Heat Transfer热传作业4解答

工程热力学课后作业答案chapter4

工程热力学课后作业答案chapter4

1 p73 4-1 1kg 空气在可逆多变过程中吸热40kJ ，其容积增大为1102v v =，压力降低为8/12p p =，设比热为定值，求过程中内能的变化、膨胀功、轴功以及焓和熵的变化。 解：热力系是1kg 空气 过程特征：多变过程 ) 10/1ln()8/1ln()2/1ln()1/2ln(==v v p p n ＝0.9 因为 T c q n ?= 内能变化为 R c v 2 5= ＝717.5)/(K kg J ? v p c R c 5 727==＝1004.5)/(K kg J ? =n c ==--v v c n k n c 51＝3587.5)/(K kg J ? n v v c qc T c u /=?=?＝8×103 J 膨胀功：u q w ?-=＝32 ×103J 轴功：==nw w s 28.8 ×103J 焓变：u k T c h p ?=?=?＝1.4×8＝11.2 ×103J

2 熵变：1 2ln 12ln p p c v v c s v p +=?＝0.82×103)/(K kg J ? 4－2 有1kg 空气、初始状态为MPa p 5.01=，1501=t ℃，进行下列过程： （1）可逆绝热膨胀到MPa p 1.02=； （2）不可逆绝热膨胀到MPa p 1.02=，K T 3002=； （3）可逆等温膨胀到MPa p 1.02=； （4）可逆多变膨胀到MPa p 1.02=，多变指数2=n ； 试求上述各过程中的膨胀功及熵的变化，并将各过程的相对位置画在同一张v p -图和s T -图上 解：热力系1kg 空气 （1） 膨胀功： ])12(1[111k k p p k RT w ---=＝111.9×103 J 熵变为0 （2）)21(T T c u w v -=?-=＝88.3×103J 1 2ln 12ln p p R T T c s p -=?＝116.8)/(K kg J ?

管理学原理Chapter4课后题答案复习进程

第三次作业 宋子瑶经济1503 41507663 2. 3.Describe in detail the six-step strategic management process. The strategy management process is a six-step process that encompasses strategy planning, implementation and evaluation. The first step is identifying the organization’ s current mission, goals and strategies. Every organization needs a mission-a statement of its purpose. Defining the mission forces managers to identify what it’ s in business to do. It’s also important for managers to identify the current goals and strategies for managers have a basis for assessing whether they need to be changed. The second step is doing an external analysis. Analyzing that environment is a critical step in the strategic management process. Once the managers’ve analyzed the environment, managers need to pinpoint opportunities that the organization can exploit and threats that it must counteract or buffer against. And opportunities are positive trends in the external environment and threats are negative trends. The third step is doing an internal analysis. The internal analysis provides important information about an organization’s specific resources and

Chapter 4 Exercises 含答案

Chapter 4 From Word to Text I. Decide whether each of the following statements is true or false: 1. Grammatical sentences are formed following a set of syntactic rules. 2. The syntactic rules of any language are finite in number, but there is no limit to the number of sentences native speakers of that language are able to produce and comprehend. 3. An endocentric construction is also known as headed construction because it has just one head 4. Constituents that can be substituted for one another without loss of grammaticality belong to the same syntactic category. 5. In English syntactic analysis, four phrasal categories are commonly recognized and discussed, namely, noun phrase, verb phrase, infinitive phrase, and auxiliary phrase. 6. Number and gender are categories of noun and pronoun. 7. Word order plays an important role in the organization of English sentences. 8. Like English, modern Chinese is a SVO language. 9. In English the subject usually precedes the verb and the direct object usually follows the verb. 10. A noun phrase must contain a noun, but other elements are optional. ( 1-5 TTFTF 6-10 TTTTT ) ＩI. There are four given choices for each statement below. Mark the choice that can best complete the statement: 1 The head of the phrase “the city Rome”is_______D___ A the city B Rome C city D the city and Rome 2. The phrase “on the half” belongs to __B______construction A endocentric B exocentric C subordinate D coordinate

会计学：企业决策的基础exercises-chapter4答案

SOLUTIONS TO EXERCISES Ex 4–1 a. Book value b. Materiality c. Matching principle d. Unrecorded revenue e. Adjusting entries f. Unearned revenue g. Prepaid expenses h. None (This is an example of ―depreciation expense.‖) Ex. 4–2 Income Statement Balance Sheet Adjusting Entry a b NE I D NE I D c I NE I I NE I d NE I D NE I D e NE I D D NE D f I NE I NE D I Ex. 4–3 a. Rent Expense....................................................................................... 240,000 Prepaid Rent ........................................................................... 240,000 To record rent expense for May ($1,200,000 ÷ 5 months = $240,000 per month). b. Unearned Ticket Revenue .................................................................. 148,800 Ticket Revenue ....................................................................... 148,800 To record earning portion of season ticket revenue relating to May home games. Ex. 4–4 a. (1) Interest Expense (375) Interest Payable (375) $50,000 x 9% annual rate x 1/12 = $375. (2) Accounts Receivable .................................................................... 10,000 Consulting Fees Earned ................................................... 10,000 To record ten days of unbilled consulting fees at $1,000 per day. b. $2,250 ($50,000 x 9% x 6/12 = $2,250) c. $15,000 ($25,000 - $10,000 earned in December, 2002)

2Radiative heat transfer between nanostructurespdf

Radiative heat transfer between nanostructures A.I.Volokitin1,2and B.N.J.Persson1 1Institut fu¨r Festko¨rperforschung,Forschungszentrum Ju¨lich,D-52425,Germany 2Samara State Technical University,443010Samara,Russia ?Received14July2000;revised manuscript received25September2000;published16April2001? We use a general theory of the?uctuating electromagnetic?eld and a generalized Kirchhoff’s law?Ref.8? to calculate the heat transfer between macroscopic and nanoscale bodies of arbitrary shape,dispersive,and absorptive dielectric properties.We study the heat transfer between:?a?two parallel semi-in?nite bodies,?b?a semi-in?nite body and a spherical body,and?c?two spherical bodies.We consider the dependence of the heat transfer on the temperature T,the shape and the separation d,and discuss the role of nonlocal and retardation effects.We?nd that for low-resistivity material the heat transfer is dominated by retardation effects even for the very short separations. DOI:10.1103/PhysRevB.63.205404PACS number?s?:65.80.?n I.INTRODUCTION It is well known that for bodies separated by d?d W ?c?/k B T the radiative heat transfer between them is de- scribed by the Stefan-Bolzman law: J? ?2k B4 60?3c2 ?T14?T24?,?1? where T1and T2are the temperatures of solid1and2,re-spectively.In this limiting case the heat transfer is connected with traveling electromagnetic waves radiated by the bodies, and does not depend on the separation d.For d?d W,the heat transfer increases by many order of magnitude,which can be explained by the existence of evanescent electromag-netic?eld that decay exponentially into the vacuum.At the present time there is an increasing number of investigations of heat transfer due to evanescent waves in connection with scanning tunneling microscopy and scanning thermal mi-croscopy?STM?under ultrahigh vacuum conditions.1–4STM can be used for local heating of the surface,resulting in local desorption or decomposition of molecular species,and this offers further possibilities for the STM to control local chem-istry on a surface. A general formalism for evaluating the heat transfer be-tween macroscopic bodies was proposed some years ago by Polder and Van Hove.1Their theory is based on the general theory of the?uctuating electromagnetic?eld developed by Rytov5and applied by Lifshitz6for studying the conservative part,and by Volokitin and Persson7for studying the dissipa-tive part of the van der Waals interaction.The formalism of Polder and Van Hove can be signi?cantly simpli?ed using a generalized Kirchhoff’s law.2,8In this approach,the calcula-tion of the correlation functions for the?uctuating electro-magnetic?eld is reduced to?nding the electromagnetic?eld created by a point dipole outside the bodies.The formalism of Polder and Van Hove requires the determination of the electromagnetic?eld for all space and for all position of a point dipole,and requires the integration of the product of the component of the electric and magnetic?eld over the volumes of two bodies.In the present paper we use a simpler formalism,which is originally due to Levin and Rytov.8This formalism requires only the evaluation of a surface integral over one of the bodies and is simpli?ed further in the non-retarded limit?small distances between bodies?,where the calculation of the heat transfer is reduced to the problem of ?nding the electrostatic potential due to a point charge.We apply the formalism to the calculation of the heat transfer between:?a?two semi-in?nite bodies,?b?a semi-in?nite body and a spherical particle,and?c?two spherical particles. Problem?a?was considered by Polder and Van Hove,1 Levin,Polevoy,and Rytov,2and more recently by Pendry.3 In comparison with other treatments,we study in detail the nonlocal and retardation effects.A striking result we?nd is that for low-resistivity metals retardation effects become cru-cial and in fact dominate the heat transfer between bodies. The problem?b?was recently studied by Pendry in a differ-ent formalism.3We shall point out the differences between our results and those obtained by Pendry,wherever appropri-ate. II.FORMALISM Following Polder and Van Hove,1to calculate the?uctu-ating electromagnetic?eld we use the general theory of Ry-tov?see Refs.5,8?.This method is based on the introduction of a?uctuating current density in the Maxwell equations ?just as,for example,the introduction of a‘‘random’’force in the theory of Brownian motion of a particle?.For a mono- chromatic?eld?time factor exp(?i?t)?in a dielectric,non-magnetic medium,these equations are “?E?i?c B,?2? “?H??i?c D?4?c j f,?3? where,according to Rytov,we have introduced a?uctuating current density j f associated with thermal and quantum?uc-tuations.E,D,H,and B are the electric and the electric-displacement?eld,and the magnetic and the magnetic-induction?elds,respectively.For non-magnetic media B ?H and D??E,where?is the dielectric constant of the PHYSICAL REVIEW B,VOLUME63,205404

第4章 作业

Chapter4 homework 一、Select the best answer based on this couese ( 选择题). 1.The computer memory system refers to _________ A.RAM B. ROM C. Main memory D.Register , main memory, cache, external memory 2.If the word of memory is 16 bits, which the following answer is right? A. The address width is 16 bits B. The address width is related with 16 bits C.The address width is not related with 16 bits D. The address width is not less than 16 bits 3.The characteristics of internal memory compared to external memory A. Big capacity, high speed, low cost B. Big capacity, low speed, high cost C. small capacity, high speed, high cost D. small capacity, high speed, low cost 4．On address mapping of cache, any block of main memory can be mapped to any line of cache, it is ___________ . A. Associative Mapping B. Direct Mapping C. Set Associative Mapping D. Random Mapping 5. Cache’s write-through polity means write operation to main memory _______. A. as well as to cache B. only when the cache is replaced C. when the difference between cache and main memory is found D. only when direct mapping is used 6. Cache’s write-back polity means write operation to main memory ______________. A. as well as to cache B. only when the relative cache is replaced C. when the difference between cache and main memory is found D. only when using direct mapping 7. On address mapping of cache, the data in any block of main memory can be mapped to fixed line of cache, it is _________________. A. associative mapping B. direct mapping C. set associative mapping D. random mapping 8. On address mapping of cache, the data in any block of main memory can be mapped to fixed set any line(way) of cache, it is _________________. A. associative mapping B. direct mapping C. set associative mapping D. random mapping 9. Computer memory is organized into a hierarchy. At the highest level are the ___________. A. registers B. cache C. main memory D. external memory 10. On address mapping of cache, the data in any block of main memory can be mapped to __________ of cache, it is direct mapping . A. any line B. fixed line C. fixed set any line D. A and B 11. The characteristics of external memory compared to internal memory are _______ . A.Big capacity, high speed, low cost B. Big capacity, low speed, low cost C. small capacity, high speed, high cost D. small capacity, high speed, low cost 12. Write _____ policy can result in memory write bottle-neck. A. back B. through C. from D. to 13.A 16KByte cache has a line size of four 32-bit words, the number of line is . A 210 B 10 C 28 D 8 14. If the address-length of memory is 16 bits, which the following answer is NOT right ? . A.The addressable unit is 16 bits B.The addressing range is 216 C.The maxmum possible memory capacity is determined D.The addressable unit is not related with 16 bits 15. Suppose that the word from location X in memory can be mapped any line in the cache, this mapping function is called _____________. A. associative Mapping B. direct Mapping C. set Associative Mapping D. random Mapping 16. The simplest technique is called _____________.Using this technique, all write operations are made to main memory as well as to the cache. A. write-back B. write-through C. write-invalidate D. write-update

Chapter 4 练习题

第四章练习题 I. Define the following terms (名词解释) ： 1. Culture 6. Factual knowledge 2. Cultural sensitivity 7. Interpretive knowledge 3. Linguistic distance 8. Cultural values 4. Strategy of unplanned change 9. Cultural borrowing 5. Social institutions 10. Material culture II. Multiple Choice Questions (单项选择)： 1.__________ is pervasive in all marketing activities; the marketer's efforts actually become a part of the __________. A. Resistance; opposition to change B. Culture; fabric of culture C. Acceptance; new global context D. Public relations; culture E. Change; marketing strategy 2._____________ is the human-made part of human environment. A. Sociology B. Psychology C. Culture D. Reference groups E. Cohort groups 3.The sum total of knowledge, beliefs, art, morals, law, customs, and any other capabilities and habits acquired by humans as members of society is called: A. Sociology B. Psychology C. Culture D. Reference groups E. Cohort groups 4.When marketers first introduced the PDA to the American consumer they performed the role of being a(n) _____________ because the cultural impact of the product became

chapter4习题

DC-AC变换器（无源逆变电路） 一、学习目的： 通过本章的学习，学者可以了解逆变器的电路结构、分类、特点及主要性能 指标；对三种基本变换方式——方波变换、阶梯波变换、正弦波变换，有一定的 认识；可以理解采用各种变换方式的逆变器的工作原理；了解空间矢量PWM控制 的基本原理。 二、主要内容： 1、基本概念 DC-AC变换器是指能将一定幅值的直流输入电压（或电流）变换成一定幅值、 一定频率的交流输出电压（或电流），并向无源负载（如电机、电炉、或其它用 电器等）供电的电力电子装置，又称为无源逆变电路，常简称作逆变器（Inverter）。 完成直流电压变换的逆变器称为电压型逆变器，而完成直流电流变换的逆变 器则称为电流型逆变器。 2、变换方式的分类 （1）方波变换方式 逆变器的交流输出有两种基本调制方式：脉冲幅值调制（PAM－Pluse Amplitude Modulation）和单脉冲调制（SPM－Single Pluse Modulation）。 所谓脉冲幅值调制（PAM）是指：逆变器的输出频率可由180°方波或120°方波（如图4-3b 所示）的周期来控制，而逆变器输出基波的幅值则由输出方波的幅值即逆变器直流侧电压（或电流）的幅值来控制。显然，采用PAM控制方式时，其方波的导通角恒定（180°方波或120°方波）。 所谓的单脉冲调制（SPM）是指：逆变器的输出频率仍由方波的周期来控制，而逆变器输出基波的幅值则由逆变器输出方波的导通角进行控制，即可使导通角在0°～180°范围调节。显然，采用SPM控制方式时，逆变器输出方波的幅值即逆变器直流侧电压（或电流）的幅值恒定。 （2）阶梯波变换方式 （3）斩控调制方式：是指逆变器输出的调制脉冲幅值固定不变，而逆变器中的功率管 以一定的控制规律进行调制。斩控调制方式主要有以下二类即：①脉冲宽度调制（PWM ）；②脉冲频率调制（PFM） 3、逆变器的分类 （1）按直流侧储能元件的性质，逆变器可分为电压型逆变器（VSI－Voltage Source Inverter）和电流型逆变器（CSI－Current Source Inverter）。 （2）按逆变器输出波形的不同，逆变器可分为方波逆变器、阶梯波逆变器、以及正弦 波逆变器等。 （3）按逆变器功率电路结构形式的不同，逆变器可分为半桥逆变器、全桥逆变器、推 挽式逆变器等。 （4）按逆变器功率电路的功率器件的不同，逆变器可分为半控型逆变器和全控型逆变 器。 （5）按逆变器输出频率的不同，逆变器可分为工频逆变器、中频逆变器以及高频逆变 器。

Abaqus 热传递Heat Transfer

ENGI 7706/7934: Finite Element Analysis Abaqus CAE Tutorial 4: Heat Transfer ___________________________________________________________ Problem Description The thin plate (7035) shown below is exposed to a temperature of 25 degree. When the temperature reaches 150 degree, the plate will have expansion. A fixed boundary condition of the top plate will cause changes in stress field. The thermal expansion coefficient is . Plot the stress filed at 150 degree. r = 7.5 10

ABAQUS Analysis Steps 1. Start Abaqus and choose to create a new model database 2. In the model tree double click on the “Parts” node (or right click on “parts” and select Create) 3. In the Create Part dialog box name the part and a. Select “2D Planar” b. Select “Deformable” c. Select “Shell” d. Set approximate size = 100 e. Click “Continue…” 4. Create the geometry shown below (not discussed here)

工程热力学课后作业标准答案chapter4

p73 4-1 1kg 空气在可逆多变过程中吸热４0kJ ，其容积增大为1102v v =，压力降低为8/12p p =,设比热为定值，求过程中内能的变化、膨胀功、轴功以及焓和熵的变化。 解:热力系是1kg 空气 过程特征：多变过程 ) 10/1ln()8/1ln()2/1ln()1/2ln(==v v p p n =0.９ 因为 T c q n ?= 内能变化为 R c v 2 5= ＝717.5)/(K kg J ? v p c R c 5 727==＝100４.5)/(K kg J ? =n c ==--v v c n k n c 51＝358７.5)/(K kg J ? n v v c qc T c u /=?=?＝8×103J 膨胀功:u q w ?-==3２ ×１0３J 轴功：==nw w s ２８．8 ×103J 焓变:u k T c h p ?=?=?=1.4×８=11.2 ×103J

熵变:1 2ln 12ln p p c v v c s v p +=?=０.82×1０３)/(K kg J ? 4－2 ?有1ｋg 空气、初始状态为MPa p 5.01=,1501=t ℃，进行下列过程： （1）可逆绝热膨胀到MPa p 1.02=； （2)不可逆绝热膨胀到MPa p 1.02=,K T 3002=; (3）可逆等温膨胀到MPa p 1.02=； （4)可逆多变膨胀到MPa p 1.02=，多变指数2=n ； 试求上述各过程中的膨胀功及熵的变化，并将各过程的相对位置画在同一张v p -图和s T -图上 解:热力系1kg 空气 （1） 膨胀功： ])12(1[111k k p p k RT w ---==１11．9×１03J 熵变为0 （2）)21(T T c u w v -=?-=＝88.３×103J 1 2ln 12ln p p R T T c s p -=?＝116.8)/(K kg J ?

语言学chapter4课后练习答案

Chapter 4 Revision Exercises 1. What is syntax? Syntax is a branch of linguistics that studies how words are combined to form sentences and the rules that govern the formation of sentences. 2. What is phrase structure rule? The grammatical mechanism that regulates the arrangement of elements . specifiers, heads, and complements) that make up a phrase is called a phrase structure rule. The phrase structural rule for NP, VP, AP, and PP can be written as follows: NP→(Det) N (PP) ... VP→(Qual) V (NP) ... AP→(Deg) A (PP) ... PP→(Deg) P (NP) ... We can formulate a single general phrasal structural rule in which X stands for the head N, V, A or P. The XP rule: XP→(specifier) X (complement) 3. What is category? How to determine a word’s category? Category refers to a group of linguistic items which fulfill the same or similar functions in a particular language such as a sentence, a noun phrase or a verb. To determine a word's category, three criteria are usually employed, namely meaning, inflection and distribution. The most reliable of determining a word’s category is its distribution. 4. What is coordinate structure and what properties does it have? The structure formed by joining two or more elements of the same type with the help of a conjunction is called coordinate structure. It has four important properties: 1)there is no limit on the number of coordinated categories that can appear prior to the conjunction. 2) a category at any level a head or an entire XP can be coordinated. 3)coordinated categories must be of the same type. 4)the category type of the coordinate phrase is identical to the category type of the elements being conjoined. 5. What elements does a phrase contain and what role does each element play?