Chapter 4 Defects
材料科学基础--第2章晶体缺陷PPT课件
12
2.1.5点缺陷与材料行为
Or, there should be 2.00 – 1.9971 = 0.0029 vacancies per unit cell. The number of vacancies per cm3 is:
17
Other Point Defects
Interstitialcy - A point defect caused when a ‘‘normal’’ atom occupies an interstitial site in the crystal.
11
2.1.4 过饱和点缺陷
晶体中的点缺陷浓度可能高于平衡浓度，称为过饱和点 缺陷，或非平衡点缺陷。获得的方法：
高温淬火：将晶体加热到高温，然后迅速冷却（淬火 ），则高温时形成的空位来不及扩散消失，使晶体在低 温状态仍然保留高温状态的空位浓度，即过饱和空位。
冷加工：金属在室温下进行冷加工塑性变形也会产生 大量的过饱和空位，其原因是由于位错交割所形成的割 阶发生攀移。
6
2.1.1 分类
3.置换原子（Substitutional atom） 异类原子代换了原有晶体中的原子，而处于晶体点阵的结 点位置，称为置换原子，亦称代位原子。 各种点缺陷，都破坏了原有晶体的完整性。它们从电学
和力学这两个方面，使近邻原子失去了平衡。空位和直 径较小的置换原子，使周围原子向点缺陷的方向松弛， 间隙原子及直径较大的置换原子，把周围原子挤开一定 位置。因而在点缺陷的周围，就出现了一定范围的点阵 畸变区，或称弹性应变区。距点缺陷越远，其影响越小 。因而在每个点缺陷的周围，都会产生一个弹性应力场 。
OSHA现场作业手册说明书
DIRECTIVE NUMBER: CPL 02-00-150 EFFECTIVE DATE: April 22, 2011 SUBJECT: Field Operations Manual (FOM)ABSTRACTPurpose: This instruction cancels and replaces OSHA Instruction CPL 02-00-148,Field Operations Manual (FOM), issued November 9, 2009, whichreplaced the September 26, 1994 Instruction that implemented the FieldInspection Reference Manual (FIRM). The FOM is a revision of OSHA’senforcement policies and procedures manual that provides the field officesa reference document for identifying the responsibilities associated withthe majority of their inspection duties. This Instruction also cancels OSHAInstruction FAP 01-00-003 Federal Agency Safety and Health Programs,May 17, 1996 and Chapter 13 of OSHA Instruction CPL 02-00-045,Revised Field Operations Manual, June 15, 1989.Scope: OSHA-wide.References: Title 29 Code of Federal Regulations §1903.6, Advance Notice ofInspections; 29 Code of Federal Regulations §1903.14, Policy RegardingEmployee Rescue Activities; 29 Code of Federal Regulations §1903.19,Abatement Verification; 29 Code of Federal Regulations §1904.39,Reporting Fatalities and Multiple Hospitalizations to OSHA; and Housingfor Agricultural Workers: Final Rule, Federal Register, March 4, 1980 (45FR 14180).Cancellations: OSHA Instruction CPL 02-00-148, Field Operations Manual, November9, 2009.OSHA Instruction FAP 01-00-003, Federal Agency Safety and HealthPrograms, May 17, 1996.Chapter 13 of OSHA Instruction CPL 02-00-045, Revised FieldOperations Manual, June 15, 1989.State Impact: Notice of Intent and Adoption required. See paragraph VI.Action Offices: National, Regional, and Area OfficesOriginating Office: Directorate of Enforcement Programs Contact: Directorate of Enforcement ProgramsOffice of General Industry Enforcement200 Constitution Avenue, NW, N3 119Washington, DC 20210202-693-1850By and Under the Authority ofDavid Michaels, PhD, MPHAssistant SecretaryExecutive SummaryThis instruction cancels and replaces OSHA Instruction CPL 02-00-148, Field Operations Manual (FOM), issued November 9, 2009. The one remaining part of the prior Field Operations Manual, the chapter on Disclosure, will be added at a later date. This Instruction also cancels OSHA Instruction FAP 01-00-003 Federal Agency Safety and Health Programs, May 17, 1996 and Chapter 13 of OSHA Instruction CPL 02-00-045, Revised Field Operations Manual, June 15, 1989. This Instruction constitutes OSHA’s general enforcement policies and procedures manual for use by the field offices in conducting inspections, issuing citations and proposing penalties.Significant Changes∙A new Table of Contents for the entire FOM is added.∙ A new References section for the entire FOM is added∙ A new Cancellations section for the entire FOM is added.∙Adds a Maritime Industry Sector to Section III of Chapter 10, Industry Sectors.∙Revises sections referring to the Enhanced Enforcement Program (EEP) replacing the information with the Severe Violator Enforcement Program (SVEP).∙Adds Chapter 13, Federal Agency Field Activities.∙Cancels OSHA Instruction FAP 01-00-003, Federal Agency Safety and Health Programs, May 17, 1996.DisclaimerThis manual is intended to provide instruction regarding some of the internal operations of the Occupational Safety and Health Administration (OSHA), and is solely for the benefit of the Government. No duties, rights, or benefits, substantive or procedural, are created or implied by this manual. The contents of this manual are not enforceable by any person or entity against the Department of Labor or the United States. Statements which reflect current Occupational Safety and Health Review Commission or court precedents do not necessarily indicate acquiescence with those precedents.Table of ContentsCHAPTER 1INTRODUCTIONI.PURPOSE. ........................................................................................................... 1-1 II.SCOPE. ................................................................................................................ 1-1 III.REFERENCES .................................................................................................... 1-1 IV.CANCELLATIONS............................................................................................. 1-8 V. ACTION INFORMATION ................................................................................. 1-8A.R ESPONSIBLE O FFICE.......................................................................................................................................... 1-8B.A CTION O FFICES. .................................................................................................................... 1-8C. I NFORMATION O FFICES............................................................................................................ 1-8 VI. STATE IMPACT. ................................................................................................ 1-8 VII.SIGNIFICANT CHANGES. ............................................................................... 1-9 VIII.BACKGROUND. ................................................................................................. 1-9 IX. DEFINITIONS AND TERMINOLOGY. ........................................................ 1-10A.T HE A CT................................................................................................................................................................. 1-10B. C OMPLIANCE S AFETY AND H EALTH O FFICER (CSHO). ...........................................................1-10B.H E/S HE AND H IS/H ERS ..................................................................................................................................... 1-10C.P ROFESSIONAL J UDGMENT............................................................................................................................... 1-10E. W ORKPLACE AND W ORKSITE ......................................................................................................................... 1-10CHAPTER 2PROGRAM PLANNINGI.INTRODUCTION ............................................................................................... 2-1 II.AREA OFFICE RESPONSIBILITIES. .............................................................. 2-1A.P ROVIDING A SSISTANCE TO S MALL E MPLOYERS. ...................................................................................... 2-1B.A REA O FFICE O UTREACH P ROGRAM. ............................................................................................................. 2-1C. R ESPONDING TO R EQUESTS FOR A SSISTANCE. ............................................................................................ 2-2 III. OSHA COOPERATIVE PROGRAMS OVERVIEW. ...................................... 2-2A.V OLUNTARY P ROTECTION P ROGRAM (VPP). ........................................................................... 2-2B.O NSITE C ONSULTATION P ROGRAM. ................................................................................................................ 2-2C.S TRATEGIC P ARTNERSHIPS................................................................................................................................. 2-3D.A LLIANCE P ROGRAM ........................................................................................................................................... 2-3 IV. ENFORCEMENT PROGRAM SCHEDULING. ................................................ 2-4A.G ENERAL ................................................................................................................................................................. 2-4B.I NSPECTION P RIORITY C RITERIA. ..................................................................................................................... 2-4C.E FFECT OF C ONTEST ............................................................................................................................................ 2-5D.E NFORCEMENT E XEMPTIONS AND L IMITATIONS. ....................................................................................... 2-6E.P REEMPTION BY A NOTHER F EDERAL A GENCY ........................................................................................... 2-6F.U NITED S TATES P OSTAL S ERVICE. .................................................................................................................. 2-7G.H OME-B ASED W ORKSITES. ................................................................................................................................ 2-8H.I NSPECTION/I NVESTIGATION T YPES. ............................................................................................................... 2-8 V.UNPROGRAMMED ACTIVITY – HAZARD EVALUATION AND INSPECTION SCHEDULING ............................................................................ 2-9 VI.PROGRAMMED INSPECTIONS. ................................................................... 2-10A.S ITE-S PECIFIC T ARGETING (SST) P ROGRAM. ............................................................................................. 2-10B.S CHEDULING FOR C ONSTRUCTION I NSPECTIONS. ..................................................................................... 2-10C.S CHEDULING FOR M ARITIME I NSPECTIONS. ............................................................................. 2-11D.S PECIAL E MPHASIS P ROGRAMS (SEP S). ................................................................................... 2-12E.N ATIONAL E MPHASIS P ROGRAMS (NEP S) ............................................................................... 2-13F.L OCAL E MPHASIS P ROGRAMS (LEP S) AND R EGIONAL E MPHASIS P ROGRAMS (REP S) ............ 2-13G.O THER S PECIAL P ROGRAMS. ............................................................................................................................ 2-13H.I NSPECTION S CHEDULING AND I NTERFACE WITH C OOPERATIVE P ROGRAM P ARTICIPANTS ....... 2-13CHAPTER 3INSPECTION PROCEDURESI.INSPECTION PREPARATION. .......................................................................... 3-1 II.INSPECTION PLANNING. .................................................................................. 3-1A.R EVIEW OF I NSPECTION H ISTORY .................................................................................................................... 3-1B.R EVIEW OF C OOPERATIVE P ROGRAM P ARTICIPATION .............................................................................. 3-1C.OSHA D ATA I NITIATIVE (ODI) D ATA R EVIEW .......................................................................................... 3-2D.S AFETY AND H EALTH I SSUES R ELATING TO CSHO S.................................................................. 3-2E.A DVANCE N OTICE. ................................................................................................................................................ 3-3F.P RE-I NSPECTION C OMPULSORY P ROCESS ...................................................................................................... 3-5G.P ERSONAL S ECURITY C LEARANCE. ................................................................................................................. 3-5H.E XPERT A SSISTANCE. ........................................................................................................................................... 3-5 III. INSPECTION SCOPE. ......................................................................................... 3-6A.C OMPREHENSIVE ................................................................................................................................................... 3-6B.P ARTIAL. ................................................................................................................................................................... 3-6 IV. CONDUCT OF INSPECTION .............................................................................. 3-6A.T IME OF I NSPECTION............................................................................................................................................. 3-6B.P RESENTING C REDENTIALS. ............................................................................................................................... 3-6C.R EFUSAL TO P ERMIT I NSPECTION AND I NTERFERENCE ............................................................................. 3-7D.E MPLOYEE P ARTICIPATION. ............................................................................................................................... 3-9E.R ELEASE FOR E NTRY ............................................................................................................................................ 3-9F.B ANKRUPT OR O UT OF B USINESS. .................................................................................................................... 3-9G.E MPLOYEE R ESPONSIBILITIES. ................................................................................................. 3-10H.S TRIKE OR L ABOR D ISPUTE ............................................................................................................................. 3-10I. V ARIANCES. .......................................................................................................................................................... 3-11 V. OPENING CONFERENCE. ................................................................................ 3-11A.G ENERAL ................................................................................................................................................................ 3-11B.R EVIEW OF A PPROPRIATION A CT E XEMPTIONS AND L IMITATION. ..................................................... 3-13C.R EVIEW S CREENING FOR P ROCESS S AFETY M ANAGEMENT (PSM) C OVERAGE............................. 3-13D.R EVIEW OF V OLUNTARY C OMPLIANCE P ROGRAMS. ................................................................................ 3-14E.D ISRUPTIVE C ONDUCT. ...................................................................................................................................... 3-15F.C LASSIFIED A REAS ............................................................................................................................................. 3-16VI. REVIEW OF RECORDS. ................................................................................... 3-16A.I NJURY AND I LLNESS R ECORDS...................................................................................................................... 3-16B.R ECORDING C RITERIA. ...................................................................................................................................... 3-18C. R ECORDKEEPING D EFICIENCIES. .................................................................................................................. 3-18 VII. WALKAROUND INSPECTION. ....................................................................... 3-19A.W ALKAROUND R EPRESENTATIVES ............................................................................................................... 3-19B.E VALUATION OF S AFETY AND H EALTH M ANAGEMENT S YSTEM. ....................................................... 3-20C.R ECORD A LL F ACTS P ERTINENT TO A V IOLATION. ................................................................................. 3-20D.T ESTIFYING IN H EARINGS ................................................................................................................................ 3-21E.T RADE S ECRETS. ................................................................................................................................................. 3-21F.C OLLECTING S AMPLES. ..................................................................................................................................... 3-22G.P HOTOGRAPHS AND V IDEOTAPES.................................................................................................................. 3-22H.V IOLATIONS OF O THER L AWS. ....................................................................................................................... 3-23I.I NTERVIEWS OF N ON-M ANAGERIAL E MPLOYEES .................................................................................... 3-23J.M ULTI-E MPLOYER W ORKSITES ..................................................................................................................... 3-27 K.A DMINISTRATIVE S UBPOENA.......................................................................................................................... 3-27 L.E MPLOYER A BATEMENT A SSISTANCE. ........................................................................................................ 3-27 VIII. CLOSING CONFERENCE. .............................................................................. 3-28A.P ARTICIPANTS. ..................................................................................................................................................... 3-28B.D ISCUSSION I TEMS. ............................................................................................................................................ 3-28C.A DVICE TO A TTENDEES .................................................................................................................................... 3-29D.P ENALTIES............................................................................................................................................................. 3-30E.F EASIBLE A DMINISTRATIVE, W ORK P RACTICE AND E NGINEERING C ONTROLS. ............................ 3-30F.R EDUCING E MPLOYEE E XPOSURE. ................................................................................................................ 3-32G.A BATEMENT V ERIFICATION. ........................................................................................................................... 3-32H.E MPLOYEE D ISCRIMINATION .......................................................................................................................... 3-33 IX. SPECIAL INSPECTION PROCEDURES. ...................................................... 3-33A.F OLLOW-UP AND M ONITORING I NSPECTIONS............................................................................................ 3-33B.C ONSTRUCTION I NSPECTIONS ......................................................................................................................... 3-34C. F EDERAL A GENCY I NSPECTIONS. ................................................................................................................. 3-35CHAPTER 4VIOLATIONSI. BASIS OF VIOLATIONS ..................................................................................... 4-1A.S TANDARDS AND R EGULATIONS. .................................................................................................................... 4-1B.E MPLOYEE E XPOSURE. ........................................................................................................................................ 4-3C.R EGULATORY R EQUIREMENTS. ........................................................................................................................ 4-6D.H AZARD C OMMUNICATION. .............................................................................................................................. 4-6E. E MPLOYER/E MPLOYEE R ESPONSIBILITIES ................................................................................................... 4-6 II. SERIOUS VIOLATIONS. .................................................................................... 4-8A.S ECTION 17(K). ......................................................................................................................... 4-8B.E STABLISHING S ERIOUS V IOLATIONS ............................................................................................................ 4-8C. F OUR S TEPS TO BE D OCUMENTED. ................................................................................................................... 4-8 III. GENERAL DUTY REQUIREMENTS ............................................................. 4-14A.E VALUATION OF G ENERAL D UTY R EQUIREMENTS ................................................................................. 4-14B.E LEMENTS OF A G ENERAL D UTY R EQUIREMENT V IOLATION.............................................................. 4-14C. U SE OF THE G ENERAL D UTY C LAUSE ........................................................................................................ 4-23D.L IMITATIONS OF U SE OF THE G ENERAL D UTY C LAUSE. ..............................................................E.C LASSIFICATION OF V IOLATIONS C ITED U NDER THE G ENERAL D UTY C LAUSE. ..................F. P ROCEDURES FOR I MPLEMENTATION OF S ECTION 5(A)(1) E NFORCEMENT ............................ 4-25 4-27 4-27IV.OTHER-THAN-SERIOUS VIOLATIONS ............................................... 4-28 V.WILLFUL VIOLATIONS. ......................................................................... 4-28A.I NTENTIONAL D ISREGARD V IOLATIONS. ..........................................................................................4-28B.P LAIN I NDIFFERENCE V IOLATIONS. ...................................................................................................4-29 VI. CRIMINAL/WILLFUL VIOLATIONS. ................................................... 4-30A.A REA D IRECTOR C OORDINATION ....................................................................................................... 4-31B.C RITERIA FOR I NVESTIGATING P OSSIBLE C RIMINAL/W ILLFUL V IOLATIONS ........................ 4-31C. W ILLFUL V IOLATIONS R ELATED TO A F ATALITY .......................................................................... 4-32 VII. REPEATED VIOLATIONS. ...................................................................... 4-32A.F EDERAL AND S TATE P LAN V IOLATIONS. ........................................................................................4-32B.I DENTICAL S TANDARDS. .......................................................................................................................4-32C.D IFFERENT S TANDARDS. .......................................................................................................................4-33D.O BTAINING I NSPECTION H ISTORY. .....................................................................................................4-33E.T IME L IMITATIONS..................................................................................................................................4-34F.R EPEATED V. F AILURE TO A BATE....................................................................................................... 4-34G. A REA D IRECTOR R ESPONSIBILITIES. .............................................................................. 4-35 VIII. DE MINIMIS CONDITIONS. ................................................................... 4-36A.C RITERIA ................................................................................................................................................... 4-36B.P ROFESSIONAL J UDGMENT. ..................................................................................................................4-37C. A REA D IRECTOR R ESPONSIBILITIES. .............................................................................. 4-37 IX. CITING IN THE ALTERNATIVE ............................................................ 4-37 X. COMBINING AND GROUPING VIOLATIONS. ................................... 4-37A.C OMBINING. ..............................................................................................................................................4-37B.G ROUPING. ................................................................................................................................................4-38C. W HEN N OT TO G ROUP OR C OMBINE. ................................................................................................4-38 XI. HEALTH STANDARD VIOLATIONS ....................................................... 4-39A.C ITATION OF V ENTILATION S TANDARDS ......................................................................................... 4-39B.V IOLATIONS OF THE N OISE S TANDARD. ...........................................................................................4-40 XII. VIOLATIONS OF THE RESPIRATORY PROTECTION STANDARD(§1910.134). ....................................................................................................... XIII. VIOLATIONS OF AIR CONTAMINANT STANDARDS (§1910.1000) ... 4-43 4-43A.R EQUIREMENTS UNDER THE STANDARD: .................................................................................................. 4-43B.C LASSIFICATION OF V IOLATIONS OF A IR C ONTAMINANT S TANDARDS. ......................................... 4-43 XIV. CITING IMPROPER PERSONAL HYGIENE PRACTICES. ................... 4-45A.I NGESTION H AZARDS. .................................................................................................................................... 4-45B.A BSORPTION H AZARDS. ................................................................................................................................ 4-46C.W IPE S AMPLING. ............................................................................................................................................. 4-46D.C ITATION P OLICY ............................................................................................................................................ 4-46 XV. BIOLOGICAL MONITORING. ...................................................................... 4-47CHAPTER 5CASE FILE PREPARATION AND DOCUMENTATIONI.INTRODUCTION ............................................................................................... 5-1 II.INSPECTION CONDUCTED, CITATIONS BEING ISSUED. .................... 5-1A.OSHA-1 ................................................................................................................................... 5-1B.OSHA-1A. ............................................................................................................................... 5-1C. OSHA-1B. ................................................................................................................................ 5-2 III.INSPECTION CONDUCTED BUT NO CITATIONS ISSUED .................... 5-5 IV.NO INSPECTION ............................................................................................... 5-5 V. HEALTH INSPECTIONS. ................................................................................. 5-6A.D OCUMENT P OTENTIAL E XPOSURE. ............................................................................................................... 5-6B.E MPLOYER’S O CCUPATIONAL S AFETY AND H EALTH S YSTEM. ............................................................. 5-6 VI. AFFIRMATIVE DEFENSES............................................................................. 5-8A.B URDEN OF P ROOF. .............................................................................................................................................. 5-8B.E XPLANATIONS. ..................................................................................................................................................... 5-8 VII. INTERVIEW STATEMENTS. ........................................................................ 5-10A.G ENERALLY. ......................................................................................................................................................... 5-10B.CSHO S SHALL OBTAIN WRITTEN STATEMENTS WHEN: .......................................................................... 5-10C.L ANGUAGE AND W ORDING OF S TATEMENT. ............................................................................................. 5-11D.R EFUSAL TO S IGN S TATEMENT ...................................................................................................................... 5-11E.V IDEO AND A UDIOTAPED S TATEMENTS. ..................................................................................................... 5-11F.A DMINISTRATIVE D EPOSITIONS. .............................................................................................5-11 VIII. PAPERWORK AND WRITTEN PROGRAM REQUIREMENTS. .......... 5-12 IX.GUIDELINES FOR CASE FILE DOCUMENTATION FOR USE WITH VIDEOTAPES AND AUDIOTAPES .............................................................. 5-12 X.CASE FILE ACTIVITY DIARY SHEET. ..................................................... 5-12 XI. CITATIONS. ..................................................................................................... 5-12A.S TATUTE OF L IMITATIONS. .............................................................................................................................. 5-13B.I SSUING C ITATIONS. ........................................................................................................................................... 5-13C.A MENDING/W ITHDRAWING C ITATIONS AND N OTIFICATION OF P ENALTIES. .................................. 5-13D.P ROCEDURES FOR A MENDING OR W ITHDRAWING C ITATIONS ............................................................ 5-14 XII. INSPECTION RECORDS. ............................................................................... 5-15A.G ENERALLY. ......................................................................................................................................................... 5-15B.R ELEASE OF I NSPECTION I NFORMATION ..................................................................................................... 5-15C. C LASSIFIED AND T RADE S ECRET I NFORMATION ...................................................................................... 5-16。
SMT-04WaferPrepCD
Table 4.2
© 2001 by Prentice Hall
Float Zone Crystal Growth
Gas inlet (inert)
Chuck Polycrystalline
rod (silicon)
RF
Molten zone
Traveling RF coil
Seed crystal Inert gas out
by Michael Quirk 001 by Prentice Hall
Semiconductor-Grade Silicon
Steps to Obtaining Semiconductor Grade Silicon (SGS)
Step Description of Process
• CZ Method
– CZ Crystal Puller – Doping – Impurity Control
• Float-Zone Method • Reasons for Larger Ingot Diameters
Semiconductor Manufacturing Technology
5. State and discuss seven quality measures for wafer suppliers.
6. Explain what is epitaxy and why it is important for wafers.
Semiconductor Manufacturing Technology
2
chemical reaction to produce a silicon-bearing gas of
Si (s) + 3HCl (g) SiHCl3 (g) + H2 (g) + heat
summary 2013
若干个问题
缺陷的类型。 位错的类型。 刃型位错、螺旋位错位错线、滑 移方向、柏氏矢量的几何关系 柏氏矢量的确定方法，不同类型 柏氏矢量的区别。 位错的分解，增殖以及反应
重点与难点
位错的概念及位错理论。
Chapter 3 Binary alloys phase diagrams
3.1 Phases constituent of alloys
• 1.2 Crystal structures of pure metals
基本概念
晶体 空间点阵 阵点 晶胞 晶格常数 晶面指数 晶向指数 晶系 布拉菲点阵 晶面族 晶向族 晶带 面心立方结构 体心立方结构 密排六方结构 致密度 配位数
重点与难点： 晶面指数与晶向指数的确定 fcc,bcc,hcp晶格结构特征
5.1 The representing way of ternary phase diagrams 5.2 Ternary isomorphous phase diagrams 5.3 Ternary eutectic phase diagrams
• • • • • • • • • • •
若干问题 浓度三角形 等比例规则 背向规则 等含量规则 重心规则 杠杆规则 等温截面图 投影图 三元匀晶相图 三元共晶相图（组元不互 溶，组元有限互溶）
Chapter 2 Defects in crystals (crystalline imperfections)
• • • • •
• • • •
2.1 2.2 2.3 2.4 2.5
2.5.1 2.5.2 2.5.3 2.5.4
Point defects Fundamentals of dislocations The movement of dislocations Elastic properties associated with dislocations Dislocation in real crystals
铸造不良英语手册defects handbook
VIII.Metal Composition 金属成份 1) Improper metal composition for the application 使用不合适的金属成份 2) A composition having too high a shrinkage characteristic 有高缩松特性的成分存在 IX. Melting Practice 熔解 1) Excessive carbide stabilizers which promote high shrinkage 过多的碳稳定成份促进缩松 2) Composition having too high a shrinkage characteristic 含有高缩松特性的成分存在 3) Gaseous or over oxidized metal 过多氧化铁 4) Contamination 杂物
IV. Gating and Risering 浇冒口 1) Gating and risering do not, by themselves, create a crush although mounting gate runners too close to the pattern can result in an inadequate sand bearing surface. V. Molding Sand 型砂 1) Weak sand 2) Low green tensile strength may permit the cope to sag 较低的湿压强度可能导致上箱塌落 3) Low green deformation causes the sand to crush rather than give during closing of flask 湿剪变形量低，导致合箱时砂子的破碎 4) Low dry strength in a dry sand mold may fail to support the normal load 干砂型中烘干强度低可能导致不能支撑正常的载荷
传输原理课件 chapter4b位错基本概念
电子显微镜下的位错
透射电镜下钛合金中的位错线(黑线)
高分辨率电镜下的刃位错 （白点为原子）
1926年，苏联物理学家雅科夫∙弗兰克尔（Jacov Frenkel）假定材料发生塑性变形时晶体各部分作为刚 体相对滑动，连接滑移面两边的原子结合键同时断裂 ，这样计算得到常用金属的临界分切应力（见6.2节） 值约为0.1G，G为剪切模量
塑性变形试验中，测得的这些金属的临界分切应力仅 为10−4~10−8G，比理论强度低了整整3个数量级。
Semicoherent interface between Ag-richen and Cu-richen phases
J M HOWE. Interface in Materials, 1997
K.L. Merkle, M.I. Buckett, and Y. Gao 1992
位错理论的发展
1934年，埃贡·奥罗万（Egon Orowan），迈克尔·波拉 尼（Michael Polanyi）和 G. I. 泰勒（G. I. Taylor）三 位科学家几乎同时提出了塑性变形的位错机制理论， 解决了上述理论预测与实际测试结果相矛盾的问题。 位
b a
’ ’
=b
S.P. x
x
• 把滑移设想为刚性整体滑动所需的理论 临界切应力值比实际测量临界切应力值 大3-4个数量级。
滑移不是刚性滑动
多脚 虫 的 爬 行
9
位错（Dislocation）理论的发展
1934年 Taylor、Orowan、Polanyi提出“位错模 型”，滑移是通过称为位错的运动而进行的
材料工程专业英语4晶格缺陷Imperfections in solids
Ordinarily alloying is used in metals to improve mechanical strength and corrosion resistance.
➢Point defects (点缺陷) —zero dimension ➢Linear defects (线缺陷) —one dimension ➢Interfacial defects (面缺陷) —two dimensions
4.2 Point defects in metals
Vacancy or vacant lattice site(空位): the simplest of the point defects, one normally occupied from which an atom is missing.
For a given quantity of material, the number of the vacancies increases with temperature according to
Nv=Nexp(-Qv/kT)
Self-interstitial(填隙子) is another kind of point defects, it is such that an extra atom is crowded into an interstitial site，a small void space that under ordinary circumstances is not occupied.
By “crystalline defect” is meant a lattice irregularity having one or more of its dimensions on the order of an atomic diameter. Classification of crystalline imperfections is frequently made according to geometry or dimensionality of the defect. Several different imperfections are discussed in this Chapter including point detects (those associated with one or two atomic positions)，linear (or one-dimensional) defects，as well as interfacial defects，or boundaries ，which are two dimensional．Impurities in solids are also discussed，since impurity atoms may exist as point defects. Finally, techniques for the microscopic examination of defects and the structure of materials are briefly described．
chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案
chemical-reaction-engineeri ng-3ed-edition作者-octave-Levenspiel-课后习题答案Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (20)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (23)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (27)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (34)CHAPTER 11 BASICS OF NON-IDEAL FLOW (36)CHAPTER 18 SOLID CATALYZED REACTIONS (45)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1Solution:)/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day molgm L mg g L mg day day m dayday m VdtdN r A A ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within theWaste Waste Clean200 mgMean residenZerobeds, based on the oxygen used.Solution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hrkgckgcoal kgc hr coal t N c ⋅-=⨯-=⨯⨯⨯-=∆∆ )/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯= )/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the ratesof formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C 0.5 A C BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance ofsubstrate is given by-r A =A06]][[1760C E A + , mol/m 3·sWhat are the units of the two constants? Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and withsecond-order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C 2 A , mol/cm 3·min If the concentration is to be expressed in mol/liter and time in hours, what wouldbe the value and units of the rate constant?Solution:min)()(3'⋅⨯-=⋅⨯-cm molr hr L mol r A A 22443'300005.0106610)(minAA A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ AA A A A C C cmmol mol L C cmmolC L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(AA A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as -dtdp A= 3.66 p 2 A , atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = - dtdN V A1 = k C2 A , mol/m 3·s Solution:(a) The unit of the rate constant is ]/1[hr atm ⋅ (b) dtdN V r AA 1-=-Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=-22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃?Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150160230295370Temperatu re, ℃13 16 22 24 28 What activation energy represents this change in bustliness? Solution:RTE RTE RTE ek eak t cons ion concentrat f let ion concentrat f ek r ---=⋅⋅=⋅='00tan )()(RET Lnk Lnr A 1'-=∴ Suppose Tx Lnr y A 1,==, so ,REslope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230 295 370 A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752CT o / 13 16 22 24 28 3101-⨯T3.4947 3.4584 3.3881 3.3653 3.3206-y = 5417.9x - 15.686R2 = 0.9712340.00330.003350.00340.003450.00351/T-L n r-y = -5147.9 x + 15.686Also K REslope 9.5147-=-=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter? Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C LnAAo=, 11kt C C LnA Ao =, 22kt C CLn A Ao = Ao A C C 5.01=, Ao A C C 25.02=So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd -order reaction, kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C AoAo Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run?Solution: In a-21order reaction: 5.0AA A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao AoA Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , whichis impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =-And k C C Lnoomin)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Lnk X Ln k t t A A , dissatisfied.In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao Ao Ao Ao Ao Ao A Ao A C C C C C C C C k C C k t t, satisfied.According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n LnAAo=, ''')()(tn n Ln t n n Ln AAo A Ao= 3135213180'Ao n Ln Ln =, 28'=An3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction. Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or-r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hrt ,AC ,mmol /L A Ao AAo C C C C Ln-AAo C C t -1 0.84 1.0897 6.25 20.681.20526.25C A , millimol /liter0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 1 2 3 4 5 6 7 8 9 10 113 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 110.00256.006511.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graphy = 0.9879x + 5.0497R 2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t /(C a o -C a )9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A −−→−enzyme R, -r A =min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.0100)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A tC C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9℃:H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20 t, minC 2H 5SO 4H , mol/li ter t, minC 2H 5SO 4H , mol/li ter1804.1141 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t,mi nLmol C R /,Lmol C A /,AXAAe AAe Ae X X X X X Ln---)12()1(AeAX X Ln -0 0 5.5 0 0 041 1.18 4.91 0.10730.2163 -0.227548 1.38 4.81 0.12540.2587 -0.271755 1.63 4.685 0.14820.3145 -0.329975 2.24 4.38 0.20360.4668 -0.488196 2.75 4.125 0.25 0.6165 -0.642712 7 3.31 3.8450.30090.8140 -0.845614 6 3.76 3.620.34181.0089 -1.044916 2 3.81 3.5950.34641.0332 -1.069718 0 4.11 3.4450.37361.1937 -1.233119 4 4.31 3.3450.39181.3177 -1.359121 2 4.45 3.2750.40451.4150 -1.4578267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:y = 0.0067x - 0.0276R 2= 0.998800.511.522.530100200300400500tL n0067.0)11(21=-=Ao AeC X k Slope ,min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line: -y = 0.0068x - 0.0156R 2 = 0.998600.511.522.530100200300400500x-L n0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter1 2 4 6 7 9 12-r A , mol/liter·hr0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002 4 68 10 12 14Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃ 508427 393 356 283k,cm 3/mol·s0.10590.003100.00058880.9×10-60.942×10-6Find the complete rate equation for this reaction. Use units of joules, moles, cm 3,and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R 2= 0.987904812160.0010.0020.0030.0041/T-L n (k )From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.567E = 60851 J/mol k 0 = 105556 cm 3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e -60851/R T ·C A 2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400K, π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-literpopcorn to be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volumegoes from 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a spacevelocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min.What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be cert ain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double thebolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC 1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) isto be converted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a givenenzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R.Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flowreactor at a flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exitstream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC 2 A1 ,-r A2=kC 2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC C C - So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle ofthe tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offeredabout the first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①。
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OBSERVING EQUIL. VACANCY CONC.
increases via atom motion from the crystal to the surface, where they join the island.
Reprinted with permission from Nature (K.F. McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies in Solids and the Stability of Surface Morphology", Nature, Vol. 412, pp. 622-625 (2001). Image is 5.75 µm by 5.75 µm.) Copyright (2001) Macmillan Publishers, Ltd.
Crystal Structure
Thermomechanical Processing
defects introduction and manipulation
Microstructure
University of Tennessee, Dept. of Materials Science and Engineering 4
Ns =
(6.023 ×10
23
atoms
× 8 .4 g 3 mol cm = 8 × 10 22 atoms 3 g cm 63.5 mol
)
N v = 8 ×1022
atoms 0.9 eV atom = exp − 0.026 eV atom cm3
= 7.4 ×107 vacancies cm3
Introduction To Materials Science, Chapter 4, Imperfections in solids
Types of Defects
Defects may be classified into four categories depending on their dimension: ¾ 0D, Point defects: atoms missing or in irregular places in the lattice (vacancies, interstitials, impurities) ¾ 1D, Linear defects: groups of atoms in irregular positions (e.g. screw and edge dislocations) ¾ 2D, Planar defects: the interfaces between homogeneous regions of the material (grain boundaries, external surfaces) ¾ 3D, Volume defects: extended defects (pores, cracks)
The equilibrium number of vacancies formed as a result of thermal vibrations may be calculated from thermodynamics:
Q N v = N s exp − v k T B
where Ns is the number of regular lattice sites, kB is the Boltzmann constant, Qv is the energy needed to form a vacant lattice site in a perfect crystal, and T the temperature in Kelvin (note, not in oC or oF). Using this equation we can estimate that at room temperature in copper there is one vacancy per 1015 lattice atoms, whereas at high temperature, just below the melting point there is one vacancy for every 10,000 atoms.
Introduction To Materials Science, Chapter 4, Imperfections in solids
Chapter Outline “Crystals are like people, it is the defects in them which tend to make them interesting!” - Colin Humphreys. • Defects in Solids
3
Introduction To Materials Science, Chapter 4, Imperfections in solids
Defects – Introduction (III)
The processing determines the defects
Composition
Bonding
University of Tennessee, Dept. of Materials Science and Engineering
9
Introduction To Materials Science, Chapter 4, Imperfections in solids
Self-interstitials Vacancy
Vacancy - a lattice position that is vacant because the atom is missing. Interstitial - an atom that occupies a place outside the normal lattice position. It may be the same type of atom as the others (self interstitial) or an impurity interstitial atom.
Q N v = N s exp − v k T B
The Boltzmann’s constant kB = 1.38 × 10-23 J/atom-K = 8.62 × 10-5 eV/atom-K The temperature in Kelvin T = 27o C + 273 = 300 K. kBT = 300 K × 8.62 × 10-5 eV/K = 0.026 eV The energy for vacancy formation Qv = 0.9 eV/atom The number of regular lattice sites Ns = NAρ/Acu NA = 6.023 × 1023 atoms/mol ρ = 8.4 g/cm3 Acu = 63.5 g/mol
¾ 0D, Point defects 9 vacancies 9 interstitials 9 impurities, weight and atomic composition ¾ 1D, Dislocations 9 edge 9 screw ¾ 2D, Grain boundaries 9 tilt 9 twist ¾ 3D, Bulk or Volume defects ¾ Atomic vibrations 4.9 - 4.10 Microscopy & Grain size determination – Not Covered / Not Tested
University of Tennessee, Dept. of Materials Science and Engineering
5
Introduction To Materials Science, Chapter 4, Imperfections in solids
Point defects: vacancies & self-interstitials
University of Tennessee, Dept. of Materials Science and Engineering 1
Introduction To Materials Science, Chapter 4, Imperfections in solids
Defects – Introduction (I)
University of Tennessee, Dept. of Materials Science and Engineering
2
Introduction To Materials Science, Chapter 4, Imperfections in solids
Defects – Introduction (II)
University of Tennessee, Dept. of Materials Science and Engineering 7
Introduction To Materials Science, Chapter 4, Imperfections in solids
Let’s estimate the number of vacancies in Cu at room T
Note, that the above equation gives the lower end estimation of the number of vacancies, a large numbers of additional (nonequilibrium) vacancies can be introduced in a growth process or as a result of further treatment (plastic deformation, quenching from high temperature to the ambient one, etc.)