs in wheel paths while weathering is found in non traffic zones and it extends over all surface. Raveling is caused by lack of HMA compaction, construction of thin lift during the cold weather, dirty or disintegrating aggregates, too little asphalt in the mix or over heating of the asphalt mix. Raveling almost always requires the presence of both water and traffic to occur. Potholes Potholes are bowl shaped holes resulting from the localized disintegration. Most potholes occur in the pavements having thin asphalt concrete surface on an untreated aggregate base. Thin surfaces showing severe alligator cracking begin to lose the pieces of the asphalt out of the cracked area creating potholes. Skid Hazards One of the most common cause of the skid hazards in the asphalt pavement is a thin film of water on the pavement surface another is the thick film of water on the pavement surface that causes a high speed vehicle to hydro plane. Slipperiness may also develop from the surface contamination such as from oil spillage or certain type of clay etc. Bleeding or flushing Bleeding or flushing is the upward movement in the asphalt pavement. This results in the formation of film of asphalt on the surface. Bleeding is identified by the pavement surface with a stick, glassy appearance that may be sticky to touch and usually occurs in hot weather .The most common cause of bleeding is excess asphalt in one or more of the pavement courses .Also traffic may cause the over compaction of the asphalt layers, forcing the binder to the surface. Polished aggregate These are the aggregate particles on the surface of the pavement that have been polished smooth. Some aggregates, particularly lime stone become polished rather quickly under traffic. Some type of gravel are naturally polished and if they are used in the pavement surface without crushing they will be a skid hazard. These polished aggregates are quite slippery when they are wet. Surface Treatment Distresses Because of the construction procedures being used, surface treatments may develop some defects that dont occur in other type of pavement surfaces. These include loss of aggregate cover and streaking. Some of the asphalt pavement distresses such as corrugations, depressions, up heave, potholes and raveling occur most frequently in the pavement constructed with surface treatments. Loss of cover aggregate This distress is identified by the whipping off of aggregate by traffic from a surface treated pavement. Several things can cause loss of aggregate cover including weather too cool, fast traffic permitted on the new surface treatment too soon, a surface absorbing part of the asphalt, aggregates that are too dusty or too dry etc. Longitudinal / Transverse Streaking Longitudinal streaking is alternate lean and heavy lines of asphalt and/or aggregate running parallel to the center line of the road .Transverse streaking is the same phenomenon except that the direction is running transverse across the road way. Several things can cause longitudinal streaking including: improper height of the spray bar, incorrect asphalt pump speed, asphalt too cold, incorrect pump pressure etc. Transverse cracking is caused by spurts in the asphalt spray from the distributor spray bar. These spurs may be produced by improper pump speed, pulsation of the asphalt pump etc.

Extinction On Dinosaurs :: essays research papers

Theories of the Extinctions of the Dionsaurs: Dinosaurs became extincted 65 million years ago, at the end of the Cretaceous period, something so devastating that it altered the course of life on earth. It seems like it happened so sudden, as geologic time goes, that almost all the dinosaurs living on earth disappeared. So how did these dominant creatures just die off? Was it a slow extinction, or did it happen all of the sudden? These questions bring rise to many different beliefs on how the dinosaur disappeared over 65 million years ago. Extinction is when the birth rate fails to keep up with the death rate, it is called extinction. But, the definition does not answer the question about the nature or causes of extinction. Paleontologists generally divide extinctions into two types, for that of different causes arose. The first is called background extinctions, isolated extinctions of species due to a variety of causes. Included is out competition, depletion of resources in a habitat, changes in climate, the development or destruction of a mountain range, river channel migration, the eruption of a volcano, the drying of a lake, or the destruction of a forest, grassland, or wetland habitat. The second type of extinction is called mass extinctions. Large numbers of species go extinct; many types of species go extinct; the effects must be global, and the effects must occur in a geologically short period of time.1 The dinosaur could not have lived for ever. No creatures, no plants, no tiny bacteria are forever, not even Homo sapiens. Extinction is the fate of all species. One theory on how the dinosaurs became extinct is that of carbon dioxide, and the greenhouse effect. Volcanoes produced the proposed conditions. A massive volcanic eruption could have saturated the atmosphere with carbon dioxide so that it caused a sharp rise in temperatures worldwide. The excessive carbon dioxide would have permitted solar energy to enter the atmosphere but would have blocked the radiation of most surface heat back out into space, therefore causing the greenhouse effect. Rising temperatures could have killed off or reduced the activity of plankton, disrupting food chains and also messing up the plankton's normal role in converting carbon dioxide to oxygen through photosynthesis. From there it would not have been long for all the dinosaurs to have been suffering, and then to become extinct. My theory of the extinction of the dinosaurs is the theory of the comet

Death Penalty :: essays research papers

This paper will fallow the process of a capital trial from arrest to execution. It will discus the aspects of federal and state law, trial, appeal, and executions. It will go into further detail on arraignment and the trail details of defense and sentencing. The federal law on capital punishment begins with the constitution, which states in the eighth amendment of the bill of rights that, no person shall be subject to cruel or unusual punishment. Despite this and for the reason that it is the government that decides what is cruel and unusual, capital punishment is still federally legal. Under the united states code, title eighteen there are certain crimes that can be punished by death. Section thirty-four of the said title and code says that any crime that results in the death of any person can be punished by death. Section 1512 deals with witnesses, victims, or informants. It states that anyone who kills or atemps to kill another person with the intent to prevent the attendance or t estimony at trail may be punished by death. Section 2332 states that who ever kills a national of the united states while the national is outside the united states is subject to death if the killing is murder as it is defined. Section 36 states that participants in any continuing criminal enterprise dealing with controlled substances may be punished by death. Section 1992 states that whoever willfully derails, disables, or recks any train used in interstate or foreign commerce can be punished by death. Finally section 831 states that anyone involved in prohibited transactions involving nuclear material can be subject to the death penalty. State laws in capital punishment defer from state to state and vary in a wide range of crimes for which it can be imposed. This range usually contains one or more of the fallowing, murder of a law enforcement officer, vehicular homicide while under the influence, contract killings, felony murder, first degree murder, or any murder. No matter the la ws of the state are certain states have and will always use their own discretion in handing down a death sentence. This means that for what ever reason, be it social make up, religious make up, or the simple fact that a death sentence may inhibit the prosecution, in that the jury may be hesitant to take a life no matter what the crime, the death sentence is not always used in all cases that it is allowed in.

Metalworks Case

| 2012| | KLU | Metalworks case study| Students: | Introduction : Metalwork is a company supplying cabinets and safety boxes. At the moment it has two plants and two warehouses which are â€Å"Des Moines† and â€Å"Dover†. Metalwork also uses an external supplier in case they can’t meet the demand. However in the case Metalwork has to buy products from supplier it doesn’t make any profit since the selling price $75 for the cabinet and $107 equals the buying price.Regarding this situation Metalwork has decided to improve its logistic efficiency by either increasing its capacity in â€Å"Des Moines† or investing in Juarez, Mexico by building up a new factory. Our job in this condition is to analyze the best option, to optimize the logistic efficiency and help Metalwork make a decision. At first we will see and optimize the current logistic system and then we will try to run the two solutions. And see which one is the best in order to give Metalwork th e best possible answer. Baseline scenario 1: without distance constrains and with direct shipment from supplier to customers.The first job we had to do was to actualize every data and verify everything was correct. This work consisted in adding the data for the 3 time periods missing (2011-2012-2013). So we added the data regarding warehouse capacity, production capacity, production costs, and customers demand. We allowed direct shipment from the supplier to customers. We also checked the flows between every actor of the logistic process (exhibit 1). We use the Rail Warehouses Midwest between warehouses and ABC fleet carrier from warehouses to customers. Then we made the software run for a first try and saw the results.So as a result we can see that first the scenario is feasible. Hopefully by the way since it is how the Metalwork is supposed to work. We can see that the total cost (which contains manufacturing, transportation, warehousing, variables, and holding costs) is $321  7 41  907 75 and total profit of $2  001  748  527 44. So the situation is pretty good since the company realizes profits. We can also see that in this baseline scenario we order 629  398 84 units to our supplier. Units on which we don’t make any profit. So we can clearly see that there is a need to invest in order to meet the demand thanks to our products.Baseline scenario 2 without distance constrains and without direct shipment from supplier to customers. In this scenario we will try to see if it is possible work without direct shipment from the supplier to the customers. That is why the lane visual changes compared to the first one. The answer in this case is that this scenario is not feasible, because of warehouses capacity. Indeed we begin to see that it is necessary to work with more than 2 warehouses. Baseline scenario with direct shipment and with distance constrains for warehouse to customer. In this scenario we will introduce constraints regarding the maxim um distance to customers.Indeed high quality service and responsiveness is highly important to Metalwork that is why we add a distance constraint of maximum 800 miles to Tier 1 customers and 1000 miles to Normal customers. We keep the same data and the same lanes and logistic system. Except that we allow this time delivery from the supplier to the customers. Because if not we would have add the same problem as in the second scenario since the situation is even harder regarding the distance constraints. That is why we decided to run it with this scenario. So we can see that the two warehouses supply the customers in within the 1000 miles away.That is why the supplier has to send directly to the other customers and also to supply the rest that is needed by the customers. We can see that only having two warehouses is really not enough. In this situation the supplier has much more importance and that has huge effects on the costs. Dual supply On the diagram bellow we can see that the su pplier has a huge importance on the production since it produces even more than the Plant in Dover. So in this situation we can see that the total costs have dramatically increased from $341  741  907 75 to $887  796  558, 11 and the profit has gone down to $1  435  693  607, 07.The manufacturing cost is really high: 749  750  420 $ compared to 138  145  001$ in the first scenario. Regarding the new constraint we really realize that Metalwork needs to invest in new plant and in new warehouses. Baseline scenario with direct shipment and with distance constrains for warehouse to customer and supplier to customer. The scenario is not feasible because the warehouse capacity is limited and the distance from supplier to some customers is greater than the distance restriction. We decided to apply this restriction because we fought it was necessary to apply the distance constraints to the supplier also.Indeed the service level must be equal for every product to every c ustomer. Indeed the customer doesn’t need to know and don’t care if the product is from the supplier plant or Metalwork plant. Increased demand scenario in Des Moines We increase capacity of Des Moines. The production capacity of the safety boxes increases by 25% and cabinets by 50%. We set the additional $1  250 000 operation costs. And we decrease the cost for each unit produced in Des Moines by 50 cents. We apply these directives into our data base for every time period. We also add the possibility to go from 2 warehouses to 4 warehouses for the 4 time period.Two of the warehouses are fixed. One in Des Moines and one in Dover. So in this scenario we can see that the total cost has decreased compared to the first baseline scenario. Now we have a total cost of $248  104  881 97 and a profit of $2  075  385  283 22 which is much higher than in the first baseline scenario. We can also notice that with these investments the supplier is no longer needed. So th e objective achieved. Plus we have 100% of the demand met. We clearly see that the Plant in Des Moines is the main motor of Metalwork. Plus even with the distance constraint we see that the scenario is feasible.This is able thanks to the 2 more warehouses available. So as we see on the map bellow, all the customers are supplied and the 4 warehouses are dispatched on every side of the United States. Plus there are only a few customers supplied by two warehouses, because of warehouse capacity. So clearly the result is conclusive. The demand is met the costs go down and the profit rises. Double supply Mexican Plant scenario In this scenario the objective of Metalwork is to improve the network of its plants and also to relocate its investment into a low labor cost country.The opening cost of this plant opening is $5  000  000. In order to make this scenario work we had to reset the data for the 4 time period, that is to say cancel the downsize in costs and production capacity in Des Moines. We also forbid the direct shipment from supplier to customer if needed. And used the railway West transportation for Railway warehouses West to supply our products to the warehouses. But we still have the same problem as in the previous scenario that is to say that some customers are supplied by two warehouses. Double supplyAs we can see can see in the chart below, our total cost is $ 243  950  541 68 which is smaller than in the previous scenario. The profit is also a little bit higher, with $2  079  539  623 51. 100% of the demand is met so we can say that the objective is met. And with better results on every side whether it is financial or quality we can say that we would recommend this investment rather than the first one. However this also depends on the image the brand wants to have, something such as made in America and avoid social problems in the plants.When the activity is relocated. We can see on the last chart that the production almost equally balanc ed between Des Moines and Juarez. The good thing is that the supplier is no longer needed. So every unit sold makes the company make profit. Conclusion: 1. After comparing the different scenarios based on the guidelines we had, we found that the scenario that involved building a plant in Juarez, Mexico turns out to be the best one. 2. While comparing the increased capacity scenario and the â€Å"Mexico† scenario we found that they are nearly the same.In both of these scenarios we find that we don’t need an external supplier, we would be able to supply the demand on our own. 3. Compared to the baseline scenario we can see that increasing the number of warehouses decreases the overall total costs. 4. While optimizing the solution, we figured out that having to use two given warehouses might not be the optimal choice, those two warehouses should be catalogued as potential, the maximum number of warehouses should be increased. 5. We observed baseline scenarios can’t compete with the improved capacity scenario or the â€Å"Mexico† scenario since the cost of buying from a supplier is too high.

Lamarsh Solution Chap7

LAMARSH SOLUTIONS CHAPTER-7 PART-1 7. 1 Look at example 7. 1 in the textbook,only the moderator materials are different Since the reactor is critical, k ? ? ? T f ? 1 ?T ? 2. 065 from table 6. 3 so f ? 0. 484 We will use t d ? t dM (1 ? f ) and t dM from table 7. 1 t dM,D2O ? 4. 3e ? 2; t dM,Be ? 3. 9e ? 3; t dM,C ? 0. 017 Then, t d,D2O =0. 022188sec;t d,Be =2. 0124e-3sec;t d,C ? 8. 772e ? 3sec 7. 5 One? delayed? neutron group reactivity equation; ?lp 1 ? ?lp ? ? ? where ? ? 0. 0065; ? ? 0. 1sec? 1 1 ? ?lp ? ? ? For lp ? 0. 0sec For lp ? 0. 0001sec For lp ? 0. 001sec Note:In this question examine the figure 7. and see that to give a constant period value ,say 1 sec,you should give much more reactivity as p. neutron lifet ime increases. And it is strongl recommended that before exam,study figure 7. 1 . 7. 8 ? ? 2e ? 4 from figure 7. 2 so you can ignore jump in power(flux) in this positive reactivity insertion situation t P Pf ? Pi e T then t=ln f ? T ? 3. 456hr Pi 7. 10 In eq 7. 19 p rompt neutrons:(1-? )k ? ? a ? T delayed neutrons:p? C ? in a critical reactor(from 7. 21) ?k ? ? dC ? 0 ? C ? ? a T ? p? C ? ? k ? ? a ? T dt p? ? s T ? (1-? )k ? ? a ? T ? ? k ? ? a ? T ? ? ? prompt delayedNow you can compare their values prompt (1-? ) ? delayed ? LAMARSH SOLUTIONS CHAPTER-7 PART-2 7. 12 P0? t 1 P(t) ? e in here ? ? then, and ? ? T t P0 T P(t) ? e in here take T=-80sec ? 1? ? t ? P0 P0 ? 10 ? e 80 ? t ? 25. 24 min . 1 ? (? 5) ?9 7. 14 k ? ,0 ? pf 0 ,critical state k ? ,1 ? pf1 ,original state k ? ,1 ? 1 k ? ,1 ? k ? ,1 ? k ? ,0 k ? ,1 ? pf1 ? pf 0 f ? 1? 0 pf1 f1 ?a1F ?a 0 F f1 ? F f0 ? and we know ? a1F =0. 95 ? a 0 F and finally, M F M ? a1 ? ? a ?a 0 ? ?a f0 1 0. 95? a 0 F ? ?a M 1? ? 1? ( ) f1 0. 95 ? a 0 F ? ?a M 7. 16 20 min? 60sec/ min ? 1731. 6sec. ln 2 )From fig 7. 2 rectivity is small so small reactivity assumption can be used as, 1 1 T= ? ?i t i ? ? 0. 0848(from table 7. 3)=4. 89e-5=4. 89e-3% ?i 1731. 6 4. 89e-5 also in dollars= ? 7. 52e ? 3$ ? 0. 752cents 0. 0065(U235) t T a)2P0 ? P0e ? T ? 7. 17 8hr ? 60 min? 60sec 8hr ? 60 min? 60sec ?T? ? 6253. 8sec(very large) T ln100 b)We will make small reactivity insertion approximation using the insight given by figure 7. 2 for U-235 so, 1 1 T= ? ?i t i ? ? 0. 0324(from table 7. 3)=5. 18e-6 ?i 6253. 8 a)100MW ? 1MWe 7. 18 a)From fig 7. 1 when ? ? 0 ? 1 ? 0 so T= 1 ?T ?1 b)Use prompt jump approximation, t tP0? T P0 T 10watts (300? 100)sec P(t)= e? e? e 100sec ? 82watts ? 0. 099 1? 1? ? 1 c)Use T=-80sec. 300)sec t t P0? T P0 T 82watts ? (t ? 80sec P(t)= e? e? e ? 8 1? 1 ? (? ) ? 1 LAMARSH SOLUTIONS CHAPTER-7 PART-3 7. 20 Insert 7. 56 into 7. 57 and plot reactivity vs rod radius Using eq. 7. 57 and 7. 56 we plotted and found the radius value for 10% reactivity=3. 9 cm reactivity vs rod radius(a) 0. 14 0. 12 X: 3. 9 Y: 0. 1004 reactivity 0. 1 0. 08 0. 06 0. 04 0. 02 0 0 0. 5 1 1. 5 2 2. 5 rod radius 3 3. 5 4 4. 5 5 7. 23 a)For a slab this equation is solved you know as, x xq ?T (x) ? A1 sinh( ) ? A 2 cosh( ) ?T then to find the constants you must introduce L L ? a 2 boundary conditions 1 d? T 1 d? T 1 B. C. 1: ? 0 @ x=0 and B. C. 2: ? ? @ x=(m/2)-a ?T dx ?T dx d Introducing B. C. 1 you find A1 ? 0 and B. C. 2 x ? ? cosh( ) ? ? q L A2=- T ? 1 ? ? d ?a ? sinh((m ? 2a) / 2L) ? cosh((m ? 2a) / 2L) ? ?L ? So finally, x ? ? cosh( ) ? ? qT L ?T (x) ? ?1 ? ? d ?a ? sinh((m ? 2a) / 2L) ? cosh((m ? 2a) / 2L) ? ?L ? b) Neutron current density at the blade surface, d? L J @(m/2)-a ? ? D T ? d dx @(m/2)-a ? coth((m ? 2a) / 2L) L Let ‘s follow the instructions in the question Multiply the n. current density by the area of the blades in the cell†¦ –What is the area of the blades in the cell: From fig 7. 9,assume unit depth into the page so the cross sectional area of one of four blades, A=(l-a) ? 1 Divide by the total number of neutrons thermalizing per second in the cell —What is the volume of the cell: From fig 7. 9,assume unit depth into t he page so V=(m-2a) ? (m ? 2a) ? 1 So as in page 358 4(l ? a) 1 fR ? 2 (m ? 2a) d ? coth((m ? 2a) / 2L) L 7. 25 You should find the B-10 average atom density in the reactor Total mass of B-10=50rods ? 500g=25 ? 103g 25e3 N? ? 0. 6022e24 ? 1. 39e27atoms 10. 8 Atom density averaged over whole reactor volume, 1. 39e27 NB ? ? 2. e21 atoms/cm3 ? ? aB ? 2. 9e21? 0. 27b ? 7. 8e ? 4cm ? 1 4 ?(48. 5)3 3 7. 8e ? 4 ? use eq. 7. 62 then find,? w ? ? 0. 0938 ? 9. 4% 0. 00833 ? 0. 000019 7. 27 H ? 100cm and ? ? 0. [email  protected] x ? H a) For x ? 3H / 4 ? 75cm 1 ?x ? ? Sin(2? x / H ) ? ? (3H / 4) ? ?0. 4545$ ? H 2? ? so the positive reactivity insertion is -0. 4545$-(-0. 5$)=0. 04545$ ( x) ? ( H ) ? b) The rate of reactivity per cm can be found by differentiating the reactivity equation over the distance. ?1 1 ? d ( x) d ? 1 ?x ? ? ( H ) ? ? Sin(2? x / H ) ? ? ? ( H ) ? ? Cos(2? x / H ) ? dx dx ? ?H H ? ? H 2? ? d ( x) ? 0. 005$ / cm ? 0. cent / cm dx x ? 3H / 4 7. 31 There is a de crease in T so let’s examine the effects of sign of temperature coefficients, If ? T ? (? ) decrease in T ? decrease in k ? reduces P ? gives further dec. in k ? shut down(unstable) If ? T ? (? ) decrease in T ? increase in k ? increase in P ? inc. in T and finally reactor returns to its original state! (stable) 7. 33 ? N FVF I ? p ? exp ? ? ? ? ? M ? sM VM ? I: Resonance Integral ? sM : Scattering Cross-Section of Moderator ? M : Constant 2a ? 1. 5 ? a ? 0. 75 (rod radius) dI I (300 K ) ? 1 ? ? I (T ) ? I (300 K )(1 ? ?1 ( T ? T0 )) dT 2T I (T ) ? ? ? sM ? M VM ln p N FVF T ? T0 ?I (T ) ? I (T0 ) ? ?k ln 0. 912 ? 0. 0921k where k ? ? sM ? M VM N FVF For slightly enriched uranium dioxide reactor take ? ? 10. 5 g / cm3 (See Chapter 6). ?1 ? A? ? C? / a? where A? ? 61? 10? 4 and C? ? 2. 68 ? 10? 2 (Table 7. 4) ? ?1 ? 0. 009503 T ? 665? C (? 938K ) ? I (T ) ? I (T0 )(1 ? 13. 31* ? 1 ) ? 1. 1264I (T0 ) ? I (T ) ? 0. 0921? 1. 1264 ? k ? 0. 1037k ?1 ? ?k ? [email  protected] 665o C ? exp ? ? I (T ) ? ? exp ? ? 0. 1037 ? ? 0. 9014 ? k ? ?k ? 7. 34 70 F ? 210C 550 F ? 287 0C d ? ?T ? ? ? ? (287 ? 21) ? ?2 ? 10? 5 0C dT ? T where ? =0. 0065 ?1 ? ? 5. 32e ? 3 ? ?0. 532% ? ?0. 81$ 7. 37 First you should solve problem 7. 6 to find the fraction of expelled water, 575F ? 301 0 C 585F ? 307 C 0 Vvessel ? 6 0 C increase in T ?D 2 ? ? 6. 5m3 ? Vwater ? v 0 ? 3. 25m3 4 ?v ? ? v ? T ? ?v ? 3. 25m3 ? 3e ? 3 ? 6 0 C ? 5. 85e ? 2m3 v0 ?v ? 0. 018 v0 Then find f after expelling, k ? ,0 ? pf 0 ,critical state k ? ,1 ? pf1 ,original state k ? ,1 ? 1 k ? ,1 ? k ? ,1 ? k ? ,0 k ? ,1 ? pf1 ? pf 0 f ? 1? 0 pf1 f1 ? a1F ?a 0 F f0 ? and we know ? a1F =0. 95 ? a 0 F and finally, F M F M ? a1 ? ? a ?a 0 ? ?a f1 ? f0 1 0. 95? a 0 F ? ? a M 1? ? 1? ( ) f1 0. 95 ? a 0 F ? ? a M f0 ? ?a F ?a F ? ?a M f? in here f 0 ? 0. 682 so ?a F ? a F ? 1 ? ?)? a M ?a M 1 ? ? 1 ?a F f0 so f? 1 1 1 ? 0. 0982 ? ( ? 1) f0 ? 0. 956 f-f 0 ? 0. 287 f 0. 287 Finally, ? T (f ) ? ? 0 ? 0. 0478per 0 C ?T 6C Then = LAMARSH SOLUTIONS CHAPTER-7 PART-4 7. 39 The reactivity equivalent of equilibrium xenon is to be; ? ? I ? ? X ? T where ? X ? 0. 770 ? 1013 / cm2 ? sec and ? X ? 0. 00237 and ? I ? 0. 0639 ? p? ?X ? ?T ? ? 2. 42 and p ? ? ? 1 0 -0. 005 reactivity -0. 01 -0. 015 -0. 02 X: 4. 8 Y: -0. 02695 -0. 025 -0. 03 0 0. 5 1 1. 5 Note the convergence †¦.. 2 2. 5 3 thermal flux x 1e14 3. 5 4 4. 5 5 7. 42 For Xenon using eq. 7. 94 X? ? (? I ? ? X )? f ? T ?X ? ? aX ? T here ? I ? 6. 39e ? 2 and ? X ? 2. 37e ? 3 (from table 7. 5) ? X ? 2. 09e ? 5 (from table 7. 6) You should make a correction to the thermal absorption cross section as follows, ? 20 0. 5 ) 2 200 ? aX (200? C ) ? 0. 886 ? 1. 236 ? 2. 65e6 ? 1e ? 24 ? 0. 316 ? a,X ? ? g aXe (200 0C ) ? ? a,X (20 0C ) ? ( ? aX (200? C ) ? 9. 17e ? 19cm 2 ? 9. 17e5b finally, X? ? 0. 06627 ? ? f ? 1e13 2. 09e ? 5 ? 9. 17e5b ? 1e13 For Samarium using eq. 7. 94 S? ? ? P ?f ? aX where ? P ? 0. 01071 ? 20 0. 5 ) 2 200 ? aX (200? C ) ? 0. 886 ? 2. 093 ? 41e3 ? 1e ? 24 ? 0. 316 ? a,S ? S ? g a (200 0C ) ? ? a,S (20 0C ) ? ( ? aX (200? C ) ? 2. 9e4b finally, S? ? 0. 01071 ?f 2. 39e4b Note:When finding fission cross sections you should find the atom density of uranium 235 for this infinite thermal reactor. To do this ,refer to example 6. 5 on page 294 taking buckling zero and find a relation between moderator number density and fuel density. 7. 43 Using eq. 7. 98 0. 06627 1e13 ? 2. 42 1e13 ? 0. 773e13 where p=? =1 0. 01071 2. 42 ? Xe ? ? ? Sm 7. 44 First of all, we must write the rate equations for each element; dN Sm ? Sm N Sm ? ? a Sm N Sm? T ? ? Sm ? f ? T dt dN Eu ? ? Sm N Sm ? ? Eu N Eu ? ? a Eu N Eu? T dt dN Gd ? ? Eu N Eu ? ? a Gd N Gd? T dt ) For equilibrium reactivity; N (t ) ? N (t ? dt ) ? Xi Xi and ignore ? a Sm N Sm? T & ? a Eu N Eu? T Inserted into all rate equations; N Sm ? Sm ? f ? T ? ? Sm dN X i (t ) ?0 dt ? Sm N Sm ? ? Eu N Eu ?a N Gd Gd ? Eu N Eu ? ?T Reactivity equation is found as below; where ? a Gd / ? f p ? Sm p ? Sm ? 7 ? 10? 5 and ? ? 2. 42 and ? ? p ? 1 ? ? ? ? 2. 893 ? 10? 5 b) 157 Sm decays rapidly relative to 157 Eu and half-life of the 157 Sm is too small so, dN Sm ? 0 ? Sm N Sm ? ? Sm? f ? T ? ? Sm N Sm ? ? Sm? f ? T dt This equation is inserted into rate equation of 157 Eu and 157 Gd ; dN Eu ? ? Sm ? f ? T ? Eu N Eu dt dN Gd (t ) ? ? Eu N Eu ? ? a Gd N Gd? T dt Gd At shutdown ? N0Eu & N0 are equal to equilibrium concentration for 157 Eu and 157Gd . ? No fission & no absorption is observed. From rate equation of From rate equation of Eu ? N 157 157 Gd Eu ?N Eu ? ? Eu t 0 (t ) ? N e Gd (t ) ? N Gd 0 ? Sm ? f ? T Eu t ? e ? Eu ? Sm ? f ? T Eu ? (1 ? e t ) ? Eu From equilibrium of Gd ? N 157 Gd 0 ? Sm ? f ? ? a Gd ? Sm ? f ? Sm ? f ? T Eu ? N (t ) ? ? (1 ? e t ) ? a Gd ? Eu Gd Maximum reactivity is reached at time goes to infinity! Gd ? N max (t ? ?) ? ? Sm? f ( ? a Gd / ? f p 1 ?a ? ?T ) ? Eu Gd Sm where ? a ? ? f (1 ? ?T ? a Gd ? ? ? (1 ? ) /? ? Eu Sm Gd where ?T ? a Gd ) ? Eu ? Eu ? 1. 162 ? 10? 5 s ? ? ? ? 4. 386 ? 10? 5 ? ?0. 675cents 7. 47 a) For constant power; P ? ER ? ? fF (r , t )? T (r , t )dV V So as N decreases ,flux should increase to keep power constant, dN F (t ) ? ? N F (t )? aF ? T (t ) (1) dt P ? ER ? fF (t )? T (t ), ? fF (t ) ? N F (t )? aF N F (t )? T (t ) ? N F (0)? T (0) ? constant integrating (1) between 0,t we get, N F (t ) ? N F (0) ? ? N F (0)? aF ? T (0)t ? N F (t ) ? N F (0)[1 ? ? aF ? T (0)t ] b) P ? ER ? fF (t )? T (t ) ?T (t ) ? P ER? fF 1 P 1 ? N F (t ) ER? fF N F (0)[1 ? ? aF ? T (0)t ]