自动化专业英语课后习题答案

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自动化专业英语第三版王宏文

UNIT 1Electrical NetworksA 电路An electrical circuit or network is composed of elements such as resistors, inductors, and capacito rs connected together in some manner. If the network contains no energy sources, such as batteri es or electrical generators, it is known as a passive network. On the other hand, if one or more en ergy sources are present, the resultant combination is an active network. In studying the behavior of an electrical network, we are interested in determining the voltages and currents that exist wit hin the circuit. Since a network is composed of passive circuit elements, we must first define the electrical characteristics of these elements.电路或电网络由以某种方式连接的电阻器、电感器和电容器等元件组成。

如果网络不包含能源，如电池或发电机，那么就被称作无源网络。

换句话说，如果存在一个或多个能源，那么组合的结果为有源网络。

在研究电网络的特性时，我们感兴趣的是确定电路中的电压和电流。

因为网络由无源电路元件组成，所以必须首先定义这些元件的电特性.In the case of a resistor, the voltage-current relationship is given by Ohm's law, which states that t he voltage across the resistor is equal to the current through the resistor multiplied by the value of the resistance. Mathematically, this is expressed as就电阻来说，电压-电流的关系由欧姆定律给出，欧姆定律指出：电阻两端的电压等于电阻上流过的电流乘以电阻值。

自动控制专业英语习题参考答案.doc

自动控制专业英语习题参考答案Lesson 1 Introduction to Control Systems1.Translate the following into Chinese.(1)In 1922 Minorsky worked on automatic controllers for steering ships and showed how stability could be determined from the differential equations describing the system.1922年，Minorsky开发了用于轮船驾驶的自动控制器，并指出根据描述系统的差分方程确定系统稳定性的方法。

(2) A home heating system in which a thermostat is the controller is an example of an automatic regulating system.家用供暖系统是自动调节系统的实例，其中的温度调节装置就是控制器。

(3)An engine which rejected no heat and which converted all the heat absorbed to mechanical work would therefore be perfectly consistent with the first law of thermodynamics.一台不散热并且把吸收的所有热量都转换为机械功的机器就与热力学第一定律完全一致。

(4)In short, a robot can do the dirty work —the dull, repetitious, dehumanizing and sometimes dangerous work that humans won"t or shouldn't do.简言之，机器人能干脏活，即那些人们不愿意或不该做的枯燥的、重复性的、呆板且有时有危险的活。

电气工程及其自动化专业英语苏小林课后答案

电气工程及其自动化专业英语苏小林课后答案【篇一：电气工程及其自动化准耶英语】/p> characterize描绘…的特征，塑造人物，具有….的特征property 性质，财产equal in magnitude to 在数量（数量级）上等同于 convert 转换converter 转换器time rate 时间变化率mathematically 从数学上来讲differentiatev 区分，区别in honor of 为纪念某人 name in honor of为纪念某人而以他命名electromotive force （ e m f ）电动势voltaic battery 伏打电池，化学电池an element 一个电器元件interpret 口译，解释，说明potential difference/voltage 电势差/电压 expend 花费，消耗instantaneous 瞬时的，促发的passive sign convention 关联参考方向the law of conservation of energy 能量守恒定律 reference polarity 参考极性electron 电子 electronic 电子的 electric 电的，电动的 time-varying 时变的 constant-valued 常量的metallic 金属的be due to 是因为，由于，归功于building block 模块coulomb库伦，ampere安培，joule焦耳，volt伏特，watt瓦特，work 功变量u（t），i(t)是电路中最基本的概念。

他们描述了电路中的各种关系。

电荷量的概念是解释电现象的基本原理，电荷量也是电路中最基本的量。

电荷也是构成物质的原子的电器属性，量纲是库伦。

我们从初等物理可以得知所有物质是由基本组成部分原子组成，而原子又包括电子（electron），质子（proton）和中子（neutron）我们都知道电荷e是带负电的电子，在数量上等于1.60210*1019 c, 而质子携带同等电荷量的正电荷，相同数量的质子，电子使原子呈现电中性（neutrally charged）。

机械设计制造及其自动化专业英语课后题

Aliuminum 铝copper 铜nicke 镍titanium 钛structural strength 结构强度deep drawing 拉伸加工hardenability 硬化性machinability 可加工性cold drawn冷拔steel sheet钢板percent reduction in area 断面收缩率endurance limit 疲劳极限rolled-steel shapes 轧制钢板corrosion resistance 抗腐蚀性rupture 断裂non-ferrous 非钢的stress-strain curve 应力应变曲线yield point 屈服点percentage elongation 伸长率necking 颈缩sensitivity 灵敏性Kinematic elements运动员素External appearance外观Sound judgment准确判断Fatigue strenghth结构强度Enviroment damage环境损害Ductile or brittle韧性或脆性Blow out吹息Interference fit joint干涉配合关节定义 defintion力 f orce轴axle非金属nometal结构structure载荷load用途use性质properties低碳钢low-carbon高强度钢hinger-strengt steel热处理heat treatment屈服强度yield strength弹性模量elastic modulus伸长率percentage elongation韧性toughness内应力internal stresses应变硬化strain hardening横截面cross-sectional area断面收缩率reduction in area比例极限limit of proportionality屈服极限yield limit延性ductiliy机械性质mechanical propertiece 用..除..divide byT he distinction between a mental and nonmetal is not away clear cutEngineers would not be particularly interest in such a metal if absolute pure metal were to be producedOf the 50 or so metallic elements,only a few produce and used in large quantities in engineer practices.In the elastic range,the deformation of the specimen disappeared after the load was removed.Logically speaking,once the elastic limit is exceeded,the metal should start to yield,and finally break,without any increase in the value of stressThe purpose of design calculations is to predict the stress or deformation in the part in order that it may safely carry the load which will be imposed on it and that it my last for the expected life of machineDynamic loads are generally more dangerous than strain loads and fatigue strength must be consideredWhen loaded the material deforms and the amount of deformation depends on the size of the load。

2012机械设计制造及其自动化专业外语参考答案

一词汇a 英译汉Dimension 尺寸Machine tool Adhesiveness Compound rest复合刀具台机床粘性Apron 挡板，Climb-cut Peripheral milling Slab milling阔面铣削同向铣削Centerless grinding无心轮磨Machinery Cylindrical hole Snag 障碍机械圆柱孔Coarseness 粗糙度Up-milling逆铣Abrasive belt磨带Profile Machine table剖面机表Rectilinear Protecctive guards 直线防护String milling连续铣削b 汉译英刚度，硬度hardess 圆柱形端铣刀cylindric end mill 进给速度feed speed 托架bracket软刚，低碳钢mild steel沉头螺钉sunk screw螺线spiral 平面磨surface grind镗床boring machine 摩擦离合器friction clutch多刃刀具multipoint tool 牛头刨床shaping machine被动皮带轮Passive pulley润滑油lubricating oil燕尾槽dovetail groove 卡盘chuck材料去除装备Material removal equipment插入式磨削Plug-in grinding冷却液coolant 床头箱headstock二翻译句子1) The slides a nd slideways of a machine tool locate and guide members which move relative to each other,usually changing the position of the tool relative to the workpiece.车刀机床的滑块和滑道s使定位和导向构件之间将对移动，常常用于改变工件与工作台的相对位置2）In order to maintain a constant angular velocity, the individual tooth profile must obey the fundamental law of gearing: for a pair of gears to transmit a constant angular velocity ratio, the shape of their contacting profile must be such that the common normal passes through a fixed point on the line of the centers为了维持一个恒定的角速度,单个齿廓必须遵守啮合基本定律:对于一对以固定角速度传动的齿轮,他们的传动比两轮连心线被齿廓接触点公共法线分割的两顿先吨的反比3）Plain milling cutters are adapted to cut by teeth with cutting edges situatued on the surface of a cylinder which can be circumscribed on cutter. face mills are adapted to cutting by teeth with cutting edges situated on the surface of the mill and partially on its cylindrial surface平铣刀适用于使用位于圆柱面上的齿上的切削刃工作的铣削，平面铣刀适用于使用位于铣刀表面和其部分圆柱面的齿上的切削刃做的铣削4）Pressures angles for spur gear are usually 14.5 or 20 degrees, although other values can be used. Meshing gears must have the same pressure angles.虽然其他值也也可以使用，但齿轮的压力角经常为14.5度或20度，相啮合的齿轮必须有相同的压力角5）The cross section of the milled surface corresponds to the outline or contour of the milling cutter or combination of cutters used.加工表面的界面要与使用的刀具或组合刀具的外轮廓一致6）Lets us now discuss the different concepts associated with the manufacturing accuracy required modern mass-production technologies.让我们来讨论有关现代大规模生产技术所要求的制造精度的不同概念。

机械设计制造及其自动化专业英语课后题

机械设计制造及其自动化专业英语课后题第一单元3. Aliuminum铝copper 铜nicke镍titanium 钛structural strength结构强度deep drawing拉伸加工4. 定义defition 力torce 轴axle(roller) 非金属nometal 结构structure 载荷load 用途use(application)性质properties(nature)(character)第二单元4.hardenability硬化性machinability可加工性cold drawn 冷拔steel sheet钢板percent reduction in area断面收缩率endurance limit疲劳极限rolled-steel shapes 轧制钢板corrosion resistance 抗腐蚀性rupture断裂5.低碳钢low-carbon 高强度钢hinger-strengt steel热处理heat treatment屈服强度yield strength弹性模量elastic modulus伸长率percentage elongation韧性toughness内应力internal stresses第三单元4.non-ferrous 非钢的stress-strain curve应力应变曲线yield point屈服点percentage elongation伸长率necking 颈缩sensitivity 灵敏性5.应变硬化strain hardening横截面cross-sectional area断面收缩率reduction in area比例极限limit of proportionality 屈服极限yield limit 延性ductiliy机械性质mechannicalpropertiece用..除..divide…by…第六单元3.tangential notes肤浅的事情flexible manufacturingsystem 柔性制造系统machine instruction机器指令economy of scale规模经济Hardwireyd logic controller硬固线逻辑控制transfer-line运输线，流水线numerically control(NC)数字控制direct numerical control(DNC)直接数字控制computer numerical control(CNC)计算机数字控制4.计算机辅助制造computer-aided manufacturing数控机床手工、半自动化或全自动化manal semiautomatic or full automation 尽管机械制造业一直在持续发展，但知道20世纪50年代才出现又一个重大发展。

自动化课后答案ch13 intro mfg sys-3e-si

Chapter 13INTRODUCTION TO MANUFACTURING SYSTEMSREVIEW QUESTIONS13.1What is a manufacturing system?Answer: The definition given in the text is the following: A manufacturing system is a collection of integrated equipment and human resources, whose function is to perform one or more processing and/or assemblyoperations on a starting raw material, part, or set of parts.13.2Name the four components of a manufacturing system.Answer: As listed in the text, the four components are (1) production machines plus tools, fixtures, and other related hardware, (2) a material handling system, (3) a computer system to coordinate and/or control thepreceding components, and (4) human workers to operate and manage the system.13.3What are the three classifications of production machines, in terms of worker participation?Answer: In terms of worker participation, the machines can be classified as (1) manually operated, (2) semi-automated or (3) fully automated.13.4What are the five material handling functions that must be provided in a manufacturing system?Answer: The five material handling functions that must be provided in a manufacturing system are (1) loading work units at each station, (2) positioning the work units at the station, (3) unloading the work units from the station, (4) transporting work units between stations in manufacturing systems comprised of multipleworkstations, and (5) temporary storage of work units to prevent starving of workstations.13.5What is the difference between fixed routing and variable routing in manufacturing systems consisting ofmultiple workstations?Answer: In fixed routing, the work units always flow through the same sequence of workstations. This means that the work units are identical or similar enough that the processing sequence is identical. In variable routing, work units are transported through a variety of different station sequences. This means that the manufacturing system processes or assembles different types of work units.13.6What is a pallet fixture in work transport in a manufacturing system?Answer: A pallet fixture is a workholder that is designed to be transported by the material handling system.The part is accurately attached to the fixture on the upper face of the pallet, and the under portion of the pallet is designed to be moved, located, and clamped in position at each workstation in the system.13.7 A computer system is an integral component in a modern manufacturing system. Name four of the eight functionsof the computer system listed in the text.Answer: The eight functions of the computer system identified in the text are (1) communicate instructions to workers, (2) download part programs to computer-controlled machines, (3) control the material handlingsystem, (4) schedule production, (5) failure diagnosis, (6) safety monitoring, (7) quality control, and (8)operations management.13.8What are the five factors that can be used to distinguish manufacturing systems in the classification schemeproposed in the chapter?Answer: The five factors identified in the text are (1) types of operations performed, (2) number ofworkstations, (3) system layout, (4) automation and manning level, and (5) part or product variety.13.9Why is manning level inversely correlated with automation level in a manufacturing system?Answer: Manning level is inversely correlated with automation level in a manufacturing system because thenumber of workers required to operate the system tends to be reduced as the level of automation increases.13.10Name the three cases of part or product variety in manufacturing systems. Briefly define each of the three cases.Answer: The three cases of part or product variety in manufacturing systems are (1) single model, (2) batch model, and (3) mixed model. In the single-model case, all parts or products made by the manufacturingsystem are identical. In the batch-model case, different parts or products are made by the system, but they are made in batches because the physical setup and/or equipment programming must be changed over between models. In the mixed-model case, different parts or products are made by the system, but the differences are not significant, so the system is able to handle them without the need for time-consuming changeovers insetup or program.13.11What is flexibility in a manufacturing system?Answer: Flexibility is the attribute that allows a mixed-model manufacturing system to cope with a certain level of variation in part or product style without interruptions in production for changeovers between models.13.12What are the three capabilities that a manufacturing system must possess in order to be flexible?Answer: As identified in the text, the three capabilities are (1) identification of the different work units, (2)quick changeover of operating instructions, and (3) quick changeover of the physical setup.。

自动化课后答案ch17 automated asby-3e-si

Chapter 17AUTOMATED ASSEMBLY SYSTEMSREVIEW QUESTIONS17.1Name three of the four conditions under which automated assembly technology should be considered.Answer: The four conditions named in the text are (1) high product demand, so that millions of units areproduced, (2) stable product design, because design changes require tooling changes in the assembly system, (3) the assembly consists of no more than a limited number of components, say around a dozen or fewer, and (4) the product is designed for automated assembly.17.2What are the four automated assembly system configurations listed in the text?Answer: The four configurations are (a) in-line assembly machine, (b) dial-type assembly machine, (c) carousel assembly system, and (d) single-station assembly machine.17.3What are the typical hardware components of a workstation parts delivery system?Answer: The typical components are (1) hopper, (2) parts feeder, (3) selector and/or orientor, (4) feed track, and(5) escapement and placement device.17.4What is a programmable parts feeder?Answer: As defined in the text, a programmable parts feeder is a feeder that is capable of feeding components of varying geometries with only a few minutes required to make the adjustments (change the program) for thedifferences.17.5Name six typical products that are made by automated assembly.Answer: The products identified in the text are alarm clocks, audio tape cassettes, ball bearings, ball point pens, cigarette lighters, computer disks, electrical plugs and sockets, fuel injectors, gear boxes, light bulbs, locks, mechanical pens and pencils, printed circuit board assemblies, pumps for household appliances, small electric motors, spark plugs, video tape cassettes, and wrist watches.17.6Considering the assembly machine as a game of chance, what are the three possible events that might occur whenthe feed mechanism attempts to feed the next component to the assembly workhead at a given workstation in a multi-station system?Answer: The three possible events are (1) the component is defective and causes a station jam, (2) thecomponent is defective but does not cause a station jam, and (3) the component is not defective.17.7Name some of the important performance measures for an automated assembly system.Answer: The performance measures considered in the text include yield (proportion of assemblies produced with no defective components), production rate, proportion uptime (a.k.a. line efficiency), and unit cost per assembly.17.8Why is the production rate inherently lower on a single-station assembly system than on a multi-station assemblysystem?Answer: Because all of the work elements are performed sequentially at one station in the single station system, whereas the elements are performed simultaneously at multiple workstations in a multi-station system.17.9What are two reasons for the existence of partially automated production lines?Answer: The two reasons given in the text are (1) automation is introduced gradually on an existing manual line and (2) certain manual operations are too difficult or too costly to automate.17.10What are the effects of poor quality parts, as represented by the fraction defect rate, on the performance of anautomated assembly system?Answer: The two effects given in the text are (1) jams at stations that stop the entire assembly system toadversely affect production rate, uptime proportion, and cost per unit produced; or (2) assembly of defective parts in the product to adversely affect yield of good assemblies and product cost.17.11Why are storage buffers used on partially automated production lines?Answer: Storage buffers are used on partially automated production lines to isolate the manual stations from the breakdowns of the automated stations. Thus, workers do not have to stop working when an automated station breaks down.PROBLEMSParts Feeding17.1 A feeder-selector device at one of the stations of an automated assembly machine has a feed rate of 25 parts perminute and provides a throughput of one part in four. The ideal cycle time of the assembly machine is 10 sec.The low level sensor on the feed track is set at 10 parts, and the high level sensor is set at 20 parts. (a) How long will it take for the supply of parts to be depleted from the high level sensor to the low level sensor once thefeeder-selector device is turned off? (b) How long will it take for the parts to be resupplied from the low level sensor to the high level sensor, on average, after the feeder-selector device is turned on? (c) What proportion of the time that the assembly machine is operating will the feeder-selector device be turned on? Turned off?Solution: (a) Time to deplete from n f2 to n f1Rate of depletion = cycle rate R c = 60/10 = 6 parts/minTime to deplete = (20 - 10)/6 = 10/6 = 1.667 min(b) Time to resupply from n f1 to n f2Rate of resupply = fθ - R c = 25(0.25) - 6 = 0.25 parts/minTime to resupply = (20 - 10)/0.25 = 10/0.25 = 40 min(c) Total cycle of depletion and resupply = 41.667 minProportion of time feeder-selector is on = 40/41.667 = 0.96Proportion of time feeder-selector is off = 1.667/41.667 = 0.0417.2Solve Problem 17.1 but use a feed rate of 32 parts per minute. Note the importance of tuning the feeder-selectorrate to the cycle rate of the assembly machine.Solution: (a) Time to deplete = (20 - 10)/6 = 1.667 min(b) Rate of resupply = 32(.25) - 6 = 8 - 6 = 2 parts/minTime to resupply = (20 - 10)/2 = 10/2 = 5 min(c) Total cycle time = 6.667 minProportion of time feeder-selector turned on = 5/6.667 = 0.75Proportion of time feeder-selector turned off = 1.667/6.667 = 0.2517.3 A synchronous assembly machine has 8 stations and must produce at an average rate of 480 completedassemblies per hour. Average downtime per jam is 2.5 min. When a breakdown occurs, all subsystems (including the feeder) stop. The frequency of breakdowns of the machine is once every 50 parts. One of the eight stations is an automatic assembly operation that uses a feeder-selector. The components fed into the selector can have any of six possible orientations, each with equal probability, but only one of which is correct for passage into the feed track to the assembly workhead. Parts rejected by the selector are fed back into the hopper. What minimum rate must the feeder deliver components to the selector during system uptime in order to keep up with the assembly machine?Solution: T p = 60/480 = 0.125 min/asbyT p = T c + FT d = T c +15025(.)= T c + 0.05T c = 0.125 - 0.05 = 0.075 min/asbyR c = 110.075cT== 13.333 asbys/minMin fθ = 0.1667 f = 13.333 asbys/min Feeder rate f = 1333301667..= 80 parts/minMulti-Station Assembly Systems17.4 A dial indexing machine has six stations that perform assembly operations on a base part. The operations,element times, q and m values for components added are given in the table below (NA means q and m are not applicable to the operation). The indexing time for the dial table is 2 sec. When a jam occurs, it requires 1.5 min to release the jam and put the machine back in operation. Determine (a) production rate for the assemblymachine, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system.Station Operation Element time q m1 Add part A 4 sec 0.015 0.62 Fasten part A3 sec NA NA3 Assemble part B 5 sec 0.01 0.84 Add part C 4 sec 0.02 1.05 Fasten part C 3 sec NA NA6 Assemble part D 6 sec 0.01 0.5Solution: (a) (mq) = 0.6(0.015) + 0.8(0.01) + 1(0.02) + .5(0.01) = .042T p = 0.1333 + 0.042(1.5) = 0.19633 min/asbyR p = 60/0.19633 = 305.6 asbys/hr(b) P ap = (1-0.015+0.6x.015)(1-0.01+0.8x.01)(1-0.02+1x0.02)(1-0.01+0.5x0.01)= (0.994)(0.998)(1.0)(0.995) = 0.98705(c) E = 0.1333/0.19633 = 0.679 = 67.9%17.5An eight-station assembly machine has an ideal cycle time of 6 sec. The fraction defect rate at each of the 8stations is q = 0.015 and a defect always jams the affected station. When a breakdown occurs, it takes 1 minute, on average, for the system to be put back into operation. Determine the production rate for the assemblymachine, the yield of good product (final assemblies containing no defective components), and proportionuptime of the system.Solution: T p = 0.1 + 8(1.0)(0.015)(1.0) = 0.22 min/asby.R p = 60/0.22 = 272.7 asbys/hrIf defects always jam the affected station, then m = 1.0P ap = (1 - 0.015 + 1x0.015)8 = 1.0 = yieldE = 0.1/0.22 = 0.4545 = 45.45%17.6Solve Problem 17.5 but assume that defects never jam the workstations. Other data are the same.Solution: T p = 0.1 + 8(0)(0.015)(1.0) = 0.10 min/asby.R p = 60/0.10 = 600 asbys/hrIf defects never jam, then m = 0P ap = (1 - 0.015 + 0x0.015)8 = 0.8861 = yieldE = 0.1/0.1 = 100%17.7Solve Problem 17.5 but assume that m = 0.6 for all stations. Other data are the same.Solution: T p = 0.1 + 8(0.6)(0.015)(1.0) = 0.172 min/asbyR p = 60/0.172 = 348.8 asbys/hrP ap = (1 - 0.015 + 0.6x0.015)8 = 0.953 = yieldE = 0.1/0.172 = 0.5814 = 58.14%17.8 A six-station automatic assembly line has an ideal cycle time of 12 sec. Downtime occurs for two reasons. First,mechanical and electrical failures cause line stops that occur with a frequency of once per 50 cycles. Average downtime for these causes is 3 min. Second, defective components also result in downtime. The fraction defect rate of each of the six components added to the base part at the six stations is 2%. The probability that a defective component will cause a station jam is 0.5 for all stations. Downtime per occurrence for defective parts is 2 min.Determine (a) yield of assemblies that are free of defective components, (b) proportion of assemblies that contain at least one defective component, (c) average production rate of good product, and (d) uptime efficiency.Solution: (a) P ap = (1 - 0.02 + 0.5x0.02)6 = (0.99)6 = 0.9415(b) P qp = 1 - 0.9415 = 0.0585(c) T p = 12/60 + 0.02(3) + 6(0.5)(0.02)(2) = 0.38 minR p = 60/0.38 = 157.6 cycles/hrR ap = (0.9415)(157.9) = 148.66 good asbys/hr(d) E = 0.2/0.38 = 0.526 = 52.6%17.9An eight-station automatic assembly machine has an ideal cycle time of 10 sec. Downtime is caused by defectiveparts jamming at the individual assembly stations. The average downtime per occurrence is 2.0 min. The fraction defect rate is 1.0% and the probability that a defective part will jam at a given station is 0.6 for all stations. The cost to operate the assembly machine is $90.00 per hour and the cost of components being assembled is $.60 per unit assembly. Ignore other costs. Determine (a) yield of good assemblies, (b) average production rate of good assemblies, (c) proportion of assemblies with at least one defective component, and (d) unit cost of the assembled product.Solution: (a) P ap = (1 - 0.01 + 0.6x0.01)8 = (0.996)8 = 0.9684(b) T p = 0.1667 + 8(0.6)(0.01)(2) = 0.2627 min/asbyR p = 60/0.2627 = 228.4 asbys/hrR ap = (0.9684)(228.4) = 221.2 good asbys/hr(c) P qp = 1 - 0.9684 = 0.0316(d) C pc = [0.60+1.50(0.2627)]/0.9684 = $1.0265/asby17.10An automated assembly machine has four workstations. The first station presents the base part, and the otherthree stations add parts to the base. The ideal cycle time for the machine is 3 sec, and the average downtimewhen a jam results from a defective part is 1.5 min. The fraction defective rates (q) and probabilities that adefective part will jam the station (m) are given in the table below. Quantities of 100,000 for each of the bases, brackets, pins, and retainers are used to stock the assembly line for operation. Determine (a) proportion of good product to total product coming off the line, (b) production rate of good product coming off the line, (c) totalnumber of final assemblies produced, given the starting component quantities. Of the total, how many are good product, and how many are products that contain at least one defective component? (d) Of the number ofdefective assemblies determined in above part (c), how many will have defective base parts? How many willhave defective brackets? How many will have defective pins? How many will have defective retainers?Station Part identification q m1 Base 0.01 1.02 Bracket 0.02 1.03 Pin 0.03 1.04 Retainer 0.04 0.5Solution: (a) P ap = (1-0.01+1x0.01)(1-0.02+1x0.02)(1-0.03+1x0.03)(1-0.04+0.5x0.04)= (1.0)(1.0)(1.0)(0.98) = 0.98(b) T p = 3/60 + (0.01+0.02+0.03+0.04x0.5)(1.5) = 0.17 min/cycle.R ap = 0.98(60/0.17) = 345.9 good asbys/hr(c) The diagram below shows quantities of components at the four workstations in the assembly machine:base 100,000 bracket100,000pin100,000retainer100,000→→→→95060 good asbys+ 1940 with defective 1000 def 2000 def 3000 def 1940 def retainers Total number produced = 95,060 + 1,940 = 97,000Number of units of good product = 95,060Number of units containing at least one defect = 1,940 (d) Number of products containing defective base parts = 0 Number of products containing defective brackets = 0 Number of products containing defective pins = 0Number of products containing defective retainers = 1,94017.11 A six-station automatic assembly machine has an ideal cycle time of 6 sec. At stations 2 through 6, parts feedersdeliver components to be assembled to a base part that is added at the first station. Each of stations 2 through 6 is identical and the five components are identical. That is, the completed product consists of the base part plus the five components. The base parts have zero defects, but the other components are defective at a rate q . When an attempt is made to assemble a defective component to the base part, the machine stops (m = 1.0). It takes an average of 2.0 min to make repairs and start the machine up after each stoppage. Since all components areidentical, they are purchased from a supplier who can control the fraction defect rate very closely. However, the supplier charges a premium for better quality. The cost per component is determined by the following equation:Cost per component = 0.1 +0.0012qwhere q = the fraction defect rate. Cost of the base part is 20 cents. Accordingly, the total cost of the base part and the five components is:Product material cost = 0.70 +0.006qThe cost to operate the automatic assembly machine is $150.00 per hour. The problem facing the production manager is this: As the component quality decreases (q increases), the downtime increases which drives production costs up. As the quality improves (q decreases), the material cost increases because of the priceformula used by the supplier. To minimize total cost, the optimum value of q must be determined. Determine by analytical methods (rather than trial-and-error) the value of q that minimizes the total cost per assembly. Also, determine the associated cost per assembly and production rate. (Ignore other costs). Solution : Product material cost C m = 0.20 + 5(0.1 +0.0012q ) = 0.70 + 0.006qT p = 0.1 + 5(1)(q )(2.0) = 0.1 + 10qC pc = C m + C L T p = 0.70 + 0.006q + 2.50(0.1 + 10q ) = 0.70 + 0.006q + 0.25 + 25q = 0.95 + 0.006q + 25qTaking the derivative of the cost equation with respect to q : 20.006250pc dC dqq -=+= 20.006q = 25 q 2 = 0.006/25 = 0.00024q = 0.0155Using this value in the preceding cost equation,C pc = 0.70 + 0.006/0.0155 + 2.50(0.1 + 10x0.0155) = $1.725/asbyT p = 0.1 + 5(10)(0.0155)(2.0) = 0.255 min/asby R p = 60/0.255 = 235.3 asbys/hr17.12 A six-station dial indexing machine is designed to perform four assembly operations at stations 2 through 5 aftera base part has been manually loaded at station 1. Station 6 is the unload station. Each assembly operation involves the attachment of a component to the existing base. At each of the four assembly stations, ahopper-feeder is used to deliver components to a selector device that separates components that are improperly oriented and drops them back into the hopper. The system was designed with the operating parameters for stations 2 through 5 as given in the table below. It takes 2 sec to index the dial from one station position to the next. When a component jam occurs, it takes an average of 2 min to release the jam and restart the system. Line stops due to mechanical and electrical failures of the assembly machine are not significant and can be neglected. The foreman says the system was designed to produce at a certain hourly rate, which takes into account the jams resulting from defective components. However, the actual delivery of finished assemblies is far below thatdesigned production rate. Analyze the problem and determine the following: (a) the designed average production rate that the foreman alluded to, (b) the proportion of assemblies coming off the system that contain one or moredefective components, (c) the problem that limits the assembly system from achieving the expected production rate, and (d) the production rate that the system is actually achieving. State any assumptions that you make in determining your answer.Station Assembly time Feed rate f Selector θq m2 4 sec 32/min 0.25 0.01 1.03 7 sec 20/min 0.50 0.005 0.64 5 sec 20/min 0.25 0.02 1.05 3 sec 15/min 1.0 0.01 0.7Solution: T d = 2 min, T c = 7 + 2 = 9 sec = 0.15 min Cycle rate R c = 1/0.15 = 6.667 cycles/min(a) ∑(mq) = 1(0.01) + 0.6(0.005) + 1(0.02) + 0.7(0.01) = 0.04T p = 0.15 + 0.04(2) = 0.23 minR p = 60/0.23 = 260.9 asbys/hr(b) P ap = (1-0.01+1x0.01)(1-0.005+0.6x0.005)(1-0.02+1x0.02)(1-0.01+0.7x0.01)= (1)(0.998)(1)(0.997) = 0.995P qp= 1 - 0.995 = 0.005R ap = 260.9(0.995) = 259.6 good asbys/hr(c) Station 2: fθ = 32(.25) = 8 components/minStation 3: fθ = 20(.50) = 10 components/minStation 4: fθ = 20(.25) = 5 components/minStation 5: fθ = 15(1.0) = 15 components/minThe problem is that the feeder for station 4 is slower than the machine's cycle rate of 6.667 cycles/min(d) If the machine operates at the cycle rate that is consistent with the feed rate of Station 4, then T c = 12 sec =0.20 minT p = 0.20 + 0.04(2) = 0.28 minR p = 60/0.28 = 214.3 asbys/hrR ap = 214.3(0.995) = 213.2 good asbys/hr17.13For Example 17.4 in the text, dealing with a single-station assembly system, suppose that the sequence ofassembly elements were to be accomplished on a seven-station assembly system with synchronous parts transfer.Each element is performed at a separate station (stations 2 through 6) and the assembly time at each respective station is the same as the element time given in Example 17.4. Assume that the handling time is divided evenly(3.5 sec each) between a load station (station 1) and an unload station (station 7). The transfer time is 2 sec, andthe average downtime per downtime occurrence is 2.0 min. Determine (a) production rate of all completed units,(b) yield, (c) production rate of good quality completed units, and (d) uptime efficiency.Solution: (a) T c = 7 + 2 = 9.0 sec = 0.15 minF = 0.02(1.0) + 0.01(0.6) + 0.015(0.8) + 0.02(1.0) + 0.012 = 0.070T p = 0.15 + 0.070(2.0) = 0.15 + 0.14 = 0.29 minR p = 1/0.29 = 3.45 asbys/min = 206.9 asbys/hr(b) P ap = (1.0)(0.996)(0.997)(1.0) = 0.993(c) R ap = 206.9(0.993) = 205.5 good asbys/hr(d) E = 0.15/0.29 = 0.5172 = 51.7%Comment: Comparing the values with those in Example 17.4, production rate is more than double for the multi-station system, yield is the same, and line efficiency is greatly reduced because of the much faster cycle time. Single Station Assembly Systems17.14 A single-station assembly machine is to be considered as an alternative to the dial-indexing machine in Problem17.4. Use the data given in the table for that problem to determine (a) production rate, (b) yield of good product(final assemblies containing no defective components), and (c) proportion uptime of the system. Handling time to load the base part and unload the finished assembly is 7 sec and the downtime averages 1.5 min every time acomponent jams. Why is the proportion uptime so much higher than in the case of the dial-indexing machine in Problem 17.4?Solution: (a) T c = 7 + (4 + 3 + 5 + 4 + 3 + 6) = 7 + 25 = 32 sec = 0.5333 min∑(mq) = 0.6(0.015) + 0.8(0.01) + 1(0.02) + 0.5(0.01) = 0.042 (same as for Problem 17.4)T p = 0.5333 + 0.042(1.5) = 0.59633 minR p = 60/0.59633 = 100.6 asbys/hr(b) P ap = (1-0.015+0.6x0.015)(1-0.01+0.8x0.01)(1-0.02+1x0.02)(1-0.01+0.5x0.01) = 0.98705 (same as forProblem 17.4)(c) E = .5333/.59633 = 0.8943 = 89.43%Comment: Proportion of uptime E is so much higher than in Problem 17.4 because the cycle time is much longer for the single-station machine than for the six-station dial-indexing machine (32 sec vs 6 sec). The averagedowntime per cycle is the same for both machines, but it is a much lower proportion of the longer cycle time in the single station case.17.15 A single station robotic assembly system performs a series of five assembly elements, each of which adds adifferent component to a base part. Each element takes 4.5 sec. In addition, the handling time needed to move the base part into and out of position is 4 sec. For identification, the components, as well as the elements thatassemble them, are numbered 1, 2, 3, 4, and 5. The fraction defect rate is 0.005 for all components, and theprobability of a jam by a defective component is 0.7. Average downtime per occurrence = 2.0 min. Determine (a) production rate, (b) yield of good product in the output, (c) uptime efficiency, and (d) proportion of the output that contains a defective type 3 component.Solution: (a) T p = T c + nmqT dT c = 4 + 5(4.5) = 26.5 sec = 0.44167 minT p = 0.44167 + 5(0.7)(0.005)(2.0) = 0.47667 minR p = 1/0.47667 = 2.098 asbys/min = 125.9 asbys/hr(b) P ap = (1 - 0.005 + 0.7(0.005))5 = (0.9985)5 = 0.9925(c) E = 0.44167/0.47667 = 0.9266 = 92.66%(d) Type 3 defect = 1 - (1 - 0.005 + 0.7(0.005)) = 0.005 - 0.7(0.005) = 0.3(0.005) = 0.001517.16 A robotic assembly cell uses an industrial robot to perform a series of assembly operations. The base part andparts 2 and 3 are delivered by vibratory bowl feeders that use selectors to insure that only properly oriented parts are delivered to the robot for assembly. The robot cell performs the elements in the table below (also given are feeder rates, selector proportion θ, element times, fraction defect rate q, and probability of jam m, and, for the last element, the frequency of downtime incidents p). In addition to the times given in the table, the time required to unload the completed subassembly takes 4 sec. When a linestop occurs, it takes an average of 1.8 min to make repairs and restart the cell. Determine (a) yield of good product, (b) average production rate of good product, and(c) uptime efficiency for the cell? State any assumptions you must make about the operation of the cell in orderto solve the problem.Element Feed rate f Selector θElement Time T e q m p1 15 pc/min 0.30 Load base part 4 sec 0.01 0.62 12 pc/min 0.25 Add part 23 sec 0.02 0.33 25 pc/min 0.10 Add part 34 sec 0.03 0.84 Fasten 3 sec 0.02Solution: Assumptions: (1) Feeders continue to operate and deliver parts into the feed track even when a jam occurs during assembly. (2) Low level quantity n f1 is sufficient to eliminate possibility of a stockout.T c = T h + ∑T e = 4 + 4 + 3 + 4 + 3 = 18 sec = 0.3 minT p = T c + (∑qm + p)T d = 0.3 + (0.01 x 0.6 + 0.02 x 0.3 + 0.03 x 0.8 + 0.02)(1.8)= 0.3 + 0.056)(1.8) = 0.4008 min/asbyCheck feed rates:f1θ1 = 15(0.3) = 4.5 pc/min or 0.2222 min/pcf2θ2 = 12(0.25) = 3.0 pc/min or 0.3333 min/pcf3 3 = 25(0.1) = 2.5 pc/min or 0.40 min/pcEach of these can be completed within the average cycle time of the robot assembly operation.(a) P ap = (1 - 0.01 + 0.6 x 0.01)(1 - 0.02 + 0.3 x 0.02)(1 - 0.03 + 0.8 x 0.03)= (0.996)(0.986)(0.994) = 0.976(b) R p = 1/0.4008 = 2.495 asbys/min = 149.7 asbys/hrR ap = 149.7(0.9762) = 146.1 good asbys/hr(c) E = 0.3/0.4008 = 0.7485 = 74.85%Alternative assumption: (1) Feeder stops when jam occurs. Hence, cycle time is limited by feeder #3.T c = 0.4 min T p = 0.4 + (0.056)(1.8) = 0.5008 min(a) R p = 1/0.5008 = 1.9968 asbys/min = 119.8 asbys/hr P ap = 0.976 (same as above)(b) R ap = 119.8(0.9762) = 116.9 good asbys/hr(c) E = 0.400/0.5008 = 0.7987 = 79.87%Partial Automation17.17 A partially automated production line has a mixture of three mechanized and three manual workstations. Thereare a total of six stations, and the ideal cycle time of 1.0 min, which includes a transfer time of 6 sec. Data on the six stations are listed in the accompanying table. Cost of the transfer mechanism C at = $0.10/min, cost to run each automated station C as = $0.12/min, and labor cost to operate each manual station C w = $0.17/min. It has been proposed to substitute an automated station in place of station 5. The cost of this station is estimated at C as5 = $0.25/min and its breakdown rate p5 = 0.02, but its process time would be only 30 sec, thus reducing the overall cycle time of the line from 1.0 min to 36 sec. Average downtime per breakdown of the current line, as well as for the proposed configuration, is 3.5 min. Determine the following for the current line and the proposed line: (a) production rate, (b) proportion uptime, and (c) cost per unit. Assume the line operates without storage buffers, so when an automated station stops, the whole line stops, including the manual stations. Also, in computing costs, neglect material and tooling costs.Station Type Process time p i1 Manual 36 sec 02 Automatic 15 sec 0.013 Automatic 20 sec 0.024 Automatic 25 sec 0.015 Manual 54 sec 06 Manual 33 sec 0Solution: For the current line,(a) T c = 1.0 min, F = 0.01 + 0.02 + 0.01 = 0.04T p = 1.0 + 0.04(3.5) = 1.0 + 0.14 = 1.14 min/unit, R p = 1/1.14 = 0.877 units/min = 52.6 units/hr(b) E = 1.0/1.14 = 0.877 = 87.7%(c) C o = 0.10 + 3(0.12) + 3(0.17) = $0.97/min. C pc = (0.97)(1.14) = $1.106/unit.For the proposed line in which station 5 is automated,(a) T c = 36 sec = 0.6 min F = 0.01 + 0.02 + 0.01 + 0.02 = 0.06T p = 0.6 + 0.06(3.5) = 0.6 + 0.21 = 0.81 min/unit, R p = 1/0.81 = 1.235 units/min = 74.1 units/hr(b) E = 0.6/0.81 = 0.7407 = 74.1%(c) C o = 0.10 + 3(0.12) + 0.25 + 2(0.17) = $1.05/min C pc = (1.05)(0.81) = $0.851/unit.17.18Reconsider Problem 17.17 except that both the current line and the proposed line will have storage buffers beforeand after the manual stations. The storage buffers will be of sufficient capacity to allow these manual stations to operate independently of the automated portions of the line. Determine (a) production rate, (b) proportionuptime, and (c) cost per unit for the current line and the proposed line.Solution: For the current line,(a) Automated section of line: T c = 25 + 6 = 31 sec = 0.5167 minF = 0.01 + 0.02 + 0.01 = 0.04 T p = 0.5167 + 0.04(3.5) = 0.6567 min/unit.Manual section of line: T c = 54 + 6 = 60 sec = 1.0 minNo breakdowns, so F = 0. T p = 1.0 + 0(3.5) = 1.0 min/unit.Using the manual section as the limiting section, R p = 1/1.0 = 1.0 units/min = 60.0 units/hr(b) E = 0.5167/1.0 = 0.5167 = 51.7%Comment: The automated sections of the line could operate with an efficiency = 0.5167/0.6567 = 0.787, but production rate is ultimately limited by the manual cycle time of 1.0 min.(c) C o = 0.10 + 3(0.12) + 3(0.17) = $0.97/min C pc = (0.97)(1.0) = $0.97/unit.For the proposed line in which station 5 is automated,(a) Automated section of line: T c = 36 sec = 0.6 minF = 0.01 + 0.02 + 0.01 + 0.02 = 0.06 T p = 0.6 + 0.06(3.5) = 0.6 + 0.21 = 0.81 min/unit.Manual section of line: T c = 36 + 6 = 42 sec = 0.7 minNo breakdowns, so F = 0. T p = 0.7 + 0(3.5) = 0.7 min/unit.Using the automated section as the limiting section, R p = 1/0.81 = 1.235 units/min = 74.1 units/hr(b) E = 0.6/0.81 = 0.7407 = 74.1%(c) C o = 0.10 + 3(0.12) + 0.25 + 2(0.17) = $1.05/min C pc = (1.05)(0.81) = $0.851/unit.17.19 A manual assembly line has six stations. The assembly time at each manual station is 60 sec. Parts are transferredby hand from one station to the next, and the lack of discipline in this method adds 12 sec (T r = 12 sec) to the cycle time. Hence, the current cycle time is 72 sec. The following two proposals have been made: (1) Install a mechanized transfer system to pace the line; and (2) automate one or more of the manual stations using robots that would perform the same tasks as humans only faster. The second proposal requires the mechanized transfer system of the first proposal and would result in a partially or fully automated assembly line. The transfer system would have a transfer time of 6 sec, thus reducing the cycle time on the manual line to 66 sec. Regarding thesecond proposal, all six stations are candidates for automation. Each automated station would have an assembly time of 30 sec. Thus if all six stations were automated the cycle time for the line would be 36 sec. There aredifferences in the quality of parts added at the stations; these data are given in the accompanying table for each station (q = fraction defect rate, m = probability that a defect will jam the station). Average downtime per station jam at the automated stations is 3.0 min. Assume that the manual stations do not experience line stops due todefective components. Cost data: C at = $0.10/min; C w = $0.20/min; and C as = $0.15/min. Determine if either or both of the proposals should be accepted. If the second proposal is accepted, how many stations should beautomated and which ones? Use cost per piece as the criterion of your decision. Assume for all cases considered that the line operates without storage buffers, so when an automated station stops, the whole line stops, including the manual stations.Station q i m i Station q i m i1 0.005 1.0 4 0.020 1.02 0.010 1.0 5 0.025 1.03 0.015 1.0 6 0.030 1.0Solution: Proposal 1: Current operation: T c = 1.2 min C o = 6(0.20) = $1.20/minC pc = 1.20(1.2) = $1.44/unit.Proposal: T c = 1.1 min C o = 0.10 + 6(0.20) = 1.30/minC pc = 1.30(1.1) = $1.43/unit.Conclusion: Accept Proposal 1.Proposal 2: T c = 36 sec = 0.6 min if all six stations are automated.F = 0.005(1.0) + 0.01(1.0) + 0.015(1.0) + 0.02(1.0) + 0.025(1.0) + 0.03(1.0) = 0.105T p = 0.6 + 0.105(3.0) = 0.6 + 0.315 = 0.915 min/unitC o = 0.10 + 6(0.15) = 1.00/minC pc = 1.00(0.915) = $0.915/unitConclusion: Accept Proposal 2.。

自动化专业英语课后答案

自动化专业英语课后答案【篇一：自动化专业英语试卷及答案】translate the following words or phrases into chinese ①metric度量收敛方差随机等价动态方程判据、评价标准测量模型偏微分方程在概率意义上，依概率② convergence ③ covariance④stochastic ⑤ equivalence ⑥ dynamic equation ⑦ criteria ⑧measurement model ⑨ partial derivative equation ⑩ in the sense of probability2. explain the following symbols in english?? ① bm the second order derivative of b sub m, b double prime sub m② a?b③ xi?a④ ?yx a is equivalent to b xi approaches a the partial derivative of y with respective to x⑤ ?abintegral between limits b to a ⑥ a‖b ⑦ a is parallel to b the cube root of a set a is contained in x a ⑧ a?x⑨ ?a?b?c?d??e?fdivided by e equals f a plus b minus c multiplied d, all⑩ ??,?n,s.t ai?a as i??for any special ?, there is a respected n, such that ai approaches a as i approaches ?二、translate the following paragraphs into chineseparagraph a[8 points]a subset a?xis said to be bounded if there exists m?rsuch that d(x,y)?mfor allx,y?a. the diameter of a is defined to besup?d(x,y)x,y?a?. a bounded metric space is one for which x itself is bounded.bounded intervals in r are bounded sets. a discrete metric space is bounded (take m?1).译文：一个子集a?x被称为是有界的，如果存在m?r使得对所有x,y?a有d(x,y)?m。

电气工程及其自动化专业英语苏小林课后答案

他们描述了电路中的各种关系。

电荷量的概念是解释电现象的基本原理，电荷量也是电路中最基本的量。

电荷也是构成物质的原子的电器属性，量纲是库伦。

自动化专业英语教程

自动化专业英语教程
《自动化专业英语教程》是一本面向高等工科院校自动化专业学生的教材，涵盖了自动化专业的各个发展方向。

以下是该教程的相关介绍：
全书共包括电气与电子工程基础、控制理论、计算机控制技术、过程控制系统、网络化与信息化控制、自动化技术的综合应用6部分共30个单元。

新增了智能控制综述、DSP、嵌入式系统、电力系统自动化、智能电网、大数据应用、知识自动化、云计算、智慧城市和智慧企业技术等诸多内容，覆盖9万余字的专业词汇量。

该教程可作为自动化专业本科生及研究生专业英语课程的教材，也可供有关工程技术人员参考。

自动化专业英语答案

自动化专业英语答案【篇一：自动化专业英语考试答案】>二。

句子翻译： 40分，8题1:p2a common method of analyzing an electrical network is mesh or loop analysis. the fundamentallaw that is applied in this method is kirchhoffs first law, which states that the algebraic sun of thevoltages around a closed loop is 0 ,or , in any closed loop , the sum of the voltage rises must equal the sum of the voltage drops. mesh analysis consists of assuming thatcurrents~termed loop currents~flow in each loop of anetwork ,algebraically summing the voltage drops around each loop ,and setting each sum equal to 0.分析电网络的一般方法是网孔分析法或回路分析法。

应用于此方法的基本定律是基尔霍夫第一定律，基尔霍夫第一定律指出：一个闭合回路中的电压代数和为0，换句话说，任一闭合回路中的电压升等于电压降。

网孔分析指的是：假设有一个电流——即所谓的回路电流——流过电路中的每一个回路，求每一个回路电压降的代数和，并令其为零。

2:p17alternatively, suppose that there had been attached to each pilot’s seat an electronic device that provided an output voltage which is v1 when the seat is occupied and v2 when the seat is not occupied. let us attach the designation “true” to the voltage level v2 so that the le vel v1 is “false”. let us further construct an electric circuit with two sets of input terminals and one set of output terminals. the circuit is to have the property that the output voltage will be v2 if and only if both inputs, i.e., one input and simultaneously the other, are at the level v2. otherwise the output is v1 finally let us connect the inputs to the devices on the chairs of pilots a and b and arrange that an alarm bell, connected to the output z, respond when the output is v2 (“true”) and not o therwise. we have then constructed a circuit which performs the and operation and is capable of making the logical deduction that the plane is unpiloted when, indeed, both pilots leave the cockpit换句话说，假设每一位飞行员座位下面有一个电子装置，当座位上有人时，其输出电压为v1，当座位上无人时，其输出电压为v2。

2电气自动化专业英语翻译中文译文课后答案（李久胜马洪飞陈宏钧

电气自动化专业英语（翻译1-3）指示电压，内阻相对高一些时，这个表可以用来测量电压值。

第一部分：电子技术注意：不管如何设计，指针移动的距离取决于线圈的电流值。

第一章电子测量仪表为了让这类表用在交流电中，在设计时必须作微小的改电子技术人员使用许多不同类型的测量仪器。

一些工作需要动。

整流器可以把交流变成直流电。

整流器合并进仪表中并且精确测量面另一些工作只需粗略估计。

有些仪器被使用仅仅是量程要指示出正确的交流电压值。

整流器类型的仪表不能用于确定线路是否完整。

最常用的测量测试仪表有：电压测试仪，直流电中并且它一般被设计成电压表。

电压表，欧姆表，连续性测试仪，兆欧表，瓦特表还有瓦特小如图1。

2，电测力计是另一种能用于交流电的既能作安培时表。

表也能作电压表的仪器。

它由两个固定线圈和一个移动线圈构所有测量电值的表基本上都是电流表。

他们测量或是比成。

这三个线圈通过两个螺旋型弹簧串联在一起。

这个弹簧支较通过他们的电流值。

这些仪表可以被校准并且设计了不同的撑住移动线圈。

当电流流行性过线圈时移动线圈顺时针方向移量程，以便读出期望的数值。

动。

1．1安全预防电测力计因为属永久磁铁型仪表，量程不是均匀分布的。

仪表的正确连接对于使用者的安全预防和仪表的正确维作用在动线圈上的力根据流过该线圈的电流平方来变化。

有必护是非常重要的。

仪表的结构和操作的基本知识能帮助使用者要在量程开始比量程结束分割的密一点。

分割点之间距离越大，按安全工作程序来对他们正确连接和维护。

许多仪表被设计的仪表的读数越精确。

争取精确的读值是重要的。

只能用于直流或只能用于交流，而其它的则可交替使用。

注意：移动叶片结构是仪表的另一种类型。

电流流过线圈引起两每种仪表只能用来测量符合设计要求的电流类型。

如果用在不个铁片（叶片）磁化。

一个叶片是可动的，另一个是固定的。

正确的电流类型中可能对仪表有危险并且可能对使用者引起伤在两个叶片间的磁的作用引起可动叶片扭转。

移动的数值取决害。

自动化专业英语教程机械工业出版社

自动化专业英语教程机械工业出版社PART 1Electrical and Electronic Engineering BasicsUNIT 1A Electrical Networks ————————————3B Three-phase CircuitsUNIT 2A The Operational Amplifier ———————————5B TransistorsUNIT 3A Logical Variables and Flip-flop ——————————8B Binary Number SystemUNIT 4A Power Semiconductor Devices ——————————11B Power Electronic ConvertersUNIT 5A Types of DC Motors —————————————15B Closed-loop Control of DC DriversUNIT 6A AC Machines ———————————————19B Induction Motor DriveUNIT 7A Electric Power System ————————————22B Power System AutomationPART 2Control TheoryUNIT 1A The World of Control ————————————27B The Transfer Function and the Laplace Transformation —————29UNIT 2A Stability and the Time Response —————————30B Steady State—————————————————31UNIT 3A The Root Locus —————————————32B The Frequency Response Methods: Nyquist Diagrams —————33UNIT 4A The Frequency Response Methods: Bode Piots —————34B Nonlinear Control System 37UNIT 5 A Introduction to Modern Control Theory 38B State Equations 40UNIT 6 A Controllability, Observability, and StabilityB Optimum Control SystemsUNIT 7 A Conventional and Intelligent ControlB Artificial Neural NetworkPART 3 Computer Control TechnologyUNIT 1 A Computer Structure and Function 42B Fundamentals of Computer and Networks 43UNIT 2 A Interfaces to External Signals and Devices 44B The Applications of Computers 46UNIT 3 A PLC OverviewB PACs for Industrial Control, the Future of ControlUNIT 4 A Fundamentals of Single-chip Microcomputer 49B Understanding DSP and Its UsesUNIT 5 A A First Look at Embedded SystemsB Embedded Systems DesignPART 4 Process ControlUNIT 1 A A Process Control System 50B Fundamentals of Process Control 52UNIT 2 A Sensors and Transmitters 53B Final Control Elements and ControllersUNIT 3 A P Controllers and PI ControllersB PID Controllers and Other ControllersUNIT 4 A Indicating InstrumentsB Control PanelsPART 5 Control Based on Network and InformationUNIT 1 A Automation Networking Application AreasB Evolution of Control System ArchitectureUNIT 2 A Fundamental Issues in Networked Control SystemsB Stability of NCSs with Network-induced DelayUNIT 3 A Fundamentals of the Database SystemB Virtual Manufacturing—A Growing Trend in AutomationUNIT 4 A Concepts of Computer Integrated ManufacturingB Enterprise Resources Planning and BeyondPART 6 Synthetic Applications of Automatic TechnologyUNIT 1 A Recent Advances and Future Trends in Electrical Machine DriversB System Evolution in Intelligent BuildingsUNIT 2 A Industrial RobotB A General Introduction to Pattern RecognitionUNIT 3 A Renewable EnergyB Electric VehiclesUNIT 1A 电路电路或电网络由以某种方式连接的电阻器、电感器和电容器等元件组成。

机械设计制造及其自动化专业英语第三版第二章课后答案

机械设计制造及其自动化专业英语第三版第二章课后答案1、Location is the first thing customers consider when_____to buy a house. [单选题] *A.planning(正确答案)B.plannedC.having plannedD.to plan2、It was _____the policeman came_____the parents knew what had happened to their son. [单选题] *A.before…asB. until…whenC. not until…that(正确答案)D.until…that3、A?pen _______ writing. [单选题] *A. is used toB. used toC. is used for(正确答案)D. used for4、It is reported（）three people were badly injured in the traffic accident. [单选题] *A. whichB. that(正确答案)C.whileD.what5、It _____ us a lot of time to do this job. [单选题] *A. spentB. madeC. took(正确答案)D. cost6、78．—Welcome to China. I hope you'll enjoy the ________.—Thank you. [单选题] * A．tour(正确答案)B．sizeC．nameD．colour7、As I know, his salary as a doctor is much higher_____. [单选题] *A. than that of a teacher(正确答案)B. than a teacherC. to that of a teacherD. to a teacher8、It’s raining outside. Take an _______ with you. [单选题] *A. cashB. life ringC. cameraD. umbrella(正确答案)9、In order to find the missing child, villagers _______ all they can over the past five hours. [单选题] *A. didB. doC. had doneD. have been doing(正确答案)10、He often comes to work early and he is _______ late for work. [单选题] *A. usuallyB. never(正确答案)C. oftenD. sometimes11、Do not _______ me to help you unless you work harder. [单选题] *A. expect(正确答案)B. hopeC. dependD. think12、She talks too much; you’ll be glad when you’re free of her. [单选题] *A. 与她自由交谈B. 离开她(正确答案)C. 受她的控制D. 与她在一起13、We have ______ homework today. （）[单选题] *A. too manyB. too much(正确答案)C. much tooD. very much14、( ) What she is worried __ is ____ her daughter is always addicted to chatting online./; that [单选题] *A /; thatB of thatC about that(正确答案)D about what15、The students in that university are not fewer than（）in our university. [单选题] *A. the oneB. thatC. themD. those(正确答案)16、Kate has a cat _______ Mimi. [单选题] *A. called(正确答案)B. callC. to callD. calling17、There was a time（）I wondered why I would like to do this boring job. [单选题] *A. whichB. whyC. whereD. when(正确答案)18、Do you know what（）the change in his attitude? [单选题] *A. got throughB. brought about(正确答案)C. turned intoD. resulted from19、_______ your parents at home last week? [单选题] *A. IsB. WasC. AreD. Were(正确答案)20、The early Americans wanted the King to respect their rights. [单选题] *A. 统治B. 满足C. 尊重(正确答案)D. 知道21、一Mary wants to invite you to see the movie today. 一I would rather she（B）me tomorrow. [单选题] *A.tellsB. told (正确答案)C. would tellD. had told22、We’re proud that China _______ stronger and stronger these years. [单选题] *A. will becomeB. becameC. is becoming(正确答案)D. was becoming23、The relationship between employers and employees has been studied（）. [单选题] *A. originallyB. extremelyC. violentlyD. intensively(正确答案)24、The young man had decided to give up the chance of studying abroad, _____ surprised his parents a lot. [单选题] *A. whenB. whereC. which(正确答案)D. that25、The children are playing wildly and making a lot of?_______. [单选题] *A. cryB. voicesC. noises(正确答案)D. music26、_____from far away, the 600-meter tower is stretching into the sky. [单选题] *A. SeeB. SeeingC. To seeD. Seen(正确答案)27、—What do you think of Animal World? —______. I watch it every day.（）[单选题] *A. I don’t mind it.B. I like it.(正确答案)C. I can’t stand it.D. I don’t like it.28、I hope to see you again _______. [单选题] *A. long long agoB. long beforeC. before long(正确答案)D. long29、Chinese is one of ____ most widely used languages in ____ world. [单选题] *A. a, theB. /, theC. the, the(正确答案)D. a, /30、In winter, animals have a hard time_____anything to eat. [单选题] *A.to findB.finding(正确答案)C.foundD.to finding。

15544+自动化专业英语(习题参考解答)

自动化专业英语姜书艳主编张昌华徐心皓何芳编著习题参考解答Unit 1A. Basic laws of Electrical Networks[EX.1] Comprehension1. KCL:The algebraic sum of the currents entering any node is zero.KVL:The algebraic sum of the voltage around any closed path is zero.2. Node: A point at which two or more elements have a common connection is calleda node.Branches: a single path in a network composed of one simple element and the node at each end of that element.Path: If no node was encountered more than once, then the set of nodes and elements that we have passed through is defined as a path.Loop: If the node at which we started is the same as the node on which we ended, then the path is, by definition, a closed path or a loop. a path is a particular collection of branches.3. 4, 5, We can form a path but not a loop.4. v R2=32V, V x=6V[EX.2] Translation from English to Chinese1. 如果定义具有最大连接支路数的节点为参考节点，那么得到的方程相对来说比较简单。

自动化专业英语第三版王宏文

如果网络不包含能源，如电池或发电机，那么就被称作无源网络。

换句话说，如果存在一个或多个能源，那么组合的结果为有源网络。

在研究电网络的特性时，我们感兴趣的是确定电路中的电压和电流。

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• 比例加积分加微分 proportional plus integral plus
derivative (PID)
济南大学控制科学与工程学院
自动化专业英语
5
1.3 Stability of closed-loop control
systems
• 2. Put the following into chinese
• 判据 criteria 拉氏变换 Laplace transform • 零点zero-point 极点 pole-point • 特征方程 characteristic equation • 系数 parameter/coefficient • 偏差变量 deviation variable 比较器 comparator • 伺服问题 servo problem • 给定变化 given changes • 负载变化 load fluctuation/variation in load
• linear quadratic regulator 线性二次调节器
• Linear quadratic 线性二次型 controller 控制器
• Simulation 仿真
temperature 温度
• Flow rate 流速
pressure 压力
• Concentration 浓度 level 液位，液面
• 控制系统设计 control system design • 设计模型 design model • 炼油 oil refinery 化工 chemical (engineering) • 电力 (electric) power 造纸 paper (making) • 水泥 cement 纺织 textile • 塑料 plastic 钢铁 steel (iron and steel) • 航空航天 aerospace (Aeronautics and Astronautics) • 再沸器 reboiler
济南大学控制科学与工程学院
自动化专业英语
9
1.5Controller tunning
• 2. Put the following into chinese
• Setting 设置
tunning 整定，调节
• Trial and error 试差法 guideline 方针，指南
• Averaging control 平均控制 residence time 滞留时间
• observable output 可观测输出
• Measurable disturbance 可测量扰动
济南大学控制科学与工程学院
自动化专业英语
8
1.4 The design process of control system
• 3 Put the following into English
济南大学控制科学与工程学院
自动化专业英语
3
1.2 What is feedback and what are its effects?
• 2. Put the following into chinese
• Stability 稳定性
sensitivity 灵敏度
• Actuator 执行器 sensor 传感器 Transducer 传感器
• Pole-zero cancellation 零极相消
• Self-regulating 自调节
济南大学控制科学与工程学院
自动化专业英语
6
1.3 Stability of closed-loop control systems
• 3 Put the following into English
• Investment 投资
fundamental 基础的
• Evaluating 评价
benifit 利润，益处
• Distillation column 分裂蒸馏塔
• Overhead 在头上的, 高架的 process dynamic 过程动态
• Manipulatable input 可操作输入
• Oscillatory 振动的
damp 衰减
• Unstable 不稳定地 complex number 复数
• Controlled variable 被控变量
• Manipulated variable 操纵变量
• Load variable 负载变量
error 误差
• Process transfer function 过程传递函数
济南大学控制科学与工程学院
自动化专业英语
7
1.4 The design process of control system
• 2. Put the following into chinese
• Design 设计 decision 决定，决策
• Methodology 方法论 productivity 生产率，生产力
济南大学控制科学与工程学院
济南大学控制科学与工程学院
自动化专业英语
14
2.2 Level measurements
• 2. Put the following into English
• 密度 density 浮力式液面计 buoyancy level meter • 浮球 float 标尺 scale • 电位计 potentiometer 导电电极 conductive probe • 电容电极 capacitive probe 超声波 ultrasonic wave • 张力式仪表 tension gauge 应变仪 strain gauge • 荷重传感器 load sensor
自动化专业英语
课后作业答案
1.1 Introduction to Process Control
• 2. Put the following into chinese
• performance 性能 process plant 过程系统 • process control 过程控制 reference 参考 • operating conditions 运行条件 • disturbance 扰动，干扰 • Automatic control 自动控制 • Manual control 手动控制，人工控制 • Perfect control 完美控制 • Feedback control 反馈控制 • Set point 设定值 • Controlled variable 被控变量
济南大学控制科学与工程学院
自动化专业英语
11
2.1 Pressure measurements
• 1. Put the following into chinese
• Atmospheric pressure 大气压 • absolute pressure 绝压 • Gauge pressure 表压 • differential pressure 差压 • Total vacuum 完全真空 diaphragm 膜片 • Capsule 膜盒 bellow 波纹管 • Bourdon tube 波登管 silicon diaphragm 硅膜片
济南大学控制科学与工程学院
自动化专业英语
15
2.3 Flow measurements
• 1. Put the following into chinese
• Flow rate 流速 accounting 统计，计算 • Laminar flow 层流 energy losses 能量损失 • Total flow 总体流量 pressure energy 压能 • Kinetic energy 动能 potential energy 势能 • Restriction 约束，限制 orifice plate 孔板 • Venturi tube 文丘里管 flow nozzle 测流嘴 • Elbow 肘，弯管接头 rotameter 转子流量计 • Mass flow 质量流
济南大学控制科学与工程学院
自动化专业英语
10
1.5Controller tunning
• 3 Put the following into English
• 流量控制 flow control 液面控制 level control • 压力控制 pressure control 温度控制 temperature control • 成分控制 composition control • 临界增益法 ultimate gain method • 临界周期 ultimate cycling 超调 overshoot • 热交换器 heat exchanger 化学反应器 chemical reactor • 汽化器 vaporization (carburettor)
pressure gauges)
济南大学控制科学与工程学院
自动化专业英语
13
2.2 Level measurements
• 1. Put the following into chinese
• Level measurement 液位测量 container 容器 • Interface 界面 buoyancy 浮力 • Immersed 浸没 radium 镭 • Capacitive probe 电容电极 nonconductive 不导电的 • Dielectric constant 介电常数 sight glass 玻璃液面计 • Float 浮体，浮球，浮标 level formula 液位公式
• 可编程逻辑控制器programmable logic controller (PLC)
• 模型 model
黑箱 black-box

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