GLOBAL KNOWLEDGE ARK

Engineering & Technology · Electrical Power, Conversion & Machines

Protection, power quality, reliability and lifecycle trade-offs

Adding a second supply can improve a simplified reliability model. It cannot help if both depend on the same failed upstream path or if neither can carry the required load. Reliability is a statement about a defined service, time and set of failure mechanisms.

Electrical Circuits supplies current/voltage, power, rms, phase and ideal storage relations. Engineering Thermodynamics supplies energy balances, losses and system boundaries. This chapter introduces magnetic flux and pole count before using them, derives the simple machine/converter relations and labels every invented parameter. All diagrams and numerical cases are original teaching models. They do not specify equipment construction, protective settings, installation, switching or qualification.

Back to the chapter

After reading, you can

Evaluate protection coverage, common causes and lifecycle trade-offs without turning a redundancy or efficiency number into a guarantee.

01

Identify the function being protected

A protection system observes relevant electrical conditions and initiates action to limit consequences. A protective relay can detect a condition and command interruption; a breaker must have appropriate interrupting capability and a functioning control/energy path. Detection, decision and physical action are distinct links.

Coordination aims to remove the relevant faulty part while retaining appropriate service elsewhere, with backup for failures to act. It depends on system topology, fault behaviour and actual device characteristics. A one-line drawing or ideal switch does not establish clearing time, selectivity or safety. This chapter gives no settings, wiring or switching procedure.

Power quality also asks whether delivered voltage, frequency, waveform and interruptions suit the equipment’s function. Adequate average kWh does not prove that a controller tolerates a brief interruption or a machine tolerates unbalance. A solution to one disturbance can introduce losses, control complexity or another dependency.

02

Make independence and capacity visible

For a defined mission, let each of two hypothetical paths have survival probability 0.99. Assume either path can carry the full service, the paths fail independently and switching/control never fails. Both fail with probability 0.01²=0.0001, so system mission survival is 0.9999. This is a constructed probability model, not an observed availability or reliability figure.

Now introduce a separate common-cause event of probability p that disables both. Conditional on its absence, retain the independent path model. Survival becomes (1−p)×0.9999. The figure shows how strongly a common dependency can dominate the optimistic independent result. Shared cooling, controls, routing or environment can make independence unjustified; their actual probabilities require evidence.

Mission reliability concerns surviving the specified interval without the defined failure. Availability additionally involves restoration/repair and whether service is present at a chosen time. Do not convert one into the other without a suitable time/repair model. Redundancy also requires the remaining path’s capacity and transition behaviour to be adequate.

A shared mechanism changes the redundant-service calculationHorizontal common-cause event probabilityp 0–0.05 at 10000 units/probability, labelled 0–5%. Vertical total mission failure 0–0.05 at 6500 units/probability, labelled 0%,2%,4%. Conditional path survival 0.99 each; independently both fail 0.0001 absent common event. Total failure 1−(1−p)0.9999=0.0001+0.9999 p. At p 0.02 failure 0.020098; survival 0.979902. Either path assumed sufficient full capacity, perfect transfer/control absent common event. No measured failure rates, lifetime curve or availability implied.Original mission model: common cause dominates independent paths0%2%4%0%1%2%3%4%5%System mission failure probabilityTwo conditionally independent paths: each survives with 0.99At p = 2%: system failure = 2.0098%Probability p of shared disabling eventFailure = 1 − (1 − p) × 0.9999. Either path must carry full service.Invented mission probabilities; no repair model or observed reliability data.
Horizontal common-cause event probabilityp 0–0.05 at 10000 units/probability, labelled 0–5%. Vertical total mission failure 0–0.05 at 6500 units/probability, labelled 0%,2%,4%. Conditional path survival 0.99 each; independently both fail 0.0001 absent common event. Total failure 1−(1−p)0.9999=0.0001+0.9999 p. At p 0.02 failure 0.020098; survival 0.979902. Either path assumed sufficient full capacity, perfect transfer/control absent common event. No measured failure rates, lifetime curve or availability implied.
03

Compare a whole service over its intended life

Lifecycle comparison includes manufacture, operating energy, maintenance, replacement, recovery and the consequences of lost service. A more efficient component may have higher acquisition impacts, different repairability or a changed duty cycle. Establish a common functional output and boundary before ranking options.

A calculation can isolate one contribution while recording what it omits. Price, discounting, future load, material recovery and failure probabilities are assumptions, not constants supplied by an efficiency percentage. Keep uncertainties and sensitivity visible; no universal winner follows from one operating case.

Test an optimistic survival claim and an energy-only premium

With common-cause probability p=0.02, the declared mission survival is 0.98×0.9999=0.979902, or about 97.99%. Failure probability is 0.020098. The near-perfect independent figure would hide the dominant shared mechanism. This does not compare two real products or establish how often either fails.

Separately, compare two hypothetical converters delivering the same constant 1 kW for 4000 h/year over five years. At assumed efficiencies 90% and 95%, total input energies are 20,000/0.90≈22,222.2 kWh and 20,000/0.95≈21,052.6 kWh. Difference is about 1169.6 kWh.

At an explicitly invented constant energy price €0.20/kWh, this saves about €233.92. If acquisition premium is €250, that energy-only five-year comparison leaves €16.08 unrecovered. Maintenance, discounting, changed duty, downtime and environmental impacts are excluded; adding them could change the decision. This is arithmetic for a stated scenario, not a tariff forecast or purchasing recommendation.

Try the reasoning

  1. Does two-path independence follow from drawing two separate boxes?

    Show the answer and reasoning

    No. Shared physical, control or environmental causes can couple their failures.

  2. Why must each path’s capacity be checked in the survival model?

    Show the answer and reasoning

    The model requires either surviving path to provide the whole defined service; survival without sufficient capacity is not success.

  3. Does the lower-input-energy option necessarily have the best lifecycle outcome?

    Show the answer and reasoning

    No. It is one contribution. Other costs, impacts, service conditions and uncertainty require a common-boundary comparison.

What to carry forward

Define success, time, dependencies and capacity. Compare protection and lifecycle evidence at the level of the complete service.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. US Department of Energy2 sources
    1. DOE Fundamentals Handbook Electrical Science, Volume 4 of 4 — DOE-HDBK-1011/4-92 (archived training handbook)US Department of EnergyTeaching resource · PDFenergy.govRead source
    2. Improving Motor and Drive System Performance — A Sourcebook for Industry, second editionUS Department of EnergyTeaching resource · PDFenergy.govRead source
  2. e-Handbook of Statistical Methods — 8.1.8.4 R out of N modelNIST/SEMATECHTeaching resourceitl.nist.govRead source