下载文档原格式
附着强度英语Adhesive StrengthThe concept of adhesive strength is a fundamental aspect of material science and engineering, with far-reaching implications in various industries. Adhesives are materials that facilitate the bonding of two or more surfaces, creating a strong and durable connection. The strength of this bond, known as adhesive strength, is a critical factor in determining the overall performance and reliability of the joined components.Adhesive strength is influenced by a multitude of factors, including the chemical composition of the adhesive, the surface properties of the materials being bonded, the environmental conditions, and the applied stresses. Understanding and optimizing these factors is crucial for ensuring the integrity and longevity of adhesive-based assemblies.One of the primary determinants of adhesive strength is the chemical composition of the adhesive itself. Adhesives can be classified into various categories, such as epoxies, acrylics, silicones, and polyurethanes, each with its own unique properties andperformance characteristics. The choice of adhesive depends on the specific application, the materials being bonded, and the desired level of strength and durability.Epoxy adhesives, for instance, are known for their exceptional strength and resistance to environmental factors, making them a popular choice for applications in the aerospace, automotive, and construction industries. These adhesives form a strong covalent bond with the substrate, resulting in a high-strength connection that can withstand significant stresses and loads.Acrylic adhesives, on the other hand, offer a more versatile and flexible bonding solution. They are commonly used in the electronics and consumer goods industries, where the ability to bond a wide range of materials, including plastics and metals, is essential. Acrylic adhesives typically exhibit good impact resistance and can accommodate slight movements or deformations in the bonded components.Silicone adhesives, in contrast, are prized for their excellent resistance to high temperatures, weathering, and chemical exposure. They are often employed in applications where these environmental factors are a concern, such as in the automotive, aerospace, and construction industries. Silicone adhesives can form strong bonds with a variety of substrates, including glass, metal, and plastic.Polyurethane adhesives, meanwhile, are known for their superior flexibility and impact resistance. They are commonly used in the construction and transportation industries, where the ability to accommodate movement and vibration is crucial. Polyurethane adhesives can form strong bonds with a wide range of materials, including wood, concrete, and various plastics.The surface properties of the materials being bonded also play a significant role in determining the adhesive strength. The surface roughness, wettability, and chemical composition of the substrates can influence the adhesive's ability to form a strong and durable bond. Surface preparation techniques, such as cleaning, etching, or priming, can be employed to enhance the adhesion properties of the materials, thereby improving the overall adhesive strength.Environmental factors, such as temperature, humidity, and exposure to chemicals or UV radiation, can also have a significant impact on the adhesive strength. Adhesives may experience degradation or weakening over time due to these environmental stresses, leading to a reduction in the overall bond strength. Understanding and mitigating these environmental factors is essential for ensuring the long-term performance and reliability of adhesive-based assemblies.The applied stresses on the bonded assembly are another criticalfactor in determining the adhesive strength. Adhesives can experience various types of stresses, including tensile, shear, peel, and cleavage stresses, each of which can have a different effect on the bond strength. Designing the bonded assembly to minimize the impact of these stresses, through the use of appropriate joint geometries and load-bearing configurations, can help maximize the adhesive strength and ensure the overall integrity of the assembly.In addition to these fundamental factors, advances in adhesive technology have led to the development of specialized adhesives that can further enhance the adhesive strength. For example, structural adhesives, which are designed to withstand high loads and stresses, are commonly used in the aerospace and automotive industries. These adhesives can provide superior strength and durability, often outperforming traditional mechanical fasteners in certain applications.Another example of advanced adhesive technology is the use of nanoparticle-reinforced adhesives. These adhesives incorporate nanoscale reinforcements, such as carbon nanotubes or graphene, which can significantly improve the adhesive strength, toughness, and thermal stability of the bond. The incorporation of these nanomaterials can lead to enhanced interfacial interactions between the adhesive and the substrate, resulting in a stronger and more resilient bond.In conclusion, adhesive strength is a critical factor in the design and performance of a wide range of products and structures. Understanding the factors that influence adhesive strength, such as the chemical composition of the adhesive, the surface properties of the materials being bonded, environmental conditions, and applied stresses, is essential for ensuring the reliability and durability of adhesive-based assemblies. Advances in adhesive technology, including the development of specialized structural adhesives and nanoparticle-reinforced adhesives, have further expanded the capabilities and applications of adhesive bonding in various industries.。
刘容旭，李春雨，王语聪，等. 超高压辅助酶解法改性汉麻分离蛋白及其理化性质的研究[J]. 食品工业科技，2023，44（19）：99−107.doi: 10.13386/j.issn1002-0306.2023010016LIU Rongxu, LI Chunyu, WANG Yucong, et al. Study on the Modification and Physicochemical Properties of Hemp Protein Isolate by Ultra-High Pressure Assisted Enzymatic Hydrolysis[J]. Science and Technology of Food Industry, 2023, 44(19): 99−107. (in Chinese with English abstract). doi: 10.13386/j.issn1002-0306.2023010016· 研究与探讨 ·超高压辅助酶解法改性汉麻分离蛋白及其理化性质的研究刘容旭1，李春雨2，王语聪2，谢智鑫2，谢宜桐2，李双鹏2，刘丹怡1, *，韩建春2,*（1.黑龙江省绿色食品科学研究院，黑龙江哈尔滨 150028；2.东北农业大学 食品学院，黑龙江哈尔滨 150030）摘 要：本研究以汉麻分离蛋白（Hemp Protein Isolate ，HPI ）为原料，通过超高压辅助酶解反应对HPI 进行改性，以溶解度和水解度为判定指标筛选酶解改性反应最佳条件，并探究超高压辅助酶解反应对酶解产物溶解性、起泡性、乳化性、持水性、持油性的影响。
结果表明，HPI 酶解反应最适条件为：加酶量（复合蛋白酶）5000 U/g 、酶解改性pH8.0、酶解改性温度55 ℃、酶解改性时间50 min 。
以HPI 为对照，当压力为200 MPa 时，酶解产物的溶解度、起泡性、乳化性、持油性最高，压力为100 MPa 时，泡沫稳定性最好，酶解后的乳化稳定性存在不同程度的下降，压力为0.1 MPa 时其持水性达到最大值。
如何解决疲劳效应英语作文Title: Strategies to Combat Fatigue Effect。
Fatigue effect, the depletion of mental and physical energy over time, poses a significant challenge to individuals in various aspects of life, be it academic, professional, or personal. Addressing this issue requires a multifaceted approach encompassing lifestyle adjustments, cognitive strategies, and self-care practices. In this essay, we will delve into effective methods to tackle fatigue effect and enhance overall well-being.Firstly, establishing a balanced lifestyle is paramount in combating fatigue. Adequate sleep is fundamental for replenishing energy levels and sustaining cognitive function. Research suggests that adults should aim for 7-9 hours of sleep per night for optimal health. Additionally, maintaining a consistent sleep schedule, even on weekends, helps regulate the body's internal clock, promoting better sleep quality and reducing daytime fatigue.Moreover, incorporating regular physical activity into one's routine is beneficial for combating fatigue. Exercise stimulates the release of endorphins, neurotransmittersthat elevate mood and energy levels. Engaging in activities such as jogging, yoga, or swimming not only enhances physical fitness but also boosts mental clarity and alertness. Even short bouts of exercise throughout the day can mitigate feelings of fatigue and improve overall productivity.In conjunction with lifestyle modifications, adopting cognitive strategies can mitigate the impact of fatigue on cognitive function. Time management techniques, such as the Pomodoro Technique, involve breaking tasks into manageable intervals separated by short breaks. This method prevents burnout and enhances focus by capitalizing on the brain's natural rhythm of attention.Furthermore, implementing mindfulness practices can counteract the cognitive effects of fatigue. Mindfulness meditation, characterized by non-judgmental awareness ofthe present moment, promotes mental clarity and resilience to stress. Studies have shown that regular meditation reduces fatigue and enhances cognitive flexibility, enabling individuals to navigate challenges with greater ease.In addition to lifestyle and cognitive interventions, self-care practices play a crucial role in combatingfatigue effect. Nutrition plays a pivotal role in sustaining energy levels throughout the day. Consuming a balanced diet rich in whole grains, lean proteins, fruits, and vegetables provides essential nutrients that support optimal brain function and mitigate fatigue.Moreover, practicing self-compassion and setting realistic expectations are vital components of self-care. Perfectionism and excessive self-criticism can contribute to burnout and exacerbate feelings of fatigue. Cultivating self-compassion involves treating oneself with kindness and understanding, particularly during periods of heightened stress or fatigue.Furthermore, establishing boundaries and prioritizing self-care activities is essential for preventing burnout. Carving out time for leisure activities, social connections, and relaxation fosters resilience and replenishes depleted energy stores. Whether it's reading a book, spending time with loved ones, or engaging in hobbies, prioritizing activities that bring joy and fulfillment is essential for maintaining well-being.In conclusion, addressing fatigue effect necessitates a comprehensive approach encompassing lifestyle adjustments, cognitive strategies, and self-care practices. Byprioritizing sleep, exercise, mindfulness, nutrition, and self-compassion, individuals can mitigate the impact of fatigue and cultivate resilience in the face of challenges. Empowering oneself with effective coping mechanisms is keyto sustaining energy levels and optimizing overall well-being.。
第44卷第6期2021年6月V ol.44，No.6June2021核技术NUCLEAR TECHNIQUES熔盐堆低功率工况下反应性引入事故初始条件敏感性探讨焦小伟王凯王超群杨群何兆忠（中国科学院上海应用物理研究所上海201800）摘要熔盐堆低功率工况反应性引入事故中，不同的反应性引入速率将触发不同的停堆信号。
同时反应堆初始功率和反应性温度系数等初始条件影响事故的进程，引起事故后果的差异。
本文选取了7个反应性引入速率工况、25个初始功率水平和反应性温度系数的参数组合初始工况，分别讨论了这三个参数对事故后果的影响。
分析结果表明：熔盐堆低功率工况反应性引入事故的后果对反应性引入速率的变化较敏感，在其他初始条件一定的情况下，存在特定的反应性引入速率会导致最不利的事故后果；事故后果对反应堆初始功率和反应性温度系数的变化不敏感，由初始功率和反应性温度系数差异造成的事故后果差异较小。
关键词熔盐堆，低功率，反应性引入事故，敏感性中图分类号TL36DOI:10.11889/j.0253-3219.2021.hjs.44.060602Study on sensitivity of initial conditions of reactivity initiated accident under low powerconditions of molten salt reactorJIAO Xiaowei WANG Kai WANG Chaoqun YANG Qun HE Zhaozhong(Shanghai Institute of Applied Physics,Chinese Academy of Sciences,Shanghai201800,China)Abstract[Background]In the reactivity initiated accidents under low power operating conditions of molten salt reactor(MSR),different reactivity insertion rates will trigger different emergency shutdown signals.At the same time,the initial conditions such as the initial reactor power and the temperature coefficients of reactivity affect the accident process and cause differences in accident consequences.[Purpose]The study aims to conduct a sensitivity analysis of the impact of the reactivity insertion rate,the initial reactor power,and the reactivity temperature coefficient on transient consequences.[Methods]First of all,7reactivity insertion rate conditions were selected and simulated through RELAP5-TMSR.Then,25combinations of the initial reactor power and the temperature coefficients of reactivity were assumed as initial conditions.Finally,the effects of these three parameters on the consequences of the accident were discussed separately by using local sensitivity analysis method.[Results]The insertion rate that causes a concurrent trigger of the high outlet temperature and the high-power shutdown signal leads to the most unfavorable consequence.The difference between the peak temperatures of the fuel salt and structural materials and their respective initial values under the worst reactivity insertion rate condition is negatively correlated with initial power.However,the temperature difference of each parameter caused by different initial power does not中国科学院青年创新促进会项目(No.Y929022031)资助第一作者：焦小伟，男，1989年出生，2019年于中国科学院大学获博士学位，副研究员，主要从事反应堆事故分析通信作者：杨群，E-mail：收稿日期：2021-01-14，修回日期：2021-03-29Supported by the Project of Youth Innovation Promotion Association of Chinese Academy of Sciences(No.Y929022031)First author:JIAO Xiaowei,male,born in1989,graduated from University of Chinese Academy of Sciences with a doctoral degree in2019,associate professor,focusing on reactor safetyCorresponding author:YANG Qun,E-mail:Received date:2021-01-14,revised date:2021-03-29焦小伟等：熔盐堆低功率工况下反应性引入事故初始条件敏感性探讨exceed3℃.The difference between the peak temperatures decrease first and then increases with the increase of the temperature coefficients of reactivity,but the maximum difference does not exceed0.5℃.[Conclusions]Under low power operating conditions of MSR,the consequences of reactivity introduced events are highly sensitive to the reactivity insertion rate and low sensitivity to the initial power and temperature coefficients of reactivity.Key words Molten salt reactor,Low power,Reactive initiated accident,Sensitivity熔盐堆（Molten Salt Reactor，MSR）是第四代核能系统候选堆型之一。
The Effect of Temperature on ProteinConformationProteins are essential components of living organisms and are responsible for carrying out various cellular functions. They are composed of long chains of amino acids that are folded into intricate 3-dimensional structures. The specific shape of a protein, or its conformation, plays a critical role in its function. Temperature is one of the key factors that can influence protein conformation. In this article, we will explore the effect of temperature on protein conformation and how it impacts their function.Temperature-induced protein denaturationProtein denaturation is a process in which the protein loses its native conformation and unfolds into a linear or random coil structure. This process can be triggered by several factors, including pH, salts, mechanical stress, and temperature. Among these, temperature is the most commonly studied factor that can induce protein denaturation.When proteins are exposed to high temperatures, the thermal energy causes the bonds that hold the protein structure together to break. Hydrogen bonds, which are weaker than covalent bonds, are the first to be broken. As the temperature continues to rise, the more significant covalent bonds that hold the protein together begin to break, further destabilizing the structure. Ultimately, the protein loses its native conformation, and its function is impaired.The effect of temperature on protein stabilityThe stability of a protein refers to its ability to maintain its native conformation in the face of various environmental conditions, including temperature. The stability of a protein is influenced by several factors, including the amino acid sequence, solvent conditions, and the presence of ligands or cofactors. Temperature can disrupt the stability of a protein by altering its structure and causing it to denature.Proteins have a range of thermal stability that depends on their amino acid sequence and their specific structure. Generally, proteins that are stable at higher temperatures have a higher content of hydrophobic amino acids, which can help to stabilize the structure through hydrophobic interactions. In contrast, proteins that are stable at lower temperatures tend to have more polar amino acids and a lower content of hydrophobic amino acids.The temperature at which a protein denatures is known as its melting temperature or Tm. The Tm of a protein is influenced by its intrinsic stability as well as the specific conditions under which it is studied. For example, the pH, salt concentration, and presence of other molecules can all affect the Tm of a protein.The effect of temperature on protein functionThe specific conformation of a protein plays a critical role in its function. Therefore, changes in protein conformation due to temperature can have a significant impact on their function. The effect of temperature on protein function can vary depending on the specific protein and the conditions under which it is studied.Some proteins are more sensitive to changes in temperature than others. For example, enzymes, which catalyze chemical reactions in the cell, have a specific optimal temperature range at which they function best. Outside of this range, the reaction rate can slow down or even stop altogether due to changes in protein conformation.Other proteins, such as transporters and receptors, are also sensitive to changes in temperature. Changes in protein conformation due to temperature can affect the ability of these proteins to bind to their ligands and carry out their function.ConclusionIn conclusion, temperature has a significant impact on protein conformation. High temperatures can cause proteins to denature, while changes in temperature can alter their stability and affect their function. Understanding the effect of temperature on protein conformation and function is essential for designing experiments and developing new drugs and therapies that target specific proteins.。
压力交变试验英语Stress Alternating TestStress is an integral part of our daily lives, and it can have a significant impact on our physical and mental well-being. The ability to manage stress effectively is crucial for maintaining a healthy and balanced lifestyle. One method of assessing and understanding the effects of stress is through the use of a stress alternating test.The stress alternating test is a technique used to evaluate an individual's response to varying levels of stress. The test involves exposing the subject to a series of controlled stress-inducing situations, alternating between high and low-stress conditions. This approach allows researchers and clinicians to observe how the body and mind react to the fluctuations in stress levels.During the stress alternating test, the subject may be asked to perform a variety of tasks or engage in different scenarios that are designed to elicit specific stress responses. For example, the subject may be required to solve complex mathematical problems under time pressure, engage in public speaking, or participate in simulated high-stakes decision-making scenarios. The test may also involveexposing the subject to environmental stressors, such as loud noises, extreme temperatures, or challenging physical activities.As the subject navigates through these varying stress conditions, researchers closely monitor a range of physiological and psychological parameters. These may include heart rate, blood pressure, cortisol levels, skin conductance, and self-reported measures of anxiety, mood, and cognitive performance. By analyzing these data points, researchers can gain insights into how the individual's body and mind respond to the fluctuations in stress levels.One of the key benefits of the stress alternating test is its ability to provide a comprehensive understanding of an individual's stress resilience. By observing how the subject's physiological and psychological responses change in response to the alternating stress levels, researchers can identify patterns and identify potential areas of vulnerability or strength. This information can then be used to develop personalized stress management strategies and interventions.Moreover, the stress alternating test can also be valuable in the context of occupational health and safety. Many professions, such as emergency services, military, and high-stakes decision-making roles, often require individuals to navigate high-stress situations regularly.The stress alternating test can be used to assess the suitability and resilience of individuals for these demanding roles, ensuring that they are equipped to handle the challenges they may face.In addition to its practical applications, the stress alternating test also contributes to our scientific understanding of the human stress response. By studying the physiological and psychological reactions to varying stress levels, researchers can gain insights into the underlying mechanisms that govern the stress response. This knowledge can then be used to develop more effective strategies for stress management and prevention, ultimately improving overall health and well-being.In conclusion, the stress alternating test is a valuable tool for assessing and understanding the effects of stress on the human body and mind. By exposing individuals to controlled stress-inducing situations and monitoring their responses, researchers and clinicians can gain valuable insights that can be used to develop personalized stress management strategies and interventions. As we continue to navigate the challenges of modern life, the stress alternating test will likely play an increasingly important role in promoting health, well-being, and resilience.。
Unit 3 Science and nature Section Ⅲ Word power，Task & ProjectⅠ品句填词1．Use your good ____________(判定) before you decide.答案：judgement2．The ____________ (多数) of students find it hard to accept the new theory.答案：majority3．From his ____________(可怕的) look，I guess he must have found something horrible.答案：frightened4．No one is to leave the building without my ____________(许诺)．答案：permission5．Please reenter your password to____________(确认) it is correct.答案：confirmⅡ单句改错1．I’d say she is pretty rich，judge from her →judging2．The project，conducting by our headmaster，has won the students’→conducted3．There was frost on the ground，confirmed that fall had arrived in →confirming4．This frighten boy whose mother was lost in the disaster is looking for her →frightened 5．We view the holi day for a time for recreation，but she has a different idea.第一个for→as Ⅲ完成句子1．就我个人而言，这部电影糟糕透顶。
鲶鱼效应可以有效促进竞争英语作文The Catfish Effect stimulates competition in various fields, propelling individuals to achieve greater accomplishments. This phenomenon describes the notion that if a small entity enters a larger competitive environment, it can motivate the established players to work harder and achieve better results. This effect can be observed in different domains such as sports, education, technology, and business.In sports, the Catfish Effect is often witnessed when a talented newcomer emerges and challenges the established athletes. The presence of this new talent motivates the existing players to train harder, improve their skills, and maintain their position at the top. The competition created by the Catfish Effect drives athletes to push beyond their limits, breaking records and achieving remarkable feats.Similarly, in the field of education, the Catfish Effect can enhance competition among students. When a new student with exceptional abilities joins a class, it inspires others to strive for excellence. The existing students realize the need to work harder, study more rigorously, and actively participate in academic activities to maintain their academic standing. This competitive environment fosters a culture of continuous improvement and enhances overall academic performances.The Catfish Effect is also evident in the realm of technology. When a small startup with an innovative idea enters an industry dominated by established companies, it often disrupts the market. The established companies recognize the potential of the new entrant and are motivated to innovate and develop better products or services in order to maintain their market share. This healthy competition drives technological advancements, benefiting consumers with improved offerings.Moreover, in the world of business, the Catfish Effect fuels healthy competition among companies. When a new player emerges with a unique value proposition or a disruptive business model, it compels existing companies to reassess their strategies and find ways to stay competitive. Established companies may invest in research and development, improve their products or services, or enhance customer experiences to successfully counter the competition andretain their market position.In conclusion, the Catfish Effect is a catalyst for competition in various domains. It motivates individuals and entities to step up their game, work harder, and strive for excellence. This effect has a positive impact on sports, education, technology, and business, fostering innovation and enhancing overall performances. The presence of new players creates an environment of healthy competition, pushing everyone to reach their full potential.。
琴弦不能太紧也不能太松英语作文The Delicate Balance of String Tension.In the realm of musical instruments, the strings that produce the enchanting melodies and harmonies are finely tuned to a specific tension. This tension plays a crucial role in determining the instrument's tone, intonation, and overall playability.The Effects of String Tension on Tone.String tension has a direct impact on the pitch of a musical note. When a string is tighter, it vibrates at a higher frequency, producing a higher sound. Conversely, a looser string vibrates at a lower frequency, resulting in a lower pitch. This principle is evident across all stringed instruments, from violins to guitars to pianos.Moreover, string tension affects the timbre, or tonal quality, of the sound produced. Tighter strings tend toproduce a brighter, more piercing tone, while looserstrings often yield a warmer, more mellow sound. This variation in timbre is due to the different overtone series generated by strings of different tensions.The Importance of Intonation.String tension also plays a vital role in intonation, the accuracy of pitch intervals. When the strings are tuned to the correct tension, they will produce notes that are perfectly in tune with each other. However, if the strings are too tight or too loose, the notes will be out of tune, making it difficult to produce harmonious intervals and melodies.Playability and String Tension.The tension of the strings also affects theinstrument's playability. Tighter strings require more force to press down and produce a clear sound, which can be physically demanding for musicians, especially in extended playing sessions. On the other hand, looser strings areeasier to play but may lack the responsiveness and articulation required for certain musical styles.Finding the Optimal Balance.The optimal string tension for a musical instrument depends on various factors, including the size and construction of the instrument, the type of music being played, and the musician's personal preferences. Finding the right balance requires experimentation and careful consideration.Generally, strings that are too tight can cause excessive stress on the instrument, leading to damage or premature wear. They can also result in a sharp, metallic sound that lacks warmth and depth. Conversely, strings that are too loose may produce a dull, lifeless tone and poor intonation.Conclusion.The tension of musical strings is a delicate balancethat must be carefully maintained to achieve the desired tone, intonation, and playability. Musicians must strive to find the optimal string tension for their instruments, taking into account the specific characteristics of the instrument, the musical genre, and their own individual needs. By mastering this aspect of their craft, they can unlock the full potential of their musical endeavors and create truly captivating performances.。
New1H-Pyrazole-Containing Polyamine Receptors Able ToComplex L-Glutamate in Water at Physiological pH ValuesCarlos Miranda,†Francisco Escartı´,‡Laurent Lamarque,†Marı´a J.R.Yunta,§Pilar Navarro,*,†Enrique Garcı´a-Espan˜a,*,‡and M.Luisa Jimeno†Contribution from the Instituto de Quı´mica Me´dica,Centro de Quı´mica Orga´nica Manuel Lora Tamayo,CSIC,C/Juan de la Cier V a3,28006Madrid,Spain,Departamento de Quı´mica Inorga´nica,Facultad de Quı´mica,Uni V ersidad de Valencia,c/Doctor Moliner50, 46100Burjassot(Valencia),Spain,and Departamento de Quı´mica Orga´nica,Facultad deQuı´mica,Uni V ersidad Complutense de Madrid,A V plutense s/n,28040Madrid,SpainReceived April16,2003;E-mail:enrique.garcia-es@uv.esAbstract:The interaction of the pyrazole-containing macrocyclic receptors3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene1[L1],13,26-dibenzyl-3,6,9,12,13,16,-19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene2[L2],3,9,12,13,16,22,-25,26-octaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene3[L3],6,19-dibenzyl-3,6,9,12,13,-16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene4[L4],6,19-diphenethyl-3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene5[L5],and 6,19-dioctyl-3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetra-ene6[L6]with L-glutamate in aqueous solution has been studied by potentiometric techniques.The synthesis of receptors3-6[L3-L6]is described for the first time.The potentiometric results show that4[L4]containing benzyl groups in the central nitrogens of the polyamine side chains is the receptor displaying the larger interaction at pH7.4(K eff)2.04×104).The presence of phenethyl5[L5]or octyl groups6[L6]instead of benzyl groups4[L4]in the central nitrogens of the chains produces a drastic decrease in the stability[K eff )3.51×102(5),K eff)3.64×102(6)].The studies show the relevance of the central polyaminic nitrogen in the interaction with glutamate.1[L1]and2[L2]with secondary nitrogens in this position present significantly larger interactions than3[L3],which lacks an amino group in the center of the chains.The NMR and modeling studies suggest the important contribution of hydrogen bonding andπ-cation interaction to adduct formation.IntroductionThe search for the L-glutamate receptor field has been andcontinues to be in a state of almost explosive development.1 L-Glutamate(Glu)is thought to be the predominant excitatory transmitter in the central nervous system(CNS)acting at a rangeof excitatory amino acid receptors.It is well-known that it playsa vital role mediating a great part of the synaptic transmission.2However,there is an increasing amount of experimentalevidence that metabolic defects and glutamatergic abnormalitiescan exacerbate or induce glutamate-mediated excitotoxic damageand consequently neurological disorders.3,4Overactivation ofionotropic(NMDA,AMPA,and Kainate)receptors(iGluRs)by Glu yields an excessive Ca2+influx that produces irreversible loss of neurons of specific areas of the brain.5There is much evidence that these processes induce,at least in part,neuro-degenerative illnesses such as Parkinson,Alzheimer,Huntington, AIDS,dementia,and amyotrophic lateral sclerosis(ALS).6In particular,ALS is one of the neurodegenerative disorders for which there is more evidence that excitotoxicity due to an increase in Glu concentration may contribute to the pathology of the disease.7Memantine,a drug able to antagonize the pathological effects of sustained,but relatively small,increases in extracellular glutamate concentration,has been recently received for the treatment of Alzheimer disease.8However,there is not an effective treatment for ALS.Therefore,the preparation of adequately functionalized synthetic receptors for L-glutamate seems to be an important target in finding new routes for controlling abnormal excitatory processes.However,effective recognition in water of aminocarboxylic acids is not an easy task due to its zwitterionic character at physiological pH values and to the strong competition that it finds in its own solvent.9†Centro de Quı´mica Orga´nica Manuel Lora Tamayo.‡Universidad de Valencia.§Universidad Complutense de Madrid.(1)Jane,D.E.In Medicinal Chemistry into the Millenium;Campbell,M.M.,Blagbrough,I.S.,Eds.;Royal Society of Chemistry:Cambridge,2001;pp67-84.(2)(a)Standaert,D.G.;Young,A.B.In The Pharmacological Basis ofTherapeutics;Hardman,J.G.,Goodman Gilman,A.,Limbird,L.E.,Eds.;McGraw-Hill:New York,1996;Chapter22,p503.(b)Fletcher,E.J.;Loge,D.In An Introduction to Neurotransmission in Health and Disease;Riederer,P.,Kopp,N.,Pearson,J.,Eds.;Oxford University Press:New York,1990;Chapter7,p79.(3)Michaelis,E.K.Prog.Neurobiol.1998,54,369-415.(4)Olney,J.W.Science1969,164,719-721.(5)Green,J.G.;Greenamyre,J.T.Prog.Neurobiol.1996,48,613-63.(6)Bra¨un-Osborne,H.;Egebjerg,J.;Nielsen,E.O.;Madsen,U.;Krogsgaard-Larsen,P.J.Med.Chem.2000,43,2609-2645and references therein.(7)(a)Shaw,P.J.;Ince,P.G.J.Neurol.1997,244(Suppl2),S3-S14.(b)Plaitakis,A.;Fesdjian,C.O.;Shashidharan,S Drugs1996,5,437-456.(8)Frantz,A.;Smith,A.Nat.Re V.Drug Dico V ery2003,2,9.Published on Web12/30/200310.1021/ja035671m CCC:$27.50©2004American Chemical Society J.AM.CHEM.SOC.2004,126,823-8339823There are many types of receptors able to interact with carboxylic acids and amino acids in organic solvents,10-13yielding selective complexation in some instances.However,the number of reported receptors of glutamate in aqueous solution is very scarce.In this sense,one of the few reports concerns an optical sensor based on a Zn(II)complex of a 2,2′:6′,2′′-terpyridine derivative in which L -aspartate and L -glutamate were efficiently bound as axial ligands (K s )104-105M -1)in 50/50water/methanol mixtures.14Among the receptors employed for carboxylic acid recogni-tion,the polyamine macrocycles I -IV in Chart 1are of particular relevance to this work.In a seminal paper,Lehn et al.15showed that saturated polyamines I and II could exert chain-length discrimination between different R ,ω-dicarboxylic acids as a function of the number of methylene groups between the two triamine units of the receptor.Such compounds were also able to interact with a glutamic acid derivative which has the ammonium group protected with an acyl moiety.15,16Compounds III and IV reported by Gotor and Lehn interact in their protonated forms in aqueous solution with protected N -acetyl-L -glutamate and N -acetyl-D -glutamate,showing a higher stability for the interaction with the D -isomer.17In both reports,the interaction with protected N -acetyl-L -glutamate at physiological pH yields constants of ca.3logarithmic units.Recently,we have shown that 1H -pyrazole-containing mac-rocycles present desirable properties for the binding of dopam-ine.18These polyaza macrocycles,apart from having a highpositive charge at neutral pH values,can form hydrogen bonds not only through the ammonium or amine groups but also through the pyrazole nitrogens that can behave as hydrogen bond donors or acceptors.In fact,Elguero et al.19have recently shown the ability of the pyrazole rings to form hydrogen bonds with carboxylic and carboxylate functions.These features can be used to recognize the functionalities of glutamic acid,the carboxylic and/or carboxylate functions and the ammonium group.Apart from this,the introduction of aromatic donor groups appropriately arranged within the macrocyclic framework or appended to it through arms of adequate length may contribute to the recognition event through π-cation interactions with the ammonium group of L -glutamate.π-Cation interactions are a key feature in many enzymatic centers,a classical example being acetylcholine esterase.20The role of such an interaction in abiotic systems was very well illustrated several years ago in a seminal work carried out by Dougherty and Stauffer.21Since then,many other examples have been reported both in biotic and in abiotic systems.22Taking into account all of these considerations,here we report on the ability of receptors 1[L 1]-6[L 6](Chart 2)to interact with L -glutamic acid.These receptors display structures which differ from one another in only one feature,which helps to obtain clear-cut relations between structure and interaction(9)Rebek,J.,Jr.;Askew,B.;Nemeth,D.;Parris,K.J.Am.Chem.Soc.1987,109,2432-2434.(10)Seel,C.;de Mendoza,J.In Comprehensi V e Supramolecular Chemistry ;Vogtle,F.,Ed.;Elsevier Science:New York,1996;Vol.2,p 519.(11)(a)Sessler,J.L.;Sanson,P.I.;Andrievesky,A.;Kral,V.In SupramolecularChemistry of Anions ;Bianchi,A.,Bowman-James,K.,Garcı´a-Espan ˜a,E.,Eds.;John Wiley &Sons:New York,1997;Chapter 10,pp 369-375.(b)Sessler,J.L.;Andrievsky,A.;Kra ´l,V.;Lynch,V.J.Am.Chem.Soc.1997,119,9385-9392.(12)Fitzmaurice,R.J.;Kyne,G.M.;Douheret,D.;Kilburn,J.D.J.Chem.Soc.,Perkin Trans.12002,7,841-864and references therein.(13)Rossi,S.;Kyne,G.M.;Turner,D.L.;Wells,N.J.;Kilburn,J.D.Angew.Chem.,Int.Ed.2002,41,4233-4236.(14)Aı¨t-Haddou,H.;Wiskur,S.L.;Lynch,V.M.;Anslyn,E.V.J.Am.Chem.Soc.2001,123,11296-11297.(15)Hosseini,M.W.;Lehn,J.-M.J.Am.Chem.Soc.1982,104,3525-3527.(16)(a)Hosseini,M.W.;Lehn,J.-M.Hel V .Chim.Acta 1986,69,587-603.(b)Heyer,D.;Lehn,J.-M.Tetrahedron Lett.1986,27,5869-5872.(17)(a)Alfonso,I.;Dietrich,B.;Rebolledo,F.;Gotor,V.;Lehn,J.-M.Hel V .Chim.Acta 2001,84,280-295.(b)Alfonso,I.;Rebolledo,F.;Gotor,V.Chem.-Eur.J.2000,6,3331-3338.(18)Lamarque,L.;Navarro,P.;Miranda,C.;Ara ´n,V.J.;Ochoa,C.;Escartı´,F.;Garcı´a-Espan ˜a,E.;Latorre,J.;Luis,S.V.;Miravet,J.F.J.Am.Chem.Soc .2001,123,10560-10570.(19)Foces-Foces,C.;Echevarria,A.;Jagerovic,N.;Alkorta,I.;Elguero,J.;Langer,U.;Klein,O.;Minguet-Bonvehı´,H.-H.J.Am.Chem.Soc.2001,123,7898-7906.(20)Sussman,J.L.;Harel,M.;Frolow,F.;Oefner,C.;Goldman,A.;Toker,L.;Silman,I.Science 1991,253,872-879.(21)Dougherty,D.A.;Stauffer,D.A.Science 1990,250,1558-1560.(22)(a)Sutcliffe,M.J.;Smeeton,A.H.;Wo,Z.G.;Oswald,R.E.FaradayDiscuss.1998,111,259-272.(b)Kearney,P.C.;Mizoue,L.S.;Kumpf,R.A.;Forman,J.E.;McCurdy,A.;Dougherty,D.A.J.Am.Chem.Soc.1993,115,9907-9919.(c)Bra ¨uner-Osborne,H.;Egebjerg,J.;Nielsen,E.;Madsen,U.;Krogsgaard-Larsen,P.J.Med.Chem.2000,43,2609-2645.(d)Zacharias,N.;Dougherty,D.A.Trends Pharmacol.Sci.2002,23,281-287.(e)Hu,J.;Barbour,L.J.;Gokel,G.W.J.Am.Chem.Soc.2002,124,10940-10941.Chart 1.Some Receptors Employed for Dicarboxylic Acid and N -AcetylglutamateRecognitionChart 2.New 1H -Pyrazole-Containing Polyamine Receptors Able To Complex L -Glutamate inWaterA R T I C L E SMiranda et al.824J.AM.CHEM.SOC.9VOL.126,NO.3,2004strengths.1[L1]and2[L2]differ in the N-benzylation of the pyrazole moiety,and1[L1]and3[L3]differ in the presence in the center of the polyamine side chains of an amino group or of a methylene group.The receptors4[L4]and5[L5]present the central nitrogens of the chain N-functionalized with benzyl or phenethyl groups,and6[L6]has large hydrophobic octyl groups.Results and DiscussionSynthesis of3-6.Macrocycles3-6have been obtained following the procedure previously reported for the preparation of1and2.23The method includes a first dipodal(2+2) condensation of the1H-pyrazol-3,5-dicarbaldehyde7with the corresponding R,ω-diamine,followed by hydrogenation of the resulting Schiff base imine bonds.In the case of receptor3,the Schiff base formed by condensation with1,5-pentanediamine is a stable solid(8,mp208-210°C)which precipitated in68% yield from the reaction mixture.Further reduction with NaBH4 in absolute ethanol gave the expected tetraazamacrocycle3, which after crystallization from toluene was isolated as a pure compound(mp184-186°C).In the cases of receptors4-6, the precursor R,ω-diamines(11a-11c)(Scheme1B)were obtained,by using a procedure previously described for11a.24 This procedure is based on the previous protection of the primary amino groups of1,5-diamino-3-azapentane by treatment with phthalic anhydride,followed by alkylation of the secondary amino group of1,5-diphthalimido-3-azapentane9with benzyl, phenethyl,or octyl bromide.Finally,the phthalimido groups of the N-alkyl substituted intermediates10a-10c were removed by treatment with hydrazine to afford the desired amines11a-11c,which were obtained in moderate yield(54-63%).In contrast with the behavior previously observed in the synthesis of3,in the(2+2)dipodal condensations of7with 3-benzyl-,3-phenethyl-,and3-octyl-substituted3-aza-1,5-pentanediamine11a,11b,and11c,respectively,there was not precipitation of the expected Schiff bases(Scheme1A). Consequently,the reaction mixtures were directly reduced in situ with NaBH4to obtain the desired hexaamines4-6,which after being carefully purified by chromatography afforded purecolorless oils in51%,63%,and31%yield,respectively.The structures of all of these new cyclic polyamines have been established from the analytical and spectroscopic data(MS(ES+), 1H and13C NMR)of both the free ligands3-6and their corresponding hydrochloride salts[3‚4HCl,4‚6HCl,5‚6HCl, and6‚6HCl],which were obtained as stable solids following the same procedure previously reported18for1‚6HCl and2‚6HCl.As usually occurs for3,5-disubstituted1H-pyrazole deriva-tives,either the free ligands3-6or their hydrochlorides show very simple1H and13C NMR spectra,in which signals indicate that,because of the prototropic equilibrium of the pyrazole ring, all of these compounds present average4-fold symmetry on the NMR scale.The quaternary C3and C5carbons appear together,and the pairs of methylene carbons C6,C7,and C8are magnetically equivalent(see Experimental Section).In the13C NMR spectra registered in CDCl3solution, significant differences can be observed between ligand3,without an amino group in the center of the side chain,and the N-substituted ligands4-6.In3,the C3,5signal appears as a broad singlet.However,in4-6,it almost disappears within the baseline of the spectra,and the methylene carbon atoms C6and C8experience a significant broadening.Additionally,a remark-able line-broadening is also observed in the C1′carbon signals belonging to the phenethyl and octyl groups of L5and L6, respectively.All of these data suggest that as the N-substituents located in the middle of the side chains of4-6are larger,the dynamic exchange rate of the pyrazole prototropic equilibrium is gradually lower,probably due to a relation between proto-tropic and conformational equilibria.Acid-Base Behavior.To follow the complexation of L-glutamate(hereafter abbreviated as Glu2-)and its protonated forms(HGlu-,H2Glu,and H3Glu+)by the receptors L1-L6, the acid-base behavior of L-glutamate has to be revisited under the experimental conditions of this work,298K and0.15mol dm-3.The protonation constants obtained,included in the first column of Table1,agree with the literature25and show that the zwitterionic HGlu-species is the only species present in aqueous solution at physiological pH values(Scheme2and Figure S1of Supporting Information).Therefore,receptors for(23)Ara´n,V.J.;Kumar,M.;Molina,J.;Lamarque,L.;Navarro,P.;Garcı´a-Espan˜a,E.;Ramı´rez,J.A.;Luis,S.V.;Escuder,.Chem.1999, 64,6137-6146.(24)(a)Yuen Ng,C.;Motekaitis,R.J.;Martell,A.E.Inorg.Chem.1979,18,2982-2986.(b)Anelli,P.L.;Lunazzi,L.;Montanari,F.;Quici,.Chem.1984,49,4197-4203.Scheme1.Synthesis of the Pyrazole-Containing MacrocyclicReceptorsNew1H-Pyrazole-Containing Polyamine Receptors A R T I C L E SJ.AM.CHEM.SOC.9VOL.126,NO.3,2004825glutamate recognition able to address both the negative charges of the carboxylate groups and the positive charge of ammonium are highly relevant.The protonation constants of L 3-L 6are included in Table 1,together with those we have previously reported for receptors L 1and L 2.23A comparison of the constants of L 4-L 6with those of the nonfunctionalized receptor L 1shows a reduced basicity of the receptors L 4-L 6with tertiary nitrogens at the middle of the polyamine bridges.Such a reduction in basicity prevented the potentiometric detection of the last protonation for these ligands in aqueous solution.A similar reduction in basicity was previously reported for the macrocycle with the N -benzylated pyrazole spacers (L 2).23These diminished basicities are related to the lower probability of the tertiary nitrogens for stabilizing the positive charges through hydrogen bond formation either with adjacent nonprotonated amino groups of the molecule or with water molecules.Also,the increase in the hydrophobicity of these molecules will contribute to their lower basicity.The stepwise basicity constants are relatively high for the first four protonation steps,which is attributable to the fact that these protons can bind to the nitrogen atoms adjacent to the pyrazole groups leaving the central nitrogen free,the electrostatic repulsions between them being therefore of little significance.The remaining protonation steps will occur in the central nitrogen atom,which will produce an important increase in the electrostatic repulsion in the molecule and therefore a reduction in basicity.As stated above,the tertiary nitrogen atoms present in L 4-L 6will also contribute to this diminished basicity.To analyze the interaction with glutamic acid,it is important to know the protonation degree of the ligands at physiological pH values.In Table 2,we have calculated the percentages ofthe different protonated species existing in solution at pH 7.4for receptors L 1-L 6.As can be seen,except for the receptor with the pentamethylenic chains L 3in which the tetraprotonated species prevails,all of the other systems show that the di-and triprotonated species prevail,although to different extents.Interaction with Glutamate.The stepwise constants for the interaction of the receptors L 1-L 6with glutamate are shown in Table 3,and selected distribution diagrams are plotted in Figure 1A -C.All of the studied receptors interact with glutamate forming adduct species with protonation degrees (j )which vary between 8and 0depending on the system (see Table 3).The stepwise constants have been derived from the overall association constants (L +Glu 2-+j H +)H j LGlu (j -2)+,log j )provided by the fitting of the pH-metric titration curves.This takes into account the basicities of the receptors and glutamate (vide supra)and the pH range in which a given species prevails in solution.In this respect,except below pH ca.4and above pH 9,HGlu -can be chosen as the protonated form of glutamate involved in the formation of the different adducts.Below pH 4,the participation of H 2Glu in the equilibria has also to be considered (entries 9and 10in Table 3).For instance,the formation of the H 6LGlu 4+species can proceed through the equilibria HGlu -+H 5L 5+)H 6LGlu 4+(entry 8,Table 3),and H 2Glu +H 4L 4+)H 6LGlu 4(entry 9Table 3),with percentages of participation that depend on pH.One of the effects of the interaction is to render somewhat more basic the receptor,and somewhat more acidic glutamic acid,facilitating the attraction between op-positely charged partners.A first inspection of Table 3and of the diagrams A,B,and C in Figure 1shows that the interaction strengths differ markedly from one system to another depending on the structural features of the receptors involved.L 4is the receptor that presents the highest capacity for interacting with glutamate throughout all of the pH range explored.It must also be remarked that there are not clear-cut trends in the values of the stepwise constants as a function of the protonation degree of the receptors.This suggests that charge -charge attractions do not play the most(25)(a)Martell,E.;Smith,R.M.Critical Stability Constants ;Plenum:NewYork,1975.(b)Motekaitis,R.J.NIST Critically Selected Stability Constants of Metal Complexes Database ;NIST Standard Reference Database,version 4,1997.Table 1.Protonation Constants of Glutamic Acid and Receptors L 1-L 6Determined in NaCl 0.15mol dm -3at 298.1KreactionGluL 1aL 2aL 3bL 4L 5L 6L +H )L H c 9.574(2)d 9.74(2)8.90(3)9.56(1)9.25(3)9.49(4)9.34(5)L H +H )L H 2 4.165(3)8.86(2)8.27(2)8.939(7)8.38(3)8.11(5)8.13(5)L H 2+H )L H 3 2.18(2)7.96(2) 6.62(3)8.02(1) 6.89(5)7.17(6)7.46(7)L H 3+H )L H 4 6.83(2) 5.85(4)7.63(1) 6.32(5) 6.35(6) 5.97(8)L H 4+H )L H 5 4.57(3) 3.37(4) 2.72(8) 2.84(9) 3.23(9)L H 5+H )L H 6 3.18(3) 2.27(6)∑log K H n L41.135.334.233.634.034.1aTaken from ref 23.b These data were previously cited in a short communication (ref 26).c Charges omitted for clarity.d Values in parentheses are the standard deviations in the last significant figure.Scheme 2.L -Glutamate Acid -BaseBehaviorTable 2.Percentages of the Different Protonated Species at pH 7.4H 1L aH 2LH 3LH 4LL 11186417L 21077130L 3083458L 4083458L 51154323L 6842482aCharges omitted for clarity.A R T I C L E SMiranda et al.826J.AM.CHEM.SOC.9VOL.126,NO.3,2004outstanding role and that other forces contribute very importantly to these processes.26However,in systems such as these,which present overlapping equilibria,it is convenient to use conditional constants because they provide a clearer picture of the selectivity trends.27These constants are defined as the quotient between the overall amounts of complexed species and those of free receptor and substrate at a given pH[eq1].In Figure2are presented the logarithms of the effective constants versus pH for all of the studied systems.Receptors L1and L2with a nonfunctionalized secondary amino group in the side chains display opposite trend from all other receptors. While the stability of the L1and L2adducts tends to increase with pH,the other ligands show a decreasing interaction. Additionally,L1and L2present a close interaction over the entire pH range under study.The tetraaminic macrocycle L3is a better(26)Escartı´,F.;Miranda,C.;Lamarque,L.;Latorre,J.;Garcı´a-Espan˜a,E.;Kumar,M.;Ara´n,V.J.;Navarro,mun.2002,9,936-937.(27)(a)Bianchi,A.;Garcı´a-Espan˜a,c.1999,12,1725-1732.(b)Aguilar,J.A.;Celda,B.;Garcı´a-Espan˜a,E.;Luis,S.V.;Martı´nez,M.;Ramı´rez,J.A.;Soriano,C.;Tejero,B.J.Chem.Soc.,Perkin Trans.22000, 7,1323-1328.Table3.Stability Constants for the Interaction of L1-L6with the Different Protonated Forms of Glutamate(Glu) entry reaction a L1L2L3L4L5L6 1Glu+L)Glu L 3.30(2)b 4.11(1)2HGlu+L)HGlu L 3.65(2) 4.11(1) 3.68(2) 3.38(4) 3Glu+H L)HGlu L 3.89(2) 4.48(1) 3.96(2) 3.57(4) 4HGlu+H L)H2Glu L 3.49(2) 3.89(1) 2.37(4) 3.71(2)5HGlu+H2L)H3Glu L 3.44(2) 3.73(1) 2.34(3) 4.14(2) 2.46(4) 2.61(7) 6HGlu+H3L)H4Glu L 3.33(2) 3.56(2) 2.66(3) 4.65(2) 2.74(3) 2.55(7) 7HGlu+H4L)H5Glu L 3.02(2) 3.26(2) 2.58(3) 4.77(2) 2.87(3) 2.91(5) 8HGlu+H5L)H6Glu L 3.11(3) 3.54(2) 6.76(3) 4.96(3) 4.47(3) 9H2Glu+H4L)H6Glu L 2.54(3) 3.05(2) 3.88(2) 5.35(3) 3.66(4) 3.56(3) 10H2Glu+H5L)H7Glu L 2.61(6) 2.73(4) 5.51(3) 3.57(4) 3.22(8) 11H3Glu+H4L)H7Glu L 4.82(2) 4.12(9)a Charges omitted for clarity.b Values in parentheses are standard deviations in the last significantfigure.Figure1.Distribution diagrams for the systems(A)L1-glutamic acid, (B)L4-glutamic acid,and(C)L5-glutamicacid.Figure2.Representation of the variation of K cond(M-1)for the interaction of glutamic acid with(A)L1and L3,(B)L2,L4,L5,and L6.Initial concentrations of glutamate and receptors are10-3mol dm-3.Kcond)∑[(H i L)‚(H j Glu)]/{∑[H i L]∑[H j Glu]}(1)New1H-Pyrazole-Containing Polyamine Receptors A R T I C L E SJ.AM.CHEM.SOC.9VOL.126,NO.3,2004827receptor at acidic pH,but its interaction markedly decreases on raising the pH.These results strongly suggest the implication of the central nitrogens of the lateral polyamine chains in the stabilization of the adducts.Among the N-functionalized receptors,L4presents the largest interaction with glutamate.Interestingly enough,L5,which differs from L4only in having a phenethyl group instead of a benzyl one,presents much lower stability of its adducts.Since the basicity and thereby the protonation states that L4and L5 present with pH are very close,the reason for the larger stability of the L4adducts could reside on a better spatial disposition for formingπ-cation interactions with the ammonium group of the amino acid.In addition,as already pointed out,L4presents the highest affinity for glutamic acid in a wide pH range,being overcome only by L1and L2at pH values over9.This observation again supports the contribution ofπ-cation inter-actions in the system L4-glutamic because at these pH values the ammonium functionality will start to deprotonate(see Scheme2and Figure1B).Table4gathers the percentages of the species existing in equilibria at pH7.4together with the values of the conditional constant at this pH.In correspondence with Figure1A,1C and Figure S2(Supporting Information),it can be seen that for L1, L2,L5,and L6the prevailing species are[H2L‚HGlu]+and[H3L‚HGlu]2+(protonation degrees3and4,respectively),while for L3the main species are[H3L‚HGlu]+and[H4L‚HGlu]2+ (protonation degrees4and5,respectively).The most effective receptor at this pH would be L4which joins hydrogen bonding, charge-charge,andπ-cation contributions for the stabilization of the adducts.To check the selectivity of this receptor,we have also studied its interaction with L-aspartate,which is a competitor of L-glutamate in the biologic receptors.The conditional constant at pH7.4has a value of3.1logarithmic units for the system Asp-L4.Therefore,the selectivity of L4 for glutamate over aspartate(K cond(L4-glu)/K cond(L4-asp))will be of ca.15.It is interesting to remark that the affinity of L4 for zwiterionic L-glutamate at pH7.4is even larger than that displayed by receptors III and IV(Chart1)with the protected dianion N-acetyl-L-glutamate lacking the zwitterionic charac-teristics.Applying eq1and the stability constants reported in ref17,conditional constants at pH7.4of 3.24and 2.96 logarithmic units can be derived for the systems III-L-Glu and IV-L-Glu,respectively.Molecular Modeling Studies.Molecular mechanics-based methods involving docking studies have been used to study the binding orientations and affinities for the complexation of glutamate by L1-L6receptors.The quality of a computer simulation depends on two factors:accuracy of the force field that describes intra-and intermolecular interactions,and an adequate sampling of the conformational and configuration space of the system.28The additive AMBER force field is appropriate for describing the complexation processes of our compounds,as it is one of the best methods29in reproducing H-bonding and stacking stabiliza-tion energies.The experimental data show that at pH7.4,L1-L6exist in different protonation states.So,a theoretical study of the protonation of these ligands was done,including all of the species shown in5%or more abundance in the potentiometric measurements(Table4).In each case,the more favored positions of protons were calculated for mono-,di-,tri-,and tetraprotonated species.Molecular dynamics studies were performed to find the minimum energy conformations with simulated solvent effects.Molecular modeling studies were carried out using the AMBER30method implemented in the Hyperchem6.0pack-age,31modified by the inclusion of appropriate parameters. Where available,the parameters came from analogous ones used in the literature.32All others were developed following Koll-man33and Hopfinger34procedures.The equilibrium bond length and angle values came from experimental values of reasonable reference compounds.All of the compounds were constructed using standard geometry and standard bond lengths.To develop suitable parameters for NH‚‚‚N hydrogen bonding,ab initio calculations at the STO-3G level35were used to calculate atomic charges compatible with the AMBER force field charges,as they gave excellent results,and,at the same time,this method allows the study of aryl-amine interactions.In all cases,full geometry optimizations with the Polak-Ribiere algorithm were carried out,with no restraints.Ions are separated far away and well solvated in water due to the fact that water has a high dielectric constant and hydrogen bond network.Consequently,there is no need to use counteri-ons36in the modelization studies.In the absence of explicit solvent molecules,a distance-dependent dielectric factor quali-tatively simulates the presence of water,as it takes into account the fact that the intermolecular electrostatic interactions should vanish more rapidly with distance than in the gas phase.The same results can be obtained using a constant dielectric factor greater than1.We have chosen to use a distance-dependent dielectric constant( )4R ij)as this was the method used by Weiner et al.37to develop the AMBER force field.Table8 shows the theoretical differences in protonation energy(∆E p) of mono-,bi-,and triprotonated hexaamine ligands,for the (28)Urban,J.J.;Cronin,C.W.;Roberts,R.R.;Famini,G.R.J.Am.Chem.Soc.1997,119,12292-12299.(29)Hobza,P.;Kabelac,M.;Sponer,J.;Mejzlik,P.;Vondrasek,put.Chem.1997,18,1136-1150.(30)Cornell,W.D.;Cieplak,P.;Bayly,C.I.;Gould,I.R.;Merz,K.M.,Jr.;Ferguson,D.M.;Spelmeyer,D.C.;Fox,T.;Caldwell,J.W.;Kollman,P.A.J.Am.Chem.Soc.1995,117,5179-5197.(31)Hyperchem6.0(Hypercube Inc.).(32)(a)Fox,T.;Scanlan,T.S.;Kollman,P.A.J.Am.Chem.Soc.1997,119,11571-11577.(b)Grootenhuis,P.D.;Kollman,P.A.J.Am.Chem.Soc.1989,111,2152-2158.(c)Moyna,G.;Hernandez,G.;Williams,H.J.;Nachman,R.J.;Scott,put.Sci.1997,37,951-956.(d)Boden,C.D.J.;Patenden,put.-Aided Mol.Des.1999, 13,153-166.(33)/amber.(34)Hopfinger,A.J.;Pearlstein,put.Chem.1984,5,486-499.(35)Glennon,T.M.;Zheng,Y.-J.;Le Grand,S.M.;Shutzberg,B.A.;Merz,K.M.,put.Chem.1994,15,1019-1040.(36)Wang,J.;Kollman,P.A.J.Am.Chem.Soc.1998,120,11106-11114.Table4.Percentages of the Different Protonated Adducts[HGlu‚H j L](j-1)+,Overall Percentages of Complexation,andConditional Constants(K Cond)at pH7.4for the Interaction ofGlutamate(HGlu-)with Receptors L1-L6at Physiological pH[H n L‚HGlu]an)1n)2n)3n)4∑{[H n L‚HGlu]}K cond(M-1)L13272353 2.44×103L2947763 4.12×103L31101324 3.99×102L423737581 2.04×104L51010222 3.51×102L6121224 3.64×102a Charges omitted for clarity.A R T I C L E S Miranda et al. 828J.AM.CHEM.SOC.9VOL.126,NO.3,2004。
应力对电化学的滞后作用英语Electrochemical Lag Effects of Stress.Stress can have a significant impact on the electrochemical processes that occur in materials. These effects can be either positive or negative, depending on the specific material and the type of stress applied. In some cases, stress can lead to an increase in electrochemical activity, while in other cases it can lead to a decrease.One of the most common ways that stress affects electrochemical processes is by altering the microstructure of the material. When a material is subjected to stress,its atoms and molecules are forced to move and rearrange themselves. This can lead to changes in the material's grain size, crystal structure, and other microstructural features. These changes can, in turn, affect the material's electrochemical properties.For example, a study by the University of California, Berkeley found that stress can increase the corrosion rate of steel. The researchers found that when steel was subjected to tensile stress, its grain boundaries became more active and more susceptible to corrosion. This led to an increase in the overall corrosion rate of the steel.In other cases, stress can have the opposite effect and decrease electrochemical activity. A study by theUniversity of Tokyo found that stress can decrease the rate of hydrogen evolution from platinum. The researchers found that when platinum was subjected to compressive stress, its surface became less active and less likely to react with hydrogen ions. This led to a decrease in the rate of hydrogen evolution.The effects of stress on electrochemical processes can be complex and depend on a number of factors, including the type of material, the type of stress applied, and the environmental conditions. However, it is clear that stress can have a significant impact on the electrochemical properties of materials.Applications of Electrochemical Lag Effects of Stress.The electrochemical lag effects of stress can be usedin a variety of applications. One common application is in the field of corrosion control. By understanding how stress affects the corrosion rate of materials, engineers can design materials and structures that are more resistant to corrosion.Another application of the electrochemical lag effects of stress is in the field of energy storage. By understanding how stress affects the electrochemicalactivity of materials, scientists can develop new and more efficient energy storage devices.Conclusion.The electrochemical lag effects of stress are a complex and fascinating phenomenon. By understanding these effects, scientists and engineers can develop new materials and technologies that are more resistant to corrosion, moreefficient at energy storage, and more responsive to external stimuli.。
《材料科学与工程基础》英文习题及思考题及答案第二章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l ＝0 corresponds to an s subshelll ＝1 corresponds to a p subshelll ＝2 corresponds to a d subshelll ＝3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell: 3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible,explain why.10.20 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heatedfrom a temperature of 1300_C .(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to completemelting?10.28 .Is it possible to have a copper–silver alloy of composition 50 wt% Ag–50 wt%Cu, which, at equilibrium, consists of _ and _ phases having mass fractions W_ _0.60 and W_ _ 0.40? If so, what will be the approximate temperature of the alloy?If such an alloy is not possible, explain why.10.30 At 700_C , what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?第三章习题和思考题3.3If the atomic radius of aluminum is 0.143nm, calculate the volume of its unitcell in cubic meters.3.8 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomicweight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover.3.9 Calculate the radius of an iridium atom given that Ir has an FCC crystal structure,a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.3.13 Using atomic weight, crystal structure, and atomic radius data tabulated insidethe front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.3.15 Below are listed the atomic weight, density, and atomic radius for threehypothetical alloys. For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. A simple cubic unitcell is shown in Figure 3.40.3.21 This is a unit cell for a hypotheticalmetal:(a) To which crystal system doesthis unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its atomic weight is 141g/mol.3.25 For a ceramic compound, what are the two characteristics of the component ionsthat determine the crystal structure?3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for thefollowing materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.3.35 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.(a) Determine the unit cell edge length. (b) How does this result compare withthe edge length as determined from the radii in Table 3.4, assuming that theMg2_ and O2_ ions just touch each other along the edges?3.36 Compute the theoretical density of diamond given that the CUC distance andbond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density?3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it isknown that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3 , how many Cd 2+ and S 2—ions are there per unit cell?3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively.On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).3.42 The unit cell for Mg Fe2O3 (MgO-Fe2O3) has cubic symmetry with a unit celledge length of 0.836 nm. If the density of this material is 4.52 g/cm 3 , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.3.44 Compute the atomic packing factor for the diamond cubic crystal structure(Figure 3.16). Assume that bonding atoms touch one another, that the angle between adjacent bonds is 109.5°, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the unit cell edge length).3.45 Compute the atomic packing factor for cesium chloride using the ionic radii inTable 3.4 and assuming that the ions touch along the cube diagonals.3.46 In terms of bonding, explain why silicate materials have relatively low densities.3.47 Determine the angle between covalent bonds in an SiO44—tetrahedron.3.63 For each of the following crystal structures, represent the indicated plane in themanner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100)plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.3.66 The zinc blende crystal structure is one that may be generated from close-packedplanes of anions.(a) Will the stacking sequence for this structure be FCC or HCP? Why?(b) Will cations fill tetrahedral or octahedral positions? Why?(c) What fraction of the positions will be occupied?3.81* The metal iridium has an FCC crystal structure. If the angle of diffraction forthe (220) set of planes occurs at 69.22°(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute(a) the interplanar spacing for this set of planes, and (b) the atomic radius for aniridium atom.4.10 What is the difference between configuration and conformation in relation topolymer chains? vinyl chloride).4.22 (a) Determine the ratio of butadiene to styrene mers in a copolymer having aweight-average molecular weight of 350,000 g/mol and weight-average degree of polymerization of 4425.(b) Which type(s) of copolymer(s) will this copolymer be, considering thefollowing possibilities: random, alternating, graft, and block? Why?4.23 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylenemay have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both mer types.4.25 (a) Compare the crystalline state in metals and polymers.(b) Compare thenoncrystalline state as it applies to polymers and ceramic glasses.4.26 Explain briefly why the tendency of a polymer to crystallize decreases withincreasing molecular weight.4.27* For each of the following pairs of polymers, do the following: (1) state whetheror not it is possible to determine if one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.(a) Linear and syndiotactic polyvinyl chloride; linear and isotactic polystyrene.(b) Network phenol-formaldehyde; linear and heavily crosslinked ci s-isoprene.(c) Linear polyethylene; lightly branched isotactic polypropylene.(d) Alternating poly(styrene-ethylene) copolymer; randompoly(vinylchloride-tetrafluoroethylene) copolymer.4.28 Compute the density of totally crystalline polyethylene. The orthorhombic unitcell for polyethylene is shown in Figure 4.10; also, the equivalent of two ethylene mer units is contained within each unit cell.5.11 What point defects are possible for MgO as an impurity in Al2O3? How manyMg 2+ ions must be added to form each of these defects?5.13 What is the composition, in weight percent, of an alloy that consists of 6 at% Pband 94 at% Sn?5.14 Calculate the composition, in weight per-cent, of an alloy that contains 218.0 kgtitanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.5.23 Gold forms a substitutional solid solution with silver. Compute the number ofgold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 , respectively.8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite)will not be an elastomer.10.50 Compute the mass fractions of αferrite and cementite in pearlite. assumingthat pressure is held constant.10.52 (a) What is the distinction between hypoeutectoid and hypereutectoid steels?(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explainthe difference between them. What will be the carbon concentration in each?10.56 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727_C(a) What is the proeutectoid phase?(b) How many kilograms each of total ferrite and cementite form?(c) How many kilograms each of pearlite and the proeutectoid phase form?(d) Schematically sketch and label the resulting microstructure.10.60 The mass fractions of total ferrite and total cementite in an iron–carbon alloyare 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?Why?10.64 Is it possible to have an iron–carbon alloy for which the mass fractions of totalferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why orwhy not?第四章习题和思考题7.3 A specimen of aluminum having a rectangular cross section 10 mm _ 12.7 mmis pulled in tension with 35,500 N force, producing only elastic deformation. 7.5 A steel bar 100 mm long and having a square cross section 20 mm on an edge ispulled in tension with a load of 89,000 N , and experiences an elongation of 0.10 mm . Assuming that the deformation is entirely elastic, calculate the elasticmodulus of the steel.7.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa ,and the modulus of elasticity is 115 Gpa .(a) What is the maximum load that may be applied to a specimen with across-sectional area of 325mm, without plastic deformation?(b) If the original specimen length is 115 mm , what is the maximum length towhich it may be stretched without causing plastic deformation?7.8 A cylindrical rod of copper (E _ 110 GPa, Stress (MPa) ) having a yield strengthof 240Mpa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an elongation of 0.50 mm?7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10mm in diameterand 75 mm long that is pulled in tension. Determine its elongation when a load of 23,500 N is applied.7.16 A cylindrical specimen of some alloy 8 mm in diameter is stressed elasticallyin tension. A force of 15,700 N produces a reduction in specimen diameter of 5 _ 10_3 mm. Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa .7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression.If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 Gpa and 39.7 GPa,respectively.7.19 A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of380 MPa, and an elastic modulus of 103 GPa . A cylindrical specimen of thisalloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm . On the basis of the information given, is it possible tocompute the magnitude of the load that is necessary to produce this change inlength? If so, calculate the load. If not, explain why.7.20A cylindrical metal specimen 15.0mmin diameter and 150mm long is to besubjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic.(a)If the elongation must be less than 0.072mm,which of the metals in Tabla7.1are suitable candidates? Why ?(b)If, in addition, the maximum permissible diameter decrease is 2.3×10-3mm,which of the metals in Table 7.1may be used ? Why?7.22 Cite the primary differences between elastic, anelastic, and plastic deformationbehaviors.7.23 diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It mustnot experience either plastic deformation or a diameter reduction of more than7.5×10-3 mm. Of the materials listed as follows, which are possible candidates?Justify your choice(s).7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to besubjected to a tensile load. If the rod is to experience neither plastic deformationnor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy.(a) What is the modulus of elasticity?(b) What is the proportional limit?(c) What is the yield strength at a strain offset of 0.002?(d) What is the tensile strength?7.27 A load of 44,500 N is applied to a cylindrical specimen of steel (displaying thestress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm .(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm), how much will it increase inlength when t his load is applied?7.29 A cylindrical specimen of aluminumhaving a diameter of 12.8 mm and a gaugelength of 50.800 mm is pulled in tension. Usethe load–elongation characteristics tabulatedbelow to complete problems a through f.(a)Plot the data as engineering stressversusengineering strain.(b) Compute the modulus of elasticity.(c) Determine the yield strength at astrainoffset of 0.002.(d) Determine the tensile strength of thisalloy.(e) What is the approximate ductility, in percent elongation?(f ) Compute the modulus of resilience.7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for atypical metal alloy.(b) Superimpose on this plot a schematic curve for the compressive truestress–strain behavior for the same alloy. Explain any difference between thiscurve and the one in part a.(c) Now superimpose a schematic curve for the compressive engineeringstress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b.7.39 A tensile test is performed on a metal specimen, and it is found that a true plasticstrain of 0.20 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the true strain that results from the application of a true stress of 600 Mpa.7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of0.475. How much will a specimen of this material elongate when a true stress of325 MPa is applied if the original length is 300 mm ? Assume a value of 0.25 for the strain-hardening exponent n.7.43 Find the toughness (or energy to cause fracture) for a metal that experiences bothelastic and plastic deformation. Assume Equation 7.5 for elastic deformation,that the modulus of elasticity is 172 GPa , and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 Mpa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2mm (0.75in×0.125in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500lbf ), then(a) Determine the elastic and plastic strain values.(b) If its original length is 460 mm (18 in.), what will be its final length after theload in part a is applied and then released?7.50 A three-point bending test was performed on an aluminum oxide specimenhaving a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm . Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm ?7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimenof aluminum oxide having a reported flexural strength of 390 MPa . If the speci- men radius is 2.5 mm and the support point separation distance is 30 mm ,predict whether or not you would expect the specimen to fracture when a load of 620 N is applied. Justify your prediction.(b) Would you be 100% certain of the prediction in part a? Why or why not?7.57 When citing the ductility as percent elongation for semicrystalline polymers, it isnot necessary to specify the specimen gauge length, as is the case with metals.Why is this so?7.66 Using the data represented in Figure 7.31, specify equations relating tensilestrength and Brinell hardness for brass and nodular cast iron, similar toEquations 7.25a and 7.25b for steels.8.4 For each of edge, screw, and mixed dislocations, cite the relationship between thedirection of the applied shear stress and the direction of dislocation line motion.8.5 (a) Define a slip system.(b) Do all metals have the same slip system? Why or why not?8.7. One slip system for theBCCcrystal structure is _110__111_. In a manner similarto Figure 8.6b sketch a _110_-type plane for the BCC structure, representingatom positions with circles. Now, using arrows, indicate two different _111_ slip directions within this plane.8.15* List four major differences between deformation by twinning and deformationby slip relative to mechanism, conditions of occurrence, and final result.8.18 Describe in your own words the three strengthening mechanisms discussed inthis chapter (i.e., grain size reduction, solid solution strengthening, and strainhardening). Be sure to explain how dislocations are involved in each of thestrengthening techniques.8.19 (a) From the plot of yield strength versus (grain diameter)_1/2 for a 70 Cu–30 Zncartridge brass, Figure 8.15, determine values for the constants _0 and ky inEquation 8.5.(b) Now predict the yield strength of this alloy when the average grain diameteris 1.0 _ 10_3 mm.8.20. The lower yield point for an iron that has an average grain diameter of 5 _ 10_2mm is 135 MPa . At a grain diameter of 8 _ 10_3 mm, the yield point increases to 260MPa. At what grain diameter will the lower yield point be 205 Mpa ?8.24 (a) Show, for a tensile test, thatif there is no change in specimen volume during the deformation process (i.e., A0 l0 _Ad ld).(b) Using the result of part a, compute the percent cold work experienced bynaval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied.8.25 Two previously undeformed cylindrical specimens of an alloy are to be strainhardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and11 mm, respectively. The second specimen, with an initial radius of 12 mm, musthave the same deformed hardness as the first specimen; compute the secondspecimen’s radius after deformation.8.26 Two previously undeformed specimens of the same metal are to be plasticallydeformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular is to remain as such. Their original and deformeddimensions are as follows:Which of these specimens will be the hardest after plastic deformation, and why?8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. Ifits coldworked radius is 10 mm (0.40 in.), what was its radius beforedeformation?8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strengthof 275 MPa ?(b) What is the approximate Brinell hardness of a 1040 steel having a yieldstrength of 690 MPa?8.41 In your own words, describe the mechanisms by which semicrystalline polymers(a) elasticallydeform and (b) plastically deform, and (c) by which elastomerselastically deform.8.42 Briefly explain how each of the following influences the tensile modulus of asemicrystallinepolymer and why:(a) molecular weight;(b) degree of crystallinity;(c) deformation by drawing;(d) annealing of an undeformed material;(e) annealing of a drawn material.8.43* Briefly explain how each of the following influences the tensile or yieldstrength of a semicrystalline polymer and why:(a) molecular weight;。
分析测试新成果 (39 ~ 46)惰气熔融-红外吸收/热导法同时测定无烟煤中氮和氢王 琳1，王 楠1，沈峰满2（1. 东北大学 分析测试中心，辽宁 沈阳 110819；2. 东北大学 冶金学院，辽宁 沈阳 110819）摘要：首次使用惰气熔融-红外吸收/热导法实现无烟煤中氮、氢元素的同时、快速、准确测定. 探究分析条件，发现当称样量为0.030 0 g ，分析功率为5 500 W ，氮元素的积分延迟时间为15 s ，集成时间为55 s ，氢元素的积分延迟时间为5 s ，集成时间为85 s ，且使用石墨套埚时，氮氢元素的释放最完全、合理. 方法中氮、氢校准曲线的相关系数分别为0.994 9、0.994 0，检出限分别为0.321%、0.189%，定量限分别为0.326%、0.194%，精密度分别为3.60%、0.63%，满足线性关系及方法要求. 惰气熔融-红外吸收/热导法重复性好、高效便捷、操作和维护简单，可用于无烟煤中氮、氢元素的定量检测.关键词：惰气熔融；红外吸收/热导法；无烟煤；氮；氢中图分类号：O657. 3 文献标志码：B 文章编号：1006-3757（2024）01-0039-08DOI ：10.16495/j.1006-3757.2024.01.007Simultaneous Determination of Nitrogen and Hydrogen in Anthracite by Inert Gas Melting-Infrared Absorption/Thermal Conductivity MethodWANG Lin 1， WANG Nan 1， SHEN Fengman2（1. Analysis and Measurement Centre , Northeastern University , Shenyang 110819, China ；2. School ofMetallurgy , Northeastern University , Shenyang 110819, China ）Abstract ：The contents of nitrogen and hydrogen in anthracite were simultaneously, rapidly and accurately determined by the inert gas melting-infrared absorption/thermal conductivity method. A series of experiments were studied. The results indicated that the most complete and reasonable release of nitrogen and hydrogen was achieved when the sample was 0.030 0 g, the analysis power was 5 500 W, the integration delay time of nitrogen was 15 s, the integration time of nitrogen was 55 s, the integration delay time of hydrogen was 5 s, the integration time of hydrogen was 85 s, and the graphite sleeve crucible was used. The correlation coefficients of calibration curves of nitrogen and hydrogen were 0.994 9and 0.994 0, respectively. The limits of detection were 0.321% and 0.189%, the limits of quantification were 0.326% and 0.194%, and the precision were 3.60% and 0.63%, respectively, which met the requirements of linearity and method. The inert gas melting-infrared absorption/thermal conductivity method is reproducible, efficient and convenient, easy to operate and maintain, and can be used for the quantitative determination of nitrogen and hydrogen in anthracite.Key words ：inert gas melting ；infrared absorption/thermal conductivity method ；anthracite ；nitrogen ；hydrogen自2020年我国提出碳达峰、碳中和的发展目标以来[1]，我国的能源、经济等发展始终围绕碳排放、绿色清洁等话题. 煤是工业原料之一，素来被称为“工业之母”，是世界工业、制造业、经济、民生等的重要支撑，其用途广泛，在新材料制备、化工生产、生活供暖、交通出行、发电等方面有着不可替代的作用. 我国属于煤矿矿产丰富的国家[2]，煤、石油、天然气是重要的能源，特点是“富煤、贫油、少气”[3].收稿日期：2023−10−11； 修订日期：2023−12−18.基金项目：国家自然科学基金资助项目 (51974073) [National Natural Science Foundation of China (51974073)]作者简介：王琳（1990−），女，实验师，主要从事气体成分分析等化学分析，E-mail ：****************.第 30 卷第 1 期分析测试技术与仪器Volume 30 Number 12024年1月ANALYSIS AND TESTING TECHNOLOGY AND INSTRUMENTS Jan. 2024煤根据品种及品质的不同，分为烟煤、无烟煤、焦炭等，并应用于不同行业，其中无烟煤因其燃烧无烟、煤化程度高、含碳量高、热值高、挥发分低等特点，普遍用于燃料及燃料电池、先进碳材料[4-7]、催化剂[8]、吸附剂[9-10]、滤料、民用煤等. 而据统计显示，我国空气污染源中的粉尘、PM2.5、SO2及NO x等大部分来自于民用煤燃烧的排放[11]，因此加强对无烟煤的质量监测，是提升煤炭质量、发展低碳与绿色能源的重要环节.煤炭的检测标准溯源到上世纪60年代，检测指标一般包括工业分析[12]（水分、灰分、挥发分、固定碳）、元素分析[13-15]（C、S、O、N、H）、有价元素分析[16-17]（As、Ga、Se、Ge等）、阴离子[18]（氟等）等. 其中无烟煤中的氮元素在燃烧后会形成NO x，对人类及居住环境污染影响较大[11]. 无烟煤中氢元素含量的多少，代表了热值的大小. 因此准确快速测定无烟煤中氮、氢含量对煤炭质量控制，煤炭行业的检验检测、标准制定、能源开发及环境保护等均具有重要意义.对于无烟煤中氮、氢元素的检测，通常使用半微量开氏法和半微量蒸汽法[19]、高温燃烧-检测器测定法[14, 20]测定无烟煤中的氮含量，采用三节炉法、二节炉法[13]、电量-重量法[21]、高温燃烧-检测器测定法[14]测定无烟煤中的氢含量. 其中三节炉法、二节炉法、电量-重量法均存在硫、氯等元素的干扰，需使用铬酸铅、银丝、二氧化锰等试剂消除干扰，污染较大且成本高. 随着科技的进步，仪器法逐渐被用于测定无烟煤中的氮、氢元素含量，现有的仪器法[22]原理是将无烟煤在氧气下燃烧，对燃烧生成的H2O、N2气体进行检测. 但该法存在燃烧炉/管升降温时间长、分析时间长、维护复杂、耗材昂贵等缺点. 而以惰气熔融-红外吸收/热导法为分析原理设计的氧氮氢分析仪通常用于陶瓷、粉末[23]、钢铁[24]等无机材料中氧、氮、氢元素的测定，并以快速、精准的优势成为冶金、材料等领域以及检验检测机构在气体元素分析方面的常用仪器. 但目前为止，未见其应用于无烟煤类产品的检测工作中，其在使用中无需强酸、重金属等试剂，具有无需等待升降温、分析时间短、样品前处理简易、维护相对简单等优势，满足绿色、安全、快速、准确分析的要求，因此本文首次尝试将惰气熔融-红外吸收/热导法应用于无烟煤中氮、氢元素的检测.1 试验部分1.1 仪器与试剂氧氮氢分析仪：美国力可公司，ONH836；天平：赛多利斯，SQP；石墨套埚（内坩埚加外坩埚）、石墨标准坩埚、镍嚢，LECO公司；有机元素分析仪：德国元素公司，Vario MACRO cube.氦气（99.999%），氮气（99.5%），沈阳顺泰特种气体有限公司；无烟煤标准物质：ZBM093、ZBW112A、ZBM095A，济南众标科技有限公司生产；GBW11104j，国家煤炭质量监督检验中心；GBW11108o，山东省冶金科学研究院. 对氨基苯磺酰胺（C6H8N2O2S）、WO3，德国元素公司；未知样品为某学生客户日常送检的无烟煤样品.1.2 试验原理在惰性气体氦气保护下，样品置于上下电极间的石墨坩埚中，经过坩埚脱气、吹扫、脉冲炉通电，上、下电极及石墨坩埚形成电路并加热，使待测样品完全熔融，N、H元素分别以N2、H2分子形式释放，随载气氦气流经热的氧化铜催化剂，H2被完全氧化成H2O，N2、H2O一起进入红外检测池，根据H2O的特征红外吸收波长，检测得到氢元素的含量，之后H2O被高氯酸镁等过滤试剂吸收，N2进入热导检测池完成氮元素的测定，其原理图如图1所示.样品上电级红外检测池检测 H2O热导检测池检测 N2坩埚下电极脉冲熔融炉N2N2H2催化剂H2OH2O图1 氧氮氢分析仪测定氮、氢的工作原理图Fig. 1 Working principle diagram ofOxygen/Nitrogen/Hydrogen Analyzer determined nitrogenand hydrogen1.3 试验方法1.3.1 准备工作将标准物质、待测样品置于110 ℃洁净的烘箱中烘干2 h，保证粒度在0.074 mm以下，然后再置40分析测试技术与仪器第 30 卷于干燥器中冷却备用.对氧氮氢分析仪进行彻底维护，包括上电极、下电极、投样口的清扫清洁，催化剂、过滤试剂等试剂的更换，并通过漏气检查，保证仪器的气密性.1.3.2 试验步骤打开稳压电源、氧氮氢分析仪主机及软件，将下电极升高，在氦气保护模式下进行仪器预热至少1 h，预热完成后打开氦气至流速为450 mL/min，开通冷却水，使检测器保持在稳定的工作温度. 本方法以镍嚢及空白石墨套锅作为空白，分别称取0.010 0~0.100 0 g（精确到±0.000 3 g）的样品，小心倾倒于镍嚢内，等待投样，设置4 500~6 000 W的分析功率，对比石墨套埚与石墨标准坩埚的分析效果，分别设置0~15 s的分析延迟时间、50~85 s数据集成时间等仪器参数. 开始测试后进行投放样品、取下坩埚、更换新的内坩埚、脱气、吹扫等操作，依次进行空白、标准物质及未知样品的测试，建立标准曲线，并对方法进行检出限、定量限、精密度等试验验证.1.3.3 未知样品对比试验本文使用有机元素分析仪作为未知样品测试的对比方法，并命名为方法1. 对有机元素分析仪（CHNS模式）的燃烧管进行清理并更换试剂及灰分坩埚，还原管内铜及银丝重新装填，酒精擦拭干净后放回到炉子内，通高纯氦气，流速为600 mL/min，室温检漏通过后，分别升至1 150、850 ℃工作温度下吹扫4 h后进行试验. 使用仪器自带标准曲线，以75 mg的锡纸包裹，称取25 mg的对氨基苯磺酰胺作为“run”和漂移标准物质进行曲线校正，待测样品称样量为50 mg，加入WO3助熔，75 mg锡纸包裹，使用工具压除空气后置于自动进样器中进样，试样在1 150 ℃下通高纯氧气燃烧，850 ℃下催化还原，释放出N2和H2O，进入相应检测池分析检测，经过“吹扫-捕集”吸附解析的分离过程，得到氮、氢的分析数据，完成检测.2 结果与讨论2.1 进样方式的确定本试验采用直投法进样，对于粉末类样品以此方式进样时，会造成进样系统污染、进样量减少、分析数据偏低等问题，为避免因进样造成的分析误差，需采用镍嚢作为样品包裹体，保证进样量的准确性及释放完全性.2.2 进样量的确定样品的进样量会影响熔融效果，使用标准物质ZBM095A作为待测样品，对比0.010 0、0.020 0、0.030 0、0.040 0、0.050 0、0.060 0、0.080 0、0.100 0 g 进样量对氮、氢元素释放效果的影响. 由图2可见，随着进样量的增加，氮质量比在进样量为0.010 0~ 0.030 0 g时的测定结果变化不大，而在0.0300 g时出现拐点呈下降趋势，随着进样量的继续增加，由于释放条件不足，氮质量比下降，因此氮的最佳进样量为0.0300 g. 氢质量比随进样量增加，先呈明显上升趋势，在进样量为0.030 0 g时，氢质量比达到了最高点，而随着进样量的继续增大，氢质量比缓慢降低，在进样量大于0.060 0 g时，氢质量比迅速下降. 由此可见，0.0300 g是其最佳进样量. 产生该现象的原因可能是进样量较低时，样品分析浓度不够，导致氢元素质量比偏低，而进样量过高时，样品的分析条件不足以使氢完全释放，氢元素质量比降低，且就仪器本身的检测范围而言，氢的测量上限绝对质量为0.002 5 g，因此对于标准物质ZBM095A 的氢元素质量比的测定，当进样量超过0.050 0 g时，检测池处于饱和状态，无法正常检测. 因此，0.030 0 g 为该方法的最佳进样质量.4.54.03.53.02.52.01.51.00.500.020 00.040 00.060 0NH0.080 00.100 0m/g质量比/%图2 不同进样量下氮、氢的测试结果Fig. 2 Test results of nitrogen and hydrogen underdifferent sample masses2.3 分析功率的确定在氮、氢元素分析中，分析功率是决定样品释放的重要参数. 本试验依次设置4 500、5 000、5 500、6 000 W的功率梯度，观察功率对于无烟煤中氮、氢元素检测的影响. 图3为氮、氢的测试值随功率变化的关系图. 由图3可见，当功率较低，在4 500、5 000 W时，氮、氢元素质量比偏低，说明过低的功第 1 期王琳，等：惰气熔融-红外吸收/热导法同时测定无烟煤中氮和氢41率不足以使无烟煤完全熔融释放，这与无烟煤本身含碳量高、燃点高的特性一致. 但当功率为6 000 W 时，质量比再次下降，这是因为功率过高，导致氮、氢元素过早溢出，数据捕捉不及时，导致数据偏低.当分析功率为5 500 W 时，氮、氢元素的释放最完全，测定值最高. 由此可见，无烟煤的最佳分析功率为5 500 W.2.4 分析坩埚的对比氮、氢元素分析的样品载体一般分为石墨套埚（外坩埚加内坩埚）和标准坩埚. 本试验对比二者的分析效果，观察图4（a ）的氮元素及图4（b ）的氢元素在使用不同坩埚时的测定谱图，可发现氮、氢元素在使用石墨套埚得到的测定值明显高于标准坩埚，说明石墨套埚的分析效果优于标准坩埚. 究其原因，标准坩埚对比石墨套埚来说相对单薄，在5 500 W 的高功率下其承压能力小，甚至存在标准坩埚被烧漏或者断裂的情况，因而标准坩埚的使用会导致数据偏低，对于无烟煤这类燃点高、熔融产生热量大的样品来说，双层结构的套埚更适用. 因此，本试验选用石墨套埚作为分析坩埚.2.5 分析参数的设定（包括分析延迟时间、数据集成时间）本方法对仪器分析参数（分析延迟时间、数据集成时间）进行了探究. 对比了15、10、5、0 s 四种延迟时间，观察图5（a ）可见，15 、10 s 时氢的出峰过早、不完整且峰形不佳，导致氢元素的数据捕集不完全，测试数据偏低. 当调整为5 s 时，氢峰的前端有平缓的基线，0 s 时出峰过缓. 因此，5 s 是合理的延迟时间. 由图5（b ）可见，氮的测试值随延迟时间的增加而增大，其延迟时间设置为15 s 较合理.对于出峰不完全的问题，本试验采用将数据集成时间延长的方式，分别设置为55、65、75、80、85 s ，观察图6（a ）发现，当集成时间为55、65、75 s 时，氢峰的末端均未回到基线的位置，数据偏低. 80 s 时谱线回到基线，85 s 时形成相对完整的正态分布峰，与图6（b ）的数据趋势吻合. 同时观察图6（b ）发现，氮的集成时间为55s 数据更合理. 因此本方法选择氮的延迟时间为15 s 、集成时间为55 s ，氢的延迟时间为5 s 、集成时间为85 s 为最佳分析参数.2.6 标准曲线建立及检出限测定无烟煤中的氮、氢元素含量范围较宽泛，单点校准的方式并不适用. 本文采用建立标准曲线的校准方式，在称样质量为0.030 0 g 、分析功率为5 500W ，氮、氢元素延迟时间分别为15、5 s ，捕集时间分别为55、85 s ，使用石墨套埚的试验条件下，选择有证标准物质ZBM093、GBW11104j 、GBW11108o 、2.754.34.24.14.03.92.702.652.602.554 5005 000N H5 5006 000P /W质量比/%质量比/%图3 分析功率的探究试验Fig. 3 Test results of nitrogen and hydrogen underdifferent analytical powers100(a)608040积分强度石墨套锅标准坩埚2000102030t /s405060100(b)608040积分强度石墨坩埚标准坩埚2000102030t /s405060图4 石墨套埚与标准坩埚的确定试验(a)不同坩埚对氮元素的测试谱图，(b)不同坩埚对氢元素的测试谱图Fig. 4 Comparison of test results between graphite sleeve pote and standard crucible (a) spectra of nitrogen in different crucibles, (b) spectra of hydrogen in different crucibles42分析测试技术与仪器第 30 卷ZBW112A 建立标准曲线，其认定值及测量值结果如表1所列. 氮、氢元素的线性方程分别为：Y =2.098 404 22X −0.000 200 66、Y =0.789 376 46X −0.000 044 57，相关系数分别为0.994 9、0.994 0，满足线性关系. 对空白坩埚连续测试11次，得到氮、氢元素的平均值分别为0.318 9%、0.186 9%，以该结果与3倍标准偏差之和作为检出限，分别为0.321%、0.189%，以平均值与10倍标准偏差之和作为定量限，分别为0.326%、0.194%，结果如表2所列，表明该方法检测范围较宽，适用于无烟煤中氮、氢元素的定量检测.2.7 方法的准确度、精密度测试精密度测试是验证方法可靠性的重要指标，本试验使用有证无烟煤标准物质ZBM095A 进行精密度测试，平行测定7次，并计算其精密度. 如表3所列，其氮、氢元素的测定平均值分别为1.30%、3.30%，由表1可知，其认证值分别为1.31%±0.07%、3.23%±0.10%，因此该方法准确度较好. 经计算，氮、氢的精密度分别为3.60%、0.63%，满足方法精密度要求. 由此可见该方法准确可靠.表 1 标准物质及其认证值、测量值Table 1 Certified and measured values of standardsubstances/%标准物质NH 认证值测量值认证值测量值ZBM0930.56±0.060.563 3.01±0.12 2.92GBW11104j 0.94±0.070.929 2.64±0.15 2.71GBW11108o 1.30±0.06 1.30 4.58±0.13 4.59ZBW112A 1.10±0.06 1.12 3.78±0.10 3.79ZBM095A1.31±0.071.303.23±0.103.3010015 s 10 s 5 s 0 s(a)8060积分强度402005101520253035t /s 404550556065702.655.04.03.02.01.00N H(b)2.602.552.50质量比/%质量比/%2.452.402.3551015t /s图5 氮、氢的分析延迟时间对比试验(a) 不同延迟时间下氢的测试谱图， (b)延迟时间对氮、氢的影响Fig. 5 Comparison test of analysis delay times of nitrogen and hydrogen(a) spectra of hydrogen in different delay times, (b) effect of delay times on nitrogen and hydrogen100 2.705.04.94.84.74.62.682.662.642.622.6055606570758085909585 s 80 s 75 s 65 s 55 s806040积分强度质量比/%质量比/%20002040t /st /s6080100(a)(b)图6 氮、氢的集成时间对比试验(a)不同集成时间下氢的测试谱图， (b)集成时间对氮、氢的影响Fig. 6 Comparison test of integration times of nitrogen and hydrogen(a) spectra of hydrogen in different integration times, (b) effect of integration times on nitrogen and hydrogen第 1 期王琳，等：惰气熔融-红外吸收/热导法同时测定无烟煤中氮和氢432.8 未知样品测试对日常送检的无烟煤样品进行抽检，并标号为样品1、样品2，使用方法1与本方法进行对比，随试验进行ZBM095A的测试. 分别平行测定7次，其测试结果如表4所列. 由表可见，方法1测得样品1、样品2、ZBM095A中氮的平均值分别为0.096 6%、1.086%、1.30%，相对标准偏差（RSD）分别为2.67%、1.75%、3.60%. 氢的平均值分别为2.899%、3.312%、3.30%，RSD分别为1.90%、1.50%、0.63%. 本方法测得样品1、样品2、ZBM095A中氮的平均值分别为0.094 6%、1.067%、1.25%，RSD分别为2.99%、1.69%、3.90%. 氢的平均值分别为2.927%、3.300%、3.20%，RSD分别为1.87%、1.56%、0.72%. 对比两种方法，准确度与精密度均能够满足试验要求，再次证实本文建立的方法适用于无烟煤中的氮、氢两种元素的定量测定.表 3 ZBM095A的精密度试验Table 3 Precision test of ZBM095A/%元素测定值平均值RSDN 1.28、1.26、1.34、1.35、1.36、1.30、1.24 1.30 3.60H 3.30、3.32、3.33、3.29、3.29、3.33、3.28 3.300.63表 4 两种方法测试未知样品的对比试验Table 4 Comparison of two methods for testing unknown samples/%样品方法1平均值方法1 RSD本方法平均值本方法RSD N H N H N H N H样品10.096 6 2.899 2.67 1.900.094 6 2.927 2.99 1.87样品 2 1.086 3.312 1.75 1.50 1.067 3.300 1.69 1.56 ZBM095A 1.30 3.30 3.600.63 1.25 3.20 3.900.723 结论（1）本文首次将惰性气体熔融-红外吸收/热导法应用于无烟煤类产品的检测中，该方法满足同时、快速、准确的特点，减少了强酸化学试剂的使用，体现了绿色化学宗旨.（2）建立了无烟煤中氮、氢元素定量测试的方法，为煤炭行业的检验检测、标准制定、贸易等提供参考.（3）拓展了氧氮氢分析仪的使用范围，在有色金属、高温合金、难熔金属、稀土、陶瓷、矿石等材料的使用范围之外，增加了无烟煤类产品的使用.参考文献：习近平. 在第七十五届联合国大会一般性辩论上的讲话[N]. 人民日报, 2020-09-23(3).［ 1 ］元雪芳, 任恒星, 郭鑫, 等. 不同物质对无烟煤生物转化的影响研究[J].煤化工，2022，50（5）：79-82.[YUAN Xuefang, REN Hengxing, GUO Xin, et al.Study on impact of adding different substances on bio-transformation of anthracite[J]. Coal Chemical In-dustry，2022，50 （5）：79-82.]［ 2 ］吕俊复, 蒋苓, 柯希玮, 等. 碳中和背景下循环流化床燃烧技术在中国的发展前景[J]. 煤炭科学技术，2023，51（1）：514-522. [LV Junfu, JIANG Ling, KEXiwei, et al. Future of circulating fluidized bed com-［ 3 ］表 2 方法的线性与检出限Table 2 Linearity and limits of detection元素线性方程线性相关系数（R2）平均值/(%，n=11)检出限/%定量限/% N Y=2.098 404 22X − 0.000 200 660.994 90.318 90.3210.326H Y=0.789 376 46X − 0.000 044 570.994 00.186 90.1890.19444分析测试技术与仪器第 30 卷bustion technology in China for carbon neutralization [J ]. Coal Science and Technology ，2023，51 （1）：514-522.]于昭仪, 谢卫宁, 邱钿, 等. 添加剂对煤基石墨微观结构的影响[J ]. 煤炭科学技术，2023，51（5）：302-308.[YU Zhaoyi, XIE Weining, QIU Tian, et al. Effect of additives on microstructure of coal-based graphite [J ].Coal Science and Technology ，2023，51 （5）：302-308.]［ 4 ］传秀云, 鲍莹. 煤制备新型先进炭材料的应用研究[J ]. 煤炭学报，2013，38（S1）：187-194. [CHUAN Xiuyun, BAO Ying. Application of coal as raw materi-als in preparing new advanced carbon materials [J ].Journal of China Coal Society ，2013，38 （S1）：187-194.]［ 5 ］唐跃刚, 陈鹏翔, 李瑞青, 等. 京西煤制备氧化石墨烯分子结构模型的构建与优化[J ]. 煤炭科学技术，2021，49（6）：126-134. [TANG Yuegang, CHEN Pengxiang, LI Ruiqing, et al. Model construction and optimization of molecule structure of coal-based grapheme oxide from Jingxi coal [J ]. Coal Science and Technology ，2021，49 （6）：126-134.]［ 6 ］Zhang C, Xie Y C, Zhang C, et al. Upgrading coal tomultifunctional graphene-based materials by direct laser scribing [J ]. Carbon ，2019，153 ：585-591.［ 7 ］杨丽, 刘帅, 辛春梅, 等. 炭黑负载增加活性炭缺陷位点催化甲烷裂解制氢[J/OL ]. 煤炭科学技术, 2023:1-11[2023-12-08]. https:///kcms/detail/11.2402.td.20230706.1933.007.html. [YANG Li, LIU Shuai, XIN Chunmei, et al. The loading of carbon black increased the defect site of activated carbon Catalytic pyrolysis of methane for hydrogen produc-tion [J/OL ]. Coal science and technology, 2023: 1-11[2023-12-08]. https:///kcms/detail/11.2402.td.20230706.1933.007.html.]［ 8 ］张昆, 孟召平, 金毅, 等. 不同煤体结构煤的孔隙结构分形特征及其研究意义[J ]. 煤炭科学技术，2023，51（10）：198-206. [ZHANG Kun, MENG Zhaoping,JIN Yi, et al. Fractal characteristics of pore structures on different coal structures and its research signific-ance [J ]. Coal Science and Technology ，2023，51 （10）：198-206.]［ 9 ］狄军贞, 曹洋, 赵文琦. 巯基改性褐煤的制备及其对Fe 2+、Mn 2+的吸附特性研究[J ]. 煤炭科学技术，2023，51（3）：261-270. [DI Junzhen, CAO Yang, ZHAO Wenqi. Preparation of mercapto modified lignite and its adsorption characteristics for Fe 2+、Mn 2+[J ]. Coal［ 10 ］Science and Technology ，2023，51 （3）：261-270.]何绪文. 民用燃煤大气污染物控排技术对策[J ]. 洁净煤技术，2017，23（4）：12-17. [HE Xuwen. Counter-measure of air pollutant controlled pollutation for ci-vilian coal [J ]. Clean Coal Technology ，2017，23 （4）：12-17.]［ 11 ］国家质量监督检验检疫总局, 中国国家标准化管理委员会. 煤的工业分析方法 仪器法: GB/T 30732—2014[S ]. 北京: 中国标准出版社, 2014. [General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of China,Standardization Administration of the People's Re-public of China. Proximate analysis of coal-instru-mental method: GB/T 30732—2014[S ]. Beijing:Standards Press of China, 2014.]［ 12 ］国家质量监督检验检疫总局. 煤的元素分析方法:GB/T 476—2001[S ]. 北京: 中国标准出版社, 2001.[General Administration of Quality Supervision, In-spection and Quarantine of the People's Republic of China. Ultimate analysis of coal: GB/T476—2001[S ]. Beijing: Standards Press of China,2001.]［ 13 ］国家质量监督检验检疫总局, 中国国家标准化管理委员会. 煤中碳氢氮的测定 仪器法: GB/T 30733—2014[S ]. 北京: 中国标准出版社, 2014. [General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of China,Standardization Administration of the People's Re-public of China. Determination of total carbon, hy-drogen and nitrogen content in coal-instrumental method: GB/T 30733—2014[S ]. Beijing: Standards Press of China, 2014.]［ 14 ］杜强, 王喜武. 煤直接液化工艺硫氮元素分布及其影响研究[J ]. 中国煤炭，2022，48（8）：104-108. [DU Qiang, WANG Xiwu. Research on the distribution of sulfur and nitrogen and its influence in direct coal li-quefaction process [J ]. China Coal ，2022，48 （8）：104-108.]［ 15 ］国家市场监督管理总局, 国家标准化管理委员会.煤中有价元素含量分级及应用导则: GB/T 41042—2021[S ]. 北京: 中国标准出版社, 2021.[State Administration for Market Regulation, Na-tional Standardization Administration. Guidance for utilization and classification of content of valuable elements in coal: GB/T 41042—2021[S ]. Beijing:Standards Press of China, 2021.]［ 16 ］姬晓燕, 张志峰, 祁风华, 等. 任家庄井田晚古生代煤［ 17 ］第 1 期王琳，等：惰气熔融-红外吸收/热导法同时测定无烟煤中氮和氢45系战略性金属元素富集特征[J/OL ]. 煤炭科学技,2023: 1-15 [2023-12-08]. https://doi. org/10.13199/ki.cst.2023-0146. [JI Xiaoyan, ZHANG Zhifeng, QI Fenghua, et al. Characteristics of strategic metal ele-ment enrichment in late Paleozoic coal measures in Renjiazhuang Jingtian [J/OL ]. Coal Science and Technology, 2023: 1-15[2023-12-08]. https:///10.13199/ki.cst.2023-0146.]唐佳伟, 张锁, 刘兆峰, 等. 吸附法去除水中F -进展及其在矿井水处理中发展方向[J/OL ]. 煤炭科学技术,2023: 1-16[2023-12-08]. https: //doi. org/10.13199/ki. cst. 2022-1835. [TANG Jiawei, ZHANG Suo,LIU Zhaofeng, et al. Progress in removal of F -in wa-ter by adsorption method and its development direc-tion in mine water treatment [J/OL ]. Coal science and technology, 2023: 1-16[2023-12-08]. https:///10.13199/ki.cst.2022-1835.]［ 18 ］国家质量监督检验检疫总局, 中国国家标准化管理委员会. 煤中氮的测定方法: GB/T 19227—2008[S ]. 北京: 中国标准出版社, 2009. [General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of China,Standardization Administration of the People's Re-public of China. Determination of nitrogen in coal:GB/T 19227—2008[S ]. Beijing: Standards Press of China, 2009.]［ 19 ］龚婉莉. 采用元素分析仪测定煤中碳氢氮含量的应用研究[J ]. 煤质技术，2018（1）：38-41, 49. [GONG Wanli. The application study on determination of car-bon, hydrogen and nitrogen content in coal by ele-mental analyzer [J ]. Goal Quality Technology ，2018（1）：38-41, 49.]［ 20 ］国家质量监督检验检疫总局, 中国国家标准化管理委员会. 煤中碳和氢的测定方法: GB/T 476—2008[S ]. 北京: 中国标准出版社, 2009. [General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of China,Standardization Administration of the People's Re-public of China. Determination of carbon and hydro-gen in coal: GB/T 476—2008[S ]. Beijing: Standards Press of China, 2009.]［ 21 ］孙玉芳, 王曦. 元素分析仪测定土壤氮、碳含量的不确定度评定[J ]. 分析测试技术与仪器, 2016, 22(4):240-245. [SUN Yufang, WANG Xi. Uncertainty eval-uation of measurement results for determination of total carbon and nitrogen in soil samples using ele-mental analyzer [J ]. Analysis and testing technology and instruments. 2016, 22(4): 240-245.]［ 22 ］朱春要, 秦建, 赵希文, 等. 脉冲熔融-红外/热导法测定钛合金粉末微注射成形脱脂坯中氧氮氢[J ]. 中国无机分析化学，2023，13（5）：499-504. [ZHU Chunyao, QIN Jian, ZHAO Xiwen, et al. Determina-tion of oxygen, nitrogen and hydrogen in micro injec-tion molding of titanium alloy powder about degrease billets by impulse fusion-infrared absorption and thermal conductivity method [J ]. Chinese Jorunal of Inorganic Analytical Chemistry ，2023，13 （5）：499-504.]［ 23 ］侯桂臣, 张琳, 李一懿, 等. “直投法” 应用于金属粉末样品中氧氮氢分析[J ]. 冶金分析，2019，39（6）：7-13. [HOU Guichen, ZHANG Lin, LI Yiyi, et al. Ap-plication of "direct-dropping" for analysis of oxygen,nitrogen and hydrogen in metal powder sample [J ].Metallurgical Analysis ，2019，39 （6）：7-13.]［ 24 ］46分析测试技术与仪器第 30 卷。
最后译文：压力限制肿瘤生长法国物理学家发现了简单的力压在医学上可应用于降低肿瘤的生长速度并限制其生长大小。
通过使用老鼠细胞来完成这项工作的研究者说这个结果可以引生出更好的癌症诊断工具并很可能最终实现用药物治疗癌症。
众所周知当生长细胞中的DNA发生突变时就会形成肿瘤并发展为癌症, .但是这种发展是如何受到肿瘤周围环境的影响仍是一个需要讨论的课题。
由巴黎居里学院的让.弗朗斯科乔尼和其他一些院校进行了一项新的调查，研究肿瘤的生长是如何受到它所经受的压力的限制的，如同按压周围的健康组织一样。
很难把基因学、生物化学和力学在生物机体内的肿瘤中所扮演的角色分离出来。
为了解释这一问题，乔尼的团队用老鼠细胞中的一个直径十余毫米的类似肿瘤的球在实验台上进行了这项工作，工作者们把这个模拟肿瘤放入一个由半渗透聚合物制成的几毫米长的袋子中，这之后就进入到一个滋生细胞的包含营养物的研究方案中。
肿瘤在这种自由的状态下会继续生长两周或者三周, 直到达到细胞的死亡和分裂刚好平衡的稳态。
糖分的严厉打击为了找出在这个生长过程中是什么影响到了压力, 小组在此方案中加入了很多糖分这些糖分由于颗粒太大而无法穿过袋子的微小孔洞所以仍在袋子外面，造成了一种浓度的不平衡，而使其迫切的要解决掉袋子外的溶液以努力恢复其浓度的平衡，袋子外较大浓度的溶液随即对袋子产生了力度的压迫，并且这种压迫被里面的肿瘤所感应到。
这种方法被重复用于同样的肿瘤上，每个不同袋子中的肿瘤被不同浓度的糖分溶液所浸透，因此揭示出每个肿瘤都受到了不同的压力。
该小组发现压力越大，肿瘤生长越慢并且最终尺寸越小。
比如施加500帕的压力，仅仅百分之两点五的气压），便可将肿瘤的增长率和稳态量减半。
为了精准地确立压力是如何减弱增长的，乔恩和他的同事将肿瘤冰冻起来，将其切成非常薄的薄片，.并在薄片上覆盖两种抗体，这个方法显示出了在每个肿瘤上已死亡而被分离的细胞----这两种细胞发出的荧光波长不同-。
The effect of surfactants on surfacetensionSurface tension is a characteristic of liquids that arises from the cohesive forces between the molecules of the liquid. It is the tension that forms between the liquid surface and the air or the interface between two liquids. Surface tension is expressed in units of force per unit length, and it is measured in dynes per centimeter (dyn/cm).Surfactants are compounds that reduce the surface tension of liquids. They are amphiphilic molecules that have both hydrophilic (water-loving) and hydrophobic (water-hating) regions. Because of their amphiphilic nature, surfactants are able to dissolve in both water and oil, which makes them useful in many different applications.The effect of surfactants on surface tension is due to their ability to reduce the cohesive forces between the liquid molecules. Surfactants work by adsorbing onto the liquid surface, which causes the molecules to rearrange themselves. This rearrangement of the liquid molecules reduces the cohesive forces between them, which in turn lowers the surface tension of the liquid.The reduction in surface tension caused by surfactants has many important applications. One example is in cleaning products, where surfactants are used to remove dirt and grime from surfaces. The ability of surfactants to lower the surface tension of water allows them to penetrate into tight spaces and remove dirt and oil from surfaces.Surfactants are also used in the food industry, where they are added to liquids to form stable emulsions. An emulsion is a mixture of two immiscible liquids, such as oil and water, where one liquid is dispersed in the other. Surfactants are added to emulsions to prevent the droplets from coalescing, which would cause the emulsion to separate. The ability of surfactants to lower the surface tension of liquids makes them effective at stabilizing emulsions.Surfactants also have applications in the pharmaceutical industry. They are used to improve the solubility and bioavailability of drugs, which makes them more effective. Surfactants can also be used to control the release of drugs, which allows for more precise dosing. The ability of surfactants to lower the surface tension of liquids makes them effective at improving the solubility and bioavailability of drugs.In conclusion, surfactants have a significant effect on surface tension. The ability of surfactants to reduce the surface tension of liquids has many important applications in a wide range of industries, including cleaning products, food, and pharmaceuticals. Understanding the effect of surfactants on surface tension is crucial for the development of new and innovative products that utilize surfactants.。
Accepted ManuscriptEffect of Sn addition on stress hysteresis and superelastic properties of aTi-15Nb-3Mo alloyMuhammad Farzik Ijaz, Hee Young Kim, Hideki Hosoda, Shuichi MiyazakiPII:S1359-6462(13)00505-8DOI:/10.1016/j.scriptamat.2013.10.007Reference:SMM 10081To appear in:Scripta MaterialiaReceived Date:12 September 2013Revised Date:8 October 2013Accepted Date:10 October 2013Please cite this article as: M.F. Ijaz, H.Y. Kim, H. Hosoda, S. Miyazaki, Effect of Sn addition on stress hysteresis and superelastic properties of a Ti-15Nb-3Mo alloy, Scripta Materialia(2013), doi: /10.1016/ j.scriptamat.2013.10.007This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.Effect of Sn addition on stress hysteresis and superelasticproperties of a Ti-15Nb-3Mo alloyMuhammad Farzik Ijaz a , Hee Young Kim a,*, Hideki Hosoda b and Shuichi Miyazaki a,c,** a Division of Materials Science, University of Tsukuba, Tsukuba, Ibaraki 305-8573,Japan b Precision and Intelligence Laboratory, Tokyo Institute of Technology, Yokohama226-8503, Japanc Center of Excellence for Advanced Materials Research, King Abdulaziz University,PO Box 80203, Jeddah 21589, Saudi Arabia* Corresponding author. Tel./Fax: +81 29 853 6942.**Corresponding author. Tel./Fax: +81 29 853 5283. E-mail addresses: heeykim@ims.tsukuba.ac.jp (H. Y. Kim),miyazaki@ims.tsukuba.ac.jp (S. Miyazaki).The effects of Sn content on stress hysteresis and superelastic properties of Ti-15Nb-3Mo-(0-1.5)Sn were investigated. The stress hysteresis decreased with increasing Sn content due to the suppression of athermal omega phase formation. The addition of Sn was also very effective for increasing superelastic recovery strain. Dueto a two-fold role of Sn, i.e. on the martensitic transformation temperature and omega phase, the stress for inducing martensitic transformation decreased with increasing Sn content up to 1at.%, then increased by further addition.Keywords: Titanium alloys; Shape memory alloys; SuperelasticityTi-Ni alloys have been successfully utilized in various biomedical devices due to their exceptional superelastic properties. The major biomedical applications of Ti-Ni alloys have been found in devices such as stents, guide wires and dental arch wires [1]. However, despite their excellent superelastic performance and successful safe usage for medical applications, there are still concerns over the high concentration of Ni in various Ti-Ni-based superelastic alloys. Therefore efforts have been devoted to develop new classes of Ni-free Ti-based superelastic alloys [2-8]. Among them, Ti-Nb-based alloys have been able to get significant attention [7-13]. Superelasticity in Ti-Nb-based alloys is attributed to the reversible stress-induced martensitic transformation between parent βphase (bcc) and martensite α" phase (orthorhombic structure) [14, 15]. Superelastic recovery strain is primarily dependent on the transformation strain which is determined by the crystal structures of parent phase and martensite phase. Recently our group reported that the addition of Mo to Ti-Nb alloys is very effective to increase superelastic recovery strain due to the solid solution strengthening effect of Mo and the increase in transformation strain [16]. However, excess addition of Mo for the replacement of Nb in Ti-Nb-Mo alloy system deteriorates superelastic properties mainly due to the formation of a large amount of athermal ω phase [16]. It was also reported that the increase in Mo content significantly increases the stress hysteresis of Ti-Nb-Moalloys [17]. A small stress hysteresis is preferable for superelastic alloys to take advantage of a high recovery force. In this study, we show that the addition of Sn is very effective not only to reduce stress hysteresis but also to improve superelastic recovery strain of a Ti-Nb-Mo alloy. The unique Sn content dependence of the critical stress for inducing martensitic transformation in Ti-15Nb-3Mo alloys is analyzed by microstructural observation.Ti-15Nb-3Mo-(0-1.5)Sn (at.%) (all compositions are hereafter described in atomic per cent) ingots were prepared by arc-melting in an Ar atmosphere. The ingots were homogenized at 1273 K for 7.2 ks followed by cold-rolling up to 98.5% reduction in thickness without intermediate annealing. Specimens for tensile tests, X-ray diffraction (XRD) measurements and transmission electron microscope (TEM) observation were cut from the as-rolled strips using an electric discharge cutting machine. All the specimens were heat-treated at 973 K for 0.6 ks in an Ar filled quartz tube, followed by quenching in water. Superelastic properties were characterized by loading-unloading tensile tests at room temperature (298 K). The dimensions of the tensile specimens were 0.14 mm in thickness and 1.5 mm in width, with a length of 40 mm. Tensile tests were performed using a Shimadzu Autograph 5KN tensile testing machine at a cross-head speed of 2.5 х 10-4 s-1. Both ends of the tensile specimens were fixed with in two chucks so that gage length becomes 20 mm and the strain of the specimens was determined by measuring the distance between the chucks using an extensometer. Constituent phases were investigated using an XRD machine with a Cu Kαradiation source. Microstructural analysis was conducted by TEM, using a JEOL 2010F microscope operated at 200 kV.Figure 1a shows stress–strain curves of Ti-15Nb-3Mo-(0-1.5)Sn alloys obtained at room temperature. The specimen was elongated until reaching 2.5% strain and then unloaded. All the alloys exhibited superelastic recovery. A single headed arrow points at the critical stress required for inducing martensitic transformation (σβ→α"), whereas a double head arrow points at the stress where the reverse transformation finishes (σα"→β) upon unloading. Stress hysteresis (∆σ) was defined as the difference between σβ→α" and σα"→β.The Sn content dependences of σβ→α" and σα"→βare shown in Figure 1b. So far, it has been reported that Sn decreases M s of Ti-Nb based alloys [18, 19]. Therefore, it is expected that σβ→α"increases with increasing Sn content, since the stress for inducing martensitic transformation at a fixed test temperature increases with decreasing M s of the alloy. However, Figure 1b shows a peculiar Sn content dependence of σβ→α": it decreased by the addition of Sn up to 1 at.%, and then increased by further addition. This implies that the addition of Sn up to 1 at.% raised M s of the Ti-15Nb-3Mo alloy, but further addition caused the decrease of M s. On the other hand, σα"→β exhibited a monotonic increasing tendency with the increase in Sn content as shown in Figure 1b. It is important to note that the addition of Sn causes a significant decrease in ∆σof the Ti-15Nb-3Mo alloy. It is apparent that the decreasing trend of ∆σ is more pronounced particularly up to 1 at.% Sn addition as shown in Figure 1c.In order to investigate superelastic properties of the Ti-15Nb-3Mo-(0-1.5)Sn alloys, cyclic loading–unloading tensile tests were carried out and the results are shownin Figure 2. In the first cycle, the tensile stress was applied until the strain reached 1.5% and then unloaded. The similar measurement was repeated by increasing maximum strain by the interval of 0.5% (such as 1.5 %, 2.0% for each successive cycle of loading) using the same specimen. In order to evaluate superelastic properties, two types of strain, i.e. recovery strain (εr) and remained plastic strain (εp) were measured at each cycle. It is seen that the strain was almost totally recovered upon unloading up to the first cycle for Ti-15Nb-3Mo, up to the third cycle for Ti-15Nb-3Mo-0.5Sn and Ti-15Nb-3Mo-0.75Sn, up to the fourth cycle for Ti-15Nb-3Mo-1Sn and Ti-15Nb-3Mo-1.25Sn, and up to the first cycle for Ti-15Nb-3Mo-1.5Sn, respectively. The magnitudes of maximum recovery strain εr max, σβ→α" and ∆σ are listed in Table 1. Interestingly, these properties, i.e. εr max, σβ→α"and ∆σshow different Sn content dependences. As mentioned above, with increasing Sn content σβ→α"decreased to minimum at Ti-15Nb-3Mo-1Sn and it was then increased by further addition of Sn. On the other hand, ∆σ exhibited a monotonic decreasing tendency with increasing Sn content. The decrease in the stress hysteresis caused the increase in the superelastic recovery strain; εr max increased to a maximum value of 3.7% at Ti-15Nb-3Mo-1.25Sn. However Ti-15Nb-3Mo-1.5Sn exhibited a small recovery strain of 2.1% in spite of its smallest stress hysteresis. It is suggested that the small recovery strain of Ti-15Nb-3Mo-1.5Sn is due to its high σβ→α". A high σβ→α"reduces the difference between the critical stress for permanent plastic deformation and the stress for inducing martensitic transformation, resulting in that the permanent plastic deformation occurs during the martensitic transformation.In order to discuss the effect of Sn content on superelastic properties, XRD measurement and TEM observation were carried out. Figure 3a shows XRD profiles of the Ti-15Nb-3Mo-(0-1.5)Sn alloys obtained at room temperature. Within the measured 2θ (deg.) range, β phase was identified by four major reflections from (110)β, (200)β, (211)β, and (222)βwhereas ωphase was identified by two major reflections from (001)ω and (002)ω. The reflections from the ω phase are found to be sensitive to the Sn content. It is clearly seen that the Ti-15Nb-3Mo alloy reveals the strongest intensity of the ωphase. The ωphase is suggested to be formed athermally during quenching from the annealing temperature (973 K). The peak intensities of the ω phase gradually became weaker as the Sn content increased. The peaks from the ωphase could not be detected in the XRD profiles of the Ti-15Nb-3Mo-1.25Sn and Ti-15Nb-3Mo-1.5Sn alloys. This implies that the ω phase was suppressed significantly with the increase in Sn content, which is consistent with previous works [20, 21]. The suppression of athermal omega phase (ωath) by the addition of Sn was also confirmed by TEM observation. Figure 3b shows dark-field images and the corresponding selected area diffraction patterns with zone axis of [113]βin the Ti-15Nb-3Mo and Ti-15Nb-3Mo-1Sn alloys. Dark-field images showing the ω phase were formed using the diffraction spot indicated by a white circle in each diffraction pattern. It is seen that the size and volume fraction of the ωath phase were remarkably reduced by the Sn addition.On the basis of microstructure analysis, the unique Sn content dependences of σβ→α"and stress hysteresis in the Ti-15Nb-3Mo-(0-1.5)Sn alloys can be explained by considering a two-fold role of Sn; the one hand Sn decreases M s of Ti-Nb based alloys, on the other hand it suppresses the ω phase formation. It has been confirmed that the athermal ωphase suppresses the martensitic transformation and increases in σβ→α"of Ti-Nb-Mo alloys [16, 17], implying that the suppression of the athermal ωphase increases M s. Figure 4 illustrates the schematic explanation of the effect of Sn on apparent M s and σβ→α". If we only consider the compositional effect, σβ→α"should increase monotonically with increasing Sn content because the difference between M s and test temperature (RT) increases. However, a large amount of athermal ω phase in the Ti-15Nb-3Mo alloy decreases M s and increases σβ→α"of the alloy. As mentioned above, the volume fraction of athermal ω phase decreases with increasing Sn content, indicating that the decrease in M s due to the ωphase is reduced as the Sn content increases. Consequently the decrease in σβ→α" with increasing Sn content up to 1 at.% in the Ti-15Nb-3Mo-(0-1.5)Sn alloys implies the fact that the effect of Sn on the suppression of athermal ωphase is stronger than the compositional effect which decreases M s. As the Sn content increased the effect of Sn on the suppression of athermal ω phase becomes weaker, hence the compositional effect becomes dominant. This explains why σβ→α" increases with increasing Sn content from 1 at.% to 1.5 at.%.On the other hand, the monotonic increase in σα"→β with increasing Sn content suggests that the intrinsic compositional effect of Sn is dominant for the reverse transformation. It has been proposed that the athermal ω phase also transforms into the α" phase when the β phase transforms [17, 22]. As a result, according to the current understanding, it is believed that the monotonic change in σα"→β is mainly attributed to the absence of the ωphase in the α" phase while unloading. This explains not only a large stress hysteresis loop in the ternary alloy but also the decrease of stress hysteresis with increasing Sn content.In conclusion, we are able to reduce the stress hysteresis and increase the superelastic recovery strain of a Ti-Nb-Mo alloy by the addition of Sn. The decrease in the stress hysteresis with increasing Sn content is due to the suppression of the athermal ωphase. The unique Sn content dependence of the critical stress for inducing martensitic transformation in Ti-15Nb-3Mo-(0-1.5)Sn alloys is due to a two-fold role of Sn, i.e. the decrease of M s and the suppression of the athermal ω phase.AcknowledgmentsThis work was partially supported by JSPS KAKENHI Grant Number 23360300 and 25289247, MEXT KAKENHI Grant Number 23102503 and 25102704.References[1] T. Yoneyama, S. Miyazaki, Shape Memory Alloys for Biomedical Applications, Woodhead Publishing, Cambridge, 2009.[2] S. Miyazaki, H.Y. Kim, H. Hosoda, Mater. Sci. Eng. A 438-440 (2006) 18.[3] H.Y. Kim, S. Hashimoto, J.I. Kim, T. Inamura, H. Hosoda, S. Miyazaki, Mater. Sci. Eng. A 417 (2006) 120.[4] T.W. Duerig, J. Albrecht, D. Richter, P. Fischer, Acta Metall. 30 (1982) 2161.[5] L.C. Zhang, T. Zhou, S.P. Alpay, M. Aindow, Appl. Phys. Lett. 87 (2005) 241909.[6]T. Maeshima, M. Nishida, Mater. Trans. 45 (2004) 1101.[7]M. Niinomi, T. Akahori, S. Katsura, K. Yamauchi, M. Ogawa, Mater. Sci. Eng. C 7 (2007) 154.[8] D.C. Zhang, Y.F. Mao, M. Yan, J.J. Li, E.L. Su, Y.L. Li, S.W. Tan, J.G. Lin, J. Mech. Behav. Biomed. Mater. 13 (2012) 156.[9] V. Brailovski, S. Prokoshkin, M. Gautheir, K. Inaekyan, S. Dubinskiy, M. Petrzhik, M. Filonov, Mater. Sci. Eng. C 31 (2011) 643.[10] F. Sun, S. Nowak, T. Gloriant, P. Laheurte, A. Eberhardt, F. Prima, Scripta Mater.63 (2010) 1053.[11] S. Cai, M.R. Daymond, Y. Ren, D.M. Bailey, L.E. Kay, Mater. Sci. Eng. A 562 (2013) 172.[12] E.G. Obbard, Y.L. Hao, R.J. Talling, S.J. Li, Y.W. Zhang, D. Dye, R. Yang, Acta Mater. 59 (2011) 112.[13] H. Hosoda, Y. Kinoshita, Y. Fukui, T. Inamura, K. Wakashima, H.Y. Kim, S. Miyazaki, Mater. Sci. Eng. A 438-440 (2006) 870.[14] H.Y. Kim, T. Sasaki, J.I. Kim, T. Inamura, H. Hasoda, S. Miyazaki, Acta Mater. 54 (2006) 423.[15] H.Y. Kim, T. Sasaki, K. Okutsu, J.I. Kim, T. Inamura, H. Hosoda, S. Miyazaki, Acta Mater. 54 (2006) 2419.[16] Y. Al-Zain, H.Y. Kim, H. Hosoda, T.H. Nam, S. Miyazaki, Acta Mater. 58 (2010) 4212.[17] Y. Al-Zain, H.Y. Kim, T. Koyano, H. Hosoda, T.H. Nam, S. Miyazaki, Acta Mater.59 (2011) 1464.[18] E. Takahashi, T. Sakurai, S. Watanabe, N. Masahashi, S. Hanada, Mater. Trans. 43 (2002) 2978.[19] Y. L. Hao, S.J. Li, S.Y. Sun, R. Yang, Mater. Sci. Eng. A 441 (2006) 112 .[20] P.J.S. Buenconsejo, H.Y. Kim, S. Miyazaki, Scripta Mater. 64 (2011) 1114.[21] Y. Al-Zain, Y. Sato, H.Y. Kim, H. Hosoda, T.H. Nam, S. Miyazaki, Acta Mater.60 (2012) 2437.[22] T. Inamura, H. Hosoda, H.Y. Kim, S. Miyazaki, Phil. Mag. 90 (2010) 3475.List of Figure captions.Figure 1. (a) Stress-strain curves of Ti-15Nb-3Mo-(0-1.5) Sn alloys, (b) effect of Sn content on σβ→α" and σα"→β and (c) effect of Sn content on stress hysteresis ∆σ.Figure 2. Stress-strain curves obtained by cyclic loading–unloading tensile tests for Ti-15Nb-3Mo-(0-1.5) Sn alloys.Figure 3. (a) XRD profiles for Ti-15Nb-3Mo-(0-1.5)Sn alloys and (b) dark field images and the corresponding selected area diffraction patterns for Ti-15Nb-3Mo and Ti-15Nb-3Mo-1Sn alloys.Figure 4. A schematic explanation on the effect of Sn on Ms and σβ→α".Figure 1.(a) Stress-strain curves of Ti-15Nb-3Mo-(0-1.5) Sn alloys, (b) effect of Sn content on σβ→α" and σα"→β and (c) effect of Sn content on stress hysteresis ∆σ.Figure 2.Stress-strain curves obtained by cyclic loading–unloading tensile tests for Ti-15Nb-3Mo-(0-1.5) Snalloys.alloys and (b) dark field images and the corresponding selectedarea diffraction patterns of Ti-15Nb-3Mo and Ti-15Nb-3Mo-1Sn alloys.and σβ→α".Table 1.Summary of superelastic properties obtained by cyclic loading-unloading tensile tests of Ti-15Nb-3Mo-(0-1.5)Sn alloys.Alloy (at.%)σβ-α″(MPa)∆σ(MPa)εr max(%)Ti-15Nb-3Mo-0Sn297 253 2.1 Ti-15Nb-3Mo-0.5Sn259 212 3.1 Ti-15Nb-3Mo-0.75Sn252 172 3.2 Ti-15Nb-3Mo-1Sn212 123 3.3 Ti-15Nb-3Mo-1.25Sn271 121 3.7 Ti-15Nb-3Mo-1.5Sn344 78 2.1。
