Reliability Design of Mechanical Systems Such As Compressor Subjected to Repetitive Stresses †

metals

Article Reliability Design of Mechanical Systems Such as Compressor Subjected to Repetitive Stresses †

Seongwoo Woo 1,* and Dennis L. O’Neal 2

1 Department of Mechanical Engineering, College of Electrical and Mechanical Engineering, Addis Ababa Science and Technology University, Addis Ababa P.O. Box 16417, Ethiopia 2 Department of Mechanical Engineering, School of Engineering and Computer Science, Baylor University, Waco, TX 76798-7356, USA; [email protected] * Correspondence: [email protected]; Tel.: +251-90-047-6711 † Presented at the First International Electronic Conference on Metallurgy and Metals (IEC2M 2021).

Abstract: This study demonstrates the use of parametric accelerated life testing (ALT) as a way to recognize design defects in mechanical products in creating a reliable quantitative (RQ) specification. It covers: (1) a system BX lifetime that X% of a product population fails, created on the parametric ALT scheme, (2) fatigue and redesign, (3) adapted ALTs with design alternations, and (4) an evaluation of whether the system design(s) acquires the objective BX lifetime. A life-stress model and a sample size formulation, therefore, are suggested. A refrigerator compressor is used to demonstrate this method. Compressors subjected to repetitive impact loading were failing in the field. To analyze the pressure loading of the compressor and carry out parametric ALT, a mass/energy balance on the vapor-compression cycle was examined. At the first ALT, the compressor failed due to a cracked or Citation: Woo, S.; O’Neal, D.L. fractured suction reed valve made of Sandvik 20C carbon steel (1 wt% C, 0.25 wt% Si, 0.45 wt% Mn). Reliability Design of Mechanical The failure modes of the suction reed valves were similar to those valves returned from the field. Systems Such as Compressor The fatigue failure of the suction reed valves came from an overlap between the suction reed valve Subjected to Repetitive Stresses. Metals 2021, 11, 1261. https:// and the valve plate in combination with the repeated pressure loading. The problematic design was doi.org/10.3390/met11081261 modified by the trespan dimensions, tumbling process, a ball peening, and brushing process for the valve plate. At the second ALT, the compressor locked due to the intrusion between the crankshaft Academic Editors: João and thrust washer. The corrective action plan specified to perform the heat treatment to the exterior Pedro Oliveira, Belén Díaz Fernández, of the crankshaft made of cast iron (0.45 wt% C, 0.25 wt% Si, 0.8 wt% Mn, 0.03 wt% P). After these Leszek Adam Dobrzański, design modifications, there were no troubles during the third ALT. The lifetime of the compressor Jae-chun Lee, Alex Lanzutti, Eric was secured to have a B1 life of 10 years. D. van Hullebusch and Filippo Berto Keywords: mechanical product; fatigue; design defects; parametric ALT; compressor Received: 2 July 2021 Accepted: 6 August 2021 Published: 10 August 2021 1. Introduction Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in Because of demands in the global market, refrigerators must be designed to have low published maps and institutional affil- energy usage and reliability. For those aimed functions, the compressor in the refrigerator iations. often needs to be redesigned to enhance its comprehensive energy efficiency on a yearly basis. It involves recently designed features for the system, which should be rapidly brought to the end-users. With insufficient testing or no assumption of how the new attributes may be utilized, their launching may increase product failures in the marketplace and negatively influence the manufacturer’s brand name. These attached features for Copyright: © 2021 by the authors. Licensee MDPI, Basel, Switzerland. the product should be totally assessed in the design phase before being released into the This article is an open access article marketplace. Therefore, reliability quantitative (RQ) specifications utilizing an established distributed under the terms and system of method should be presented [1,2]. conditions of the Creative Commons A compressor is designed to increase the refrigerant pressure in a refrigeration cycle Attribution (CC BY) license (https:// by several mechanical compression mechanisms. One of the key components of the creativecommons.org/licenses/by/ compressor is the crankshaft that converts the rotational motion into reciprocating motion 4.0/). by a crank mechanism. A compressor is subjected to repeated stresses due to internal

Metals 2021, 11, 1261. https://doi.org/10.3390/met11081261 https://www.mdpi.com/journal/metals Metals 2021, 11, 1261 2 of 21

pressure loadings over the course of its lifetime. New rotary compressors introduced in 1987 in refrigerators were experiencing large recalls due to the locking of the compressor in the field [3]. Oil metal sludge formed during normal refrigeration operations was separating from the sintering crankshaft and plugging the capillary tubes, which forced the refrigerator to no longer function. To stop a compressor recall, any flawed components needed to be identified and altered using testing methodology such as parametric ALT, which can produce reliability quantitative (RQ) specifications before the system launches. The procedures of robust design, such as the Taguchi approach [4,5] and design of experiments (DOE) [6], were developed to help identify the most advantageous designs for products. Especially, Taguchi’s method employs design parameters to put it in the right location where “noise” parameters do not have any effect on the output. As a result, the right designs of mechanical products can be selected. However, without identifying failure mechanisms such as fatigue, this methodology can only pursue system optimization. If there is a design fault, the product may fail during its lifetime as loads are exerted on it. Many parameters must be considered to identify an optimal design of a mechanical structure. However, the large number of parameters may require huge computations unless some options are neglected to reduce computational requirements. Material faults, such as extremely small voids and contacts when subjected to repeated loads, may begin to fail because of fatigue. Fatigue is the chief source of destruction in metallic elements, explaining roughly 80–95% of all constructional failures [7–9]. Fatigue in ductile metals appears itself in the shape of cracks that grow in stress accumulation, such as holes, grooves, etc. Those failures can affect the reliability of mechanical systems such as moving automobiles, airplane wings, marine ships, turbo engines, and atomic reactors. The fatigue procedure covers three fluctuating stress/time modes: (1) symmetrical about zero stresses, (2) asymmetrical about zero stresses, and (3) random stress cycles. The fatigue may also rely on the parameters such as the cyclic stress amplitude, mean stress or stress ratio, R (=σmin/σmax), which can be defined as s the proportion of the minimum cyclic stress to the maximum cyclic stress [10]. In other words, in periodical shapes, the peaks on both the maximum (high side) and the minimum (low side) are crucial. When employing an elevated load which can be stated as an accelerated factor (AF), accelerated life testing (ALT) can be examined to discover the design defects such as stress raiser in the structure. The ALT combined on the reliability block diagram was investigated as another method [11]. It included a test plan for the system, identifying failure mechanics such as fatigue, and using sample size equation, accelerated loads, etc. Elsayed [12] categorized statistical, physics/statistics, and physics/experimental-established prototypes for examination. Meeker [13] suggested numerous practical ways to organize an ALT. Carrying out an ALT [14,15] necessitates numerous notions such as the BX life for the system test scheme, a simplified life-stress description, sample size formulation, and fracture mechanics [16–18] because failure may happen suddenly from the fragile components in a product. Contem- porary test techniques [19–26] might be hard to replicate the design flaws of components in a multi-module system because those procedures evaluate insufficient component samples and do not identify the fatigue(s) which actually occur in the marketplace. To attain the sound design of a mechanical product, designers have employed traditional techniques such as fracture mechanics and strength of materials [27]. As quantum as one of the branches of mechanics has developed, engineers identified which failures came from micro-void coalescence (MVC) and discovered many metallic alloys or some engineering plastics. To find the fatigue source of a mechanical product, a life-stress model can be incorporated with conventional design methods and relevant techniques to help identify the failure of electronic parts due to present material flaws or small cracks when the parts are subjected to (mechanical) stresses. Finite element methods (FEMs) may not identify the source of failure [28]. To show the effectiveness for identifying and altering the design defects of a mechanical product, this research proposes using a parametric ALT as a systematic reliability method that can create the RQ specifications—mission cycles. It covers: (1) a system BX Metals 2021, 11, x FOR PEER REVIEW 3 of 22

Metals 2021, 11, 1261 To show the effectiveness for identifying and altering the design defects of 3a ofmechan- 21 ical product, this research proposes using a parametric ALT as a systematic reliability method that can create the RQ specifications—mission cycles. It covers: (1) a system BX lifetime created on the ALT scheme, (2) a load examination for ALT, (3) tailored ALTs lifetimewith created the alterations, on the ALT and scheme, (4) an (2)estimation a load examination of whether forthe ALT,product (3) tailoreddesign(s) ALTs accomplishes with the alterations, and (4) an estimation of whether the product design(s) accomplishes the the objective BX lifetime. The derivation of the sample size equation, time-to-failure, and objective BX lifetime. The derivation of the sample size equation, time-to-failure, and BX BX lifetime is provided. To confirm ALT results in real life and compare the current design lifetime is provided. To conﬁrm ALT results in real life and compare the current design with with previous one, it would then be necessary to monitor the introduction of the new previous one, it would then be necessary to monitor the introduction of the new design in design in the market to ensure it reached the targeted lifetime and failure rate. A newly the market to ensure it reached the targeted lifetime and failure rate. A newly designed designed compressor in a domestic refrigerator subjected to repeated pressure impact compressor in a domestic refrigerator subjected to repeated pressure impact loading is loading is used as an example of this methodology. used as an example of this methodology.

2. Parametric2. Parametric ALT ALT for Mechanical for Mechanical Product Product 2.1. BX2.1. Lifetime BX Lifetime of a Mechanical of a Mechanical System System

To performTo perform an ALT, an ALT, the BX the life, BX Llife,B, isLB needed, is needed as a as measure a measure of theof the product product lifetime. lifetime. It It cancan identify identify the the accumulative accumulative failure failure rate rate and and meet meet the the market market needs needs for for reliability reliability re- requirementsquirements of theof the product. product. BX BX life life is theis the elapsed elapsed time time for for which which X% X of% aof population a population of of a a productproduct fails. fails. Thus, Thus, a “BX a “BX life Ylife years” Y years” is a wayis a way for expressing for expressing the system the system lifetime. lifetime. For For example,example, if a system if a system lifetime lifetime has a has B20 a life B20 of life 10 of years, 10 years, then 20%then of20% the of population the population will failwill fail duringduring 10 years 10 years of operation. of operation. On the On other the other hand, hand, the mean the mean time time to failure to failure (MTTF)—the (MTTF)—the B60 lifetime—isB60 lifetime—is not acceptable not acceptable for identifying for identifying the system the system lifetime lifetime because because it takes it takes too long too long for 60%for of60% the of population the population of the of product the product to fail. to Thefail. BXThe life BX is life a more is a more suitable suitable measure measure of of a lifetime.a lifetime. In a mechanicalIn a mechanical product, product, movements movements of power of power for acquiring for acquiring mechanical mechanical advantages advantages are utilizedare utilized in functions in functions that that require require forces forces and and movement movement by by accommodating accommodating certain certain sys- systemtem mechanisms. mechanisms. As As products products are are subjected subjected to repeatedto repeated stresses, stresses, they they should should have have a a properproper lifetime lifetime for those for those stresses. stresses. For instance, For instance, to maintain to maintain the freshness the freshness of food of food products, products, a refrigeratora refrigerator is designed is designed to provide to provide cooled cooled air from air a heatfrom exchanger a heat exchanger such as ansuch evaporator as an evapo- to therator refrigerator to the refrigerator (or freezer) (or sections. freezer) Thesections. refrigerator The refrigerator is made up is made of different up of different modules: mod- the cabinet,ules: the door, cabinet, internal door, ﬁxtures internal (shelves fixtures and (she drawers),lves and controls drawers), and controls instruments, and instruments, electric motor,electric compressor, motor, heat compressor, exchangers heat (evaporator exchange andrs (evaporator condenser), and water condenser), supply device, water and supply numerousdevice, dissimilar and numerous parts. A dissimilar refrigerator parts. has approximatelyA refrigerator 2000has approximately components (Figure 2,0001 a)compo-. A targetnents of (Figure the refrigerator’s 1a). A target lifetime of the refrigerator’s is set to have lifetime a B20 life is set 10 to years. have Ifa B20 a refrigerator life 10 years. If consistsa refrigerator of 20 units consists and each of unit 20 units has 100 and components, each unit has the 100 lifetime components, of each unitthe lifetime should beof each designedunit toshould have be a B1 designed life 10 years. to have The a requiredB1 life 10 system years. lifetimeThe required of a refrigerator system lifetime relies onof a re- the manyfrigerator components, relies on including the many the components, compressor. including If the compressor the compressor. has design If the faults, compressor it impactshas thedesign potential faults, life it impacts of the refrigerator the potential (Figure life of1b). the refrigerator (Figure 1b).

(a) (b)

FigureFigure 1. Product 1. Product lifetime lifetime decided decided by newly by newly designed designed module module (a) categorization (a) categorization of multi-module of multi-module refrigerator; refrigerator; (b) product (b) product lifetime LB and failure rate λs. lifetime LB and failure rate λs. Metals 2021, 11, x FOR PEER REVIEW 4 of 22 Metals 2021, 11, 1261 4 of 21

2.2. Placing an Entire Parametric ALT Scheme 2.2. Placing an Entire Parametric ALT Scheme For a stated period of time, the reliability for a compressor can be explained as the For a stated period of time, the reliability for a compressor can be explained as the capacity needed to continue performing the intended function under specified environ- capacity needed to continue performing the intended function under specified environmental/operational conditions [29]. It can be explained using the traditional “bathtub mental/operational conditions [29]. It can be explained using the traditional “bathtub curve”, which is composed of three regions, as seen in the top curve of Figure 2. In the curve”, which is composed of three regions, as seen in the top curve of Figure2. In the firstfirst region, region, during during the the early early product product life, life, there there is some is some reduction reduction in the in failure the failure rate. In rate. theIn secondthe second region, region, during during its middle its middle life, there life, is there a relatively is a relatively constant constant failure failurerate. In rate.the third In the region,third there region, is a there growing is a growing failure rate failure until rate the untilfinal life the of final the life product. of the If product. a product If afollows product thisfollows classic this pattern, classic it pattern,might not it might be successful not be successful in the marketplace in the marketplace due to the due tall to initial the tall failureinitial rates failure and rates the relatively and the relatively short lifetime short of lifetime the product of the productdue to built-in due to design built-in flaws. design Companiesflaws. Companies need to upgrade need to upgradethe system the design system by design setting by its setting reliability its reliability targets to targets (1) re- to move(1) remove premature premature failures, failures, (2) reduce (2) reduce random random failures failures over the over product the product lifetime, lifetime, and (3) and enlarge(3) enlarge product product lifetime. lifetime. As the As design the design of a mechanical of a mechanical product product improves, improves, its failure its failure rate in ratethe infield the decreases, field decreases, and the and system the system lifetime lifetime can canbe increased. be increased. For For such such situations, situations, the the traditionaltraditional bathtub bathtub curve curve might might be bealtered altered to tothe the bottom bottom curve curve in inFigure Figure 2,2 where, where failure failure ratesrates are are low low throughout throughout the the life life of ofthe the produc productt except except toward toward the the end end of ofits its expected expected life. life.

FigureFigure 2. Bathtub 2. Bathtub curve curve and and straight straight line. line. The failure rate on the bathtub in Figure2 can be stated as The failure rate on the bathtub in Figure 2 can be stated as 0 f dF/dt (1 − R)′ −R′0 λ = f = dF dt = ()1− R =− R (1) λ = R = R = R = R (1) R R R R where λ is the failure rate, f is the failure density function, R is the reliability, and F is the where λ is the failure rate, f is the failure density function, R is the reliability, and F is the unreliability of the product. unreliability of the product. If Equation (1) is integrated over time, the X% cumulative failure F(LB) at BX life, LB, If Equation (1) is integrated over time, the X% cumulative failure F(LB) at BX life, LB, can be obtained: can be obtained: Z λdt = − ln R (2) λdt = −ln R (2) That is, it can be stated as:  That is, it can be stated as: Z LB L B ∼ A = hλλi⋅·LB == λλ()(t)⋅·dt == − ln R ()(LB) ==−− ln ((1−− F ))≅= F ()()(LB)(== X) (3) A LB t dt ln R LB ln 1 F F LB X (3) 0 Metals 2021, 11, x FOR PEER REVIEW 5 of 22 Metals 2021, 11, 1261 5 of 21

As a product has a lower failure rate of the straight line that follows an exponential As a product has a lower failure rate of the straight line that follows an exponential distribution, the reliability of a mechanical system could be stated as the product lifetime distribution, the reliability of a mechanical system could be stated as the product lifetime LB and failure rate λ: LB and failure rate λ: − -λLL R(L()) = =1 −−F(L ()) ==e λ BB =∼≅ 1−λL (4) R BLB 1 F BLB e 1- LBB (4) ThisThis relationship relationship is is suitable suitable below below about about 20% 20% of of the cumulative failure [30]. [30]. After After targetingtargeting the the lifetime, lifetime, LLBB, ,and and failure failure rate, rate, λλ, ,the the mechanical mechanical product product can can be be redesigned redesigned by by identifyingidentifying the the troublesome troublesome structures structures and and alte alteringring them them by by a a parametric parametric ALT ALT (Figure 33).).

FigureFigure 3. 3. ParameterParameter diagram diagram of of compressor compressor (example). (example).

ToTo target thethe lifetimelifetime of of a a mechanical mechanical system system through through a parametric a parametric ALT, ALT, there there are three are threepossible possible system system modules: modules: (1) modified (1) modified independent independent units, (2)units, newly (2) newly designed designed independent inde- pendentunits, and units, (3) similarand (3) independentsimilar independent units to theunit earliers to the design earlier base design on requestbase on inrequest the field. in The modified compressor in the refrigerator inspected here is used as a case study. It the field. The modified compressor in the refrigerator inspected here is used as a case was originally redesigned to enhance its energy efficiency and make the refrigerator more study. It was originally redesigned to enhance its energy efficiency and make the refrig- competitive in the marketplace. However, the redesigned compressor had design faults erator more competitive in the marketplace. However, the redesigned compressor had that needed to be altered because of failures in the field. design faults that needed to be altered because of failures in the field. The altered independent units D from the field data manifested in Table1 had a failure The altered independent units D from the field data manifested in Table 1 had a fail- rate of 0.31% per year and a B1 life of 3.2 years. The lifetime of compressors, based on ure rate of 0.31% per year and a B1 life of 3.2 years. The lifetime of compressors, based on data from the field, indicated they had an expected B1 life of 1.6 years because they had data from the field, indicated they had an expected B1 life of 1.6 years because they had a a failure rate of 0.62% per year. To satisfy consumer demands, a new lifetime target for failure rate of 0.62% per year. To satisfy consumer demands, a new lifetime target for the the mechanical system such as compressor was put to have a B1 life of 10 years with 0.1 mechanical system such as compressor was put to have a B1 life of 10 years with 0.1 per- percent per year. cent per year. Table 1. Whole ALT scheme of mechanical modules in a refrigerator. Table 1. Whole ALT scheme of mechanical modules in a refrigerator. Field Data Expected Reliability Aimed Reliability Field Data Expected Reliability Aimed Reliability Mod-Modules Failure Rate per BX Life, BX Life, Failure Rate per BX Life, Failure Rate Per Year, BX Life, FailureFailure RateRate per Per Year, Year, %/Year BX Life, Failure Rate Per Year, BX Life, ules Year, %/Year Year Year Year, %/Year Year A%/Year 0.35 2.9Year Similar %/Year×1 0.35Year 2.9%/Year 0.10 10(BXYear = 1.0) A B0.35 0.24 4.2 2.9 NewSimilar ×5×1 0.35 1.202.9 0.83 0.10 0.10 10(BX 10(BX == 1.0)1.0) C 0.30 3.3 Similar ×1 0.30 3.33 0.10 10(BX = 1.0) B D0.24 0.31 3.2 4.2 AlteredNew ×2×5 1.20 0.620.83 1.610.10 0.10 10(BX 10(BX == 1.0)1.0) C E0.30 0.15 6.7 3.3 AlteredSimilar ×2×1 0.30 0.303.33 3.33 0.10 0.10 10(BX 10(BX == 1.0)1.0) DOthers 0.31 0.50 10.0 3.2 SimilarAltered ×1×2 0.62 0.501.61 10.0 0.10 0.50 10(BX 10(BX == 5.0)1.0) Product 1.9 2.9 - - 3.27 0.83 1.00 10(BX = 10) E 0.15 6.7 Altered ×2 0.30 3.33 0.10 10(BX = 1.0) Others 0.50 10.0 Similar ×1 0.50 10.0 0.50 10(BX = 5.0) Product 1.9 2.3. Failure 2.9 Mechanics and - Parametric- 3.27 ALT for Redesign0.83 1.00 10(BX = 10) As mentioned in Section 2.1, mechanical products usually transfer power (or energy) 2.3.from Failure one position Mechanics to anotherand Parametric by adapting ALT for proper Redesign mechanisms. For example, the compressor in a refrigerator, by utilizing the vapor-compression cycle, increases the refrigerant pressure As mentioned in Section 2.1, mechanical products usually transfer power (or energy) using a crankshaft mechanism. It is subjected to repeated stresses because of pressure from one position to another by adapting proper mechanisms. For example, the compres- loading. If there is a design defect in the structure that creates an insufficient strength (or sor in a refrigerator, by utilizing the vapor-compression cycle, increases the refrigerant stiffness) when the loads are applied, the mechanical product may abruptly collapse before pressure using a crankshaft mechanism. It is subjected to repeated stresses because of its expected lifetime. pressure loading. If there is a design defect in the structure that creates an insufficient Metals 2021, 11, x FOR PEER REVIEW 6 of 22

strength (or stiffness) when the loads are applied, the mechanical product may abruptly Metals 2021, 11, 1261 collapse before its expected lifetime. 6 of 21 Metal fatigue is the familiar word that can be used to express the unanticipated failure of product components that develop fractures in their lifetime. It is connected to the numberMetal of stress fatigue cycles is the experienced familiar word by thata comp canonent be used and to the express amount the unanticipatedof stress exerted failure on theof component. product components The S-N curve that shows develop that fractures an infinite in life their is lifetime.possible for It a is metal connected component to the ifnumber the stresses of stress in the cycles component experienced remain by abelow component a clearly and defined the amount lower of stress stress boundary. exerted on Metalthe component. fatigue grows The if S-N there curve is a showsstress raiser that an such infinite as notches life is possible in a component. for a metal An component engineer shouldif the stressesidentify inthese the componentdesign flaws remain using appropriate below a clearly analysis defined or testing. lower stressThere boundary. is also a correlationMetal fatigue between grows a if metal there component’s is a stress raiser ultimate such as tensile notches strength in a component. and hardness An engineerand its capabilityshould identify to endure these fatigue design loads. flaws The using loftie appropriater the tensile analysis strength or and testing. hardness, There the is more also a probablycorrelation it will between fatigue a metalif subjected component’s to higher ultimate fluctuating tensile loads. strength and hardness and its capabilityFatigue to failure endure due fatigue to design loads. defects The loftier can be the characterized tensile strength by two and components: hardness, the (1) more the stressprobably due itto will loads fatigue on the if subjectedconstructional to higher system fluctuating and (2) the loads. type of materials (or shape) utilizedFatigue in the failurestructure. due In to recognizing design defects the can product be characterized failure by a by parametric two components: ALT, an (1)engi- the neerstress could due tooptimally loads on design the constructional parts with good system forms and and (2) theproper type materials. of materials The (or system shape) couldutilized bear in repeated the structure. loads In over recognizing its lifetime the so product that it failurecould byfulfill a parametric the reliability ALT, target an engineer (Fig- urecould 4). optimally design parts with good forms and proper materials. The system could bear repeated loads over its lifetime so that it could fulfill the reliability target (Figure4).

Figure 4. Fatigue failure on the structure created by repetitive loading and design flaws. Figure 4. Fatigue failure on the structure created by repetitive loading and design flaws. The main problem for reliability testing is to determine how quickly the potential failureThe mode main canproblem be identified. for reliability To complete testing it,is itto is determine necessary how to successfully quickly the formulate potential a failuresimplified mode failure can be description identified. and To complete determine it, the it is correct necessary coefficients to successfully for the model. formulate A life- a simplifiedstress (LS) failure model description can be developed, and determine which incorporates the correct coefficients stresses and for reaction the model. parameters. A life- stressThis model(LS) model can explain can be variousdeveloped, mechanical which inco failuresrporates such stresses as fatigue and in reaction the structure. parameters. Fatigue Thisfailures model can can occur explain not due various to stresses mechanical in a perfect failures part such but ratheras fatigue due toin thethe presentstructure. defects Fa- tigueor very failures small can cracks occur on not the due exterior to stresses of a component. in a perfect part but rather due to the present defectsAs or fatigue very small originates cracks from on the material exterior flaws of a created component. on a macro or microscopic scale level, the life-stressAs fatigue model originates can be from defined material from flaw suchs acreated point of on view. a macro For instance, or microscopic we can utilizescale level,the procedures the life-stress for model solid-state can be diffusion defined of from impurities such a point in silicon, of view. which For is instance, commonly we usedcan utilizeas semi-conduct the procedures material, for solid-state such as in diffusion the following of impurities procedure: in silicon, electro-migration-induced which is commonly usedvoiding, as semi-conduct growth of chloride material, ions, such and as catching in the following of electrons procedure: or holes. As electro-migration-in- an electric magneto- ducedmotive voiding, force, ξ growth, is exerted, of chloride the impurities ions, and such catc ashing voids of inelectrons materials, or holes. created As by an electronic electric magneto-motivemovement, are effortlesslyforce, ξ, is exerted, migrated the because impurities the barriers such as ofvoids junction in materials, energy are created dropped by electronicand distorted/phase-shifted. movement, are effortlessly For solid-state migrated diffusion because of the impurities barriers of in junction silicon, theenergy junction are droppedequation and J might distorted/phase-shifted. be expressed as follows For solid-state [31,32] (Figure diffusion5): of impurities in silicon, the junction equation J might be expressed as follows [31,32] (Figure 5): Metals 2021, 11, x FOR PEER REVIEW 7 of 22

 q  1  J = []aC()x − a ⋅exp − w − aξ  ⋅ v (5)  kT 2    

Metals 2021, 11, 1261 [Density/Area]·[Jump Probability]·[Jump Frequency] 7 of 21

− qaξ ∂C − = −[]a2ve qw kT ⋅cosh +[]2ave qw kT Csin 2kT ∂x h q 1 i J = [aC(x − a)] · exp − kT w − 2 aξ ·v [Density/Area]·[Jump Probability]·[Jump Frequency ]Q  = Φ()()x,t,T sinh aξ exp − h i h i  2 −qw/kT qaξ ∂C −qw/kT  kTqaξ (5) = − a ve · cosh 2kT ∂x + 2ave Csinh 2kT Q = Φ(x, t, T)sinh(aξ) exp − kT  Q  = Bsinh()aξ exp−  = Bsinh(aξ) exp − Q kT kT   where C Cis is the the concentration, concentration,q is q theis the magnitude magnitude of electric of electric charge, charge,ν is the ν is frequency, the frequency,Φ() is Φ() isa constant,a constant,B isB ais constant, a constant,a is a the is the interval interval between between atoms, atoms,ξ is ξ the is appliedthe applied ﬁeld, field,k is thek is the Boltzmann’s constant,constant,T Tis is the the temperature, temperature, and andQ isQ theis the energy. energy.

Figure 5.5. PotentialPotential exchange exchange in in material material such such as as silicon silicon as electricalas electrical ﬁeld field is exerted. is exerted. The reaction process, which depends on speed, could be stated as The reaction process, which depends on speed, could be stated as − ∆E−aSΔE−aS − ∆E+aS ΔE+aS + − kT kT− kT kT − K = K −+ K =− a h ekT − a h e kT = ∆E − = kT − kT (6) KkT −KkT K aS a e a ∆Ee = a h e sinh( kT ) = Bsinh(aS) exp − kT h h Δ where K is the reaction rate, SkTis the− (chemical)E aS effect, T is the temperature, ΔEkis Boltzmann’s(6) constant, E is the (activation)= a energy,e kT andsinh(∆ is the) = difference.Bsinh()aS exp−  h kT kT The reaction rate K from Equations (5) and (6) could be simpliﬁed as  where K is the reaction rate, S is the (chemical) effect, T is the temperature, k is Boltzmann’s Ea constant, E is the (activation) Kenergy,= Bsinh and(aS Δ) expis the− difference. (7) kT The reaction rate K from Equations (5) and (6) could be simplified as If Equation (7) takes an inverse function, the life-stress (LS) model might be stated as  E a  K = B sinh( aS ) exp  −  (7) − E kT  TF = A[sinh(aS)] 1 exp a (8) If Equation (7) takes an inverse function, the life-stresskT (LS) model might be stated as

−1 −1  E a  The sine hyperbolic expressionTF [=sinhA[](sinh(aS)] aSin) Equationexp  (8) can be stated as (Figure6): (8) − kT 1. (S) 1 at ﬁrst has a little linear result;   − 2. The(S) sinen has hyperbolic what is regarded expression as a medium sinhaS result; in Equation (8) can be stated as (Figure −1 6):3. eaS at end is large. 1. S at first has a little linear result; Metals 2021, 11, x FOR PEER REVIEW 8 of 22

Metals 2021, 11, 1261 8 of 21 2. S has what is regarded as a medium result; 3. e at end is large.

FigureFigure 6. 6. HyperbolicHyperbolic sine sine stress stress term term versus versus S-N S-N curve curve from from a standpoint a standpoint of stress of stress range. range.

AnAn ALT ALT is is usually usually conducted conducted in in the the medium medium scope, scope, and and Equation Equation (5) (5) might might be bestated stated as as −n Ea TF = A(S)−n exp Ea  (9) TF = A()S exp kT (9)  kT  Because the stress level in a mechanical system may be hard to be computed during Because the stress level in a mechanical system may be hard to be computed during ALT, Equation (9) must be restated. As the power is stated as the product of ﬂows and ALT, Equation (9) must be restated. As the power is stated as the product of flows and effort,effort, stresses stresses may may originate originate from from effort effort in ina multi-port a multi-port product product (Table (Table 2) [33].2)[33 ].

TableTable 2. 2. PowerPower concept concept in ina multi-port a multi-port product product expressed expressed as effort as effort and andflow. ﬂow.

SystemSystem Effort, Effort, e(t)e (t)Flow, Flow, f(t) f(t) TranslationTranslation system system Force, Force, F(t)F (t)Velocity, Velocity, V(t)V (t) RotationRotation system system Torque, Torque, τ(t)τ (t)Angular Angular velocity, velocity, ω(t)ω (t) Pump,Pump, compressor compressor Pressure Pressure difference, difference, ΔP(t)∆P( t) Volume Volume flow ﬂow rate, rate, Q(t)Q (t) ElectricElectric system system Voltage, Voltage, V(t)V (t)Current, Current, i(t) i(t) Magnetic Magneto-motive force, em Magnetic ﬂux, ϕ Magnetic Magneto-motive force, em Magnetic flux, φ

StressStress is is a amaterial material quantity quantity that that specifies speciﬁes the theinner inner forces forces which which adjoining adjoining particles particles ofof a acontinuum continuum material material exert exert on each on eachother. other. For a mechanical For a mechanical system, because system, stress because orig- stress inatesoriginates from fromeffort, effort, Equation Equation (9) might (9) mightbe stated be as stated as

−n E  −λ  E  = ()− Eaa = () − aEa TFTF = A(SS) nexpexp  =BBe(e) expλ exp  (10) (10)  kTkT  kTkT

wherewhere AA andand BB areare constants. constants. ToTo obtain obtain the the acceleration acceleration factor factor (AF), (AF), which which can can mostly mostly affect affect the the evaluation evaluation of of fa- fatigue tiguestrength strength in a product,in a product,AF mightAF might be statedbe stated as theas the ratio ratio between between the the proper proper elevated elevated stress stress levels and normal working conditions. AF might be modified to integrate the effort levels and normal working conditions. AF might be modiﬁed to integrate the effort notion: notion: nn λλ S1  Ea 1 1  e1  Ea 1 1 AF =  S1  Ea  1 − 1  = e1  Ea  1 −1  (11) AF =  S   k  T − T  =  e  k T− T (11)  S 0  k T0 T 1  e 0 k  T 0 T 1  0    0 1   0    0 1  2.4. Parametric ALT for Mechanical Systems To acquire the mission cycle of parametric ALTs from the targeted BX lifetime on the test plan in Table1, the sample size equation combined with the AF should be attained [1] (see AppendixA). Metals 2021, 11, 1261 9 of 21

1 L∗ β n ≥ (r + 1) × × B + r (12) x AF·ha If the lifetime of a mechanical system, such as the domestic compressor, is targeted to have a B1 life of 10 years, the mission cycles might be acquired for an assigned set of samples subjected to (impact) loading. In parametric ALTs, the design defects of the new product might be identiﬁed to help satisfy the lifetime target.

2.5. Case Study—Design of a Recently Designed Domestic Compressor in a Refrigerator Operating with unspecified consumer usage conditions, the suction reed valves in compressors used in the field were unsuccessful. The fractured suction reed valve caused the domestic compressor to lock and quit functioning. As the primary purpose of the refrigerator, including the compressor, was lost, consumers would solicit to have the product exchanged. To address the problem, it was crucial to reproduce the failure mode(s) of the compressor in simulated conditions in a room or building equipped for engineering Metals 2021, 11, x FOR PEER REVIEW 10 of 22 experiments. Seemingly, the troublesome compressors that came from the field had two evident design flaws: (1) the suction reed valve had a quantity of overlap with the valve plate, and (2) the valve plate had a sharp edge. As the suction reed valve impacted the valve plate, it would fail before its expected lifetime (Figure7).

(a)

(b) Figure 7. Failed suction reed valve: (a) suction reed valve and valve plate in a compressor. (b) Frac- Figure 7. Failed suction reed valve: (a) suction reed valve and valve plate in a compressor. (b) Frac- tured suction reed valve from the marketplace or parametric ALT. tured suction reed valve from the marketplace or parametric ALT. To chill the stored products in a refrigerator, the refrigerator supplies cooled air from the evaporator heat exchanger to the freezer (or refrigerator) departments. The vapor- compression cycle in a refrigerator covers a compressor, condenser, capillary tube, and evaporator. Electrical energy in the compressor motor is converted to work energy in the compressor that is used to elevate the refrigerant pressure. With refrigerant flowing in the system, heat energy absorbed by the evaporator is moved to the condenser, where it is rejected to the surrounding air. A capillary tube manages the refrigerant flow and controls the flow of refrigerant from the high-pressure refrigerant in the condenser to the low- pressure refrigerant in the evaporator. In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor (Processes 1–2) and is cooled to the saturated liquid state in the condenser (Processes 2–3). It then drops in pressure as it goes through the expansion device to the lower pressure evaporator (Processes 3–4). In the evaporator, the refrigerant is vaporized as it absorbs heat from the refrigerated space (Processes 4–1) (Figure 8). Metals 2021, 11, 1261 10 of 21

To chill the stored products in a refrigerator, the refrigerator supplies cooled air from the evaporator heat exchanger to the freezer (or refrigerator) departments. The vapor- compression cycle in a refrigerator covers a compressor, condenser, capillary tube, and evaporator. Electrical energy in the compressor motor is converted to work energy in the compressor that is used to elevate the refrigerant pressure. With refrigerant flowing in the system, heat energy absorbed by the evaporator is moved to the condenser, where it is rejected to the surrounding air. A capillary tube manages the refrigerant flow and controls the flow of refrigerant from the high-pressure refrigerant in the condenser to the low- pressure refrigerant in the evaporator. In an ideal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor (Processes 1–2) and is cooled to the saturated liquid state in the condenser (Processes 2–3). It then drops in pressure as it Metals 2021, 11, x FOR PEER REVIEW goes through the expansion device to the lower pressure evaporator (Processes11 of 3–4). 22 In

the evaporator, the refrigerant is vaporized as it absorbs heat from the refrigerated space (Processes 4–1) (Figure8).

(a) (b)

FigureFigure 8. A domestic 8. A domestic compressor compressor in the in refrigeration the refrigeration cycle: cycle: (a) vapor-compression (a) vapor-compression refrigerator refrigerator cycle, cycle, (b) P h diagram. (b) P h diagram.

DuringDuring normal normal operation, operation, refrigerator refrigerator compressors areare subjected subjected to repetitiveto repetitive stresses stressesdue due to pressure to pressure loads loads differences differences in the in compressor. the compressor. If there If is there a design is a faultdesign in fault a component, in a component,such as ansuch insufficiency as an insufficiency of strength, of strength, when the when loads the are loads applied are applied in the compressor,in the com- the pressor,component the component may abruptly may abruptly fail and notfail meetand not its expectedmeet its expected lifetime. Bylifetime. identifying By identify- the system ing thefailure system by afailure parametric by a parametric ALT, an engineer ALT, an can engineer select can proper select material proper andmaterial redesign and the redesigncomponent the component using the using best orthe most best favorableor most favorable way so the way compressor so the compressor can support can repeatedsup- portloads repeated and loads its lifetime and its canlifetime be increased. can be increased. It is necessary It is necessary to analyze to analyze the pressure the pressure loads in loadsthe in compressor.the compressor. In a Inrepresentative a representative refrigeration refrigeration cycle, cycle, to toassess assess its its design, design, it it was was required required to determinedeter- mineboth both the the condensing condensing temperature, temperature,Tc, andTc, and evaporating evaporating temperature, temperature,Te. The Te. massThe mass flow rate flowof rate refrigerant of refrigerant in a compressor in a compressor can be can modeled be modeled as as η . v ηv m m= =PDPD× × (13)(13) v vsuc suc wherewhere PD isPD theis piston the piston displacement, displacement, ην isη theν is compressor’s the compressor’s volumetric volumetric efficiency, efficiency, and andνsuc vsuc is theis specific the specific volume volume at compressor at compressor suction suction port. port. The mass flow rate of the refrigerant at the capillary tube can be modeled as [34]. 0.5   P  − 4 ρdP  P =  3  m cap A (14)  2  ρ   f ΔL + ln 3  m  ρ   D 4     By conservation of mass, the mass flow rate can be stated as: m = m (15) cap The energy conservation at the condenser can be stated as Q = m ()()h − h = T − T / R (16) c 2 3 c o c Metals 2021, 11, 1261 11 of 21

The mass ﬂow rate of the refrigerant at the capillary tube can be modeled as [34].

 0.5 −R P4 ρdP . P3 mcap = A  (14) 2 ρ3 fm∆L + ln D ρ4

By conservation of mass, the mass ﬂow rate can be stated as:

. . m = mcap (15)

The energy conservation at the condenser can be stated as

. Qc = m(h2 − h3) = (Tc − To)/Rc (16)

The energy conservation at the evaporator can be stated as

. Qe = m(h1 − h4) = (Ti − Te)/Re (17)

Using Equations (15) through (17), it is possible to obtain estimates of the mass ﬂow . rate, m, evaporator temperature, Te, and condenser temperature, Tc. Because the saturation pressure, Psat, is a function of temperature, we can obtain the evaporator pressure, Pe (or condenser pressure Pc). Pe = f (Te) or Pc = f (Tc) (18)

Both the condensing pressure, Pc, and evaporating pressure, Pe, are important when examining the load on the compressor. These pressures depend on the environmental conditions, heat exchanger size, and customer usage conditions in the design stage. During normal compressor operations, it will be subjected to repeated stresses due to the pressure differences between the suction and discharge. The internal stress of the compressor relies on the pressure difference between suction pressure, Psuc, and discharge pressure, Pdis. ∼ ∆P = Pdis − Psuc = Pc − Pe (19) Under accelerated conditions, the life-stress model (LS model) in Equation (10) can be stated as: − Ea − Ea TF = A(S) n exp = A(∆P) λ exp (20) kT kT

where A is constant, k is Boltzman’s constant, Ea is the activation energy, T is the absolute temperature, n is the quotient, and λ is the cumulative damage exponent in Palmgren– Miner’s rule. So the acceleration factor (AF) might be stated as:

S n E 1 1 ∆P λ E 1 1 AF = 1 a − = 1 a − (21) S0 k T0 T1 ∆P0 k T0 T1

where S1 (or P1) is mechanical stress (or pressure difference) under accelerated conditions, and S0 (or P0) is mechanical stress (or pressure difference) under representative conditions. For a domestic compressor in a refrigerator, the typical operating conditions were for a customer span from 0 ◦C to 43 ◦C with a humidity ﬂuctuating from 0% to 95%. Vibration conditions presumed for working the compressor were subjected to 0.2 to 0.24 g of acceleration. When the compressor operates, the suction reed valve opens to allow refrigerant to stream into the compressor. A compressor is expected to cycle on and off 22 cycles per day. A worst-case scenario was also duplicated with on and off of 98 cycles per day. Under the worst cases, the compressor working for 10 years may incur approximately 357,700 usage cycles. Metals 2021, 11, 1261 12 of 21

The compressor is usually made of (carbon, cast, stainless, alloy, etc.) steel. The allowable stresses are stated as a function of the yield stress (Fy) or tensile stress (Fu) of the structural material. For steel, the ranges of yield strength, Fy, and ultimate or tensile strength, Fu, ordinarily used are 248–345 MPa and 400–483 MPa, separately. Refrigerant R134a is the refrigerant utilized in the refrigerant cycle. It uses synthetic refrigeration compressor oils that have a high Viscosity Index (VI). They have slight viscosity changes in relation to temperature changes. Therefore, the slope of the viscosity of a synthetic lubricant with a high VI is ﬂatter with respect to temperature. The viscosity thus remains stable across a wide temperature usage range. Metals 2021, 11, x FOR PEER REVIEW 13 of 22 In the worst case, the pressure difference was 1.27 MPa, and the compressor dome temperature was 90 ◦C. For a parametric ALT, the pressure difference was elevated to be 2.94 MPa, and the compressor dome temperature was also elevated to be 120 ◦C, with an accumulativeaccumulative damage damage exponent,exponent, λλ, ,of of 2. 2. The The total total AFAF computedcomputed from from Equation Equation (21) was (21) was 7.37.3 (Table (Table3). 3).

Table 3. CompressorTable 3. CompressorALT conditions. ALT conditions. System states Worst Case ALT AF System States Worst Case ALT AF High-side 1.27 2.94 5.36 ① Pressure High-side(MPa) Low-side 0.01.27 0.0 2.94 - 5.36 1 Pressure (MPa) Low-side ΔP 1.270.0 2.94 0.0 - - Temperature∆P 1.27 2.94 - ◦ Dome 90 120 1.37 ② Temperature ( C) (°C) Dome 90 120 1.37 2 Total AF (=( 1 × 2 )) Total AF - - - 7.32 - - - 7.32 (=(① × ②)) The test cycles of the ALTs calculated from Equation (12) were 49,000 cycles for 100 sam- The test cycles of the ALTs calculated from Equation (12) were 49,000 cycles for 100 ples if the shape parameter, β, was presumed to be 2.0. The parametric ALT was designed to samples if the shape parameter, β, was presumed to be 2.0. The parametric ALT was de- reassure a lifetime target—B1 life of 10 years—with an approximate 60% level of confidence signed to reassure a lifetime target—B1 life of 10 years—with an approximate 60% level thatof it confidence would fail that less it than would once fail during less than 49,000 once cycles.during We49,000 also cycles. exerted We the also duty exerted cycles the of the pressureduty cycles difference of the betweenpressure suctiondifference pressure, betweenP suctionsuc, and pressure, discharge Psuc pressure,, and dischargePdis. pressure,To P assessdis. the design of the compressor, a straightforward refrigeration cycle was established.To assess It contained the design aof compressor,the compressor, condenser, a straightforward capillary refrigeration tube, and cycle evaporator. was es- The temperaturetablished. It in contained the enveloped a compressor, ﬁberglass conden packageser, capillary is controlled tube, and by evaporator. two 60 W The lamps tem- and a fan.perature A thermal in the switch enveloped attached fiberglass to the packag covere is of controlled the compressor by two 60 managed W lamps a and 51 ma 3fan./h axial fan.A Thethermal test switch conditions attached and to their the cover borders of the were compressor set up on managed the control a 51 m board.3/h axial As fan. the test started,The test the conditions high-side and and their low-side borders pressureswere set up were on the displayed control board. on the As the pressure test started, gauges or apparatusthe high-side screen and (Figure low-side9). pressures were displayed on the pressure gauges or apparatus screen (Figure 9).

Figure 9. Equipment for the ALTs. Figure 9. Equipment for the ALTs. 3. Results and Discussion A sample in the first ALT (n = 100) locked at 3,500 cycles. Another sample continued working but had a partially broken suction reed at 7,500 cycles (Figure 10). The shape parameter, β, formed from the first ALT, was 2.0 (see Figure 11). The forms and positions of the failure in the samples obtained from the first ALT and the field were alike (Figure 10). The fractured suction reed valve originated from three design faults: (1) it had a quantity of overlap with the valve plate, (2) it had a sharp edge on the valve plate, and (3) it used an insufficiently strong material (see Figures 6 and 12). Metals 2021, 11, 1261 13 of 21

3. Results and Discussion A sample in the first ALT (n = 100) locked at 3500 cycles. Another sample continued working but had a partially broken suction reed at 7500 cycles (Figure 10). The shape parameter, β, formed from the first ALT, was 2.0 (see Figure 11). The forms and positions of the failure in the samples obtained from the first ALT and the field were alike (Figure 10). The fractured suction reed valve originated from three design faults: (1) it had a quantity Metals 2021, 11, x FOR PEER REVIEW of overlap with the valve plate, (2) it had a sharp edge on the valve plate, and (3) it used14 an of 22 insufficiently strong material (see Figures6 and 12).

(a) (b)

FigureFigure 10. 10. FailedFailed suction suction reed reed valves valves fr fromom the the field field and firstfirst ALT:ALT: ( a(a)) failed failed products products from from the the field, field, (b()b outcome) outcome after after first first parametric parametric ALT, ALT, ( (c) partially brokenbroken suctionsuction reed reed valve valve from from the the first first ALT, ALT, (d) shattered valve plate. (d) shattered valve plate.

When the suction reed valve frequently striked the valve plate, it eventually fractured before its expected lifetime. The main failure mode of the compressor was locking due to the failed suction reed valve. The ALT methodology was well-suited for recognizing the fatigue failure attained from the units in the field. First, the shape of the failed suction reed valves from the marketplace and those in the first ALT were very similar in appearance. The first ALT failure and market failure data manifested a similar pattern on a Weibull plot (Figure 11). As the data for the two models had similar gradients on the plot, each loading condition of the first ALT and that from the field over the product lifetime were similar. Therefore, it might be anticipated that the test samples in the laboratory would break in a similar way to those in the field. For the shape parameter, β, the last shape parameter from the chart was confirmed to be 2.0, compared with the approximated value—2.0. Based on both test outcomes in the Weibull plot, the parametric ALT was successful because it recognized the design defects which were responsible for the market failures. In other words, as proven by two things—the visual likeness in both photos and similar slopes in the Weibull plot—these ALTs were justified in recognizing the design flaws that were judged as the failures from the market. These failures determined the product lifetime. MetalsMetals 20212021, ,1111, ,x 1261 FOR PEER REVIEW 1514 of of 22 21

FigureFigure 11. 11. MarketMarket statistics statistics and and consequences consequences of of ALT ALT on on Weibull Weibull chart. chart.

RefrigeratorsWhen the suction that came reed valve from frequentlythe field had striked a primary the valve failure plate, mode it eventually with no fractured cooling becausebefore its the expected compressor lifetime. did not The function. main failure Field mode data suggested of the compressor that the wastroublesome locking due com- to pressorsthe failed may suction have reed had valve.design The flaws. ALT Due methodology to these defects, was well-suited the repetitive for recognizingpressure loads the couldfatigue produce failure attainedunanticipated from the stresses units inon the the field. suction First, reed the valve, shape ofcausing the failed it to suction crack and reed propagatevalves from the the crack marketplace to its end. andTo reproduce those in the the first dominant ALT were failure very mode similar of the in appearance. compressor, The first ALT failure and market failure data manifested a similar pattern on a Weibull parametric ALTs were carried out. Based on the first ALT and market data, we realized plot (Figure 11). As the data for the two models had similar gradients on the plot, each that the AF and β values were 7.3 and 2.0 (see Figure 11). For assigned test samples, the loading condition of the first ALT and that from the field over the product lifetime were test cycles were computed in Equation (12) if the product lifetime was secured to have a similar. Therefore, it might be anticipated that the test samples in the laboratory would B1 life of 10 years. break in a similar way to those in the field. For the shape parameter, β, the last shape To investigate a compressor that failed at 3,500 cycles in the first ALT, the trouble- parameter from the chart was confirmed to be 2.0, compared with the approximated value— some compressors came from the marketplace and the first ALT were compared to deter- 2.0. Based on both test outcomes in the Weibull plot, the parametric ALT was successful mine the potential design defects. The mode of the compressor failure in the first ALT was because it recognized the design defects which were responsible for the market failures. In very similar to the ones from the field. The suction reed valves failed in areas where they other words, as proven by two things—the visual likeness in both photos and similar slopes were partly covered by the valve plate. The tests stated that the compressor was improp- in the Weibull plot—these ALTs were justified in recognizing the design flaws that were erly designed to operate with this suction reed valve. judged as the failures from the market. These failures determined the product lifetime. The failed suction reed valve originated from the unsuitable design flaws: (1) an quantity of overlap with the valve plate; (2) a sharp edge on the valve plate; and (3) insufficiently strong material (0.178t) utilized in the design of suction reed valve. These defects might cause the compressor system to fracture abruptly when subjected to repetitive pressure loads (see Figures 7 and 12). Metals 2021, 11Metals, x FOR2021 PEER, 11 ,REVIEW 1261 16 of 22 15 of 21

(a)

(b)

Figure 12. StructuralFigure 12. designStructural defects design of suction defects reed of suctionand valve reed plate and in valve a compressor: plate in a compressor:(a) overlapped (a) overlapped suction reed valve and valve plate with sharp edge; (b) impact load in combination of design defects suction reed valve and valve plate with sharp edge; (b) impact load in combination of design defects when compressor is worked. when compressor is worked.

To fix the failedRefrigerators suction thatreed camevalve from failure the that field was had due a primary to the repetitive failure mode pressure with no cooling stresses in thebecause compressor’s the compressor lifetime, the did va notlve function. plate and Fieldsuction data reed suggested valve was that modified the troublesome as follows: compressors(1) trespan size may from have 0.73 had mm design to 1.25 flaws. mm; Due (2) to adding these defects, ball peening the repetitive and brush pressure loads process (seecould Figure produce 13); (3) unanticipated thickening the stresses suction on reed the valve suction from reed 0.178 valve, t to causing 0.203 t it(4) to crack and expanding tumblingpropagate process. the crack to its end. To reproduce the dominant failure mode of the compressor, For theparametric second ALT, ALTs three were samples carried locked out. Basednear 17,000 on the cycles. first ALT The and problematic market data, com- we realized pressor systemthat came the AF as andfollows:β values (1) erosion were 7.3of the and crankshaft 2.0 (see Figure and (2) 11 the). For intrusion assigned between test samples, the crank shaft testand cycles thrust were washer. computed The design in Equation modification (12) if the was product provided lifetime to the was heat secured treat- to have a B1 ment on thelife surface of 10 of years. crank shaft that can modify the physical, and sometimes chemical, properties of a materialTo investigate to achieve a compressor the desire thatd result, failed such at 3500 as hardening cycles in the of firsta weak ALT, crank- the troublesome ≥ % ≤ % ≤ % shaft materialcompressors (Ductile Iron came FCD450 from the– C marketplace 2.5 wt , S and 0.02 the wt first, Mg ALT 0.09 were wt compared). to determine the potential design defects. The mode of the compressor failure in the first ALT was very similar to the ones from the field. The suction reed valves failed in areas where they were partly covered by the valve plate. The tests stated that the compressor was improperly designed to operate with this suction reed valve. The failed suction reed valve originated from the unsuitable design flaws: (1) an quantity of overlap with the valve plate; (2) a sharp edge on the valve plate; and (3) Metals 2021, 11, 1261 16 of 21

insufficiently strong material (0.178 t) utilized in the design of suction reed valve. These defects might cause the compressor system to fracture abruptly when subjected to repetitive pressure loads (see Figures7 and 12). To fix the failed suction reed valve failure that was due to the repetitive pressure stresses in the compressor’s lifetime, the valve plate and suction reed valve was modified as follows: (1) trespan size from 0.73 mm to 1.25 mm; (2) adding ball peening and Metals 2021, 11, x FOR PEER REVIEWbrush process (see Figure 13); (3) thickening the suction reed valve from 0.178 t to17 0.203 of 22 t

(4) expanding tumbling process.

Metals 2021, 11, x FOR PEER REVIEW 17 of 22

(a) (b)

FigureFigure 13.13. Redesigned valve plate: ( (aa)) trespan trespan in in valve valve plate; plate; ( (bb) )ball ball peening peening process. process.

ForIn the the third second ALT, ALT, there three were samples no design locked problems near 17,000 in the cycles. compressor The problematic until the para- com- pressormetric(a) ALT system was came carried as follows:out to 49,000 (1) erosion cycles. of We the( btherefore)crankshaft concluded and (2) thethat intrusion the design between mod- Figure 13. Redesignedcrankifications shaft valve attained and plate: thrust from (a) washer. trespan the first in The andvalve design second plate; modification (ALTsb) ball were peening waseffective. process. provided Table to 4 the summarizes heat treatment the onparametric the surface ALT of results. crank shaft With that the can modified modify design the physical, parameters, and sometimes the compressor chemical, samples prop- In the thirdertieswere ALT,ensured of a material there to attainwere to achieve theno designlifetime the pr desiredtarget—B1oblems result, in life the of such compressor 10 years as hardening with until about ofthe aa weakpara-60% confidence crankshaft metric ALT materialwaslevel. carried (Ductile out to Iron 49,000 FCD500 cycles. – CWe≥ therefore2.5 wt%, concluded S ≤ 0.02 wt%, that Mg the≤ design0.09 wt%). modifications attainedIn from the third the first ALT, and there second were noALTs design were problems effective. in Table the compressor 4 summarizes until the the parametric Table 4. Compressor ALT outcomes. parametric ALT wasresults. carried With out the to 49,000modified cycles. design We thereforeparameters, concluded the compressor that the designsamples modifications were ensuredattained to attain from the the lifetime firstFi andr sttarget—B1 ALT second ALTslife of were 10 years effective. Secondwith about Table ALT4 a summarizes 60% confidence theThird parametric ALT Parametric ALT level. ALT results. WithOriginal the modified Design design parameters, Designthe compressor samples Final were Design ensured to attain the lifetime3,500 cycles: target—B1 1/100 locking life of 10 years with about a 60% confidence level. In 49,000 cycles, there are no issues in the 49,000 cycles: Table 4. Compressor7,500 cycles: ALT 1/100 outcomes. broken 17,000 cycles: 3/100 locking compressor 100/100 OK Tablesuction 4. Compressor reed valve ALT outcomes. First ALT Second ALT Third ALT Parametric ALT First ALT Second ALT Third ALT Parametric ALT Original Design Design Final Design 3,500 cycles:Original 1/100 Design locking Design Final Design In 49,000 cycles, there are no issuesStructure in the 49,000 cycles: In 49,000 cycles, there are no 7,5003500 cycles: cycles: 1/100 1/100 broken locking 17,000 cycles: 3/100 locking compressor 17,000 cycles: 3/100 locking100/100 49,000 OK cycles: 100/100 OK issues in the compressor 7500 cycles:suction 1/100 reed broken valve suction reed valve

C1: Trepan size: 0.73 mm→1.25 mm StructureStructure C2: attaching ball peening C5: FCD500 + no heat treatment → and brush process Action plans FCD500 + heat treatment on the C3: SANDVIK 20C: crankshaft C1: TrepanC1: Trepan size:0.178 size: t→0.203 t → 0.73 mm0.73→ mmC4:1.25 expanding1.25 mm mm tumbling: C2: attaching ball peening and brush process C5: FCD500 + no heat C2: attaching ball peening4 h→ 14 h Action plans C3: SANDVIK 20C: C5: FCD500treatment + no heat→ treatmentFCD500 + → heat and brush0.178 process t→0.203 t treatment on the crankshaft Action plans FCD500 + heat treatment on the 4. ConclusionsC3:C4: SANDVIK expanding 20C: tumbling: 4 h→14 h crankshaft To0.178 increase t→0.203 the t lifespan of a new domestic compressor in a refrigerator, the following reliabilityC4: expanding methodology tumbling: was developed: (1) the system BX lifetime created the total para- → metric ALT4 h scheme,14 h (2) the parametric ALT with design modifications, and (3) determine if the product design attains the mission cycles. To show the effectiveness of this method- 4. Conclusionsology for identifying and altering the design defects of a mechanical product, we derived To increasethe sample the lifespan size equation, of a new domestictime-to-failure compressor equation in aand refrigerator, acceleration the method, following and BX life- reliability methodologytime. As a test was case, developed: we inspected (1) the the system redesign BX lifetimeof a compressor created thethat total came para- from the mar- metric ALT scheme,ketplace. (2) the parametric ALT with design modifications, and (3) determine if the product. designIn the attains marketplace the mission and thecycles first. To ALT, show we the found effectiveness that the failed of this suction method- reed valve in ology for identifyinga compressor and altering originated the design from defects the following of a mechanical design flaws: product, (1) a smallwe derived overlap between the sample size equation, time-to-failure equation and acceleration method, and BX lifetime. As a test case, we inspected the redesign of a compressor that came from the marketplace. . In the marketplace and the first ALT, we found that the failed suction reed valve in a compressor originated from the following design flaws: (1) a small overlap between Metals 2021, 11, 1261 17 of 21

4. Conclusions To increase the lifespan of a new domestic compressor in a refrigerator, the following reliability methodology was developed: (1) the system BX lifetime created the total parametric ALT scheme, (2) the parametric ALT with design modifications, and (3) determine if the product design attains the mission cycles. To show the effectiveness of this methodology for identifying and altering the design defects of a mechanical product, we derived the sample size equation, time-to-failure equation and acceleration method, and BX lifetime. As a test case, we inspected the redesign of a compressor that came from the marketplace. • In the marketplace and the first ALT, we found that the failed suction reed valve in a compressor originated from the following design flaws: (1) a small overlap between the valve with the valve plate, (2) a sharp edge on the valve plate, and (3) inadequate width of the suction reed valve (SANDVIK20 C0.178t). As design modifications, the trepan on the valve plate was extended from 0.73 mm to 1.25 mm, and a ball peening technique was secured to eliminate the sharp edge. For the second ALT, we found that three samples failed near 17,000 cycles because of the intrusion between the crankshaft and thrust washer. As an action plan, the crankshaft was heat-treated. • During the third ALT, no problems were found. The compressor systems should attain the lifetime target—B1 life of 10 years with about a 60% confidence level. • By examining problematic products returned from the marketplace and performing parametric ALTs with design modifications, it was possible to improve the expected lifetime of the compressor. This reliability methodology can be applied to other mechanical products such as airplane, automobiles, and construction machines. To use this methodology, designers should understand why products fail during their lifetime. That is to say, if there are design flaws in the structure that are subjected to repetitive loads, the system will fail before its expected lifetime. Engineers would need to identify the load characteristics of a mechanical system so that the parametric ALT can be carried out until the required mission cycles under accelerated stress conditions. Finally, engineers can use parametric ALT to identify and alter the design problems of a mechanical product.

Author Contributions: S.W. managed the concept forming, methodology, examination and testing, and wrote the manuscript. D.L.O. examined the analysis and writing for the original manuscript and edited it. All authors have read and agreed to the published version of the manuscript. Funding: This research accepted no external finance. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: The data provided in this research can be obtained on request from the corresponding author. Conflicts of Interest: The authors express no conflict of interest.

Abbreviations

A Cross-sectional area of the capillary tube: cm2 BX Time which is a cumulated failure rate of X%: durability index Ea Activation energy, eV e Effort f Flow F Impact force, kN F(t) Unreliability h Testing cycles (or cycles) ∗ h* Non-dimensional testing cycles, h = h/LB ≥ 1 J Junction function equation Metals 2021, 11, 1261 18 of 21

K Boltzmann’s constant, 8.62 × 10−5 eV/deg ∆L Capillary tube distance in the two-phase LB Target BX life and x = 0.01X, on the condition that x ≤ 0.2 n Number of test samples ∆P Pressure difference between the condenser and evaporator, MPa PD Volume flow rate in compressor, m3/s Pc Pressure in the condenser, MPa Pe Pressure in the evaporator, MPa Psuc Pressure at compressor suction, MPa Pdis Pressure at compressor discharge, MPa Amount of energy absorbed or released during the reaction. For the Q semiconductor, total number of dopants per unit area Qc Heat transfer by temperature difference in the condenser, kW Qe Heat transfer by temperature difference in the evaporator, kW R Ratio for minimum stress to maximum stress in stress cycle, σmin/σmax r Failed numbers Rc Thermal resistance in the condenser, K/kW Re Thermal resistance in the evaporator, K/kW S Stress T Temperature, K Tc Absolute temperature in the condenser, K Te Absolute temperature in the evaporator, K ti Test time for each sample TF Time to failure X Accumulated failure rate, % x x = 0.01X, on condition that x ≤ 0.2 Wc Compressor power, kW Greek symbols ξ Electrical field exerted η Characteristic life λ Cumulative damage exponent in Palmgren–Miner’s rule χ2 Chi-square distribution α Confidence level 3 νsuc Specific volume of refrigerant at compressor suction, m /kg ρ Refrigerant density, kg/m3 ηv Volumetric efficiency ω Angular velocity, rad/s Superscripts β Shape parameter in Weibull distribution h ∂ ln(T ) i n Stress dependence, n = − f ∂ ln(S) T Subscripts 0 Normal stress conditions 1 Accelerated stress conditions

Appendix A. Derivation of Sample Size Equation for Parametric ALT Presently, many methods have been proposed to resolve issues on sample size. The Weibayes model by using Weibull examination is a widely acknowledged method of inspecting reliability data. However, it is hard to be straightly used due to the complexity of the mathematical formulation. The entire instances as failures (r ≥ 1) and no failures (r = 0) need to be split. As a result, it is feasible to attain a predictable sample size equation which might supply the mission cycle after correct presumptions. In selecting the model parameters to maximize the likelihood function, the maximum likelihood estimation (MLE) statistic is a general way of approximating the parameters of a model. The characteristic life ηMLE would be stated as:

n β β ti ηMLE = ∑ (A1) i=1 r Metals 2021, 11, 1261 19 of 21

where ηMLE is the maximum likelihood estimate for the characteristic life, n is the entire number of samples, ti is the experimental time for each sample, and r is the number of failures. If the failure numbers, r, greater than or equal to 1 and the conﬁdence level is 100 (1 − α), then the characteristic life, ηα, can be estimated from Equation (A1).

n β 2r β 2 β η = × η = × t for r ≥ 1 (A2) α 2 ( + ) MLE 2 ( + ) ∑ i χα 2r 2 χα 2r 2 i=1

2 where χα() is the chi-square distribution when the p-value is α. Presuming there are no failure numbers, ln (1/α) is similar to the chi-square value, 2 χα(2) 2 [35].

− x ν −1 ! − x ν −1 ! Z ∝ e 2 x 2 Z ∝ e 2 x 2 p − value : α = ν dx = ν dxfor x ≥ 0 (A3) χ2 (2) 2 ν 2 ln α−1 2 ν α 2 Γ 2 2 Γ 2 where ν is the shape parameter and Γ is the gamma function. For r = 0, the characteristic life ηα from Equation (A2) might be stated as:

n n β 2 β 1 β η = × t = × t (A4) α χ2 (2) ∑ i 1 ∑ i α i=1 ln α i=1

As Equation (A2) is demonstrated for all cases r ≥ 0, characteristic life, ηα, might be stated as: n β 2 β η = × t for r ≥ 0 (A5) α 2 ( + ) ∑ i χα 2r 2 i=1 If the logarithm in the Weibull distribution is used, the relationship between characteristic life and BX life, LB, might be stated as: β 1 β L = ln × η (A6) B 1 − x α

If the estimated characteristic life of the p-value α, ηα, in Equation (A5), is substituted into Equation (A6), we obtain the BX life equation:

n β 1 2 β L = ln × × t (A7) B − 2 ( + ) ∑ i 1 x χα 2r 2 i=1

As whole reliability testing has insufﬁcient sample numbers to assess the lifetime for the designated failures, which might be less than that of the sample size, the experiment plan can be stated as: β β β nh ≥ ∑ ti ≥ (n − r) × h (A8) If Equation (A8) is changed with Equation (A7), the BX life equation might be restated as: 1 2 1 2 ∗ β ∼ × · β ≥ × × ( − ) β ≥ β LB = ln 2 nh ln 2 n r h LB (A9) 1 − x χα(2r + 2) 1 − x χα(2r + 2) If Equation (A9) is reorganized, the sample size equation with the failures can be stated as: 2 ∗ β χα(2r + 2) 1 LB n ≥ × × + r (A10) 2 1 h ln 1−x Metals 2021, 11, 1261 20 of 21

2 2 3 χα(2r+2) ∼ −1 x x ∼ Because 2 = (r + 1) for α = 0.6 and ln(1 − x) = x + 2 + 3 + ... = x, the sample size Equation (A10) can be straightforwardly adjacent to:

1 L∗ β n ≥ (r + 1) × × B + r (A11) x h

where the sample size equation can be stated as n ~ (failure numbers + 1)·(1/cumulative failure rate)·((target lifetime)/(plan testing time)) ˆ β + r.

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