Prevention is better than cure

Prevention is better than cure Prevention is better than cure The current raging debate on healthcare should give us time to ponder over several issues. The need of the hour is to decrease the rapidly escalating cost of health care, expand its coverage and ensure that Medicare and Medicaid are viable future propositions. We should put our faith in the old adage that prevention is better than cure. Primary care should be easily accessible to USs uninsured and underinsured people. People should also be offered incentives in order to bring about much needed change in the ways they take care of their health. The Congress has a major role to play in making prevention the key and basic principal of health care reforms. According to the U.S. Centers for Disease Control and Prevention (CDC) 133 million people in the country are victims of at least one chronic disease. It is estimated that more than 75% of the two trillion dollars allocated for health are being utilized to fight chronic diseases. The figures are alarming. One out of every four Americans may be victims of two or more chronic diseases by the year 2025. The majority of Medicares funds are spent on treating patients suffering from chronic illnesses. It is estimated that Medicaid spends 80% of its funds in treating chronic diseases which are increasing rapidly. The Congressional Budget Office prophesizes that unless drastic action is taken health care expenditure is likely to increase from 17 percent of GDP today to 49 percent in 2082. Federal expenditure on Medicare and Medicaid is expected to shoot up from 4 percent of GDP to a staggering   20 %. The silver lining on this ominous cloud is the fact that most chronic diseases can be prevented. Preventing a few can lead to a multiplier effect. If one were to launch a campaign of national awareness on how to decrease diabetes and obesity it will also lead to lesser number of   people suffering from heart disease and stroke besides reducing medical costs in addition to the trauma and suffering that a patient undergoes. These four chronic diseases alone consume 75% of one trillion dollars. A few skeptics claim that there would be negligible or very little savings from preventive measures. This fear however is unfounded as facts prove otherwise. A study published in the Annals of Internal Medicine in 2005 reveal that a federally sponsored program to prevent diabetes in pre diabetic patients generated remarkable cost savings. Several studies conducted prove conclusively that educating consumers about life style management and health care resulted in cutting down medical services by 20%. Academic health centers such as the Johns Hopkins University and the University of Maryland in Baltimore can play a pivotal role in conducting incentive based prevention programs. This will enable people at high risk to receive treatment at an early stage which can reduce medical costs drastically. A major drive for preventive health care is the need of the hour to keep health care costs under control otherwise courting disaster will be the only answer. Third article JAMA Application of Comparative Utilization Data in Managed care Organizations One of the main problems facing managed care organizations all over the US is data management. One of the key elements contributing to successful running of MCOs relies on the way they utilize data. A multi dimensional approach to data management is necessary for efficient working. One of the main contributing factors is the effective application of comparative utilization data. One of the chief advantages of using comparative utilization data is that it helps in cost cutting measures in the MCOs and thereby helps them to function efficiently. The MCOs can integrate comparative data into their UM process which will ensure better functioning. External benchmarking is of vital importance. It is of primary importance that MCO managers should make comparisons between their internal utilization patterns and accepted benchmarks from other communities. In performing this task MCO managers face problems like conformity in data-collection techniques as well as comparability between the subject and the benchmark data cost In order to get a clearer picture let us deal with the problems. Conformity of data is a challenging task. The healthcare industry tries its best to compile benchmark data. Several surveys are conducted for this purpose. If a survey is conducted regarding Cardiology costs $PMPM, should sub-capitate invasive cardiology be included? Should the EKGs done by internists who are specializing in cardiology be considered? Such questions demonstrate the pitfalls of collecting data through surveys. Data can prove to be of use if it contains claims-level detail. It should also be consistent Comparability is another hurdle which needs to be overcome. Most feel   â€Å"Data from other markets does not apply here in my City, U.S.A.† Many physicians are averse to considering data from other locales. They may go by the logic that they have higher imaging costs in so and so area as they have seven outpatient CT scanners serving 14000 patients. Such logic does not hold water. If the source of comparative actual data is understood, then it will be able to illuminate the MCOs data by comparison. The amount of services required by people residing in a particular locality should be predictable to a large extent though prices may vary. However there are solutions which will ensure efficient functioning if they are followed. Start benchmarking using good data which pass the conformity and comparability tests. The MCOs data should not have any weakness or loopholes. MCO should also learn to handle benchmark data with care. It should not be considered the last word. . It is to be used to â€Å"uncover clues that might lead to utilization management opportunities.† It is necessary to analyze all variations. If the analyst finds abnormal patterns he should find out a reason for it. Sometimes variations occur because of data errors, fee schedule differences random variation, population differences, regional health trends, benefit differences. Efficient benchmarking does not work without the involvement of physicians at every level. Successful intervention can take place only if there is a high level of communication. If MCOs are able to lay their hands on reliable and superior quality benchmarking data it will make them function more efficiently,   and benefit them in the long run. Fourth article Generic Drugs Many people are not aware of what the term generic drugs actually mean. Generic drugs are basically copies of brand name drugs even though it lacks patent protection.   New innovator medicines are marketed under brand names. According to the U.S. Food and Drug Administration (FDA) generic drugs are identical in dose, strength, route of administration, safety, efficacy, and intended use as their brand name counterpart. These are as safe to use as their brand name drug from which they have been copied as the same ingredients are used in their preparation and they have the same effect on our body. The main advantage that generic drugs enjoy over brand name drugs is their cost. They are much cheaper. One of the main reasons behind this is that producers of generic drugs do not have to bear the investment cost like the company developing new drugs. They help companies like Medicare to reduce costs and save patients a lot of money. As these drugs do not enjoy patent protection they face stiff competition, hence they try to keep the price low. Most developing countries are going in for genetic drugs in a big way. Generic drugs can be produced when patents of brand name drugs are near expiration. Companies interested in manufacturing the generic drug can send an application to the FDA. Once approval is granted these drugs can be floated in the market. These drugs also enjoy the fruits of the marketing effort made by the patent drug company. Today generic drugs are widely used. They find their way into half of the prescriptions that qualified physicians write. The facilities used to manufacture both the brand name drug and generic drugs are the same as the two are identical. Generic drugs boast of the same quality as brand name drugs. The FDA prohibits drugs to be manufactured in sub standard facilities. Around 3,500 inspections are conducted by the FDA to ensure that the requisite standards are fulfilled. About half of the brand name firms also produce generic drugs. Many of them even make their own copies. Generic drugs however look different from the brand name drugs they have been copied from. In US trademark laws prohibits the generic drug to look identical to the brand name drug. However even if looks differ the constituents are the same. Colors and flavors are allowed to vary. Contrary to popular opinion all brand name drugs do not possess a generic version. Brand name drugs enjoy patent protection for a period of two decades. Other companies can bring out the generic version only after the patent expires and they gain FDA approval. The use of generic drugs is on the increase. According to IMS health the global sales of generic drugs have shot up from $29 billion in 2003 to $78 billion in 2008

Process Safety And Loss Prevention Plant Engineering Essay

The system in figure 1 schematic of a nomadic incineration unit. The equipment is arranged as a skid mounted bundle, recess and out pipes have been disconnected from unit.for the care purpose unit can be skiding out to open infinite and accessing needed constituents straight, or subsequently taking constituents from the unit in order to derive the entree. All supply and waste connexion are from the unit. Because of cramped conditions. Figure 2 it shows the forepart and side positions of the unit is 2.5m tallness, 5m deep, 2m broad. [ 1 ] Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) The kiln, heat money changer, and scrubber are each secured to border by 6 bolts and there are 4 connexions to each of the motors. The whole unit can be slid out to let care utilizing raising cogwheel and this requires 20 proceedingss to hale out and 40 proceedingss to return. The clip takes to take nuts and bolts 2 proceedingss and the clip takes to replace 5 proceedingss [ 1 ] MTTR ( Average Time To Repair ) is besides known as Mean Corrective Tim – Mct, or TC. is colored norm of the fix times for the system. ( a ) ( I ) Calculation of MTTR when the unit is slid out for fix: Here failure constituents are removed from unit and it will be repaired and replaced to unit. Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) Formula for MTTR: TE† c = [ a?‘ni=1 ( I »i.Tc ( I ) ) ] / a?‘ni=1 ( I »i ) Where: TE† c ( I ) is the disciplinary clip for the i'th unit. I »i is the failure rate of the i'th unit. N is the figure of unit. [ 2 ] Failure informations ( I » ) : Heat exchanger failure rate ( I » ) = 40 ( failure per 10^6hours ) or 40A-10^-6hours [ 3 ] Rotary kiln ( I » ) basic constituents of a rotary kiln are the shell, the furnace lining liner, support tyres, rollers, driven cogwheel and internal heat money changer. So rotary kiln failure rate we may gauge amount of all constituents which are utilizing to do rotary kiln. Under technology premise rotary kiln failure rate ( I » ) = 30 ( failures per 106hours ) or 30A-10-6 hours Under technology premise Scrubbing unit failure rate ( I » ) = 45 ( failures per 106hours ) or 45A-10-6hours Under technology premise fan failure rate ( I » ) = 57 ( failures per 106 hours ) or 57A-10-6 Corrective clip for constituents ( Tc ) : ( Tc ) = Tdet + Tloc + Tpla + Tsel + ( Tpre / Tlog ) + ( [ Trem + Trep ] /Trip ) + Tver + Tstu Tdet = observing mistake Tlo = placement failure Tpla = be aftering the work Ts = select the failed point Tpre = shutdown & A ; readying Tlog = logistics clip Trem = remotion of failed point Trep = replacing of failed point Trip = repair-in-place Tver = verify the repaired point Tstu = re-start [ 4 ] Corrective clip for heat money changer ( Tc ) Heat money changer has four connexions in the unit and heat money changer framed by 6 bolts and nuts so clip to take take that constituent ( heat money changer ) Entire nuts and bolts for the heat money changer in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take heat exchanger = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace heat money changer = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug pipe line the unit line from whole unit Unpluging pipe line from temperature accountant it will take clip = 20 proceedingss Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from rotary kiln it will take clip = 40 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to temperature accountant it will take clip = 25 proceedingss Connecting pipe line to fan it will take clip = 35 proceedingss Connecting pipes line to rotary kiln it will take clip = 45 proceedingss Connecting pipe line to another connexion it will take clip = 30 proceedingss Corrective clip for heat money changer ( Tc ) = 12+30+20+25+40+20+25+35+45+30 =282 proceedingss or 4.7 hours Corrective clip for rotary kiln ( Tc ) Rotary kiln has four connexions connexions in the unit and rotary kiln framed by 6 bolts and nuts so clip to take take that constituent ( rotary kiln ) Entire nuts and bolts for the rotary kiln in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take rotary kiln = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace rotary kiln = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug the unit line from whole unit Unpluging pipe line from prison guard motor it will take clip = 23 proceedingss Unpluging pipe line from heat money changer it will take clip = 30 proceedingss Unpluging pipe line from another connexion it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to sleep together motor it will take clip = 28 proceedingss Connecting pipe line to heat exchanger it will take clip = 35 proceedingss Connecting pipe line to another connexion it will take clip = 25 proceedingss Connecting pipe line to another connexion it will take clip = 40 proceedingss Corrective clip for rotary kiln ( Tc ) = 12+30+23+30+25+20+28+35+25+40 = 268 proceedingss or 4.46 hours Scrubing unit has four connexions in the unit and framed by 6 bolts and nuts so clip to take take that constituent ( scouring unit ) Entire nuts and bolts for the scouring unit in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take scouring unit = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace scouring unit = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug the unit line from whole unit Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 30 Unpluging pipe line from another connexion it will take clip = 35 Unpluging pipe line from another connexion it will take clip = 25 Connecting pipe line to fan it will take clip = 30 proceedingss Connecting pipe line to another connexion it will take clip = 33 Connecting pipe line to another connexion it will take clip = 38 Connecting pipe line to another connexion it will take clip = 30 Corrective clip for scouring unit ( Tc ) = 12+30+25+30+35+25+30+33+38+30= 288 proceedingss or 4.80 hours Fan has besides four connexions with whole unit Unpluging pipe line from heat money changer it will take clip = 25 proceedingss Unpluging pipe line from temperature accountant it will take clip = 30 Unpluging pipe line from scouring unit it will take clip = 33 Unpluging pipe line from fan motor it will take clip = 27 Connecting pipe line to heat exchanger it will take clip = 30 proceedingss Connecting pipe line to temperature accountant it will take clip = 33 Connecting pipe line to scouring unit it will take clip = 38 Connecting pipe line to fan motor it will take clip = 30 Corrective clip for fan unit ( Tc ) = 25+30+33+27+30+33+38+30= 246 proceedingss or 4.10 hours Table 1: Technetium for the when the unit is slid out for fix Component I » ( failures per 106or A-10-6hours ) Tc ( hours ) I » . Tc Heat money changer 40 4.70 188 Rotary kiln 30 4.46 133.8 Scrubing unit 45 4.80 216 Fan 57 4.10 233.7 a?‘I »= 172 a?‘I »Tc= 771.5 Tc = a?‘I »Tc / a?‘I » = 771.5 /172 = 4.48 hours The MTTR ( Average Time To Repair ) when the unit is slid out for fix = 4.48 hours ( a ) ( two ) Calculation of MTTR when the unit is repaired in topographic point: Here we have to cipher MTTR ( Average Time To Repair ) whole unit in topographic point Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) Formula for MTTR: TE† c = [ a?‘ni=1 ( I »i.Tc ( I ) ) ] / a?‘ni=1 ( I »i ) Where: TE† c ( I ) is the disciplinary clip for the i'th unit. I »i is the failure rate of the i'th unit. N is the figure of unit. [ 5 ] Failure informations ( I » ) : Heat exchanger failure rate ( I » ) = 40 ( failure per 10^6hours ) or 40A-10^-6hours [ 6 ] Rotary kiln ( I » ) basic constituents of a rotary kiln are the shell, the furnace lining liner, support tyres, rollers, driven cogwheel and internal heat money changer. So rotary kiln failure rate we may gauge amount of all constituents which are utilizing to do rotary kiln. Under technology premise rotary kiln failure rate ( I » ) = 30 ( failures per 106hours ) or 30A-10-6 hours Under technology premise Scrubbing unit failure rate ( I » ) = 45 ( failures per 106hours ) or 45A-10-6hours Under technology premise fan failure rate ( I » ) = 57 ( failures per 106 hours ) or 57A-10-6 Corrective clip for constituents ( Tc ) : ( Tc ) = Tdet + Tloc + Tpla + Tsel + ( Tpre / Tlog ) + ( [ Trem + Trep ] /Trip ) + Tver + Tstu Tdet = observing mistake Tlo = placement failure Tpla = be aftering the work Ts = select the failed point Tpre = shutdown & A ; readying Tlog = logistics clip Trem = remotion of failed point Trep = replacing of failed point Trip = repair-in-place Tver = verify the repaired point Tstu = re-start [ 7 ] here we do n't necessitate to take constituents from unit for fix Corrective clip for heat money changer ( Tc ) : Heat money changer has four connexion in the whole unit Time taking to unpluging the unit line from whole unit Unpluging pipe line from temperature accountant it will take clip = 20 proceedingss Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from rotary kiln it will take clip = 40 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to temperature accountant it will take clip = 25 proceedingss Connecting pipe line to fan it will take clip = 35 proceedingss Connecting pipes line to rotary kiln it will take clip = 45 proceedingss Connecting pipe line to another connexion it will take clip = 30 proceedingss Corrective clip for heat money changer unit ( Tc ) = 20+25+40+20+25+35+45+30 = 240 minute or 4 hours Corrective clip for rotary kiln ( Tc ) : Unpluging pipe line from prison guard motor it will take clip = 23 proceedingss Unpluging pipe line from heat money changer it will take clip = 30 proceedingss Unpluging pipe line from another connexion it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to sleep together motor it will take clip = 28 proceedingss Connecting pipe line to heat exchanger it will take clip = 35 proceedingss Connecting pipe line to another connexion it will take clip = 25 proceedingss Connecting pipe line to another connexion it will take clip = 40 proceedingss Corrective clip for rotary kiln ( Tc ) = 23+30+25+20+28+35+25+40 = 226 minute or 3.76 hours Corrective clip for scouring unit ( Tc ) : Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 30 Unpluging pipe line from another connexion it will take clip = 35 Unpluging pipe line from another connexion it will take clip = 25 Connecting pipe line to fan it will take clip = 30 proceedingss Connecting pipe line to another connexion it will take clip = 33 Connecting pipe line to another connexion it will take clip = 38 Connecting pipe line to another connexion it will take clip = 30 Corrective clip for scouring unit ( Tc ) = 25+30+35+25+30+33+38+30 = 246 proceedingss or 4.10 hours Corrective clip for fan ( Tc ) : Unpluging pipe line from heat money changer it will take clip = 25 proceedingss Unpluging pipe line from temperature accountant it will take clip = 30 Unpluging pipe line from scouring unit it will take clip = 33 Unpluging pipe line from fan motor it will take clip = 27 Connecting pipe line to heat exchanger it will take clip = 30 proceedingss Connecting pipe line to temperature accountant it will take clip = 33 Connecting pipe line to scouring unit it will take clip = 38 Connecting pipe line to fan motor it will take clip = 30 Corrective clip for fan unit ( Tc ) = 25+30+33+27+30+33+38+30= 246 proceedingss or 4.10 hours So based on computations and observation MTTR ( Mean To Time Repair ) for unit is slid out for fix is significantly more than unit is repaired in topographic point. Table 2: Technetium for the when the unit is repaired in topographic point Component I » ( failures per 106or A-10-6hours ) Tc ( hours ) I » . Tc Heat money changer 40 4.0 160 Rotary kiln 30 3.76 112.8 Scrubing unit 45 4.10 184.5 Fan 57 4.10 233.7 a?‘I »= 172 a?‘I »Tc= 691.0 Tc = a?‘I »Tc / a?‘I » = 691 /172 = 4.01 hours The MTTR ( Average Time To Repair ) when the unit is slid out for fix = 4.01 hours Mentions: ( 1 ) ( a ( I ) ) ( a ( two ) ) [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] [ 4 ] [ 5 ] [ 7 ] Cris Whetton, ility technology. Maintainability. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 3 ] [ 6 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. 1b ) Design alterations to cut down Mean Time To Repair ( MTTR ) : To accomplish optimal MTTR the undermentioned design consideration are recommended: The heat exchanger stuff must be considered based on the operating temperature of the liquid More dependable and maintainable stuff used in the rotary kiln Better we have one more scouring unit to cut down the Mean Time To mend MTTR Motor capacity must designed based on chilling demands All the pipe parametric quantities must be based on the operating temperature of the liquid throwing it Material which is utilizing to do all constituents should be defy all status The temperature accountant must be calibrated for the liquid temperature 1c ) Instrumentality which has system is utile to find the mistakes.so instrumentality in this system temperature accountant ( TC ) : Here TC maps to modulate the temperature of the liquid come ining the heat money changer that is, it pre-controls the liquid come ining the heat money changer. As shown in the figure, the temperature accountant modulate the temperature of the liquid released from the heat money changer and before being cooled by the fan which is control by fan motor. So temperature accountant is utile to observing the mistake which may happen in the heat money changer. Based on the given figure it can be likely assume that degree index may be used for the rotary kiln. a flat index is placed at the top of the rotary kiln. This is used is indicate the maximal degree of the mixture that can be accommodated in a rotary kiln. So this may be indicated the mistakes if anything occur. A flow rate valve is placed in the scrubber unit, so as to command the flow rate alkalic solution into the scouring unit. This flow rate valve allows merely the coveted sum of solution in to the scouring unit. Once the coveted degree is reached the valve will automatically close off the flow of liquid into the unit. And we have some detector dismay at the fan and fan motor and screw motor why because if these have any jobs will gives the signals so we can easy find the mistakes. Due to the incorporation of these instrumentality into the chief system the opportunities of failure is significantly reduced 2 ) Question description: Procedure works to respond liquid A and liquid B to bring forth merchandise C. liquid A passing into storage A utilizing liquid accountant. From storage it will pump to reactor. Liquid B go throughing into storage B utilizing liquid accountant from storage B to pumping to reactor. From reactor merchandise C coming out. Acid gas from reactor pumping to scouring unit. In scouring unit acid gas is cleaned utilizing alkalic solution which is go throughing into scouring unit. Scrubing unit leaves impersonal waste watercourse. Liquids ever available at the recesss to the procedure. There is at least two scouring units working right for the procedure. Stand-by pumps switch over automatically. Pipe work failures can be ignored. [ 1 ] Available informations: The computing machine system has a dependability of 0.9997 over one twelvemonth The operator dependability over one twelvemonth is 0.85 for indicated mistakes and 0.95 for mistakes which raise an dismay Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 Reactor failures can affect the fomenter which has two failure manners. Shaft break failure rate = 0.1/year Motor failure rate = 0.3/year [ 1 ] 2 ( a ( I ) ) Fault tree analysis here merchandise fails to run into specification is the top event Alarm failure Liquid control LAL fails Liquid control Low degree High degree Agitator failure Coking job Motor failure Shaft break High degree Low degree Excess flow of liquid Angstrom Excess flow of liquid B Reactor Pump failure 2 ( a ( two ) ) Fault tree analysis here liquid waste watercourse composing outside bounds is the top event Low degree High degree Internal mal maps failure Connection fails between scrubbers Improper cleansing temperature Improper alkaline solution pumping to scrubber unit Scrubber unit failure Improper flow reactor to scrubber High degree Low degree Low degree High degree 2a ) computation of dependability of parts of the system Here parts of the system: Storages Reactor Agitator Pumps Scrubing unit Dependability of reactor: Here reactor failure can affect the fomenter failure. First one is shaft break and 2nd one is motor failure Failure rate of shaft break = 0.1/year Failure rate of the motor = 0.3/year Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 [ 1 ] Failure rate of pump ( I » ) = 13A-10-6hours [ 2 ] Dependability of shaft break: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 [ 3 ] Failure rate of shaft break = 0.1/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.1/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.1/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 90.11 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.995 So dependability for shaft break = 0.995 Dependability of motor: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 Failure rate of the motor = 0.3/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.3/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.3/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 177.29 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.948 So dependability for motor = 0.948 Dependability for scouring unit: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? Here we have the T = 133.6 yearss Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Z ( T ) = ( 1.8/ ( 600 ) 1.8 ) A- ( 133.6-60 ) 1.8-1 Z ( T ) = 0.2/year Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? = 0.996 So dependability for scouring unit R ( T ) = 0.996 Dependability of pump: Failure rate of pump ( I » ) = 13A-10-6hours Dependability of pump R ( T ) = e-I »t Surviving clip t = 70 yearss One twenty-four hours = 24 hours Surviving clip T = 1680 hours Dependability of pump R ( T ) = e-I »t = vitamin E ( -13A-10^-6A-1680 ) Dependability of pump R ( T ) = 0.978 Mentions: [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. [ 3 ] Cris Whetton, ility technology. Failure information analysis. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. 2b ) Reliability block diagram for the complete system Pump 1 Storage A Pump2 Scrubing unit Reactor Pump Storage B computation of dependability of the complete system over one twelvemonth: Here parts of the system: Storages Reactor Agitator Pumps Scrubing unit Dependability of reactor: Here reactor failure can affect the fomenter failure. First one is shaft break and 2nd one is motor failure Failure rate of shaft break = 0.1/year Failure rate of the motor = 0.3/year Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 [ 1 ] Failure rate of pump ( I » ) = 13A-10-6hours Failure rate of fan ( I » ) = 57A-10-6hours [ 2 ] Dependability of shaft break: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 [ 3 ] Failure rate of shaft break = 0.1/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.1/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.1/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 90.11 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.995 So dependability for shaft break = 0.995 Dependability of motor: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 Failure rate of the motor = 0.3/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.3/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.3/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 177.29 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.948 So dependability for motor = 0.948 Dependability for scouring unit: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? Here we have the T = 133.6 yearss Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Z ( T ) = ( 1.8/ ( 600 ) 1.8 ) A- ( 133.6-60 ) 1.8-1 Z ( T ) = 0.2/year Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? = 0.996 So dependability for scouring unit R ( T ) = 0.996 Dependability of pump: Failure rate of pump ( I » ) = 13A-10-6hours Dependability of pump R ( T ) = e-I »t Surviving clip t = 70 yearss One twenty-four hours = 24 hours Surviving clip T = 1680 hours Dependability of pump R ( T ) = e-I »t = vitamin E ( -13A-10^-6A-1680 ) Dependability of pump R ( T ) = 0.978 Dependability of the complete system over twelvemonth R ( T ) = norm of system parts dependability = ( 0.995+0.948+0.996+0.978 ) /4 = 0.979 Therefore dependability of the complete system over twelvemonth = 0.979 Mentions: [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. [ 3 ] Cris Whetton, ility technology. Failure information analysis. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. 2c ) To accomplish a mark dependability of 0.90 over one twelvemonth: Reliability mark is a nothing failure mark. This is an of import mark implied for those low acting workss, such workss does non accomplish certain ends designed by applied scientists. So we have to put appropriate mark to accomplish works design. the dependability of the system must be improved to accomplish the mark. to accomplish the dependability mark or to better dependability three basic ways must be employed. By system design By component specification By preventative care By system design: – The basic regulation of our system design is to maintain the design has simple as possible. the system is more dependable if the system is simple. Some of the stairss include, System simplification: To cut down the complexnesss in procedure works at the design phase its ego Decrease in the usage of Complex parts by replacing them with more cardinal parts The design should be made simple and easy to under base Decrease in constituent count: The figure of constituents used in the works must be reduced. complex constituents must be avoided for the simpleness of the design. Mistake tolerance: The basic features of mistake tolerance require: No individual point of failure No individual point of repair- the system must run without any break during the procedure of fix when the system experiences any jobs. Mistake isolation to the neglecting component- in instance of failures the failed portion of the system must be isolated from the pained system. This requires necessary failure sensing mechanism. Fault containment to forestall extension of the failure Handiness of reversion modes- some failures may do cripples to the full system, to avoid the full procedure system must force to the safe manner By component specification: For the dependability of a constituent it must be adequately specified for their full length of service. Extra dependability can be provided by runing the constituents at lower emphasis so their operating emphasiss. By making so early failures of the constituents can be reduced. in a procedure industry it is really hard to better dependability merely by specification. This is attributed to the deficit of necessary informations sing the affect of emphasiss on the constituents. Components of high quality can non be used ever for economic grounds. Normally the parametric quantities required to better dependability frequently contradict with procedure demands. Some of the dependability betterments include: Use of disciplinary maintenance- it is defined as the care which is required to mend and convey merchandise after the fix is carried out. it is carried out in constituents who is failures does n't impact of the overall working of the procedure system significantly. This activity chiefly involves fix, Restoration or replacing of constituents. Design improvement-the design of any high quality procedure works is based on the design parametric quantities and proficient specifications. the reactor design must be improved for high rates of efficiency. Temperature, force per unit area and other external considerations must be included in the design of reactor and storage armored combat vehicles. Quality control-Quality control assures conformity to specifications. quality control checks whether measurings of the constituents like reactors, storage armored combat vehicle, scrub units as in this instance conform to the demands. Preventive care: Is defined as a care carried out to forestall failure or warring out of constituents in the procedure works. This is carried out by supplying systematic review, sensing and bar of inchoate failure. The preventive care attempts are aimed at continuing the utile life of equipment and avoiding premature equipment failures, minimising any impact on operational demands. In add-on to the everyday facets of cleansing, adjusting, lubricating and proving. it is carried out merely on those points where a failure would hold expensive or unacceptable effects e.g. reactors, storage armored combat vehicles, scouring units. Many of these points are besides capable to a statutory demand for review and preventative care. [ 1 ]

A Monster Cannot Be Created Essay

Do Humans have freewill or are they products of their environment? Everyone is different, everyone lives a different life, which factor cause a person to be the way they are nature or nurture? In Mary Shelly’s Frankenstein human nature causes Victor Frankenstein to become the real monster of the novel.  First of all let’s start out by analyzing Victor’s childhood. Victor came from a renowned family in Geneva. â€Å"I am by birth a Genevese, and my family is one of the most distinguished of that republic† (page 17). Also along with this he was born into a family full of love, in fact he was given almost all of the attention and love from his parents. â€Å"Much as they were attached to each other, they seemed to draw inexhaustible stores of affection from a very mine of love to bestow them upon me.† (page 19). So Nurture can not be held responsible for Victor’s monstrous persona, it must be because of nature. This shows that people have a predetermined personality, but this also shows certain parts of person’s personality can be reveled and or amplified due to changes in their environment. An example of this would be Alphonse Frankenstein’s disapproval of Victor’s scientific interests. Consequently forcing Victor into solitude, and revealing his obsessive tendencies. This is because Victor feels the need to measure the onward race for achievement against the yardstick of benefit to the human community. As a result of this human nature causes Victor to become a monster. Various factors influenced the development of Victor Frankenstein. For example, Victors unsupervised reading and education. This illustrates Victor’s natural obsession for knowledge. This shows the value of the human fulfillment of the pursuit of knowledge, and the importance of individual creative effort. Another example would be the immoral desertion of his creation. This shows that Victor cares little about anyone else’s interests and wellbeing. This is because Victor is extremely narcissistic and fails to assume responsibility for his relationship with others. Another case in point is the actual creation of a living human being. This shows that Victor has an innate passion for pushing the envelope of modern sciences. This helps illustrate the value of the need to measure the onward race for achievement against the yardstick of benefit to the human community. As a result Victor has a fascination for bending the laws of human nature. It seems a part of Victor’s personality goes into his creation. If you look closely at the novel you can see that there is a direct link between Victor Frankenstein and his monster, almost as if the two are doubles. For example, all of the murders Victor’s creation commits, could it be that Victor actually wants these people to be killed? Could Victor’s creation actually be another other side of him, much like a Dr. Jekyll and Mr. Hyde? â€Å"I considered the being whom I had cast among mankind, and endowed with the will and power to effect purposes of horror, such as the deed which he had now done, nearly in the light of my own vampire, my own spirit let loose from the grave, and forced to destroy all that was dear to me.† (page 61) Look at people that were murdered, Henry Clerval, Victor’s closest friend since childhood. He is Victors intellectual opposite, and quite possibly his rival for success. William, the more charming and spirited sibling, â€Å"when he smiles [William] two little dimples appear on each cheek, which are rosy with health. He has already had one or two little wives†¦Ã¢â‚¬  (page 56).  And what else could explain Victor’s judgment for not protecting Elizabeth from the monster. Victor is entirely aware that the monster will only kill the people close to him, and he is clearly warned that the monster will be with him on his wedding night, â€Å"It is well. I go; but remember, I shall be with you on your wedding-night.† (page 153 ). Does he tell Elizabeth, does he tell anyone what he knows? Absolutely not. The only time he tells anyone about his monster and its correlation with the murders is after everyone is already murdered. Coincidence? Some might argue that Victor’s creation is the real monster in the story because of free will. This is a valid argument but not necessarily true. Yes, it’s true that the monster had free will and that it was his decision to murder those people, but you have to understand the position the monster was in. He was abandoned by Victor as soon as he was created, and he was looked at as a hideous beast by anyone who laid eyes upon him. He was not given an equal opportunity to find acceptance among people because of his appearance, not because of his actions. An example of this is when he tries to save a young girl from drowning. He rescues the young girl, but as soon as someone else sees what is happening he automatically assumes that he is trying to murder the young girl. â€Å"The whole village was mused; some fled, some attacked me, until, grievously bruised by stones and many other kinds of missile weapons, I escaped to the open country, and fearfully took refuge in a low hovel, quite bare, and making a wretched appearance after the palaces I had beheld in the village.† (page 91). What people forget is that the monster was created by Victor which is not a natural occurrence; therefore the rules of nature do not apply to him. The monster is the way he is because of how Victor created him. The only difference between the two is that the creation was not accepted in his environment, thus bringing out a hidden persona.  A monster can not just be created; a monster comes from within the creator, and in the end Victor’s creation becomes his death. â€Å"Blasted as thou wert, my agony was still superior to thine; for the bitter sting of remorse will not cease to rankle in my wounds until death shall close them for ever.† (page 205).

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