Globalization And Workers Developing Countries - 2349 Words

Globalization and Workers in Developing Countries Table of Content Introduction 2 Part 1: Does globalization harm workers’ interests in developing countries? 3 Condition of Sweatshop Workers in Developing Countries 3 Average Wages and Hours 4 Job Destruction and Job Creation 4 Unemployment 5 Child Labour 5 Working Conditions, safety and security issues 5 Benefits of Globalization in the Labour Market 6 Part 2: Suggestions for overcoming sweatshops in developing countries: 6 Trade Policy Reforms 7 Capital Market Reforms 7 Labour Market Reforms 7 The Role of International Community 7 Conclusion 8 Reference 8 Introduction Globalization has become an international dynamics of today’s fast growing economy around the world. Multinational Companies (MNCs) get benefits which is important for the economic development because it is the international integration of world trade, and resources, labour, and technology of financial market (IMF, 2014). The labour market is the key distribution network to influence the developing counties through globalization (Rama, 2003). In the first part of the paper, to answer the question â€Å"does globalization harm workers’ interests?†, the form of globalization as average wages and hours, job destruction and creation, unemployment, child labour, working conditions, safety and security issues are discussed to understand the condition of sweatshop workers in developing countries; and benefits of globalization in the labour marketShow MoreRelatedImpact Of Globalization On The Social Consciousness Of The Late 20th Century Essay1667 Words   |  7 PagesI ntroduction Globalization is a term that profoundly characterized the social consciousness of the late 20th century, and still continues to shape 21st century social discourse. But what exactly does globalization imply? Unfortunately, despite being referred to almost incessantly, the true meaning of globalization is often obscure (Reich, 1998). 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Process Safety And Loss Prevention Plant Engineering Essay Free Essays

string(1087) " 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\." 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. We will write a custom essay sample on Process Safety And Loss Prevention Plant Engineering Essay or any similar topic only for you Order Now 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 ] How to cite Process Safety And Loss Prevention Plant Engineering Essay, Essay examples

Brisbane Power Centre

Questions: 1. What are the current purchasing and inventory management processes being used by the company?2. What are the comparative advantages and disadvantages of the currently used processes?3.What supply-chain and inventory management concepts would help the company increase efficiency and reduce investment whilst maintaining adequate stock levels?4. What recommendations would you make to Ms Green with respect to restructuring the purchasing and inventory functions of the company? Answers: Introduction Brisbane outdoor Power Centre is started with 1 store now has 3 retail stores. All the stores are profitable and operate quite independently in managing their inventory and purchasing systems. As all the stores are in profits, Management did not care about the details of the operations of each store and how they are handling various operations and all the stores are operated in fairy loose and decentralized fashion with no common policies or rules. As the founding member decided to retire, branch manager of one of the store, Ms Green, overtook this business. She was not convinced with the existing way of things and wanted to make some improvements. The objective of this report is analyze the current system and sport the inefficiencies and thereby suggest or recommends approaches to improve the current system and rationalize the procurement and inventory processes of all the 3 stores. 1. Brisbane Outdoor Power Centre as evolved from a single retail lawnmower business to a full-fledged thriving group of 3 retail outlets that supplies possibly all the equipment used in gardening. Donald who is the founder of this business manages 1 store initially and then acquired the other 2 stores which are both running in profits. As the other groups are running in profits, Donald didnt interfere much in the operations of the other 2 stores and manages his own store at Mt. Gravatt. Present state is that all the 3 stores are operated independently manages their own sourcing and inventory. Each retail stores are having their own set of suppliers from which they generally purchases. As a result, it may be the case in case of the same suppliers it is approached by the 3 different purchasing departments of each of the retail store for different requirements and orders. This also means that Brisbane outdoor is making much more additional efforts in negotiation with suppliers than what is required if they could have negotiate with the suppliers collectively (Leuschner, 2013). Not only this, it may be possible that same supplier is supplying same materiel at different prices to different retail stores of Brisbane Outdoor power Centre. Thus, there could be more potential of saving the cost, which is not actually been realized. Also, Donalds style of non-interference as long as the stores are making profits have given rises to some odd sourcing and inventory management practices. Currently, different retail stores stocked products of different brands at different retail stores. This will lead to different experience for customer when they visit different stores. In case of shortages of any particular item, they will be transported from different retail stores but there was no integrated approach. The company operated in a complete decentralized fashion with each retail store stocking products of its preferable brand, buys from supplier of its own choice and kept inventory and safety stock based on its own decision and having goals and values of its own. Though this approach worked very fine earlier but as the Brisbane outdoor power is growing at a rapid rate, it becomes important for the organization to make appropriate changes in its production and inventory management system to explore further potential of cost savings and provide customers uniform experience across all of its retail stores. The inventory management system at the company was not good. The company was using the old and traditional methods of inventory management. There was not any integration between the inventory management of the three stores and each store used to have its own stocks and inventory. The three inventory managers used to have their own calculations about the stock in hand and the order to be placed. This was not a good inventory management practiced as it lead to completely different, independent and exclusive inventory management systems of three stores. 2. Advantages of Current Process Each retail store has full independent to manage its operation and hence are very agile in response as there are no approvals required for its decision making (Knoppen, Saenz, 2015). Another advantage of having decentralized purchasing and inventory management is that each store is having information about the upcoming events in their area and aware of the local factors and thus are stocking more products based on their own requirements (Chung, Talluri, Narasimhan, 2015). For instance, Ipswich store is having more range of ride-on mowers and other gardening equipment that are required for maintaining large properties and hobby farms. As each retail store is stocking products of its own chosen brands, it will help the store to customers in an effective way. Each store can independently decide which brands are selling more and liked more by its consumers and can stock the same leading to higher customer satisfaction. More control with the local management lead to more sense of ownership, which is very much required to run the retail store effectively. Also, local management is much more sensitive to the demands of its local customers and hence can be able to connect with the customer in a better way (Stadtler, 2015). Customers will feel more connected with the store. Disadvantages of Current Process There are many disadvantages of the current purchasing and inventory management systems. Few of them are as follows: One of the downside of the existing system is that each retail store is dealing with the same suppliers individually which means that it is possible that different stores have negotiated different rates with same suppliers (Rushton, Croucher, Baker, 2014). All this leads to 2 things, first there is additional effort organization is spending on buying. Second, there is potential of saving that is lost because each store is buying at different rates Each retail store is creating separate purchase order a supplier and thus unable to take the benefits of bulk discounts if all the requirements of different stores can be combines (Chopra, Sodhi, 2004). Another scenario here is that it is possible that one retail store is getting products at a very cheap price from the supplier based on his relationship while another retail is getting same materiel from different suppliers (Konur, Geunes, 2016).Different stores are having same products of different brands. This will lead to different customer experience in different stores (Jorgensen, Kort, 2002). There are incidents of over and under purchasing of materials. Also, each store is maintaining its own safety stock which lead to the higher inventory levels for the while organization. Same capital can be divested from inventory and can be invested to reap better returns (Yano, Gilbert, 2005). There will no opportunity of discounts on bulk purchases. 3. Safety Stock Reduction Currently, all the retails stores are maintaining their own inventory and safety stock. Thus, there is high level of inventory it is maintaining and there is huge scope of reduction. Brisbane Outdoor Power should maintain all the safety stock at only one of its retail stores whichever is having large area and enough space (Wisner, 2015). All the stores should be replenished from the safety stock whenever required. This will help in reducing the capital tied up in the inventory, which can be invested or used at some other place to earn greater returns. Exploit Economies of Scale Brisbane outdoor power should exploit economies of scale. Management should collect the data from the different retail stores and analyze it and identifies the different scenarios like below and the action plan for each of the scenario: Scenario 1: Different retail stores are buying from same supplier. In such cases, lowest prices paid by the store should be noted and further negotiated for bulk discounts while ordering the material for all the 3 stores. Scenario 2: Different retail stores are buying same material from different suppliers. In such cases, it should be analyzed which store is paying the least amount to the supplier, the quality of the items supplied by different suppliers, their lead time and appropriate supplier should be selected after prioritizing the needs between price, quality and delivery. In this case, also discounts can be availed on ordering bulk quantity for all the stores (Fearne, Hughes, 2013). Scenario 3: Different retail stores are buying different materials from different suppliers. In all such cases, best supplier should be identified for all the products considering various options. Using EOQ (Economic Order Quantity) for Items with Fairly Constant Demand and Very less Lead Time It can be defined the economic order quantity so as to minimize the ordering and holding costs of the inventory. If the organization orders fewer inventories, less holding costs will be incurred however it has to order quite frequently and every time incurred a ordering costs. At the same time, if it will order too much quantity, it needed space and other associated inventory holding costs. Thus, EOQ is a quantity, which is defined by minimizing the holding and ordering costs (LEv, 2013). Reorder Point Strategy for Items with High Lead Time In this strategy, automatic order is places when the quantity reaches the certain level. That certain level is defined by the safety stock and lead-time. An order will be placed when the quantity is left enough to be consumed during the lead-time and reaches threshold safety stock. According to this strategy, inventory will be replenished as soon as the level just reaches the safety stock level (Rossin, 2012). Therefore, this approach will allow storing the safety inventory but not allowing the accumulation of excess inventory in the system. 4. There are many things that can be done by Ms Green to restructure the purchasing inventory functions of the company. The company should centralize this function. It can purchase the software that can be installed in all the 3 factories and maintains the inventory levels of each of the store. The same software can be used to purchase the material from the same set of suppliers but with different ship to and bill to locations. Therefore, the purchase order will be one for all the material and but material will be delivered at different stores. This helps the organization to take advantage of discounts using bulk quantity (Sridharan, 2013). If some items are in demand in particular store, then it can be delivered at same store. The best approach here is to keep the safety stock inventory at 1 store only and not at all the stores. And regular supplies should me order by common Purchase order at all stores. There are much Software like Xero, QuickBooks online and MYOB that offers software services on the cloud and charges based on the usage basis (Ma, 2015). At this time, there is definitely requirement of software to manage the operations effectively and this software provides value of money. Brisbane power does not have to invest in infrastructure and consulting fees to use this software on cloud. Once it grows and expands, it considers to use the full fledge ERP systems like Microsoft dynamics. Conclusion To conclude, Brisbane outdoor power centre new management has spotted very valid observations in the current system that all the stores are having very unconnected purchasing and inventory systems and operates quite independently. It could be possible that all the 3 stores are negotiating separately with same suppliers and paying different rates. This report suggested several measures to improve the system like centralizing the purchasing function using software. This will increase the visibility of the material at all the stores as well as help organization to save money on bulk orders. References Chopra, S., Sodhi, M. S. (2004). Managing risk to avoid supply-chain breakdown.MIT Sloan management review,46(1), 53. Chung, W., Talluri, S., Narasimhan, R. (2015). Optimal pricing and inventory strategies with multiple price markdowns over time.European Journal of Operational Research,243(1), 130-141. Fearne, A., Hughes, D. (2013). 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Push-Pull Boundary Location, Information Quality, and Supply Chain Performance: An Exploratory Analysis.Journal of Global Business Issues,6(1), 7. Rushton, A., Croucher, P., Baker, P. (2014).The handbook of logistics and distribution management: Understanding the supply chain. Kogan Page Publishers. Sridharan, R., Simatupang, T. M. (2013). Power and trust in supply chain collaboration.International Journal of Value Chain Management,7(1), 76-96. Stadtler, H. (2015). Supply chain management: An overview. InSupply chain management and advanced planning(pp. 3-28). Springer Berlin Heidelberg. Wisner, J., Tan, K. C., Leong, G. (2015).Principles of supply chain management: a balanced approach. Cengage Learning. Yano, C. A., Gilbert, S. M. (2005). Coordinated pricing and production/procurement decisions: A review. InManaging Business Interfaces(pp. 65-103). Springer US.