CH-1 Introduction – Automation, Production Systems, and Computer-Integrated Manufacturing : Book Guide

This document contains Chapters 1 to 6 Chapter 1 INTRODUCTION REVIEW QUESTIONS 1.1 What is a production system? Answer: As defined in the text, a production system is a collection of people, equipment, and procedures organized to perform the manufacturing operations of a company. 1.2 Production systems consist of two major components. Name and briefly define them. Answer: The two major components given in the text are (1) facilities, which consist of the factory, the equipment in the factory, and the way the equipment is organized; and (2) manufacturing support systems, which are the procedures used by the company to manage production and to solve the technical and logistics problems encountered in ordering materials, moving the work through the factory, and ensuring that products meet quality standards. Product design and certain business functions are included among the manufacturing support systems. 1.3 What are manufacturing systems, and how are they distinguished from production systems? Answer: A manufacturing system is a logical grouping of equipment in the factory and the worker(s) who operate(s) it. Examples include worker-machine systems, production lines, and machine cells. A production system is a larger system that includes a collection of manufacturing systems and the support systems used to manage them. A manufacturing system is a subset of the production system. 1.4 Manufacturing systems are divided into three categories, according to worker participation. Name the three categories. Answer: The three categories are (1) manual work systems, (2) worker-machine systems, and (3) automated systems. 1.5 What are the four functions included within the scope of manufacturing support systems? Answer: As identified in the text, the four functions are (1) business functions, (2) product design, (3) manufacturing planning, and (4) manufacturing control. 1.6 Three basic types of automation are defined in the text. What is fixed automation and what are some of its features? Answer: Fixed automation is a system in which the sequence of processing (or assembly) operations is fixed by the equipment configuration. Each operation in the sequence is usually simple, but the integration and coordination of many such operations in one piece of equipment makes the system complex. Typical features of fixed automation are (1) high initial investment for custom-engineered equipment, (2) high production rates, and (3) relatively inflexible in accommodating product variety. 1.7 What is programmable automation and what are some of its features? Answer: In programmable automation, the production equipment is designed with the capability to change the sequence of operations to accommodate different part or product configurations. The operation sequence is controlled by a program, which is a set of instructions coded so that they can be read and interpreted by the system. Some of the features of programmable automation are (1) high investment in general purpose equipment, (2) lost production time due to changeovers of physical setup and reprogramming, (3) lower production rates than fixed automation, (4) flexibility to deal with variations and changes in product configuration, and (5) most suitable for batch production. 1.8 What is flexible automation and what are some of its features? Answer: Flexible automation is an extension of programmable automation. A flexible automated system is capable of producing a variety of parts (or products) with virtually no time lost for changeovers from one part style to the next. There is no lost production time while reprogramming the system and altering the physical setup. Accordingly, the system can produce various mixes and schedules of parts or products instead of requiring that they be made in batches. Features of flexible automation are (1) high investment for a customengineered system, (2) continuous production of variable mixtures of products, (3) medium production rates, and (4) flexibility to deal with product design variations 1.9 What is computer-integrated manufacturing? Answer: As defined in the text, computer-integrated manufacturing (CIM) denotes the pervasive use of computer systems to design the products, plan the production, control the operations, and perform the various information-processing functions needed in a manufacturing firm. True CIM involves integrating all of these functions in one system that operates throughout the enterprise. 1.10 What are some of the reasons why companies automate their operations? Answer: The reasons give in the text are (1) increase labor productivity, (2) reduce labor cost, (3) mitigate the effects of labor shortages, (4) reduce or eliminate routine manual and clerical tasks, (5) improve worker safety, (6) improve product quality, (7) reduce manufacturing lead time, (8) accomplish processes that cannot be done manually, and (9) avoid the high cost of not automating. 1.11 Identify three situations in which manual labor is preferred over automation. Answer: The five situations listed in the text are the following: (1) The task is technologically too difficult to automate. (2) Short product life cycle. (3) Customized product. (4) To cope with ups and downs in demand. (5) To reduce risk of product failure. 1.12 Human workers will be needed in factory operations, even in the most highly automated operations. The text identifies at least four types of work for which humans will be needed. Name them. Answer: The four types of work identified in the text are (1) equipment maintenance, (2) programming and computer operations, (3) engineering project work, and (4) plant management. 1.13 What is the USA Principle? What does each of the letters stand for? Answer: The USA Principle is a common sense approach to automation and process improvement projects. U means “understand the existing process,” S stands for “simplify the process,” and A stands for “automate the process.” 1.14 The text lists ten strategies for automation and process improvement. Identify five of these strategies. Answer: The ten strategies listed in the text are (1) specialization of operations, (2) combined operations, (3) simultaneous operations, (4) integration of operations, (5) increased flexibility, (6) improved material handling and storage, (7) on-line inspection, (8) process control and optimization, (9) plant operations control, and (10) computer integrated manufacturing (CIM). 1.15 What is an automation migration strategy? Answer: As defined in the text, an automation migration strategy is a formalized plan for evolving the manufacturing systems used to produce new products as demand grows. 1.16 What are the three phases of a typical automation migration strategy? Answer: As defined in the text, the three typical phases are the following: Phase 1: Manual production using single-station manned cells operating independently. Phase 2: Automated production using single-station automated cells operating independently. Phase 3: Automated integrated production using a multi-station automated system with serial operations and automated transfer of work units between stations. Chapter 2 MANUFACTURING OPERATIONS REVIEW QUESTIONS 2.1 What is manufacturing? Answer: Two definitions are given in the text. The technological definition is the following: Manufacturing is the application of physical and chemical processes to alter the geometry, properties, and/or appearance of a given starting material to make parts or products. Manufacturing also includes the joining of multiple parts to make assembled products. The economic definition is the following: Manufacturing is the transformation of materials into items of greater value by means of one or more processing and/or assembly operations. 2.2 What are the three basic industry categories? Answer: The three basic industry categories are (1) primary industries, which are those that cultivate and exploit natural resources, such as agriculture and mining; (2) secondary industries, which convert the outputs of the primary industries into products; they include manufacturing, construction, and power generation; and (3) tertiary industries, which constitute the service sector of the economy; examples include banking, retail, transportation, education, government. 2.3 What is the difference between consumer goods and capital goods? Answer: Consumer goods are products that are purchased directly by consumers, such as cars, personal computers, TVs, tires, toys, and tennis rackets. Capital goods are products purchased by other companies to produce goods and supply services. Examples include commercial aircraft, mainframe computers, machine tools, railroad equipment, and construction machinery. 2.4 What is the difference between a processing operation and an assembly operation? Answer: A processing operation transforms a work material from one state of completion to a more advanced state that is closer to the final desired part or product. It adds value by changing the geometry, properties, or appearance of the starting material. An assembly operation joins two or more components to create a new entity, called an assembly, subassembly, or some other term that refers to the joining process. 2.5 Name the four categories of part-shaping operations, based on the state of the starting work material. Answer: The four categories are (1) solidification processes, (2) particulate processing, (3) deformation processes, and (4) material removal processes. 2.6 Assembly operations can be classified as permanent joining methods and mechanical assembly. What are the four types of permanent joining methods? Answer: The joining processes are (1) welding, (2) brazing, (3) soldering, and (4) adhesive bonding. 2.7 What is the difference between hard product variety and soft product variety? Answer: Hard product variety is when the products differ substantially. In an assembled product, hard variety is characterized by a low proportion of common parts among the products; in many cases, there are no common parts. Soft product variety is when there are only small differences between products. There is a high proportion of common parts among assembled products whose variety is soft. 2.8 What type of production does a job shop perform? Answer: Low production of specialized and customized products. The products are typically complex, such as experimental aircraft and special machinery. 2.9 Flow line production is associated with which one of the following layout types: (a) cellular layout, (b) fixed-position layout, (c) process layout, or (d) product layout? Answer: (d) Product layout. 2.10 What is the difference between a single-model production line and a mixed-model production line? Answer: A single-model production line makes products that are all identical. A mixed model production line makes products that have model variations characterized as soft product variety. 2.11 What is meant by the term technological processing capability? Answer: Technological processing capability of a plant (or company) is its available set of manufacturing processes. It includes not only the physical processes, but also the expertise possessed by plant personnel in these processing technologies. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. 2.1 (A) A manufacturing plant produces three product lines in one of its plants: A, B, and C. Each product line has multiple models: 3 models within product line A, 5 models within B, and 7 within C. Average annual production quantities of model A is 400 units, 800 units for model B, and 500 units for model C. Determine the number of (a) different product models and (b) total quantity of products produced annually in this plant. Answer: (a) The total number of different product models produced is P = 3 + 5 + 7 = 15 different models (b) The total production quantity of all products made in the factory is Qf = 3(400) + 5(800) + 7(500) = 1200 + 4000 + 3500 = 8700 units annually 2.2 Consider product line A in preceding Problem 2.1. Its three models have an average of 46 components each, and the average number of operations needed to produce each component is 3.5. All components are made in the same plant. Determine the total number of (a) components produced and (b) operations performed in the plant annually. Answer: (a) The total number of components produced is given by Eq. (2.7). npf = PQnp = 3(400)(46) = 55,200 components (b) The total number of operations performed annually in the plant is given by Eq. (2.9). nof = PQnpno = 3(400)(46)(3.5) = 193,200 operations 2.3 A company produces two products in one of its plants: A and B. Annual production of Product A is 3600 units and of Product B is 2500 units. Product A has 47 components and Product B has 52 components. For Product A, 40% of the components are made in the plant, while 60% are purchased parts. For Product B, 30% of the components are made in the plant, while 70% are purchased. For these two products taken together, what is the total number of (a) components made in the plant and (b) components purchased? Answer: (a) The total number of components produced in the plant can be determined using Eq. (2.3), adjusting it for the proportions of each part made in the factory: npf = 3600(47)(0.40) + 2500(52)(0.30) = 67,680 + 39,000 = 106,680 parts made in plant (b) Let npp = the total number of parts purchased: npp = 3600(47)(0.60) + 2500(52)(0.70) = 101,520 + 91,000 = 192,520 purchased parts 2.4 (A) A product line has two models: X and Y. Model X consists of 4 components: a, b, c, and d. The number of processing operations required to produce these four components are 2, 3, 4, and 5, respectively. Model Y consists of 3 components: e, f, and g. The number of processing operations required to produce these three components are, 6, 7, and 8 respectively. The annual quantity of Model X is 1000 units and of Model Y is 1500 units. Determine the total number of (a) components and (b) processing operations associated with these two models. Answer: (a) The total number of components can be determined using Eq. (2.3): npf = 1000(4) + 1500(3) = 4000 + 4500 = 8500 components Alternatively, Eq. (2.7) can be used, first computing the average values for Q and np using Eqs. (2.6) and (2.8). Q = (1000 + 1500)/2 = 1250 units np = (1000(4) + 1500(3))/(2 × 1250) = 3.4 components per unit product npf = 2(1250)(3.4) = 8500 components (b) The total number of processing operations can be determined using Eq. (2.4): nof = 1000(2 + 3 + 4 + 5) + 1500(6 + 7 + 8) = 1000(14) + 1500(21) = 45,500 operations Alternatively, Eq. (2.9) can be used, first computing the average values for np and no using Eqs. (2.8) and (2.10). The value of np was calculated above: np = 3.4 components per unit product no = (1000(2 + 3 + 4 + 5) + 1500(6 + 7 + 8))/3.4 = 5.353 operations per component nof = 2(1250)(3.4)(5.353) = 45,500 operations 2.5 The ABC Company is planning a new product line and a new plant to produce the parts for the line. The product line will include 8 different models. Annual production of each model is expected to be 900 units. Each product will be assembled of 180 components. All processing of parts will be accomplished in the new plant. On average, 6 processing operations are required to produce each component, and each operation takes an average of 1.0 min (including an allowance for setup time and part handling). All processing operations are performed at workstations, each of which includes a production machine and a human worker. The plant operates one shift. Determine the number of (a) components, (b) processing operations, and (c) workers that will be needed to accomplish the processing operations if each worker works 2000 hr/yr. Answer: (a) Number of components produced in the plant: npf = PQnp = 8(900)(180) = 1,296,000 components Number of operations performed in the plant: nof = PQnpno = 8(900)(180)(6) = 7,776,000 operations in the plant per year Total operation time TT = nofTp, where Tp = time for one processing operation. TT = 7,776,000(1.0) = 7,776,000 min = 129,600 hr of processing time At 2000 hours/yr per worker, number of workers w = 129,600/2000 = 64.8 workers This should be rounded up to 65 workers. 2.6 The XYZ Company is planning a new product line and a new factory to produce the parts and assembly the final products. The product line will include 10 different models. Annual production of each model is expected to be 1000 units. Each product will be assembled of 300 components, but 65% of these will be purchased parts (not made in the new factory). There are an average of 8 processing operations required to produce each component, and each processing step takes 30 sec (including an allowance for setup time and part handling). Each final unit of product takes 48 min to assemble. All processing operations are performed at work cells that include a production machine and a human worker. Products are assembled at single workstations consisting of one worker each plus assembly fixtures and tooling. Each work cell and each workstation require 25 m2 of floor space and an additional allowance of 45% must be added to the total production area for aisles, work-in-process storage, shipping and receiving, rest rooms, and other utility space. The factory will operate one shift (the day shift, 2000 hr/yr). Determine (a) how many processing and assembly operations, (b) how many workers (direct labor only), and (c) how much total floor space will be required in the plant. Answer: (a) The number of products is Qf = PQ = 10(1000) = 10,000 products/yr Therefore, the number of final assembly operations = 10,000 asby ops/yr Total number of parts npf = 10(1000)(300) = 3,000,000 components, but 65% of these are purchased, so the number made in the plant will be 0.35(3,000,000) = 1,050,000 Number of processing operations nof = 1,050,000(8) = 8,400,000 proc ops/yr Total processing time TTp = nofTp, where Tp = time for one processing operation. TTp = 8,400,000(0.50) = 4,200,000 min = 70,000 hr/yr Total assembly time TTa = QTa, where Ta = assembly time for each product. TTa = 10,000(48) = 480,000 min/yr = 8000 hr/yr Number of workers w = (70,000 + 8000)/2000 = 39 workers With 1 worker per workstation for processing operations and 1 worker per assembly workstation, n = w = 39 workstations. Total floor space TA = nAw(1 + AL), where Aw = area of each work cell or workstation, and AL = allowance for aisles, storage, etc. TA = 39(25)(1 + 0.45) = 1413.75 m2 (~15,217 ft2) 2.7 Suppose the company in Problem 2.6 were to operate two shifts (a day shift and an evening shift, a total of 4000 hr/yr) instead of one shift to accomplish the processing operations. The assembly of the product would still be accomplished on the day shift. Determine (a) how many processing and assembly operations, (b) how many workers on each shift (direct labor only), and (c) how much total floor space will be required in the plant. Answer: (a) Number of final assembly operations PQ = 10(1000) = 10,000 asby ops/yr Total number of parts npf = 10(1000)(300) = 3,000,000 components, but 65% of these are purchased, so the number made in the plant will be 0.35(3,000,000) = 1,050,000 Number of processing operations nof = 1,050,000(8) = 8,400,000 proc ops/yr (b) Total processing time TTp = nofTp, where Tp = time for one processing operation. TTp = 8,400,000(0.50 min) = 4,200,000 min = 70,000 hr/yr total for two shifts TTp = 70,000/2 = 35,000 hr/yr total for each shift Number of processing operation workers per shift wp = 35,000/2000 = 17.5 rounded up to 18 parts production workers per shift Total assembly time TTa = QTa, where Ta = assembly time for each product. TTa = 10,000(48) = 480,000 min/yr = 8000 hr/yr Number of assembly workers wa = 8000/2000 = 4 assembly workers Number of workers on day shift w = 18 + 4 = 22 workers Number of workers on evening shift w = 18 workers (c) The floor space must be based on the number of day shift operations, which includes processing and assembly operations. Total floor space TA = nAw(1 + AL), where Aw = area of each work cell or workstation, and AL = allowance for aisles, etc. TA = 22(25)(1 + 0.45) = 797.5 m2 (~8584 ft2) Comment: This is a savings in floor space of ~44% compared to the one-shift operation in the previous problem. 2-5 Chapter 3 MANUFACTURING METRICS REVIEW QUESTIONS 3.1 What is the cycle time in a manufacturing operation? Answer: As defined in the text, the cycle time Tc is the time that one work unit spends being processed or assembled. It is the time between when one work unit begins processing (or assembly) and when the next unit begins. 3.2 What is a bottleneck station? Answer: The bottleneck station is the slowest workstation in a production line, and therefore it limits the pace of the entire line. 3.3 What is production capacity? Answer: Production capacity is the maximum rate of output that a production facility (or production line, work center, or group of work centers) is able to produce under a given set of assumed operating conditions. 3.4 How can plant capacity be increased or decreased in the short term? Answer: As listed in the text, the two ways that plant capacity can be increased or decreased in the short term are (1) add workers, (2) change the number of work shifts per week Sw, and (3) change the number of hours worked per shift Hsh. 3.5 How can plant capacity be increased or decreased in the intermediate or long term? Answer: As listed in the text, the two ways that plant capacity can be increased or decreased in the intermediate or long term are (1) increase or decrease the number of work centers in the plant or (2) increase the production rate of the work centers by making methods improvements or using more productive processing technologies. 3.6 What is utilization in a manufacturing plant? Provide a definition. Answer: Utilization is the proportion of time that a productive resource (e.g., a work center) is used relative to the time available under the definition of capacity. Expressing this as an equation, where Ui = utilization of machine i, and fij = the fraction of time during the available hours that machine i is processing part style j. An overall utilization for the plant is determined by averaging the Ui values over the number of work centers: n 3.7 What is availability and how is it defined? Answer: Availability is a reliability metric that indicates the proportion of time that a piece of equipment is up and working properly. It is defined as follows: A = (MTBF – MTTR)/MTBF where A = availability, MTBF = mean time between failures, and MTTR = mean time to repair. 3.8 What is manufacturing lead time? Answer: Manufacturing lead time is the total time required to process a given part or product through the plant, including any lost time due to delays, time spent in storage, reliability problems, and so on. 3.9 What is work-in-process? Answer: Work-in-process (WIP) is the quantity of parts or products currently located in the factory that are either being processed or are between processing operations. WIP is inventory that is in the state of being transformed from raw material to finished product. 3.10 How are fixed costs distinguished from variable costs in manufacturing? Answer: Fixed costs remain constant for any level of production output. Examples include the cost of the factory building and production equipment, insurance, and property taxes. Variable costs vary in proportion to the level of production output. As output increases, variable costs increase. Examples include direct labor, raw materials, and electric power to operate the production equipment. 3.11 Name five typical factory overhead expenses? Answer: Table 3.2 in the text lists the following examples of factory overhead expenses: plant supervision, applicable taxes, factory depreciation, line foremen, insurance, equipment depreciation, maintenance, heat and air conditioning, fringe benefits, custodial services, light, material handling, security personnel, power for machinery, shipping and receiving, tool crib attendant, payroll services, and clerical support. 3.12 Name five typical corporate overhead expenses? Answer: Table 3.3 in the text lists the following examples of corporate overhead expenses: corporate executives, engineering, applicable taxes, sales and marketing, research and development, cost of office space, accounting department, support personnel, security personnel, finance department, insurance, heat and air conditioning, legal counsel, fringe benefits, and lighting. 3.13 Why should factory overhead expenses be separated from corporate overhead expenses? Answer: A manufacturing company may operate more than one factory, and each factory has its own overhead expenses that are different from the expenses at other factories. On the other hand corporate overhead expenses are applied to all factories operated by the company. Also, in matters of analyzing costs, corporate overhead would simply inflate the operating costs so they should not be included in the cost analyses, but at least some of the factory overhead costs should be included. In pricing decisions, both factory and corporate expenses must be included. 3.14 What is the capital recovery factor in cost analysis? Answer: The capital recovery factor (A/P, i, N) converts an initial cost of an investment at year 0 into a series of equivalent uniform annual year-end values, where i = annual interest rate and N = number of years in the service life of the equipment. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Cycle Time and Production Rate 3.1 (A) A batch of parts is produced on a semi-automated production machine. Batch size is 200 units. Setup requires 55 min. A worker loads and unloads the machine each cycle, which takes 0.44 min. Machine processing time is 2.86 min/cycle, and tool handling time is negligible. One part is produced each cycle. Determine (a) average cycle time, (b) time to complete the batch, and (c) average production rate. Answer: (a) Th = 0.44 min, To = 2.86 min, and Tt = 0 Cycle time Tc = 0.44 + 2.86 = 3.30 min Batch time Tb = 55 + 200(3.30) = 715 min = 11.92 hr Average production time Tp = 715/200 = 3.575 min/pc Production rate Rp = 60/3.575 = 16.78 pc/hr 3.2 In a batch machining operation, setup time is 1.5 hours and batch size is 80 units. The cycle time consists of part handling time of 30 sec and processing time of 1.37 min. One part is produced each cycle. Tool changes must be performed every 10 parts and this takes 2.0 min. Determine (a) average cycle time, (b) time to complete the batch, and (c) average production rate. Answer: (a) Th = 0.50 min, To = 1.37 min, and Tt = 2.0/10 = 0.20 min Cycle time Tc = 0.50 + 1.37 + 0.20 = 2.07 min Batch time Tb = 1.5(60) + 80(2.07) = 255.6 min = 4.26 hr Average production time Tp = 255.6/80 = 3.195 min/pc Production rate Rp = 60/3.195 = 18.78 pc/hr 3.3 A batch production operation has a machine setup time of 3.0 hours and a processing time of 1.60 min per cycle. Two parts are produced each cycle. No tool handling time is included in the cycle. Part handling time each cycle is 45 sec. It consists of the worker obtaining two starting work units from a parts tray, loading them into the machine, and then after processing, unloading the completed units and placing them into the same tray. Each tray holds 24 work units. When all of the starting work units have been replaced with completed units, the tray of completed parts is moved aside and a new tray of starting parts is moved into position at the machine. This irregular work element takes 3.0 min. Batch quantity is 2400 units. Determine (a) average cycle time, (b) time to complete the batch, and (c) average production rate. Answer: (a) Th = 0.75 min/cycle, To = 1.60 min, and Tt = 0. The irregular work element is an additional component of work handling time: Th2 = 3.0/(24/2) = 0.25 min Cycle time Tc = 0.75 + 1.60 + 0.25 = 2.60 min Batch time Tb = 3.0(60) + (2400/2)(2.60) = 180 + 3120 = 3300 min = 55 hr Average production time Tp = 3300/1200 = 2.75 min/cycle = 1.375 min/pc Production rate Rp = 60/1.375 = 43.64 pc/hr 3.4 A flow line mass production operation consists of eight manual workstations. Work units are moved synchronously and automatically between stations, with a transfer time of 15 sec. The manual processing operations performed at the eight stations take 40 sec, 52 sec, 43 sec, 48 sec, 30 sec, 57 sec, 53 sec, and 49 sec, respectively. Determine (a) cycle time for the line, (b) time to process one work unit through the eight workstations, (c) average production rate, and (d) time to produce 10,000 units. Answer: (a) The longest processing operation time, Max To = 57 sec at station 6. It sets the pace for the rest of the line. Tc = 15 + 57 = 72 sec = 1.2 min Time to process one work unit through the eight workstations = 8(1.2) = 9.6 min Production rate Rp = 60/1.2 = 50 pc/hr Time to produce 10,000 units = 10,000/50 = 200 hr 3.5 Average setup time on a certain production machine is 4.0 hours. Average batch size is 48 parts, and average operation cycle time is 4.2 min. The reliability of this machine is characterized by mean time between failures of 37 hours and a mean time to repair of 55 min. (a) If availability is ignored, what is the average hourly production rate of the machine. (b) Taking into account the availability of the machine, determine its average hourly production rate. (c) Suppose that availability only applied during the actual run time of the machine and not the setup time. Determine the average hourly production rate of the machine under this scenario. Answer: (a) Batch time Tb = 4(60) + 48(4.2) = 240 + 201.6 = 441.6 min Tp = 441.6/48 = 9.2 min and production rate Rp = 60/9.2 = 6.52 pc/hr Availability A = (37 – 55/60)/37 = 0.975 = 97.5% Production rate including effect of availability = ARp = 0.975(6.52) = 6.36 pc/hr If availability only applies during run time, then Tb = 4(60) + 48(4.2/0.975) = 240 + 206.77 = 446.77 min Tp = 446.77/48 = 9.31 min and production rate Rp = 6.45 pc/hr Plant Capacity and Utilization 3.6 (A) A mass-production plant has six machines and currently operates one 8-hour shift per day, 5 days per week, 50 weeks per year. The six machines produce the same part each at a rate of 12 parts per hour. (a) Determine the annual production capacity of this plant. (b) If the plant were to operate three 8-hour shifts per day, 7 days per week, 52 weeks per year, determine the annual percentage increase in plant capacity? Answer: (a) From Table 3.1, Hpc = 2000 hr Annual plant capacity using 50 weeks/yr PCy = 2000(6)(12) = 144,000 pc/yr (b) Working three 8-hr shifts/day, 7 days/wk, Hpc = 8736 hr. Annual plant capacity PCy = 8736(6)(12) = 628,992 pc/wk This is an increase of 628,992/144,000 – 1 = 3.368 = 336.8% Comment: This problem illustrates the significant effect that hours of operation can have on plant capacity. 3.7 One million units of a certain product are to be manufactured annually on dedicated production machines that run 16 hours per day, five days per week, 50 weeks per year. (a) If the cycle time of a machine to produce one part is 1.2 minute, how many of the dedicated machines will be required to keep up with demand? Assume that availability and utilization are 100%, and that setup time can be neglected. (b) Solve part (a) except that availability = 90%. Answer: (a) Total workload WL = 1,000,000(1.2 min/60) = 20,000 hr/yr From Table 3.1, hours available/machine = 4000 hr Number of machines n = (20,000 hr)/(4000 hr/machine) = 5 machines (b) At A = 90%, n = (20,000 hr)/(4000 × 0.90) = 5.55 rounded up to 6 machines 3.8 A job shop has four machines and operates 40 hours per week. During the most recent week, machine 1 processed part A for 25 hours at a production rate of 10 parts per hour, and part B for 12 hours at a rate of 7 parts per hour. Machine 2 processed part C for 37 hours at a rate of 14 parts per hour, and was idle 3 hours. Machine 3 processed part D for 15 hours at a rate of 20 parts per hour, and part E for 25 hours at a rate of 15 parts per hour. Machine 4 processed part F for 13 hours at a rate of 9 parts per hour, part G for 12 hours at a rate of 18 parts per hour, and was idle the rest of the week. Determine (a) weekly production output of the shop and (b) average utilization of equipment. Answer: (a) For machine 1, f1A = 25/40 = 0.625 and f1B = 12/40 = 0.30. For machine 2, f2C = 37/40 = 0.925. For machine 3, f3D = 15/40 = 0.375 and f3E = 25/40 = 0.625. For machine 4, f4F = 13/40 = 0.325 and f4G = 12/40 = 0.30. Average weekly production rate is Rph = 0.625(10) + 0.3(7) + 0.925(14) + 0.375(20) + 0.625(15) + 0.325(9) + 0.3(18) Rph = 46.5 pc/hr Weekly production rate Rpw = 40(46.5) = 1860 pc/wk (b) U = (0.625 + 0.30 + 0.925 + 0.375 + 0.625 + 0.325 + 0.30)/4 = 0.869 = 86.9% 3.9 A batch production plant works 40 hours per week and has three machines. In a typical week, five batches of parts are processed through these machines. Production rates, batch times, and operation sequences for the parts are given in the table below for one week. (a) Determine the weekly production rate for the shop. (b) Is this weekly production rate equal to the plant capacity? If not, determine what the output would be if all three machines could be operated up to 40 hours per week, given the constraint that no reductions in weekly production rates are allowed for any part. Use of a spreadsheet calculator is recommended for this problem. Part Machine 1 Machine 2 Machine 3 Rp1 (pc/hr) Tb1 (hr) Rp2 (pc/hr) Tb2 (hr) Rp3 (pc/hr) Tb3 (hr) A 10 15 15 10 18.75 8 B 15 14 10 21 C 8 21 D 20 5 25 4 E 16 5 10 8 Answer: (a) Current weekly output can be determined by multiplying the Rp values by the Tb values in each cell of the table. This is done in the following spreadsheet. Part Rp1Tb1 Rp2Tb2 Rp3Tb3 Batch size A 150 150 150 150 B 210 210 210 C 168 168 D 100 100 100 E 80 80 80 Total 708 Total weekly output of the three machines is 708 pc/wk Alternative solution: Use Equation (3.13) to find the time fractions out of 40 hr for each part and machine combination. This is done in the following spreadsheet, which also calculates the fijRpij/noj terms in Equation (3.13). Part f1j f2j f3j Rp1j Rp2j Rp3j noj f1jRp1j/noj f2jRp2j/noj f3jRp3j/noj A 0.375 0.25 0.2 10 15 18.75 3 1.25 1.25 1.25 B 0.35 0.525 15 10 2 2.625 2.625 C 0.525 8 1 4.20 D 0.125 0.1 20 25 2 1.25 1.25 E 0.125 0.2 16 10 2 1.00 1.00 Total 0.85 1.0 0.925 5.125 7.70 4.875 Weekly production rate is Rppw = (40)(5.125 + 7.7 + 4.875) = 40(17.7) = 708 pc/wk (b) Machine 2 is already operating 40 hr/wk, so no increases can be made in its weekly production output. It produces parts A, C, D, and E, so the weekly quantities of these parts remain unchanged. The only part whose output can be increased is therefore part B, produced on machines 1 and 3. Machine 1 operates 34 hr/wk (85% utilization), so the hours devoted to the production of part B could be increased from 14 to 20 hr/wk. At a production rate of 15 pc/hr, the quantity produced by Machine 1 could be increased from 210 to 300. Machine 3 operates 37 hr/wk (92.5% utilization), so the hours devoted to the production of part B could be increased from 21 to 24 hr/wk. At a production rate of 10 pc/hr, the quantity produced by Machine 3 could be increased from 210 to 240. Machine 3 turns out to be the bottleneck in the operation sequence for Part B, so that means that Machine 1 cannot be used the full 40 hr/wk. Any output beyond 240 pc/wk could not be processed by Machine 3 within the 40-hour week. Thus, 16 hours of time on Machine 1 will be devoted to Part B, thus increasing the hours of operation for Machine 1 from 34 to 36 (90% utilization). That still leaves 4 hours of unutilized time per week on Machine 1. Perhaps those 4 hours could be used to produce a sixth part (Part F). Plant capacity has been increased by 30 parts per week (output of Part B increases from 210 pc/wk to 240 pc/wk). Thus, PCw = 150 + 240 + 168 + 100 + 80 = 738 pc/wk 3.10 There are ten machines in the automatic lathe section of a certain machine shop. The setup time on an automatic lathe averages 5 hours. Average batch size for parts processed through the section is 100. Average operation time = 9.0 minutes. Under shop rules, an operator can be assigned to run one or two machines. Accordingly, there are five operators in the section for the ten lathes. In addition to the lathe operators, there are two setup workers who perform only machine setups. These setup workers are busy the full shift. The section runs one 8-hour shift per day, 5 days per week. Scrap losses are negligible and availability = 100%. The production control manager claims that the capacity of the section should be 2000 parts per week. However, the actual output averages only 1600 units per week. What is the problem? Recommend a solution. Answer: Hours/week = 40 hr Tp = (5 + 1000 × 9/60)/100 = 20 hr/100 pc = 0.20 hr/pc, Rp = 5 pc/hr Production capacity of automatic lathe section: PC = (40 hr/wk)(5 pc/hr)(10 machines) = 2000 pc/wk But the actual output = 1600 pc/wk. Why? Consider the workload of the setup workers. Number of setups per week = (2 setup workers)(40 hr/wk)/(5 hr/setup) = 16 setups/wk = 16 batches/wk. At 100 pc/batch, total pc/week = 16 × 100 = 1600 pc/week The problem is that the setup workers represent a bottleneck. To solve the problem, hire one more setup worker. With three setup workers, number of batches per week = 3(40)/5 = 24 batches/wk. At 100 pc/batch, total pc/week = 24 x 100 = 2400 pc/week Comments: (1) There are 10 machines but with 2 setup workers each setting up during the 40-hr week, this means that only 8 machines are producing parts at any given moment. Thus two of the five machine operators only have one machine to tend at any given moment. To maximize worker utilization, each machine operator should be tending two machines at all times. An ideal situation would be to have 12 machines. With two machines being set up at any given time, that leaves 10 machines producing, or two machine for each of the five machine operators. (2) In order to increase output and balance the workloads between the machine operators and setup workers, the shop should consider purchasing two additional automatic lathes. Three setup workers would be capable of (3)(40 hr/wk)/(5 hr/setup) = 24 setups/wk. Each batch takes 20 hr, so each machine can complete 2 batches/wk. Total batches with 12 machines is 24/wk or 24 setups/wk. Manufacturing Lead Time and Work-In-Process 3.11 (A) A certain batch of parts is routed through six machines in a batch production plant. The setup and operation times for each machine are given in the table below. Batch size is 100 and the average nonoperation time per machine is 12 hours. Determine (a) manufacturing lead time and (b) hourly production rate for operation 3. Machine 1 2 3 4 5 6 Setup time (hours) 4 2 8 3 3 4 Operation time (min) 5.0 3.5 6.2 1.9 4.1 2.5 Answer: Average Tsu = (4 + 2 + 8 + 3 + 3 + 4)/6 = 24/6 = 4.0 hr Average Tc = (5 + 3.5 + 6.2 + 1.9 + 4.1 + 2.5)/6 = 23.2/6 = 3.867 min MLT = 6(4.0 + 100(3.867/60) + 12) = 6(22.444) = 134.67 hr Rp for operation 3: Tp = [8.0 + 100(6.2/60)]/100 = 18.33/100 = 0.1833 hr/pc Production rate Rp = 1/0.1833 = 5.455 pc/hr 3.12 Suppose the part in the previous problem is made in very large quantities on a production line in which an automated work handling system is used to transfer parts between machines. Transfer time between stations is 15 sec. Total time required to set up the entire line is 150 hours. Assume that the operation times at the individual machines remain the same as in the previous problem. Determine (a) manufacturing lead time for a part coming off the line, (b) production rate for operation 3, (c) theoretical production rate for the entire production line. (d) How long would it take to produce 10,000 parts after the setup has been completed? Answer: (a) Given that operation 3 is the bottleneck station, Tc = 6.2 + 0.25 = 6.45 min For first part including setup, MLT = 6(6.45)/60 + 150 = 150.645 hr For each part after steady state operation is achieved, MLT = 6(6.45) = 38.7 min Tp for operation 3 = 6.45 min Production rate for operation 3 is Rp = 60/6.45 = 9.302 pc/hr Theoretical production rate for line = 9.302 pc/hr since station 3 is the bottleneck station on the line. Time to produce 10,000 parts = 10,000/9.302 = 1075 hr 3.13 A batch production plant processes an average of 35 batches of parts per week. Five of those part styles have been selected as a sample for analysis to assess the production performance of the plant. The five parts are considered to be representative of the population of parts processed by the plant. Any machine in the plant can be set up for any type of batch processed in the plant. Batch quantities, numbers of operations in the processing sequence, setup times, and cycle times for the parts are listed in the table below. The average nonoperation time per batch per operation is 11.5 hours. The plant operates 40 hours per week. Use this sample to determine the following measures of plant performance: (a) manufacturing lead time for an average part, (b) plant capacity if all machines could be operated at 100% utilization, and (c) work-in-process (number of parts-in-process). Use of a spreadsheet calculator is recommended for this problem. Part j Qj noj Tsuij (hr) Tcij (min) Tno (hr) A 100 6 5.0 9.3 11.5 B 240 3 3.8 8.4 11.5 C 85 4 4.0 12.0 11.5 D 250 8 6.1 5.7 11.5 E 140 7 3.5 3.2 11.5 * In the next-to-last column, the WLj values are computed as follows: WLj = nojQjTp. They represent a weekly workload for each part. In the last column, Rpph = WLjRpj/40noj. Manufacturing lead time MLT = 956.8/5 = 191.35 hr An alternative calculation of MLT using Eq. (3.20) can be based on average values computed from the formulas in Table 3.1: Q = 163 pc, no = 5.6 ops, Tsu = 4.67 hr, Tc = 6.63 min, and Tno = 11.50 hr: MLT = 5.6(4.67 + 163 × 6.63/60 + 11.50) = 191.35 hr For the five parts in the sample, Rppw = HpwRpph = 40(20.375) = 815 pc/wk. For 35 batches launched per week, Rppw = (35/5)(815) = 5705 pc/wk on average An alternative way to arrive at this figure is to take the 35 batches launched each week and multiply by the average batch size of 163 pc: Rppw = 35(163) = 5705 parts launched per week. From the last column of the table, hourly production rate for the five machines in the sample Rpph = 20.375 pc/hr. For the plant with 35 machines, Rpph = (35/5)(20.375) = 142.625 pc/hr Using Eq. (3.24), WIP = 142.625(191.35) = 27,291 pc 3.14 The table below presents production data for three batches of parts processed through a batch production plant. Production rates (Rp) are given in parts per hour. Utilization fractions (f) are the fractions of time during the 40-hour week that the machine is devoted to the production of these parts. The parts do not proceed through the machines in the same order. Determine (a) weekly production rate, (b) workload, and (c) average utilization of this set of equipment. A spreadsheet calculator is recommended for this problem. Machine 1 Machine 2 Machine 3 Machine 4 Part Rp1 f1 Rp2 f2 Rp3 f3 Rp4 f4 A 15 0.3 22.5 0.2 30 0.15 7.5 0.6 B 10 0.5 12.5 0.4 8 0.625 C 30 0.2 15 0.4 20 0.3 Answer: (a) Equation (3.13) was used with a spreadsheet calculator to solve for the weekly production rate, noting that the number of operations per part differs for the three parts: Rppw = 40(4.79167 + 4.79167 + 2.79167 + 3.125) = 620 pc/wk Workload WL = 40(1.0 + 1.0 + .775 + .9) = 147.0 hr Average utilization for these 4 machines U = (1.0 + 1.0 + .775 + .9)/4 = 0.9188 3.15 The table below shows production data for five batches of parts processed through a batch production plant. Setup times (Tsu) are given in hours. Operation cycle times (Tc) are given in min per cycle. Utilization fractions (f) are the fractions of time during the 40-hour week that the machine is devoted to the production of these parts. The parts do not proceed through the machines in the same order. Determine (a) weekly production rate, (b) workload, and (c) average utilization of this set of equipment. A spreadsheet calculator is recommended for this problem. Machine 1 Machine 2 Machine 3 Part Tsu Tc f Tsu Tc f Tsu Tc f A 2.50 1.50 0.125 2.25 2.01 0.140 5.00 1.56 0.190 B 3.00 0.78 0.140 2.90 1.53 0.200 1.25 2.39 0.230 C 1.50 1.50 0.150 1.80 4.00 0.345 2.30 1.50 0.170 D 4.00 7.20 0.250 1.75 5.10 0.150 1.60 2.88 0.100 E 2.00 3.20 0.150 2.55 1.80 0.120 2.55 1.80 0.120 Answer: (a) Equation (3.13) was used with a spreadsheet calculator to solve for the weekly production rate, noting that the number of operations per part is the same (no = 3) for the five parts: Tp1 Tp2 Tp3 Rp1 Rp2 Rp3 Rpph1 Rpph2 Rpph3 3.00 3.36 4.56 20 17.86 13.16 0.8333 0.8333 0.8333 1.68 2.40 2.76 35.71 25 21.74 1.6667 1.6667 1.6667 2.00 4.60 2.27 30 13.04 26.47 1.5 1.5 1.5 12.00 7.20 4.80 5 8.33 12.5 0.4167 0.4167 0.4167 4.80 3.84 3.84 12.5 15.63 15.63 0.625 0.625 5.04167 5.04167 5.04167 The Rpph values for each machine were calculated as Rpph = fRp/no. Rppw = 40(3)Rpph = 40(3)(5.04167)) = 605 pc/wk Utilization for each machine: U1 = Σf1 = 0.815, U2 = Σf2 = 0.955, U3 = Σf3 = 0.810 Workload WL = 40(0.815 + 0.955 + 0.810) = 103.2 hr Average utilization for these 3 machines U = (0.815 + 0.955 + 0.81)/3 = 0.860 3.16 The average part produced in a certain batch manufacturing plant must be processed sequentially through an average of eight operations. Twenty (20) new batches of parts are launched each week. Average operation time is 6 min, average setup time is 5 hours, average batch size is 25 parts, and average nonoperation time per batch is 10 hours per machine. There are 18 machines in the plant. Any machine can be set up for any type of batch processed in the plant. The plant operates 75 production hours per week. Determine (a) manufacturing lead time for an average part, (b) plant capacity if all machines could be operated at 100% utilization, (c) plant utilization, and (d) work-in-process (number of parts-in-process). (e) How would you expect the nonoperation time to be affected by plant utilization? Answer: (a) Manufacturing lead time MLT = 8(5 + 25(6/60) + 10) = 140 hr Average production time Tp = (5 + 25 × 6/60)/25 = 0.30 hr/pc Average production rate Rp = 1/0.30 = 3.333 pc/hr Plant capacity at 100% utilization PCw = 75(18)(3.333)/6 = 750 pc/week Parts launched per week = 20 × 25 = 500 pc/week Utilization U = 500/750 = 0.667 = 66.7% If the plant produces 500 pc/wk in 75 hr, then Rpph = 6.667 pc/hr WIP = Rpph(MLT) = 6.667(140) = 933 pc As utilization increases towards 100%, one would expect the nonoperation time to increase, because queues of parts are allowed to develop in front of each machine to assure that the machine does not run out of work. When the workload in the plant grows, the shop becomes busier, and it usually takes longer to get the jobs completed. As utilization decreases, one would expect the nonoperation time to decrease, because the queues in front of each machine are smaller. 3.17 On average 16 new batches of parts are started through a certain plant each week. Average batch quantity is 50 parts that are processed through a sequence of seven machines. Setup time per machine per batch averages 4 hours, and average operation time per machine for each part is 12 min. Nonoperation time per batch at each machine averages 8 hours. There are 37 machines in the factory. Any machine can be set up for any type of batch processed in the plant. The plant operates 40 hours per week. The plant manager complains that almost 65 hours of overtime must be authorized each week to keep up with the production schedule. Determine (a) manufacturing lead time for an average order, (b) weekly production capacity of the plant if all machines were operated at 100% utilization, (c) current utilization of the plant, and (d) average work-in-process (number of parts) in the plant. (e) Why must overtime be authorized to achieve the desired output? Answer: (a) Manufacturing lead time MLT = 7(4 + 50(12/60) + 8) = 154 hr/batch Average production time Tp = (4 + 50 × 12/60)/50 = 14/50 = 0.28 hr/op Average production rate for each machine Rp = 1/0.28 = 3.57 pc/hr Monthly production capacity PCw = 37(40)(3.57)/7 = 755 pc/wk Parts launched per month = 16(50) = 800 pc/wk Schedule exceeds plant capacity by 800 - 755 = 45 pc/wk This requires overtime hours = (45 pc × 7 ops/pc)(0.28 hr/op) = 88 hr. Utilization U = (800 pc)/(755 pc) = 1.0596 = 105.96% If the plant produces 800 pc/wk in 40 hr, then Rpph = 20 pc/hr WIP = (20 pc/hr)(154 hr) = 3080 pc Overtime must be authorized because utilization is greater than 100%. Given that normal operating hours for plant capacity is 40, at U = 105.96%, the number of hours actually worked in the plant = 40(1.0596) = 42.38 hr. Thus the number of overtime hours in the plant = 2.38 hr × 37 machines = 88 hr. 3.18 A certain job shop specializes in one-of-a-kind orders dealing with parts of medium-to-high complexity. A typical part is processed sequentially through ten machines in batch sizes of one. The shop contains a total of eight conventional machine tools and operates 40 hours per week of production time. Average time values on each part per machine are: machining time = 0.5 hour, work handling time = 0.3 hour, tool change time = 0.2 hour, setup time = 3 hours, and nonoperation time = 12 hours. A new programmable machine is being considered that can perform all ten operations in a single setup. The programming of the machine for this part will require 20 hours; however, the programming can be done off-line, without tying up the machine. Setup time will be just 2 hours because simpler fixtures will be used. Total machining time will be reduced to 80% of its previous value due to advanced tool control algorithms; work handling time will be the same as for one machine; and total tool change time will be reduced by 50% because tools will be changed automatically under program control. For the one machine, nonoperation time is expected to be 12 hours, same as for each conventional machine. (a) Determine the manufacturing lead time for the conventional machines and for the new programmable machine. (b) Compute the plant capacity for the following alternatives: (i) a job shop containing the eight traditional machines, and (ii) a job shop containing two of the new programmable machines. Assume the typical jobs are represented by the data given above. (c) Determine the average level of work-in-process for the two alternatives in part (b), if the alternative shops operate at full capacity. Answer: (a) Present method: MLT = 10(3 + (0.3 + 0.5 + 0.2) + 12) = 160 hr New machine: MLT = 1(2 + (0.3 + 0.8 × 10 × 0.5 + 0.5 × 10 × 0.2) + 12) = 19.3 hr (b) (i) Present method: For 1 machine, Tp = (3 + 1)/1 = 4 hr, Rp = 1/7 = 0.25 pc/hr For 8 machines, plant capacity PCw = (8 machines)(40 hr)(0.25 pc/hr)/(10 ops/pc) = 8 orders/week = 8 parts/wk (ii) New machine: For one machine, Tp = (2 + 5.3)/1 = 7.3 hr, Rp = 1/7.3 = 0.137 pc/hr For 2 machines, plant capacity PCw = (2 machines)(40 hr)(0.137 pc/hr)/(1 op/pc) = 10.96 orders/week = 10.96 parts/week (c) Present method: WIP = (8 orders/week)(160 hr/order)/(40 hr/wk) = 32 parts New machines: WIP = (10.96 orders/week)(19.3 hr/order)/(40 hr/wk) = 5.29 parts (on average) 3.19 A factory produces cardboard boxes. The production sequence consists of three operations: (1) cutting, (2) indenting, and (3) printing. There are three machines in the factory, one for each operation. The machines are 100% reliable and operate as follows when operating at 100% utilization: (1) In cutting, large rolls of cardboard are fed into the cutting machine and cut into blanks. Each large roll contains enough material for 4,000 blanks. Production cycle time = 0.03 minute per blank during a production run, but it takes 35 minutes to change rolls between runs. (2) In indenting, indentation lines are pressed into the blanks to allow the blanks to later be bent into boxes. The blanks from the previous cutting operation are divided and consolidated into batches whose starting quantity = 2,000 blanks. Indenting is performed at 4.5 minutes per 100 blanks. Time to change dies on the indentation machine = 30 min. (3) In printing, the indented blanks are printed with labels for a particular customer. The blanks from the previous indenting operation are divided and consolidated into batches whose starting quantity = 1,000 blanks. Printing cycle rate = 30 blanks per min. Between batches, changeover of the printing plates is required, which takes 20 minutes. In-process inventory is allowed to build up between machines 1 and 2, and between machines 2 and 3, so that the machines can operate independently as much as possible. Determine the maximum possible output of this factory during a 40-hour week, in completed blanks per week (completed blanks have been cut, indented, and printed)? Assume steady state operation, not startup. Answer: Determine maximum production rate Rp for each of the three operations: Operation (1) - cutting: Tb = 35 min. + 4000 pc × 0.03 min/pc = 35 + 120 = 155 min/batch Rp = 4000 pc/batch/(155 min/batch) = 25.806 pc/min = 1548.4 pc/hr Operation (2) - indenting: Tb = 30 min. + 2000 pc(4.5/100 min./pc) = 30 + 90 = 120 min/batch Rp = 2000 pc/batch/(120 min/batch) = 16.667 pc/min = 1000 pc/hr Operation (3) - printing: Tb = 20 min. + 1000 pc/(30 pc/min) = 20 + 33.33 = 53.33 min/batch Rp = 1000 pc/(53.33 min/batch) = 18.751 = 1125 pc/hr Bottleneck process is operation (2). Weekly output = (40 hr/wk)(1000 pc/hr) = 40,000 blanks/wk Manufacturing Costs 3.20 (A) The break-even point is to be determined for two production methods, one manual and the other automated. The manual method requires two workers at $16.50 per hour each. Together, their production rate is 30 units per hour. The automated method has an initial cost of $125,000, a 4-year service life, no salvage value, and annual maintenance costs = $3000. No labor (except for maintenance) is required for the machine, but the power to operate it is 50 kW (when running). Cost of electric power is $0.05 per kWh. The production rate for the automated machine is 55 units per hour. (a) Determine the break-even point for the two methods, using a rate of return = 25%. (b) How many hours of operation per year would be required for each method to reach the breakeven point? Answer: (a) Manual method: variable cost = (2 workers)($16.50/hr)/(30 pc/hr) = $1.10/pc Total cost as a function of Q is TC = 1.10 Q assuming no fixed costs. Automated method: (A/P,25%,4) = 0.25(1 + 0.25)4/1 + 0.25)4 – 1 = (0.4234) UAC = 125,000(A/P, 25%, 4) + 3000 = 125,000(0.4234) + 3000 = $55,930/yr Variable cost = (50 kWh/hr × $0.13/kWh)/(55 pc/hr) = $0.182/pc Total cost as a function of Q = 55,930 + 0.182 Q Break-even point: 1.10 Q = 55,930 + 0.182 Q 0.918Q = 55,930 Q = 60,913 pc/yr (b) Hours of manual method per year: H = 60,913/30 = 2030 hr/yr For the automated method: H = 60,913/55 = 1108 hr/yr 3.21 Theoretically, any given production plant has an optimum output level. Suppose a certain production plant has annual fixed costs = $2,000,000. Variable cost is functionally related to annual output Q in a manner that can be described by the function Cv = $12 + $0.005Q. Total annual cost is given by TC = Cf + CvQ. The unit sales price for one production unit P = $250 (a) Determine the value of Q that minimizes unit cost UC, where UC = TC/Q; and compute the annual profit earned by the plant at this quantity. (b) Determine the value of Q that maximizes the annual profit earned by the plant; and compute the annual profit earned by the plant at this quantity. Answer: (a) TC = 2,000,000 + (12 + 0.005Q)Q = 2,000,000 + 12 Q + 0.005 Q2 Unit cost UC = TC/Q = 2,000,000/Q + 12 + 0.005 Q D(UC)/dQ = -2,000,000/Q2 + 0.005 = 0 0.005Q2 = 2,000,000 Q2 = 400 x 106 Q = 20 x 103 = 20,000 pc Profit = 250(20,000) - (2,000,000 + 12(20,000) + 0.005(20,000)2) = $760,000/yr (b) Profit = 250 Q -(2,000,000 + 12 Q + 0.005 Q2) = 238 Q - 2,000,000 - 0.005 Q2 d/dQ = 238 – 2(0.005Q) = 238 - 0.010 Q = 0 Q = 238/0.010 = 23,800 pc dQ Profit = 250(23,800) - (2,000,000 + 12(23,800) + 0.005(23,800)2) = $832,200/yr 3.22 Costs have been compiled for a certain manufacturing company for the most recent year. The summary is shown in the table below. The company operates two different manufacturing plants, plus a corporate headquarters. Determine (a) the factory overhead rate for each plant, and (b) the corporate overhead rate. The firm will use these rates in the following year. Expense category Plant 1 Plant 2 Corporate headquarters Direct labor $1,000,000 $1,750,000 Materials $3,500,000 $4,000,000 Factory expense $1,300,000 $2,300,000 Corporate expense $5,000,000 Answer: (a) Plant 1: Factory overhead rate FOHR1 = 1,300,000/1,000,000 = 1.30 = 130% Plant 2: Factory overhead rate FOHR2 = 2,300,000/1,750,000 = 1.3143 = 131.43% (b) Corporate overhead rate COHR = 5,000,000/1,000,000 + 1,750,000 = 1.8182 = 181.82% 3.23 (A) The hourly rate for a certain work center is to be determined based on the following data: direct labor rate = $15.00 per hour; applicable factory overhead rate on labor = 35%; capital investment in machine = $200,000; service life of the machine = 5 years; rate of return = 15%; salvage value in five years = zero; and applicable factory overhead rate on machine = 40%. The work center will be operated two 8-hour shifts, 250 days per year. (a) Determine the appropriate hourly rate for the work center. (b) If the workload for the cell can only justify a one shift operation, determine the appropriate hourly rate for the work center. Answer: (a) (A/P,i,n) = (A/P,15%,5) = 0.15(1 + 0.15)5/(1 + 0.15)5 - 1 = 0.2983 0.15) − UAC = $200,000(0.2983) = $59,663/yr Hours/yr = (16 hr/day)(250 days/yr) = 4000 hr/yr, so Cm = 59,663/4000 = $14.92/hr Given CL = $15.00/hr, Co = 15.00(1 + 0.35) + 14.92(1 + 0.40) = $41.13/hr (b) Same UAC as in (a): UAC = $59,663/yr, Cm = 59,663/2000 = $29.83/hr Co = 15.00(1 + 0.35) + 29.83(1 + 0.40) = $62.01/hr 3.24 In the operation of a certain production machine, one worker is required at a direct labor rate = $10 per hour. Applicable labor factory overhead rate = 50%. Capital investment in the machine = $250,000, expected service life = 10 years, with no salvage value at the end of that period. Applicable machine factory overhead rate = 30%. The work cell will operate 2000 hours per year. Rate of return is 25%. (a) Determine the appropriate hourly rate for this work cell. (b) Suppose that the machine were operated three shifts, or 6000 hours per year, instead of 2000 hours per year. Determine the effect of increased machine utilization on the hourly rate compared to the rate determined in (a). Answer: (a) (A/P, 25%, 10) = 0.25(1 + 0.25)10/1 + 0.25)10 – 1 = 0.2801 UAC = 250,000(0.2801) = $70,025/year Machine rate Cm = 70,025/2000 = $35.01 Co = 10.00(1 + 0.50) + 35.01(1 + 0.30) = $60.52/hr (b) (A/P, 25%,10) = 0.2801 and UAC = $70,025/year as in (a) Machine rate Cm = $70,025/6000 = $11.67 Co = 10.00(1 + 0.50) + 11.67(1 + 0.30) = $30.17/hr 3.25 A customer has requested a quotation for a machining job consisting of 80 parts. The starting work part is a casting that will cost $17.00 per casting. The average production time of the job is 13.80 min on an automatic machine whose equipment cost rate is $66.00 per hour. This rate does not include any overhead costs. Tooling cost is $0.35 per part. The factory overhead rate is 128% and the corporate overhead rate is 230%. These rates are applied only to time and toooling costs, not starting material costs. The company uses a 15% markup on total cost for its price quotes. What is the quoted price for this job? Answer: Cpc = 17.00 + (66.00/60)(13.80) + 0.35 = 17.00 + 15.53 = $32.53/pc For 80 parts, cost = 80(17.00) + 80(15.53) = 1360 + 1242.40 = $2602.40 The factory and corporate overhead charges should be applied only to the labor and tooling costs, but not to the raw material cost ($17.00 cost per piece of the castings). Factory overhead charge = 1242.40(1.28) = 1590.27 Corporate overhead charge = 1242.40(2.30) = 2857.52 Total cost including overheads = 2602.40 + 1590.27 + 2857.52 = $7,050.19 With 15% markup, price quote = 7,050.19(1.15) = $8,107.72 3.26 A part is processed through six operations in a batch production plant. Cost of the starting material for each unit is $5.85. Batch quantity = 40 parts. The table below presents time and cost data for the six operations: Tsu = setup time, Th = part handling time, To = operation processing time, Tt = tool handling time where applicable, Tno = nonoperation time, Co = operating cost of the work center, and Ct = tool costs where applicable. Determine the total production cost for this part. A spreadsheet calculator is recommended for this problem. Operation Tsu (hr) Th (min/pc) To (min/pc) Tt (min/pc) Tno (hr) Co ($/hr) Ct ($/pc) 1 2.5 0.25 1.25 0 10 32.88 0 2 1.3 0.22 2.50 0.25 10 65.50 0.22 3 4.1 0.30 1.75 0 10 48.25 0 4 1.7 0.25 0.85 0.10 10 72.15 0.16 5 1.4 0.18 1.67 0 10 45.50 0 6 0.8 0.33 0.95 0.15 10 29.75 0.18 Answer: Tc = Th + To + Tt and Tp = Tsu/Q + Tc, the calculations for each operation are presented in the spreadsheet below: Chapter 4 INTRODUCTION TO AUTOMATION REVIEW QUESTIONS 4.1 What is automation? Answer: Several definitions of automation are given throughout the book. The definition given in this chapter is the following: Automation can be defined as the technology by which a process or procedure is accomplished without human assistance. 4.2 An automated system consists of what three basic elements? Answer: The three basic elements of an automated system are (1) power to accomplish the process and operate the system, (2) a program of instructions to direct the process, and (3) a control system to actuate the instructions. 4.3 What is the difference between a process parameter and a process variable? Answer: Process parameters are inputs to the process, such as temperature setting of a furnace, coordinate axis value in a positioning system, and motor on or off. Process variables are outputs from the process; for example, the actual temperature of the furnace, the actual position of the axis, and the rotational speed of the motor. 4.4 What are the five categories of work cycle programs, as listed in the text? Briefly describe each. Answer: The five categories listed in the text are: (1) Set point control in which the process parameter value is constant during the work cycle; (2) logic control, in which the process parameter value depends on the values of other variables in the process; (3) sequence control, in which the value of the process parameter changes as a function of time; (4) interactive program, in which interaction occurs between a human operator and the control system during the work cycle; and (5) intelligent program, in which the control system exhibits aspects of human intelligence (e.g., logic, decision making, cognition, learning) as a result of the work cycle program. 4.5 What are three reasons why decision-making is required in a programmed work cycle? Answer: Reasons given in the text are (1) operator interaction is required, so the operator must enter the data, (2) different part or product styles are processed by the system, and the system must decide how to process the current work unit, and (3) there are variations in the starting work units, and decisions must be made about adjustments in the work cycle to compensate. 4.6 What is the difference between a closed-loop control system and an open-loop control system? Answer: A closed loop control system is one in which the output variable is compared with an input parameter using a feedback loop, and any difference between the output and input is used to drive the output into agreement with the input. By contrast, an open loop control system operates without the feedback loop, and so there is no verification that the control action has been correctly carried out. 4.7 What is safety monitoring in an automated system? Answer: Safety monitoring in an automated system involves the use of sensors to track the system’s operation and identify conditions and events that are unsafe or potentially unsafe. The system is programmed to respond to unsafe conditions in some appropriate way, such as stopping the system or sounding an alarm. 4.8 What is error detection and recovery in an automated system? Answer: Error detection and recovery refers to the capability of an automated system to both diagnose a malfunction when it occurs and take some form of corrective action to restore the system to normal operation. In the ideal recovery mode, the system recovers from the malfunction on its own, without human assistance. 4.9 Name three of the four possible strategies in error recovery. Answer: The four possible strategies identified in the text are (1) make adjustments at the end of the current work cycle to recover from the malfunction, (2) make adjustments during the current cycle, (3) stop the process to invoke corrective action, and (4) stop the process and call for help. 4.10 Identify the five levels of automation in a production plant. Answer: The five levels of automation defined in the text are (1) device level, (2) machine level, (3) cell or system level, (4) plant level, and (5) enterprise level. Chapter 5 INDUSTRIAL CONTROL SYSTEMS REVIEW QUESTIONS 5.1 What is industrial control? Answer: As defined in the text, industrial control is the automatic regulation of unit operations and their associated equipment as well as the integration and coordination of the unit operations in the larger production system. 5.2 What is the difference between a continuous variable and a discrete variable? Answer: A continuous variable is one that is uninterrupted as time proceeds, and it is generally considered to be analog, which means it can take on any value within a certain range. A discrete variable is one that can take on only certain values within a given range, such as on or off. 5.3 Name and briefly define each of the three types of discrete variables. Answer: The three types of discrete variables are (1) binary, (2) discrete other than binary, and (3) pulse data. Binary means the variable can take on either of two possible values, ON or OFF, open or closed, and so on. Discrete variables other than binary are variables that can take on more than two possible values but less than an infinite number. Pulse data consist of a series of pulses and each pulse can be counted. 5.4 What is the difference between a continuous control system and a discrete control system? Answer: A continuous control system is one in which the variables and parameters are continuous and analog. A discrete control system is one in which the variables and parameters are discrete, mostly binary discrete. 5.5 What is feedforward control? Answer: Feedforward control is a means of control that anticipates the effect of disturbances that will upset the process by sensing them and compensating for them before they affect the process. The feedforward control elements sense the presence of a disturbance and take corrective action by adjusting a process parameter that compensates for any effect the disturbance will have on the process. 5.6 What is adaptive control? Answer: Adaptive control combines feedback control and optimal control by measuring the relevant process variables during operation (as in feedback control) and using a control algorithm that attempts to optimize some index of performance (as in optimal control). Adaptive control is distinguished from feedback control and steady-state optimal control by its unique capability to cope with a time-varying environment. 5.7 What are the three functions of adaptive control? Answer: The three functions of adaptive control are the following: (1) Identification, in which the current value of the index of performance of the system is determined, based on measurements collected from the process; (2) decision, which consists of deciding what changes should be made to improve system performance, including decisions to change one or more input parameters to the process, alter some of the internal parameters of the controller, or make other changes; (3) modification, which means implementing the decision. Whereas decision is a logic function, modification is concerned with physical changes in the system, such as changing the system parameters or process inputs to drive the system toward a more optimal state. 5.8 What is the difference between an event-driven change and a time-driven change in discrete control? Answer: An event-driven change means that some event has occurred to cause the state of the system to be altered. A time-driven change refers to a change that occurs at a specific point in time or after a certain time lapse has occurred. 5.9 What are the two basic requirements that must be managed by a computer controller to achieve real-time control? Answer: The two basic requirements are: (1) it must be able to respond to process-initiated interrupts and (2) it must be able to execute certain actions at specified points in time. These two requirements correspond to the two types of changes encountered in discrete control systems: (1) event-driven changes and (2) time-driven changes. 5.10 What is polling in computer process control? Answer: In computer process control, polling refers to the periodic sampling of data that indicates the status of the process. 5.11 What is an interlock? Answer: An interlock is a capability by which the controller is able to sequence the activities in a work cell, ensuring that the actions of one piece of equipment are completed before the next piece of equipment begins its activity. 5.12 What are the two types of interlocks in industrial control? Answer: The two types of interlocks are (1) input interlocks, which are signals originating from external devices and (2) output interlocks, which are signals sent by the controller to external devices. 5.13 What is an interrupt system in computer process control? Answer: According to the text, an interrupt system is a computer control feature that permits the execution of the current program to be suspended in order to execute another program or subroutine in response to an incoming signal indicating a higher priority event. Upon receipt of an interrupt signal, the computer system transfers to a predetermined subroutine designed to deal with the specific interrupt. 5.14 What is computer process monitoring? Answer: Computer process monitoring involves the use of the computer to observe the process and associated equipment and to collect and record data from the operation, but the computer is not used to directly control the process. 5.15 What is direct digital control (DDC), and why is it no longer used in industrial process control applications? Answer: DDC is a computer process control system in which certain components in a conventional analog control system are replaced by the digital computer, and the regulation of the process is accomplished by the digital computer on a time-shared, sampled-data basis rather than by the many individual analog components working in a dedicated continuous manner. The reason why DDC is no longer used in industrial process control applications is that today’s control computers are capable of much more than simply imitating the control mode of analog devices. 5.16 What is a programmable logic controller (PLC)? Answer: As defined in the text, a programmable logic controller is a microprocessor-based controller that uses stored instructions in programmable memory to implement logic, sequencing, timing, counting, and arithmetic control functions for controlling machines and processes. 5.17 Are programmable logic controllers (PLCs) more closely associated with the process industries or the discrete manufacturing industries? Answer: The discrete manufacturing industries. They replaced the electromechanical relays previously used to control on-off type control actions. 5.18 What is a programmable automation controller (PAC)? Answer: As defined in the text, a programmable automation controller is a digital controller that combines the capabilities of a personal computer with those of a conventional PLC; specifically, the input/output capabilities of a PLC together with the data processing, network connectivity, and enterprise data integration features of a PC. 5.19 What is a remote terminal unit? Answer: As defined in the text, a remote terminal unit is a microprocessor-based device that is connected to the process, receiving electrical signals from sensors and converting them into digital data for use by a central control computer; in some cases it also performs a control function for local sections of the process. 5.20 What does SCADA stand for, and what is it? Answer: SCADA stands for supervisory control and data acquisition. Supervisory control is superimposed on the process-level control systems such as CNC and PLCs. The term supervisory control and data acquisition emphasizes the fact that SCADA systems also collect data from the process, which often includes multiple sites distributed over large distances. 5.21 What is a distributed control system? Answer: A distributed control system is one consisting of multiple microcomputers connected together to share and distribute the process control workload. 5.22 What does open architecture mean in control systems design? Answer: Open architecture in control systems design means that vendors of control hardware and software agree to comply with published standards that allow their products to be interoperable, thus permitting components from different vendors to be interconnected in the same system. Chapter 6 HARDWARE COMPONENTS FOR AUTOMATION AND PROCESS CONTROL REVIEW QUESTIONS 6.1 What is a sensor? Answer: As defined in the text, a sensor is a device that converts a physical stimulus or variable of interest (such as temperature, force, pressure, or displacement) into a more convenient form (usually an electrical quantity such as voltage) for the purpose of measuring the stimulus. 6.2 What is the difference between an analog sensor and a discrete sensor? Answer: An analog measuring device produces a continuous analog signal such as electrical voltage, whose value varies in an analogous manner with the variable being measured. A discrete measuring device produces an output that can have only certain values. Discrete sensor devices divide into two categories: (1) binary, in which the measuring device produces an on/off signal, and (2) digital, in which the measuring device produces either a set of parallel status bits or a series of pulses that can be counted. 6.3 What is the difference between an active sensor and a passive sensor? Answer: An active sensor is one that responds to a stimulus without the need for any external power. A passive sensor is one that requires an external source of power in order to operate. 6.4 What is the transfer function of a sensor? Answer: The transfer function of a sensor is the relationship between the value of the physical stimulus and the value of the signal produced by the sensor in response to the stimulus. It is the input/output relationship of the sensor. 6.5 What is an actuator? Answer: An actuator is a hardware device that converts a controller command signal into a change in a physical parameter. An actuator is a transducer, because it changes one type of physical quantity, such as electric current, into another type of physical quantity, such as rotational speed of an electric motor. 6.6 Nearly all actuators can be classified into one of three categories, according to type of drive power. Name the three categories. Answer: The three categories are (1) electrical, (2) hydraulic, and (3) pneumatic. 6.7 Name the two main components of an electric motor. Answer: The two components are the stator, which is the stationary component, and the rotor, which rotates inside the stator. 6.8 In a DC motor, what is a commutator? Answer: A commutator is a rotary switching device that rotates with the rotor and picks up current from a set of carbon brushes that are components of the stator assembly. Its function is to continually change the relative polarity between the rotor and the stator, so that the magnetic field produces a torque to continuously turn the rotor. 6.9 What are the two important disadvantages of DC electric motors that make the AC motor relatively attractive? Answer: According to the text, the two important disadvantages of DC motors are (1) the commutator and brushes used to conduct current between the stator assembly and the rotor result in maintenance problems, and (2) the most common electrical power source in industry is alternating current, not direct current. In order to use AC power to drive a DC motor, a rectifier must be added to convert the alternating current to direct current. 6.10 How is the operation of a stepper motor different from the operation of conventional DC or AC motors? Answer: Conventional DC and AC motors rotate continuously based on a continuous DC or AC power source. A stepper motor rotates in discrete angular displacements, called step angles. Each angular step is actuated by a discrete electrical pulse. The total rotation of the motor shaft is determined by the number of pulses received by the motor, and rotational speed is determined by the frequency of the pulses. 6.11 What are three mechanical ways to convert a rotary motion into a linear motion? Answer: The three conversion mechanisms described in the text are (1) lead screws and ball screws, (2) pulley systems, and (3) rack and pinion. 6.12 What is a linear electric motor? Answer: A linear electric motor provides a linear motion without the need for rotary-tolinear conversion. Its operation is similar to that of rotary electric motors, except that the ring-shaped stator and cylindrical-shaped rotor are straight rather than circular. The rotor is called the forcer in a linear motor, and it moves along a straight track. 6.13 What is a solenoid? Answer: A solenoid is an actuator that consists of a movable plunger inside a stationary wire coil. When current is applied to the coil, it acts as a magnet, drawing the plunger into the coil. When current is switched off, a spring returns the plunger to its previous position. 6.14 What is the difference between a hydraulic actuator and a pneumatic actuator? Answer: Oil is used in hydraulic actuators, whereas compressed air is used in pneumatic actuators. 6.15 Briefly describe the three steps of the analog-to-digital conversion process? Answer: The three steps of the A/D conversion process are (1) sampling, which consists of converting the continuous signal into a series of discrete analog signals at periodic intervals; (2) quantization, in which each discrete analog signal is assigned to one of a finite number of previously defined amplitude levels, which are discrete values of voltage ranging over the full scale of the ADC; and (3) encoding, in which the discrete amplitude levels obtained during quantization are converted into digital code, representing the amplitude level as a sequence of binary digits. 6.16 What is the resolution of an analog-to-digital converter? Answer: The resolution of an ADC is the precision with which the analog signal is evaluated. Since the signal is represented in binary form, precision is determined by the number of quantization levels, which in turn is determined by the bit capacity of the ADC and the computer. In equation form, resolution RADC = L/(2n-1), where L = full scale range of the ADC and n = number of bits. 6.17 Briefly describe the two steps in the digital-to-analog conversion process? Answer: The two steps in the D/A conversion process are (1) decoding, in which the digital output of the computer is converted into a series of analog values at discrete moments in time, and (2) data holding, in which each successive value is changed into a continuous signal (usually electrical voltage) used to drive the analog actuator during the sampling interval. 6.18 What is the difference between a contact input interface and a contact output interface? Answer: A contact input interface is a device by which binary data are read into the computer from some external source (e.g., the process). It consists of a series of simple contacts that can be either closed or open (on or off) to indicate the status of binary devices connected to the process. A contact output interface is a device that communicates on/off signals from the computer to the process. 6.19 What is a pulse counter? Answer: A pulse counter is a device used to convert a series of pulses into a digital value. 6.20 What is a pulse generator? Answer: A pulse generator is a device that produces a series of electrical pulses whose total number and frequency are determined and sent by the control computer.. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Sensors 6.1 (A) During calibration, an Iron/Constantan thermocouple emits a voltage of 1.02 mV at 20°C and 27.39 mv at 500°C. The reference temperature is to be set to emit a zero voltage at 0°C. Assume the transfer function is a linear relationship between 0°C and 500°C. Determine (a) the transfer function of the thermocouple and (b) the temperature corresponding to a voltage output of 24.0 mV. Answer: (a) E = mT (27.39 – 1.02) mV = m(500 – 20)°C m = 26.37/480 = 0.05494 mV/°C Transfer function: E = 0.05494 T (b) At E = 24.0 mV, T = V/m = 24.0/0.05494 = 437°C 6.2 A digital tachometer will be used to determine the surface speed of a rotating workpiece in surface meters per sec. The tachometer is designed to read rotational speed in rev/sec, but in this case the shaft of the tachometer is directly coupled to a wheel whose outside rim is made of rubber. When the wheel rim is pressed against the surface of the rotating workpiece, the tachometer should provide a direct reading of surface speed in m/sec. What is the diameter of the wheel rim that will provide a direct reading of surface speed in m/sec? Answer: Circumference of wheel C = πD Surface speed v = NπD Let v = 1 m/sec. For the tachometer to read this value, N = 1 rev/sec D = v/Nπ = (1 m/sec)/((1 rev/sec)π) = 1/π = 0.31831 m = 31.831 cm 6.3 A rotary encoder is connected directly to the spindle of a machine tool to measure its rotational speed. The encoder generates 72 pulses for each revolution of the spindle. In one reading, the encoder generated 237 pulses in a period of 0.25 sec. What was the rotational speed of the spindle in (a) rev/min and (b) rad/sec? Answer: (a) N = (237 pulses)/(72 pulses/rev × 0.25 sec) = 13.167 rev/sec = 790 rev/min (b) ω = 2π(13.167 rev/sec) = 82.73 rad/sec 6.4 A digital flow meter operates by emitting a pulse for each unit volume of fluid flowing through it. The particular flow meter of interest has a unit volume of 50 cm3 per pulse. In a certain process control application, the flow meter emitted 3688 pulses during a period of 2.5 min. Determine (a) the total volume of fluid that flowed through the meter and (b) the flow rate of fluid flow. (c) What is the pulse frequency (Hz) corresponding to a flow rate of 60,000 cm3/min? Answer: (a) V = 3688(50) = 184,400 cm3 Q = 184,400/2.5 = 73,760 cm3/min Q = 60,000 cm3/min = 1000 cm3/sec fp = (1000 cm3/sec)/(50 cm3/pulse) = 20 pulse/sec = 20 Hz 6.5 A tool-chip thermocouple is used to measure cutting temperature in a turning operation. The two dissimilar metals in a tool-chip thermocouple are the tool material and the workpiece metal. During the turning operation, the chip from the work metal forms a junction with the rake face of the tool to create the thermocouple at exactly the location where it is desired to measure temperature: the interface between the tool and the chip. A separate calibration procedure must be performed for each combination of tool material and work metal. In the combination of interest here, the calibration curve (inverse transfer function) for a particular grade of cemented carbide tool when used to turn C1040 steel is the following: T = 48.94Etc – 53, where T = temperature, °C; and Etc = the emf output of the thermocouple, mV. (a) Revise the temperature equation so that it is in the form of a transfer function similar to that given in Equation (6.3). What is the sensitivity of this toolchip thermocouple? (b) During a straight turning operation, the emf output of the thermocouple was 9.25 mV. What was the corresponding cutting temperature? Answer: (a) T = 48.94 Etc − 53 S = Etc s = T s = C + m s Manipulating the temperature equation into the form of Equation (6.3), T + 53 = 48.94 Etc Etc = (T + 53)/48.94 = 0.02043 T + 1.083 In Equation (6.3), C = 1.083 and m = 0.02043, where m = sensitivity (b) T = 48.94(9.25) – 53 = 453 – 53 = 400°C Actuators 6.6 (A) A DC servomotor has a torque constant of 0.075 N-m/A and a voltage constant of 0.12 V/(rad/sec). The armature resistance is 2.5 Ω. A terminal voltage of 24 V is used to operate the motor. Determine (a) the starting torque generated by the motor just as the voltage is applied, (b) the maximum speed at a torque of zero, and (c) the operating point of the motor when it is connected to a load whose torque characteristic is proportional to speed with a constant of proportionality = 0.0125 N-m/(rad/sec). Answer: (a) Ia = 24/2.5 = 9.6 A T = 0.075(9.6) = 0.72 N-m Max ω occurs at Kvω = Vin 0.12ω = 24 ω = 24/0.12 = 200 rad/sec TL = 0.0125 ω 0.72 – 0.0036ω = 0.0125ω 0.72 = (0.0125 + 0.0036) ω = 0.0161 ω ω = 44.7 rad/sec 6.7 In the previous problem, what is the power delivered by the motor at the operating point in units of (a) Watts and (b) horsepower? Answer: (a) P = T ω ω = 44.7 rad/sec from previous problem T = 0.72 – 0.0036ω = 0.72 – 0.0036(44.7) = 0.56 N-m P = 0.56(44.7) = 25.0 W (b) HP = 25.0/745.7 = 0.034 hp 6.8 A DC servomotor is used to actuate one of the axes of an x-y positioner. The motor has a torque constant of 10.0 in-lb/A and a voltage constant of 12.0 V/(1000 rev/min). The armature resistance is 3.0 Ω. At a given moment, the positioning table is not moving and a voltage of 48 V is applied to the motor terminals. Determine the torque (a) immediately after the voltage is applied and (b) at a rotational speed of 500 rev/min. (c) What is the maximum theoretical speed of the motor? Answer: (a) Ia = 48/3 = 16 A T = (10.0 in-lb/A)(16 A) = 160 in-lb Ia = 48 -12(500/1000)/3.0 = (48 – 6)/3 = 14 A T = 10(14) = 140 in-lb Max N occurs at Vin = KvN Set KvN = 48 = (12/1000) N = 0.012 N N = 4000 rev/min 6.9 A DC servomotor generates 50 W of mechanical power in an application in which the constant of proportionality between the load and angular velocity = 0.022 N-m/(rad/sec). The motor has a torque constant of 0.10 N-m/A and a voltage constant of 0.15 V/(rad/sec). A voltage of 36 V is applied to the motor terminals. Determine the armature resistance of the motor. Answer: Power P = 50 W = Tω. Rearranging, T = 50/ ω The load torque TL = KL ω = 0.022 ω Setting T = TL, 0.022 ω = 50/ ω 0.02 ω2 = 50, ω2 = 50/0.022 = 2273, ω = 47.67 rad/sec T = 0.022(47.67) = 1.049 N-m Check: T = 50/47.67 = 1.049 N-m 3.6 − 0.715 = 2.885 = 1.049 Ra Armature resistance Ra = 2.885/1.049 = 2.75 Ω 6.10 A voltage of 24 V is applied to a DC motor whose torque constant = 0.115 N-m/A and voltage constant = 0.097 V/(rad/sec). Armature resistance = 1.9 Ω. The motor is directly coupled to a blower shaft for an industrial process. (a) What is the stall torque of the motor? (b) Determine the operating point of the motor if the torque-speed characteristic of the blower is given by the following equation: TL = KL1 + KL2 ω2, where TL = load torque, N-m; ω = angular velocity, rad/sec; KL1 = 0.005 N-m/(rad/sec), and KL2 = 0.00033 Nm/(rad/sec)2. (c) What horsepower is being generated by the motor at the operating point? Answer: (a) Ia = 24/1.9 = 12.63 A T = 0.115(12.63) = 1.45 N-m (b) Load torque (given): TL = 0.005 ω + 0.00033 ω2 T = 1.452 – 0.00587ω Set T = TL 1.452 – 0.00587 ω = 0.005ω + 0.00033 ω2 1.452 – 0.01087ω - 0.00033 ω2 = 0 Rearranging, 0.00033ω2 + 0.01087 ω - 1.452 = 0 Solving the quadratic equation, ω = -16.47 ± 68.35 = 51.88 rad/sec (negative value not feasible). The corresponding torque T = 1.452 – 0.00587(51.88) = 1.1475 N-m Check: TL = 0.005(51.88) + 0.00033(51.88)2 = 1.476 N-m (Close enough!) (c) T = 1.452 – 0.00587(51.88) = 1.452 – 0.304 = 1.148 HP = 1.148(51.88)/745.7 = 0.08 hp 6.11 The input voltage to a DC motor is 12 V. The motor rotates at 2200 rev/min at no load (maximum speed). Stall torque is 0.44 N-m, and the corresponding current is 9.0 A. Operating at 1600 rev/min, the torque is 0.12 N-m, and the current is 2.7 A. Based on these values, determine (a) the torque constant, (b) voltage constant, and (c) armature resistance of the motor. (d) How much current does the motor draw operating at 1600 rev/min? Answer: (a) Given stall torque T = 0.44 N-m, and corresponding current Ia = 9.0 A, Kt = 0.44/9.0 = 0.0489 N-m/A (b) At no load, T = 0, and N = 2200 rev/min. Converting this to rad/sec, ω = 2200(2π/60) = 230 rad/sec Whatever the value of Ra, 0.587 – 11.25Kv = 0 Kv = 0.587/11.25 = 0.0522 V/(rad/sec) Torque T = 0.12 N-m at N = 1600 rev/min, and ω = 167.5 rad/sec Ra = 0.159/0.12 = 1.325 Ω 6.12 The step angle of a stepper motor = 1.8°. The motor shaft is to rotate through 15 complete revolutions at an angular velocity of 7.5 rad/sec. Determine (a) the required number of pulses and (b) the pulse frequency to achieve the specified rotation. (c) How much time is required to complete the 15 revolutions? Answer: α = 1.8° ns = 360/1.8 = 200 step angles To rotate 15 revolutions, Am = 15(360) = 5400° np = 5400/1.8 = 3000 pulses To rotate at 7.5 rad/sec, fp = 7.5(200)/2π = 238.7 Hz Time T = (3000 pulses)/(238.7 pulses/sec) = 12.57 sec 6.13 (A) A stepper motor has a step angle = 3.6°. (a) How many pulses are required for the motor to rotate through five complete revolutions? (b) What pulse frequency is required for the motor to rotate at a speed of 180 rev/min? Answer: α = 3.6° ns = 360/3.6 = 100 step angles To rotate 5 revolutions, Am = 5(360) = 1800° np = 1800/3.6 = 500 pulses To rotate at 180 rev/min, fp = Nns/60 = 180(100)/60 = 300 Hz 6.14 The shaft of a stepper motor is directly connected to a lead screw that drives a worktable in an x-y positioning system. The motor has a step angle = 5°. The pitch of the lead screw is 6 mm, which means that the worktable moves in the direction of the lead screw axis by a distance of 6 mm for each complete revolution of the screw. It is desired to move the worktable a distance of 275 mm at a top speed of 20 mm/sec. Determine (a) the number of pulses and (b) the pulse frequency required to achieve this movement. (c) How much time is required to move the table the desired distance at the desired speed, assuming there are no delays due to inertia? Answer: α = 5°/step ns = 360/5 = 72 step angles Pitch p = 6 mm/rev x = 275 mm at v = 20m/sec Number of revolutions = 275/6 = 45.8333 revolutions of the motor shaft N = (20 mm/sec)(60 sec/min)(1 rev/6 mm) = 200 rev/min = 3.3333 rev/sec Am = (45.8333 rev)(360°/rev) = 16,500° np = 16,500°/5° = 3300 pulses fp = Nns/60 = 3.333(72) = 240 Hz Move time T = (3300 pulses)/(240 pulses/sec) = 13.75 sec 6.15 A single-acting hydraulic cylinder with spring return has an inside diameter of 95 mm. Its application is to push pallets off of a conveyor into a storage area. The hydraulic power source can generate up to 2.5 MPa of pressure at a flow rate of 100,000 mm3/sec to drive the piston. Determine (a) the maximum possible velocity of the piston and (b) the maximum force that can be applied by the apparatus. (c) Is this a good application for a hydraulic cylinder, or would a pneumatic cylinder be better? Answer: Area of cylinder A = 0.25π(95)2 = 7088 mm2 V = Q/A = (100,000 mm3/sec)/7088 mm2 = 14.11 mm/sec F = pA = (2.5 N/mm2)(7088) = 17,720 N (that’s almost 4000 lbf) The speed seems slow for material handling, and the force seems very high. A pneumatic cylinder would have a higher speed but lower force, which would be appropriate for this material handling application. 6.16 (A) A double-acting hydraulic cylinder has an inside diameter of 80 mm. The piston rod has a diameter of 15 mm. The hydraulic power source can generate up to 4.0 MPa of pressure at a flow rate of 125,000 mm3/sec to drive the piston. (a) What are the maximum possible velocity of the piston and the maximum force that can be applied in the forward stroke? (b) What are the maximum possible velocity of the piston and the maximum force that can be applied in the reverse stroke? Answer: Forward stroke area A = 0.25π(80)2 = 5026 mm2 Reverse stroke area A = 5026 – 0.25π(15)2 = 4850 mm2 Forward stroke v = 125,000/5026 = 24.9 mm/sec F = 4(5026) = 20,104 N Reverse stroke v = 125,000/4850 = 25.8 mm/sec F = 4(4850) = 19,400 N 6.17 A double-acting hydraulic cylinder is used to actuate a linear joint of an industrial robot. The inside diameter of the cylinder is 3.5 in. The piston rod has a diameter of 0.5 in. The hydraulic power source can generate up to 500 lb/in2 of pressure at a flow rate of 1200 in3/min to drive the piston. (a) Determine the maximum velocity of the piston and the maximum force that can be applied in the forward stroke. (b) Determine the maximum velocity of the piston and the maximum force that can be applied in the reverse stroke. Answer: Forward stroke area A = 0.25π(3.5)2 = 9.62 in2 Reverse stroke area A = 9.62 – 0.25π(0.5)2 = 9.42 in2 Forward stroke v = 1200/9.62 = 124 in/min F = 500(9.62) = 4810 lb Reverse stroke v = 1200/9.42 = 127.4 in/min F = 500(9.42) = 4710 lb Analog-Digital Conversion 6.18 (A) A continuous voltage signal is to be converted into its digital counterpart using an analog-to-digital converter. The maximum voltage range is ±30 V. The ADC has a 12-bit capacity. Determine (a) number of quantization levels, (b) resolution, and (c) the quantization error for this ADC. Answer: (a) Number of quantization levels Nq = 212 = 4096 (b) RADC =60/4096 - 1 = 0.01465 V (c) Quantization error = ±(0.01465)/2 = ±0.00732 V 6.19 A voltage signal with a range of zero to 115 V. is to be converted by means of an ADC. Determine the minimum number of bits required to obtain a quantization error of (a) ±5 V maximum, (b) ±1 V maximum, (c) ±0.1 V maximum. Answer: 6.20 A digital-to-analog converter uses a reference voltage of 120 V dc and has eight binary digit precision. In one of the sampling instants, the data contained in the binary register = 01010101. If a zero-order hold is used to generate the output signal, determine the voltage level of that signal. Answer: Vo = 120{0.5(0) + 0.25(1) + 0.125(0) + 0.0625 (1) + 0.03125(0) + 0.015625(1) + 0.007812(0) + 0.003906(1)} Vo = 39.84 V 6.21 A DAC uses a reference voltage of 80 V and has 6-bit precision. In four successive sampling periods, each 1 second long, the binary data contained in the output register were 100000, 011111, 011101, and 011010. Determine the equation for the voltage as a function of time between sampling instants 3 and 4 using (a) a zero-order hold, and (b) a first-order hold. Answer: First sampling instant: 100000, Vo = 80(0.5) = 40.0 V Second sampling instant: 011111, Vo = 80(0.25 + 0.125 + 0.0625 + 0.03125 + 0.015625) = 38.75 V Third sampling instant: 011101, Vo = 80(0.25 + 0.125 + 0.0625 + 0.015625) = 36.25 V Fourth sampling instant: 011001, Vo = 80(0.25 + 0.125 + 0.015625) = 31.25 V Zero order hold: V(t) = 36.25 between instants 3 and 4 First order hold: V(t) = 36.25 + α t between instants 3 and 4 α = (36.25 - 38.75)/1 = -2.5 V(t) = 36.25 - 2.5t 6.22 In the previous problem, suppose that a second order hold were to be used to generate the output signal. The equation for the second-order hold is the following: E(t) = E0 + αt + βt2 , where E0 = starting voltage at the beginning of the time interval. (a) For the binary data given in the previous problem, determine the values of α and β that would be used in the equation for the time interval between sampling instants 3 and 4. (b) Compare the first order and second-order holds in anticipating the voltage at the 4th instant. Answer: t = 0: V(t) = 36.25 = 36.25 + α (0) + β (0) t = -1: V(t) = 38.75 = 36.25 + α (-1) + β (1) t = -2: V(t) = 40.0 = 36.25 + α (-2) + β (4) Simultaneous solution yields α = -3.125 and β = - 0.625 V(t) = 36.25 − 3.125t − 0.625t2 At the fourth instant, the second order hold yields V(t) = 36.25 − 3.125(1) - 0.625(1)2 = 32.5 V At the fourth instant, the first order hold yields V(t) = 36.25 − 2.5(1) = 33.75 V The actual voltage at the fourth instant is 31.25 V from the previous problem. Hence, the second-order hold more accurately predicts the voltage. Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463