THE USE OF COMPUTERS IN MECHANICAL ENGINEERING EDUCATION Franklin H. Westervelt Department of Mechanical Engineering The University of Michigan June 1, 1962 This material is distributed by The Ford Foundation Project on the Use of Computers in Engineering Education at The University of Michigan. This report appears in the library edition of the Final Report of the Project and is also issued as a separate booklet. Similar "Curriculum Reports" for other engineering disciplines are also available on request.

ABSTRACT The Mechanical Engineering Department at The University of Michigan has taken an active part in computing work at the University with a total of nine faculty members serving as participants in the Project on the Use of Computers in Engineering Education during the past two-year period. In addition, there have been twenty-six Mechanical Engineering professors from other universities who participated in Project activities for periods varying from one week to a full semester. All undergraduate Mechanical Engineering students at the University are required to take an introductory sophomore level digital computer course taught by Computing Center and Mathematics Department personnel. Digital and analog computer work has been assigned in 17 departmental courses during the past year, giving students an opportunity to gain practice in the application of computers in the solution of their engineering problems. The Mechanical Engineering curriculum, the areas where computer activity has been strongest, and a sampling of opinions of faculty members as to the effectiveness of computer usage in engineering instruction are described. The Department is responsible for the operation of an analog computer laboratory and students are required to become familiar with these computers as well as with the digital machine. A selected set of eleven example problems prepared by faculty Project participants and other faculty members in the Department is also included. These may be considered as a supplement to the 64 example engineering problems, including several related to Mechanical Engineering subject areas, which have been published previously by the Project. -B -

Table of Contents Page I. Introduction B3 II. The Undergraduate Program B3 III. The Graduate Program B6 IV. Computers in Mechanical Engineering B8 V. Application of Computers to Specific Subject Areas B9 a) Thermodynamics B9 b) Heat Transfer and Fluid Flow Bll c) Kinematics, Dynamics and Machine Design B13 VI. Available Computing Facilities B16 VII. Conclusions B16 VIII. Sample Problems B18 65 Isentropic Process for Ideal Gas with Variable Specific Heat B19 66 Compressibility Factors Using the Beattie-Bridgeman Equation of State B25 67 The Effect of Pressure and Propellant Ratio on HydrogenOxygen Rocket Performance B28 68 Determination of the Composition of the Products of Combustion B45 69 Flame Temperature for Combustion of Air and Methane B50 70 Transient Temperature Calculation for a Jacketed Kettle B60 71 Surge System Oscillations B69 72 Analog Analysis of a Sinusoidally Excited Spring-MassDashpot System B76 73 Bevel Gear Speed Reducer Force Analysis B80 74 Cam Design Proposal Analysis B81 75 Use of the Electronic Differential Analyzer to Study the Dynamics of Machinery B83 TABLES IB Undergraduate Program in Mechanical Engineering, 1962-1963 B4 IIB Graduate Courses in Mechanical Engineering B7 FIGURES IB Mechanical Engineering Program, 1962-1963 B5 B2

USE OF COMPUTERS IN MECHANICAL ENGINEERING EDUCATION I. INTRODUCTION The Department of Mechanical Engineering at the University of Michigan has participated extensively in the study of the educational utility of computers. Many of the faculty have learned to use the computer in the classroom and the number of new uses increases in breadth and depth continually. Both analog and digital computer techniques have been employed successfully. It is the purpose of this report to show how these techniques have been introduced and to present the general philosophy surrounding the application of computers in the departmental courses. In order to orient the computer and its educational value to the degree programs of the Department of Mechanical Engineering, the structure of the undergraduate and graduate degree programs will be considered first. II THE UNDERGRADUATE PROGRAM The program in Mechanical Engineering leading to the degree of Bachelor of Science in Engineering may be thought of as two sections. The first is technical and devoted to the preparation of engineers for professional development through industrial service or graduate study. The second is devoted to the humanities and social sciences and is designed to prepare the student for an effective life in society. The technical section emphasizes the basic sciences and engineering sciences. Courses in these areas include associated laboratories that are the direct responsibility of the teachers in the corresponding lecture courses. To this foundation is added a group of courses in analysis and design to foster the association of theory, experiment and practice that is vital to good engineering practice. Flexibility to meet the varied needs and aims of the students is provided by substitutions, combined programs with other departments, and electives in the area of the student's interest. Variation in the abilities of the students is accounted for in a level of attainment requirement for graduation, which permits gifted students to acquire the degree in less than the usual number of credit hours by taking special sequences of courses that cover the same material in fewer class hours. Hence the program may vary from one student to another, but they are all variations on a theme, which is the recommended program. The current recommended program in Mechanical Engineering is listed below, and followed by a flow chart of the program. All students are required to complete this program or its equivalent. B3

The program as laid out in the flow diagram (Fig. 1B) indicates the prerequisite and corequisite courses and the order in which the courses should be taken to complete the program in eight semesters and a summer session. The rows on the chart give the semester schedule and credit hour loads. The columns show the various area sequences involved. For example, the first column shows the sequence in mathematics running from the first semester through the fifth, TABLE IB Undergraduate Program in Mechanical Engineering, 1962-63 Course Number* Elective English I English II English III Econ. 401 Math. 233, 234, 371, 372 Math. 373 Math. 404 Physics 145 Physics 146 Chem. 103 or 104 Chem. 105 or 106 Chem/Met. Eng. 250 Chem/Met. Eng. 270 Eng. Graphics 101 Eng. Graphics 104 Eng. Mech. 208 Eng. Mech. 210 Eng. Mech. 212 Eng. Mech. 324 Eng. Mech. 343 Mech. Eng. 251 Mech. Eng. 324 Mech. Eng. 335 Mech. Eng. 336 Mech. Eng. 340 Mech. Eng. 362 Mech. Eng. 371 Mech. Eng. 381 Mech. Eng. 460 Elec. Eng. 315 Elec. Eng. 442 Technical Electives Subject Humanities, Arts & Social Sciences English Composition and Speech English Composition, Speech or Literature Literature Modern Economic Society Analytic Geometry and Calculus Elementary Computer Techniques Differential Equations Mechanics, Sound, and Heat Electricity and Light General or Inorganic Chemistry General and Inorganic Chemistry Principles of Engineering Materials Metals and Alloys Engineering Drawing Introduction to Descriptive Geometry Statics Mechanics of Materials Laboratory in Strength of Materials Fluid Mechanics Dynamics Manufacturing Processes I Fundamentals of Fluids Machinery Thermodynamics I Thermodynamics II Dynamics of Machinery Machine Design I Heat Transfer I Manufacturing Processes II Machine Design II D. C. and A. C. Apparatus and Circuits Motor Control and Electronics Hours 11 6 2 2 3 16 1 3 5 5 4 4 3 3 3 2 3 4 1 3 3 3 4 3 4 4 3 4 3 3 4 4 12 Total 138 *Descriptions of these courses may be found in the University of Michigan College of Engineering Announcement of 1962-63. B4

Undergraduate Mechanical Engineering Program 1962-63 Math & EE Phys & ME Grfx & ME EM & ME Chem & CM Humanities English ition Speech English /; LL'" KU I C Heat rk y Chemistry Composition Statics of E3 Light " Materials Literature I ----- Computer Descript. Elasticity' Elast. Engineering IV "IV " G aterials of o0: Techniques Ge om. Lab Materials a) Thermo- Dynamics Summer r ~rl*H~~~ ~dynamics FuDiffer)ential ids Dynamics M) Diff er enti Metallurgy I) EatoII (E. M.).> ci) J)~quations Circuits Fluids Mechanical Manufact. Humanities VI & (A) J - (M. E.) Desin Process Electronics Process Electronics Heat Mechanical Technical Humanities English VII & 4 (?) Hun e 0 Control Transfer Design IP Elective Literature Technical Technical Technical VIII l 3 T Economics Humanities Elective Elective Elective 0 Q Figure IB Total Credits D o (D 0 r0; (D H' S CD I I oq 11 P0 138 P~\ ~ s ~t ~. ~

In the area bounded by rows five through seven and columns two through four the flow of thermodynamics, fluid mechanics and heat transfer becomes evident. The sequence in engineering mechanics occurs in column four, starting with Statics, and flows downward and to the right through Dynamics to Dynamics of Machinery which initiates a mechanical design sequence. The design sequence is also fed by the sequence in Chemistry, Materials and Processing that occurs in the sixth column. The diagram indicates clearly the interrelations of the various areas of the program and should be helpful in relating the use of computers to the fundamental purposes of such a program. Training in the use of computers is introduced to the program in two ways. First, a specific course entitled Elementary Computer Techniques (Math 573) occurs in the second column in the fourth semester, as shown in the flow chart. In that position this course may be used as a prerequisite for any or all junior technical subjects. It is at present a specific prerequisite for the second course in thermodynamics (M.E. 336) and the course in mechanical design (M.E. 562) and is used in that manner by others, fluid mechanics (M.E. 324), and heat transfer (M.E. 371) in particular. Secondly, training in the techniques of analog computers is an integral part of the course content of Dynamics of Machinery (M.E. 540). Experience shows that it is usually better to separate the introduction to the digital computer because of the dominance of the language aspects of the problem in the early stages. Analog computers may be introduced either way depending upon the complexity of the problems to be handled. Coupled with an introduction to mechanical vibrations,it is most effective to integrate the computer training. In Dynamics of Machinery (M.E. 340) the analog computer is used in both classroom and laboratory experiences. It should be observed that the early occurence of the Computer Techniques course makes possible a large number of classroom applications throughout the upper class years. Only the most general of these applications will be reported here. However, the successes in these areas are continually germinating new applications in other course areas. III THE GRADUATE PROGRAM The graduate program of the Mechanical Engineering Department is structured in six major areas. These are: 1) Fluid Flow 2) Heat Transfer 5) Thermodynamics 4) Stress Analysis 5) Dynamics and Vibration 6) Materials and Manufacturing B6

Use of Computers in Mechanical Engineering Education The requirements for the master's degree include at least one course in four of the engineering science areas cited above, and also advanced calculus. No thesis is required but the majority of the students gain some experience in research either through work on one of the research projects or through study or research in selected Mechanical Engineering Topics (ME 600), a three credit individual-research-type course. The requirements for the doctorate involve thirty credits of course work beyond the Master of Science degree, two languages, the successful passing of written and oral qualifying examinations, and the doctoral thesis. Courses at the 600, 700, and 800 level are primarily for doctoral students. Table IIB lists the graduate Mechanical Engineering courses available to show the breadth and depth of the offerings. The Table is divided somewhat arbitrarily into Basic and Professional courses. The intent is to indicate those course es whose content is taken primarily from the fundamental aspects of the science and those extending the student into the frontiers of current practice. It should be apparent that such a separation is most difficult because of the ever changing state of technology. However, the reader may grasp some feeling of current emphasis from this tabulation. Table IIB Graduate Courses in Mechanical Engineering Basic Courses ME 417 ME 421 ME 432 ME 461 ME 517 ME 524 ME 531 ME 555 ME 540 ME 541 ME 550 ME 556 ME 567 ME 571 ME 583 ME 625 ME 672 ME 673 ME 772 ME 855 Plastic Forming of Metals I Dynamics and Thermodynamics of Compressible Flow Combustion Automatic Control Plastic Forming of Metals II Fluid Mechanics II Statistical Thermodynamics Advanced Thermodynamics Mechanical Vibrations Synthesis of Mechanisms Physical Behavior of Materials Stress Considerations in Design Reliability Consideration in Design Heat Transfer II (Conduction) Machinability Introduction to Viscous Flow Theory Heat Transfer III (Convection) Heat Transfer IV (Radiation) Heat Transfer V Seminar in Thermodynamics B7

Professional Courses ME 422 ME 437 ME 463 ME 465 ME 467 ME 480 ME 482 ME 486 ME 491 ME 492 ME 493 ME 494 ME 495 ME 496 ME 497 ME 498 ME 539 ME 594 Design Theory of Fluid Machinery Applied Energy Conversion Wear Considerations in Design Mechanical Analysis Laboratory Lubrication and Bearing Analysis Design of Manufacturing Equipment Manufacturing Engineering Manufacturing Considerations in Design Heating and Air Conditioning Design of Heating and Air Conditioning Systems Gas Turbine Engines Design of Gas Turbine Engines Analysis and Design of Rocket Engines Internal Combustion Engines Automotive Laboratory Automotive Chassis Cryogenics and Refrigeration Advanced Theory of Internal Combustion Engines IV COMPUTERS IN MECHANICAL ENGINEERING The philosophy of the use of computers in Mechanical Engineering is related to the program structure just presented. The fundamental conviction is that Mechanical Engineering students should be introduced to the use of computers, both analog and digital, as an integral part of their educational experience. Further, the computer experience must be a graduated experience, repeated in many meaningful situations in the various subject areas of the student's degree program. Within the course work several valuable attributes of the introduction and application of computers are evident. First, the computer challenges the student with a new degree of discipline in his work. The solution of problems on the computer forces the student to think of the logical structure of the problem. The various alternatives presented by the typical engineering problem are better treated by the creation of procedures of generality for the computer than by the usual set of restricted cases considered without the availability of the computer. Second, the student may extend his problem solving experience by considering more complex and realistic problems. The student has the satisfaction of obtaining the results of quite involved procedures more frequently. Less often must he be content with a superficial treatment of analytical procedures. Third, the student is extended in his mathematical experience, through the association with numerical analysis. Even "trial and error" methods are made more rigorous in the consideration of Newton-Raphson methods, half-interval methods and other methods applicable to the solution of the non-linear expressions common to engineering practice. Vector and matrix analysis may be applied to appropriate problems in realistic situations with little more effort than was previously expended in the consideration of situations constructed so as to be simple enough to be solved by hand. In addition, it should B8

Use of Computers in Mechanical Engineering Education be observed that graduate students, at the doctoral level, may now undertake many problems that were unreasonable or even impossible heretofore. Finally, it has been observed that the solution of problems on the computer frequently produces a greater interaction between the student and the faculty. Possibly, this has been due to the fact that both the faculty and the students have been sharing the learning of the computer techniques. However, it is more reasonable to suppose that the ability to engage problems of more realistic proportions than the typical academic excercise is stimulating to both faculty and students,thus creating an atmosphere of mutual interest. V APPLICATION OF COMPUTERS IN SPECIFIC SUBJECT AREAS Because of the dynamic growth of computer applications it is certain that any presentation of the applications is, at once, obsolete. Therefore, a set of representative subject areas have been selected to illustrate the nature of activities at one point in time. The subject areas chosen include Thermodynamics, Heat Transfer, Fluid Flow, Kinematics and Dynamics. Thus it is possible, in most instances, to display problem materials that are of varying complexity and thus suitable for courses ranging from the undergraduate through the graduate level. The comments quoted concerning the problem materials and the applications of computers have been extracted from written commentaries submitted by the faculty concerned with the subject areas. It is hoped that their candid frankness will helpe the reader to orient the experiences of one faculty body to the context in which he may find himself. a) Thermodynamics In this area the student has a large number of opportunities to meet situations that are suitable for computer solution. While a few problems have been assigned in Thermodynamics I (ME 33555, 3 credits), it is more likely that the first computer problems in Thermodynamics will be encountered in Thermodynamics II (ME 336, 4 credits). This is because of the nature of the material presented in the first course and because of the extreme diversity of students present in the course. Many students from departments other than Mechanical Engineering are found in Thermodynamics I. The second course is taught by a number of faculty and thus provides an opportunity to observe the reaction of several staff members to the presentation of a common problem. The problem presented in all these classes was the computation of the compressibility factors of gases from the Beattie-Bridgeman equation. (See example problem No. 15 of the First Annual Report of the Ford Foundation Computer Project). The comments of the staff reflect a general feeling that the problem was of value to the student. "The problem was of considerable benefit in understanding the use of equations of state and their limitations." B9

"I think the problem was helpful and should be continued." "This computer problem served as a good demonstration as to the use of digital computers as an engineering tool".'"The assignments have been useful as far as the objectives go". The comments also indicate the value of previous course work in computers (e.g. Elementary Computer Techniques) prior to courses given in the professional departments. At the time this problem was given relatively few Mechanical Engineering students had had the opportunity to take such a course. (This course, Math 373, is now a prerequisite for ME 336 and ME 362.) "However, it has been very time consuming for both students and myself when the students have not had any previous experience...". "More sophisticated problems can be covered when the students have had a formal and more thorough training program such as is now required in the curriculum". One staff member mentioned the "drag-out" of the problem "when service at the Computing Center is poor". This comment is mentioned because it emphasizes the incompatibility of effective education and long queues. The Computing Center currently processes more than 6,000 problem runs per month and is anticipating delivery of equipment of much higher capacity. The comment may thus be interpreted as a warning against the diversification of resources into small low-capacity machines and a strong argument for support of the very largest capacity facility that can be afforded. The use of the digital computer as a demonstrating device was made in Advanced Thermodynamic (ME 555, 3 credits, graduate level). Here the problem of determining the effects of pressure and percent theoretical air on the adiabatic flame temperature of methane,including the effects of dissociation of the products of combustion,was presented. The students developed the solution but did not program or run the problem. Instead the program and results were given to the class. The objective was to evaluate the combustion process with quantitative evidence to support the discussion. (It should be observed that the lack of previous computer experience is even more noticeable among the upper level students. In a few semesters it may become commonplace to expect and receive student programs for more complicated problems than this one.) The Seminar in Thermodynamics (ME 835, 3 credits, upper graduate level) considered the problem of the analysis of rocket engine performance as a function of propellant mixture ratio and chamber pressure. This was programmed individually by the students and "was beneficial to the students' understanding of chemical equilibrium and to the approach to the solution of advanced problems in Thermodynamics". The problem statements and solutions for a set of problems of graduated complexity in the area of Thermodynamics are given in the problem section of this report. B10

Use of Computers in Mechanical Engineering Education b) Heat Transfer and Fluid Flow These areas are representative of many that are strongly founded in engineering principles and build from these foundations into engineering practice. Accordingly, the computer finds considerable application and utility in developing these materials. Each area will be treated separately but the level of the material covered is comparable, the first encounter coming in the junior year. Heat Transfer I (ME 371, 4 credits, junior level) is concerned with the study of the mechanisms of heat transfer processes. For the past two semesters, a problem in two-dimensional heat conduction using finite difference techniques has been programmed for the digital computer. The students themselves have prepared this program. Their instructor comments.... "In assigning problems for programming by the students the primary objective was to give the students an experience in setting up an algorithm for the solution of a problem. In so doing they must think in terms of unambiguous logical operations, an experience which has value in all phases of engineering. A secondary objective was to instill in the students confidence in solving specific heat transfer problems with a digital computer." A comparison of analytic, graphical, analog computer and numerical methods with actual experimental evidence is readily possible. In this course the students "make physical measurements of a transient heat conduction phenomenon and compare these with the results of an analytic solution, a numerical solution using hand computations, a graphical solution, the analog computer solution, and a numerical digital computer solution. The advantages and limitations of each of these is discussed." An apparent benefit of such an approach is the opportunity to teach the limitations, as well as the advantages. This aspect is sometimes ignored in an attempt to "prove a point". "This program demonstrates interaction of errors due to discretization, or grid-size spacing, and due to round-off, inherent in the iteration techniques. The objective was to demonstrate potential pitfalls in the use of numerical solutions, and also to demonstrate their value with complex geometries and boundary conditions." A representative set of problems to be treated are the following: I. A steady-state two dimensional conduction problem with heat transfer coefficients and fluid temperatures specified. (Example 3-4 in "Principles of Heat Transfer" by Frank Kreith.) II. Transient one dimensional heat conduction in an infinite slab with a step change in one surface temperature. (Both analog and digital computer solutions obtained for this problem.) III. Transient one dimensional heat conduction in an infinite slab with a step change in one fluid temperature. The comments concerning the difficulties associated with the computer solutions are also interesting. Bll

"In first presenting a digital computer problem some apathy on the part of students was detected. It was even found necessary to provide a certain amount of incentive. However, once having gained some experience and success in programming the attitude in most cases was changed to one of enthusiasm." "It has been my personal experience that it is unwise to give a student problems in learning the language and learning the problem at the same time. For this reason I think computers can be very nicely used in the classroom because most of the students are familiar with the problems being discussed, and hence are not only oriented but motivated to their solutior on a high speed computer." It might have been observed that the foregoing remarks apply equally to faculty as well as students. In fact one rather expects that the use of the computer in the classroom must "prove itself" with many faculty. These men are highly qualified in their respective areas and quite naturally may hesitate to venture into an area in which they are much less certain of themselves initially. A staff member comments.... "In order to introduce the staff to the use of digital computers, a program (such as the Project on the Use of Computers in Engineering Education) is invaluable in the early stages". The computer has also found much application in the graduate level Heat Transfer courses. A partial list of problems programmed in these courses is given to indicate a little of the scope of the work done. I. Evaluation of the double integral to obtain the geometric view factor for radiation between two surfaces of arbitrary shape and orientation. (Radiative Heat Transfer, ME 673, 3 credits.) II. Solution of the radiation between a multiplicity of gray surfaces. (ME 673) III. Linear problems of diffusion with simultaneous convective flow (Convective Heat Transfer, ME 672, 3 credits). IV. Convective heat flow in a duct with simultaneous axial conduction. (ME 672) The application of computers in fluid flow is first encountered in Fundamentals of Fluid Machinery (ME 324, 4 credits). This course was using the IBM 650 computer in classroom activities as early as 1957-58. It was recognized that the availability of a compiler was essential to effective classroom use. The IT (Internal Translator) and later the GAT (General Algebraic Translator) compilers were employed to allow the solution of the non-linear energy dissipation equations in turbulent flow by the students. In this connection an approach was developed that seems effective when it is necessary, as it was at that time, to teach programming to the students as well as the problem. "First, select a problem that may be solved a small piece at a time, each piece leading into a larger problem. In this case, the first problem was to solve the Colebrook equation for turbulent B12

Use of Computers in Mechanical Engineering Education flow at Reynolds numbers greater than 4000 and with general relative roughness. This function has the form: = 1.14 - 2.*log0 [ e/D + 9.35 OIoT 10 NR/ This is solved as a rather simple but meaningful program using the Newton-Raphson method, and allows the student to put most of his attention on the computer solution after a brief introduction to the fluid mechanics. It is sometimes desirable to assign one solution to be done by hand so that the method is made completely clear (and so that the student can verify his computer solution). Next, this program is converted into a subroutine and the problem is generalized to allow any Reynolds number and a general set of conduit and fluid parameters to solve for pressure drop given the flow,or flow given the pressure drop. This causes the student to consider the action to be taken in the "transition zone"., a topic frequently avoided otherwise. Finally, this program is converted into a subroutine and applied to the solution of the flow in a network of inter-connected conduits." In this way, the student may begin with a fairly simple exercise, gain familiarity and confidence but retain the early work as a valuable part of later, more complicated and meaningful programs. Other problems that have been treated successfully and effectively by junior level students are: I. Supersonic flow along a wall using the method of characteristics. II. Representation of characteristics of two phase flow with variable temperature, pressure and fluid parameters. III. Surge system oscillations. IV. Dynamic behavior of positive displacement fluid machinery with associated external systems. Representative problems in Heat Transfer and Fluid Flow are included in the problem section of this report. c) Kinematics, Dynamics and Machine Design Considerable activity has occurred using analog computers in the course/s in Machine Design, specifically in Dynamics of Machinery, (ME 540, 4 credits). "One objective in the use of the analog in ME 540 is simply to introduce the student to the use of the analog computer. A second is to demonstrate the validity of derived relationships of parameters for the systems involved. Important is the ability to observe visually the frequency response effect, and the response of these simple systems to various input functions." B13

An important use of the analog computer occurs in the laboratory. Here the student simulates the mechanism used in a torsional vibration experiment. The physical model and the simulated model are compared under the excitation of various frequencies. "The analog computer is a stimulant to the student learning the dynamics of mechanisms. Students better understand dynamics and are better equipped to solve problems in that area as a result of their classroom and laboratory experiences." Another faculty member states..."Many vibrations problems are quite abstract and in solving them the student is often so far from physical reality that it is merely a mathematical exercise. Next to an actual laboratory demonstration the EDA (Electronic Differential Analyzer) is the best tool to establish a connection between the physics and the mathematics of a problem". The course in Mechanical Vibrations (ME 540, 3 credits, graduate level) further extends the use of analog computers. In this course, the student is brought into contact with vibrating systems with several degrees of freedom. In addition certain non-linear problems may also be treated very effectively. The Mechanical Engineering Department analog computer has the equipment to allow function generation and square law multiplication. The multipliers may be extended in usefulness by using the built-in Zener diodes to obtain the appropriate sign on the results of the multiplication. In general, the results with the analog equipment have been regarded as successful. The feeling is strong that additional work will be incorporated into the courses using these machines. The Machine Design sequence also uses the digital computer in many varied applications. The first point at which contact is made with the digital computer is in Design of Machine Elements (ME 362, 3 credits, junior level). Typical problems have included a Bevel Gear speed reducer force analysis and variation in Hertz contact stresses in cam systems. The kinematic analysis has also been treated successfully.... "To me, however, the problem of kinematic analysis, as distinguished from synthesis, has been reduced almost at one stroke to pure routine. In other words, the act of programming the computer has produced a set of rules which can be followed by any technician (if computer time is too expensive) to analyze any mechanism problem". (It should be noted that "too expensive" was clarified later in a set of solutions using the IBM 704 in which the cost per solution was found to be of the order of 5 cents per point.) Other applications in Experimental Research in Mechanical Engineering (ME 408, 3 credits, senior level) included: Deflections in shafts with complex loading and discontinuous cross sections; Determination of transverse shaft vibration frequencies for multiple degrees of freedom by the Stodola iterative method. An evaluation of the use of computers in engineering education may be found in the following remarks submitted by one of the faculty:'"1. It contains an automatic feedback control on the learning experience of the B14

Use of Computers in Mechanical Engineering Education student. Problem solution by the student with subsequent correction by the instructor has long been a basis of our approach, but the insistence that the student correct his errors and resubmit until correct has not been part of it. The use of the computer furnishes the discipline that insists upon this complete approach to the learning process. 2. The element of reward for success is an inherent part of dealing with computers. This is also a vital part of the learning process. It is very interesting to observe the stimulation of interest, and the enthusiasm which is generated by an experience with the computer. 3. It induces conscious, careful, deliberate organization of analytical approach, and thereby brings about a deeper understanding of the problem at hand. (It seems to me that this is probably the most important virtue of the computer at this point in the program (ME 562). The introduction of the problem to go on the computer should be related to a method of approach to complex analytical problems. Up to this point students have relied a great deal upon intuitive, mathematical approaches, and have tended to deal with solutions which could be seen as a whole. At this point they should be dealing with several variables simultaneously, and should learn to derive what they want from the several equations involved.) 4. A rational approach to problems coupled with the use of applied mathematics is encouraged. 5. The temptation to generalize, and the ease with which it can be done on computers is helpful in studying alternative solutions to design problems. 6. The complexity and length of problems that can be handled is increased. Like all tools, the computer has its faults as well as virtues. It does absorb time at this stage, cutting down the number of problems that can be handled in a semester. This will be compensated for by practice, of course, and by the fact that fewer more complex problems may serve just as well educationally, any way. Sometimes students become more fascinated by the tool than by the problem. The search for punctuation marks is sometimes as tedious and more frustrating than the job of hand computation." Others are more "restrained" but show definite tendencies toward being "sold" within a reasonable period. The following two paragraphs are taken, intact, from a report submitted by one of the faculty to illustrate this point: "Of course it is a natural consequence of the Computer Project, but I think the digital computer's role in engineering education has been somewhat overdone at Michigan. I have noted B15

particularly that industrialists take a long hard look at a problem before considering a computer solution. Not often are they looking for the exact answers the computer will provide. Many times they will be perfectly satisfied if they can find something about the general trend. Hard justification is necessary before they will consider using a computer. I think this same justification should be employed in education. On the other hand problems exist today, and even more are foreseen for the future, in which computer solutions will be the only available means. In general these seem to be in the design area where optimization theory, reliability, and statistical information, are important factors. These will be dealt with, for the most part, at the Senior and the Senior-graduate level." Problems in the machine design area, including both analog and digital applications, are included in the problem section of this report. VI. AVAILABLE COMPUTING FACILITIES The digital computing facilities available to faculty and students in the Mechanical Engineering Department consist of the University Computing Center's IBM 709 with an associated IBM 1401 for off-line card, tape, and paper handling. The 709 will be replaced by an IBM 7090 during the coming summer. Considerable analog computing power is also available to the Department. This includes a Department-operated analog laboratory with five 8-16 amplifier Applied Dynamics (AD1) analog computers, each equipped with Sanborn hot-wire recorders. Additional equipment includes x-y plotters and square law multipliers. Diode function generators will be installed in the near future. In addition, a large Applied Dynamics computer (AD64) containing 32 amplifiers, built-in multipliers and diode function generators, is available for student use and laboratory demonstrations. The machine is equipped with hardware for repetitive operation and an oscilloscope for display of problem solutions. On occasion, students also take more comprehensive analog computer courses taught in the Instrumentation Department which has several large analog computers available for student use. VII. CONCLUSIONS The consensus, taken from the faculty reports submitted and condensed here, is that computers serve a vital function in the education of Mechanical Engineering students. In the main, the applications have grown conservatively and cautiously from small beginnings to the rather substantial uses quoted. It is useful, especially for those considering similar computer activities in engineering education, to review briefly some of the salient aspects of the use of computers as gleaned from the reports of the faculty of the Mechanical Engineering Department First of all, the learning process is considerably assisted by the proper use of computers in the class work. This benefit seems to be due to three major factors that are particularly active in the use of computers: B16

Use of Computers in Mechanical Engineering Education 1. The student must organize the solution of problems more explicitly and carefully when using the computer than when performing the solution by hand. This is because the student is always "present" when solving problems by hand. When using the computer, only the student's instructions for solving the problem can be "present" at solution time. 2. Rewards for a successful effort are very positive and immediate. This greatly enhances the learning process, stimulates interest and generates enthusiasm. 3. Discipline to submit and resubmit problems until completely correct is inherent in the use of computers by the student. It is interesting to note that the student will readily accept the flaggingof errors and termination of a computer run by the computer while he would probably complain if his instructor were to make the same correction in his work. Secondly, the courses themselves are improved by the use of computers. Some of these benefits are: 1. The extension of the student's mathematical experience through the association with numerical methods. 2. The treatment of more realistic and meaningful problems in the various course areas, with the added benefit of studying the results of these solutions under many varying conditions. This means that the courses are also extended into areas not previously touched. 3. The stimulation of greater student - faculty interactionthrough the treatment of interesting problems and solution techniques. Of course, there have been some dissatisfactions and criticisms. These are the natural byproducts of any creative process. It is significant that the reactions have almost always been stated positively with the obvious intent to correct and improve the use of computers in education. A part of the difficulties arose from the fact that both the faculty and the students were learning together. This is a little frustrating to both. As time progresses, more and more students will have had the basic computer course (now a required course for graduation in the undergraduate program). This will mean that less time will be spent in the classroom on programming details with more value extracted in the engineering training of the student. The evolution of engineering education has progressed into another stage. It is safe to say that the Mechanical Engineering curriculum will be molded for a long time to come by the forces generated during this effort. The processes of natural selection administered by an alert and conscientious faculty will insure the dynamic growth of the effective use of computers in the classroom. What has been observed and reported here is a glimpse of this process during its early, vigorous, formative period. B17

VIII. EXAMPLE PROBLEMS Several problems from the various Mechanical Engineering subject areas, many prepared by faculty participants in the Ford Project, are listed below. Complete solutions follow for most of the problems. For a few, only the problem statement is shown. All the digital computer programs are written in the MAD language which is fully described in A Computer Primer for the MAD Language by E. I. Organick. These problems may be considered as a supplement to Problems 1 through 45 published in the First Annual Report of the Project, Problems 46 through 56 published in the Second Annual Report, and Problems 57 through 64 in a recent Project publication, Use of Computers in Industrial Engineering Education. It should be remembered that many of these problems represent the instructor's first effort at a computer solution and may not necessarily illustrate sophisticated programming techniques. List of Problems Number 65 66 67 68 69 70 71 72 73 74 75 Title Isentropic Process for Ideal Gas with Variable Specific Heat Compressibility Factors Using the BeattieBridgeman Equation of State The Effect of Pressure and Propellant Ratio on Hydrogen-Oxygen Rocket Performance Determination of the Composition of the Products of Combustion Flame Temperature for Combustion of Air and Methane Transient Temperature Calculation for a Jacketed Kettle Surge System Oscillations Analog Analysis of a Sinusoidally Excited Spring-Mass-Dashpot System Bevel Gear Speed Reducer Force Analysis Cam Design Proposal Analysis Use of the Electronic Differential Analyzer to Study the Dynamics of Machinery Author R. D. Slonneger R. E. Sonntag R. E. Sonntag R. M. Shastri R. E. Sonntag E. T. Kirkpatrick H. P. Hale C. W. Messersmith J. R. Pearson J. R. Pearson Page B19 B25 B28 B45 B50 B60 B69 B76 B80 B81 B83 B18

Example Problem No. 65 ISENTROPIC PROCESS FOR IDEAL GAS WITH VARIABLE SPECIFIC HEAT by Robert D. Slonneger Department of Mechanical Engineering West Virginia University Course: Thermodynamics I Credit hours: 3 Level: Junior Statement of Problem The entropy change for an ideal gas can be computed with an equation as follows: 2 S = f C dT -R In ] (1) 2 1 c T " 1 Solutions for this equation are simple when the specific heat is constant, but when Cp is a function of temperature the solution is more tedious. In the case of the isentropic process, the equation can be used to compute one value of a variable if the other three values are known. For example, knowing the initial temperature T1 and the pressures P1 and P2,the final temperature T2 can be computed. When specific heats are constant the familiar equation k-l T2 [ P2 (2) can be applied. Tables of thermodynamic properties are available for certain substances (namely air) which account for variations in specific heat, but a trial and error solution of equation (1) must be used for other substances. Write and test a MAD program which will accept as data an initial temperature T1 (degrees R), an initial pressure, psia, and a final pressure, psia, for which the temperature, T2, will be computed if Cp is a function of temperature as follows: C =11.515 17 + 1 (3) P JT T Note: The above equation is for oxygen and gives the specific heat BTU/(lb mol ~R) if T is in degrees R. A solution to an accuracy of + 2 degrees will be satisfactory. Solution Since this problem can be solved easily by a simple trial and error technique the problem statement to the student can be very brief. In this problem the reason for using the computer can be justified by two arguments B19

Isentropic Process for Ideal Gas With Variable Specific Heat (1) Several solutions will be needed. (2) Although the solution is simple the arithmetic is such that the possibility of error is quite high. In solving the problem, S2-S1 in equation (1) is set equal to zero because the process to be analyzed is isentropic, and then the equation for Cp (equation (3)) is introduced: 0 = (7.515 + 1530 dT R in 2/p = (11.515-T —~Z2 + 0= 11.515 Xn [T1 + 44 [ T2 - 1 2 1/2 T /2 1 - 1550 L - T2i -1 R-OXn[2/P T1 For facility in program writing, an internal function F0 (T2,T1) = 11.515in 1 + 544 [T22 = - 15 50 2 was defined so that the following equation could be written DIFF = F0 (T2,T1) -n [P 2/P1 (5) (6) Thus if the value of DIFF is exactly 0, the exact solution is obtained. Since the tolerance of + 2 degrees permits some simplification, the solution follows a simple pattern. (1) The initial guess for T2 is made by assuming the constant specific heat case and applying equation'(2). The value thus obtained is substituted into equation (6) and the value for DIFF could theoretically be positive or negative. Note: This may not be so but a good computer program should handle either eventuality even though it is improbable that DIFF will be negative. In either case the value of T2 is changed in steps of 10 degrees until a sign change occurs, at which time the value of T2 is altered in steps of 1 in the opposite direction until the sign changes again. When the second sign change occurs the current value of T2 is called the answer. List of Symbols Tl = T1, Initial temperature, OR (Data). T2 = T2 = Final temperature, OR. P1 = P1 = Initial pressure, psia (data). P2 = P = Final pressure, psia (data). TISEN = Temperature calculated with equation (2), R. DIFF = Difference calculated from equation (6). COUNT = A counter to count iterations. DNIO,UPIO = A switch to indicate path taken. TEMP = A temporary storage location. B20

Example Problem No. 65 Flow Diagram IOb1OI T2 = Ta,- 0-A oD'JoI= 1 0o631 DIo / T? - T2 +/10' UPloi - O i FB W i 9 T2- * - ) Lz — -, B21

Isentropic Process for Ideal Gas With Variable Specific Heat MAD Program and Data iROIRT SLOMEGER D044N 1 0.0 050 SCONPILE MADvEXECUTE PRINT OBJECT*DUHP R THIS PROGRAM WILL CALCULATE THE TEMPERATURE AFTER R AN ISENTROPIC CHANGE OF STATE WHEN THE SPECIFIC R HEATS ARE A FUNCTION OF TEMPERATURE) INTEGER UP10Q DN10, COUNT START READ FOMAT INPUTe P1 P2- Tl PRINT FORMAT HEAD, P Pt P2 Tl INTERNAL FUNCTION F.(S2-*.S) l 1.*515*ELOG (S2/Si)+344.* 1 l1.*SQRT &Sf2 H)-t 1 ~/:SQRT-IS 1 ).::-1530*t t 1 /S2 ) -!( 1i/S i EXECTE ZERO.( UP10O DN10O COUNT) T2 * T1*f4P2/PI) *P* *266) TISEN * T2 608AIK 0FF a F. {T2tTl) - l. tG*ELOGC (P2/P1) COUNT * COUNT + 1 PRINT FORMAT CHECK. COUNT. D.FF. T2 WHENEVER DFW *F eG 0o0.AND.O COUNT *E. 1l tRAMSFER TO DOWN10 WHENEVER DI.-FF*,L.0. &AD.*COUNT*.,*1 *TRANSFER TO ADO10 WHENEVER DIFF*G0i.*O0AND*lDNlO.*1, TRANSFER TO DOWN10 WHENEVER DIFF:*L*O.AND*UP1 *E:*1 TRANSFER TO AD010 WHENEVER DIFF *6E..OOANDOUP1O.E1,TRANSFER TO LESS1 WHENEVER DIFF*L.O., AND*DN1 *El.tTRANSFER TO LARG1 DOWNl10 T2 a T2-10.0 DN1 " 1 TRANSFER TO GOBAK AO10 T2 * T2+10.0 UP1 a 1 TRANSFER TO 60BAK LESS1 TEMP I DIFF THROUGH ALPHA. FOR T2aT2-i~ 1TEMP-DIF.FaGETEMP 1DFF' F. TZT1) - 1.96*ELOG (P2/P1) COUNT COUNT+1 PRINT FORMAT CHECK. COUNT.DIFFvT2 ALPHA CONTINUE TRANSFER TO FINI LAR61 TEMP DI0FF THROUGf BETA*FOR T2T2+1.. 1 DIFF-TEMP *GE.-TEMP DOFF * F. T2*T1) - 1~986*ELOG6 (P2/P1) COUNT * COUNT +1 PRINT FORMAT CHECK. COUNT Di.FF T2 BETA CONTINUE FINI PRINT FORMAT ANS.*T2T ISEN TRANSFER TO START VECTOR VALUES INPUT $SF10.3,F10o3,F1O.O*S VECTOR VALUES HEAOSSI1HlI74HTEMPERATURE AFTER ISENTROPIC CHANG 1E OF STATE WITH VARIABLE SPECIFIC HEAT //19H INITIAL PRESSUR 2E * F 1S.5 SHPSIA*/17H FINAL PRESSURE = F15.5,5HPSIA./ 322N INITIAL TEMPERATURE a F5.0, 15HDEGREES RANKINE*S VECTOR VALUES CHECK*S8H COUNT - 15 S22 12HDIFFERENCE = F10*5 1*S2' 19HFINAL TEMPERATURE a F5.0*S VECTOR VALUES ANS * S20MOFINAL TEMPERATURE FS.o0 * 1XH5DEGREES RAMKIME/44HOIF SPECIFIC HEAT CONSTANT THE TEMPERAT 2URE a F50 15iHDiEGREES RANKINE*S END OF PROGRAM $DATA 14 7 100. 520. 10. 500. 600. B22

Example Problem No. 65 Computer Output TEh'MPER.TLURE AFTER I-SENTROFIC CHA.ING-F OIF STRTE WIiTH'VRF.:IRBLE SFEC IFIIC: HEPT INITI L PRESSURE = C T NtCI D tPP'.I lD1 - 14. 7000iOPSI A. tI l I )l 1 ~ n' l ~,r' T Ti *_J,- _.LF~;L'L _4_______ g: - V-'_ —-J- -I- I ilUr w;J-LU M IL-L - -- -- -- -- ---— ___ ---— _ --- INITITL TEMPERATURE = 520. DEGREES Rr-NKINE — _ JlUL I_- _ D I F F E R E"i C EIL -F,3- FI I- LTIE I - E I F. T R ECOL1JNT = 2 DIFFERENCE = 0. 0473 "- FINFL TEMPrERATLURE - 891. CCIil,i T -. T FFFFPFti- I - -n f-l.fQ PFT. TFM PFPQTU;pF I - fiiO COUNT = 4 DIFFERENCE0 = -0.02847 FINfIHL TEMFERiTUIFRE =:S1. __~:W'.iI _!...........5. IEEERE H 1 F-. _ -... I 0 L__-E'."l -: i L _Et.4EJF.FIU E. -;E.o —. COIUNT = 6 DIFFERENCE = -0.0115 F I RL TEMPFER TURE.= 8 —:3. __- LJ -'.__. =.....IR....... 0-_.....DEEREE __=__F'.i-.LE_ _ l ___. COLU-NT = S DIFFEREtNCE = 0.00532 FItRL TEMPERRiTURE = 8l5. FINHL TEMPERRTURE = 886. DEGREES RFANKINE IF SFECIFIC HEFT CON-STRNT THE TEMPERFTURFE - 900.DEGREES RfAKI.INE TE1ERPT. __FTR ~_SENTROIC CHlGE OF ST flTE.ITH'RIi.....S..EC C... HE.T TEMP:-ERATURE'`FTE-.. ISEI'tTR I 6:HnrGE OF STAT'E J.iT P R I Ai RIrLE -SPECIFIC HEAT INITIFL PRESSURE = FT NIi F PRF'-;! IF - 10. OOO0,FSI P..,i-in - n n 1 nnp — T Q INITIRL TEMFPEFRTURE = -600..DEGREES* FF.NKIK4NE _F:0 U MT... —.. C.:1 U;HT = C:!_.1!i r-4 T F,: i'OJT - C0 U lNT C 0~, U.i T - C 0 liU T C O 1'-liT = C: 0 U- r T C IAI! Ur"'T - C 0 O LiH T =: i0i U! T CIUNT C 0 i!_i iT - I0 11'-I=T -:, 1 1 I T = C: 0 LU1N,'T = C: CI N T = C O'! Ul T = CO: I Il'. T = C 0._I N T _:UNT = C 0 UNHT -: O! ii tT = LN:!Ii T = 1::: C1-.I T =.........___ -D]_E:iE E __.. =... _Q,_-._ -- "!/. LL..- r':, r___ LF lu t 4 5 7 8'. 10 11 12 13 16 20 21 22 2 3 24 25 26 ~ 7 DIFFERENCE = DIFFERENCE - DIFFERENCE = _DIF.EEREN' -=, DIFFERENCE = ICJIF-ENI tE_DIFFERENCE = DIFFERENCE = DIFFrRE.NE - DIFFEFRENCE DIFFEREE -E DI F F E r' E NrCE D I F FEEI-I:E CE D I F F E E Nf C E DIFFERENCE DI FFERE;NI"ICE -,1- I F F ~E..E t D I FFE F.:E, "IC:C E DlI FFEr.'EtC:E = D I F F Ett E ": E EDIFFER EN CE _ I Frr]E Eti-H _= I F F Err.E l:rE - E C I F' r E " 4 C: E FI'T F ~~F'~.- FI' -. F:F O..-7, -.,: - 7,,;-,,'-:. 13.0-,76 5.'!. 0. C9 I F.'. L: i 4..1 4' II.ll r'* Fi 4Q 4 4 [i. 0J O.':,...: 13 i 4.' 0. 0t i - 2 iO., -,.i I -0. 0344_ -0.0_2 4 5., _ i3 t.!... 4. i: l _. i'l 1 i"' _' -~-:. j-l i |'.!.8_.. -i_'-'~i. " i _,,8,.. I- 1 )3i FIHAL F I' t l F T 4A I - F lFi L F I NAL.JE I EiL.. -- LNiL. FINAL F f! i A L FINAL F Li FI NfAi. L F I Ai AL FIN F L F I AHRL F I HAL.t FINAL F I H L FI,-lhL F I HAL. F I HAL F' t.-I cc I T,:... - I - " - _-:.... - T__EJ- P.E:i-A 1 -'-.._' I TEMPERFiTURE =1'87. TFMp,r F',TSIS!.'F -1 -1'i7. T' r: F.: R T i R E - 1' -'. TENFMPERATURIE =17 T E " IE-IU, I.S — -. _. T tEF..T!F. 1',. TEMPERRTURi E -1787 T -EiiP F:EER; a ItE -- - 1.... _'1 FF.:l:._TII,_.,'~tE -__1 -,',7. T F I'RE R.R T E - 1 7 CJ - TEPF'E.RATURE =19 7. T E l F E R H T U R E = 1 I. _T E tPF EF,-T. - t_. --- E TEM1ERTURE.' _ZF-L 1 6 67. T E tiF FE.'Ri T I R, E -! (,.~ TEMFPERRTURE = 16'. TEMPERAfTURE =1 6:-. TEMPE. RTURE =-16,.TEMPERRTLIRE =16i4. Tr F,EP F:ii T! I E' -T 1 E r;,',. TEMPERA-TURE 1 4 iTE PI P E R i i R F' I r i.; F;, - F:.IH H.L TEr 1PiFER FTURF E, 66. D- EG1 R,[, -!-F t-... J-' I -- --- -- IF:FE: E IFI C HE A T' irTRN T T.HE -TE. L F:Tl. L- -:-DLF:E TLL.E.. ~ ~ ~ ~ ~ I.!;....-..r.~.~.....,... -* * RLL D CAT FTF: PRCE SSED. B23

Isentropic Process for Ideal Gas With Variable Specific Heat Critique The selection of a trial and error technique was deliberate to emphasize to the student the approach that can be used to such a problem. Since this problem is intended for first semester junior students it was felt that more elegant numerical techniques should not be considered. It is an example of an early computer problem which is easy to solve, but the student should realize that the computer can solve routine, tedious problems as well as the very elegant problems. It seems to this writer that this type of problem, solvable by conventional means but tedious and filled with chances for error, are excellent problems for the junior level engineering student. Moreover it is possible that a more sophisticated problem might discourage a beginning student. The selection of 10 degree steps for successive approximations was quite arbitrary. In the original instructor's solution some conditional print statements were included and for fairly low pressures (P2) the 10 degree guess was excellent. For higher pressures machine time could be reduced by specifying larger steps based on the magnitude of the term DIFF. B24

Example Problem No. 66 COMPRESSIBILITY FACTORS USING THE BEATTIE-BRIDGEMAN EQUATION by Richard E. Sonntag Department of Mechanical Engineering The University of Michigan Course: Thermodynamics II Credit hours: 3 Level: Junior Statement of Problem Write a program for the I.B.M. 709 Computer to calculate compressibility factor as a function of reduced temperature and pressure for any pure substance following the BeattieBridgeman equation of state. For any Tr, consider values of Pr from 0.2 to an arbitrary Pr max in steps of 0.2. Plot the resulting reduced isotherms, compare with the generalized charts, and discuss the correlation. Test the program using data for nitrogen with values of Tr = 1.0, 1.1, 1.3, 1.6, and 1.8. Let P be 3.0 for all values of T. max Solution The Beattie-Bridgeman Equation of State is P RT 2 + v+ 2 3 +v v v where = BoRT - A - cR/T2 = -BbRT + A a - B cR/T2 = Bobc R/T2 Newton's method of solving iteratively for v is as follows. f(v) = Pv4 - RTv3 - pv2 - Yv - = f'(v) = 4Pv3 - 3RTv2 - 2v - f(vi) v = vi f'(v ) = vi(l - Ei) where f(vi)/f'(vi) E = __ i Vi For the first trial, ~ / RT + + + RT + + P v = - + + - 1 Pv PvR (RT) 2 B25

Compressibility Factors Using the Beattie-Bridgeman Equation Flow Diagram MAD Program and Data $COMPILE MAD, EXECUTE, DUMP READ DATA PRINT RESULTS TC,PC,AO,BO,A,B,C S1 READ DATA T=TR*TC RT=.08206*T RDT2=. 08206/(T*T) BETA=BO*RT-AO-C*RDT2 GAMMA=AO*A-BO*B*RT-BO*C*RDT2 DELTA=BO*B*C *RDT2 PR=O. S2 I=0 INTEGER I PR=PR+0.2 P=PR*PC V=RT/P+BETA/RT+GAMMA*P/(RT*RT) S3 I=I+1 WHENEVER I.GE. 25, TRANSFER TO S4 V2=V*V V3=V*V2 F=P*V*V3-RT*V3-BETA*V2-GAMMA*V-DELTA DF=4. *P*V3-3. *RT*V2-2. *BETA*V-GAMMA E=F/(V*DF) V=V*(l.-E) WHENEVER.ABS.E.G.0.0001, TRANSFER TO S3 B26

Example Problem No. 66 MAD Program and Data (continued) Z=P*V/RT PRINT RESULTS T,P,V,E,I,TR,PR,Z WHENEVER PR.GE.PRMAX, TRANSFER TO Sl TRANSFER TO S2 S4 PRINT COMMENT $O.NO SOLUTION $ PRINT RESULTS T,P,V,E,TR,PR,F,DF,I TRANSFER TO S1 END OF PROGRAM $ DATA TC=126.1,PC=33.5,AO=1.3445,BO=0. 05046,A=0.02617,B=-0.00691,C=0.42E+05* TR=1.0, PRMAX=3. 0* TR=. 1, PRMAX=3.0* TR=1.3, PRMAX=3.0* TR=1.6, PRMAX=3.0* TR=1.8,PRMAX=3.0* Computer Results for this program are not shown. B27

Example Problem No. 67 THE EFFECT OF PRESSURE AND PROPELLANT RATIO ON HYDROGEN-OXYGEN ROCKET PERFORMANCE by R. E. Sonntag Course: Seminar in Thermodynamics Credit Hours: 3 Level: Graduate Statement of the Problem Consider the following rocket motor. Liquid hydrogen at -423~F and liquid oxygen at -290~F are fed to the combustion chamber in a molar ratio of m:l/2, where m may range from 1.0 to 2.0. The combustion chamber pressure may range from 200 - 300 psia. a) Assuming no heat transfer from the chamber, determine the flame temperature and composition, considering the products to consist of H20, H2, 02, H, O, OH. b) These products are then expanded through a reversible adiabatic nozzle to 14.7 psia. Consider the following two possible situations. 1) frozen equilibrium; i.e., there is no change in composition through the nozzle, although specific heat varies with temperature. 2) changing equilibrium; i.e., at each point in the nozzle there is a condition of equilibrium among the six constituents. For each of these cases, determine the nozzle exit temperature and velocity, and in 2), the composition at the nozzle exit. Changing Equilibrium or () Frozen Equilibrium Combustion Liq H2 V v3 or v4 Data Liquid 02 at -290~F Liquid H2 at -423~F T-Eo = -5400 BTU liq 0 Lb-Mole;-O, = -3380 BTU liq H2 Lb-Mole B28

Both relative to 0 for gaseous 02 and H2 at 537 R. Cp = A + BT + CT2 + DT3 MT T = ~R P Mole-R A B'103 C.106 D.109 H20: 6.970 1.925 -.149 - H2: 6.424.576 -.0241 02: 6.732.835 -.0554 - OH: 7.1663 -.3365.310 -.0358 0: 5.3621 -.360.1046 -.00942 H: 4.968 — o BTU -o BTUo BTU 537 Mole g537 Mole 537 Mole-R H20: -104,071 -98,344 45.106 H2: 0 0 31.191 02 0 0 49.003 OH: 18,100 16,080 43.888 0: 106,500 98,900 38.469 H: 94,000 87,500 27.393 Solution Consider the combustion mH2 + 2 - H20 + (m - 1) H2, m 1 (1) subject to the dissociation reactions 1. H20 = H2 + 02 (2) 1 2. H20 H + OH (3) 3. 2 H H (4) 4. 1 0 0 (5) The four simultaneous equilibrium equations are x x 1/2 K H2 02 1/2 (6) 1 XH20 X 1/2 2 OH XH1/2 B29

Effect of Pressure and Propellant Ratio XH p1/2 K p 3 -= 1/2 H2 xo 0 p1/2 K4 = x0-/2- The mass balance ratio of H to 0 is The mass balance ratio of H to 0 is (8) (9) GM-ATOMS H 2m GM-ATOMS 0 1 2XH 0 + 2XH + XOH + H XH20 + 2x0 + XOH + X0 (10) (11) and the sum of the mole fractions must equal unity, or XH20 + XH + + XH + x + XH = 1 2 2 2 Substituting equations (6), (7), (8), (9) into (10) and (11), (A4) xH x0l/2 + 4m x t (A5) x1/2 + (A6) xH /2 /2 H2 2 2 2 2 2 - 2 x - (A7) xH/2 = H2 ( 2 and (Al) x2 x/2 + XH + + (A2) x 1/2 1/2 + (A7) xH/2 H2+ (A3) x12 /2 2 1=0 + (A3) 1/2 _ 1 = 0 2 (12) (13) where (Al),...,(A7) are constants for a given temperature, pressure, and m, given by Al = P1/2/K A2 = K2/K1 A3 = K4/P1/2 A4 = 2(m-l)(Al) (14) A5 = 2m(A3) A6 = (2m-l)(A2) A7 = K3/P1/2 The equilibrium constants are functions of temperature only, and can be evaluated from the relation AH T - T n KT = n K + + (v) dT dT (15) o R T0'o R TLT 0 T / T Prod- RT To o React. The resulting four equations are those listed in the program for Kl, K2, K3, and K4 in terms of temperature T. The procedure for solution for a given pressure P and ratio m is to assume a temperature, evaluate the four equilibrium constants, and then the coefficients Al through A7 according to (14). The set of equations (12), (13) must then be solved for xH and x 2 2 B30

Example Problem No. 67 The Newton-Raphson iterative method was used for this solution, in terms of the variables Y = 1/2 XH2 (16) Z = 1/2 02 Once these values have been found, all the mole fractions can be calculated from equations (6) through (9). To check the assumed temperature, it will be necessary to make use of the First Law of Thermodynamics at state 2 and the Second Law for either state 3 or 4. Thus, for Part a) of the problem, assuming negligible velocities, H2 H = O (17) 1 liq H2 2 liq 2 (18) and H2 nHOh hH20 n02 h0 + nOH h0 + nOH hH (19) 22n h2 n2 2 O2 The problem introduced here is that it is necessary to know the number of moles of each component, rather than the mole fractions, which have been determined. Consider that a moles of H20 dissociate according to reaction 1 (Eq. 2), thereby forming a moles of H2 and a/2 moles of 02. Similarly, let b moles of H20, c/2 moles of H2, and d/2 moles of 02 dissociate according to the other three reactions. The composition at equilibrium, then is nHO = 1 - a - b n m 1 + a + b/2 - c/2 2 nO = a/2 - d/2 (20) ~2 \ (20) nOH nH = c n = d Therefore nT n. =m+ 2 (a + b + c +d) (21) nT n i - and n. i n- (22) T Since the mole fractions have already been determined at the assumed. temperature, a, b, c, and d can be evaluated by reduction of the six equations (22), resulting in X2 [1 - m + X3 + X4 2m - 1- X1 X + X3 [l + X+ [l - X - X [l + Xl] 2 2 1~ [1 XLx B31

Effect of Pressure and Propellant Ratio X -[X t a x4 -[ x4 + a (24) b = ----— (24) X4 - X2/2 c = bX1 (25) d = bX2 (26) where xH X1 x0 H XOH X2 x XOH XH (27) 2 X = 3 XH20 X o2 4 xH ) -H20 These values are then substituted into (20) to obtain the number of moles of each substance. Assuming ideal gases, the partial enthalpy of each component is equal to its enthalpy at one atmosphere pressure, so that for each component T h. h. + Cp dT (28) Z ~ o i' P~i T T0 i 0 The resulting six equations, to be found in the program, are given according to the nomenclature HH20 for hH2o, etc. If, at the assumed temperature, the First Law (17) is not satisfied, then the temperature must be incremented by some AT, the direction depending on the sign of (17). The entire procedure is then repeated for the new temperature. Whenever the sign of (17) changes from one trial to another, the AT is cut in half and the temperature incremented in the opposite direction the procedure being repeated until (17) is as close to zero as desired. For part b), the frozen equilibrium calculation is performed first, because the composition is the same as that just determined at state 2. The half-interval method of assuming a temperature and incrementing is again used, the condition to be satisfied being the Second Law, S4 - s2 = 0 (29) where S nH20 SH20 + nH2 SH+ + n + n OH SOH HH + nOSO (30) 2 2 2 2 2 2 and S = S - R n(xP) (31) i i'-T T o TO + i dT s - + I P T (32)'T0 ^ o(32) 0 B32

Example Problem No. 67 S2 is determined from the temperature, pressure and composition at state 2. Values for T4 are assumed until Eq. (29) is as close to zero as desired. Upon solution, the velocity can be calculated from the First Law, -2 V4 H1 = H + (2m + 16) - (33) 2gc by using the set of enthalpy equations (28) at the temperature T4. In evaluating the changing equilibrium part, the requirement to be met is that S3 - S2 = O (34) where equations (30), (31), (32) give the entropies. The difference between this part and the preceding is that the composition no longer is the same as at state 2, but instead must satisfy the equilibrium equations (6) through (9). Thus, the procedure is similar to that of the first part of the problem except that Eq. (34) must be satisfied. Upon determination of the temperature T3 and composition satisfying these equations, the velocity is found from the First Law, -2 H = H3 + (2m + 16) - (35) 2gc where the enthalpy equations are now used at T3. It should be pointed out that the six equations corresponding to (22) could be substituted into the equilibrium equations (6) through (9), resulting in a set of four nonlinear equations in the four unknowns a, b, c, d, which could be solved simultaneously. The problem was not solved by this procedure due to the additional computational time required for such a solution. Flow Diagram B33

Effect of Pressure and Propellant Ratio Flow Diagram, Continued,3 T2-T2-AT2 IAH/ I-* PUNCH (NO (}- SOLUTION) M, P2,T,I,J,L - T3=T3-AT3 AS/ ASI B34

Example Problem No. 67 Flow Diagram, Continued CALCULATE PUNCH MP2,T4, T4,I, (vY4^ -*J,V4,XH20,XH2,X02, S=1.0 T3=T4+lO00/M 8 XOH,XO,XH'PUNCH M,P2,T3,AT3,I, /\X CALCULATE _^ J,V3,XH20,XH2,X02, V3 XOH,XO,XH NOTE: FA(Y,Z), FB(Y,Z) APPEARING AFTER DELTA ARE THE SET OF EQUATIONS (12), (13) IN TERMS OF Y, Z, TO BE SOLVED BY THE NEWTON-RAPHSON ITERATIVE METHOD. B35

Effect of Pressure and Propellant Ratio MAD Program and Data R. E. SOINNTflG::.. 1:: nn innn nn *COMPILE MRD, EXECUTEPUNCH OB.JECT, DUMP VECTOR VRLLUES DRTR=-7F10..5*,$ *002 ----- _ P3=P —-- P=l14.7- - - - - - - -ln.Z DELH=1.0 *004 I=0 *!P=P2...'14.7 *006 L=2 *_/.00l7 BET' I=I+1 * 008 T=T2 *009 IE1 —------------------------------------------------------------------ 009 GAMMfl ~ ~VY=SQRT. CXH2i}' -01 0 K1 =EXP. (-4 7=0 + 1.42*ELOG. T - H0-II].?lT +. LCKE1 6,10.. P. - 0012 16*T*T - 51 702.. -T':'.01 2 K2=EXP. (C-4.G94 + 1.71 6*ELOG. (T-) - i0. t00497#T + I0.0375*10..P.,013 1-6*T*T - i0.001510... P.-'-.*T*T*T - 6:]:1.T,n13.' —! —F'"-'-' —— 6-K3E: + 1 ~:-.:-4*ELOG. (T - L00CIOO2=#T +. c001 005.*1'0.' *01 4 1P.-G *T*T - 46896T......:: 1 4 " ~ ""-'" K4=EXP. C-.02 ii + 1.1005*ELOG.3. T} -. i000 1i *36T + 0.-01 1 1 1 10.. F. -iS# 015 1T*T - 0.003'955*i0..,.-9#TTT -9*T*T*T 53132.2.T: 01.5 P1=SQRT. CP).'-K1 01' A2=K2."K 1 01 7 FA3=K4.-"-SQRT.:p:) 018 R4=2.* - 1}:*Ri *019'.: —-—. -, — - -m,..J -0 #n2..-:'*'~T —I.T... Fih^vJz-_J-lILz-Lf^ A + 1 6*02n FRA=*'A4*Y* -.'* +. + R.*M + R *Y*Z - 2.*'.':*Y - R 7,'.,.'.02., -- -"F"F...,'i..','.. "+*i1* + Z#Z + R*2*"# + it7*'.,.' + P.3*Z - 1. *0 02, DFHDY'=2. *Q4l*Y*Y*Z +'. A6","Z -4..'#A - it7#'. 7,027*' DFRDZ R4#*#,'''2 +:-:.*MZ* + l.5-Z + Hit*'-,Z #02" DFEBD[= RlI-"'*'' *Z + Z +','Z + R3*Z *i.0 -DErNOM=DFRAD'YDFEiDZ -. DFADDF1 01 —--------- ------ ------- DELY'= CFB*DFRDZ - Fi #DFBDZ: E..-DEuOM #032 DELZ.=CFIA*DFBDY - FB.DFi:DY.F':"..... DEHrOMi #033 WHENEyER J.E. 50, TRANSFER TO EF' IL *034 WHENEVERF.. RE'.. CDEL'.-:. LE. 0. 0001.R' AND.. AE. C,::ELZ::,. LE. *035 1 0.00001:. TRFNSFER TO MU *035 Y =,' *EFP'. CDEL V::,, ]3 Z=Z*E:P.:DELZ) *037 TRAfNSFER TO DELTA *0.37_MUXH;=.:<:2 03'::-: 02=Z-*z *4C:XH2-'0=XH2*Z*R 1 #041 X:H= ", * A 3 XO_____ =Z*fi..":,_________________________4 044 X1 =H/X-OH,045 X::<:2=^:':0/Xc-iH _Li- -4' —---------------------— *046 ----- -"" —---- = X2."-H2 - n_047 A=:'::2*.C1. - M +.3':, + 44*:C2.*M - 1. -:-:':,.:.::2 +:3::: 1. + X2 i04'. 1 + 0.5*C1. -::XI: -I 4 1 - 1 +':.: 1",# E:= X:4 - R*::4 +.5:::..: 4 -' 0..5:4.X2 n. *050 i D=B*X2 052 -. —-f —.H2_ J —-. —--- _ —----—.- - - ----------------------— _..-._JL5 3 —----— _ NH2=M - 1 + P + 0. 5*B - i.# *054 Nr02=0.5*(R -;>.:0). NH=B *056 NO=D *058' IdHEhE VER I. E. 1 ClJi; TFRAHN FER TgI EP SIL *!E;-!._..................... ------- Ij.IHENEVER L.E. 3: TFAPNSFER TO IETP *0ci60 DELHPR=DELH *Or:. B36

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Effect of Pressure and Propellant Ratio Computer Output M = 1.00 P2 = 200.0 XH20 = 0.6703460 XOH " 0.0936007 M= 1.00 P2= 200.0 T4= XH20 = 0.6703460 XOH a 0.0936007 M= 1.00 02= 200.0 T3= XH20 = 0.8103833 XOH = 0.0475383 M = 1.20 P2 = 200.0 XH20 = 0.6532267 XOH a 0.0673720 M= 1.20 P2= 200,0 T4XH20 = 0.6532267 XOH = 0.0673720 M= 1.20 P2= 200.0 T3= XH20 = 0.7765779 XOH = 0.0226282 M = 1.40 P2 = 200.0 XH20 = 0.6179716 XOH = 0.0416415 M= 1.40 )2= 200.0 T4= XH20 = 0.6179716 XOH = 0.0416415 M= 1.40 P2= 200.0 T3= XH20 = 0.7002676 XOH = 0.0058107 M - 1.60 P2 = 200.0 XH2C = 0.5739799 XOH = 0.0231559 M= 1.60 P2= 200.0 T4= XH20 = 0.5739799 XOH = 0.0231559 M= 1.60 P2= 200.0 T3= XH20 = 0.6217212 XOH - 0.0011077 M = 1.80 P2 = 200.0 XH20 = 0.5287101 XOH = 0.0120117 M= 1.80 P2= 200.0 T4= XH20 = 0.5287101 XOH = 0,0120117 M= 1.80 P2= 200.0 T3= XH20 - 0.5548105 XOH = 0.0001862 M = 2.00 P2 = 200.0 XH20 = 0.4859722 XOH = 0.0059508 M= 2.00 P2= 200.0 T4= XH20 = 0.4859722 XOH = 0.0059508 M= 2.00 P2= 200.0 T3= XH20 = 0.4998301 XOH x 0.0000299 M = 1.00 P2 = 225.0 XH20 = 0.6730543 XOH = 0.0934934 M" 1.00 P2" 225.0 T4" XH20 = 0.6730543 XOH = 0.0934934 M= 1.00 P2= 225.0 T3= XH20 = 0.8188794 XOH = 0.0454017 M = 1.20 P2 = 225.0 XH20 = 0.6555651 XOH a 0.0670443 M= 1.20 P2= 225.0 T4= XH20 = 0.6555651 XOH = 0.0670443 M= 1.20 P2= 225.0 T3mXH20 = 0.7828021 XOH = 0.0204497 M = 1.40 P2 = 225.0 XH20 = 0.6197767 XOH = 0.0411674 T2 = 6086.5 DELT2 = 0.50 I = 19 J = 2 XH2 = 0.1243038 X02 = 0.0397145 XO = 0.0227281 XH = 0.0493072 3827.0 DELT4= 8.00 1= 9 J= 2 V4= 9011.5 XH2 = 0.1243038 X02 = 0.0397145 XO a 0.0227281 XH = 0.0493072 4954.0 DELT3= 1.00 1= 11 J= 2 V3= 9418.6 XH2 = 0.0808551 X02 = 0.0296413 XO = 0.0090634 XH = 0.0225200 T2 = 6042.2 DELT2 = 0.25 1 = 22 J = 2 XH2 = 0.1955084 X02 = 0.0133436 XO = 0.0123252 XH = 0.0582241 3780.7 DELT4= 4.00 1= 10 J= 2 V4= 9479.1 XH2 = 0.1955084 X02 = 0.0133436 XO = 0.0123252 XH = 0.0582241 4830.1 DELT3= 8.00 1= 10 J= 3 V3= 9891.9 XH2 = 0.1697389 X02 = 0.0034956 XO = 0.0023376 XH = 0.0252219 T2 = 5899.0 DELT2 = 1.00 I = 16 J = 2 XH2 = 0.2747403 X02 = 0.0038769 XO = 0.0053182 XH = 0.0564521 3663.7 DELT4= 4.00 1= 8 J= 2 V4= 9831.4 XH2 = 0.2747403 X02 = 0.0038769 XO = 0.0053182 XH = 0.0564521 4486.0 DELT3= 4.00 1= 7 J= 3 V3= 10210.0 XH2 = 0,2788168 X02 = 0.0001843 XO = 0.0002233 XH = 0.0146972 T2 = 5694.0 DELT2 = 2.00 I = 12 J = 2 XH2 = 0.3527253 X02 = 0.0010338 XO = 0.0-019589 XH = 0.0471494 3502.1 DELT4= 4.00 1= 7 J= 2 V4= 10088.3 XH2 = 0.3527253 X02 = 0.0010338 XO = 0.0019589 XH = 0.0471491 4083.1 DELT3= 4.00 1= 11 J= 3 V3= 10408.3 XH2 = 0.3714413 X02 = 0.0000074 XO = 0.000013-2 XH = 0.005-7092 T2 = 5457.7 DELT2 = 0.25 I = 18 J = 2 XH2 = 0.4230364 X02 = 0.0002635 XO = 0.0006494 XH = 0.0353290 3320.0 DELT4= 2.00 1= 10 J= 2 V4= 10273.8 XH2 = 0.4230364 X02 = 0.0002635 XO = 0.0006494 XH = 00.353290 3705.5 DELT3= 2.00 1= 14 J= 3 V3= 10524.3 XH2 = 0.4431828 X02 = 0,0000003 XO = 0.0000006 XH = 0.0018196 T2 = 5211.0 DELT2 = 1.00 I = 15 J = 2 XH2 = 0.4833094 X02 = 0.0000656 XO a 0.0002004 XH = 0.0245026 3133.6 DELT4= 2.00 1= 6 J= 2 V4= 10398.9 XH2 = 0.4833094 X02 = 0.0000656 XO = 0.0002004 XH = 0.0245026 3376.6 DELT3= 1.00 1= 14 J= 3 V3= 10587.4 XH2 = 0.4996101 X02 = 0.0000000 XO = 0.0000000 XH = 0.0005299 T2 = 6116.2 DELT2 = 0.25 I = 21 J = 2 XH2 = C.1235838 X02 = 0.0393344 XO = 0.0222903 XH = 0.0482439 3764.2 DELT4s 4.00 1= 10 J= 2 V4= 9190.4 XH2 = 0.1235838 X02 = 0.0393344 XO = 0.0222903 XH = 0.0482439 4925.2 DELT3= 1.00 1= 12 J= 2 V3= 9610.9 XH2 = 0.0778967 X02 = 0.0286360 XO = 0.0083451 XH = 0.0208426 T2 z 6071.0 DELT2 = 1.00 I = 17 J = 2 XH2 = 0.1952693 X02 = 0.0130606 XO = 0.0120062 XH = 0.0570548 3719.9 DELT4= 2.00 1= 10 J= 2 V4= 9658.2 XH2 = 0.1952693 X02 = 0.0130606 XO = 0.0120062 XH = 0.0570548 4789.3 DELT3= 4.00 I= 9 J= 3 V3= 10092.2 XH2 = 0.1687971 X02 = 0.0029589 XO = 0.0019510 XH = 0.0230412 T2 = 5923.7 DELT2 = 0.25 I = 19 J = 2 XH2 = 0,2750221 X02 = 0.0037410 XO = 0.0051223 XH = 0.0551705 B38

Example Problem No. 67 Computer Output, Continued M 1.40 P2= 225.0..- T4= —.36 A....ELT04............-.-....=l XH20 = 0.6197767 XH2 = 0.2750221 X02 = 0.0037410 XOH = 0.0411674..XO-.0-.0051.22.3. JH =........................170... M= 1.40 P2= 225.0 T3= 4422.9 DELT3= 4.00 I= 7 J= 3 V3= 10408.4 XH20 = 0.7027159 XH2 = 0.2796673 —- X02 = 0.0001302 XOH = 0.0047577 XO = 0.0001574 XH = 0.0125715 M = 1.60 P2 = 225.0 T2 -. 513.5 _DE-I23 =-050 __L.-4 18 i = 2 XH20 = 0.5752619 XH2 = 0.3533399 X02 = 0.0009828 XOH = 0*0227106 XO = 0.0018614 XH =.0.i458.5436 M= 1.60 P2= 225.0 T4= 3436.8 DELT4= 2.00 I= 9 J= 2 V4= 10276.2 XH20 = 0.5752619 XH2 = 0*3533399 XO2 = 0.0009828. XOH = 0.0227106 XO s 0.0018614 XH 0 0.0458436 M= 1.60 P2= 225,.0 T.3-4QQ7.8 4-.8- ELT3= 200 1=1- 10 J 1= _3 V3 1060;-.5 XH20 = 0.6224668 XH2 = 0.3721416 X02 = 0*0000044 XOH = 0.0008276.XO -, 0.0000080.XH = 0.-004-5516.. M = 1.80 P2 = 225.0 T2 = 5473.0 DELT2 = 1*00 I = 13 J = 2 XH20 = 0.5295160 XH2 = 0.4237242 X02 = 0.0002478 XOH = 0.0117066 XO = 0.0006106 XH = 0.0341956 M= 1.80 P2" 225.0 T4- 3253'4- EL-Ts —-- -r04 — l- 10 J 2 V'4 10459;6 XH20 = 0.5295160 XH2 = 0.4237242 X02 = 0.0002478 XOH = 0.0117066 XO = 0.0006106 XH = 0.0341956 - M= 1.80 P2= 225.0 T3= 3625.0 DELT3= 8.00 1= 9 J= 3 V3= 10715.9 XH20 = 0.5550157 XH2 = 0.4435018 X02 = 0.0-0000001 - XOH = 0.0001274 XO = 0.0000003 XH = 0.0013547 M = 2.00 P2 = 225.0 T2 -= =.222.0- l - - -..0 —4 -44- J - -2 XH20 = 0.4864771 XH2 = 0.4839343 X02 = 0.0000609 XOH = 0.0057587 XO - 0.0001862.-.XH- 0.0235864- ---- M= 2.00 P2= 225.0 T4= 3066.0 DELT4= 8.00 I= 4 J= 2 V4= 10584.6 XH20 = 0.4864771 XH2 = 0.4839343 X02 =.- 00000609...................... XOH = 0.0057587 XO = 0*0001862 XH = 0.0235864 M= 2.00 P2= 225.0 T3- 3Z98.0 — ELT3= 4.00 -I-L —1 -J-= -.-3 — V-107737 —2 XH20 = 0.4998824 XH2 = 0.4997238 X02 = 0.0000000 XOH = 0.0000192 XO = 0.0000000 XH = 0.0003747 - M = 1.00 P2 = 250.0 T2 = 6143.0 DELT2 = 1.00 I = 20 J = 2 XH20 = 0.6754867 XH2 = 0.1229302 X02 = 0.03a9939 - - XOH = 0.0933852 XO = 0.0219012 XH = 0.0473030 M= 1.00 P2= 250.0 T4= 3711-.4 DELT4= 2.00 1=!0 — J —-2 — 4= 9340.5 XH20 = 0.6754867 XH2 = 0.1229302 X02 = 0.0389939 XOH = 0.0933852 XO = 0.0219012 XH = 0.0473030 -- M= 1.00 P2= 250.0 T3= 4898.4 DELT3= 1.00 I= 11 J= 2 V3= 9780.2 X.H-= —-e * 826Mi4 —— H2 — Q 0 --- X751 --.01.5 2- -71-752,:, XOH = 0*0434714 XO = 0.0077222 XH = 0.0193809 M = 1*20 P2 = 250.0 T2 6 6096,5 DELT2' 0.50 I = 2-3 — J 2XH20 = 0*6577448 XH2 = 0*1950338 X02 = 0.0128015 XOH = 0.0667130 XO = 0.-01-17151 XH-= 0.05599'1-9 _.... M= 1*20 P2= 250.0 T4= 3665*0 DELT4= 2*00 1= 11 J= 2 V4= 9817.7 - XH20 = 066577448 ----—.-XH2 * —0- -. 0r-30 —.-.... —-— 1 XOH = 0.0667130 XO = 0.0117151 XH = 0.0559919 Mu 1*20 P2= 250*0 T3= 4751.4 DE-LT3= 1.00- 1= 14 J= 2 V3= 10264.7 - - XH20 = 0.7880678 XH2 = 0.1680439 X02 = 0.0025201 XOrH 0.0185633 XO a 0.0016422 XH = 0.0211647 M = 1.40 P2 = 250.0 T2 = 5946.0 DELT2 = 2.00 I = 15 J = 2 XH20 = 0.621-3724 7 —5XH+2-.. 0.2752699 X02 = 0.00362-2 6-. ------ XOH = 0.0407417 XO = 0.0049522 XH = 0.0540435 Mu 1*40 P2= 250.0 T4= 3545,7 DELT4 4.00 1= 7 J= -2. V..4= —-1-1-7 XH20 = 0.6213724 XH2 = 0.2752699 X02 = 0.0036226 XOH w 0.0407417 XO = 0.0049522 XH = 0.0540435-. —--- - M= 1.40 P2= 250.0 T3= 4363.9 DELT3= 8.00 1= 6 J= 3 V3= 10581.0 XH20 -- 0.7046486 X$22 = 0.280400Q4- -.X02 =0.00 -0-9 ------ 1 XOH = 0.0039253 XO = 0.0001125 XH = 0.0108201 M = 1.60 P2 = 250.0 T2 = 5731.0 DELT2 = 100 =13 -................... XH20 = 0.5763867 XH2 = 0.3538786 X02 = 0.0009391 XOH = 0.0223164 XO = 0.0017780 XH = 0.0447019 -- M= 1.60'2= 250.0 T4= 3379.7 DELT4= 64.00 1= 2 J= 2 V4= 10437.7 XH20 = —.5763.67 — X2 — -0.353-A&6 —-— X2. 939 XOH = 0*0223164 XO = 0*0017780 XH = 0*0447019 Mm 1.60 P2= 250.0 T3= 3-936.7 DELT3= 4.00 I= 8 J- — 3...VY3 1Q7 3Q —--- --- XH20 = 0.6230291 XH2 = 0.3726952 X02 = 0.0000027 XOH = 0.0006221 - XO-9 XH = 0.0000 36.46Q0- - M 1.80 P2 = 250.0 T2 = 5486.5 DELT2 = 0O50 I = 16 J = 2 XH20 0*-.5302249 XH2 = 0/423?8l X_?= O0.OO2343 XOH = 0.0114350 XO = 0.0005775 XH = 0.03320UQ M= 1.80?2= 250.0 T4=-3194..9- DEL4-.2.00 -1=. 10 -J. 2- V4=-.620.-9-. XH20 = 0.5302249 XH2 = 0.4243281 X02 = 0.0002343 XOH = 0.0114350 X..Q000575 XH -0.0332003 -..-............ —. —. -—. M= 1.80 P2= 250.0 T3= 3554.5 DELT3= 4.00 I= 9 J= 3 V3= 10877.8 XH20 — 0.55514 —-4XH2 - Q.44434722341 — X02 - - n)00nnnn0 XOH = 0.0000901 XO = 0*0000002 XH = 0.0010352 B39

Effect of Pressure and Propellant Ratio Computer Output, Continued.... -— 20..P2 —P —2 5 --. -— =-22 -- DELT2 = 4.00 1 12 J. -.-=. —-....: ___.0 XH20 = 0*4869036 XH2 = 0*4844709 X02 = 0.0000570 H- _ — =- — ^ 0-59 3 -.... —- -XH — O0.022.Q05 -.............. Ma 2.00 P2= 250*0 T4= 3006.5 DELT4= 2.00 1= 8 J= 3 V4= 10745.5 -XX4- X20 — -0-.4869036 XH2 = 0-4844709 XQ2-= 0_QO00570 XOH = 0.0055936 XO = 0*0001745 XH = 0.0228005 — M -2.-00. P2 -250.0~ T3 322-8-5 --— DELT-3 2-O-..- -= 11 -l J= 3 -V3- 10934-1 ---. XH20 = 0.4999161 XH2 = 0.4997991 X02 = 0.0000000....... XOH —- + 00-1o2-7 - o —-- - -XH =0 002721 -.-.- —...- -- ------ --.-. M = 1.00 P2 = 275.0 T2 = 6167.5 DELT2 = 0.50 I = 22 J = 2 -; —— X — 4. 6..7763.3.XH2 -.122 3475 2 —= —-- XOd = 0.0932926 XO = 0,0215587 XH = 0.0464750 M= 1.00 P2= 275.0 T4= 3665.3 —- DELT4= 8.00 1= 7 J= 2 V4= 9470.0XH20 = 0.6776363 XH2 = 0.1223475 X02 = 0.0386900 -O XOH- -0.0932926 XO = 0.0215587 -XH = 0*.Q064750. Ma 1.00 P2= 275.0 T3= 4873.3 DELT3= 16.00 1= 8 J= 3 V3= 9932.5...._-........4 —.8-3-34940- XH2 - 00726958 X02 = 0.-268582 XOH = 0.0416983 XO = 0.0071714 XH = 0.0180824 M = 1.20 P2 = 275.0 T2 = 6119.5 DELT2 = 0.50 I = 19 J = 2 XH20 = 0.6597564 XH2 = 0.1948089 X02 = 0.0125652 XOH - 0.0663916 XO - 0.0114506 XH - 0.05502-3Ma 1.20 P2= 275.0 T4= 3616.5 DELT4= 2.00 I= 12 J= 2 V4= 9957.9 ---— XH2 —4-4,4597564 — XH2 —a-0.-94-8069 - X02 =0.0125652 XOH = 0.0663916 XO = 0.0114506 XH = 0.0550273 * —M —_20 —-2-=a-2-7-5-0 -.-T-38-4.74-3-9 —--..-DL -3 80 - I —10 - J - 3 -~V3a — -10423o7 ---- - — _ - -. XH20 = 0.7928145 XH2 = 0*1674065 X02 = 0.0021382 XOH * 0.0168233 XO - 0.0013790 XH - 0.01943-85 M = 1.40 P2 = 275.0 T2 = 5965.5 DELT2 a 0.50 I = 21 J = 2 --- -- 2O —-— 0-= — -06229082 —---— XH2 - -0*-2754911 -- X42- -0.0035114 --- ---- XOH = 0.0403108 XO = 0.0047923 XH = 0.0529862 -M 3-4 —-P-2 —- 2-5. —- -T-4 —3494.8 —-LDFLT4= 2.00 I= — 9 J 2 V4a 1039.44 —- -.. XH20 = 0.6229082 XH2 = 0.2754911 X02 = 0.0035114 XOH - 0.0'43108 XO - -Q.047923 XH = 0Q0529862 M= 1.40 P2= 275.0 T3= 4309.1 DELT3= 4.00 I= 8 J= 3 V3= 10732.2 -X... -— XH20. 070 817....2 — =- 007 -X02 — — 0 *0000677 -.... XOH = 0.0032658 XO = 0.0000816 XH = 0.0093752 M -- 1.6- P2 27-5 —0- - -T-2-5-0 7-47-,0 —D~_T2a — 140 -— 2 —1.1 J 2_. XH20 = 0.5773730 XH2 = 0.3543539 X02 = 0.0009014 XOH - 0Q0219682 XC a 0-.0017062 XH a-0436980 M= 1.60 P2= 275.0 T4= 3327.2 DELT4= 64.00 1= 2 J= 2 V4= 10581.6 -.....-.XH20 — = -O0+577371. - X2- - Q..35-43-539.-._ —-XQ2 - =- 0.000901A -.. —--- - —. XOH = 0.0219682 XO = 0.0017062 XH = 0.0436980 M= 1.60 P2- 275o0 T3= 3874,2 DELT3= 2.00 I= 10 J= 3 V3= 10918..... XH20 = 0.6234293 XH2 = 0.3731059 X02 = 0.0000017 _ XH = 0-0004799 XO = 0O000031, XH = 0.Q00298n M = 1.80 P2 = 275*0 T2 = 5498.5 DELT2 = 0.50 I = 16 J = 2 - XH2C - 0.5308623 -- H2.-. =Q4268.............. X02 a= 0-..QQ2225 - - - -...................XOH = 0.0111884 XO = 0.0005486 XH = 0.0323104 _ M..... P2= 275.0 - lT4_3=42... JlEL.TA-... Z a.QQ -10Q... Z Y4=-. l4Q76..i..... XH20 = 0.5308623 XH2 = 0.4248680 X02 = 0.0002225 XOH a e00Tl11884 Xf) = 0.0Q05486 XH = 0.0323104 M= 1*80 P2= 275*0 T3= 3490.4 DELT3= 16.00 1= 6 J= 3 V3= 11020.9 ---- XHN-20 = — 5S-25.-1._..XH2 -.-_ 4._38,Z-.... Q2_ —0O.QOO-....... XOH = 0.0000650 XO = 0.0000001 XH = 0.0008030 -. M -2.OQ- - - P2- =-2-.75 I2-524 _ DELIZ-__=__I-215 _.. 1 -=__=______ _____________ XH20 = 0.4872964 XH2 = 0.4849561 X02 = 0.0000536 XOH 00054411 XO = 0-0001641 XH = 0.0220888 Ma 2.00 P2= 275.0 T4= 2953.2 DELT4= 4.00 I= 6 J= 2 V4= 10886.6..-..XHf2-= -0a872964/- - - --— 488/5L61 -..XO — =-O-0.l05-36 -— _ ---—. XOH a 0.0054411 XO = 0.0001641 XH = 0.0220888 — 4 -— 2,OO -1-P2 =-27 t0.-. — T3^-3146 70-.DELT3 —— Z.ICL —I —IL2 —J= —:_ —— 3 — 1 --— 1J:ZXH20 a 0.4999384 XH2 = 0.4998501 X02 = 0.0000000 -— XH =0A0000087 X - _.0OOQ00 XH = 0.Iu02280 — M = 1.00 P2 = 300.0 T2 = 6189.5 DELT2 = 0.50 I = 23 J = 2 *.- XH20- =-0.6-97450-.XH2- -= —0-12-1797-.2-. —0 —.0 0 384016.. —-. —------ -_ —- -— __ —-_-.. —-- XOH = 0.0931626 XO = 0.0212306 XH = 0.0456907 -M -— l00 —-2-= —300.0 — T.4 - -34 —-4 EL-. —4s. — 40Q-0-4 — - -- 8 --— = —-2.-V4-= -95Q9 XH20 = 0.6797450 XH2 = 0.1217697 X02 = 0.0384016.... -_ —-.H' a- 0-rG-9'3-1626 AX - O.Q2123Q6 XH - a "456907 M= 1.00 P2= 300.0 T3= 4850.1 DELT3= 4.00 I= 10 J= 3 V3= 10067.0 XH20 = 0.&397476 --— X2 —40-744265 - — X0-2 = 0.-02607-3 ---- XOH = 0.0401037 XO = 0.0066934 XH = 0*0169506 M = 1 —20 P2- 300-0 -.T2 -—.6140.7 —- -— DELl-2 -=.-.-25 -- I = 23 -_J -J. 2. _ XH20 = 0.6615463 XH2 = 0.1946086 X02 = 0.0123545 -..-..... - -= 0661-0-4-7-7 XO = 0)0112165. XH = 0e0541693 M= 1.20 P2= 300.0 T4= 3573.4 DELT4= 2.00 I1 13 J= 2 V4= 10080.7 XH20 = 0.6615463 - -— ^ XH2 = 0.1946086 — X02 -0.-0123545 —.-.-. XOH = 0.0661047 XO = 0.0112165 XH = 0.0541693 B40

Example Problem No. 67 Computer Output, Continued M- 1.20 P2- 300.0 T3: 4679.7 DELT3- 1.00 In 136 d 2 V3 10559.6 XH20 = 0.7967626 XH2 = 0.1669135 X02 = 0.0018328 -------- -X- w- --— 2+ — - _H —= —. — -- 5 —--— X0e -5 +-2- -4 —--— O- X-1~ - 0739 —------------------- M - 1.40 P2 = 300.0 T2 = 5983.5 DELT2 = 0.50 I = 18 J a 2 ------- — 6-2*274.4. —— e —76 -- -------------------— 3 —XOH = 0.0399251 XO = 0.0046520 XH = 0*0520451 M-m 140 P2 300.*0 T4" 3449.4 DELT4 16.00 1 — 5 j- 2 V 1044 6.7 XH20 = 0.6242743 XH2 = 0.2756902 X02 = 0.0034135 -------— XoH — — 0-99-25 --- -Q6 ------ 05 —------------------ M= 1.40'2= 300.0 T3= 4257.7 DELT3 2 2.00 I= 11 Ja 3 V3= 10866.7 X —- )200 —74 7. —-- - - --— XH -=-281-j87 —-— O.X2 — l-0 —---------------- XOH = 0.0027359 XO = 0.0000599 XH = 0.0081678 M 1.U P2 - 300-.0 - - T2r 576. ELT2. 00-I T 18 J - 2 XH20 = 0.5782698 XH2 = 0.3547839 X02 a 0.0008678 ---- XH —------ -— 2-1473 ~- --- -012 47 —--- -— 0. 0-2 —--- -04O27&94 —------- M= 1.60 P2= 300.0 T4= 3281.9 DELT4= 2.00 In 10 J= 2 V4= 10705.3 -- XH — - -X -+ —-578a —--— 2- 3 -Z83 —X —-0.74 —X2 --- ------— A7. 768 - 3.0216473 XO = 0.0016422 OH 3 0.04278 1 M-" 160 P2 — 33000 T3-s 3869 F ELT3a 2-00 I= 10 3= V3= 11047.1 XH20 = 0.6237312 XH2 = 0.3734272 X02 0 0000011 ---------- XOH =- 0-.037-5- ----— XO-=4-0-02 —--- — 0X- Q2.629 —-------------- M = 1.80 P2 = 300.0 T2 = 5508.0 DELT2 = 2.00 I a 9 J a 2 ----- -26 4.~-0 — H- 253 —-- -X02- -0300l2114 —----------------- 3 XO = 0.0005206 XH 3 0.0314422.,80 P2- 300.0 T'= 3095:2 EDLT4" 2-00 I= 10 I= 2 VA= 10886aXH20 = 0.5315038 XH2 = 0.4253911 X02 = 0.0002111 ----------- -0-1 O09343 --------. -— O - 26005 ----— X 0 L42 —------- M: 1.80 P2= 300.0 T3= 3432.8 DELT3= 2.00 I= 11 J= 3 V3= 11146.7 -------— XH2- — 0 - 3 129 —-— X0 2 —-D ___ QOD —------------------- XOH = 0.00000479 XO = 0.00000001 XH = 0.0006340 M - 2.00Q P2 300 0 T2 - 524,8-0 DELT2 = 2-00 I 3 13 1 2 XH20 = 0.4876552 XH2 = 0.4853934 X02 = 0.0000507._ —--— x- -OH- o0-5-3020 —---- 4a —. X.1Q_55-0- 1 ---..- _X __= _L_- -iD 2 -z - __ M= 2,00 P2= 300.0 T4= 2906*0 DELT4= 1.00 1= 8 J= 2 V4: 11009.1 --- ____ —32a0= —848J-_65-52 ---—.XH12_-_=0 4 _85349 — XQ2-_- — QD07 ___ —-----— __, - XOH = 0.0053020 XO = 0.0001550 XH = 0.0214474 M- 2_00 P2"-'300.0 T'= 3117-0 r1ELT3'-=.900 I= 13 1= JS3 V~= 111_.~O0 XH20 = 0.4999538 XH2 = 0.4998859 X02 = 0.0000000 _ ____=__o0aQO1. —-— X- 0 -.Q0QQ00.0I-000 _ —-X_-_A — 0QQl1-1 -_- ------ - Discussion of Results The program output gives the values of chamber pressure in psia, combustion temperature in OR, and the various mole fractions at state 2. The second set lists tne temperature and velocity, ft/sec, at the nozzle exit state 4, resulting from a frozen equilibrium calculation. The composition here is necessarily the same as at state 2. Finally, the temperature, velocity and composition at the nozzle exit state 3, resulting from a changing equilibrium calculation, are given. Accuracy of the calculated temperatures in the half-interval method is indicated in the output by the smallest AT values, although the true accuracy of solution is somewhat less, due to inaccuracies in the specific heat equations. The combustion temperature and both exit velocities have been plotted versus propellant ratio at three pressures. Of particular interest is the increase in exit velocity with increasing m, even though the combustion temperature decreases, this being due to the decreased molecular weight of the products from the excess hydrogen. Eventually the velocity would reach B41

Effect of Pressure and Propellant Ratio a maximum and then decrease for higher values of m, but this range was not studied since the nozzle exit temperatures become so low as to make the dissociation reactions negligible and consequently introduce convergence problems in the solution. It should also be mentioned that the actual nozzle process, although not isentropic, should lie somewhere between the two extreme conditions considered here. ~~6300 rnri 8 W f lii -It 11200 P2 = 200. 6200 g 11000 6100oo 10800 6000 ~ ~ 10600 o 3 59800 10200 o 5800 10200 o -p tH 57 loi|j! lg0 >O r Tr~ ~N ~553~~~00 p. X~0 o z6o 98o0leee ^'"fgalE -- 54900 9 10400 1. 1.2 1.4 1.6 18 2.0 t 5200 10200 Propellant Ratio M (lb-moles Hydrogen per one-half lb-mole Oxygen) (Ib-moles Hydrogen per one-ha~#!~:~lf Ib-mole Oxygen B42

6300 6200 6100 6000 0 o 0 cf o (D P cD TO ) r (D 5900 5800 5700 5600;:pi =:::250, t i,...f,I:Ld:++HH''iffil I Ji ^1 ":',;1 1 -- -+1|......t XN I- i,,'- i;!Wi ~tI|:i-l: - l,,'11 1 " i':, 11|,11 I IIIH-1:H-t+H1tl I I Nl 1.0 1.2 1.4 1.6 i 1.8 2.0 Propellant Ratio M 1.0 1.2 1.4 1.6 1.8 2.0 Propellant Ratio M 11200 11000 10800 6100 6300. t. I;;1i-tl;0;ll tll,, tii 1,11+ i t'i! -t: T t tu'm tt iy t 4' i t j | T t t' I tr' 4 i T ttatl:ilT'ali i|| fi it llt 1, 4: I. TilTi:0$tJli-t td,,44ltt2+X V4 44S0 ~g-1ja, s- X | W liit if 41 rfi iif-4tt ti r-+t;i:n + t+:!~:tt;i bf,!i-t~tiittitit itti t+'t+ti ii~tiicWiift~it~fit t g i;: j I, i ^ f'itt'' itt i i I Jit tt|i 4L F+ ++i -' | P.2 = 3~~00 Li+. I it +,t.. I';, t... i.,,,.. 11 ll-M, ni,,...,.,.,,, 10600 6000 o 10400 N N (D Li 10200 ct CD 0 * 10000 c 9800 c< CD 0 0 0 o 0' ct (D H1 0;3'' tU t' t i r tj l-t I t tli 0 At 6200 t IIIII-JiiI i 144ii~4~ - Lf 0 oi 5900 5800 5700 5600 5500 5400 5300 5200 11200 11000 10800 10600 It I'T i',i~i t4 41, -r; 4 t i 4IMt-' T'44F1 liFTF 1 -, W,':;T~ m -tffr':if~~~~ittl t~in~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~iL-H 4 4- t::~~. n + c;4f:Li4+ llf -+H tt~ Un T 0 N 10400 N (D NI ct 10200 CD o 0 10000 c 9800 9800 0 CD 0 a (D o 0 - I 5500 5400 5300 9600 9400 9200 9000, 9600 9400 9200 5200 9000 1.0 1.2 1.4 1.6 1.8 2.0 Propellant Ratio M (lb-moles Hydrogen per one-half lb-mole Oxygen) (lb-moles Hydrogen per one-half lb-mole Oxygen)

Effect of Pressure and Propellant Ratio Critique The problem solved is a realistic one, although simplified as the interest was in the thermodynamics of the problem rather than in rocket design. The project was experimental, that of assigning a very difficult problem to a class with no programming experience. The class consisted of only five students, thereby permitting more individual attention than would normally be possible, and the students had previously studied the basic subject matter, enabling them to begin the solution fairly early in the semester. The class showed unusual interest in the problem and progressed extremely well. The students attended the evening lectures, which were very beneficial to them. Problems common to all the students were discussed frequently in class, and each benefited from the others' mistakes. The MAD language proved to be ideal, due to its simplicity and freedom from unnatural statements and conventions. By the semester's end, all five students had working programs, two obtaining complete results, the others having only partial success because of convergence problems. All were hindered by the relatively long processing time. Complete success could have been achieved if the processing time had been less, or if the instructor had pushed the class harder early in the semester. The students' knowledge of the computer, programming, use of trial and error methods, and interest in progression to other problems exceeded the instructor's expectations, although there were more difficulties in convergence than anticipated. The solution of a problem of this complexity also aided materially in their overall understanding of the fundamental principles involved. B44

Example Problem No. 68 DETERMINATION OF THE COMPOSITION OF THE PRODUCTS OF COMBUSTION by R. M. Shastri School of Mechanical Engineering Purdue University Course: Thermodynamics or Combustion Credit Hours: 3 Level: Senior Statement of the Problem Calculate the composition of the products of combustion of a stoichiometric ethylene and oxygen flame at 2700~C. The following products are believed to exist: C02, CO, H20, 02, H2, OH, 0 and H. It is further assumed that the total pressure is one atmosphere and that chemical equilibrium exists in the burnt gases. All gases are considered to be ideal gases. Solution The total pressure P is given by, In ger The fi P = PCO2 + PCO + PH20 + + + POH + PO + PH 2 2 2 2 ieral we know, NC = PCO + PCO 2 N = 2pCO + 2P + PH20+ 2 + POH + PO NH = PH20 + 2PH + POH +PH Lve equilibrium constants are given by, K1 = CO\/2 = 0.297 PC02 PH2 \Po K2 = = 0.0435 PH20 POH/PH2 K3 = H2 = 0.0465 PH20 (1) (2) (3) (4) (5) (6) (7) PH K4 = H = 0.154 PH K5 = = 0.110 (8) (9) B45

Determination of the Composition of the Products of Combustion From the above equations the eight unknown partial pressures of Equation 1 are calculated. We also know that, NC/NH = 0.5 and NO/NH = 1.5 The method of Damkohler and Edse (1)* is used to solve the system of equations. The values of equilibrium constants are obtained from Gaydon and Wolfhard (2). Flow Diagram Nomenclature /~S ^\~~TART ~PC02CO = Pco2/PCO PH20 = PH 0 2 PC02CO, PH20 NCNH = NC/NH NCNH, NONHA. -f*~ ^~~~.NONHA = NO/NH (Actual) PC02CO, PH20 NCNH, NONHA P02 = p0 P02 = (.297*PC02CO).P.2 PH2 = p i -w~~~~~~~~H2 |PH2 =.0435*PH20/SQRT4PO)I POH = POH POH =.0465*PH20/SQRT.(PH2)| PH = PH PH =.154*SQRT.(PH2)| PO = P PO =.lOSQRT.(PO NH = NH |NH = 2*PH20+2*PH2+PH+POH NC = NC NC = NH*(NCNH) PCO = PCO PCO NC/(l+PC02CO) PC02 = PC PC02 = PCO*PC02CO NO = N0 NO = 2*PC02+PCO+PH20+2*P02 NONH = NO/NH (Cal.) +POH+PO NONH = NO/NH P = P02+PH2+POH+PO+PCO+PC02 +PH+PH20 PRINT FORMAT ______ -___c=~ P02, PH2, POH, P0, START PRIIN FORMAT PH, NH, NC, PCO, PCO2, TABLE NO, NONH, P L —,/~ ~ t " ~~~o~~o * The numbers in parentheses refer to the references at the end.

Example Problem No. 68 MAD Program and Data R M SHASTRI Du054N..___ 001 00... 6.___025__. R M SHASTRI DU54N 001 006 025 * COMPILE MAD, EXECUTE, PRINT OBJECT, DUMP RDETERMINATION OF THE COMPOSITIONOF THE PRODUCTS OF COMBUSTION ROF STOCHIOMETRIC ETHYLENE AND OXYGEN MIXTURE, WHEN CHEMICAL REQUILIBRIUM EXISTS AT HIGH TEMPERATURES S___START READ__QRMAT DATA, PCO2COPH20,NCNHNONHA PUNCH FORMAT DDATA,PC02CO,PH20,NCNH,NONHA PRINT FORMAT DDATA, PC02CO,PH20,NCNH,NONHA P02=(.297*PC02CO).P.2 PH2=.0435*PH20/SQRT.(P02) POH=.0465*PH20/SQRT.(PH2) ---------- -— PH=. —---------- ---— 54 —--- ------— *SQRT(PH2) —------ PO=.110*SQRT.(P02) NH=2*PH20+2*PH2+PH+POH NC=NH*NCNH PCO=Nr/(1+PC02CO)....... - C-'() 2-=~-c —-------—' —--- -c' c 20) —--------- PC02=PCO*(PCO2CO) NO=2*PC02+PCO+PH20+2*PO2+ POHP+PO NONH=NO/NH P=P02+PH2+POH+PO+PCO+PC02+PH+PH20 PUNCH FORMAT TABLE PRINT FORMAT TABLE PUNCH FORMAT CALCP02,PHH 2, POHP POPHNH, NCPCO)PC02, NONONHP PRINT FORMAT CALC,PO2,PH2,POH,PO,PH,NH,NCPCO,PCO2,NONONHP _-. _-.-"`~ ~~ ~ ~'~ -, -.,... _I -`- - I ~-~'- ~ - ~-_- - -- ~ I^ -- ~ ~- -..^ ^^ ^^ ^ ^- I I -~- - ---- - - - - - - - - - - - - - - - - - - - - - TRANSFER TO START VECTORVALUESDATA=$4F10 3*$ VECTORVALUESDDATA=$1H1 S55,6HPC02CO 56,4HPH20 S5,4HNCNH S4,5HN 1ONHA/1H S5,F7.4,S3,F7.4,S2_,F7.4S2, F7.4*$.,..-..-...-VECTORVALUESTABLE':H.. S1'S3HP02 S4,3HPH2 3,3HPOH S3,2HPO S4,2 1HPH S4,2HNH S4,2HNC S4,3HPCO S2,4HPC02 S3,2HNO S3,4HNONH S3,1 — _-~..~.~._...-.~^- _.-~1 —~ —" 1_^p-' ^ 1. —. —. —. -. — c — ^- ---- ------ - - --- - - 1HP*$ VECTORVALUESCALC=$1H 12F6.4*$ END OF PROGRAM'DA, T A ------------- 1.5 0.35 0.5 1.5 0.75 0.35 0.5 1.5.. 0.75 0.25 0.5 1. __ 1.-3 0.307 0.5 1.5 Computer Output (Rearranged) Results from-the first three data points: PC02CO PH20 NCNH NONHA. -Y500-0-' 0..3500 0'5000 1.5000.............0 03'o...500 o, i. oo1.5ooo 0.7500 0.2500 0.5000 1.5000 -P02 PH2 POH P0 PH - NH NC7 PCO PC02 NO NONH P 0.198 o.03420.0P80.049'CO:02850.884S 0.442 0.177a0.26551.r,9191.799 1.19 6 O.O 4~i9T.0684 0060.024.0400.939 0.469 268 0 20 1.2~06 1.285C1.0646 0.0490*048t 0.05260.024 0.034U0.o84 0.342.019560.146 0.9151.337 30.8017; Results from the fourth data point: PC02CO PH20.NCNH NONHA 1.0300 0.3070 0.50o0 1.50j0 P0...:....."~Pr-:.....D —'-~ —....p P..' — PO C2 "N? D ANONH P 0.0936n.04370.06830.03370.032220.6018.48009O.1395o02u341l.?0051.49 720.Q9793 B47

Determination of the Composition of the Products of Combustion Discussion of Results Initial trial assumptions are made for PH 0 and PC02/PCO Now all the remaining partial pressures are calculated. A check is made to see whether the total pressure and the ratio NOINH will have the correct values. If they do not check,new assumptions are made and the procedure is repeated. After 3 trial calculations a plot is made of PC /PCO vs. PH20 as shown in Figure 1. Using this figure,better approximations are made. 1.5 1.3 1.2 PC02 1.1 PCO 1.0 0.9 0.8 0.7 1 I A I II.... m X I _ _ 3 2 W I I L! 1 t L I I l W w n 0.2 0.3 0.4 PH20 Figure 1. Plot of Pc02/PCO vs. PH20 B48

Example Problem No. 68 Calculations showed a total pressure of more than one atmosphere at Point 1 and less than 1 atmosphere at Point 3. A linear variation of pressure is assumed. The Point'm' is determined where the total pressure is approximately one atmosphere. Similarly, a point'n' is established between Points 2 and 3. The whole procedure is repeated for N /NH = 1.5 and points'x' and'y' are established. The values of pC o/Pco and pH20 are given by the point 4the point of intersection of the lines m-n and a-b will be the values for the 4th assumption. The procedure is now repeated using Points 1, 2, 4 to get Point 5. The procedure is repeated until the calculations give the required values for P and NO/NH. This, then, will also give the values of the correct composition. Critique This problem has been taken from Gaydon and Wolfhard (2). The method could be applied when there are any number of products of combustion. The equilibrium constants for the products at the desired temperature must be known. Here the total pressure has been assumed a constant before and after the reaction. With slight changes the above method could also be used for constant volume processes. References 1. Damkohler, G. and Edse, R., Z. Elektrochem 49, 178 (1943). 2. Gaydon, A. G. and Wolfhard, H. G. "Flames," Chapman and Hall Ltd., London (1953). B49

Example Problem No. 69 ADIABATIC FLAME TEMPERATURE FOR COMBUSTION OF AIR AND METHANE by Richard E. Sonntag Department of Mechanical Engineering The University of Michigan Course: Advanced Thermodynamics Credit hours: 3 Level: Graduate Statement of Problem It is desired to determine the effects of pressure and percent theoretical air on the adiabatic flame temperature and product composition for the combustion of methane with air over the following range: 1 to 50 atmospheres pressure 100 to 150 percent theoretical air Assume all gases behave ideally, and that the reactants enter at 77~F. Let 100 X a = percent theoretical air where a > 1.0 Then, for the combustion at 77~F, CH4 + (2a) 02 + (7.52a) N2 - C0 + 2H20 + 2 (a-l) 0 + (7.52a) N2 As the temperature of the products is increased to the flame temperature, consider that only two dissociation reactions occur: A: C2 = CO + 1 2 (1) B: H20 = H2 + 02 (2) In the solution, use the following specific heat equation and data: C = a +. 10-T + 10-6T2 Btu p0 Lb-Mole-OR where T = ~R a _ _ _ CO2 6.214 5.776 -1.094 CO 6.420 0.925 -0.060 02 6.148 1.722 -0.285 H20 7.256 1.277 0.087 H2 6.947 -0.111 0.148 N2 6.524 0.694 -0.001 B50

The resulting specific heat equations will not prove to be particularly accurate at temperatures above 3500~R, but are sufficient for this illustration. At 77~F, CH4 C02 H20 CO -o h -32,200 -169,293 -104,071 -47,548 g~ (Btu/lb-Mole) -169,667 -98,344 -59,054 Solution Let y and z be defined as the moles involved in reactions A and B respectively, so that the change in each component is 1 1 A: C02 = CO + - 02 - y + 2 B: H H2 = H + 2 O -z + + z 2 2 2 2 2 (1) (2) Since the initial moles equation, the number of nco O2 of each component are given in the product side of the combustion moles at equilibrium (flame temperature) is = 1 - y nH O = 2- z 2 n02 02 = 2(a-l) + 1 y + z y2 nCO nH nN2 n N2 = z = 7.52 a total=+ 9.52a+y+z ntota1 = 1 + 9.52a + 7 y + 7 z Note that although the nitrogen is inert insofar as the reactions are concerned, it must be considered in the total number of moles present at equilibrium. The two equilibrium equations are now written as 1 2 XCO x02 K = KA(T) =X cA02 1 P (1 -(y) 1 2 2a-2 + zy + - z \1+9.52a + l y + l z / 1 P (3) B51

Adiabatic Flame Temperature for Combustion of Air and Methane 1 1 2 2 KB = KB(T) P = ) (l+ a 2 (4) 2cop (z ( 1+9.52a- + y z 1 Z Xc02 It will, therefore, be necessary to express the equilibrium constants as functions of temperatur It is known that d(ln K)p = dT (5) RT However, the standard-state enthalpy change AH~ is a function of temperature, and can be written 0 0 -O AHT = AHT + ( h o prod-react o T = AHT + P-R (p ) dT (6) 0 Substituting equation (6) into (5), and integrating between any two temperatures T1 and T2, n 4] AH TT2 T in = To 1 1 +1 Z (7) l[T1] R tT1 T2 + K P-R Cpo T2 () T1 LT0 Reaction A At 557 R, using the enthalpy and free energy of formation data, 0 o ~ Btu Btu AHA = 121,745 = Mole H1 Me 2 1 Therefore, AG0 n KA RT =-103.6 1 Also, z (VC ) = 3.28 -.00399 T +.892 10 6T2 P-R o Substituting these values into equation (7) and integrating, In K = 1.65 In T -.001 T +.075 10-6T2 60759 _ 0.41 (8) A T- T Reaction B At 557~R, AHB = 104,071 BtuG = 98,44 Btu B1 Mole H20 B1 Mole Therefore, AG ln KB1 RT - 92.2 B52

Example Problem No. 69 Also Z (-C ) = 2.765 -.000527 T -.082 10-6T2 P-R po As before, In KB = 1.393 In T -.000133T -.007 106 T2 51673 - 4.6 (9) T The third equation to be set up is the First Law, which requires that HR =Hp (10) where HR = H = 32,200 Btu (11) 4CH + n _EH h 2+ nC 00'h Hp nC02 hC02 +nH20 + n0 T + J [ nC0 Cp + nH 2 Cp + n C + nC C H57R 2 C02 H20 2 0 2 + nH CD + 2 CN ] dT (12) i2 H 2 2N 2 Substituting numerical values into equation (12), H = - 582,646 - 34,163a + 120,511y + 102,665 z + T(8.43 + 61.356 a + 3.28 y + 2.765 z) + ] (24.44 + 43.31 a - 19.96 y - 2,635 z) - os1 (0.117 + 0.1923 a - 0.297 y + 0.028 z) (13) Therefore, Equation (10) becomes T(8.43 + 61.356 a + 3.28 y + 2.765 z) + (24.44 + 4 1 a - 19.96 y - 2.635 z) -[ o] (0.117 +0.1923 a - 0.297 y + 0.028 z ) + 120,511 y + 102,665 z - 54,163 a - 350,446 = 0 (14) Thus, for given values of a and P, there are three equations (3, 4, 14) in three unknowns (T, y, z), since KA, KB are given in terms of T by equations (8), (9). For an assumed T, y and z are determined from (3) and (4), this necessitating the simultaneous solution of two non-linear equations. This could be accomplished by elimination of one of the variables in terms of the other, but a more general method will be used here as an illustration, applying as well in the case of more complex reactions or systems of reactions. B53

Adiabatic Flame Temperature for Combustion of Air and Methane Specifically, initial values of y and z are assumed, and then incremented by linearized correction equations obtained from the partial derivatives of the equations involving y and z (3 and 4). The process is repeated until the correction is sufficiently small. Once y and z have been so determined for the assumed T, all three are substituted into equation (14) to check the assumption of T. The value of T is then incremented and the procedure repeated until the error in equation (14) is less than some predetermined amount. For any given T, there are two equations FA (y,z) = 0, FB (Y,z) = 0 Assume initial values for y, z. Evaluating FA (y,z), FB (y,z) for this y, z, it is found that, in general, both are unequal to zero. Therefore y and z must be corrected by some amounts Ay andAz such that the required conditions are met. That is, FA + AFA = 0 FB +AFB = 0 where A _ 6R n y [fn zy AF FB [ FB 1] n a^ L J nz nz Note that a logarithmic form has been used here for the incrementation of y and z, this being done to prevent y or z from becoming negative during the course of the solution. Otherwise it would be very possible to obtain a physically non-real solution of the equations, with one or more of the equilibrium mole fractions having negative values. The solution desired is, of course, always the one with all positive masses. Solving the last four equations, bFA -FA 6-n z FB -FB B AAn y = n z(15) aFA aFA ntn y bnn z bFB FB bin Y b6n z B54

Example Problem No. 69 ALn z = FA -Xn y FB 3An y ^ FA 6bn y FB 3An y -FA -FB (16) FA aLn z FB 6Ln z Therefore, the new values of y and z nn y = nn y + A in to be used y, nn z in the succeeding trial are = Rnz + A n z (17) or, y = y -(e) ~n y, z = z (e) n z and the iteration process is repeated until AL n y and A tn z are both sufficiently small. For the problem at hand, the quantities to be used in evaluating equations (15) and (16) are found by rewriting equations (3) and (4) in the form FA L [L l- () - 2 A n K (18) FB n [[-z] *P - -in KB (19) where, for convenience, 1 1 (Dl) = nttal = 1 + 9.52a + y + z 1 1 (D2) = n = 2 a - 2 + y + An KA and in KB are given by equations (8) and (9) respectively. (20) (21) Thern since a FA Qln y idy I A FA i dy FA 1 J Ld n = L y J etc. 6FA 2 &6 = y ~ + (D3) ~ Y bA = ^n = (Dy) * y 6ny - (D3). y (22) (23) (24) b FB 6%n z - + (D3) * z 2-z (25) where 1.5 + 3.76 a (D3) - (D1) ~ (D2) (26) B55

Adiabatic Flame Temperature for Combustion of Air and Methane Thus, for an initial assumption of y, z, equations (18), (19), (22),..., (25) are evaluated and substituted into equations (15) and (16), thereby giving the corrections to be applied to y, z (according to equation (17)) for the next trial. It should be stated here that convergence to a solution using this method is certainly a real problem, since the successive incrementations may very well grow and cause divergence from any solution if the initial assumptions of y and z are not in the general region of the true values. There are special techniques for handling this particular problem, namely, means for predicting whether or not the assumed values are inside the area of convergence. If not, then different assumptions are generated. Finally, it should be obvious that all three equations (3,4, 14) could be solved simultaneously using the same method that is used here to solve (3) and (4) for any assumed T. This would not follow as closely the paths taken to solve such a system by hand, but is perhaps somewhat more rigorous mathematically. Such a method naturally will pose a more severe test of the convergence, requiring that the initial assumption of T also be a reasonably good one (which is not the case in our method of solution). The three equations lead to a set of three linear correction equations, the set then being solved for the corrections (Ay, Az, AT, or logarithmic counterparts) by matrix inversion. These same remarks apply also to the general case of n simultaneous equilibrium reactions, resulting in a set of n+l linear correction equations (including the First Law and T), which can be solved by the same method. Flow Diagram B56

Example Problem No. 69 Flow Diagram (continued) IT~2 T (^)a~~~~~ (6a~) r~~.- ] T=T+ AT EVALUATE PRINT xcCO, xH O.0P., a., TA,EI ( e XC02' XH20' X02' ~ J, x' MAD Program MAD Program $ COMPILE MAD, EXECUTE ALPHA READ FORMAT DATA,P,A,T, DELTAT VECTOR VALUES DATA=$4F10.2*$ J=O E=-1.0 Y=O. 1 Z=0.1 BETA J=J+l LNKA=1.65*ELOG.(T)-.001*T+.075*(T/1000.).P.2-60759./T-.41 LNKB=1.393*ELOG.(T)-.000133*T-.007*(T/1000.).P.2-51673./T-4.61 I=0 INTEGER I,J GAMMA I=I+1 D1=1.0+9.59*A+0.5*(Y+Z) D2=2.0*(A-1.0)+0.5*(Y+Z) D3=(1.5+3.76*A)/(D1*D2) FA=ELOG.(Y*Y*D2*P/(Dl*(1. -Y.P.2))-2. 0*LNKA FB=ELOG. (*Z*D2*P/(D*(2.0-Z).P.2))-2. 0*LNKB DFADZ=D3*Z DFBDY=D3*Y DFADY=2.0/( 1.0-Y)+DFBDY DFBDZ=4.0/(2.0-Z) +DFADZ DENOM=DFADY*DFBDZ -DFADZ *DFBDY DY= FB*DFADZ-FA*DFBDZ) /DENOM DZ=(FA*DFBDY-FB*DFADY)/DENOM Y=Y*EXP.(DY) Z=Z*EXP.(DZ) WHENEVER I.E.50, TRANSFER TO DELTA WHENEVER.ABS.(DY/Y).G.O.0001, TRANSFER TO GAMMA WHENEVER.ABS.(DZ/Z).G.0.0001, TRANSFER TO GAMMA EPREV=E E=(8.43+61.356*A+3.28*Y+2.765Z) *T+( 24.44+43.31*A-19.96*Y-2.635*) *(T/100. 1 ). P. 2-(.117+.192*A-. 297*Y+.028*Z)*(T/100.).P.3+120511. *Y+102665*Z2 34163.*A-350446. WHENEVER.ABS.(E).LE.100., TRANSFER TO DELTA WHENEVER (E*EPREV).L.O,DELTAT=O. 5*DELTAT WHENEVER J.E.100, TRANSFER TO DELTA WHENEVER E.G.O T=T-DELTAT OTHERWISE T=T+DELTAT END OF CONDITIONAL B57

Adiabatic Flame Temperature for Combustion of Air and Methane MAD Program (continued) TRANSFER TO BETA DELTA XC02=(1.0-Y) /D1 XH20= (2.0-Z /D1 X02=D2/D1 XCO=Y/D1 XH2=Z/D1 XN2=7.52*A/AD1 PRINT FORMAT RESULT,P,A,T,DELTAT,E,I,J,XC02 XH20,X02,XCO,XH2,XN2 VECTOR VALUES RESULT=$1HO,5F10.2,2I4,6F10.7*$ TRANSFER TO ALPHA END OF PROGRAM Computer Output The following table has been obtained from the computer output. PATM a T XC2 XH2O 0 O X X2 Xc H0 0 2 X coX H2 N2 1 1.0 4008.084702.185716.006456.009741.003171.710214 5 1.0 4056.088028.187242.004333.006617.002048.711732 10 1.0 4072.089196.187751.00599.05519.001679.712256 20 1.0 4086.090203.188178.002971.004572.001371.712705 30 1.0 4094.090714.188393.002654.004090.0.0121732 40 1.0 4098.091062.188536.002439.003763.001115.713085 50 1.0 4102.091304.188636.002289.003535.001043.713193 1 1.1 3866.083812.172937.019501.003174.001033.719543 5 1.1 3890.085438.173630.018493.001637.000520.720282 10 1.1 3897.085894.173819.018213.001206.000380.720488 20 1.1 3902.086240.173961.018001.000879.000275.720644 30 1.1 3904.086401.174027.017902.000727.000227.720716 40 1.1 3906.086496.174066.017844.000636.000199.720759 50 1.1 3906.086566.174095.017801.000570.000178.720790 1 1.2 3690.079338.160483.032907.001092.000378.725802 5 1.2 3698.079948.160748.032529.000513.000176.726086 10 1.2 3701.080101.160813.032435.000369.000126.726156 20 1.2 3702.080212.160861.032366.000263.000090.726208 30 1.2 3704.080261.160881.032336.000217.000074.726231 40 1.2 3704.080292.160894.032317.000188.0000oooo64.726245 50 1.2 3704.080312.160904.032305.000168.000057.726254 1 1.3 3516.074354.149337.045110.000387.000145.730667 5 1.3 3520.074574.149437.044973.000177.0000oooo66.730773 10 1.3 3521.074628.149462.044939.000126.000047.730798 20 1.3 3522.074667.149479.044915.000090.000033.730816 30 1.3 35522.074684.149487.044905.000073.000027.730824 40 1.3 3522.074694.149491.044898.000064.000024.730829 50 1.3 3522.074701.149494.044894.000057.000021.730833 1 1.4 3357.069643.139514.05590.000144.000058.734711 5 1.4 3358.069725.139554.055878.0000oooo65.000026.734752 10 1.4 3359.069745.139564.055865.000046.000019.734761 20 1.4 3359.069758.139570.055857.000033.000014.734768 30 1.4 3359.069765.139573.055853.000027.000011.734771 40 1.4 3359.069768.139575.055850.000024.000010.734773 50 1.4 3359.069771.139576.055849.000021.000009.734774 1 1.5 3212.065387.130861.065482.000055.000024.738191 5 1.5 3213.065419.130877.065462.000025.000011.738206 10 1.5 3213.065427.130881.065457.000017.000008.738210 20 1.5 3213.065432.130883.065454.000012.000006.738213 30 1.5 3213.065434.130885.065452.000010.000005.738214 40 1.5 3213.065436.130885.065451.000009.000004.738215 50 1.5 3213.065437.130886.065450.000008.000003.738216 B58

Example Problem No. 69 Discussion of Results The reason for limiting the range of theoretical air to 150 per cent in this example is that for greater amounts the flame temperature is decreased sufficiently that there is no appreciable dissociation. In this case, the solution is reduced to a relatively simple calculation. The results of this program are shown in graphical form in the figure below. As a point of interest, the time required to completely solve the equations for one set of P and a (i.e., determine the flame temperature) was less than one second. This would have been reduced somewhat if a good initial assumption for T had been made in each case. With the values actually used, the temperature had to be incremented an average of twelve times per solution to reduce the error in equation (14) to the equivalent of less than one degree Rankine. 4100 4000 3900 0o 0 I) aJ 4-) (1) a 0 ca r1 0 C* *H rcii 3800 3700 a=l. 0 1.3 1.4.________...... 1.5 3600 3500 3400 3300 3200 1 5 10 Pressure - Atmospheres 20 30 40 50 B59

Example Problem No. 70 TRANSIENT TEMPERATURE CALCULATION FOR A JACKETED MIXING KETTLE by E. T. Kirkpatrick Department of Mechanical Engineering The University of Toledo Course: Heat Transfer Credit Hours: 3 Level: Graduate Statement of the Problem In developing a certain chemical mixing process, the temperature of the mixture is to be maintained at fixed values which may be varied over a moderate range near 150 degrees F. A scheme for doing this is to locate a thermocouple in the mixture at some representative spot and use its output voltage as the input to an automatic controller, as indicated in Figure 1. The controller would regulate the heat to or from the mixture by controlling the flow of hot and cold water to the jacket spaces in the steel vessel. Either the hot or cold water valve will be open depending on whether a temperature increase or decrease is desired. Zero dead time is to be assumed. In such a system, how may the temperature in the mixture be expected to vary with time after the controller is suddenly reset to call for a new temperature? The nomenclature, physical constants and initial values are shown in Table 1. The jacket is assumed to be perfectly insulated. Tout F -.. Drain W2,C2,T2 W3 C3.T3 W1,C1,T1 ThermocouW.. ^....-..-.J T/ lixerO T- J ControI e Controller W4 C4,T4! " 4:,......,.,..,., I Fig. 1. Mixing Kettle with Insulated Jacket B60

Nomenclature Subscripts 1 and p 2 3 4 and w in out c h mixture vessel jacket circulating water inlet conditions outlet conditions cold water hot water Constants and W, weight, lbs C, specific hee A, area, ft HP HW TH To T 2 T 3 TCONTL H TIMAX Variables 1 2 3 4 7700 3500 1500 84000 (lbs/hr) at, BTU/lb.deg F 0.95 0.1 0.1 1.0 400 400 400 75 BTU/hr-ft2 deg F. convective heat transfer coef. for mixture 500 BTU/hr-ft2 deg F, convective heat transfer coef. for water 200 deg F, hot water temperature 50 deg F, cold water temperature 100 deg F, initial value, product temperature 100 deg F, initial value, vessel temperature 100 deg F, initial value, jacket temperature 150 deg F, controller temperature setting 0.001 hours, time increment for Runge-Kutta Integration Procedure 0.7 hours, maximum time Solution A. Derivation of Equations Apply the principle of conservation of energy to the mixture, vessel, water and the jacket. Assumptions (a) Walls sufficiently thin so that the temperatures are constant within. (b) Sufficient mixing to cause the mixture temperature to be uniform. (c) Jacket water temperature is the average of the inlet and outlet temperatures. (d) Mixing energy negligibly small. B61

Transient Temperature Calculation for a Jacketed Mixing Kettle Apply the principle of conservation of energy. Increase in internal energy of mixture = heat flow from kettle to mixture. dT1 CAW = hA2 (T-T1) () dt Increase in internal energy of kettle = heat flow from water to kettle - heat flow from kettle to product. dT2 C2W2 - h A2 (T-T) - hA2 (T2-T1) (2) dt Increase in internal energy of jacket = heat flow from water to jacket, dT C3W3 - = hA3 (Tw-T 3) dt Heat loss by water = heat flow to kettle and jacket. C4W4 (Tin-Tout) = hw A2 (Tw-T2) + hwA3 (Tw-T3) (4) Average temperature exists within water duct. Tw = (Tin + Tout)/2 (5) The five unknown temperatures are T1, T2, T3, Tw and Tout. Tout can be eliminated by combining equations (4) and (5), and constants introduced to give Tw = [Tin + K5T2 + K6T3] /K7 (6) F1 = Kl(T2-T1) (7) F2 = K2(Tw-T2) - K3(T2-T) (8) F3 K4(Tw-T3) (9) where h A hwA2 hpA2 hwA3 K =', K2 = K, K = w C11 C2W2 C2W2 C3W3 hwA2 hwA3 K5 = 6, K7 = 1 + K5 + K6 2W4C4 2W4 C4 B. Computer Program The input data is as follows: W1, W2, W3, W4, C1, C2, C3, C4, A1, A2, A3, hp, hw, initial time (0), time increment H, TIMAX, T1, T2, T3, Th, Tc, Tcontrol The numerical values are shown both in Table 1, and in the tabulated results. The constants Ki are first calculated and the plotting subroutine given the information pertaining to the form of the plotting grid (PLOT1.)i and the maximum and minimum values of the ordinate and abscissa (PLOT2.). The Runge-Kutta subroutine is then initialized with the 1) Plot Subroutine by Brice Carnahan and Larry Evans. B62

Example Problem No. 70 number of equations, names of first unknown, first derivative, a dummy variable, initial time 2 and the time increment (SETRKD.). By trial, it was found that an increment of H = 0.01 was unstable and H = 0.001 was used. However, a print-out at intervals of time equal to 0.01 is satisfactory, so that a counter N is used and the Runge-Kutta subroutine executed (RKDEQ.). On each pass, the mixture temperature T1 is checked against the control temperature setting and switched if necessary. Because of the possibility of instability, the absolute value of the jacket temperature T3 is checked against the control temperature. It was found by trial that T3 was the most sensitive to instability. If instability is sensed, the solution of the equations is stopped and the final stages of the plotting routine entered (PLOT4.). On each tenth pass all temperatures and the time are printed out, and the same information is given to the plotting routine (PLOT3.) for future plotting print-out. The final instructions of the program are to print remarks on the plot and to return to the read-in instruction. If more sets of data are available, the program will continue; if not, the program ends. Flow Diagram:', (T, F, t, 2)(z GO V^~RC~.&) RS(\) ^WDo' o, F, K I-T,-T7 GO Y\ bCz ()) I (CL, 2) Runge-Kutta Subroutine by Bruce Arden. B63

Transient Temperature Calculation for a Jacketed Mixing Kettle MAD Program and Data TED KIRKPATRICK D002N 8 002 005 150 T1NKOOOO SCOMPILE MAD, EXECUTE R R***TRANSIENT TEMPERATURE IN A MIXING VESSEL RI*****ET. KIRKPATRICK RUSE SUBROUTINES RKDEQ AND PLOT R DIMENSION T(3) F(3),0(3) DUMMY(732) INTEGER NPLOTS K INTEGER N R READ FORMAT DATAlWlIW2,W3,W4,ClC2,C3,C4,AliA2A3HPPHW.TIME 1,HKTIMAXT(1) tT(2)~ T(3),THTCTCONTL VECTOR VALUES DATA1=$4F10.3/4F10.3/3F10.3/2F10O3/3F10.3/6F10. 13*$ PRINT FORMAT HEAD1i$ W1 W2 W3 W4 $ VECTOR VALUES HEADiS1$H2~7C6*S PRINT FORMAT ECHOlW12W2#W3vW4 VECTOR VALUES ECHOl$1H,4F10O2*S PRINT FORMAT HEAD2o$ Ci C2 C3 C4 S VECTOR VALUES HEAD2ZSlHO,7C6*$ PRINT FORMAT ECH02,C1C29C3#C4 VECTOR VALUES ECHO2=$1H,4F10.2*$ PRINT FORMAT HEAD3#$ Al A2 A3 S VECTOR VALUES HEAD3$1HO*5C6*S PRINT FORMAT ECH03.A1.A29A3 VECTOR VALUES ECH03=$lH,3F10*2*$ PRINT FORMAT HEAD4#$ HP HW TIME INCRH 1 TIMAX $ VECTOR VALUES HEAD4-$1H09C6*$ PRINT FORMAT ECH049HPHWtTIMEtHTIMAX VECTOR VALUES ECH04$SlH,5F10.2*$ PRINT FORMAT HEAD5S$ T1 T2 T3 TH 1 TC TCONTL $ VECTOR VALUES HEAD5=$1HOlOC6*$ PRINT FORMAT ECH05T(i1).T(2),T(3),THTCTCONTL VECTOR VALUES ECH05OSlH,6F10*2*$ K1C(HP*A2)/(Cl*W1) K2-(HW*A2)/(C2*W2) K3 (HP*A2)/(C2*W2) K4=(HW*A3)/(C3*W3) K5 (HW*A2)/( 2.*W4*C4) K6&(HW*A3)/( 2*W4*C4) K7-1 K5+K6 PRINT FORMAT KAYS, SCONSTANTS K1 TO K75,K1.K2,K3,K4,K5,K6, 1K7 VECTOR VALUES KAYS S$1HO0 3C6/lHO,7F10.4*$ VECTOR VALUES MARGIN-S TEMPERATURESS VECTOR VALUES N1l~, *.10.1 START EXECUTE PLOTl (N6,10,7,10) EXECUTE PLOT2.(DU1MY,0e7t0.,200*,50.) PRINT FORMAT HEAD VECTOR VALUES HEAD=$1HlS14958HTABULATED SOLUTION OF SIMULTAN lEOUS DIFFERENTIAL EQUATIONS /lHOS14947H.GIVING TRANSIENT TEMP 2ERATURES IN A MIXING TANK /1HOS6,4HTIMES52HT1,S8,2HT2,S8,2 3HT3 S8 2HTWS7 3HT IN S6.4HTOUT*$ R TIN-TH HwH/ 10 EXECUTE SETRKDe (31T(1) F(1) Q( 1) TIME~H) GO1 N1 GO K=RKDEQ (O) TRANSFER TO S(K) S(1) TW=(TIN+(K5*T(2))+(K6*T(3)))/K7 F(1) Kl1*(T(2)-T(1)) F(2) "K2*(TW-T(2)) K3*(T(2) T(l)) F(3) =K4*(TW-T(3)) TRANSFER TO GO B64

Example Problem No. 70 MAD Program and Data, Continued 54 2) TWATR =TW TOUT=2*TWATR-T I N WHENEVER N-GEl10O TRANSFER TO CONT TRANSFER TO CONT2 CONT PRINT FORMAT RESULT TIME T(1) T:42) T 3) TWATR TIN TOUT VECTOR VALUES RESULT $1SH * 1F10.3, 6F10.2*S EXECUTE PLOT3 ($E$ TIME *TOUT 1 ) EXECUTE PLOT3. ( $*S T IME TT ( 1 ) 1 ) EXECUTE PLOT3 ( SVS*TIME T(2) 1 ) EXECUTE PLOT 3* ( S TIME*T(3) 1) EXECUTE PLOT3 ($WS TIME TWATR 1) N* CONT2 WHENEVER T(1).GE.TCONTL.T INTTC WHENEVER T( 1) L.TCONTLT IN=TH WHENEVER TIME G*TIMAX*TRANSFER TO OUT Ns N 1 TRANSFER TO GO OUT PRINT FORMAT TITLE VECTOR VALUES TITLE=1H1,S4 *40HTRANSIENT TEMPERATURES IN A M 1IXING TANK *S EXECUTE PLOT4 ( 32 *MARGIN) R PRINT FORMAT BOTTOM VECTOR VALUES BOTTOM'$1HOS40330HTHE INDEPENDENT VARIABLE TIM IE //1H *S2755HPLOTTING CHARACTERS. Tl(*).T2(V).T3(J),T:WATR(W 2),TOUT(E) $ END OF PROGRAM S DATA 7700 350. 1500 84000..95.1 *1 1. 400. 400. 400. 75. 500. 0.0 0.01.7 100.0 100. 100.0 200*0 50O0 150. Computer Output Wl W? W3 W4 7700.00 35o0.00 1500.00 84000.00 C1 C? C3 C4 0.95 0.10 0.10 1.00 Al A? A3 400.00 4nr.00 400.00 HP HW TIME INCR=H TIMAX 75.00 5n0.00 0.00 0.01 0.70 Tl T? T3 TH TC TCONTL 100.00 ln2.00 100.00 20000 5150.00 B65

Transient Temperature Calculation for a Jacketed Mixing Kettle Computer Output (Tabular) TABULATED SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS GIVING TRANSIENT TEMPERATURES IN A MIXING TANK TIME T1 T2 T3 TW TIN TOUT 0.010 ln1.99 171.10 183.03 183.86 200.00 167.71 0.020 1n4.90 176.01 186.85 186.92 200.00 173.84 0.030 17.78 176.95 187.42 187.46 200.00 174.91 0.040 110.58 177.66 187.82 187.84 200.00 175.69 0.050 113.29 178.34 188.18 188.21 200.00 176.42 0.060 115.91 178.99 188.54 188.57 200.00 177.14 0.070 118.46 179.63 188.89 188.92 200.00 177.83 0.080 120.93 180.25 189.23 189.25 200.00 178.50 0.090 123.33 180.85 189.55 189.58 200.00 179.15 0.100 125.65 181.43 189.87 189.89 200.00 179.79 0.110 127.90 181.99 190.18 190.20 200.00 180.40 0.120 130.09 182.53 190.47 190.50 200.00 180.99 0.130 132.21 183.06 190.76 190.78 200.00 181.57 0.140 134.26 183.58 191.04 191.06 200.00 182.13 0.150 1i6.25 184.07 191.31 191.33 200.00 182.67 0.160 138.18 184.56 191.58 191.60 200.00 183.19 0.170 140.06 185.03 191.83 191.85 200.00 183.70 0.180 141.87 185.48 192.08 192.10 200.00 184.20 0.190 143.63 185.92 192.32 192.34 200.00 184.68 0.200 145.34 186.35 192.55 192.57 200.00 185.14 0.210 147.00 186.76 192.78 192.79 200.00 185.59 0.220 148.60 187.16 193.00 193.01 200.00 186.03 0.230 150.11 162.93 149.36 123.66 50.00 197.31 0.240 150.01 176.57 184.18 186.20 200.00 172.41 0.250 150.10 153.07 141.25 117.35 50.00 184.70 0.260 150.07 134.62 117.50 103.11 50.00 156.21 0.270 149.98 120.01 102.37 92.88 50.00 135.76 0.280 149.89 133.39 135.85 154.94 200.00 109.87 0.290 149.85 145.72 154.19 165.16 200.00 130.32 0.300 149.85 155.80 165.46 172.46 200.00 144.91 0.310 149.89 163.65 173.00 177.78 200.00 155.57 0.320 149.96 169.64 178.30 181.72 200.00 163.44 0.330 150.05 174.16 182.15 184.65 200.00 169.30 0.340 1~0.11 152.94 141.14 117.27 50.00 184.53 0.350 150.08 134.62 117.49 103.10 50.00 156.21 0.360 149.99 120.02 102.37 92.88 50.00 135.76 0.370 149.90 133.39 135.85 154.94 200.00 109.88 0.380 149.86 145.72 154.19 165.16 200.00 130.33 0,390 149.86 155.80 165.46 172.46 200.00 144.92 0.400 149.90 163.65 173.00 177.78 200.00 155.57 0.410 149.97 169.64 178.30 181.72 200.00 163.44 0.420 150.06 174.16 182.15 184.65 200.00 169.30 0.430 150.12 152.94 141.14 117.27 50.00 184.53 0.440 150.09 134.62 117.49 103.10 50.00 156.21 0.450 150.00 120.02 102.37 92.88 50.00 135.76 0.460 149.90 133.39 135.85 154.94 200.00 109.88 0.470 149.86 145.73 154.19 165.16 200.00 130.33 0.480 149.65 139.28 148.77 161.01 200.00 122.02 0.490 149.89 134.88 137.07 155.89 200.00 111.78 0.500 149.72 124.93 128.50 149.42 200.00 98.84 0.510 149.98 122.02 104.02 94.16 50.00 138.32 0.520 149.87 108.88 92.09 85.44 50.00 120.88 0.530 150.08 137.30 119.70 104.82 50.00 159.64 0.540 150.02 120.17 102.49 92.98 50.00 135.95 0.550 150.10 156.52 144.11 119.56 50.00 189.13 0.560 150.12 134.82 117.65 103.23 50.00 156.46 0.570 150.03 178.95 186.15 187.73 200.00 175.46 0.580 149.82 172.36 180.69 183.50 200.00 167.01 0.590 149.68 160.50 170.40 175.77 200.00 151.54 0.600 149.67 140.15 149.48 161.56 200.00 123.13 0.610 149.88 110.31 93.27 86.36 50.00 122.71 0.620 150.11 137.38 119.77 104.87 50.00 159.75 0.630 150.05 179.09 186.27 187.82 200.00 175.64 0.640 149.84 172.37 180.70 183.51 200.00 167.03 0.650 149.70 160.50 170.41 175.77 200.00 151.54 0.660 149.70 140.16 149.48 161.57 200.00 123.13 0.670 149.90 110.32 93.28 86.36 50.00 122.72 0.680 150.13 137.39 119.77 104.88 50.00 159.75 0.690 150.07 179.10 186.27 187.82 200.00 175.65 0.700 149.86 172.38 180.70 183.52 200.00 167.03 B66

Example Problem No. 70 Computer Output (Plotted) TRfANSIENT TEMPERfTLiRES I P MIXING THNK 200.0 + + + ~+ ~ + ~ + -+ I I I E I I I I I I I I I I I I I I...____ _..I W WW,JwtI w I I_ I _I I I Will I I I I E I a I IW I t VVEE WE I IE I WE I J I I IiJ V''EEEF I T iit. T I T Il T Wi WI I..'W^~ EEE I I I I J I'. IJ I V.' _EE I II I) lj I I i J I 1 75.0 +-VEE - - - - - + - -* — --- - -+ - —:::: -- --- + -.- k'- }. - - -~t ------------------------- - + ---- -} —--;17115. 0 _ E ~ + ~+ —~ —-~ ~+ —_ — _-_ — _............... ~ —-: - -^ ^ ^^ - ^- - - - - -- - _ _ EI I _ I _ /J wj/. I I 4 IV. I I I I I I..I J I IE 1 I I E~~~I I I WJJ W~ IA, w W I I I..I I I i-E - E I — Id- W I I I I II I I/ 1.. I E'I EI E I I I _ I ___ I I _I'E I__ I - I I I E.WJ. E E W.ijiE E.iJ I I I - I ~I ___ ~ I I ~~~ It e I~~~~''JI EI I: I I. 1 U + ~~ —-~ —--— + ~ —---- ~+ —J+ —- + -*.t~ ** - "" *J.:****'.lJ*!l'***:, * i'.J*.:** E I I - -'I I IMiI ------- ------- P I _,I...I.. IJI 1 I I I E I I ^ I Jl II J IIJ l 1 II I. E~~~~~~~~~~~~~~~~~~~~ I IR__ _ _ _ v'E i\JIEI. I' I.,E [' I'I u I I \- - - ------ -----—, E 125.0"+-w + - ~r- -- -~ + -, —-~ ~ ~+-~- r —- r - tr-r I II I I I EI \ I I EI I l I I I I _ ___J _I I I I I I IiI I. I I I I I. Id I 1W W I 100.0 ------------- - --- I I.J I W.J I d.J — I -.J -- I I I I I I I I I I I I I I I II',J I I I I l I. I I I _ I.._ W I.J J I I I I I I I I I I I I I I I IW W _ I I I I I I i.hI I I I I I I I I I I I I I I I I I I I I I I I I I 7 * *5.0 +- * — "- -j- - -- - -+- - - - - -- - -- I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I _ I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 50.0 + —------- + -------— + ------— + - ----- + -------— + —----— + —---- 0.. 0 0. 1 0. 0.. 0. 5 O... 0.7 THE INDEPENDENT'..'R I BL.E TI lE PLOTTING CH RACTERS. Ti TC T.T2 3,:,..:.. T I A T. Tl O,T U T (.E) B67

Transient Temperature Calculation for a Jacketed Mixing Kettle Discussion of Results The tabulated results show the temperatures of the mixture, vessel, jacket, tank water, and inlet and outlet conditions for time increments of 0.01 hours. The plotted results show the same information in graphical form, with the characters denoting the temperatures as follows: Character Temperature * Mixture V Vessel J Jacket W Water E Exit Water A curve has been drawn through points V to show the cyclical variations of temperature within the vessel walls. Critique Probably the most instructive features of this problem are the difficulties incurred by the instability of the Runge-Kutta solutions when too large a step interval is chosen. In order to get a solution to this problem a very small interval must be taken, and this means that many iterations must be used to yield just one solution. It is true that the temperature curves could be made smoother by a different choice of the water flow rates, and the heat transfer convective coefficients. However, the values chosen for the sample problem are probably close to the actual values. The temperature changes would initially be quite rapid. It should be pointed out that an analog computer could be used to "solve" this problem with no stability problems of the type incurred with the digital computer. However, it would be difficult to study both the small temperature differences at the beginning and end of the curve, together with the intermediate results unless a fairly complicated switching circuit was used to change the scale factor at the appropriate points in the solution. B68

Example Problem No. 71 SURGE SYSTEM OSCILLATIONS by Harry P. Hale Department of Mechanical Engineering Wayne State University Course: Hydraulic Machinery Credit hours: 3 Level: Senior Statement of Problem In hydroelectric plants quite a lot of study has been given to unsteady flow in the long hydraulic tunnels from the headwater to the hydraulic turbine. The calculations are tedious since the equations involved are nonlinear differential equations. On the next page is shown a schematic of the system involved and the various dimensions. The control valve is ordinarily connected to a turbine through a speed control governor. The overall problem involving tunnels, surge tanks, turbine, and governor is quite complicated. In this problem you will not be concerned with the turbine or governor. The control valve is so adjusted that the water velocity in the tunnel is 6.85 feet/second. (a) Set up the differential equations for unsteady flow. Assume the mass of water in the surge tank is negligible, the tunnel walls are inelastic, and the fluid is incompressible. (The 6.85 ft/sec is the maximum velocity in the tunnel.) (b) Find the period of mass oscillation neglecting friction if the control valve is suddenly closed. (c) Set up the differential equation considering friction if the control valve is suddenly closed. (d) Solve the differential equation of part (c) using the Runge-Kutta routine. To evaluate the friction factor, f, use the Moody Chart and assume the tunnel is concrete with the roughness factor, C, equal to 0.005 feet. As a first approximation assume f is constant and evaluate it at a tunnel velocity of 6.85 ft/sec. Assume the water temperature is 70~ F. Carry your solution out to about 5 cycles. References: Engineering Fluid Mechanics by Jaeger Hydraulic Transients by Rich B69

Surge System Oscillations Solution Refer to the figures shown below. dWsinc = TAtdLsina = TAtdh YAdh - Atdp - YAtdh = AdL tE t t t f 9 t T 1. As- SL H 1. 2. 3. L = A = t Reservoir 4. Surge Tank Pressure Tunnel 5. Control Valve Penstock A = 41.5 sq. ft. 8500 feet VO = 6.85 ft/sec. 31.5 sq. ft. Schematic of Hydraulic System (p + dp)At Free Body Diagram B70

Example Problem No. 71 Integrating dL between 0 and L, dh between Hi - V2/2g and Ha, dp/y between Hi and H + Z, and dhf between 0 and L,the following equation is obtained: I. L dV V dV + z + g + h = 0 From the continuity equation the following is obtained: II. VA = A dZ + Q t s dt where, K = specific weight, lb/ft3 V = tunnel velocity, ft/sec hf = head loss due to friction, feet. Q = flow in penstock, cubic feet/second. (b) At t = 0-, Q = Qo; At t = 0+, Q = 0 and AtV = AsdZ/dt. Also if the V2/2g term is neglected, (L/g)(dV/dt)+ Z = 0. Combining these two equations, d2z gAt dZ + 2 + LAs dt s From this the natural frequency, X) =/gAt/LA radians/sec. The period T = 2w/ = 117 seconds. (c) Neglect the V /2g term as it will be small compared with the friction loss. Using the Darcey-Weisback formula, hf = f (L/D)(V2/2g), the following formula is obtained: a2z/dt + 0.00287Z + 0.00208 (dZ/dt) = 0 (The friction factor, f, turns out to be about 0.02) (d) The above second order differential equation will have to be broken down into two first order equations as follows: Let Z(1) = Z; let Z(2) = dZ/dt = F(1) The two first order equations are: F(1) = Z(2) F(2) = -A.Z(1), - B-Z(2)-Z(2) At t = 0, dZ/dt or F(1) = 0 At t = 0, Z or Z(1) = Z = hf = -19.6 feet. Flow Diagram B71

Surge System Oscillations Flow Diagram (continued) CALCULATE — ~-K=RKDEQ.(O) K=l _ - L - )/No TERITE MAD Program RSURGE SYSTEM OSCILLATIONS DIMENSION F(2),Q(2),Z(2) INTEGER K EXECUTE SETRKD.(.2,Z(1),F(1),Q,T,H) FIRST READ FORMAT INPUT,A,B,Z(1),Z(2),H PRINT FORMAT HEAD T=O. WRITE PRINT FORMAT RESULT,T,Z(1),Z(2) WHENEVER T.G.500., TRANSFER TO FIRST STEP K=RKDEQ.(O) WHENEVER K.E.1 F 1)=Z(2) F( 2 =-A*Z( 1)-B*.ABS.Z(2)*Z(2) TRANSFER TO STEP END OF CONDITIONAL TRANSFER TO WRITE VECTOR VALUES HEAD=$1H1,S13,47HTABULATED SOLUTION OF THE DIFF 1ERENTIAL EQUATION/lHO,S16,lHT,S18,lHZ,S18,lHV*$ VECTOR VALUES INPUT=$5F10.5*$ VECTOR VALUES RESULT=$1H,6F19.4*$ END OF PROGRAM B72

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Example Problem No. 71 Discussion of Results and Critique The results indicate that the surge tank oscillations die out at a slow rate, as might be expected. Although these results are not startling, the main purpose of such a problem is to show the student how a rather tedious problem can be solved rather easily with the aid of a computer. Also, more realistic conditions could be used such as setting up an equation for the friction factor, f, in terms of Reynolds Number instead of using a constant value for f. Also a more complicated system could be analyzed by including the turbine and governor and checking stability of the overall system. The data for this problem was taken from an actual installation as noted in the reference, Engineering Fluid Mechanics by Jaeger. It would probably be better to analyze the differential equation with an analog computer first so that the student could quickly see how changes in parameters affected the results. Then, for more precise answers, the digital computer should be used. Surge System Oscillations B75

Example Problem No. 72 ANALOG ANALYSIS OF SINUSOIDALLY EXCITED SPRING-MASS-DASHPOT SYSTEM by C. W. Messersmith School of Mechanical Engineering Purdue University Course: Mechanical Engineering Laboratory Credit Hours: 2 Level: Junior Statement of the Problem A spring-mass-dashpot system has linear characteristics. If x is the displacement of the mass (M) from its rest position, if -kx is the spring force tending to restore the mass dx to its rest position, and if the damping force is - c-d; write the differential equation for the system assuming the mass to be acted upon by a force of the form F sin Xwt. For a particular system the mass is 3.22 Ibm, the spring constant (k) is 400 lbf per ft, the dashpot factor (c) is 2 lbf sec per ft and the exciting force has a maximum amplitude of 20 lbf. Use the analog computer to determine the amplitude of vibration of the mass for frequencies of the forcing function varying from about one-half to about twice the natural frequency of the system and compare your results with analytically determined values. Solution 0.1 -2 + 2 c + 400x = 20 sin 6ot dt2 dt dt d let s = - then 4000x + 20 sx + s2x - 200 sin W)t = 0 the natural frequency = /4000 i 63 radians per sec If Xmax = 0.16, then for optimum amplifier operation, display 4000(0.16) = 640, say 600x; 600x 600x s2x, say 10 sx; and 6~2 say 6 sy ^2 bs (63) The equation for implementing the computation then becomes 4 (600x) + 2O (10 sx) + 6 200 4 (600 x) + (10 sx) + 6( 6) - 200 sin <Ot = 0 or 2 1.111(600x) + 0.333 (10 sx) + (x) - 33.3 sin Mt = 0 Computer Circuit The computer circuit to solve the equation including generation of the forcing function is shown below. The computer was operated at frequencies one-tenth that of the problem as indicated by the note on the sketch. B76

Note: The solution was obtained at a frequency, hat of /he ptotfyre y reducing he 9 0a i5 of qrnmif ers 2 3 5 Qnd 6 by a factor of /0. Computer Circuit Computer Output Fig. 1 shows a curve for the analytical steady-state solution and various points obtained from the computer solution. Figs. 2a, 2b, and 2c show records of the computer solution made at three different frequencies of the forcing function. B77

Spring-Mass-Dashpot System 01 H x 4-) Q) e pcq Q) Q) r-I Ea Hl 02 I I I I I I i I I I II I I I I I rI I I I r Frequency, Radians per Second ie Equation 0.1 dx + 2 d + 400x = 20 sin Wt dt Fig. 1. Solution of th Fig. 2a Typical Graphical Results B78

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Example Problem No. 75 BEVEL GEAR SPEED REDUCER FORCE ANALYSIS by J. Raymond Pearson Department of Mechanical Engineering The University of Michigan Course: Machine Design I Credit hours: 3 Level: Senior Statement of Problem A firm wishes to produce speed reducers of this type for various ratings, and plans to replace the tapered roller bearings with ball bearings. The ball bearings are to be mounted, as shown in Figure 12 of Machine Analysis and Design Problems by Alvord, Pearson, and Hall, so that the thrust reactions occur on one bearing at a time. Input data and symbols to be used: Input torque and external forces on the shafts Tin = 500 in-lb Fix = 100lb Gix = 2001b Fiy = 1501b Giy = 3001b Fiz = 2001b Giz = 4001b Physical dimensions: Pressure angle (() = 20~, velocity ratio = 4:1, Rp =, Rg =, A = B =, C =, D =, E =, F = Measure Figure 12 to obtain dimensions where they are not given. Label the reactions on the bearings as shown. Prepare a general force analysis to determine all bearing reactions. Write a MAD program for solution on the digital computer. Run the program, using the given input data, X to obtain the bearing reactions. I _Gix G0 LABEL REACTIONS ACCORDING TO BEARING AND DIRECTION,iy iz USE ORIGIN 0p FOR FORCE ANALYSIS OF BEARINGS A and B. USE ORIGIN 0 FOR FORCE ANALYSIS OF BEARINGS D and E. Bearings Q X QeIIB vY l ~ix jRaxP F in Bearing B80

Example Problem No. 74 CAM DESIGN PROPOSAL ANALYSIS by J. Raymond Pearson Department of Mechanical Engineering The University of Michigan Course: Machine Design II Credit hours: 3 Level: Senior Statement of the Problem Attached is the drawing of a proposed design for a cam drive follower. The cam is an eccentric circular disk. Space limitations require that the sum of the cam and roller and that the roller width be 0.5xradius of the roller. The output are negligible. The spring load S equals - 100 - kX. The following is a suggested list of symbols and significant R2= Eccentric radius R = Follower roller radius D = R/R2 ratio Ms = Follower mass = 0.01 + roller & pin mass B = Roller width A = Follower acceleration A) = Cam speed X = Disk eccentricity 6 = Cam angle I = Inertia force S = Spring force K = Spring stiffness rate = Angle of line of centers F = Contact force - = Cont+act. stress with a straight line radii be held at 2.500 inches load and the friction load constants: in. in. #s2/in. in. in/s2 rad/s in. deg # # # /in deg # # /in2 # /in2 in. in/s2 P Y Mdl Md2 C = Roller pin bearing projected pressure =Follower displacement =Modulus of Elasticity of roller material =Modulus of Elasticity of cam material =Sum of cam & roller radii B81

Cam Design Proposal Analysis It seems likely that the critical factors of this design will be either the contact stresses between the roller and cam surfaces or the follower roller bearing pressure. The type of bearing proposed here is an ordinary sleeve bearing. The effects of change in radius ratio and speed on these two factors is desired. Write out an analysis to determine the variation of stress and pressure with cam angle for the two parameters, radius ratio and speed. Prepare and run a computer program to determine the necessary values and prepare a set of curves to demonstrate these effects. Use values of 0.125, 0.250, 0.575, 0.500 for the radius ratio and 75, 100, 125, and 150 radians/sec. for the speed parameter. 1. Frame 2. Eccentric cam 3. Follower roller I 4. Follower / 5. Follower spring I I B82

Example Problem No. 75 USE OF THE ELECTRONIC DIFFERENTIAL ANALYZER TO STUDY THE DYNAMICS OF MACHINERY* Department of Mechanical Engineering The University of Michigan Course: Dynamics of Machinery Credit Hours: 4 Level: Junior A series of exercises has been prepared to introduce the student to the use of the electronic differential analyzer as a means of studying the dynamics of machinery. Typical examples will be presented of some of the material which has been prepared. The examples consist of: a typical set of homework problems for which a written solution is required, eight computer problems to be solved on the electronic differential analyzer, an experiment combined with a computer problem, and a typical, one-hour examination. Typical Set of Homework Problems Fill in Answers 1. eI 0 e0 e1 = R Ro 3. el e1 3 e1 K R e=e0 e2 C 4. e1 R e e = 1 0.1 5. el = at 0.1 0 0.1 V e0 = 6. Draw the analog circuit to simulate the following system: x x0 = 0.10 in m im = 1.0 lbf-sec /in _iC _I k~~ |c = 1.0 lbf-sec/in C kf~ k = 100 lbm/in * These exercises were prepared jointly by several members of the Mechanical Engineering faculty. B83

Use of the Electronic Differential Analyzer Typical Set of Problems to be Solved on the Electronic Differential Analyzer Analog Computer Problem Number 1: Solve the equation mx + cx + kx = 0 where m = 1 lb-sec2/in., c = 1.5 lb-sec/in., and k = 100 lb/in. 1. For starting conditions use x = 1-1/2 in, k = 0. Obtain a record of x and x on the oscillograph. Be sure that they are both positive in the same direction. How long does it require for the amplitude to decay to 1/4 in? What voltage did you use for x? For x? On the tape, a centimeter represents what displacement? What velocity? 2. Repeat item 1 for x = 15 in/sec, x = 0. What differences are noted? 3. Disconnect the velocity loop and start the motion with x = 1-1/2 in, x = 0. What is the measured frequency? The calculated frequency? 4. Change the feedback capacitors to 1 mfd each and repeat item 3. What happens? Analog Computer Problem Number 2: A mechanical vibrating system has W = 10 lb, c = 0.305 lb-sec/in, and k = 120 lb/in. It is to be excited by a step forcing function of 8 lb. 1. Draw the circuit diagram and determine suitable scales for the displacement, time, and force. This should be done before coming to the computer lab. 2. Wire up the computer and operate it so that you can check the values of your scale factors, and also to get data for item 4. The transient force should be wired up through the function switch so that you can turn the force on and off while the computer is in "operate." Take recordings of the force on one channel and the displacement on the other. Record the actual scale on the tape. 3. Disconnect the damping loop and reproduce both parts of Fig. 18-7.* 4. Measure one of the records taken in item 2 and determine the log decrement, and, from this, calculate the damping ratio. How does this damping ratio compare with the damping ratio determined from the original differential equation? Analog Computer Problem Number 3: Use the analog computer to calculate a set of points for one of the curves of Fig. 18-24.* Since a function generator will not usually be available, you should generate the forcing function on the computer itself. Draw the circuit diagram in such a manner that changing the frequency of the forcing function will have no effect on the amplitude of the forcing voltage. Note also that there is an easier way to do this than the method illustrated in your EDA**otes. Shigley, Joseph E., Dynamic Analysis of Machines, McGraw-Hill Book Co., New York (1961). or Shigley, Joseph E., Theory of Machines, McGraw-Hill Book Co., New York (1961). ** A set of notes entitled, "Introduction to the Electronic Differential Analyzer," prepared by the faculty of the Department of Mechanical Engineering at The University of Michigan. These notes are distributed as introductory material to beginning students using the analog computer. B84

Example Problem No. 75 Analog Computer Problem Number 4: A mechanical vibrating system has k = 100 lb/in, c = 0.485 lb-sec/in, and W = 10 lb. It is to be excited by a transient force of 10 lbs. 1. Draw the circuit diagram and determine suitable scales for the displacement, time, and force. The circuit should be arranged so that the force can be switched on and off. 2. Wire up the computer and operate it so that you can check the values of your scale factors. Take several recordings of amplitude and force and calculate the percentage error between the machine solution and a phase-plane solution. 3. Operate the system by turning the force to "on". After 2-1/2 cycles of vibration,turn the force off. What is the amplitude of the first peak after turning off the force? 4. Some of the problems that occur in applying constant forces to elastic systems are that the system "overshoots" its eventual equilibrium position and vibrates too much. This can be solved by introducing more damping. If too much damping is introduced, however, then the system requires excessive time to reach its position of equilibrium. A happy medium would be obtained if only a slight overshoot were obtained provided the system settled down quickly. This is really a problem in optimization. So optimize your system by finding a value for c such that the system quickly comes to the equilibrium position after the force is applied without overshooting too much. 5. Connect x to the y axis of an oscilloscope or x-y plotter, and ~/ln to the x axis and photograph or draw the phase-plane diagrams for parts 2 and 3. Analog Computer Problem Number 5: u Using the computer, calculate the amplitude ratio k corresponding to various a/ n ratios from zero to 4. Uo Plot the results with --- as the ordinate, and X/o n as the abscissa. This is called a frequency-response chart. Get enough points to construct a smooth curve. Do not use a sine generator. Instead, generate the forcing function on the computer itself. Include specimens of the computer tape containing the forcing function as one graph and the response that both charts have positive values in the same direction. k Fcos w F0cos tt J4 = 0.15 as the other graph. Be sure On the same chart draw a frequency-response diagram obtained by calculation. What is the largest error found? B85

Use of the Electronic Differential Analyzer Analog Computer Problem Number 6: For the system shown in the figure r = 0.15, using the analog uo computer, calculate the amplitude ratio Fo-/ corresponding to various W /tn ratios from zero to 4. Plot the results with as the ordinate and ~A /) n as the abscissa. This is called a frequency response chart. Get enough points to construct a smooth curve. Do not use a sine-wave generator. Instead, generate the forcing function on the computer itself. Include specimens of the computer tape containing the forcing function as one graph and the response as the other graph. Be sure that both charts have positive values in the same direction. Analog Computer Problem Number 7: Prepare the analog solution for the system shown in the schematic diagram, using the following procedure: 1. System equation 2. Check solution 3. Variables range /////// 4. Parameter range 5. Time scale adjusted equations 6. Amplitude scale adjusted equation k C 7. Circuit diagram 8. Potentiometer settings 9. Tabulation of runs 10. Curve plots and conclusions m = 10 c = 1. Determine experimentally: k =2. a. the period x(O) b. the amplitude after 2 cycles of vibration c. the value of the critical damping constant A(0) = C Nt F eij t #s/in 0 #s/in 0 #/in 1.0 in 0 % Analog Computer Problem Number 8: Prepare the analog solution for the system shown in the schematic diagram, using the following procedure: 1. System equation 2. Check solution 3. Variables range 4. Parameter range 5. Time scale adjusted equation k 6. Amplitude scale adjusted equation 7. Circuit diagram 8. Potentiometer settings 9. Tabulation of runs 10. Curve plots and conclusions Instructions: a. Determine the response ratio, xo/Xst for o/6Jn =0.5, 0.8, 1.0, 1.41, 2.0 b. Plot response ratio vs. frequency. c. Plot the computed curve and determine the discrepancy. B86 m = 1.60 #sin c = 24.0 #s/in k = 250 #/in F = 75 # 0

Example Problem No. 75 Experiment Combined with Computer Problem Purpose: To experimentally analyze a torsional vibration model, and to simulate the model on the analog computer. Apparatus: 1. Torsional vibration model with crank and lever forcing mechanism. 2. Necessary instrumentation. 3. Analog Computer. Experimental Procedure: 1. Connect motor to variable speed control, range 0-1725 rpm, and run at low speed for a short period of time to see how it operates. DO NOT EXCEED 600 RPM OF CRANKSHAFT AT ANY TIME. 2. Calibrate both potentiometers separately. If this is accurately done, they should read the same when re-installed and the model turned over by hand. 3. Display on the scope and record with the camera the displacement of each end, superimposed, at 550 rpm of the crankshaft. Do not over speed! 4. Measure the spring rate of the model, but do not stress the torsion bar beyond 20,000 psi shear. 5. Display and record the damped free vibration of the flywheel, 6. Using a 3-string torsional pendulum, determine the moment of inertia of a similar flywheel, and calculate the moment of inertia of the flywheel shaft, spacer, and coupling. Analytical Procedure: 1. Calculate the damping coefficient from the data of 5 above. 2. Write the differential equation of motion for the model, and using the measured data from 4, 5, and 6, above, simulate the model on the analog computer, reproducing as nearly as possible the displacement diagrams obtained from the model at 550 rpm. Report Should Contain: 1. A copy of these instructions. 2. Simple but neat sketch of the test set-up. 3. Photographs or copies, neatly labeled. 4. Curve showing spring rate. 5. Observed data. 6. Calculations, computer program diagram, and recorder tape from Analytical Procedure. 7. List of important equipment used. B87

Use of the Electronic Differential Analyzer Typical One-Hour Examination I. A forced-damped system of one degree of freedom consists of a weight of 38.6 pounds supported by a spring of rate 10 pounds per inch and a dashpot with a viscous damping constant of 1.0 pound-seconds/inch. The system is excited by an external harmonic force with an amplitude of 1.50 pounds and a frequency of 105.05 cycles per minute. Determine: a. Displacement amplitude ratio xO/xt. b. Displacement amplitude xo, inches. c. Damping factor, r, dimensionless. d. Spring force amplitude, pounds. e. Damping force amplitude, pounds. f. Inertia force amplitude, pounds. g. Phase angle, 0, degrees. h. Draw the force amplitude polygon to a scale of 1 inch equals 0.5 pounds. Identify the vectors. II. a. Write the differential equation of motion for the system of problem I. b. Draw a circuit diagram to represent the mechanical system of problem I on the electronic differential analyzer. Give all necessary capacitor, resistor, and potentiometer values. B88

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