StudentShare
Contact Us
Sign In / Sign Up for FREE
Search
Go to advanced search...
Free

Critical Component for a Renewal Policy - Case Study Example

Cite this document
Summary
This case study "Critical Component for a Renewal Policy" is about pumps within the system as a critical component that requires replacement. Flow occurs because of a change in pressure. Like in displacement pumps where the liquid already has some form of energy thus it is said to be primed…
Download full paper File format: .doc, available for editing
GRAB THE BEST PAPER93% of users find it useful

Extract of sample "Critical Component for a Renewal Policy"

Name: Course name:  Tutor’s name: Date: Critical component for a renewal/replacement policy The critical component that requires to renewal/replacement is pump within the system. Flow occurs because of change in pressure. Like in displacement pumps where the liquid already has some form of energy thus it is said to be primed. In pumps, where the inlet, outlet as well as the pump lies at the same height and thus is connected to a motor which provides mechanical energy to the fluid! The pump effectively changes one from of energy into another and this second form is responsible to make the liquid flow. The simple working principle of the pump involves a motor; the liquid uses this energy and flows into the impeller of the pump; this centrifugal force of the impeller helps to increase the velocity or flow rate of the fluid. In the moving fluid; mechanical energy gets transformed to kinetic energy because of which the pressure gets increased. This increased pressure coupled with the direction of movement of the impeller blades makes the fluid go out through the outlet point. In order to completely evaluate the working/performance of a pump, various different variables and characteristics associated with the pump need to be observed and analysed. These include the fluid flow rate, efficiency, mechanical power, power input, total dynamic head values, speed of the pump etc. Data Collection Various data was collected which included for Age Replacement Model, Block Replacement Model and Minimal Repair Model. This data is presented below: Age Replacement Model Weibull Parameters Shape 3.44 Cf   2000 Scale 20 Cm   1000 Location 12 Cost-Rate min 49.18 Mean 29.98 Optimum Interval 24 delta t 0.5 tp R(tp) R(tp)dt T(tp)=∑R(t)dt Cost Rate MIL-C.R tp 12 1   12 - 83.33 12 12.5 1 0.50 12.50 80.00 80.00 12.5 13 1 0.50 13.00 76.93 76.93 13 13.5 0.9999 0.50 13.50 74.08 74.08 13.5 14 0.9996 0.50 14.00 71.46 71.45 14 14.5 0.9992 0.50 14.50 69.02 69.02 14.5 15 0.9985 0.50 15.00 66.77 66.76 15 15.5 0.9975 0.50 15.50 64.68 64.68 15.5 16 0.9961 0.50 16.00 62.76 62.75 16 16.5 0.9941 0.50 16.49 60.99 60.96 16.5 17 0.9915 0.50 16.99 59.35 59.32 17 17.5 0.9883 0.49 17.49 57.86 57.81 17.5 18 0.9842 0.49 17.98 56.50 56.43 18 18.5 0.9793 0.49 18.47 55.27 55.17 18.5 19 0.9733 0.49 18.96 54.16 54.03 19 19.5 0.9663 0.48 19.44 53.17 53.01 19.5 20 0.9581 0.48 19.92 52.29 52.09 20 20.5 0.9487 0.48 20.40 51.53 51.28 20.5 21 0.9379 0.47 20.87 50.89 50.58 21 Block Replacement Model Weibull Parameters Shape 3 Cf   20000 Scale 800 Cm   2000 Location 200 Cost-Rate min 5.75 Mean 914.38 Optimum Interval 420.00 delta t 10 t R(t) F(t) H(t) est. N(t) Cost Rate MIL-STD Cost Rate t 200 1 1.56E-08 1.56E-08 1.56E-08 10.00 10.00 200 210 0.999998 1.95E-06 1.95E-06 1.95E-06 9.52 9.52 210 220 0.999984 1.56E-05 1.56E-05 1.56E-05 9.09 9.09 220 230 0.999947 5.27E-05 5.27E-05 5.27E-05 8.70 8.70 230 240 0.999875 0.000125 0.000125 0.000125 8.34 8.34 240 250 0.999756 0.000244 0.000244 0.000244 8.02 8.02 250 260 0.999578 0.000422 0.000422 0.000422 7.72 7.72 260 270 0.99933 0.00067 0.00067 0.00067 7.46 7.45 270 280 0.999 0.001 0.001 0.001 7.21 7.21 280 290 0.998577 0.001423 0.001424 0.001423 6.99 6.98 290 300 0.998049 0.001951 0.001953 0.001952 6.80 6.78 300 310 0.997404 0.002596 0.0026 0.002598 6.62 6.60 310 320 0.996631 0.003369 0.003375 0.003372 6.46 6.44 320 330 0.995718 0.004282 0.004291 0.004286 6.32 6.29 330 340 0.994655 0.005345 0.005359 0.005352 6.20 6.17 340 350 0.99343 0.00657 0.006592 0.006581 6.09 6.05 350 360 0.992032 0.007968 0.008 0.007984 6.00 5.95 360 370 0.99045 0.00955 0.009596 0.009573 5.92 5.87 370 380 0.988674 0.011326 0.011391 0.011358 5.86 5.80 380 390 0.986693 0.013307 0.013396 0.013352 5.81 5.74 390 400 0.984496 0.015504 0.015625 0.015564 5.78 5.70 400 Minimal Repair Model Weibull Parameters Shape 4 Cr (Cost of Minimal repair) 1000 Scale 500 Cm (Cost of Renewal) 3000 Location 100 Minimum Cost Rate 6.63 Mean 553.2 Optimum Replacement Interval 570 delta t 10 Starting t Value 100 tp R(tp) F(tp) H(tp) Cost Rate tp 100 1.0000 0.0000 0.000E+00 30.00 100.00 110 1.0000 0.0000 1.600E-07 27.27 110.00 120 1.0000 0.0000 2.560E-06 25.00 120.00 130 1.0000 0.0000 1.296E-05 23.08 130.00 140 1.0000 0.0000 4.096E-05 21.43 140.00 150 0.9999 0.0001 1.000E-04 20.00 150.00 160 0.9998 0.0002 2.074E-04 18.75 160.00 170 0.9996 0.0004 3.842E-04 17.65 170.00 180 0.9993 0.0007 6.554E-04 16.67 180.00 190 0.9990 0.0010 1.050E-03 15.79 190.00 200 0.9984 0.0016 1.600E-03 15.01 200.00 210 0.9977 0.0023 2.343E-03 14.30 210.00 220 0.9967 0.0033 3.318E-03 13.65 220.00 230 0.9954 0.0046 4.570E-03 13.06 230.00 240 0.9939 0.0061 6.147E-03 12.53 240.00 250 0.9919 0.0081 8.100E-03 12.03 250.00 260 0.9896 0.0104 1.049E-02 11.58 260.00 270 0.9867 0.0133 1.336E-02 11.16 270.00 280 0.9833 0.0167 1.680E-02 10.77 280.00 290 0.9794 0.0206 2.085E-02 10.42 290.00 300 0.9747 0.0253 2.560E-02 10.09 300.00 310 0.9694 0.0306 3.112E-02 9.78 310.00 320 0.9632 0.0368 3.748E-02 9.49 320.00 330 0.9562 0.0438 4.477E-02 9.23 330.00 340 0.9483 0.0517 5.308E-02 8.98 340.00 Weibull Distribution The WEIBULL analysis provides with simple and useful graphical plot. The horizontal scale represent the aging parameters such numbers or hours and vertical scale provides the cumulative cost rate. The main three characteristics of WEIBULL parameters are beta, shape and the location to determine the predicted life. Renewal/replacement strategy There are various strategies available for the component which included age replacement of individual component, Block replacement of individual components, Age replacement of all components treated as a single component, Condition monitoring or continuous monitoring, Replacement on failure and Condition based renewal/replacement. From the strategies available the best renewal/replacement strategy for the selected component is Condition based renewal/replacement. This is arrived at based regular repairs and inspection. When the condition deteriorates the pump is replaced. The following is done regularly to pump Sequence of Procedure for inspection 1. Verify the energy isolating devices’ location and the power or magnitude of energy produced by the source. 2. All affected employees must be notified that servicing (maintenance or repair) is required on the water pump machine and that the machine must be locked out or shut down to carry out the servicing. Authorized employees who will do the lockout and tagout procedure shall specifically identify as to which valve, switch or control for energy source needs to be locked out. With this, it should be noted that more than one energy source may be involved in the lockout/tagout for water pumps (electrical, mechanical, hydraulic or water energy and thermal energy sources included). The authorized employees who will perform the lockout/tagout shall clear any uncertain identification of the power/energy sources with the unit managers/supervisors. In addition, the authorized employees shall also have sufficient knowledge in terms of the hazard potential of the identified energy source and the methods for energy control. 3. If the water pump machine is currently operating, it should be shut down using the standard procedure. For standard water pumps machines (and systems), the pump is shut down at the nearest local stop/start switch closest to the water pump unit. The manual valve on the discharge line shall be closed. Move the lever to “off” position on the central power panel. 4. Next, deactivate or disable the energy isolating device(s). If a knife switch is present on the pump, the switch should be thrown to the off position. Then, the breaker is moved to the off position. Close all the inlet and outlet water valves. 5. Using specified individual locks, lock out the energy control or isolating device(s). Then, apply lock to the knife switch. If the device(s) do not have a knife switch, use a breaker lock to lock out the breaker. Residual or stored energy such as water pressure lines should be restrained or dissipated by certain methods including blocking, repositioning or grounding. In water pumps, capacitors are dissipated and volumes of residual water in lines are drained. Allow time for hot water lines to cool first before doing the dissipation. 6. Verify if the equipment is de-energized or if isolation has been complete by attempting to start or power up the water pump to its normal operating controls. Before verifying the isolation or before ensuring that the equipment is disconnected from the source of energy, make sure that no employees or personnel are exposed to ensure safety. After verifying the isolation of the water pump from the energy source, switch back the operating controls to an “off” or “neutral” position. 7. The water pump is now locked out and the servicing (repair or maintenance can now be carried out. After the water pump servicing has been completed and to restore the machine back to its normal operations, the following procedure must be followed. 1. Check the water pump and the area near the machine to make sure that the unnecessary items are already removed and that the water pump and its parts are still operationally intact. 2. Assess the work area to make sure that all personnel are already in safe positions or are removed from the area of operation. 3. Verify if all the controls and/or switches are in neutral or off positions. 4. Remove the locking devices and re-energize the power pump machines. In some cases, the removal of the lockout devices is only accomplished after the machine has been re-energized to facilitate safe removal. If fault is found then the pump will be then be replaced Quantitative analysis The net present value method determines the dollar value of the firm as a result of undertaking an investment project. The NPV is computed by discounting the projected net cash flow of each year based on the compound annual required return on investment. This is done to reflect the effects of inflation during the projection time period. These discounted projections are combined with the original investment amount subtracted to show the expected profit dollar amount over the projected time period. The method calculates the amount of the change in a firm’s value from pursuing a particular project, and is calculated by adding the projected cash flows and subtracting the initial investment amount. Replacement upon failure The second option has a higher NPV which means Optimum renewal/replacement criteria There are very important factors that are very important when considering such machines like the proper design, assembly as well as installations are some of the requirements when it comes to optimal performance. A rotating machine normally comprises of three major parts: rotor, bearing as well as the foundation inclusive of pipe connection and the structures. As such, a problem may arise when it comes to the major use of pump. However from quantative analysis the following are optimal replacement criteria total annual cost $200,000 cost rate per month $30 inspection interval Regularly Net present value 71,999 Works Cited Bergman, W.J., 1999. Reliability centered maintenance (RCM) applied to electrical switchgear. In Power Engineering Society Summer Meeting, 1999. IEEE (Vol. 2, pp. 1164-1167). IEEE. Gallagher, T., & Andrew, J., 2007. Financial Management; Principles and Practice. New York: Freeload Press. Mitchell, Larry D 2002, 'Preventive maintenance and RCM II', Manufacturing Engineer: ME, vol. 81, no. 4, pp. 153-155 Robert B .A, D A, 2004, ‘The New Weibull Handbook’, Fifth Ed, SAE International, Florida Smith, A.M., 1993. Reliability-centered maintenance (Vol. 83). New York: McGraw-Hill. de Vasconcelos, V., da Silva, E.M.P., da Costa, A.C.L., dos Reis, S.C. and Correa, D.P., 2009. A Dynamic Maintenance FMEA To Support A Reliability Centered Maintenance Implementation Of Critical Systems In Nuclear Facilities. Read More

Authorized employees who will do the lockout and tagout procedure shall specifically identify as to which valve, switch or control for energy source needs to be locked out. With this, it should be noted that more than one energy source may be involved in the lockout/tagout for water pumps (electrical, mechanical, hydraulic or water energy and thermal energy sources included). The authorized employees who will perform the lockout/tagout shall clear any uncertain identification of the power/energy sources with the unit managers/supervisors.

In addition, the authorized employees shall also have sufficient knowledge in terms of the hazard potential of the identified energy source and the methods for energy control. 3. If the water pump machine is currently operating, it should be shut down using the standard procedure. For standard water pumps machines (and systems), the pump is shut down at the nearest local stop/start switch closest to the water pump unit. The manual valve on the discharge line shall be closed. Move the lever to “off” position on the central power panel. 4. Next, deactivate or disable the energy isolating device(s).

If a knife switch is present on the pump, the switch should be thrown to the off position. Then, the breaker is moved to the off position. Close all the inlet and outlet water valves. 5. Using specified individual locks, lock out the energy control or isolating device(s). Then, apply lock to the knife switch. If the device(s) do not have a knife switch, use a breaker lock to lock out the breaker. Residual or stored energy such as water pressure lines should be restrained or dissipated by certain methods including blocking, repositioning or grounding.

In water pumps, capacitors are dissipated and volumes of residual water in lines are drained. Allow time for hot water lines to cool first before doing the dissipation. 6. Verify if the equipment is de-energized or if isolation has been complete by attempting to start or power up the water pump to its normal operating controls. Before verifying the isolation or before ensuring that the equipment is disconnected from the source of energy, make sure that no employees or personnel are exposed to ensure safety.

After verifying the isolation of the water pump from the energy source, switch back the operating controls to an “off” or “neutral” position. 7. The water pump is now locked out and the servicing (repair or maintenance can now be carried out. After the water pump servicing has been completed and to restore the machine back to its normal operations, the following procedure must be followed. 1. Check the water pump and the area near the machine to make sure that the unnecessary items are already removed and that the water pump and its parts are still operationally intact. 2. Assess the work area to make sure that all personnel are already in safe positions or are removed from the area of operation. 3. Verify if all the controls and/or switches are in neutral or off positions. 4. Remove the locking devices and re-energize the power pump machines.

In some cases, the removal of the lockout devices is only accomplished after the machine has been re-energized to facilitate safe removal. If fault is found then the pump will be then be replaced Quantitative analysis The net present value method determines the dollar value of the firm as a result of undertaking an investment project. The NPV is computed by discounting the projected net cash flow of each year based on the compound annual required return on investment. This is done to reflect the effects of inflation during the projection time period.

These discounted projections are combined with the original investment amount subtracted to show the expected profit dollar amount over the projected time period. The method calculates the amount of the change in a firm’s value from pursuing a particular project, and is calculated by adding the projected cash flows and subtracting the initial investment amount.

Read More
Cite this document
  • APA
  • MLA
  • CHICAGO
(Critical Component for a Renewal Policy Case Study Example | Topics and Well Written Essays - 1500 words, n.d.)
Critical Component for a Renewal Policy Case Study Example | Topics and Well Written Essays - 1500 words. https://studentshare.org/engineering-and-construction/2056385-interval-determination-and-compiling-of-the-program
(Critical Component for a Renewal Policy Case Study Example | Topics and Well Written Essays - 1500 Words)
Critical Component for a Renewal Policy Case Study Example | Topics and Well Written Essays - 1500 Words. https://studentshare.org/engineering-and-construction/2056385-interval-determination-and-compiling-of-the-program.
“Critical Component for a Renewal Policy Case Study Example | Topics and Well Written Essays - 1500 Words”. https://studentshare.org/engineering-and-construction/2056385-interval-determination-and-compiling-of-the-program.
  • Cited: 0 times
sponsored ads
We use cookies to create the best experience for you. Keep on browsing if you are OK with that, or find out how to manage cookies.
Contact Us