Process Safety And Loss Prevention Plant Engineering Essay

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The system in figure 1 schematic of a mobile incineration unit. The equipment is arranged as a skid mounted package, inlet and out pipes have been disconnected from unit .for the maintenance purpose unit can be sliding out to open space and accessing required components directly ,or later removing components from the unit in order to gain the access. All supply and waste connection are from the unit. Because of cramped conditions. Figure 2 it shows the front and side views of the unit is 2.5m height, 5m deep, 2m wide.[1]

Components:

Heat exchanger (EX)

Rotary kiln (RK)

Scrubbing unit (SC)

Temperature controller (TC)

Fan motor (FM)

Screw feeder (SW)

Screw motor (SM)

Feed hopper (FH)

The kiln, heat exchanger, and scrubber are each secured to frame by 6 bolts and there are 4 connections to each of the motors. The whole unit can be slid out to allow maintenance using lifting gear and this requires 20 minutes to haul out and 40 minutes to return. The time takes to remove nuts and bolts 2 minutes and the time takes to replace 5 minutes [1]

MTTR (Mean Time To Repair) is also known as Mean Corrective Tim - Mct , or TC. is biased average of the repair times for the system.

(a)(i)

Calculation of MTTR when the unit is slid out for repair:

Here failure components are removed from unit and it will be repaired and replaced to unit.

Components:

Heat exchanger (EX)

Rotary kiln (RK)

Scrubbing unit (SC)

Temperature controller (TC)

Fan motor (FM)

Screw feeder (SW)

Screw motor (SM)

Feed hopper (FH)

Formula for MTTR:

Tˆc = [ ∑ni=1(λi .Tc(i))]/ ∑ni=1(λi)

Where:

Tˆc(i) is the corrective time for the i'th unit.

λi is the failure rate of the i'th unit.

n is the number of unit.[2]

Failure data (λ):

Heat exchanger failure rate (λ) = 40(failure per 10^6hours) or 40Ã-10^-6hours[3]

Rotary kiln (λ) basic components of a rotary kiln are the shell, the refractory lining, support tires, rollers, driven gear and internal heat exchanger. So rotary kiln failure rate we may estimate sum of all components which are using to make rotary kiln.

Under engineering assumption rotary kiln failure rate (λ) = 30 (failures per 106hours) or 30Ã-10-6 hours

Under engineering assumption Scrubbing unit failure rate (λ) = 45 (failures per 106hours) or 45Ã-10-6hours

Under engineering assumption fan failure rate (λ) = 57(failures per 106 hours) or 57Ã-10-6

Corrective time for components (Tc):

(Tc) = Tdet + Tloc + Tpla + Tsel + (Tpre / Tlog) + ([Trem + Trep]/Trip) + Tver + Tstu

Tdet = detecting fault

Tlo = locating failure

Tpla = planning the work

Ts = select the failed item

Tpre = shutdown & preparation

Tlog = logistics time

Trem = removal of failed item

Trep = replacement of failed item

Trip = repair-in-place

Tver = verify the repaired item

Tstu = re-start [4]

Corrective time for heat exchanger (Tc)

Heat exchanger has four connections in the unit and heat exchanger framed by 6 bolts and nuts so time to take remove that component (heat exchanger)

Total nuts and bolts for the heat exchanger in the unit = 6

Time taking to remove bolts and nuts at each connection = 2 minutes

So time taking to remove heat exchanger = 6Ã-2 = 12 minutes

Time taking to replace bolts and nuts at each connection = 5 minutes

Time taking to replace heat exchanger = 6Ã-5 = 30 minutes

And we have to disconnect the connections here we have total 4 connection

Time taking to disconnect pipe line the unit line from whole unit

Disconnecting pipe line from temperature controller it will take time = 20 minutes

Disconnecting pipe line from fan it will take time = 25 minutes

Disconnecting pipe line from rotary kiln it will take time = 40 minutes

Disconnecting pipe line from another connection it will take time = 20 minutes

Connecting pipe line to temperature controller it will take time = 25 minutes

Connecting pipe line to fan it will take time = 35 minutes

Connecting pipes line to rotary kiln it will take time = 45 minutes

Connecting pipe line to another connection it will take time = 30 minutes

Corrective time for heat exchanger (Tc) = 12+30+20+25+40+20+25+35+45+30 =282 minutes or 4.7 hours

Corrective time for rotary kiln (Tc)

Rotary kiln has four connections connections in the unit and rotary kiln framed by 6 bolts and nuts so time to take remove that component (rotary kiln)

Total nuts and bolts for the rotary kiln in the unit = 6

Time taking to remove bolts and nuts at each connection = 2 minutes

So time taking to remove rotary kiln = 6Ã-2 = 12 minutes

Time taking to replace bolts and nuts at each connection = 5 minutes

Time taking to replace rotary kiln = 6Ã-5 = 30 minutes

And we have to disconnect the connections here we have total 4 connection

Time taking to disconnect the unit line from whole unit

Disconnecting pipe line from screw motor it will take time = 23 minutes

Disconnecting pipe line from heat exchanger it will take time = 30 minutes

Disconnecting pipe line from another connection it will take time = 25 minutes

Disconnecting pipe line from another connection it will take time = 20 minutes

Connecting pipe line to screw motor it will take time = 28 minutes

Connecting pipe line to heat exchanger it will take time = 35 minutes

Connecting pipe line to another connection it will take time = 25 minutes

Connecting pipe line to another connection it will take time = 40 minutes

Corrective time for rotary kiln (Tc) = 12+30+23+30+25+20+28+35+25+40 = 268 minutes or 4.46 hours

Scrubbing unit has four connections in the unit and framed by 6 bolts and nuts so time to take remove that component (scrubbing unit)

Total nuts and bolts for the scrubbing unit in the unit = 6

Time taking to remove bolts and nuts at each connection = 2 minutes

So time taking to remove scrubbing unit = 6Ã-2 = 12 minutes

Time taking to replace bolts and nuts at each connection = 5 minutes

Time taking to replace scrubbing unit = 6Ã-5 = 30 minutes

And we have to disconnect the connections here we have total 4 connection

Time taking to disconnect the unit line from whole unit

Disconnecting pipe line from fan it will take time = 25 minutes

Disconnecting pipe line from another connection it will take time = 30

Disconnecting pipe line from another connection it will take time = 35

Disconnecting pipe line from another connection it will take time = 25

Connecting pipe line to fan it will take time = 30 minutes

Connecting pipe line to another connection it will take time = 33

Connecting pipe line to another connection it will take time = 38

Connecting pipe line to another connection it will take time = 30

Corrective time for scrubbing unit (Tc) = 12+30+25+30+35+25+30+33+38+30= 288 minutes or 4.80 hours

Fan has also four connections with whole unit

Disconnecting pipe line from heat exchanger it will take time = 25 minutes

Disconnecting pipe line from temperature controller it will take time = 30

Disconnecting pipe line from scrubbing unit it will take time = 33

Disconnecting pipe line from fan motor it will take time = 27

Connecting pipe line to heat exchanger it will take time = 30 minutes

Connecting pipe line to temperature controller it will take time = 33

Connecting pipe line to scrubbing unit it will take time = 38

Connecting pipe line to fan motor it will take time = 30

Corrective time for fan unit (Tc) = 25+30+33+27+30+33+38+30= 246 minutes or 4.10 hours

Table 1: Tc for the when the unit is slid out for repair

Component

λ (failures per 106or Ã-10-6hours)

Tc (hours)

λ . Tc

Heat exchanger

40

4.70

188

Rotary kiln

30

4.46

133.8

Scrubbing unit

45

4.80

216

Fan

57

4.10

233.7

∑λ=

172

∑λTc=

771.5

Tc = ∑λTc / ∑λ = 771.5 /172 = 4.48 hours

The MTTR (Mean Time To Repair) when the unit is slid out for repair = 4.48 hours

(a)(ii)

Calculation of MTTR when the unit is repaired in place:

Here we have to calculate MTTR (Mean Time To Repair) whole unit in place

Components:

Heat exchanger (EX)

Rotary kiln (RK)

Scrubbing unit (SC)

Temperature controller (TC)

Fan motor (FM)

Screw feeder (SW)

Screw motor (SM)

Feed hopper (FH)

Formula for MTTR:

Tˆc = [ ∑ni=1(λi .Tc(i))]/ ∑ni=1(λi)

Where:

Tˆc(i) is the corrective time for the i'th unit.

λi is the failure rate of the i'th unit.

n is the number of unit.[5]

Failure data (λ):

Heat exchanger failure rate (λ) = 40(failure per 10^6hours) or 40Ã-10^-6hours[6]

Rotary kiln (λ) basic components of a rotary kiln are the shell, the refractory lining, support tires, rollers, driven gear and internal heat exchanger. So rotary kiln failure rate we may estimate sum of all components which are using to make rotary kiln.

Under engineering assumption rotary kiln failure rate (λ) = 30 (failures per 106hours) or 30Ã-10-6 hours

Under engineering assumption Scrubbing unit failure rate (λ) = 45 (failures per 106hours) or 45Ã-10-6hours

Under engineering assumption fan failure rate (λ) = 57(failures per 106 hours) or 57Ã-10-6

Corrective time for components (Tc):

(Tc) = Tdet + Tloc + Tpla + Tsel + (Tpre / Tlog) + ([Trem + Trep]/Trip) + Tver + Tstu

Tdet = detecting fault

Tlo = locating failure

Tpla = planning the work

Ts = select the failed item

Tpre = shutdown & preparation

Tlog = logistics time

Trem = removal of failed item

Trep = replacement of failed item

Trip = repair-in-place

Tver = verify the repaired item

Tstu = re-start[7]

here we don't need to remove components from unit for repair

Corrective time for heat exchanger (Tc):

Heat exchanger has four connection in the whole unit

Time taking to disconnecting the unit line from whole unit

Disconnecting pipe line from temperature controller it will take time = 20 minutes

Disconnecting pipe line from fan it will take time = 25 minutes

Disconnecting pipe line from rotary kiln it will take time = 40 minutes

Disconnecting pipe line from another connection it will take time = 20 minutes

Connecting pipe line to temperature controller it will take time = 25 minutes

Connecting pipe line to fan it will take time = 35 minutes

Connecting pipes line to rotary kiln it will take time = 45 minutes

Connecting pipe line to another connection it will take time = 30 minutes

Corrective time for heat exchanger unit (Tc) = 20+25+40+20+25+35+45+30 = 240 minute or 4 hours

Corrective time for rotary kiln (Tc):

Disconnecting pipe line from screw motor it will take time = 23 minutes

Disconnecting pipe line from heat exchanger it will take time = 30 minutes

Disconnecting pipe line from another connection it will take time = 25 minutes

Disconnecting pipe line from another connection it will take time = 20 minutes

Connecting pipe line to screw motor it will take time = 28 minutes

Connecting pipe line to heat exchanger it will take time = 35 minutes

Connecting pipe line to another connection it will take time = 25 minutes

Connecting pipe line to another connection it will take time = 40 minutes

Corrective time for rotary kiln (Tc) = 23+30+25+20+28+35+25+40 = 226 minute or 3.76 hours

Corrective time for scrubbing unit (Tc):

Disconnecting pipe line from fan it will take time = 25 minutes

Disconnecting pipe line from another connection it will take time = 30

Disconnecting pipe line from another connection it will take time = 35

Disconnecting pipe line from another connection it will take time = 25

Connecting pipe line to fan it will take time = 30 minutes

Connecting pipe line to another connection it will take time = 33

Connecting pipe line to another connection it will take time = 38

Connecting pipe line to another connection it will take time = 30

Corrective time for scrubbing unit (Tc) = 25+30+35+25+30+33+38+30 = 246 minutes or 4.10 hours

Corrective time for fan (Tc):

Disconnecting pipe line from heat exchanger it will take time = 25 minutes

Disconnecting pipe line from temperature controller it will take time = 30

Disconnecting pipe line from scrubbing unit it will take time = 33

Disconnecting pipe line from fan motor it will take time = 27

Connecting pipe line to heat exchanger it will take time = 30 minutes

Connecting pipe line to temperature controller it will take time = 33

Connecting pipe line to scrubbing unit it will take time = 38

Connecting pipe line to fan motor it will take time = 30

Corrective time for fan unit (Tc) = 25+30+33+27+30+33+38+30= 246 minutes or 4.10 hours

So based on calculations and observation MTTR (Mean To Time Repair) for unit is slid out for repair is significantly more than unit is repaired in place.

Table 2: Tc for the when the unit is repaired in place

Component

λ (failures per 106or Ã-10-6hours)

Tc (hours)

λ . Tc

Heat exchanger

40

4.0

160

Rotary kiln

30

3.76

112.8

Scrubbing unit

45

4.10

184.5

Fan

57

4.10

233.7

∑λ=

172

∑λTc=

691.0

Tc = ∑λTc / ∑λ = 691 /172 = 4.01 hours

The MTTR (Mean Time To Repair) when the unit is slid out for repair = 4.01 hours

References:(1)(a(i))(a(ii))

[1] Plant reliability and maintainability, assignment question paper, module (CPE6250) held on November 30 to December 3 2009.

[2][4][5][7] Cris Whetton, ility engineering. Maintainability. [Lecture handout].from plant reliability and maintainability, module (CPE6250) held on November 30 to December 3 2009.

[3][6] Frank P. Lees, 1996, Loss prevention in the process industries, second edition, volume 3.

1b)

Design changes to reduce Mean Time To Repair (MTTR):

To achieve optimum MTTR the following design consideration are recommended:

The heat exchanger material must be considered based on the operating temperature of the liquid

More reliable and maintainable material used in the rotary kiln

Better we have one more scrubbing unit to reduced the Mean Time To repair MTTR

Motor capacity must designed based on cooling requirements

All the pipe parameters must be based on the operating temperature of the liquid throwing it

Material which is using to make all components should be withstand all condition

The temperature controller must be calibrated for the liquid temperature

1c)

Instrumentation which has system is useful to determine the faults .so instrumentation in this system temperature controller (TC): Here TC functions to regulate the temperature of the liquid entering the heat exchanger that is, it pre-controls the liquid entering the heat exchanger. As shown in the figure, the temperature controller regulate the temperature of the liquid released from the heat exchanger and before being cooled by the fan which is control by fan motor. So temperature controller is useful to detecting the fault which may occur in the heat exchanger.

Based on the given figure it can be probably assume that level indicator may be used for the rotary kiln. a level indicator is placed at the top of the rotary kiln. This is used is indicate the maximum level of the mixture that can be accommodated in a rotary kiln. So this may be indicated the faults if anything occur. A flow rate valve is placed in the scrubber unit, so as to control the flow rate alkaline solution into the scrubbing unit. This flow rate valve allows only the desired amount of solution in to the scrubbing unit. Once the desired level is reached the valve will automatically shut off the flow of liquid into the unit. And we have some sensor alarm at the fan and fan motor and screw motor why because if these have any problems will gives the signals so we can easily determine the faults.

Due to the incorporation of these instrumentation into the main system the chances of failure is significantly reduced

2)

Question description:

Process plant to react liquid A and liquid B to produce product C. liquid A passing into storage A using liquid controller. From storage it will pump to reactor. Liquid B passing into storage B using liquid controller from storage B to pumping to reactor. From reactor product C coming out. Acid gas from reactor pumping to scrubbing unit. In scrubbing unit acid gas is cleaned using alkaline solution which is passing into scrubbing unit. Scrubbing unit leaves neutral waste stream. Liquids always available at the inlets to the process. There is at least two scrubbing units working correctly for the process. Stand-by pumps switch over automatically. Pipe work failures can be ignored.[1]

Available data:

The computer system has a reliability of 0.9997 over one year

The operator reliability over one year is 0.85 for indicated faults and 0.95 for faults which raise an alarm

Scrubber unit has a weilbull failure characteristic with η = 600 days, γ = 60days, and β = 1.8

Reactor failures can involve the agitator which has two failure modes.

Shaft fracture failure rate = 0.1/year

Motor failure rate = 0.3/year[1]

2(a(i))

Fault tree analysis here product fails to meet specification is the top event

Alarm failure

Liquid control

LAL fails

Liquid control

Low level

High level

Agitator failure

Coking problem

Motor failure

Shaft fracture

High level

Low level

Excess flow of liquid A

Excess flow of liquid B

Reactor

Pump failure

2(a(ii))

Fault tree analysis here liquid waste stream composition outside limits is the top event

Low level

High level

Internal mal functions failure

Connection fails between scrubbers

Improper cleaning temperature

Improper alkaline solution pumping to scrubber unit

Scrubber unit failure

Improper flow reactor to scrubber

High level

Low level

Low level

High level

2a) calculation of reliability of parts of the system

Here parts of the system:

Storages

Reactor

Agitator

Pumps

Scrubbing unit

Reliability of reactor:

Here reactor failure can involve the agitator failure. First one is shaft fracture and second one is motor failure

Failure rate of shaft fracture = 0.1/year

Failure rate of the motor = 0.3/year

Scrubber unit has a weilbull failure characteristic with η = 600 days, γ = 60days, and β = 1.8[1]

Failure rate of pump ( λ) = 13Ã-10-6hours[2]

Reliability of shaft fracture:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^6[3]

Failure rate of shaft fracture = 0.1/year

So using this we are finding t

Z(t) = β/ηβ(t-γ)β-1

0.1/year =(1.8/(600)1.8)Ã-(t-60)1.8-1

Here one year = 365 days

0.1/365 = (1.8/(600)1.8)Ã-(t-60)1.8-1

t = 90.11 days

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

= 0.995

So reliability for shaft fracture = 0.995

Reliability of motor:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

Failure rate of the motor = 0.3/year

So using this we are finding t

Z(t) = β/ηβ(t-γ)β-1

0.3/year =(1.8/(600)1.8)Ã-(t-60)1.8-1

Here one year = 365 days

0.3/365 = (1.8/(600)1.8)Ã-(t-60)1.8-1

t = 177.29 days

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

= 0.948

So reliability for motor = 0.948

Reliability for scrubbing unit:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^β

Here we have the t = 133.6 days

Z(t) = β/ηβ(t-γ)β-1

Z(t) =(1.8/(600)1.8)Ã-(133.6-60)1.8-1

Z(t) = 0.2/year

Equation for the reliability:

R(t) = e-((t-γ)/η)^β

= 0.996

So reliability for scrubbing unit R(t) = 0.996

Reliability of pump:

Failure rate of pump ( λ) = 13Ã-10-6hours

Reliability of pump R(t) = e-λt

Surviving time t = 70 days

One day = 24 hours

Surviving time t = 1680 hours

Reliability of pump R(t) = e-λt

= e(-13Ã-10^-6Ã-1680)

Reliability of pump R(t) = 0.978

References:

[1] Plant reliability and maintainability, assignment question paper, module (CPE6250) held on November 30 to December 3 2009.

[2] Frank P. Lees, 1996, Loss prevention in the process industries, second edition, volume 3.

[3] Cris Whetton, ility engineering. Failure data analysis. [Lecture handout].from plant reliability and maintainability, module (CPE6250) held on November 30 to December 3 2009.

2b)

Reliability block diagram for the complete system

Pump 1

Storage A

Pump2

Scrubbing unit

Reactor

Pump

Storage B

calculation of reliability of the complete system over one year:

Here parts of the system:

Storages

Reactor

Agitator

Pumps

Scrubbing unit

Reliability of reactor:

Here reactor failure can involve the agitator failure. First one is shaft fracture and second one is motor failure

Failure rate of shaft fracture = 0.1/year

Failure rate of the motor = 0.3/year

Scrubber unit has a weilbull failure characteristic with η = 600 days, γ = 60days, and β = 1.8[1]

Failure rate of pump ( λ) = 13Ã-10-6hours

Failure rate of fan (λ) = 57Ã-10-6hours [2]

Reliability of shaft fracture:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^6[3]

Failure rate of shaft fracture = 0.1/year

So using this we are finding t

Z(t) = β/ηβ(t-γ)β-1

0.1/year =(1.8/(600)1.8)Ã-(t-60)1.8-1

Here one year = 365 days

0.1/365 = (1.8/(600)1.8)Ã-(t-60)1.8-1

t = 90.11 days

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

= 0.995

So reliability for shaft fracture = 0.995

Reliability of motor:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

Failure rate of the motor = 0.3/year

So using this we are finding t

Z(t) = β/ηβ(t-γ)β-1

0.3/year =(1.8/(600)1.8)Ã-(t-60)1.8-1

Here one year = 365 days

0.3/365 = (1.8/(600)1.8)Ã-(t-60)1.8-1

t = 177.29 days

Equation for the reliability:

R(t) = e-((t-γ)/η)^6

= 0.948

So reliability for motor = 0.948

Reliability for scrubbing unit:

Equation for failure rate:

Z(t) = β/ηβ(t-γ)β-1

Here β = shape factor

η = characteristic life

γ = location parameter

t = surviving a time

Equation for the reliability:

R(t) = e-((t-γ)/η)^β

Here we have the t = 133.6 days

Z(t) = β/ηβ(t-γ)β-1

Z(t) =(1.8/(600)1.8)Ã-(133.6-60)1.8-1

Z(t) = 0.2/year

Equation for the reliability:

R(t) = e-((t-γ)/η)^β

= 0.996

So reliability for scrubbing unit R(t) = 0.996

Reliability of pump:

Failure rate of pump ( λ) = 13Ã-10-6hours

Reliability of pump R(t) = e-λt

Surviving time t = 70 days

One day = 24 hours

Surviving time t = 1680 hours

Reliability of pump R(t) = e-λt

= e(-13Ã-10^-6Ã-1680)

Reliability of pump R(t) = 0.978

Reliability of the complete system over year R(t) = average of system parts reliability

=(0.995+0.948+0.996+0.978)/4 = 0.979

Therefore reliability of the complete system over year = 0.979

References:

[1] Plant reliability and maintainability, assignment question paper, module (CPE6250) held on November 30 to December 3 2009.

[2] Frank P. Lees, 1996, Loss prevention in the process industries, second edition, volume 3.

[3] Cris Whetton, ility engineering. Failure data analysis. [Lecture handout].from plant reliability and maintainability, module (CPE6250) held on November 30 to December 3 2009.

2c)

To achieve a target reliability of 0.90 over one year:

Reliability target is a zero failure target. This is an important target implied for those low performing plants, such plants does not achieve certain goals designed by engineers. So we have to set appropriate target to achieve plant design. the reliability of the system must be improved to achieve the target. to achieve the reliability target or to improve reliability three basic ways must be employed.

By system design

By component specification

By preventive maintenance

By system design:-

The basic rule of our system design is to keep the design has simple as possible. the system is more reliable if the system is simple. Some of the steps include,

System simplification:

To reduce the complexities in process plant at the design stage its self

Reduction in the use of Complex parts by replacing them with more fundamental parts

The design should be made simple and easy to under stand

Decrease in component count:

The number of components used in the plant must be reduced. complex components must be avoided for the simplicity of the design.

Fault tolerance:

The basic characteristics of fault tolerance require:

No single point of failure

No single point of repair- the system must operate without any interruption during the process of repair when the system experiences any problems.

Fault isolation to the failing component- in case of failures the failed part of the system must be isolated from the offended system. This requires necessary failure detection mechanism.

Fault containment to prevent propagation of the failure

Availability of reversion modes- some failures may cause cripples to the entire system, to avoid the entire process system must pushed to the safe mode

By component specification:

For the reliability of a component it must be adequately specified for their entire length of service. Additional reliability can be provided by operating the components at lower stress then their operating stresses. By doing so early failures of the components can be reduced. in a process industry it is very difficult to improve reliability just by specification. This is attributed to the shortage of necessary data regarding the affect of stresses on the components. Components of high quality cannot be used always for economic reasons. Normally the parameters required to improve reliability often contradict with process requirements. Some of the reliability improvements include:

Use of corrective maintenance- it is defined as the maintenance which is required to repair and bring product after the repair is carried out. it is carried out in components who is failures doesn't affect of the overall working of the process system significantly. This activity mainly involves repair, restoration or replacement of components.

Design improvement-the design of any high quality process plant is based on the design parameters and technical specifications. the reactor design must be improved for high rates of efficiency. Temperature, pressure and other external considerations must be included in the design of reactor and storage tanks.

Quality control-Quality control assures conformance to specifications. quality control checks whether measurements of the components like reactors, storage tank, scrub units as in this case conform to the requirements.

Preventive maintenance:

Is defined as a maintenance carried out to prevent failure or warring out of components in the process plant. This is carried out by providing systematic inspection, detection and prevention of incipient failure.

The preventative maintenance efforts are aimed at preserving the useful life of equipment and avoiding premature equipment failures, minimizing any impact on operational requirements. In addition to the routine aspects of cleaning, adjusting, lubricating and testing. it is carried out only on those items where a failure would have expensive or unacceptable consequences e.g. reactors, storage tanks, scrubbing units. Many of these items are also subject to a statutory requirement for inspection and preventive maintenance.[1]

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