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Design and development of a high pressure CNG intensifier Touchette, Alain 1994

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DESIGN AND DEVELOPMENT OF A HIGH PRESSURE CNG INTENSIFIERbyALAIN TOUCHETfEB.Eng. in Mechanical Engineering, Royal Military College of Canada, 1992A ThESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR ThE DEGREE OFMASTER OF APPLIED SCIENCEINTHE FACULTY OF GRADUATE STUDIESMechanical Engineering DepartmentWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAApril 1994© Main Touchette 1994In presenting this thesis in partial fulfilment of the requirements for an advanceddegree at the University of British Columbia, I agree that the Library shall make itfreely available for reference and study. I further agree that permission for extensivecopying of this thesis for scholarly purposes may be granted by the head of mydepartment or by his or her representatives. It is understood that copying orpublication of this thesis for financial gain shall not be allowed without my writtenpermission.(Signature)___________________________Department ofciThe University of British ColumbiaVancouver, CanadaDate 2?- APQ dl/DE-6 (2/88)IIABSTRACTA two-Stage, variable-stroke rotary-reciprocating intensifier was built in order to providehigh pressure natural gas to a diesel engine. In this application, the intake pressure is variable (20to 200 bar), as is the mass flow (2 to 50 kg/br) and the delivery pressure is constant (200 bar).The main feature of this intensifier is that it can provide any mass flow in this range regardless ofintake pressure or operating speed.The mechanism used to provide the variable mass flow uses a variable stroke. Thismechanism, as well as the intensifier configuration, was chosen after investigating the alternativedesign concepts. This investigation showed that the variable stroke would be an energy efficientmethod of controlling the intensifier mass flow. The specific intensifier dimensions weredetermined using a graphical design optimization technique.Prototype testing was limited to speeds below its design operating range because ofcomponent failures due to inadequate cooling. In this speed range, the variable stroke proved tobe capable of controlling the intensifier mass flow. The rotary-reciprocating geometry was not asenergy efficient as originally expected. This shOrtfall is due to large frictional losses in the rotarygas seals and roller bearings. These frictional losses are essentially constant for all strokes whichresults in low efficiencies at low mass flows.It is suggested that a new intensifier design be considered: a configuration which retainsthe variable stroke capability of the rotary reciprocating design but eliminates the main sources offrictional losses by using non-rotating pistons.111TABLE OF CONTENTSABSTRACT iiTABLE OF CONTENTS iiiLIST OF TABLES viiiLIST OF FIGURES ixACKNOWLEDGMENTS xiii1 INTRODUCTION 11.1 Alternative Fuels for Diesel Engines 112 Methods of Using Natural Gas 21.2.1 Spark Ignition 21.2.2 Low Pressure Injection 31.2.3 High Pressure Injection 31.3 General Design Requirements 51.4 Previous Work 61.5 Goals and Objectives 62 DESIGN PROBLEM 82.1 Introduction 82.2 Exhaust Pressure 8iv2.3 Intake Pressure 92.4 Capacity 102.5 Capacity Control 112.6 Additional Requirements 122.7 Summary 133 DESIGN CONCEPTS 153.1 Introduction 153.2 Turbomachinery 163.3 Rotary Compressors 183.4 Reciprocating Compressors 183.4.1 Compression Cycle 193.4.2 Volumetric Efficiency 243.4.3 Capacity Control 253.4.4 Actuation 293.5 Previous Intensifier Design 383.6 Choice of Intensifier Configuration 393.7 Rotary-Reciprocating Intensifier Terminology 403.8 Rotary-Reciprocating Intensifier Operation 424 DESIGN OPTIMIZATION 454.1 Summary of Design Requirements 454.2 Description of Design Procedure 47V4.3 System Parameters 484.4 Objective Function 494.5 Design Variables 514.5.1 Number of Stages 524.5.2 Number of Pistons 524.5.3 Stroke 534.5.4 Bore 534.5.5 Rod Diameter 544.5.6 Bearing Diameter and Length 544.5.7 Outer Ring Diameter 554.6 Design Constraints 554.6.1 Roller Bearing Load and Speed 554.6.2 Linear Bearing Load and Sliding Velocity 594.6.3 Required Mass Flow Rate 614.7 Design Procedure 614.7.1 Design Tools 614.7.2 Number of Stages and Pistons 644.7.3 Design of the First and Second Stage 664.8 Auxiliary Components 884.8.1 Valves 884.8.2 Rotary Seals 895 EXPERIMENTAL APPARATUS 96vi5.1 Test Rig 965.2 Piping Plan and Instrumentation 985.3 Data Acquisition 1035.4 Testing Procedure 1056 RESULTS 1076.1 Introduction 1076.2 Low Speed Testing 1086.2.1 Mass Flow Rate 1096.2.2 Power and Specific Power 1126.2.3 Compressor Efficiency 1166.2.4 Volumetric Efficiency 1176.3 High Speed Testing 1176.3.1 Mass Flow Rate 1186.3.2 Power and Specific Power 1196.3.3 Compressor Efficiency 1216.3.4 Volumetric Efficiency 1216.4 Variable Speed Testing Results 1237 CONCLUSIONS AND RECOMMENDATIONS 1257.1 Conclusions 1257.1.1 Discussion of Objectives 1257.1.2 Problems of Rotary Configuration 126viiCentrifugal Forces 127Rotary Gas Seals 127Alignment 128Overheating 1287.2 Recommendations 1297.2.1 Centrifugal Forces 1297.2.2 Rotary Gas Seals 1307.2.3 Alignment 1317.2.4 Overheating 1327.2.5 General Recommendations 1358 REFERENCES 1369 APPENDICES 137A Engine Specifications 137B Capacity Control Example 143C Centrifugal Compressor Calculations 144D Intensifier Drawings 147E Test Rig Specifications 190F Instruments Specifications and Calibration 192G Data Acquisition Hardware Specifications 203H Performance Data Calculation Subroutine 205I Simulation Program 210viiiLIST OF TABLESTable 3.1 Comparison of Alternative Drive Systems 37Table 4.1 Design Variables 68Table 6.1 Scope of Testing 107Table A.1 Data from DDC for the 6V92-TA Coach 139Table D.1 List of Parts 147ixLIST OF FIGURESFigure 3.1 Compressor Configurations 16Figure 3.2 Simple Reciprocating Compressor 19Figure 3.3 Beginning of Cycle at BDC 20Figure 3.4 Compression Phase 20Figure 3.5 Exhaust of Compressed Gas 21Figure 3.6 Mid Cycle at TDC 21Figure 3.7 Expansion Phase 21Figure 3.8 Intake of Fresh Gas 21Figure 3.9 Typical PV Diagram for Compression Cycle 21Figure 3.10 Variable Clearance Volume - High Mass Flow Rate 27Figure 3.11 Variable Clearance Volume - Low Mass Flow Rate 27Figure 3.12 Variable Stroke Intensifier - Long Stroke 28Figure 3.13 Variable Stroke Intensifier - Short Stroke 28Figure 3.14 Camshaft Actuation 31Figure 3.15 Crankshaft Actuation 33Figure 3.16 Crankshaft Actuation with Crosshead 34Figure 3.17 Hydura PVWH open loop pump 36Figure 3.18 End View of Rotary Intensifier 40Figure 3.19 Side View of Rotary Intensifier 41xFigure 3.20 Intensifier with Zero Eccentricity 43Figure 3.21 Intensifier with Maximum Eccentricity 44Figure 4.1 Linearization of Commercial Bearing Data 57Figure 4.2 Linearization of Commercial Bearing Data 57Figure 4.3 Linearization of Commercial Bearing Data 57Figure 4.4 Stage and Piston Configuration 65Figure 4.5 View of Geometric Variable 67Figure 4.6 Design Procedure 75Figure 4.7 Torque vs Crank Angle - Stroke of 19.05 mm 77Figure 4.8 Torque vs Crank Angle - Stroke of 25.4 mm 78Figure 4.9 Torque vs Crank Angle - Stroke of 31.75 mm 78Figure 4.10 Sketch of Spreadsheet Format 79Figure 4.11 Bearing Load vs Stroke and Rod Diameter 83Figure 4.12 Bearing Load vs Stroke and Rod Diameter 83Figure 4.13 Bearing Speed vs Stroke and Rod Diameter 84Figure 4.14 Bearing Speed vs Stroke and Rod Diameter 85Figure 4.15 Bearing Diameter Optimization 86Figure 4.16 NUPRO Check Valve 88Figure4.17 Face Seal 90Figure 4.18 Mechanical Seal 91xFigure 5.1 Test Rig 97Figure 5.2 Oil System 99Figure 5.3 Gas System 101Figure 5.4 End View of Test Rig 103Figure 5.5 Data Acquisition Hardware 104Figure 6.1 109Figure 6.2 110Figure 6.3 111Figure 6.4 113Figure 6.5 114Figure 6.6 115Figure 6.7 116Figure 6.8 117Figure 6.9 118Figure 6.10119Figure 6.11 120Figure 6.12 121Figure 6.13 122Figure 6.14 123Mass Flow Rate for Full Stroke Testing at 200 RPMStage Pressure Ratio for Full Stroke Testing at 200 RPMMass Flow Rate for Various Strokes at 200 RPMRequired Power for Various Strokes at 200 RPMMeasured and Predicted Power for Full Stroke at 200 RPMSpecific Power for Various Strokes at 200 RPMCompressor Efficiency for Various Strokes at 200 RPMVolumetric Efficiency for Various Strokes at 200 RPMMass Flow for Full Stroke at 200 and 400 RPMMeasured and Predicted Power forFull Stroke at 200 and 400 RPMSpecific Power for Full Stroke at 200 and 400 RPMEfficiency for Full Stroke at 200 and 400 RPMVolumetric Efficiency for Full Stroke at 200 and 400 RPMPower Required at Various SpeedsxiiFigure 7.1 Side View of Open Configuration 133Figure 7.2 End View of Open Configuration 134Figure A.1 Power Output of the 6V92-TA Coach 140Figure A.2 Brake Specific Fuel Consumption of the 6V92-TA Coach 140Figure A.3 Fuel Consumption of the Engine in kg/hr 141Figure A.4 Fuel Consumption in mg/revolution 141Figure F.1 Loadcell Calibration 193Figure F.2 Inlet Temperature Calibration 194Figure F.3 Interstage Temperature Calibration 195Figure F.4 Outlet Temperature Calibration 195Figure F.5 Wall Temperature Calibration 196Figure F.6 Inlet Pressure Calibration 197Figure F.7 Interstage Pressure Calibration 197Figure F.8 Outlet Pressure Calibration 198Figure F.9 Tachometer Calibration 199xmACKNOWLEDGMENTSI wish to thank Professor P.G. Hill for supervising my work at UBC. The value of hisguidance in all aspects of the project was only surpassed by his accessibility and availability.Thanks are due to K. Bruce Hodgins whose wealth of technical knowledge led to manyeye-opening lessons in design, manufacturing, and testing. His experience was invaluable insurmounting the many problems encountered during the refinement stage.I am grateful to Christoph Aichinger who laid the groundwork of the intensifier projectwith his thesis. His work in assembling a good test rig and instrumentation was very useful to meand will be to any future student who works on this project.I also wish to thank Len Drakes and Anton Schreinders who transformed the intensifierfrom an idea on paper to a working prototype. They showed me the importance of includingmachining considerations in the design phase and took the time to show me how.Finally, I must thank my colleagues of the Combustion Lab: Patric Ouellette who alwayshas time to offer a helping hand and Brad Douville who joined me in many relaxing workouts andkept me sane.1Chapter 1Introduction1.1 Alternative Fuels for Diesel En2inesOne of the worst contributors to air pollution, fossil fuel burned for transport, has been thetarget of particularly severe emission regulations. More specifically, diesel engines used in busesand trucks will, in the near future, be subject to stringent new regulations which conventionaldiesel engine technology appears unlikely to meet.These regulations mostly restrict the emission of particulate matter (PM) but also placelimits on the emission of nitrogen oxides (NOx). The traditional experience in diesel enginedesign has been that measures which reduce either PM or NOx emissions generally result in anincrease in the emission of the other. Even with new engine designs, better electronic control andparticulate traps, the solution of the emission problem seems difficult and costly.An attractive alternative to solve the diesel engine emissions problem is to convert theengine to run on a cleaner burning fuel. Methanol and natural gas seem to he the most promisingalternative fuels.Methanol has been shown to be capable of meeting the emission restrictions on PM andNOx but requires large amounts of energy to produce. Methanol also has undesirable effects suchas corrosion of engine parts and in production of aldehydes, which are pollutants though notregulated at this point. Natural gas seems to be a more attractive alternative since it is relativelyinexpensive, readily available and does not have the corrosive effects of methanol.1.2 Methods of Usin2 Natural GasThere are a number of methods of converting diesel engines to natural gas fueling:• conversion of the diesel engine to spark-ignited combustion• low pressure gas injection with compression ignition• high pressure gas injection with compression ignitionThe advantages and disadvantages of these methods may be summarized as follows [1].1.2.1 Spark IgnitionIn this case, the natural gas is introduced in the combustion chamber via a carburetor orinlet port injector. The fuel-air ratio is controlled over the engine load range by means of athrottle. With close control of the fuel-air ratio of the premixed mixture, catalytic exhausttreatment can reduce emissions of NOx and PM to acceptable levels.The disadvantages of this method are that in order to prevent detonation, the enginecompression ratio must be greatly reduced from the value for diesel operation. Reducedcompression ratio and the need for throttling at part load can seriously reduce the efficiency of theengine. Thus conversion of diesel engines to spark ignition can meet emission requirements but atthe cost of reduced efficiency.31.2.2 Low Pressure InjectionMuch experience has been gained with low pressure injection of natural gas into dieselengines. Injection is done either directly into the intake manifold or at the intake port just beforeintake port closure. Operating at full load and using supplementary injection of diesel fuel forignition of the natural gas, this method of injection can produce high efficiency and reasonablylow emissions.The disadvantages of this method are associated with part load operation. The dieselengine operates without a throttle which results in an air admission rate which is essentiallyindependent of load. Because of this, the mixture fuel-air ratio inside the cylinder becomes toolow at part load to support combustion. Emissions become serious because of unburned fuel; atcertain loads, damaging auto-ignition can be encountered. Port injection produces better resultsthan manifold injection because it entails less mixing of the fuel and air prior to combustion; thisresults in higher local fuel-air ratio and more successful combustion. Also, port injectionproduces better results in two-stroke engines because it prevents waste of fuel in the scavengingperiod. However, experience with port injection to date indicate that use of low pressureinjection of natural gas will not produce satisfactory efficiency and emissions over the whole loadrange of a vehicle diesel engine.1.2.3 High Pressure InjectionThe advantage of injecting the natural gas directly into the combustion chamber is that itpermits the high efficiency of the diesel engine to be maintained over the entire load range. Noreduction in compression ratio and no throttling are required for successful operation.4The disadvantage of this method is the need for an on-board pressurization of the naturalgas. It happens that the optimum tank pressure for storage of natural gas (approximately 200 bar)is of the same order as the required injection pressure. However, as fuel is consumed from thetank, the associated pressure drop requires recompression before injection.The task of recompression can be simplified if the natural gas is stored in liquid form(LNG). In this case, the liquid is stored at low pressure in the tank, pumped as a liquid to highpressure and vaporized in a heat exchanger before injection. In certain applications, LNG will bea suitable alternative. However, the disadvantages include the need for natural gas liquefaction,cryogenic containment and gas escape during the venting needed to control the containerpressure. These disadvantages make the use of compressed natural gas (CNG) advantageous invehicle application such as bus fleets in which there is sufficient space on board the vehicle for thelarger CNG tanks. Also, most of the infrastructure relating to natural gas is equipped to deal withit in gaseous form.It is with these considerations in mind that the Department of Mechanical Engineering atthe University of British Columbia has undertaken a study of natural gas fueling of diesel engines.The purpose of this study is to evaluate the effectiveness of natural gas at reducing the emissionsof the engine. This investigation is using a Detroit Diesel 6V92-TA engine with direct injection ofthe natural gas which will be stored as CNG. This engine was chosen because is a durable andproven engine widely used in urban buses in North America. If natural gas fueling of the dieselengine can be shown to be an effective method of reducing emissions, then the conversion ofurban bus engines to run on natural gas would he an important application.As discussed earlier, the choice of direct injection of CNG requires on-boardrecompression of the natural gas. It is the design and development of a method of pressurizing5natural gas for direct injection in a vehicle diesel engine that will be described in this thesis. Theterm intensifier will be used instead of compressor because the latter is usually used when theintake pressure is constant. It will be shown that the intake pressure in this case is both elevatedand variable, hence the term intensifier1.3 General Design RequirementsAs will be shown on chapter 2, the principal design requirements for the intensifier are:• high and constant delivery pressure• inlet pressure varying over a wide range• mass flow varying over a wide range• high efficiency• small size and weightAdditional design requirements include durability, safety and low production andmaintenance costs. The preliminary design process revealed that commercially availabletechnology which could meet these requirements does not exist. It also showed the difficulty ofmaintaining high compressor efficiency over the desired range of intake pressure and mass flowrate. Thus an important goal of this work was to design the intensifier for high efficiency over theoperating range.61.4 Previous WorkThe groundwork for the intensifier project was mostly laid out by C. Aichinger and hiscontributions are contained in his master’s thesis [2]. One important contribution of this work is astudy of the different intensifier designs to examine which type of design would be suitable for thisdesign application. This study will be examined and expanded in chapter 3.Aichinger also designed and developed two prototype shaft-driven reciprocatingintensifiers. His evaluation of the performance of these intensifiers supplied valuable informationon design limitations and component performance. His results strongly emphasized the need foran intensifier design with variable capacity which is the subject of this work and which required anew intensifier concept.1.5 Goals and ObjectivesThe general goal of this work is to design and develop a natural gas intensifier that wouldmeet the requirements outlined above with special emphasis placed on achieving high efficiencyvariable capacity operation. In working towards this goal and after review of specific designrequirements and evaluation of different possible design concepts, the reasons for choosing therotary-reciprocating intensifier configuration were identified. Then followed the evaluation of thepertinent design limitations and the formulation of a computer simulation of the selected designconcept. With the simulation and additional design optimization techniques, a final design wasdeveloped for production and testing.7The specific objectives of this work are:1. To design and build a variable-stroke, multi-stage, mechanically actuated rotary-reciprocatingintensifier which would meet the design requirements for fuel flow in both design and offdesign operation of the diesel engine2. To measure the performance of this intensifier (paying close attention to its variable capacitycapabilities) in the light of the thermodynamic model used in the design process3. To assess the essential design limitations of this new class of variable stroke, high pressureintensifiers.8Chapter 2Design Problem2.1 IntroductionChapter 3 discusses intensifier design concepts and determining which intensifierconfiguration is best suited to the application of natural gas fueling of diesel engines. Beforeevaluating the different design concepts, the specific design requirements must be known. Thischapter will examine the design requirements as determined by the engine demands and theoperating conditions in which the intensifier will operate. The design requirements that have thegreatest effect on the choice of intensifier configuration are the operating pressures (intake andexhaust pressure), capacity and capacity control. Furthermore, there are practical considerationsthat will affect the choice of intensifier design concept.2.2 Exhaust PressureSince the intensifier will provide fuel to the injector, the exhaust pressure of the intensifiermust be the same as the injection pressure. The present working value for the injection pressure isapproximately 200 bar. This value comes from the need for rapid penetration of the gas acrossthe cylinder to avoid excessive combustion duration. Rapid penetration is possible when thevelocity of the gas leaving the injector nozzle is sonic or faster. To obtain sonic velocity through9a nozzle, a pressure ratio of approximately 2:1 is required. Since the cylinder pressure at the timeof injection is in the region of 80 to 100 bar, an injection pressure of approximately 200 bar isrequired. Sonic velocities also result in a choked nozzle which has the added benefit that the massflow through the nozzle is independent of combustion chamber pressure. The 200 bar injectionpressure is not a rigidly set value and it is possible that this value may be increased or lowered inthe future as testing of the injector progresses.2J Intake PressureSince the fuel will be stored as CNG in high pressure tanks, the intake pressure of theintensifier is the same as the tank pressure. The filling pressure for a CNG storage tank is 200 barwhich sets the maximum intensifier intake pressure. This 200 bar filling pressure comes from thefact that it is close to the pressure which, at ambient temperature, maximizes the ratio of mass ofgas contained to the mass of the tank required. The optimum occurs near 200 bar due to theshape of the compressibility curve of natural gas.As for the minimum tank pressure, there is no theoretical lower limit since the tanks couldbe emptied until they contain a partial vacuum. This is not feasible as it would mean a very highpressure ratio for the intensifier. Even if the tanks were only permitted to go down toatmospheric pressure (approximately 1 bar), the pressure ratio required would be 200:1. Thiswould require a very large machine due to the number of compression stages required. For thisreason, a lower limit must be arbitrarily set. This lower value must be chosen in such a way thatthe tanks are nearly empty since this would minimize the amount of tanks required on the bus (animportant consideration due to the weight of the tanks) and yet provide a reasonable pressureratio for the intensifier. A lower limit of 20 bar was chosen. This means that the tanks are10approximately 90% empty (from a simple perfect gas analysis with constant temperature). With amore precise analysis taking into account the compressibility of natural gas, the tanks are actually92.2% empty when the pressure is allowed to drop to 20 bar. This is a reasonably high valuewhich sets the maximum pressure ratio for the intensifier at 10:1, which is quite manageable.2.4 CapacityThe capacity of an intensifier is dependent on the speed at which it is operated. It is thusimportant to determine the operating speed range before any capacity considerations are made.A convenient power source for the intensifier is the engine crankshaft. The simplest wayto drive the intensifier would be to connect directly to the crankshaft and turn it at engine speed.There would be some added complexity if a gearbox was added to reduce or increase the speed,but it is an option. An important point to note is that with this option the intensifier speed isproportional to the engine speed. The addition of a hydraulic or pneumatic system to drive theintensifier would make the intensifier speed independent of engine speed.The capacity requirement is set by the engine fuel flow demand. The data from DetroitDiesel for the 6V92 lists the maximum required flow rate of natural gas is 53 kg/hr at 2100 RPMat full load. The mmimum can be found from experiments on the 6V92 and is less than 2 kg/hr atidle (600 RPM at no load). If the intensifier speed is proportional to engine speed, a moresignificant value for fuel flow is the amount of fuel required per engine revolution. The maximumflow then occurs at 1200 RPM at full load and is approximately 500 mg per engine revolution andthe minimum, at idle, is approximately 50 mg per engine revolution. A detailed calculation of themaximum values can be found in Appendix A while the values at idle were taken fromexperimental data.112.5 Capacity ControlAnother requirement related to fuel flow is capacity control. The desired situation is onewhere the intensifier capacity exactly matches the engine fuel demand at any given operatingpoint. The engine fuel demand is a function of engine load and speed. The intensifier mass flowrate is a function of its speed and intake pressure (since delivery pressure is to he kept constant),that is storage tank pressure. Since the intensifier speed is only a function of engine speed and thetank pressure simply decreases with time as fuel is consumed, some type of capacity controlsystem is required to match the intensifier flow rate to the engine requirements.It would seem simple to design the intensifier for the worst case, that is use the case wherethe required fuel flow is maximum and the tank pressure is minimum as a design point. Using theworst case as a design point would ensure that the intensifier would never fall short of enginedemand. Indeed, the intensifier could supply in the order of 20() times engine demand in a morefavorable off-design situation. Appendix B demonstrates this pOint with a simple single-stagereciprocating machine.The problem of handling this extra flow is not trivial. If it is fed from the exhaust linesback into the intake lines, the gas expands to intake pressure which results in a considerable wasteof energy. If the gas is fed back to the intake lines but is not allowed to expand to intake pressure(through the use of check valves) the result is a very large mass flow rate through the intensifierassociated with high gas velocities. These high velocities result in substantial pressure dropsacross valves which translate into another important loss. Also, high velocities result inaccelerated wear of the valves which is undesirable. Thus designing an intensifier for the worstcase and trying to control the unneeded flow at other operating points often results in large losses.12Since the engine spends a very small amount of time at maximum fuel flow with lowpressure in the tanks, the off-design operating conditions must be taken into account. The needfor some type of capacity control to reduce the intensifier capacity in the off-design cases isevident. This requirement of variable capacity will prove to be the most challenging to meet andis the primary area of work in the intensifier project.2.6 Additional ReiuirementsSome further requirements that must be met by the intensifier are set by the environmentin which the intensifier will be required to function. This environment is determined from the factthat the intensifier would be used to supply compressed natural gas to a vehicle engine.The power required to drive the intensifier should be as low as possible so losses such asfriction or uncontrolled expansion should be avoided. The power requirement should beespecially low at low load, when fuel demand is low, since the intensifier should not consume alarge percentage of engine output. Compressor efficiency is thus important since the intensifier isdrawing power from the vehicle engine.The intensifier must be placed near the engine in the engine compartment. Since space inthe engine compartment is fairly restricted, the intensifier must be as compact as possible. Also, ifthe need for additional supports for the intensifier is to be avoided, the intensifier weight must bekept low.The intensifier must be properly sealed as a leak of natural gas would be a potentialexplosion hazard, especially from a bus left overnight in a garage. If a leak in the system cannotbe avoided, it must be properly vented. One option available is to vent any leak to the engineintake. The escaped gas would then be burned in the combustion chamber.13The 6V92 TA is renowned as a very rugged engine. The intensifier, being tucked away inthe engine compartment and as accessible as the engine, should he as durable as the engine. Itscomponents should thus be designed with a long life and they should not require frequentreplacement.For the application of natural gas fueling of a bus engine to he attractive, the cost of theentire system must be reasonable. If one considers the tanks, the new gas lines, the injectors,modification to the engine control system, the intensifier and the installation of all thesecomponents, it is obvious that the intensifier should not be expensive to build. Simplicity and easeof manufacturing are thus essential. The requirement for low cost also means that the intensifiermust take advantage of any existing system (pneumatic, electric, etc.) already on the bus.2.7 SummaryIn summary, the principal design requirements are:• 200 bar exhaust pressure• 20 to 200 bar intake pressure• variable mass flow in the range of 50 to 500 mg/rev• high compressor efficiency• low size and weight• completely leak free• long life• low production and maintenance costsThe requirements listed above define the problem of constructing a suitable intensifier.The most important design consideration is to he able to match the intensifier capacity to the14engine fuel requirements and it was shown that this is not trivial. The incorporation of variablecapacity in the intensifier design is what makes this work significant. The next chapter willexamine the different intensifier design concepts and select the intensifier configuration best suitedto meet these requirements.15Chapter 3Design Concepts3.1 IntroductionThe following chapter is a study of the different intensifier design concepts. The purposeof this review is to determine, in light of the specific design requirements detailed in chapter 2, theintensifier configuration best suited to the design application. This study expands on the workdone by Aichinger [2] on this subject.The principal design alternatives are identified in Fig. 3.1. The design alternatives aredivided into two families of compressors: the intermittent flow and continuous flow machines.The intermittent flow family includes the reciprocating and the rotary compressors. Thecontinuous flow include the dynamic compressors or turbomachines. The advantages anddisadvantages of the different types will be summarized while discussing the selection of aconfiguration that will meet the design requirements.16Compressor TypesIntermvttent Flow Continos FlowTurbomchinesRotary Reciprocating II I ft Centrifugal[— Sliding Vane f— Hydrcnlic Actuationftxia1ft Straight Lobe Pneumatic Actuation LI MixedHelical LobeLMechanical ActuationFigure 3.1Compressor Configurations (Aichinger [2])3.2 TurbomachineryThe continuous flow compressors, or turbomachines, will be examined first. The onlyturbomachine that can operate at a 10:1 pressure ratio with a single stage is the centrifugalcompressor. Detailed calculations of the required diameter and angular speed of a typical highpressure ratio centrifugal compressor are described in Appendix C. The results of thesecalculations are as follows: for the given operating conditions, the required rotor diameter is 4x10cm and the angular velocity is 3x109 RPM. This extremely small rotor would be impossible tomachine. Also, because of the small size of the rotor, the clearances would be critical and therewould be unacceptable leakage and loss of efficiency. Finally, the required angular velocity isunacceptably high.17Since the angular velocity calculated above is unacceptable, a different approach will betaken. The operating speed will be set at engine speed and the specific speed, a dimensionlessparameter commonly used as a criterion for selection of compressor type, will be examined.The dimensionless parameter specific speed (N,) is defined as:N=(tpoin which is the angular velocity, Q is the inlet volumetric flow rate, Ap0 is the pressuredifference and p is the inlet density.As previously mentioned, the angular speed Q is chosen to be of the order of the enginespeed, that is 2000 RPM or 210 rad/s. The volumetric flow rate Q would be taken to be themaximum flow, 53 kg/hr rate at a density defined by the lowest tank pressure, 20 bar. Thisdensity is 13.9 kg/rn3 (20 bar, 25°C) and the volumetric flow rate works out to be 1.06 x103rn3/s. The pressure difference Apo is simply the difference between intake and exhaust pressureand is 18 MPa.The dimensionless specific speed works out to be 1.78 x104 and it has been shown that,for best efficiency, the specific speed parameter for different types of dynamic compressors are[3]:• Ns> 1.8 -Axial compressor• 1.2> Ns> 2.2 -Mixed flow pumps• 0.2 > Ns> 1.2 -Centrifugal compressor• 0.1 > Ns -Positive displacement18Of course, these values are only guidelines but the value of Ns for the urban busapplication is of the order of 0.0002 and is clearly in the range covered by positive displacementor intermittent flow compressors.3.3 Rotary CompressorsThe next family of compressors, the intermittent flow machines, can be subdivided into thereciprocating and the rotary compressors. The rotary compressors are usually limited to a 4.5:1pressure ratio and have an exhaust pressure that falls well short of the required 200 bar. Also,they typically require 25% more power than an equivalent reciprocating machine. This additionalpower requirement results in a lower efficiency. These shortfalls are due to problem with sealingthe sliding surfaces of the rotary compressors [2].3.4 Reciprocating CompressorsThe possible design concepts for the application in question has been reduced toreciprocating machinery since both turbomachines and rotary compressors proved to be unable ofmeeting the design requirements. This category of compressors includes all machines that use apiston sliding in a cylinder to produce the compression. Within this category are options open tothe designer in methods of controlling capacity and actuating the piston.Before capacity control and actuation are examined, the general reciprocating cycle, aswell as the concept of volumetric efficiency, will be described.193.4.1 Compression CycleAll reciprocating compressors function by the principle of pressure rise due to volumereduction. This volume reduction is supplied by a piston sliding inside a closed cylinder (Fig. 3.2).In the isentropic case, that is with no heat transfer or irreversibilities such as friction, thepressure rise is described as: [4](P2 (vP1)V2)3.1where k is the isentropic coefficient (ratio of the specific heats) and the subscript 1 denotes thestart of compression and 2 the end of compression. A complete compression cycle will now bedescribed.The compression cycle begins when the piston is at the lowest point in its reciprocatingmotion, that is bottom dead center (BDC) (Fig. 3.3). At this point the chamber formed by thecylinder and the piston has its largest volume and it is filled with gas at intake pressure. TheCy1incrPistonExcns± VcdvIntcke VcdvEFigure 3.2Simple Reciprocating Compressor20piston starts its upward motion with both valves closed so there is a reduction in volume and thuscompression (Fig. 3.4).When the pressure exceeds the exhaust pressure, that is the pressure that is on the otherside of the exhaust valve (which could be the pressure of an accumulator or intercooler forexample), the exhaust valve opens and the gas passes out of the cylinder (Fig. 3.5). When thepiston reaches top dead center (TDC) or its highest point in its reciprocating motion, there is nolonger any reduction in volume so the exhaust valve closes and the gas stops flowing out of thecylinder (Fig. 3.6). The piston then starts its downward travel and any gas that was left in theclearance volume (volume present in the cylinder when the piston is at TDC) begins to expand(Fig. 3.7). The pressure inside the cylinder is described by the same equation as the compressioncycle (eq 3.1). The intake valve opens when the pressure of the gas inside the cylinder dropsbelow the intake pressure (Fig. 3.8) and fresh gas enters the cylinder until the piston reachesbottom dead center (BDC) or its lowest point in its reciprocating motion. The intake valve thencloses and the cycle is repeated.r—A-———Figure 3.3 Figure 3.4Beginning of Cycle at BDC Compression Phase21ZDFigure 3.5Exhaust of Compressed GasFigure 3.7 Figure 3.8Expansion Phase Intake of Fresh GasA typical compression cycle can be plotted on a P-V diagram and would look like Fig. 3.9.Mid Cycle at TDCexhaust Pressure vs VolumePressure compressIonBDCVolumeFigure 3.9Typical P-V Diagram for Compression Cycle22The temperature of the gas in the cylinder will also vary through the compression cycle.These variations are described, if the process is isentropic, by the following equation: [4]tT2 tV3.2in which k is the isentropic coefficient.The temperature will thus rise during the compression, drop during the expansion and stayconstant during the exhaust and intake phases. Because of these variations and the fact that thetemperature of the wall of the cylinder will be fairly constant, there will often be a temperaturedifference between the gas and the wall and heat transfer either from or to the gas will occur. Theassumption of isentropic compression and expansion does not hold and equations 3.1 and 3.2need to be modified to take into account the heat transfer. This is done by replacing theisentropic coefficient with a polytropic coefficient which is different for compression andexpansion. This polytropic coefficient is found in the following way:The second law of thermodynamics requires: [4]Tds = dh — vdpand for an ideal gas, the law for polytropic compression and expansion is+ n = 0p vThe second law can be rewritten as:ds k dT_dpRk-1 T p23ds k (dpdvdpRk—1p i’) p(k_l)dskdvdpR v pCv V J3and combining with ideal gas law becomes:ds dv—= (k -n)—.Cv VIn this last equation, the only unknowns for a given compression cycle are the polytropiccoefficient n and the entropy variation ds. This equation is thus used to get an idea of the value ofn for different situations. In other words, it is expected that n be smaller than k during thecompression phase since the gas has a higher temperature than the wall for most of the phase andthus there is a net heat transfer out of the gas and ds is negative. The opposite holds for theexpansion phase, that is n is expected to be larger than k. The actual values for the n forcompression and expansion are found experimentally. Of course, these values vary depending onthe pressure ratio, the intake temperature and the wall temperature.Mention was made of the clearance volumes which play an important role during theexpansion phase of the gas since there would be no expansion phase if there were no clearancevolume. In this case, all the gas would be expelled from the cylinder when the piston would bemoving towards TDC and as soon as the piston would start its downward motion, the intake24valve would open and admit fresh gas into the cylinder. This seems like a better situation but thepresence of clearance volumes at TDC is inevitable. This volume is a result of space in the valves,space left between the top of the piston and the top of the cylinder to provide a safety margin incase of thermal expansion and diametrical clearance between the piston and the cylinder betweenthe top of the piston and the first piston ring.3.4.2 Volumetric EfficiencyThe presence of clearance volumes introduces the concept of volumetric efficiency. Thisparameter is defined as the ratio of the actual mass flow rate to the theoretical mass flow rate, thatis the mass flow if the were no clearance volumes and there were perfect filling of the cylinder justbefore the piston starts its downward motion. The volumetric efficiency would be: [2]1v=in which th is the actual mass flow, Pialet is the inlet density, n is the angular velocity and Vd is thedisplaced volume. In terms of clearance volumes, the volumetric efficiency can be expressed as:P2’—1where the clearance volume ratio ej is the ratio of the clearance volume to the displaced volumeVd and is typically between 4 and 20 %.25The pressure ratio and the clearance volumes are the most important factors indetermining the volumetric efficiency but there are also other factors that arise when dealing withreal machines such as:• poor filling of the cylinder caused by a pressure drop across the intake valve.• preheating of the incoming gas by the hot wall causing the gas to expand and reduce theamount of gas that can enter the chamber. This effect has been shown to be able to reduce thevolumetric efficiency by as much as 10% for a 5:1 pressure ratio [II.• gas leakage through the exhaust valve, permitting high pressure gas to enter the cylinder fromthe exhaust tract during the expansion phase delaying the opening of the intake valve.• leakage across piston rings and rod seals.The purpose of keeping the volumetric efficiency high is to keep the machine small. Thisis attractive since the forces tend to be lower in smaller compressors because there are smallerareas exposed to high pressure and because the masses that need to be accelerated inreciprocating motion are smaller. The use of multistaging provides an important advantage, thatof increasing the volumetric efficiency by lowering the pressure ratio of each stage.The second chapter described the need for the intensifier to have a variable capacity.Different types of capacity control for the reciprocating compressor will now be examined.3.4.3 Capacity ControlOne way to control the capacity is to vary the speed at which the intensifier is operating.This can be done continuously or discreetly. Given that the intensifier is going to be powered by avehicle engine, there are a number of methods of varying the drive speed. These include hydraulicand pneumatic actuation discussed in section 3.4.5. If the intensifier is to be directly driven by the26engine, a continuously variable drive could be possible in principle but successful prototypes havenot yet been developed. Alternatively, the intensifier could be run at engine speed when requiredand disengaged from the engine and thus not running it when the fuel flow does not require it.This is an energy efficient method since no energy is used when the intensifier is not run. Themajor problem is that this setup would cause variations in the injection pressure though thesecould be minimized with a large accumulator. Also, this method would require a clutch andcontrol system. This latter option would unfortunately require valuable engine bay space.An alternative for capacity control is a variable clearance volume arrangement. Since alarger clearance volume would reduce the volumetric efficiency, the mass flow rate would also bereduced. This can be extended to zero mass flow rate when the pressure at the end of theexpansion phase is equal to the intake pressure. This option is not as energy efficient as thevariable speed option because there is still considerable friction loss when the intensifier is runningat zero mass flow. On the other hand, this system is quite easier to implement. Fig. 3.10 and 3.11show how a simple single stage intensifier could operate with variable clearance volume. Oneproblem associated with this capacity control system is the large pressure loads on the slidingassembly containing the valves.27Figure 3.11Variable Clearance Volume - Low Mass Flow RateA third option which seems to combine the advantages of the first two is to vary thestroke of the piston. When high mass flow is required, the piston has a full stroke and acts exactlylike a conventional reciprocating compressor. When lower mass flow is needed, the stroke isshortened; this has the same effect as increasing the clearance volume. As opposed to the variableclearance volume option, there are decreasing frictional losses as the stroke is shortened. In fact,if the stroke is reduced to zero, there would no frictional losses due to the piston motion in thecylinder. Fig. 3.12 and 3.13 show the operation of such a variable stroke intensifier.Cy1indPistonExhoust VoveIntoI-<e Vo.1vFigure 3.10Variable Clearance Volume - High Mass Flow Ratez-jIIIIlIIIH28Varying the stroke of a reciprocating intensifier is difficult if mechanical actuation isconsidered. The technique used to accomplish this task is the rotary type intensifier which will bedescribed in later sections of this chapter and is the core of the work covered by this thesis.If the intensifier geometry is not to be variable, that is if the intensifier is a conventionalone with constant clearance volume and stroke with an uncontrollable speed, there exist twofurther techniques to control the capacity. One is to unload the intake valve, that is open theintake valve during the compression phase of the cycle. This causes the gas to flow back into theFigure 3.12Variable Stroke Intensifier - Long StrokeI IIIIII III IIII IFigure 3.13Variable Stroke Intensifier - Short Stroke29intake tract so that it does not get compressed. With this technique the injection pressure wouldvary unless a large accumulator were used and there is the additional disadvantage of increasedcomplexity and size of the intake valve. This modification to the intake valve usually leads to anincrease in clearance volume and thus larger components for a given mass flow rate.The second method of controlling the capacity of a fixed geometry intensifier is bypassing.In this case, the compressed gas is fed from the exhaust lines back into the intake lines of theintensifier if the mass flow required is less than the mass flow provided by the intensifier. Thepresence of a check valve prevents the expansion of the gas hack down to intake pressure(pressure of the tanks). This system was used in the first prototypes of the intensifier [2] and itwas found to consume a lot of power. Theoretically, there should be no power loss from thissystem (except for frictional losses) but at high intake pressures, the mass flow rate through thebypass system was high enough to cause excessive pressure drop across the check valves.3.4.5 ActuationThe class of reciprocating compressors can be divided into categories depending on themethod of actuation of the piston. Three options were studied in the evaluation: hydraulic,pneumatic and mechanical actuation.The hydraulic method of actuation uses a double-acting hydraulic cylinder to impart areciprocating motion to the piston. The system consists of a hydraulic pump driven by the engine,a double-acting hydraulic cylinder, a switch valve that controls the motion of the cylinder and aflow control unit that controls the frequency of the reciprocating motion.30The advantages of this type of system are as follows:• The capacity of the intensifier can be continuously varied by the hydraulic systemand thus the engine fuel requirements can be exactly matched.• Hydraulic systems have been proven to be reliable and efficient.• Hydraulic components are interchangeable.The disadvantages are:• The system is heavy because of the mass of hydraulic components required.• The hydraulic system will require a reservoir for the hydraulic fluid and this willrequire a large amount of space.• The system is expensive (as shown by Aichinger [2]).The pneumatic method of actuation is similar to the hydraulic one, the main differencebeing that the working fluid is air instead of hydraulic fluid. This system seems to have theadvantages of the hydraulic driver and seems to eliminate some of the disadvantages since theurban bus already has a pneumatic system on board used mainly for the braking system.The pneumatic system on the bus can supply air at approximately 110 psi. In order tocompress natural gas to 3000 psi, the area exposed to the air must be approximately 30 timeslarger than the area exposed to the gas. This requirement of a large piston, in addition to theintensifier, results in a large space penalty.Commercially available gas intensifiers were examined by Aichinger [2]. None could meetthe gas mass flow requirement with a typical bus pneumatic system as a source of compressed air.Considering the possibility of upgrading the bus pneumatic system to provide a larger air flow, agas intensifier manufacturer proposed a system that was composed of two gas intensifiers thatwould meet the gas mass flow requirements. This intensifier system was found to be very31expensive even before the cost of the necessary modifications to the bus pneumatic system wereincluded. Also, using two gas intensifiers would require a large amount of space due to therequirement of large air pistons.In short, the hydraulic and the pneumatic actuation systems suffer from the samedisadvantages of being bulky, heavy and expensive. On the other hand, they share the advantageof offering a method of continuously controlling the capacity of the intensifier.The mechanical actuation indicated in Fig. 3.1 encompasses all systems that use a solidlink between the prime mover (engine) and the reciprocating motion of the pistons. There aredifferent methods of giving the pistons their reciprocating motion and three were studied in theevaluation: camshaft actuation, crankshaft actuation and rotating piston.The camshaft actuation would operate in the same manner as the diesel injector, that isthat a plunger would be actuated by a lobe on the camshaft (Fig. 3.14).pr1ngCylinder Head WatiPlungerLas Inta-<e ValveNozzleFigure 3.14Camshaft Actuation32This option is attractive because it would eliminate the need for a separate intensifier.However, detailed study revealed that the size of the cylinder and piston assembly would be muchtoo large to fit in the space designed for the diesel injector. Also, there is the problem of capacitycontrol. Since the cam lift is Constant, the capacity of the camshaft-operated-intensifier is only afunction of engine speed and it would be difficult to install a system that would also make it loaddependent. The small size of the components would also make machining difficult and expensive.These problems make camshaft actuation unattractive.Crankshaft actuation would function in the same manner as the pistons in an engine in thatthe reciprocating motion is provided through a crankshaft and connecting rod assembly. Thereare two different systems that use crankshaft actuation. The first is direct transmission in whichthe compressor piston is itself connected to the connecting rod (Fig. 3.15).This allows for single action compression only. The other type is crosshead transmission(Fig. 3.16). In this configuration, a crosshead is connected to the connecting rod and is guided ina cylinder. The intensifier is then mounted on top of this arrangement and a shaft attached to thecrosshead is connected to the compression piston. This arrangement allows for both single ordouble action compression.The possibility of converting a small internal combustion engine into an intensifier hasbeen examined but it was found that the forces generated by the compression of the gas would bemuch larger than the typical forces generated by a combustion engine so extensive modification tothe engine would be required.33Exhaust ValvePistonConnecting RodFigure 3.15Crankshaft ActuationIntal-<e ValveCranI-<shaft34Ehost VutveCrossheadConnecting RodCro.nkshcxftFigure 3.16Crankshaft Actuation with CrossheadExhQust Vcx[ve-\\ Intake ValvePistonIntake Valve35Another possibility of adapting a combustion engine to become an intensifier would be toreplace the piston by a crosshead and mount an intensifier above it as shown in Fig. 3.16. The useof a double-acting intensifier greatly reduces the loads on the engine and thus makes this use ofthe engine safe.The advantages of using this type of mechanical actuation are:• The design is simple and inexpensive since it uses existing parts.• The crosshead design allows for double acting intensifier which results in a smalland light package.The disadvantages of this system are:• The danger of excessive friction causing wear and heat generation.• Need for some type of capacity control as the capacity of the intensifier is onlydependent on engine speed.The last type of mechanical actuation is the rotating piston intensifier. This intensifierwould function like a hydraulic pump (Fig. 3.17). The reciprocating motion of the pistons issupplied by the fact that the swashblock is not perpendicular to the axis of the shaft to which it isattached. As the angle of the swashblock changes, so does the stroke of the pistons. This allowsfor the attractive advantage of a variable capacity. The problem with the rotating pistonconfiguration would be with sealing. The fact that the pistons are rotating means that any inlet oroutlet for the pistons are also rotating and there has to be a transfer from these rotatingcomponents to the stationary lines or vice versa. In the hydraulic pump this problem is handled byhaving the inlet and outlet ports as slots in a stationary plate at the top of the cylinders. As thepistons and cylinders rotate, the top of the latter line up with these ports at the proper time in thecompression cycle. Sealing between the inlet and outlet port is mostly handled by small tolerances36combined with the viscosity of the hydraulic fluid but there is some leakage. The fact that thenatural gas would be in gaseous form renders this configuration unattractive as the leakage wouldbe considerable and would lead to very poor intensifier performance.Figure 3.17Hydura PVWH open ioop pump. 1-control system, 2-cylinder mounted journal bearing, 3-swashblock lubrication, 4-shaft, 5-shaft bearing, 6-swashblock with bearing, 7-plunger bearings,8-valve plate, 9-valve plate port, 10-through shaft, 11-quiet valve plate design, 12-frameAlthough this design does not seem capable of being a viable solution to the intensifierproblem, it does demonstrate the use of the variable stroke method of capacity control.Of the three mechanical actuation configurations, only the crankshaft actuation seems tobe able to meet the requirements for the intensifier. Its advantages are:• simple design• small and lightweight package• inexpensive system• high efficiency operation.37Its disadvantages are:• external flow control system required in order to have a variable capacity• possible problems with excessive wear and heat generation due to friction.In summary, Table 3.1 provides the relative advantages and disadvantages of the threetypes of actuation:Design Hydraulic Pneumatic MechanicalParameter Drive Drive DriveEfficiency 80-90% 80-90% 90-97%Flow Control variable variable control systemdisplacement displacement requiredSystem Complexity complex complex simpleCost high high lowDurability high high lowWeight and Space large large smallMaintenance Work low low highNoise low high lowTable 3.1Comparison of alternative drive systems (C. Aichinger [2])Hydraulic and pneumatic actuation configurations both have the attractive advantage ofsupplying the intensifier with variable flow but both suffer from the fact that the systems would belarge, heavy and complex. The mechanical actuation would operate at a higher efficiency andwould be lighter, smaller and simpler but would require the implementation of one of the capacitycontrol techniques discussed earlier. The following section will describe the intensifier design ofAichinger, that is the single-stage and the two-stage reciprocating designs.383.5 Previous Intensifier DesignThe first prototype built by Aichinger was a single-stage, double-acting intensifier whichcould produce the maximum required mass flow and pressure ratio. This intensifier wasmechanically actuated using the crosshead arrangement shown in Fig. 3.16. Tests of thisconfiguration showed that the temperature resulting from a 10:1 pressure ratio led to problemswith the sealing at the valves as the materials chosen could not sustain these temperatures. Otherproblems were that given the high pressure loading of the piston, maintaining alignment wasextremely critical and side loads on the crosshead were severe. Because of this, this prototypewas not tested at speeds higher than 200 RPM. Also, the valve losses were high resulting in a lowvolumetric efficiency.The second prototype, the double acting, two stage intensifier, was found to solve manyof the problems encountered with the first design. The two stages meant that intercooling waspossible and this drastically reduced the operating temperature and improved the performance oftemperature-sensitive devices such as valves and seals. The valve design was greatly improvedand this resulted in a much higher volumetric efficiency. The problems with alignment were stillpresent though and this prevented the testing at high speeds (more than 200 RPM)Both prototypes had problems when it came to capacity control. Bypassing, where theexhaust gas if fed into the intake track, was found to produce high gas velocities through thevalve and this resulted in a back pressure and a loss of efficiency. Valve unloading, in which theintake valve is forced opened during the compression part of the cycle, was found to be difficultto implement because of the added complexity and the increase in clearance volume (accompaniedby the decrease in volumetric efficiency).39The previous designs thus did not meet the requirement for a variable capacity but didestablish a good foundation of test experience on which to base a new design.3.6 Choice of Intensifier ConIiurationFrom the design requirements and the study of alternative design concepts, the firstdecision concerning intensifier configuration was to choose a reciprocating machine. From theprevious sections on volumetric efficiency, capacity control and actuation, the following can beconcluded:1. Multi-staging has beneficial effects on volumetric efficiency and thus tends to reduceintensifier size and the pressures forces felt by intensifier components.2. Varying the stroke of the intensifier is an energy efficient technique to control the capacity ofthe intensifier.3. Mechanical actuation is a space- and cost-efficient method of transferring energy from theengine to the intensifier.With these facts in mind, and with knowledge gained from previous intensifier designs, thefollowing intensifier configuration was chosen: a mechanically actuated, multi-stage, variablestroke reciprocating intensifier.The number of stages will be determined in the design optimization described in chapter 4.In the previous sections, the advantages of variable stroke operation were explained but themechanism used to vary the stroke was not described. In the following section, a new designconcept for achieving variable stroke operation in a compact mechanism is described.403.7 Rotary-Reciprocating Intensifier TerminologyFig. 3.18 and 3.19 are simplified schematic of the rotary intensifier.58Figure 3.18End View of Rotary-Reciprocating Intensifier41Figure 3.19Side View of Rotary-Reciprocating IntensifierThe basic working element of the rotary intensifier is a piston (1) sliding inside a cylinder(2). As seen in the compression cycle, each cylinder has an intake (3) and an exhaust valve (4).In the configuration shown, there are four cylinders, two of which act as a first stage (5), theothers as second stage (6). Between these two stage, there are intercooling cavities (7). Theconnecting rod (8) has at one end the piston head (1) and at the other the rollers (9). The17I.3159161342connecting rod is guided into a strictly linear motion by a linear bearing (10) All the componentslisted above are referred to as the rotor as it this assembly that spun by the engine.Attached to the rotor and rotating with it are two shafts. The first is the drive shaft (11)through which power is transmitted from the power sources to the rotor. Also on this shaft is theoil seal (12) through which oil enters the rotor. The second shaft is the gas shaft (13). On thisshaft there are gas seals (14) through which the intake gas enters the rotor and the exhaust exits.The gas and oil seals are held in place by the frame (15) which also houses two bearings (16)which hold the rotor in place. Finally, the frame guides the motion of the outer ring (17) whoseposition is controlled by the eccentricity control mechanism (18).Those are the main components of the rotary intensifier. Their functioning will beexplained in more detail in the following section.3.8 Rotary-Reciprocating Intensifier OperationThe easiest way to describe the operation of the intensifier is to follow a particle of gas asit goes through the intensifier and gets compressed. The particle first enters the intensifier at theintake gas seal. These seals are simple 0-ring type mechanical seals which seal between thestationary intake gas lines and the rotating gas shaft. The gas then enters the shaft through twoholes and follows a path through the shaft into the rotor, and more specifically, the first stageintake valves. These valves admit gas to the first stage cylinders during the intake phase of thecompression cycle. It is worthy of note that there are two paths that the gas could have takensince there are two first stage cylinders, one opposite the other in the rotor.Once compressed by the first-stage piston, the gas leaves the first stage through theexhaust valves into the intercooling cavities. After being cooled, the gas enters the second stage43cylinders and the pressure of the gas is further increased. Finally, the gas exits the second stageand follows a path through the gas shaft to the exhaust gas seals, which function in the samemanner as the intake gas seals.So far, apart from the fact that the piston and cylinders are rotating, nothing has beendescribed that makes the rotary intensifier different from a conventional reciprocating compressor.What sets this design apart is its variable stroke. The outer ring can be moved in such a way thatits center does not coincide with the axis of rotation of the rotor. The distance between the axisof rotation and the outer ring center is referred to as the eccentricity.When the eccentricity is set at zero, in other words the axis of rotation of the rotor and thecenter of the outer ring are superimposed, the pistons do not have any reciprocating motion, theysimply stay at the mid point of the cylinders and do not compress the gas at all (fig 3.20).Figure 3.20Intensifier with Zero Eccentricity44On the other hand, when the eccentricity is set at half of the maximum stroke, the pistonshave the largest reciprocating motion and offer maximum compression to the gas (fig 3.21). Ofcourse, the eccentricity can be set at any point between these two extremes and thus the amountof compression done by the intensifier is controllable.Figure 3.21Intensifier with Maximum EccentricityIt should be noted that although the rotary intensifier was always shown with two stagesand four cylinders, it is not the only possible configuration. In fact, there are many variables thatmust be considered in order to obtain the final intensifier design. The next chapter will examinethe design process and show how the final design was chosen.45Chapter 4Design Optimization4.1 Summary of Design ReiuirementsThe principal design requirements for the intensifier identified in the second chapter are:• 200 bar exhaust pressure• 20 to 200 bar intake pressure• variable mass flow in the range of 50 to 500 mg/rev• low size and weight• completely leak free• long life• low production and maintenance costsThe prototype has additional requirements that come into consideration because only oneunit will be produced. The first factor that affected the design of the prototype was speed ofconstruction. Since only one unit will be produced, it is not time efficient nor economical to havespecial tooling developed for the machining of the prototype. Also, techniques such as molding,stamping or casting were avoided since it would not be economical to use these techniques for theprototype even though they may prove to be a wise choice for the production model. This oftenresulted in components that were heavier or bulkier than required. This fact contradicts the46requirements of low size and weight but for the production model, more elegantly machinedcomponents would meet this requirementThe prototype was designed with easy assembly and disassembly in mind. Most of thepieces are bolted together and there is minimal welding of components. Also, the foreseenproblem areas were made easy to access without having to disassemble the rest of the intensifier.The safety factor also had to be considered in the design of the prototype. For this reason,some components, such as the outer ring, the frame plates and the main bearing housings, weredeliberately over designed.It was deemed economical to use the same test rig, that is the same electric motor andmounting frame, as the intensifier versions 2.1 and 1.2 and this affected the design. The overallsize, the mounting brackets and the position of the drive shaft were strongly affected by thisconsideration.In summary, the additional design requirements for the prototype version are:• speed of machining• easy assembly and disassembly• high safety factor• use of existing test rigThese are the requirements that the intensifier prototype had to meet. The techniquesused to incorporate these requirements into the design process will be examined in later sections.The design process itself will now be described and the terminology that will be used throughoutthis chapter will be defined.474.2 Description of Design ProcedureThe number, size and position of the intensifier components could not be chosen arbitrarilysince this would yield an intensifier which either does not meet the design requirements or does sowith waste of resources. The ultimate goal of the design process is to produce an intensifier thatmeets all the requirements with minimum use of resources. The optimization technique will bedescribed by examining the various elements of the process.The system parameters are the variables that are dictated by the operating conditions.These parameters cannot be changed in the optimization process. For example, the intake andexhaust pressure of the intensifier are system parameters since they do not vary with changes inthe design.The design variables are the variables that describe the number, size and relative positionof the various components of the intensifier. The design variables are the ones that are varied inan attempt to reach an optimum design. For example, the bore and stroke of a given compressionstage are design variables.The design constraints limit the range of the design variables. These constraints originatefrom the design requirements, from specific component limitations or from practicalconsiderations. The required mass flow rate is an example of a design constraint originating fromthe requirements. A maximum rolling velocity of a bearing is a constraint that arises from acomponent limitation and a minimum size for a given piston would be a practical constraint.In order to evaluate the optimality of given values of the design variables, an objectivefunction must be defined. The objective function is a statement of the general goal of theoptimization process. In the simplest case, trying to find the lightest intensifier design could be anobjective function. This function is usually in the form of a mathematical equation that links the48design variables to the quantity described in the objective function (such as weight for thisexample).In most design problem, there is not a single quantity to be maximized or minimized whilesatisfying the constraints. Trying to find the lightest intensifier with maximum efficiency would bean example of an objective function that contains two quantities. Obviously, it is highly unlikelythat both these conditions are met simultaneously at a design point. For this reason, compromisesmust be made when such a function is used. The terms minimum and maximum will still be usedbut they will not have the absolute definition they usually have. Instead, the term minimum will bedefined as “the lowest possible value for a quantity consistent with reasonable values for the otherquantities” and similarly for the term maximum.The quantities that form the objective function are often taken from non specific designrequirements such as low size and weight or long life. The four elements of the optimizationprocess defined above will now be examined in greater depth.4.3 System ParametersSome of the requirements can be transformed into system parameters. These arequantities that do not vary and are set by the operating conditions. The system parameters for thecase of the intensifier design are listed below:• exhaust pressure (200 bar)• minimum intake pressure (20 bar)• maximum required flow rate (500 mg/rev)• operating speed (engine speed)• natural gas properties49aluminum and steel propertiesIt is to be noted that system parameters such as intake pressure and mass flow rateactually vary during the operation of the intensifier but it is the extreme cases listed above thatwill be used during the design, the off-design cases being handled by the variable stroke. Thereason that the extreme case of minimum intake pressure and maximum flow rate was chosen as adesign point is that it is at this point where the intensifier requires the most power and thus thehighest torque. Furthermore, the intensifier is a full eccentricity at this point which results inmaximum contact angle between the roller bearings and the outer ring surface. This maximumcontact angle results in maximum side loads on the piston connecting rod which makes the linearbearing constraints important.From the point of roller bearing loads, the worst case occurs at the highest intakepressure, 200 bar. This point was not used as the design point since the eccentricity would bezero at this point which results in zero side loads on the piston connecting rod. The high rollerbearing loads at this point were taken into account by verifying that a given bearing size couldsustain these maximum loads.44 Objective FunctionThe objective function for the intensifier has four main quantities that are taken from thenon-specific design requirements. Since there are more than one quantity in the objectivefunction, the terms minimum and maximum will be used with the definition given in section 4.2 inmind.The first quantity in the objective function is size which will be minimized. Thisrequirement stems from the fact that the smallest intensifier that meets all the other requirements50will be the most attractive since the eventual production model will conceivably have to functionin the restrictive space of an engine bay.The design also seeks a minimum weight intensifier. This part of the objective functiongreatly affects the choice of materials used and usually works in concert with the low sizerequirement.The third quantity, that of cost, will also be minimized. The cost of the intensifier includescost of the material, machining cost, which is related to machining time, and use of existingresources such as the test rig.The last quantity in the objective function is one that will be maximized and is the life ofthe intensifier. This often translates into having a high safety factor for components such asbearings since these components have a life that decreases as they are loaded nearer to theirmaximum load.As previously mentioned, it is impossible to minimize or maximize four quantitiessimultaneously. In principle, the four quantities would be assigned a weighting factor andcombined into one equation. In practice this was not done because it was difficult to determinethe relative importance of the four quantities before starting to design. Also, the relativeimportance of the quantities varies depending on the situation. In fact the four elements of theobjective function do not always work in harmony. For example, trying to achieve long life in abearing results in opting for a larger bearing which contradicts the minimization of size, weightand cost. In these cases, a decision as to the relative importance of the various elements of theobjective function must be made. This decision is greatly affected by the situation. For example,in the case of the design of a bearing bracket which is outside of the working components of theintensifier, the minimum machining time (minimum cost) consideration would be the dominant51deciding factor and low size and weight would be temporarily ignored. On the other hand, thedesign of a piston will definitely emphasize low weight since the piston affects many othercomponents in the intensifier.The objective function then permits the designer to define the optimality of a particularconfiguration. This configuration is determined by the value of the design variables which willnow be described.4.5 Design VariablesThe design variables are the variables which have to be determined. They range fromparameters such as number of compression stages to the sizes of individual parts. The designvariables will now be listed and the considerations that come into effect when determining theirvalues will be discussed in the following sections.The design variables that have the greatest effect on the final configuration are:• number of stages• number of pistons per stage• stroke of the pistons• bore of the pistons• rod diameter• linear bearing length• roller bearing length and diameter• outer ring diameterThese design variables are interdependent, that is that a change in one of these variablesrequires a change in several others. For this reason, they have to be determined simultaneously.52Decisions as to the sizing of components such as frame supports or bolts are made after thesevariables are determined.4.5.1 Number of StagesThe compression of a gas through a 10:1 pressure ratio could conceivably be handled by asingle cylinder though earlier experiments by Aichinger [2] showed that this intensifierconfiguration led to excessively high operating temperatures. The advantage of multistaging isthat it adds the possibility of cooling the gas between the stages, which reduces the maximumoperating temperatures and compression work. This has the effect of increasing the life of mostcomponents such as seals and bushings. The disadvantage of multistaging is the added complexityand space required by the intensifier.4.5.2 Number of PistonsEach compression stage can consist of a single or multiple pistons. Increasing the numberof pistons results in a higher complexity and production cost but can have many advantages. Oneadvantage is that if one of the components of a given piston were to fail, the intensifier could stillfunction, albeit at a reduced capacity. In the vehicle application, it could mean that the vehiclewould be able to reach a service station where the intensifier could be repaired. Anotheradvantage which is particular to the rotary intensifier is ease of balancing the rotor. If pistons of asame stage are placed in an opposing manner, then there are greatly reduced forces acting on theshaft of the rotor. This can result in smaller components and reduced weight. Finally, a largernumber of pistons reduces the pressure forces felt by each individual piston so components suchas the roller bearing and the linear bearing are not as severely loaded.534.5.3 StrokeFrom preliminary design experience gained with the simulation program, the stroke of thepistons was found to be a key design variable. Because of the rotary configuration, the stroke ofall the pistons is the same. In the design phase, the term stroke is used to refer to the maximumstroke of the intensifier; the fact that it varies to fulfill the variable capacity requirement is of noconcern since most of the design centers around the worst case, that is maximum pressure ratioand maximum capacity.The stroke is key because it affects many other design variables. For example a reductionin stroke requires an increase in piston bore in order to maintain the same maximum flow rate.The larger bores result in larger pressure forces acting on the pistons. On the other hand, anincrease in stroke has the effect of moving many components away from the center of rotation andthus increasing centrifugal forces. It is desirable to find the point where the combination ofpressure and centrifugal forces is minimum since the sizing of the rod and the roller bearingsdepend on these forces. For this reason, a large part of the optimization process revolves aroundthe determination of the stroke.4.5.4 BoreThe bore of the pistons is closely related to the stroke and the number of pistons throughthe requirement of capacity. The most important effect of the bore is that pressure forces increasewith an increase in bore. Also, large bores require a large rotor and leave less space for theintercoolers. Large bores are also unattractive because they increase the clearance volume that isdue to the space left between the top of the piston and the head of the cylinder. There is a54miiiimum size for the bores which comes from machining considerations and availability of pistonseals.4.5.5 Rod DiameterThe rod diameter has two factors determining its value. The load carrying capacity of thelinear bearing that guides the rod depends on its length and the diameter of the rod. A larger rodpermits larger side loads, but has the negative effect of generating larger centrifugal forces.The second factor that affects rod diameter is the possibility of pressurizing the back sideof the piston with interstage pressure. This has the beneficial effect of counteracting the pressureand centrifugal forces. Since the back side pressure acts on an area equal to the area of the pistonminus the area of the rod, the beneficial effect is reduced with a larger rod. Also, this back sidepressurization needs a rod seal which reduces the length of the rod available to the linear bearingor increases the size of the rotor.4.5.6 Bearing Diameter and LengthThe size of the bearings is mostly determined by the pressure and centrifugal forces. Theproblem is that as the bearings become larger to accommodate larger forces, they generate largercentrifugal forces themselves. The mass of the bearings is an important factor since they are thecomponents that are situated the farthest from the center of rotation. Also, larger bearings have alower allowable rotating speed. For these reasons, a balance between the load carrying ability andthe mass and the allowable speed of the bearing must be found by studying the effect of varyingboth diameter and length.554.57 Outer Ring DiameterThe negative effect of large centrifugal forces has been noted in the preceding discussionof the design variables. Furthermore, a large outer ring results in a higher cost for the materialsand the machining. For these reasons, the outer ring diameter should be kept at a minimum.Factors that tend to increase the outer ring diameter are: longer stroke, longer required length ofthe linear bearing, larger diameter bearings and more required space in the rotor to accommodatelarge amounts of pistons.It was shown that the determination of the design variables is not trivial. The designvariables described above are all interdependent and it is not always clear if the changing of onewill have a positive or negative overall effect. The last element of the optimization process, thedesign constraints, will now be examined to see what limits are placed on the design variables.4.6 Design ConstraintsThe values for the design variables cannot be chosen arbitrarily. Not only do the values ofthe design variable affect the optimality of the design, they are also bound by the designconstraints. These constraints are linked to the requirements or originate from limitations ofparticular components. There are also many practical constraints such as the andmaximum sizes for the components but these do not need to he elaborated in detail. The fiveconstraints that most affected the design will now be discussed.4.6.1 Roller Bearing Load and SpeedThe first two constraints have to do with the roller bearings that roll on the inside of theouter ring and are connected to the piston through the rod. These bearing are loaded by the56pressure forces acting on the piston head which are transmitted down the rod. Also, thesebearings must support the centrifugal forces generated by the rotating piston, rod and bearingsthemselves.The maximum loads that these bearings can support depend on the bearing type, diameterand length. Since preliminary calculations had shown that these bearings would be heavily loaded,the type of bearing chosen was the one that would be best suited to high loads. After consultationwith INA (a bearing manufacturer) engineers, the type chosen was the unit cage needle bearing.These bearings consist of a cage holding the rolling elements in place. The races for the bearingare supplied by the shaft and casing that support them, both of which must be ground andhardened. Since other types of bearing have races made of drawn steel (which is the weakest linkof the bearing), it is this lack of built-in races that make this bearing type suitable for high loads.Also, the unit cages are small compared to full bearings which is an advantage because a smallbearing results in low centrifugal forces.The maximum load still depends on diameter and length of the bearing and it would slowthe design process considerably if commercial bearing specifications had to be consulted at everydesign iteration to verify if the bearings are overloaded. After examining commercially availablebearing specifications [5,6J, it was found that the load bearing capacity of the unit cages isapproximately proportional to both diameter and length. This can be seen in figure 4.1 and 4.2.For a first approximation, the load bearing capacity was reduced to two linear equations whichcan be seen on the graph which were used to accelerate the design process. The square points onthe graphs are data points taken from the bearing specifications and the straight lines are the linearapproximations used in the design process.57The second design constraint that is generated by the roller bearings has to do with theirFigure 4.1Linearization of Commercial Bearing Data [5,6]Figure 4.2Linearization of Commercial Bearing Data [5,6]Bearing Allowable Load vs DiameterLoad allow.per mmof length(kN) 6 10 14 18 22 26Bearing Diameter (mm)30Bearing Allowable Load vs LengthLoad allow.per mmof dia(kN)6 10 14 18 22 26Bearing Length (mm)30 34maximum allowable speed. The rollers will be kept as small as possible to reduce the centrifugal58forces and the outer ring has a diameter that is an order of magnitude larger than that of therollers. These two factors result in considerable angular speed for the bearing (of the order of 10-15 times engine speed); the maximum allowable speed depends on the type of bearing and itsdiameter. Since the unit cages were also well suited to high speeds because of their relativelysimple construction, there were no reasons to search for a different type of bearing. Afterexamination of the commercially available bearing specifications [5,6], it was found that theallowable speed decreased approximately linearly with the diameter of the bearing (figure 4.3).As with the case of allowable load, the allowable speed was reduced to a linear equation.[5]:Figure 4.3Linearization of Commercial Bearing Data [5,6]The life of the bearing is affected by both load and speed as can be seen from this formula10/16666 NLb=In2523Speed allow. 21RPM19171513Bearing Allowable Speed vs Diameter141618202224262830 32Bearing Diameter (mm)59Where Lb is the life of the bearing in hours, n is the speed in RPM, F is the load in N and Cis the basic load rating of a given bearing type and size (listed in INA catalogues [5,6]). A C/Fratio of 6 is the minimum recommended. It can be seen that if the bearing is close to themaximum load (C/F=6) and the speed is of the order of 20000 RPM, the bearing would have a lifeof approximately 300 hours. This is very short compared to the life of a typical diesel engine. Forthis reason, it is important to try to remain as far as possible from the maximum load and speedcases. In Fig. 4.10 and 4.11, a C/F ratio of 6 is used to determine the maximum allowable load.These two constraints are satisfied when the following guidelines are observed: the forcesacting along the shaft of the rod must be as small as possible, the outer ring diameter must be assmall as possible to reduce centrifugal forces and reduce the angular velocity of the rollerbearings.4.6.2 Linear Bearing Load and Sliding VelocityThe third and fourth design constraints are also supplied by a component limitation. Thiscomponent is the linear bearing that guides the rod. This linear bearing is loaded by the side loadsgenerated by the fact that the roller bearings are not rolling on a surface that is perpendicular tothe rod axis. These side loads must be supported by a linear bearing because the piston ringswould not be able to support them.The first design constraint provided by the linear bearing is a sliding velocity constraint.Most linear bearings have an upper limit on the speed that the rod they are guiding can slide pastthem. After some preliminary calculations, it was found that the Teflon bushing by Permaglide [7]would be suitable. This bushing has been used in the previous intensifier design with satisfactory60results. It also offered a load carrying capacity comparable to more complex linear bearing in avery small package.This type of bushing has three types of limitations. The first is a maximum value of thesliding velocity (v). The second is a maximum value of the ratio of the force and the projectedarea of the bushing (length times diameter) which is referred to as p. Finally, the last limitation isa maximum value of the product of the first two quantities (pv). After preliminary examination ofthe bushings, the force-area ratio (p) was found to be always well within the allowable range. Forthis reason, only the other two limitations will be considered in the analysis [7].These bushings have a maximum sliding velocity of 3 rn/s. Since the rod is traveling twicethe stroke every revolution, it can be found that this constraint limits the stroke at approximately43 mm. This was found with an intensifier speed of 2100 RPM and averaging the piston speedover the stroke. In fact, this stroke would generate a peak velocity higher that the allowablevelocity of the bushing so the stroke is limited at 40 mm.The second design constraint provided by the bushing is a combination of load per unitarea and sliding velocity. The Permaglide bushing has an allowable pv value of 3 (N/mm2)(m/s).This constraint affects the size of the bushing. It is advantageous to have a small bushing becausethis results in a shorter rod, a smaller outer ring diameter and thus lower centrifugal loads.These two constraints are best satisfied in the following cases: when the stroke is as shortas possible and when the forces acting along the rod (which translate into side loads on thebushing) are as low as possible.614.6.3 Required Mass FlowThe required capacity constraint comes directly from the design requirements. It is animportant constraint as it sets the required displaced volume of the various compression stages.The method used to determine the required displaced volume was to use the computer program torun different configurations of intensifier.4.7 Design ProcedureAll the elements of the optimization process having been identified, the following sectionswill describe the determination the design variables which will satisfy all the design constraints andyield the best objective function. First the tools used in the analysis will he examined.4.7.1 Design ToolsThe task of determining the design variables was done with two main tools. One was thespreadsheet program Lotus 123. Using this program permitted convenient graphical examinationof the effects of varying the design variables. Some choices were simple to make simply bypinpointing the minimum on a graph. The use of this software sometimes required someapproximations which will be discussed as they are used in the analysis. The insight gained by theuse of the second design tool, the simulation program, was invaluable in helping to make theseapproximations as realistic as possible.The simulation program is a Quick Basic program that simulates the functioning of arotary intensifier and can be found in Appendix I. The program calculates the pressures andtemperatures in all the stages and the intercoolers at each degree of crank shaft rotation (0). With62this information and a knowledge of how the check valves function, the mass transfers throughthe stages can be calculated.The simulation program used the following approximations. To find the pressures in thecylinders, polytropic compression and expansion are used:i1+1Pwhere the subscript i denotes a given degree of rotor rotation and the volume V is determinedfrom geometry for each degree of rotor rotation. The polytropic coefficients used were typicalcoefficients found in the previous intensifier design. The values used are n = 1.25 for compressionand n = 1.4 for expansion. This is consistent with the discussion of polytropic coefficients insection 3.4.1. Of course, these coefficients vary with pressure ratio and temperatures, but it wasbelieved that this method would yield more realistic results that the assumption of isentropiccompression and expansion. To find the temperatures from the pressures, the ideal gasrelationship was used.The simulation program functions as follows. The first step is to set all the intensifierparameters such as bores, stroke and intercooler volume. The pressures and temperaturesthroughout the intensifier are then set at intake conditions. This simulates the actual start-up ofthe real intensifier. The volumes of both stages for each angle 0 are then calculated and placed inan array. The program then starts a loop which goes through a complete revolution one degree ata time.From the volume at angles 0 and 0+1, and the initial pressures, the pressures in bothstages are calculated from polytropic compression and expansion for every angle 0. The way the63check valves are handled is that the pressures on both sides of the valve are compared and whenthere is a pressure difference in the correct direction, the valve opens and pennits gas to betransferred. The pressure in the interstage is calculated from its volume, temperature and theinflow from the first stage and outflow into the second stage. The temperatures are calculatedfrom the ideal gas law in the stages.The program also has a simple heat transfer model. This model accounts for the heatingand cooling of the gas by the rotor. A bulk temperature for the rotor is set and the temperaturesin the stages and the intercooler are modified using the following equation:T= C(TbIk—T)+Tin which T is the temperature which is modified (a stage or interstage temperature), C is aconstant and Tbjk is the chosen bulk temperature. Tblk was set at 80°C which was a typical bulktemperature of the previous intensifier versions and C was chosen so that the operatingtemperature of the intensifier were similar to the results obtained by Aichinger [2].At the end of a revolution, the mass flow through the first and second stages arecompared. Steady state is said to occur when these mass flows are the same. if they aredifferent, the program goes through another revolution. Since mass flow through the stages ismostly dependent on interstage pressure, this definition of steady state means that interstagepressure has also reached a steady state value. This technique proved to yield very good resultsas this predicted value of interstage pressure matched the measured value almost exactly.The simulation program also calculates the forces felt by the roller bearings at every angle0. This is done from knowledge of the pressure in the cylinder and the centrifugal forcesgenerated by the components such as the roller bearings and connecting rod. With a knowledge64of the contact angle between the roller bearing and the outer ring (which can be found fromgeometry), the torque felt by the rotor can be calculated. This torque was used to calculate thepower required to run the intensifier.The principal use of the program was to make possible the simulation of many differentintensifier configurations. The program also gave the design an understanding of theinterdependence of the design variables without having to build any prototypes. This was usefulin the preliminary design stage when a design strategy was formulated. The process of arriving atan optimal design will now be described4.7.2 Number of Stages and CylindersIn this section, two design variables will be determined: the number of stages and the totalnumber of cylinders. The main design constraints that come into effect when determining thesevariables are practical ones: there are no particular components which limit the choices.It is advantageous to have more than one compression stage for two reasons. First,muhistaging decreases the pressure ratio across a particular stage which has a beneficial effect onthe volumetric efficiency of that stage. Second, multistaging permits the use of intercoolers whichcool the gas between the compression stages. This results in lower operating temperatures andlower work of compression. The lower operating temperatures increase the life of componentssuch as seals and valves. The disadvantage of having more than one compression stage is theadded complexity.The number of cylinders depends on the number of stages and the number of cylinders perstage. There are two advantages of having more that one cylinder for a given stage. First, therequired displaced volume of a given cylinder is smaller which results in smaller areas exposed to65the pressure. The smaller pressure forces result in smaller components, lower centrifugal forcesand often longer predicted life of the components. Second, for the special case of the rotaryintensifier, the cylinders of a given stage can be arranged in an opposing manner which tend tobalance the forces acting on the shafts of the rotor. Also, these opposing forces tend to smoothout the torque curve of the intensifier. Again, the main disadvantage of having more than onecylinder per stage is the added complexity. Also, if there are too many cylinders in the rotor,there is little room left for the intercooling cavities which reduces their heat dissipation capabilityand one advantage of multistaging is lost.The compromise which was found between the advantages of multi staging and cylindersand the disadvantage of the added complexity was the following: two stages of two cylinders each(figure 4.4).Figure 4.4Stage and Piston ConfigurationThe four cylinders were arranged in a cross configuration with the cylinder of a givenstage opposing each other. The use of two compression stages has been proven effective in the66intensifier version 2.1 which had operating temperatures which were quite manageable. Also, aswill be shown in the next section, this choice for the number of stages and cylinders resulted incylinder which were of a size that did not present any machining difficulties. This configurationalso left plenty of room for the intercooling cavities.4.7.3 Design of the First and Second StageThis section describes the core of the design work. Sizing the various parts of the twostages proved to be an iterative process. The first step in this process was to simply find a designpoint, that is a consistent set of values for the design variables that did not violate any of thedesign constraints. Working with the computer simulation and spreadsheets, investigation of theeffects of varying the design variables around this point proved to be very valuable to obtain anunderstanding of the interaction between the design variables, the quantities of the objectivefunction and the constraints.In this section the determination of seven design variables will be discussed:• stroke• bore of the first stage• bore of the second stage• rod diameter• linear bearing length• roller bearing diameter• roller bearing length• outer ring diameter67The design constraints that will come into play are the linear bearing speed and load limits,the roller bearing speed and load limits, the required mass flow and practical constraints such asthe minimum machinable size of a piston. The list of known variables, unknown variables,equations and information extracted from the computer simulation will now be developed.The geometric variables are shown in figure 4.5 below, except for the length of the rollerbearing L3 since it is in the perpendicular direction.The variables used in the design process are classified in Table 4.1 as known, unknownand limited by the design constraints.Figure 4.5View of Geometric Variables.68VariablesKnown Unknown Limited by ConstraintsSym Description Sym Description Sym DescriptionP1 intake pressure S stroke th1 1st stage mass flowT1 intake temperature B1 1st stage bore th2 2nd stage mass flowpi intake density B2 2nd stage bore Fb 1st stage rollerbearing forceP3 exhaust pressure D1 connecting rod Fb2 2nd stage rollerdiameter bearing force0), angular velocity of the L1 connecting rod length ui, roller bearingrotor angular velocity1st stage clearance Db roller bearing PLI 1st stage linearratio diameter bearing force overareae2 2nd stage clearance I roller bearing length PL2 2nd stage linearratio bearing force overareaCi distance from top of D0 outer ring diameter V linear bearing slidingcylinder to center of velocityrotationC2 length of shaft seal L11 linear bearing lengthn isentropic coefficient 1st stage volumetricefficiencyR gas constant for NG 11v2 2nd stage volumetricefficiencyp density of aluminum P2 interstage pressurej density of steel p interstage densityTb& bulk temperature of T2 interstagerotor temperaturea contact angleTable 4.1Design Variables69The governing equations for the design process are:Mass flow through the first stagethIlvl.p1.B12.S 4.1Volumetric efficiency of the first stage (from section 3.4.2)Etp2 1T)vl = E—1j 4.2Mass flow through the second stagern2=v2 p2—B2•S 4.3Interstage densityP2p2= 4.4RT2Volumetric efficiency of the second stage (from section 3.4.2)r/P311v2 = 1 8 4.570Forces on the first stage roller bearing.2 2 ;i; 2( LrFb1=P2--B1 +w paLr--Dr iS+Ci+j+wpLD(S+C+L)4.6The first of the three terms on the right hand side is the pressure force acting on the face of thepiston. The second term is the centrifugal force generated by the rod and the third term is thecentrifugal force generated by the roller bearings. The last two terms are of the form ofw2p(volume)(distance from center of rotation).Forces on the second stage roller bearings./ Lr”Fb2P3””B2+W pQLr”” Dr2 jS+Ci+4 4 24.7+w2 ps.Lb..Db2.(S+C1+Lr)This equation has the same form as eq 4.6.Speed of the roller bearingsD0Wb=W— 4.8DbForce-area ratio for the linear bushing of the first stageFbv sin(ct)pLl= 4.9DrLth71Force-area ratio for the linear bushing of the second stageFb2 sin(a)pL2= 4.10DrLlbAverage sliding velocity on linear bushing(1)V=—2S 4.112nLength of connecting rodLrLIb+S+C2 4.12The information extracted from the simulation program can be expressed in the form of functionalrelationships forinterstage pressureP2= f(S,B1,B2,Ei,2,P1,P3,T1,Tblk), 4.13interstage temperatureT2 = f (S,B1,B2,8 i,E 2, P1,P3,T1,Tbi), 4.14contact anglea = f(Do,Lr,Db,S,C1). 4.1572This set of equations and unknowns can easily be reduced. By combining Eq 4.2 with Eq4.1 and combining Eq 4.4 and Eq 4.5 with Eq 4.3, three equations and three unknown can beeliminated. Similarly, by combining Eq 4.12 with Eq 4.6 and Eq 4.7 another equation andunknown are eliminated. Finally, the three quantities found from the computer simulation canessentially be treated as equations since the values found only depend on geometric variables(such as stroke, bores, outer ring diameter etc.). By combining these three quantities with theequations where they are used, the set is further reduced by three unknowns.The reduced set is thus 8 unknowns:S, B1, B2, D[, L, Db, Lb, D0,and 8 equations which are:mass flow for first stageth1={1_e9_1].P1.B12.S 4.16mass flow for second stage4.1773force felt by first stage roller bearings(s+1+Llb+S+C2Fb1=P2—B12+wpLlb+S+C2—D’4 4 2 ) 4.18+w2 ps.Lb..Db2.(S+C1+LIh+S÷C2)4force felt by second stage roller bearings3’:Fb2P3B2+W paLlb+S+CDr2(S+Cl L1h+52)4 4 24.19+w2 ps.Lb.Db2.(S÷C1÷Lib+S+C2)4roller bearing angular velocityD0WhW 4.20Dbforce over area felt by first stage linear bearingsFblsin(a)pLl = 4.21D1LIbforce over area felt by second stage linear bearingsFbzsin(ct)pL2 4.22DrLtbaverage sliding velocity on linear bearing.(0V=—2S 4.2323’:74These quantities given by these equations are quantities which were described in thedesign constraints. The mass flow rate design constraint makes the two mass flow rate equationstrue equalities: the mass flow rate must be a single set value for the constraint to be satisfied. Therest of the constraints make the 6 other equations inequalities. In other words, any force felt by agiven bearing that is smaller or equal to the maximum allowable force for that bearing isacceptable and the constraint is satisfied. The same is true of the sliding velocity and angularvelocity equations.Since there are 8 unknowns, 2 equality constraints and 6 inequality constraints, the designproblem has six degrees of freedom. Since six degrees of freedom makes this problem difficult tosolve graphically, one of the unknowns will be chosen before starting the optimization. The valuefor this unknown, Lb, was chosen from experience with previous intensifier design and afterrunning many simulations to get an understanding of the effect of these variables on the design.Furthermore, an additional consideration was used to reduce the problem to a four degrees offreedom problem. As will be shown, the torque curves (torque versus 0 found with the simulationprogram) were used to determine the best ratio of the first and second stage bores. Thedetermination of the 8 unknowns will now be described in detail.The procedure used to find the optimal point is shown by the flowchart in Fig. 4.6 anddescribed by the subsequent discussion of the procedure. It should be noted that the rectangleswith rounded corners in the diagram represent the points at which a value for a particular designvariable was chosen. The reason that there are only seven decision boxes while there are eightunknowns is because one of the decision boxes contains two unknowns (the two bores). All otherboxes represent steps in the calculations that were needed to make these choices. The fact that anunknown was set by practical considerations can also be seen.75Choose area ratio with torque curvjrt up spreadsheet with stroke as a variable (column][icuiate forces acting on first stage roller bearings]__1____Pressure FoJ Rod Centrifugal Forces Bearing Centrifugal Forces]Calculate bore from area ratio, I Set up rod diameter as second (Choose roller bearing length 1stroke and recuired mass flo] spreadsheet variable (row)] Lfrom practical considenJ____4____Find intercooler pressure I F Choose linear bearing length 1 [Set up bearing diameter as thirdwith computer simulatiJ L from practical considerationsJ spreadsheet variable (depth)__________4Calculate pressure forces from rElculate rod length with stroke 1ise bearing diameter from 1bore and intercooler pressure and linear bushing Pv limitatioJ Loptimal speed consideratiousjCalculate rod mass [Calculate bearingmjCalculate rod distance from [icu1ate bearing distance fromcenter of rotation from center from stroke and rod len]stroke and rod lenghtCalculate centrifugal forcesgenerated by bearingsCalculate centrifugal forcesgenerated by rodfrom angular velocityChoose rod diameter from optimal bearingload and speed considerationsChoose stroke from optimalLbearing load and speed considerationsj[The of first and second stage are set by stroke,1L bore ratio and required mass flow[ise minimum outer ring diameter formL stroke, rod lenght and bearing diameJFigure 4.6Design Procedure76The first step in determining the eight design variables was to determine the ratio of thedisplacement volumes of the two stages. Since the stroke is the same for the two stages, this isthe same as the ratio of piston areas. The effect of varying the area ratio will now he examined.A large first stage area and small second stage area results in a large portion of the workbeing done by the first stage which results in a high intercooler pressure. The high intercoolerpressure results in a higher average pressure in the first stage and this, combined with the largefirst stage area, results in large forces generated in the first stage. The average pressure in thesecond stage is also higher but the area of this stage is small. If the area of the first stage islowered and the second stage increased, the intercooler pressure drops which means that theforces in the first stage are lowered. At the same time, the second stage area has increased whilethe average pressure in this stage has decreased.It is advantageous to have similar forces in both stages, as this results in a smooth torquecurve, (i.e., torque vs crank angle). A smooth torque curve results in an intensifier which is easierto run. While this is not a major consideration if the intensifier is run by a diesel engine, it wouldbe important if the intensifier was driven by a small electric motor. Also, a smooth torque reducesthe load variations through the drive shaft and rotor which in turn increases the life on theintensifier before fatigue failure. It is thus worthwhile to study the effect of area ratio on theforces generated in the stages and thus the torque curve.It is very difficult to reduce the torque curve to an equation as the pressure in all fourcylinders varies with time. For this reason, the computer simulation was used to obtain theoptimum bore ratio. This was done by choosing three strokes that are in the expected range ofthe final stroke. These strokes are 19.05 mm, 25.40 mm and 31.75 mm. The first and second77stage bores were chosen in such a way as to provide the required mass flow. The bores were thenvaried in order to yield a different area ratio yet still provide the required mass flow.The first attempts made showed that the second stage forces were considerably larger thanthose of the first stage due to the larger pressures. This made it difficult to obtain a smoothtorque curve without having a very large first stage and a small second stage. For this reason, thechamber formed at the back of the second stage piston was linked to the intercooler and thuspressurized to intercooler pressure. This had the effect of reducing the second stage forces sincean area equal to the second stage area minus the area covered by the connecting rod was exposedto the intercooler pressure. This resulted in a force acting in the opposite direction to the pressureforces generated in the compression chamber.The graphs generated by the simulation can be seen in Fig. 4.7 to 4.9. It can be seen thatthe smoothest torque curve always occurs at approximately the same bore ratio.Torque(Nm)0 100 200 300 400Crank angle (deg)Bore Ratios: 0 2.105 + 1.689 1.167Figure 4.7Torque vs Crank Angle - Stroke of 19.05 mm78Torque(Nm)0 100 200 300 400Crank Angle (deg)Bore Ratios: 2.188 + 1.625 1.091Figure 4.8Torque vs Crank Angle - Stroke of 25.4 mmTorque(Nm)100 200 300Crank Angle (deg)Bore Ratios: 1.875 + 1.69 1.222Figure 4.9Torque vs Crank Angle - Stroke 31.75 mm79For all three strokes, the bore ratio that gave the smoothest torque curve was in the regionof 1.62-1.68. A bore ratio of 1.625 was chosen. This translates into an area ratio of 2.64 which,combined with the required mass flow rate, set the required displaced volume of both stages.These volumes were found to be 33 985 mm3 for the first stage and 12 870 mm3 for the secondstage.The next step on Fig. 4.6 consists of finding the optimum stroke. This step was doneusing a Lotus 123 spreadsheet. The spreadsheet format is sketched out in Fig. 4.10.Bearing lengthset at 100 mmBearing dia.varies form3 to 32 mmRod diameter Varies form 10 to 48 mmRod length Calculated from rod diameterRod length, axial force and Pv limitatioStroke BoreAxial Load on BearingMax1ststageVariesSum of:allow.calcu- -pressure force calculated from bore and inter-bearingfrom lated cooler pressureloadfrom-centrifugal force due to rod calculated from rodfound6mm strokedia, rod length and distance from center of rota-fromtion bearingandto req’d-centrifugal force due to rollers calculated fromdiamassbearing length, bearing diameter and diatanceand40mm flowfrom center of rotationlengthFigure 4.10Sketch of Spreadsheet Format80There were in fact two spreadsheets with the same format. The only differences are thatthe calculated value was bearing speed instead of axial load on the bearing and the last columnwas maximum allowable bearing speed.A column was set up which had the stroke varying from 6 mm to 40 mm in 2 mm steps.The reason the maximum allowable stroke is 40 mm comes from the sliding velocity limitation onthe linear bushing. A 40 mm stroke results in an average sliding velocity of approximately 3 m/s,which is the maximum allowable velocity, at 2100 RPM. The minimum stroke of 6 mm waschosen because a shorter stroke would probably present many machining difficulties as thetolerances would need to be very low to keep the volumetric efficiency high.Beside each stroke is the required bore. This bore is calculated from the requireddisplaced volume for a given stage which was found earlier. It should be noted that thespreadsheet was set up for one stage at a time. The remainder of the discussion will refer to thefirst stage spreadsheet only as the two were found to yield very similar results.Running the computer simulation gave an idea of the intercooler pressure. This pressurevaried little with stroke for a given area ratio.Knowing the intercooler pressure and the bore, the maximum pressure force acting on thefirst stage can be calculated for each stroke. To find the total force acting on the first stage, thecentrifugal forces need to be calculated.The centrifugal forces were calculated for two components of the first stage: theconnecting rod and the roller bearings. The forces for the piston head were not included as theywere found to be considerably smaller since the piston head center of gravity is close to the centerof rotation. In order to calculate the force, three quantities need to be found: mass of the81component, its distance from the center of rotation and the speed of rotation. The speed was setat 2100 RPM, or 220 s as this is the worst case.In order to find the mass of the connecting rod, its length and diameter need to bedetermined assuming that it is made of aluminum and its density is known. The diameter of therod is a design variable so it was made to vary from 10 mm to 40 mm in the row direction of thespreadsheet. In order to determine the rod length, the length of the linear bearing must be known.This length was calculated from the load limitation on the linear bushing. Since sliding velocity isknown (from the given stroke), all that is needed to find the required bushing area is the load.This load is the side load on the rod which results from the fact that the rollers are not rolling on asurface perpendicular to the rod axis. The angle between the rod axis and the inside surface of theouter ring was found (with the computer simulation) to vary little with stroke and approximately 50 To find the side load on the rod, the axial force is needed. This quantity is unknown. Indeed,it is the quantity that is presently being found. As a first approximation, the axial force was set attwice the pressure force. This approximation can be improved later as the centrifugal forces arefound which made this process an iterative one. With this approximation, a linear bushing lengthcan be calculated. The centrifugal force generated by the rod can now be calculated.At this point the spreadsheet looks like a matrix with the stroke varying from top tobottom and the rod diameter varying from left to right. At each position in the matrix, the sum ofthe pressure forces and the centrifugal forces generated by the rod are added together. The onlyforce left to add in is the centrifugal force generated by the roller bearings.It was assumed that the roller bearings were made of steel which yields the upper limit fortheir density. Both their length and their diameter were chosen as design variables but it would be82difficult to include them in the matrix as a third and fourth dimension. For this reason, the bearinglength was chosen and the bearing diameter was found graphically.The bearing length was set at 100 mm. The length was chosen after making it vary in thespreadsheet and the behavior of the design was examined. Since the load bearing capability of therollers increases linearly with length, increasing the hearing length increases the safety factorassociated with the bearing load. The effect is somewhat diminished by the fact that the bearingmass (and the centrifugal force due to this mass) also increases with length. From the standpointof bearing life, the longest possible bearing is the most attractive. From the considerations ofminimum size and weight, a longer bearing incurs a large penalty because the outer ring (which isvery massive) and the entire frame must be made larger to encase the roller bearings. For thisreason, it was judged that a maximum bearing length of 100 mm would be set.The bearing diameter was initially chosen at 20 mm. It will be shown how this value wasvaried in the spreadsheet to find the optimum bearing diameter.Now knowing the bearing length and diameter, as well as their distance from the center ofrotation (from the stroke and the rod length), the centrifugal force generated by the rollers cannow be included in the matrix. The matrix now yields the total axial force for each stroke and roddiameter combination for a given roller length and diameter. This result is shown graphically inFig. 4.11. Fig. 4.12 is an enlargement of Fig. 4.11 in the region of interest The maximumallowable bearing force was found from linear equation found in section —5045Bearing 40Forces35(kN)302520 Max allow15100 20 40 60Stroke (mm)Figure 4.11Bearing Load vs Stroke and Rod Diameter20191817 Max allowBearing 16aring loadForces 15(kN)11201.3087618 22 26 30 34 38Stroke (mm)Figure 4.12Bearing Load vs Stroke and Rod Diameter84The following points can be extracted from the graphs. First, the forces felt by the rollersincrease greatly with a decreasing stroke. Second, the rod diameter has a relatively small effecton the bearing forces. There is a noticeable improvement when the rod diameter is increased from10 nmi to 20 mm but any larger rod offers little improvement. Third, with the bearing length anddiameter chosen, it is possible to have forces smaller than the maximum allowable.By considering the bearing load limitation only, it seems that a long stroke is preferable.The only design constraint that has not yet been used is the roller bearing speed limitation. Theroller bearing speed depends on three quantities: the speed of the rotor (2100 RPM), the bearingdiameter and the outer ring diameter. The outer ring diameter is chosen by knowing the stroke,rod length and bearing diameter. The bearing speed can thus be calculated for every stroke androd diameter combination. This result is shown graphically in Fig 4.13 and 4.14.BearingSpeedRPMmax allowbearing speed40Stroke (mm)Figure 4.13Bearing Speed vs Stroke and Rod Diameter851400C -1300( Rod diaBearing 12 OO(= ::_________max allowbearing speed 86OO5 15 25 35Stroke (mm)Figure 4.14Bearing Speed vs Stroke and Rod DiameterThe following points are shown by these graphs. First, the bearing speed increases withincreasing stroke. Second, the rod diameter has an effect on speed similar to its effect on bearingforces: there is a significant improvement for bearing speed if the rod diameter is increased from10 mm to 20 mm but any further increase in diameter does not yield much improvement.The graphs showing bearing load and speed versus stroke and rod diameter weregenerated for a given roller bearing diameter. The optimum value for this design variable willnow be determined.When the roller bearing diameter was decreased from the starting point of 20 mm, the loadsituation became worse, that is to say that the axial load increased faster than the maximumallowable load. The opposite is true when the roller bearing diameter was increased: the loadsituation was better.The same is not true for the speed situation. It was found that the situation changed littlewhen the roller bearing diameter was increased or decreased slightly. When the variation was86larger, the situation became worse for both the increase and decrease of the roller bearingdiameter. This behavior points to an optimum bearing diameter for the speed considerations. Fig.4.15 shows the actual speed and the maximum allowable speed for a given stroke and roddiameter. It should be noted that the graph would have the same shape for different strokes orrod diameters but the position of the allowable line would be different.Figure 4.15Bearing Diameter OptimizationFrom Fig. 4.15, it is clear that a bearing diameter between 18 mm and 31 mm is optimumfor the speed considerations.In summary, the findings of the analysis are as follows:• the maximum stroke is 40 mm from linear bearing sliding velocity considerations• the roller bearing load limitation can be met and the safety factor increases with increasingstroke54Speed below 3allowableRPM 210 J4 18 22 26 30 34 38Bearing dia (mm)87• the roller bearing speed limitation can be met and the safety factor increases with decreasingstroke• the roller bearing load safety factor also increases with increasing roller bearing diameter• the roller bearing speed safety factor is maximum for a roller bearing diameter between 18 mmand 31 mm• the roller bearing load and speed safety factors increase with increasing rod diameter but withlittle improvement past a rod diameter of 20 mm.The above procedure yielded all the values for the design variables related to the firststage. Because of the rotary design, the stroke of the two stages must be the same. The fact thatthe area ratio was chosen from torque considerations raises questions about the optimality of thechosen stroke for the second stage. For this reason, the above procedure was repeated for thesecond stage in order to ensure that this stroke was optimal. In fact, that was relatively easy to dosince the analysis was in spreadsheet form.In view of these findings, the following choices were made. First a bearing diameter of 22mm was chosen since it is in the optimal range. The rod diameter was also chosen to be 22 mmsince it would yield the advantages of reducing bearing loads and of keeping the bearing rollingspeeds down without going to a very large rod. Also, it would be easier to mate the rod with thebearings if the sizes were the same. With the help of the load and speed graphs, a stroke of 25,4mm was chosen which yielded a first stage bore of 41,3 mm and a second stage bore of 25,4 mm.The roller bearing length was kept at 100 mm. From the stroke, rod length and bearing diameter,the outer ring diameter was set at 346.1 mm.884.8 Auxiliary ComponentsIn addition to the design variables optimized in the previous sections, there are othercomponents that need to be designed. These components do not affect directly the designvariables above so their design can be treated separately. The two main components which needto examined are the valves used in the intensifier and the rotary seals, a component that is specificto the rotary intensifier configuration.4.8.1 ValvesThe valves used in the intensifier are modified NUPRO check valves [81. This is the sametype of valves used in the previous intensifier design. A drawing of the valves can be seen in Fig.4.16.Figure 4.16NUPRO Check Valve [8JPOPPETWITHBONDEDEL.ASTOMERSEAL.0-RING BODY SEAL. RACK-UP RING89There were, in the previous intensifier versions, problems associated with these valvessuch as short life. This was mainly due to the area of the valve being too small for the requiredflow. This caused the flow velocities to be very high through the valves which cause undue wearon the bonded 0-ring. Once this 0-ring was worn, the valves permitted considerable leakage.This problem was solved in the previous design by going to a larger size of valves (CH8). It isthese larger valves that were used in the rotary intensifier design. The only place that the smallervalves (CH4) were used is for the second stage intake. This was done because the larger valvesyielded an unacceptably high clearance ratio for the small second stages.The only parts from NUPRO that were actually used in the intensifier are the poppet, thepoppet stop and the spring. The valve body was not used but rather holes of the same size weremachined directly into the rotor. After the poppet and spring were inserted, an insert of the samesize as the second half of the valve body was fitted into the hole and held all the pieces in position.The inserts were held in place by the shafts attached to the rotor. This arrangement permitted thevalves to be placed much closer to the cylinders which reduced the clearance volume penalty ofthe valves.4.8.2 Rotary SealsIn order to transfer the high pressure gas from the stationary gas lines to the rotating hubrequires some type of rotary seal. Three methods of accomplishing this task were considered forthe intensifier application: the face seal, non-contacting seals and the mechanical seal.A simplified face seal can be seen in Fig. 4.17. The stationary seal to which the gas huesare connected house a graphite ring. This ring is pressed against the rotating face of the rotor bya spring or compressed 0-ring. The disadvantage of this type of seal is that they are not well90suited to high pressure differences. Also, the graphite provides better sealing when it is hot sowhen the rotor is not rotating and there is no friction to heat the graphite ring, there is poorsealing. For the installation of the intensifier in an urban bus, this poor sealing translates intoimportant leakage when the bus is not in use.In order to improve the sealing, it is possible to add more graphite ring, thus reducing thepressure difference across a given ring. The problem with this arrangement is the limited spaceavailable on the rotor face, especially when the fact that there need to be two separate stationaryseals for the intensifier, one for intake and the other for exhaust, which need to be on the sameface (since the other face is used to drive the rotor).Frori storageStationaryseal9Figure 4.17Face Seal91The second type of rotary seal is the non-contacting seal. In this arrangement, the sealingis done by the fluid that is being transferred. The two parts of the seal, the rotating face and thestationary face, are machined in such a way that the clearances between them are very small. Theviscosity of the fluid and the sliding motion of the two faces stop the fluid from flowing outbetween the two faces. This arrangement is satisfactory when the fluid is a viscous liquid such ashydraulic fluid or oil. The problem with trying to seal a high pressure gas is that the clearancesneed to be extremely small and precise.Finally, the last type of rotary seal considered is the mechanical seal, sometimes called ashaft seal. This type of seal can be seen if Fig. 4.18.Fr’ori storagetnk5Fl Rototing//Oil chnbe r r________L__ To 1stL— Stck9eEx±eri::c:ng____Figure 4.18Mechanical SealIn this type of seal, the seal body and the exterior casing are stationary and the shaft isrotating. The exterior casing is usually attached to a larger framed and thus linked to the bearings92that guide the shaft. The seal body outside diameter is slightly smaller than the inside diameter ofthe casing which permits it to take up any eccentricity between itself and the shaft. The actualsliding is done by 0-rings which are held stationary in the seal body. As can be seen in Fig. 4.18,two 0-rings are usually used to reduce the leakage across the seal.This type of seal is used in many applications where a rotating shaft needs to be sealed.Marine propeller shafts and centrifugal compressor shafts are two important applications. For thecase of the propeller shaft, the purpose of the seal is to stop water from entering the hull. For aship, the pressure is considerable lower than the pressures encountered in the intensifier. In thecase of submarines, the pressures are comparable, but leakage is tolerated since it is possible topump out any inflowing water.The centrifugal compressor shaft seal is used to separate the high pressure gas inside thecompression chamber and the atmosphere. The pressures encountered in a centrifugalcompressor are usually considerably lower than the pressure in the intensifier and centrifugalcompressor are often used to compress air where leakage is tolerable. The state of the artcentrifugal compressor shaft seals use non-contacting surfaces that are shaped in such a way as toprovide some dynamic sealing [9]. This dynamic sealing requires high angular velocities whichare not present in the intensifier. A shaft seal manufacturer who was consulted did notrecommend the use of this type of seal for the high pressures and relatively low angular velocitiesof the intensifier.The only type of seal that seemed to be adaptable to the particular case of the intensifierwas the mechanical seal that did not use any dynamic sealing but used 0-rings for the sealing (Fig.4.18). A recent publication was found which described the design of such seals and all theequations that follow were taken from this reference [10].93One of the problems of using 0-rings is what is referred to as the Gow-Joule effect.According to this effect, an 0-ring under tensile stress which is heated tends to contract. If the0-ring is used in a conventional manner, that is slightly stretched over the shaft, it will contractwhen it is heated from friction. When the 0-ring contracts, the friction increases, as does thetendency to contract. This cycle continues until there is catastrophic failure of the 0-ring.In order to avoid the failure of the 0-ring due to the Gow-Joule effect, the 0-ring must beplaced under compressive stress. This is accomplished by designing the 0-ring groove in such away that the outside diameter of the groove is smaller that the outside diameter of the 0-ring.The procedure used to find the appropriate size of the groove and the correct 0-rings given thediameter of the shaft will now be discussed.The following values need to be determined:• H housing diameter• G 0-ring groove diameter• W width of 0-ring groove• OD 0-ring outside diameter• ID 0-ring inside diameterThe following parameters are known• S Shaft diameter• N angular velocity• P pressure difference across the sealThe shaft diameter was chosen to be 25.4 mm. It is advantageous to have a small shaftsince this reduces the sliding velocity between the shaft and the seal and thus reduces frictional94heating. The reason the shaft was not made even smaller is because both intake and exhaust gasesmust travel in this shaft and it is desirable to have large passages in order to avoid a pressure dropin the shaft.The angular velocity for the worst operating condition is 2100 RPM. This translates intoa sliding velocity of approximately 550 fpm. The worst pressure difference across a seal is 200bar.For high pressure differences and high sliding velocities, the publication recommends usinga small cross section 0-ring (0.070”). For the pressure involved, the clearance between the shaftand the housing should be smaller than 0.006” with a 90 shore A durometer hardness 0-ringmaterial in order to avoid extrusion. Since this is the maximum allowable clearance, a radialclearance of 0.003” was chosen. The housing diameter (H) must then be 1.006”.The reference then gives a table for the 0-ring groove depth and width which is applicablefor this case. The groove width (W) is given as 0.079” and the depth (measured from the shaftsurface) is given as 0.066”. The 0-ring groove diameter (0) must then be 1.132”.The last step is to find an 0-ring that will fit into the groove with the proper amount ofcompressive stress. The way this is done is to find an 0-ring with approximately 8% largeroutside diameter than the groove. For this groove, the 0-ring should have an 1.223” outsidediameter. An 2-023 0-ring would have a 5.2% smaller outside diameter and an 2-024 0-ringwould have a 10.8% smaller outside diameter. Because of the demanding operating conditions ofthe intensifier it would be preferable to use a 2-024 but there were some problem in making a hard(90 shore A durometer hardness) 0-ring fit into a groove that was much smaller than the 0-ringitself. For that reason, a 2-023 0-ring was used.95The life of the 0-ring used in the seal can be predicted from the product of the slidingvelocity and the pressure difference (Pv). For the intensifier, the sliding velocity of 550 fpm andthe pressure difference of 3000 psi result in a Pv value of 16.5 x105 fpm x psi. For indefinite life,a value of 12 x105 is recommended by an 0-ring manufacturer and a value of 2 x105 isrecommended by the reference. The elevated Pv value for the intensifier reduces the expected lifeof the 0-rings to approximately 45 mm. It should be noted that this very short life is for the worstoperating condition: maximum speed and pressure difference. The intensifier spends very littletime at this extreme operating point.In order to further extend the 0-ring life, some measures were taken to remove heat fromthe vicinity of the 0-ring. As can be seen in Fig. 4J8, there is an oil chamber between the two 0-rings of a given seal. This oil serves two purposes. First, it provides a fluid film between the shaftand the 0-ring. This has the same effect as having a smoother shaft which causes less wear on the0-ring. Second, it transports heat away from the 0-ring into the body of the seal. Anothermeasure taken to remove heat was to build the seal body out of bronze which was recommendedby the reference for good heat dissipation.This concludes the discussion of the design of the intensifier. Assembly drawings for thefinal design can be found in Appendix D. Also in this Appendix are all the drawings for theindividual pieces.96Chapter 5Experimental ApparatusThe purpose of this chapter is to describe the equipment used to measure the performanceof the intensifier. This equipment includes the test rig, the gas piping and the instrumentation. Akey part of the measurement of the performance of the intensifier is the data acquisition systemused and this will also be described. Finally, the testing procedure used will be examined.5.1 Test RigThe term test rig is used to describe the equipment used to run the intensifier. Thisincludes a power source and frame on which to attach the intensifier and all related components.Fig. 5.1 shows the major components of the test rig.The power source is an electric motor equipped with a DC motor controller. Its speed isset by the operator on the control panel. The tachometer attached to the shaft of the motormeasures the speed and sends a signal to the control panel (which displays the speed) and to thedata acquisition system. The tachometer also sends a signal to the motor controller which usesthis signal as feedback to keep the motor speed constant.A flywheel is attached to the shaft of the motor. The purpose of this arrangement is todampen out cyclic variations in power requirements of the intensifier. Also on this shaft is the belt97pulley through which the power of the motor is transmitted to the intensifier. Finally, there is anoil pump attached to the end of the shaft which takes oil from a reservoir and supplies oil to theintensifier.Figure 5.1Test RigThe motor casing is mounted on two trunnion bearing which permit it to rotate freely.Because of this arrangement, any torque developed by the motor and transmitted to the intensifiergenerates an equal and opposite torque in the motor casing. Since the casing is free to rotate, atorque arm and load cell attached to it permit the direct measurement of the torque.All these components, the motor, pulley, flywheel and oil pump are held in place withbearings mounted on the frame. This frame rests on rubber pads which dampen out vibration oftensFIer p(IeyLiiitectric ntorTrinnion Beoririgbber Pods98the rig. Also attached to this frame is the table which lies above the axis of the motor shaft. Onthis table is mounted the intensifier with its pulley. The motor pulley and intensifier pulley arelinked by a gear belt.Technical specifications for the electric motor, motor controller, belt drive and flywheelare available in Appendix E. The piping used to supply the intensifier with gas and oil, as well asthe instrumentation of the intensifier, will now be described.5.2 Piping Plan and InstrumentationThe piping plan can be divided into two parts: the oil supply system and the gas system.The oil system is shown in Fig. 5.2.99Figure 5.2Oil SystemFrom an oil reservoir (approximately 2 1 capacity), the oil is pressurized by a DDC 1-71fuel pump to approximately 70 psi. The oil is then directed through a flexible hose to the oil valveused to control the flow of oil into the intensifier. A pressure gage measures the oil pressure afterthe valve. The oil pressure is usually kept at approximately 5 psi above atmospheric. Thispressure was found to be sufficient to ensure flow of oil through the intensifier.seatpanReser’voii100The oil then enters a manifold from which it is directed through three hoses to the mainbearings and the oil seal. The seal transfers the oil into the shaft of the intensifier from which it isrouted to all four pistons. At the pistons, the oil lubricates the Permaglide hushing and also entersthe connecting rod to flow to the roller bearings.The oil is then collected at the bottom of the outer ring as it flows from the two mainbearings and the roller bearings. It then leaves the intensifier through a hole at the bottom of theouter ring and returns to the reservoir via flexible hose.It was found that the oil coming out of the roller hearings did not immediately drainthrough the hole in the outer ring. Some of this oil leaked through the sides of the outer ring andonto the table. In order to remedy this problem, 0-rings were fitted on the sides of the ring butsome leakage persisted. A pan was then fitted under the intensifier to collect any leakage of oil.This pan is emptied at the end of each test.Fig. 5.3 shows the natural gas piping. In this circuit, the following data is measured:• Inlet Pressure (P1)• Intercooler Pressure (P2)• Exhaust Pressure (P3)• Inlet Temperature (Ti)• Intercooler Temperature (T2)• Exhaust Temperature (T3)• Mass Flow Rate101The gas is provided by high pressure CNG tanks. Coming Out of the tanks, the gas isfiltered and goes through a regulator which sets intake pressure to the intensifier. The gas thengoes through a flow meter which measures the mass flow rate. The temperature of the intake gasis measured by a thermocouple (Ti) and the gas is then directed to an accumulator whosefunction is to dampen out cyclic variations in intake pressure due to the reciprocating nature ofthe intensifier. There is a similar accumulator in the exhaust tract that has the same function.From the accumulator the gas goes to the intake gas seal after the inlet pressure has beenmeasured by a strain gage pressure transducer (P1). The purpose of the gas seal is to transfer thegas to the rotating shaft of the intensifier rotor.P3 esse reliecMoss RoteFigure 5.3Gas System102After being compressed by the intensifier, it exits the latter through the exhaust gas sealwhere the exhaust temperature is measured by a thermocouple (T3). The gas pressure ismeasured by a strain gage pressure transducer (P3) before the gas goes through a pressure reliefvalve which is used to set the exhaust pressure of the intensifier. The gas is then returned toanother high pressure CNG tank to be reused in another test.It is worthy to note that there is a third gas seal. This seal is used to take some gas atintercooler conditions out of the intensifier in order to measure the temperature (T2) and thepressure (P2) of the gas after the first stage compression. This is done with a thermocouple and astrain gage pressure transducer. There is no flow through this piping except to fill it at the start ofa test.In addition to the gas properties measured at different points measured above, thefollowing variables were also measured in order to evaluate the performance of the intensifier:• Gas Seal Wall Temperature• Torque• Motor SpeedThe gas seal wall temperature thermocouple is used to ensure that the gas sealstemperature does not exceed the safe operating temperatures of the 0-rings used in the seals.The torque is measured using a load cell attached to a torque arm which is bolted to thecasing of the electric motor. Fig. 14 shows a partial end view of the rig in which this assemblycan be seen. Finally, the motor speed is measured by a tachometer attached to the end of themotor shaft. All the technical specifications of the transducers listed above, as well as detailsconcerning the calibration of the transducers, can be found in Appendix F.103Figure 5.4End View of Test Rig5.3 Data AciuisitionThe data acquisition system is shared between the intensifier and the 6V92 test rig. It iscomposed of two main components: the hardware and the software. The hardware will beexamined first and is shown in Fig. 5.5.pulleybeltElectricTrunnion beo.riiFroarmoad cell104The signals from the transducers are fed to signal conditioning modules. These modulesfilter and amplify the signals as well as providing the cold junction compensation for thethermocouples. After conditioning, the signals are sent through an interface board to a switchbox. This switch box controls which signals are sent to the PC: either those from the intensifier orfrom the 6V92. After the switch box, the signals are sent to an A/D board (PCL 818) which isinside the PC. The software then transforms the signals into useful quantities. Technicalinformation concerning the above components can be found in Appendix G.Motor’ TorqueInLet TempInterstcige TempOutLet Temp‘L( TempInlet PresInterstckge PresOutlet PresMcss FlowFigure 5.5Data Acquisition Hardware105The software first uses a calibration file to transform the voltage signals into values inengineering units. This calibration file is generated by the user and it is discussed in Appendix F.The software then accesses a configuration file which contains data such as the bore of the stagesand the speed ratio between the motor and the intensifier. This configuration file also containsinformation on which values to show on the monitor display. These values can be the data takenfrom the transducers or calculated values such as power consumed, volumetric efficiency andpressure ratio.Finally, the software permits saving data to a disk so that it can be analyzed later using aspreadsheet. The software generates two output files: one that contains the values read from thetransducers and the other that contains the calculated values. The subroutine used to calculatethese values is listed in Appendix H.5.4 Testing ProcedureThe following variables are controlled by the tester:• Intake Pressure• Exhaust Pressure• Intensifier Speed• EccentricitySince the present working value for the injection pressure is 200 bar, the exhaust pressurewill be set at this value for the testing and will not be varied. The other three variables will bevaried in such a way as to cover the range of operating conditions for which the intensifier wasdesigned.106The first series of tests was as follows:• constant speed (200 RPM)• maximum stroke (1 inch)• pressure ratio varying from 4:1 to 12:1• constant outlet pressure (200 bar)This series of tests was intended to provide data for a significant comparison to theprevious intensifier designs. In fact, there is no use of the variable stroke capability so theintensifier is in fact acting exactly like a conventional reciprocating intensifier.The second series of tests were done under the same conditions (200 RPM, 200 barexhaust pressure) but with varying strokes. Each individual test was done with a constant strokebut there were 5 different tests. The strokes for the tests varied from 0.8 to 0.0 inches in 0.2inches increments. This series of tests were intended to examine the variable capacity aspect ofthe intensifier.The third series of tests were done at a speed of 400 RPM. These tests were intended torepeat the second series of tests (five tests at different eccentricities) at a higher speed.Unfortunately, there were problems with fail tires caused by overheating which limited the series ata single test done at full eccentricity.The final series of tests was intended to examine the effect of speed on the intensifierperformance. Since testing at higher speeds is limited by operating temperatures, these test weredone at a given pressure ratio and a given eccentricity. These parameters were chosen in such away to give a chosen mass flow rate (maximum engine fuel requirement).1076.1 IntroductionAs discussed in chapter 5, there are three variables which are controlled during the testing:speed, stroke and intake pressure. The testing procedure was also described in this chapter and issummarized in Table 6.1.SpeedStroke (in) 200 RPM 400 RPM Varying from 200to 350 RPM0.2 Intake Pressure of 127bar0.4 Intake Pressure varyingfrom 53 to 100 bar0.6 Intake pressure varyingfrOm 22 to 55 bar0.8 Intake Pressure varyingfrom 16 to 66 bar1.0 Intake Pressure varying Intake Pressure varying Intake Pressure offrom 16 to 63 bar from 18 to 33 bar 20 barTable 6.1Scope of TestingChapter 6Results108As can be seen in Table 6.1, the speed was held constant for all but one test and the strokewas always held constant during a given test. As for the intake pressure, it was either heldconstant or varied with one of two methods. The first method consisted of starting at an elevatedpressure and permitting the storage tank from which the gas was taken to empty, thus decreasingthe intake pressure. The second method was to change the intake pressure with a regulator andtake several data points at each chosen pressure ratio. The first technique was used for most testwhile the second was used for the low mass flow test, that is the 0.2 inch stroke test.Data during a given test was taken as soon as the exhaust pressure was set at 200 bar(with a pressure relief valve) and the intake pressure was at the desired starting value. Steadystate was never fully achieved because the temperature of the intensifier was continuously risingdue to inadequate cooling. This cooling problem will he discussed in detail in the next chapter.The effect of overheating will be shown in some of the results of the testing. This overheatingproblem results in the fact that these tests are not completely repeatable since operatingtemperature affects intercooling and volumetric efficiency.6.2 Low Speed TestingTesting at 200 RPM consisted of running the intensifier at a given eccentricity and varyingthe intake pressure. The test was then repeated for other eccentricities. The performance datathat will be examined are mass flow rate, power requirements, power per unit mass flow,volumetric efficiency and isentropic efficiency. Most of the performance data will be examinedversus overall pressure ratio.1096.2.1 Mass Flow RateThe mass flow rate was measured in the intake line between the intensifier and the storagetank. For this reason, the measured mass flow rate also includes any leakage Out of the intensifier.This error would tend to make the measured intensifier performance better than reality. For thisreason, the leakage was always examined at the beginning of a test before the intensifier wasstarted. Any leakage above 1 % of a typical intensifier flow rate was corrected before any testingwas done. Fig. 6.1 shows a typical measured mass flow rate for a full stroke test.Figure 6.1Mass Flow Rate for Full Stroke Testing at 200 RPMIn Fig. 6.1, the solid line is the mass flow rate predicted by the computer simulation.Good agreement was achieved for the following reasons. First, the clearance volume used in thecomputer simulation is the same as the measured clearance volume in the intensifier. Second, theempirical formulation for heat transfer in the computer simulation was adjusted in such a way that32302826Mass 24Flow 22(kg/hr) 201816141210863 5 7 9 11 13Overall Pressure Ratio° measured + predicted110the interstage pressure in the simulation matched the measured interstage pressure. This matchingcan be seen in the stage pressure ratios in Fig. 6.2Figure 6.2Stage Pressure Ratio for Full Stroke Testing at 200 RPMFinally, the intake temperature for the computer simulation was set at approximately the averageintake temperature of the test. The latter would vary as expansion through regulators and checkvalves and mass flow rate would vary during the test.The stroke was varied by changing the eccentricity. The following strokes wereexamined: from 0.2” to 1.0” in 0.2” increments. This series of 5 tests gave the following massflow rate graph (Fig. 6.3) 3.4Pressure 3.2Ratio 5 7 9 11 13Overall Pressure RatioD 1st measured 2nd measured 0 1st predicted 2nd predicted111Figure 6.3Mass Flow Rate for Various Strokes at 200 RPMThe effect of the different methods of varying intake pressure can be seen in Fig. 6.3. Thehigh stroke tests show a continuous variation from high to low pressure ratio while the low stroketest only have a few discreet point.The mass flow rate graph for various strokes still shows good agreement between themeasured mass flow rate and the predicted value from the computer simulation. This graph alsoshows that the intensifier was overdesigned since the required mass flow rate of 500 mg perrevolution translates into 6 kg/hr for a speed of 200 RPM. At the worst pressure ratio (10:1) theintensifier provides a mass flow rate of 7.8 kg/hr at full stroke. This overdesign was intentionaland done to compensate for any losses in the valves or leakage across seals. This safety marginshould diminish at higher speeds as these losses increase.343230282624Mass 22Flow 20(kgr)181 3 5 7 9 11 13Overall Pressure Ratio° 1’ stroke .8 stroke 0 .6 stroke .4’ stroke X .2’ stroke1126.2.2 Power and Specific PowerThe power required to run the intensifier was calculated with the measured torque andspeed. A value for power can also be extracted from the computer simulation. This power wascalculated with the following equation:Power = Fo sin(ao) ro . dOwhere F0 is the force acting on the roller hearing due to pressure forces on the piston, a is thecontact angle between the roller bearings and the outer ring calculated form geometry, r0 is thedistance from the center of rotation to the contact point between the outer ring and roller bearingsand w is the rotational speed. Since all components in the computer simulation are frictionless andthe pressures are calculated using polytropic coefficients (1.25 for compression and 1.4 forexpansion), the value of power given by the computer simulation is the same as the polytropicpower. The difference between the measured power and the power from the computer simulationis the power loss due to friction. The power required for the intensifier can be seen in Fig. 6.4113Figure 6.4Required Power for Various Strokes at 200 RPMThis graph shows what was expected: the required power decreases with stroke. This isdue to the fact that the power required to compress the gas is lower at lower strokes becausethere is less gas to compress.To get an idea of the friction losses, the required power can be compared to the powerpredicted by the computer simulation and the isentropic power. The isentropic power iscalculated in the data acquisition program from the pressure ratio and the inlet temperature toboth stages. This comparison can be seen in Fig. 6.5 for the full stroke case.Power(kW)0.83 5 7 9 11Overall Pressure Ratio1’ stroke + .8” stroke 0 .6” stroke .4” stroke X13.2” stroke1143. 2.4(k’ 5 7 9 iiOverall Pressure Ratiomeasured + predicted isentropicFigure 6.5Measured and Predicted Power for Full Stroke at 200 RPMFrom this graph, it can be seen that the isentropic power is much less than the predictedpower from the computer simulation. There are two reasons for this difference. First, thecomputer simulation uses polytropic coefficients which are different from the isentropic one andthere is a simple heat transfer model in the program. Second, there is an important error in thecalculation of isentropic power as it is difficult to obtain a good measurement of the temperatureof the gas entering the second stage. Interstage gas is piped out of the intensifier where itspressure and temperature are measured but the gas has time to cool substantially before reachingthe thermocouple so that the temperature read is only a few degrees above room temperature.The error is worse at high pressure ratios because this is when the interstage temperature ishighest.Because of the error in the isentropic power calculation, the predicted power will be usedto calculate efficiency. This value of efficiency will be referred to as compressor efficiency The13115predicted power does take into account some heat transfer but this efficiency gives a valuableindication of power losses due to friction. As can be seen from Fig. 6.5, approximately half thepower is lost to friction. This result will be discussed in greater detail in a latter section.Another important performance characteristic is the required power per unit mass flow(specific power). The measured results can be seen if Fig. 0.4(kW*hr/kg)0.30.2 -0.1 -01 3 5 7 9 11 13Overall Pressure Ratio1” stroke .8” stroke ° .6” stroke .4’ stroke X .2” strokeFigure 6.6Specific Power for Various Strokes at 200 RPMThe fact that higher strokes require less power per unit mass flow was expected since themass flow is much higher for higher strokes and power lost to friction is expected to be similar forall stroke because of the three main sources of frictional losses (piston rings, roller bearings andgas seals), only the friction loss due to piston rings should diminish at lower strokes.aI1166.2.3 Compressor EfficiencyAs mentioned earlier, the compressor efficiency calculation will use the power predictedby the computer simulation as opposed to the isentropic power. This is not isentropic efficiencybut the efficiency calculated in the manner is still important since ii gives an indication of thefriction losses. This efficiency can be seen in Fig. 6.7 for various strokes.Figure 6.7Compressor Efficiency for Various Strokes at 20() RPMThe fact that the friction losses are approximately the same for all strokes explains the factthat the efficiency is highest at high strokes since the power required to compress the gas ishighest in this case.6050Compressor40Efficiency(%)3020101 3 5 7 9 11 13Overall Pressure Ratio° 1’ stroke .8’ stroke ° .6” stroke .4” stroke < .2” stroke1176.2.4 Volumetric EfficiencyThe volumetric efficiency is calculated in the acquisition program with knowledge of thefirst stage displaced volume and the intake conditions. The results for volumetric efficiency canbe seen in Fig. 6.8.908070VolumetricEfficiency(%)403020I I I I I I I I I1 3 5 7 9 11 13Overall Pressure Ratio1” stroke + .811 stroke 0 .6” stroke ‘ •4” stroke X .211 strokeFigure 6.8Volumetric Efficiency for Various Strokes at 200 RPMThe fact that the volumetric decreases with stroke was expected since the clearancevolume increases greatly when the stroke is decreased.6.3 High Speed TestingThis section will examine the effects of raising the speed of the intensifier. Unfortunately,the testing at higher speeds was limited to a full stroke test because of excessive overheating ofthe intensifier due to inadequate cooling. This overheating caused premature failures incomponents such as seals and piston rings.II‘C118The importance of the 400 RPM test is to give a basis for the extrapolation of theperformance data into the speed range of the engine (600 to 2100 RPM). With the presentconfiguration, testing at such speeds is impossible for two reasons. First, the test rig belt driveand electric motor limit the maximum intensifier speed at approximately 700 RPM. Second, andmore importantly, the present intensifier configuration does not permit adequate cooling of therotor and this causes failures in the rotor which makes the testing impossible.The overheating of the rotor also affects the data taken at 400 RPM and these effects willbe discussed when appropriate.6.3.1 Mass Flow RateThe mass flow rate for the intensifier at 400 RPM is compared with the predicted massflow rate from the computer simulation and also the mass flow rate from the 200 RPM test in Fig.6.9Figure 6.9Mass Flow for Full Stroke at 200 and 400 RPM3634323028Mass 26Flow 24(kgr)9 11 13 15 17 19Overall Pressure Ratio° 400 measured + 400 predicted 200 measured 200 predicted119The testing technique used is the same as for the low stroke testing at 200 RPM. Thereason for this is that at high speeds, the storage tanks pressure drops too rapidly so the intakepressure is set by a regulator in order to get some steady state data.The effect of the high temperatures can be seen in the mass flow graph. The point atapproximately 11:1 pressure ratio which is above the curve is in fact the first point to be taken.Since the intensifier was not hot at the beginning of the test, there is more heat transfer out of thecompressed gas into the intensifier body which results in lower gas density and thus a higher massflow rate. The next point to be taken is the 16.5:1 point and then the pressure ratio was loweredso that the last point to be taken before the intensifier failed was the 6:1 point.6.3.2 Power and Specific PowerThe effect of overheating is also noticeable in the power graph (Fig. 6.10). A similarbehavior to the mass flow graph is noticeable:54.5 -4Power(kW)32.521.5UI I I I I3 5 7 9 11 13 15 17 19Overall Pressure Ratio2 400 measured + 400 predicted 0 200 measured 200 predictedI*‘U’cmFigure 6.10Measured and Predicted Power for Full Stroke at 200 and 400 RPM120The 11:1 point required more power than the rest of the point of the test since it was thefirst point to be taken and had a significantly higher mass flow.The specific power requirement of the intensifier can be seen in Fig. -0.3 -0.28Specific 0.26Power 0.24(kW*hr/kg) 5 7 9 11 13 15Overall Pressure Ratio200 RPM + 400 RPMFigure 6.11Specific Power for Full Stroke at 200 and 400 RPMIt is interesting to note that the power per unit mass flow is very similar for both the 200and the 400 RPM cases. Since mass flow is proportional to intensifier speed, this graph suggeststhat the power required for the intensifier is also proportional to speed. A more detailedinvestigation of the effect of speed on the power requirement will be made in a further section.+*I I I I I I I I I I17 191216.3.3 Compressor EfficiencyThe compressor efficiency calculations for the 400 RPM cases are the same as the 200RPM cases. The predicted power from the computer program is used instead of the isentropicpower. The results for both the 200 and 400 RPM full stroke test can be seen in Fig. 6.12.Figure 6.12Efficiency for Full Stroke at 200 and 400 RPMIt is difficult to compare the two curves because of the differences in operatingtemperatures. Also, wearing in of the linear bushings and gas seals may have reduced thefrictional load on the high speed test as it was done after the set of 200 RPM tests.6.3.4 Volumetric EfficiencySince the geometry of the intensifier is the same for both 200 and 400 RPM tests, thevolumetric efficiency should not vary from one test to the other. The measured values ofvolumetric efficiency can be seen in Fig. 6.13.56555453Efficiency 52(%) 51504948474645444342413 5 7 9 11Overall Pressure Ratio° 200 RPM * 400RPM13 1512288868482Volumetnc80Efficiency(%) 7876747270686664625 7 9 11 13Overall Pressure Ratio° 200 RPM + 400 RPMFigure 6.13Volumetric Efficiency for Full Stroke at 200 and 400 RPMThe effect of operating temperature is very apparent here. The first data point to be takenin the 400 RPM test shows a higher volumetric efficiency than the equivalent 200 RPM point.This is due to the fact that the intensifier is cooler for this point. As the intensifier becomeshotter, the volumetric efficiency drops below the level of the 200 RPM tests. The differencebetween the two tests is most noticeable for the low pressure ratio region of the test when theintensifier is at its hottest in the 400 RPM tests but its coolest for the 200 RPM tests. This graphshows the important effect of cooling on the volumetric efficiency.+ +++3-If+15 17 191236.4 Variable Speed Testin2 ResultsAll previous tests were done with a set eccentricity and speed and a variable pressureratio. In order to study the effect of variable stroke, a given test was repeated with differenteccentricities. In order to study the effect of speed, it would be preferable to repeat the same testsa higher speeds but this proved to be impossible because of inadequate cooling of the rotor. Forthis reason a different testing procedure was used.The following data was taken with a constant eccentricity and a constant pressure ratio.The speed was varied from 200 RPM to 350 RPM in 20 RPM increments. The mass flow,power, specific power, efficiency and volumetric efficiency are the performance characteristic thatare examined.It proved to be difficult to maintain a constant pressure ratio for the entire test and evensmall variation in intake pressure caused important variations in mass flow and power. Fig. 6.14shows the power required by the intensifier at various speeds. IPower 2.8(kW) I200 220 240 260 280 300 320 340 360Speed (RPM)Figure 6.14Power Required at Various Speeds124The reason that the test ended at 350 RPM was because the gas seals failed due tooverheating. The points at a given speed show a wide spread because of the varying intakepressure. The fact that power is proportional to speed can be seen in this limited test range andwhen comparing the 200 and 400 RPM tests. This results in a constant compressor efficiency forfull stroke in the tested speed range.125Chapter 7Conclusions and Recommendations7.1 Conclusions7.1.1 Discussion of ObjectivesThe first objective was to design and build a variable stroke, multi-stage, mechanicallyactuated reciprocating intensifier which would meet the design requirements for fuel flow in bothdesign and off design operation of the diesel engine. This objective was met as the first rotaryintensifier prototype was built. The design optimization procedure used to arrive at the finalintensifier configuration was a graphical one which is described in chapter 4.The eccentricity of an outer ring and the axis of rotation of the rotor generates thereciprocating motion of the pistons. Two compression stages were chosen and the variable strokeis supplied by the fact that the outer ring can be moved and thus the eccentricity can be varied.The requirements for operating pressures were met as the intensifier could achieve anexhaust pressure of 200 bar with a varying intake pressure. The flow requirements at 200 RPMwere satisfied. Testing was limited to 200 RPM tests and very limited 400 RPM tests becausehigher speeds caused very rapid overheating and subsequent failure of rotor components.126The second requirement, that of measuring the performance of the intensifier in the light ofthe thermodynamic model was also met. The test rig used in previous intensifier designs wasmodified and the instrumentation adapted to the rotary prototype. The thermodynamic model,which consisted of a computer program which simulated the intensifier, proved to be aninvaluable design tool. Also, the simulation results matched the experimental results closely.The variable capacity capability of the rotary design was studied and showed that thevariable stroke method of capacity control can meet the requirements of variable fuel flow for thediesel engine. This method did not prove to be as energy efficient as had been hoped because ofthe presence of high friction losses. These friction losses proved to be fairly constant over therange of mass flow studied. This constant friction loss resulted in low compressor efficiencies atlow mass flow rates. The power requirements of the intensifier were somewhat lower at reducedmass flow because of the lower compression work required.The final objective, that of assessing the essential design limitations of this new class ofvariable stroke, high pressure intensifiers, was also met. The rotary intensifier design concept islimited by intrinsic sources of problems which are discussed in detail in the following section.7.1.2 Problems of Rotary ConfigurationThe testing of the first prototype of the rotary intensifier revealed four major sources ofproblems of the rotary configuration. Two of these problems are unavoidable because they arecaused by the rotary geometry. These problems are excessive forces due to the presence of largecentrifugal forces and the leakage due to the sealing of rotating parts. The two last problemscould be avoided in the next intensifier prototype. These problems are related to alignment andoverheating.127Centrifugal ForcesThe first major source of problems of the rotary configuration is the fact that the rotatingmachinery generates centrifugal forces. This is an important problem as at operating speeds of theorder of the engine speed, the centrifugal forces are of the same order of magnitude as thepressure forces. The components of the rotary intensifier thus have to withstand twice the forcesencountered in a non-rotating intensifier of the same capacity.The presence of the large centrifugal forces adds complexity to the design calculations.These forces also reduce the size of the feasible domain, that is the options when determining thedesign variables. Finally, and most importantly for the eventual production model, the largeforces reduce the safety factor for components such as bearing and linear bushings which in turnreduces their life.Rotary Gas SealsThe second major design limitation lies in the rotary seals. The challenge of transferringhigh pressure gas from stationary lines to the rotor or vice versa is a considerable one. Becauseof the high pressures and high sliding velocities, the seal life is very short: of the order of onehour. This is the main reason that high speed testing was limited. Also, any misalignmentbetween the gas seals and the shaft greatly reduce the seal performance. Finally, in order to sealthe high pressures, the tolerances must be very small which makes the machining difficult. Thetight fit also results in high frictional losses and high seal temperatures. The friction losses due tothe gas seals are the main cause of the low compressor efficiency at low mass flows.128AlignmentThe alignment problems originate from the fact that there are too many components thatmust be properly aligned in order to ensure proper functioning. First, the two shafts attached tothe hub must be on the same axis so that when the rotor is held by the main bearings, neither shaftoscillates. This problem is amplified by the fact that there are plates between the hub and theshafts and little room to fit in bolts. The pistons must then be aligned with the rotor in order toensure that their reciprocating motion is perpendicular to the axis of rotation. The rollers must beperpendicular to the piston rods and thus parallel to the axis of rotation of the rotor.The two main bearings must be on the same axis so that the rotor can turn freely. Thealignment of the main bearings is made difficult by the fact that it requires the alignment of thetwo frame plates and two bearing holders. All these components are attached to the base platewhich must be flat, even after being bolted to the test rig table. The frame plates also have to holdthe outer ring in such a way that its inner surface in parallel to the axis of rotation but is stillallowed to move freely in order to have a variable eccentricity.Finally, the gas cap must hold the gas seals in such a way that they are concentric to theshaft of the rotor. Any eccentricity at the gas seals caused undue wear and poor sealing.OverheatingOverheating was the main source of failures in the intensifier. High temperatures wereencountered in two areas of the intensifier: the gas seals and the rotor. The gas seals temperaturewas high because of the friction between the 0-ring and the shaft and because of the presence ofhigh temperature exhaust gases. The high gas seals temperatures would cause expansion of the0-ring which result in higher contact forces between the 0-ring and the shaft, causing even higher129heat production due to friction. This cycle would Continue until the 0-rings would seize unto theshaft and the seal would fail.High temperatures were also encountered in the rotor. The main reason for this is that therotor, being encased in the frame and outer ring, was not properly cooled. There was little flowof fresh air into the outer ring and no supplementary cooling system had been planned. The hightemperature caused failure of the Teflon piston rings. The situation is similar to the gas seal: thefriction of the piston ring against the cylinder wall would generate heat which would causeexpansion of the piston ring. The larger contact force between the piston ring and the wall causedrapid wear of the piston ring which would result in poor sealing and contamination of the oil withTeflon particles.These are the major design limitations of the rotary configuration. The following sectionwill provide modifications that can be applied to the existing intensifier or incorporated in the nextversion.7.2 RecommendationsThe recommendations that follow first deal with modifications that could be used in orderto diminish the effects of the design problems described in the previous section. The last sectiondescribes general recommendations that do not touch upon the limitations but could still improvethe intensifier performance.7.2.1 Centrifugal ForcesThe centrifugal forces cannot be eliminated. They can be minimized by attempting toreduce the mass of the components, their distance from the center of rotation or the angular130velocity. Reducing the angular velocity would require a gearbox between the intensifier and theengine. This would result in a weight and cost penalty and since the intensifier is turning at alower speed, it would require larger pistons in order to provide the same mass flow rate. Thiswould result in larger pressure forces so the problem is not entirely solved. Reducing the distancebetween the components and the center of rotation requires a more compact hub. There cannotbe much improvement over the present design as there is little wasted space inside the hub. Thefinal option of reducing the component mass could be done with a more exotic materials but thiswould result in a sharp increase in price. The centrifugal force limitation, which places limits onthe expected life of he intensifier, cannot easily be solved.7.2.2 Rotary Gas SealsPerhaps the most important design limitation is the gas seals. The need for transferringhigh pressure gas between stationary lines and the rotor cannot be avoided (unless the storagetanks and engine are also made to rotate). The three factors that affect seal life are pressure,sliding velocity and operating temperature. The pressure cannot be reduced as they are set byfactors outside the intensifier design. The sliding velocity (of the order of 2.5 mJs in the presentdesign) could be reduced by reducing the angular velocity of the intensifier or by reducing thediameter of the shaft. Reducing the angular velocity is an option that has many repercussions inthe design because of the required mass flow rate. There is a minimum shaft diameter because ofthe presence of the passages that carry the gas to the rotor.Since the pressures and the sliding velocity cannot be changed enough to make animportant difference in seal life, the last factor, that of operating temperature, must be examinedclosely. The temperature immediately around the 0-ring that is providing the sealing is131considerably higher that the surroundings because of the friction between the 0-ring and theshaft. This friction can be minimized by ensuring that the seal and the shaft are concentric. Thisresults in an even distribution of contact pressure around the 0-ring and avoids any excessivefriction in one localized area. Also, adding a lubricant close to the 0-ring and the shaft canreduce friction by creating a film between the 0-ring and the shaft. This method was already usedin the present version but the possibility of pressurizing the lubricant could be examined. Thelubricant can also be used to carry heat away from the 0-ring which reduces the operatingtemperature. The seal body design could also be improve to increase the heat transfer away fromthe 0-ring. Use of materials such as copper which have a high conductivity or the addition ofwater cooling are examples of how the heat transfer from the 0-ring could be improved.If the rotary intensifier configuration is to be pursued any further, efforts must beconcentrated on the rotary seals. Without a good rotary seal with an acceptable life, the rotaryconfiguration is worthless. It would be beneficial to set up experiments where only the seal isstudied, without any attention paid to the rest of the intensifier. A fully floating design with betterlUbrication and water cooling could prove to be long lasting.7.2.3 AlignmentThe problem of misalignment can mostly be solved with a design which takes into accountalignment in the first phases of the design process. Components that need to be aligned should beremachined as one component. For example, the two shafts attached to the rotor could bemachined as one shaft going through the hub. Also, the frame plates and the bearing holderscould be machined as one U-shaped bearing holder which would house both main bearings and inwhich would sit the outer ring. Finally, components such as the gas seals should be designed so132that they are self aligning. The shaft or the seals could be made totally free to float so that theyautomatically take up any eccentricity. The alignment problems of the present prototype could hecorrected with some modifications and could be completely avoided in the next design.7.2.4 OverheatingThe obvious solution to the overheating problem is better cooling. This could be done inseveral ways. First, the intensifier could be built in a more open configuration. In thisconfiguration, the frame would not completely enclose the rotor, thus allowing fresh air to flowaround the rotor. This would make the intensifier more dangerous because moving parts wouldbe exposed but it would make it much more easy to diagnose any problems with the rotor. Thiswould be invaluable in the refinement stage of the testing. This could he accomplished with anarrangement as seen in figure 7.1 and 7.2 where there are no frame plates but rather the use of aU-shaped bearing holder discussed earlier.133Figure 7.1Side View of Open ConfigurationU—shQped bearing holderEkrter Ring guidcince lract134This configuration would also have the advantage of simplifying the alignment of theframe. The problem of holding and locating the outer ring could be handled by having a tract inthe base plate in which the outer ring could slide back and forth. The eccentricity control couldbe done by a screw in this tract.Even with the open design, cooling may prove to be inadequate. Water or oil cooling arealso cooling option. The cooling fluid could be made to flow through the rotor by using a sealsimilar to the oil seal on each shaft of the rotor. These seals, as well as the oil seal, are similar toCDnIroI. ScrewOuter Ring gulthnce *ro-tFigure 7.2End View of Open Configuration135the gas seal but do not suffer from their short life since the pressures are much lower and there iseffective cooling due to the oil or water flow.The main problem with the open design is that the oil would tend to flow out of the side ofthe outer ring. The oil system in the present version already needs some refinement as there ismuch waste of oil because the oil collection system is inadequate and oil tends to flow out of theouter ring even with the presence of the frame walls. The oil system could be improved so that itcould function in an open intensifier design.7.2.5 General RecommendationsThese recommendations are not directly related to the rotary design limitations but arepossible modifications that would improve the intensifier performance.The next version should avoid the use of aluminum. The advantages of using steel areworth the weight penalty. Steel is usually harder that aluminum so that surfaces that need to besmooth (for 0-ring seals for example) do not become marked as easily. Steel is also stronger thanaluminum so that components that need to be flat in order to provide proper alignment do notwarp when they are bolted to other components. Finally, steel does not produce as many wearparticles that contaminate the lubrication oil.The main problems of the rotary-reciprocating configuration, the gas seals and centrifugalforces, come from the fact that the pistons are rotating. A new type on intensifier whichincorporates the design concept of variable stroke but does not use rotating pistons is presentlybeing considered. Because of patent laws, this new intensifier cannot he discussed in this thesis.136REFERENCES1. Gunawan, H., “Performance and Combustion Characteristics of a Diesel-Pilot Gas InjectionEngine”, M.A.Sc. Thesis, Department of Mechanical Engineering, The University of BritishColumbia, 1992.2. Aichinger, C., “Development of a CNG Intensifier for High Pressures”, M.A.Sc. Thesis,Department of Mechanical Engineering, The University of British Columbia, 1993.3. Csanady, G.T., “Theory of Turbomachines”, McGraw-Hiil, 1964.4. Van Wylen, G.J., Soontag, R.E., “Fundamentals of Classical Thermodynamics”, 3rd Edition,John Wiley & Sons, 1985.5. “Rolling Bearings”, INA catalogue #305, INA Walzlager Schaeffler KG, 1990.6. “Needle Roller Bearings”, INA catalogue #35 1, INA Walzlager Schaeffler KG, 1988.7. “Pennaglide Plain Bearings”, INA catalogue #703, INA Waiziager Schaeffler KG, 1991.8. Swagelok catalogue9. “Type 28 Series Dry-Running Gas Seals”, John Crane Inc., 1990.10. Martini, L.J., “Practical Seal Design”, Marcel Dekker Inc., 1984.11. Hil, P.G., Peterson, C.G., “Mechanics and Thermodynamics of Propulsion”, Addison-WesleyPublishing Company, 1992.137APPENDIX AEngine SpecificationsThe engine for which the conversion kit is being designed is for one configuration of theDetroit Diesel Corporation (DDC) 6V92-TA family. This engine is a heavy duty two strokediesel engine having 6 cylinders in a V configuration. Each cylinder displaces 92 cubic inches andthe engine is turbocharged and after cooled.The purpose of this appendix is to derive the engine fuel requirement from the data givenby DDC for the 285 hp 6V92-TA. This data is reproduced here as are the calculations and graphsfor the fuel requirement. First, some general data for the engine will be listed.General DataModel 6V92-TA CoachNumber of Cylinders 6Bore and Stroke (in) 4.84 x 5.00Displacement (in3) 552Compression Ratio 17:1Exhaust Valves per Cylinder 4Combustion System direct injectionEngine Type 63.5° VEE two strokeAspiration Turbocharged138Physical DataLength (in) 37.6Width (in) 37.6Height (in) 48.0Dry Weight (ib) 2020Fuel SystemFuel Injector Part No. 5324850Fuel Injector Timing 1.460Cart Code 0002Fuel Consumption (lb/hr) 103.5Fuel Consumption (gal/hr) 15.5Fuel Spill Rate (lb/hr) 489Fuel Spill Rate (gal/hr) 73.2Total Fuel Flow (lb/hr) 593Total Fuel Flow (gal /hr) 88.7Cooling SystemCoolant Flow (gal/mm) 160Maximum Top Tank Temperature (°F) 210Minimum Top Tank Temperature (°F) 160Minimum Coolant Fill Rate (gal/mm) 3.0Performance DataPower Output (bhp) 285Full Load Speed (RPM) 2100Peak Torque (ft ib) 870Peak Torque Speed (RPM) 1200BMEP (lbf/in2) 92139Table A1.1 shows the performance data for the 6V92-TA Coach and this data is alsoshown in graphical form below. Notice that the data had to be extrapolated for speeds below1000 RPM. This was done with a cubic fit to the data with Lotus 123.Engine Speed Power Torque BSFC(RPM) (bhp) (ft ib) Ib/bhp hr2100 285 713 .3581950 276 743 .3551800 265 773 .3501600 248 814 .3411400 228 855 .3351200 199 870 .3341000 161 846 .343Table A.lData from DDC for the 6V92-TA Coach140Figure A.1Power Output of the 6V92-TA CoachFigure A.2Engine Power vs Engine SpeedEnginePower(Bhp)3203002802602402202001801601401201008060 I I I 11111111 III600 800 1000 1200 1400 1600 1800 2000Engine speed (RPM)° Data Fitted & extrapolatedBrake Specific Fuel Consumption vs SpeedBSFC(lb/bhp* hr)600 800 1000 1200 1400 1600 1800 2000Engine Speed (RPM)° fuel cons data —fuel cons fittedBrake Specific Fuel Consumption of the 6V92-TA Coach141Figure A.3Fuel Consumption of the Engine in kg/hrFigure A.4Fuel Consumption in mg/revolution60555045Fuel 40Consumption(kgJhr) 353025201510Fuel Consumption vs Engine Speed(max power)600 800 1000 1200 1400 1600 1800 2000Engine Speed (RPM)Fuel Consumi vs Engine Speed520510500490480470Fuel 460Consumption 3(mg/rev)410400390380370360350340 800 1000 1200 1400 1600 1800 2000Engine Speed (RPM)142The above data and extrapolations were used to understand the engine fuel requirementsand this information was needed to size the intensifier. Notice that even though natural gas has aslightly higher LHV (lower heating value) than diesel fuel, the fuel requirement for natural gaswas taken to be the same as the fuel requirement for diesel. This was done to give a slightoverdesign to the intensifier that would compensate for other expected problems.143APPENDIX BCapacity Control ExampleThe following assumptions will be made: a single stage reciprocating machine compressingnatural gas (n=1.3) with a clearance ratio of 10%. This would give it a volumetric efficiency of51 % at a pressure ratio of 10:1 and 100% for a pressure ratio of 1:1 from the fact that:/P211 —1\\PI)The mass flow rate per revolution of a single stage reciprocating machine is given by:rn PiVai’n RTiThe displaced volume Va is chosen in such a way that the machine provides a mass flow of500 mg/rev an intake pressure of 20 bar (which means a 10:1 pressure ratio assuming a 200 barexhaust pressure) and intake temperature of 20°C, that is the worst case. The displaced volumefound is 7.2x105m3.Using this displaced volume for the case where the intake pressure is 200 bar (1:1 pressureratio) intake temperature is the same at 20°C, then the mass flow per revolution is an immense9800 mg/rev. This is 195 times engine requirements.144APPENDIX CCentrifugal Compressor CalculationsIn this appendix, the design of a centrifugal compressor that will meet the requirements formass flow rate and pressure ratio will be attempted. All equations and assumptions are takenfrom reference [11]. The design will start with the following known data:• 20 bar intake pressure• 300 K intake pressure• y = 1.3 for natural gas• R = 500 J/kg*K for natural gasFor high pressure ratio centrifugal compressors, the maximum attainable compressorefficiency (i’i) is approximately 80%. Also, stresses in the rotor limit the rotor tip velocity (Ut) at650 mJs. Finally, high pressure ratio compressors are typically design with blades leaning backfrom the radial direction as much as 30° (tEl’,).Since the flow in the tanks and in the injector have relatively low velocities, the pressurerise can be expressed in terms of the stagnation pressures. The stagnation pressure ratio can beexpressed as:O3 JU Y wr—=1 1+ri(y —i — j1———tan 13P01 L \,aoi,A, U146and assuming that Cp is approximately 2.8 [11], and is approximately 420 K. The outletdensity is found with ideal gas and is 95 kg/rn3. Finally, the rotor radius is found to be 2x106 rn.This is an extremely small radius for the rotor and it would be impossible to machine. Even is asmaller blade height (b) would he acceptable, the rotor would still be very small which would leadto large clearance losses. Also, with such a small radius, the angular velocity would be extremelyhigh in order to generate the 650 m/s tip velocity (of the order of 3.25x 108 rad/s or approximately3x109 RPM).147APPENDIX DIntensifier DrawingsThis Appendix contains all the drawings of the rotary intensifier. The first two pages listthe parts and part drawings numbers. This list is followed by two assembly drawings and finallythe part drawings.List of PartsNO Part Name DrwgNo1 Rotor 1-R1AI-R2AI-R3A2 1st & 2nd Stage Inserts 1-IN1A3 1st Stage End Plate 1-IP1A4 2nd Stage End Plate I-IP2A5 1st Stage Piston 1-P1A1st Stage Piston (mod) 1-P1B6 2nd Stage Piston I-P2A7 1st Stage Valve Inserts 1-VIF1A8 2nd Stage Valve Inserts 1-VIS1A9 Bearing Rod I-BR1A14810 Bearing Wheel I-BElA11 Drive Shaft I-DS1ADrive Shaft (mod) I-DS1B12 Drive Shaft Circular Plate I-CP1ADrive Shaft Circular Plate (mod) I-CP1B13 Gas Shaft I-SH1AGas Shaft mod) 1-SH1BGas Shaft (mod) I-SH1C14 Gas Shaft Circular Plate I-CP2AGas Shaft Circular Plate (mod) I-CP2B15 Gas Cap I-GC1AGas Cap (mod) I-GC1B16 Gas Seals I-GS1AGas Seal (mod) I-GS1B17 Bearing Support 1 I-BH1A18 Bearing Support 2 I-BH2A19 Oil Seal 1-OS1A20 Frame 1-FR1AI-FR2AI-FR3A1-FR4AI-FR5A21 Base Plate I-FP1A22 Outer Ring 1-RI1A23 Outer Ring Frame I-RF1AI-RF2AI-RF3A1110 9DRIVEPULLEY11716157 20TESTRIGTABLE21DRILL01/8—27NPTDEPTH3/8’2HOLESNOTE’TCLE:RANCES41.6a5’AHOLECDIAMETER’SURFACEFINISH’16RMS1/16kIntnsiferRotorcb’o,nbynoAtInToucheteI—R1Anoreqta.dAluminium6061—T4INOTE’DREAKALLEDGESUNLESSOTHERWISENOTED.ALLOUTSIDESURFACESMUSTBEMACHINEDDepurtmerrtofMecharilccLEngineerIngUBClUFEITOLERANCEON1.000’DIAHOLE’.0.002LW‘—0.000SURrACEFINISH’16RMSIwtensifierRotorCQsflbynoAtQnTouchet-eI-R2Anoreq.dAluminium6061—T41NOTEiBREAKALLEDGESUNLESSOTHERWISENOTEDALLOUTSIDESURFACESMUSTBEMACHINEDI’)DepartrentofMechnicui.EngineeringUBCTHREAD*/2’UNCDEPTH1.5’HDLESEDGEDONOTBREAKEDGESNOTE’BREAKALLEDGESUNLESSOTHERWISENOTEDALLOUTSIDESURFACESMUSTBEMACHINEDIntensifierRotor’drawnbyI’QhOAloinTouchette1—R3AhOt.rlotnoreqifr’edAtupnun6061—T4ItrTHREADØ1/2’UNCDEPTH1.5’4HOLESDRILLØS/16’DEPTH1.656HOLESDONOTBREAKEDGE-2,50\MILL011/16’DEPTH1.400’2HOLESMILLØ7/8’PTH.5802HOLESDONOTBREAKEDGE—2.625•3.081—3750DRELLØ3/8’DEPTH1.344’MILLO1/2’2HOLESEPTH1.088’2HOLESLDEPTH.468’2HOLESDeprtmen-tofMechonlco.LEngineeringUBCDONOTBREAKEDGE-dDRILLMILL013/8kDEPTH1.000’GROOVEIS.081’DEEPAND.140’WIDE1/4’1.498__Li1.500—jMILL013/8’DEPTH1.000’MILL(BALLNOSE)01/4’DEPTH.188NOTEiTOLERANCEON025000HoLEI:ggNOTEi.075’x45’CHAMFERONBOTHENDSOF025.000HOLES.+.001,NOTEiTOLERANCEONALLOUTSIDEDIAMETERS_001.DRILL0.IS.081’DEEPAND.140’WIDE.1887/16DeportmentofMechanicLEngineerIngUBC1st2ndStageInsertsdtewnbyQwfl9noAloinTouchtte1—IN1Ar,trIo1norequrdALumlnIur6061-142DRILL05/1640HOLESDepurtrerntofMechanicoi.EngineeringUBC1stStageEndPtotedrawnby)dtüWIngnoAIQInTouchette1—IP1AmoerIaLnorequredALuminiuM6061—T42NOTEITOLERANCEON025.000HDLEiNOTE,075’x45CHAMFERONBOTHENDSOF025.000HOLES..625—2.e—L_.50rw/%50It.25001/16NPT01/8’hoteI—2,0—1.0NOTETOLERANCEON025.000HOLEI1NOTE’.075’x45CHAMFERONBOTHENDSOF025.000HOLES..62501/8’holesDRILL0.999DepartmentoPMechunico].EngineeringUBC2ndStc&geEndPciterwnbyjcdruwlngnoALlnTouchetteI-IP2AnerI1norequw’edA1urInum6061—T42.625.25001/16NPTDRILL01/24HOLESNDepartmentof’MechanicalEngineeringUBC450CHAMFERDEPTH.090.750THREAD1/2’UNCDEPTH1.50-.-.522FirstStagePistondruwnbydr’wIngnoALoinToi.,iche-tte1—P1Amter’iLnorequiredAhAminkw,6061—T424,355+0.003—0,003NOTEBOTHGROOVESARE.122WIDEx.189DEEPSHAFTSURFACEMUSTBEGROUND.0625.1533,833.0625—Departmentof’MechanicalEnginee1ngUBCflrstStagePistondrawnbydruwingnoAtomToichette1—P1BruteruLnorequiredS’tomnlessS’teet2NOTEBOTHGROOVESARE122WIDEx.189DEEPSHAFTSURFACEMUSTBEGROUND4ZZ\+0.003(O))—0,00345*CHAMFERDEPTH.090THREAD1/2’UNCDEPTH1.50.5224,355.0625.1533.833.0625Depo.r’tmentof45êCHAMFERDEPTH.090THREAD1/2’UNC1,50.125—1.375F— IrDRILL01/8NOTEIBOTHGRROVESARE.083WIDEx.125DEEPSHAFTSURFACEMUSTBEGROUNDMechanco.(Engneer;ngUBCSecondStageP;stondrawnbydrawingnoAtomToche’tteI—P2AmaterialnorequiredAtuminiur6061—T424,543.522.169—10.873i1114.0210.987.0625MILL01/2’DEPTH.620DRILL03/8’MILL011/16’DEPTH.360DRILL05/16’-H.4681—0.872NOTEALL0.09’CHAMFERSARE45NOTE:TOLERANCESONALLOUTSIDEDIAMETERSARE:.268DepartmentofMechoncaJEngneerngUBC1stStageVatveInsertsdrawnbydrawingnoAtomToi..ichette1—VIF1Ar-aterlaLnorequiredArilnk.im6061—T42MILL01/2’DEPTH.620DRILL03/8’MILL07/16’DEPTH.224DRILL03/16’NDTEALL.090’CHAMFERSARE45+,001NDTETOLERANCESONALLOUTSIDEDIAMETERS’_,001.580DepartrnentofMecho.nicoi.EngineeringUBC2ndStageVaI.veInsertsdrawnbydrawingnoAtunTouchette1—VIS1Arater;a1norequiredALuniflUn6061—T42-2,625—aDRILL01/8’-2.0625--a5.250NOTESSURFACEOF0.875SHAFTMUSTBEHARDENEDANDGROUNDoi.!oom—2.000—..O.7—0,0004-.1a--.t•—5001.1251/8’-27NPTDEPTH1/4’2HOLESDRILL7/2’DepurtrriervtofMecho.nical.EngineeringUBCBearingshaftdrawnbyIdrcwIngnoAloinTouchette1—BRIAmaterialnorequiredSteeL4—2.0625(1+Q0OO5L(__)..)01.00001500 INOTE:INSIDEANDOUTSIDEDIAMETERSNEEDTOBEHARDENEDANDGROUND.Deportmentof’MechoniccdEngineeringUBCBeci.ringWheeldrawnbydrawingnoAtomTouchetteI-BElArcxteratnoreqiiredSteel812500-RINGGROOVELD.5.580085WIDEx.052DEEPMACHINEWITHDEPTH3/8’2GROOVESDRILL01/2’01/4’BALLNOSEALL0-RINGGROOVESARE.085WIDEx.052DEEPWHEREPOSSIBJ..E.LEAVEA1/16’WALLBETWEENHILEAND0-RINGGROOVE.&ewflbyfrawingnoAtcxlnTouchetteI—DS1Ano‘eq&*cISteeLI.700.620DRILL03/16’4HOLES1250SQUAREKEYHOLE.375’WIDEx,185’DEEPDepartrentofMechanIc1EngineeringUBCDriveShafttr)ALLfl—RINGGROOVESARE.085WIDEx.05kDEEPWHEREPOSSIBLE.LEAVEA1)16’WALLBETWEENHOLEAND0-RINGGROOVE.6250—3.250—SQUAREKEYHOLE.375’WIDEx.185’DEEPro.nbynoAIQIATouchetteI—DSIBnor.qL.dSteel10—RINGGROOVELD.5.580.095WIDEx.052DEEPMACHINEWITHDEPTH3/8’2GROOVES1250.70001)4’BALLIEISE1.000—.500-.DRILL01/2’06.000065.00,m 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GROOVEO.D.0.174’DEPTH0.030DepartmeritofMechanictEnglneerngUBCSeal.ShaftcbwnbyWflQflOAlcnTouchette1—SHICnoreqLr.dSteeliN0—RINGGROOVELD.2,5’.085WIDEx.052DEEP0-RINGGROOVEI.D.3/8’.085WIDEx.052DEEPMACHINEGROOVEWITH-3/8’BALLNOSEDEPTH1/2’DRILL05/16’2HOLES.3125.7001.125Depo.r-trnentof’Mechanical.EngineeringUBCGasPlotedrawnbydrawingnoALanTouchetteI—CP2AmaterkxtnorequiredAturilnium6061—T41DRILLTHROUGHHOLE01/8’.-GROOVEO.D..196DEPTH.040 MACHINEGROOVEWITH01/4’BALLNIJSEDEPTH5/16’DRILL01/4’2HOLESDepcirtmentofMechanicalEngineeMngUBCGasPI.o.tedrQwnbyAlainTouchetteN0—RINGGROOVED.D.DEPTH0.520-RINGGROOVEI.D,25’085WIDEx.052DEEP1,125.875.250.625DRILL03/8’2HOLES.328.7001.125materILALuriniur6061—T4N1/4—18NPTDEPTH3/8’WITH01/4’HDLE05.00UNCDepartrervtofMechncLEngneer1ngUBCGsSealCoverthawnbydrwmgnoAlainTouche-tteI—GC1Aiteriütnor’equredSteeL1NNOTE’THESIXINNER0—RINGGROOVESARE.085WIDEx.055DEEPANDARE1/16’FRIDOSG..ESONEACHSIDE1/4—18NPTDEPTH3)8’WITH01/4’HELEfrnwnbynoAIWTouchetteI-GCIBmetrInoSeeIIø500DepcrtmentofMechnicLEngineeringUBCGcsSeCLLCoveriiGROOVEIS.094WIDEx.050DEEPDeprtmervtofMechnIccxLEngineeringUBCGcxsSeaLs*o,nbyI0WmQnoAIQIATouchetteI—GS1AnoreqLiredBronze6NOUTSIDEDIAMETER2.747—.938—1/161/16—1/16GROOVEIS.079WIDEx.065DEEP—MACHINETHESE4HOLESWITHA1/2’ENDMILLDEPTH1/2’1/16RQEGROOVEIS.530WIDEx.250DEEP/ 1..250OUTSIDEDIAMETER2,747MACHINETHESE4HOLESWITHA1/2’ENDMILLDEPTH1/2’MACHINETHISGROOVEWITHA01/2’BALLNOSEDEPTH.750.250.4375.5165.579GROOVEIS250DEEPGROOVESARE.065DEEP45CHAMFER15CHAMFERrwnbydrQwmgnoAInTouchetteI—GSIBnoraqAredBronze330’CHAMFERDepirtewtofMechanicalEngineeringUBCGasSeatsN Nø1/8—27IPTDEPTH3/8DRILLourTIEADu/2LflC4HDESbyALInTouchetteI—BH1ASteeL45.DeprtmentofMechonIcLCngineeringUBCBottomBeuringBracketcc N1/8’—27NPTDEPTH3/8’DRILL01/8’THREAD1/2’UNC4HOLES&ownbyIdrwftnoA1anTouchetteI—BH2AmQterIoInoSteet1DRILL01/2’4HOLES04.25GROOVEIS1/8’DEEPDeprtmentofMecharcLEngineeringUBCTopDearngBracketN01/4’—18NPTDEPTH3/8’DRILL01/4’02,125NOTE:THETWOGROOVESONTHEINSIDEDIAMETERSAREBOTH.079WIDEx.056DEEPANDTHEYHAVEA1/16’WALLONBOTHSIDES.Departrentof’MechanicalEngineeringUBCOftseal,drawnbyAta;nTouchettematero.lSteetccNOTE’THEFOUR1/4’UNCHOLESARECENTEREDONPLATETHICKNESSPLATEiS1/2’THICKryA1ahTochee!—FRIAa1/4’UNCHOLES Deprtren*ofMechn$cQLEngineeringUBCFror.ePto.*e181I-flI-I-flc-flEO00j-UI1i-__8,00•1.255.500THREAD3/4’:NCDepo.rtrientof’Mechanca1EngineeringUBCFrcrnePindrawnbydrawingnoAtomTochetteI—FR3AmateriaLnor’equredSteel3.250250.250—j[..—.375.875of2.500___1.250/DRILL03/4’____.875____p.375DRILLO9/32’6,5008HOLESNDTEPLATEIS1/2’THICKBepartmentof’MechoscatEngineernguBcEccentr’icrtyControl.PLatedrcwnbydruwngnoAlainTouchetteI—FR4AmiterIxLnorequr9dSteeL1DRILL01/4’4HOLESDepcrtmentof’MechanicoJ.Engneer-inguBcSpacerdrawnbydrawIngnoAtomToiche’t-teI—FR5Amater’iutnorequIredS’teet2—‘II.’i1OObtrTHREAD1/2’UNC8HOLESdrL,nbyIdronoA1InToucheUeI—FPIAnoSteeL101/8’2HOLESDeprtrentofMechnicaI.EngineeringUBCBcLsePLteNOTE’INSIDESURrACEOFRINGMUSTBEGROUNDANDHARDENEDO.D.1’d’elflb,AtomTauchetaZ-RflA,IQWt1*0r..dTHREAD01/2’UNC3HOLESGROOVEIS1.000WIDEANDO.D.OF’14.50DeprtmentofMechQnIctEngineeringUBCOuRingN•—jm—.I><I014,50Departrentof\MechanicalEngineering\UBCNOTE’THISSURFACEMUSTBEFLATFORENTIRELENGHTOFASSEMBLYRingframeussembLykawnbyI”Q°AtInTouchette1—RIIA,oWr1li,or.qt*’eALurInium6061—14cc -415.50IIiiZRCZRADlUSlJ2 —1,750DRILL03/4’2HOLES625DRILL01/2’THREAD3/4’UNCDEPTH1.750.500r-N.I7503.001.750—Departr’ientofMechnIcoJEngineeringUBCSidesofr’;ngfro.eby110ALcinToucheteI—RF2Am11*.rIQi110IeqLa.dAluminium6061—T42NOTE’THE3/4’ -4THREAD3/4’UNCDEPTH1,250.625NOTEiA3’HIGHBARWILLGIVESUFFICIENTMATERIALTOMACHINEARCDepartmen-tofMechanical.EngineeringUBCTop8Bottomofringfrar’ec*’QwnbyIcIwmonoAkInTouchetteI—RF’3Apio1eraLnorequiredAlur’ilnlum6061—T4a190APPENDIX ETest Rig SpecificationsElectric MotorBrand name: HampstonType: variable speed, DCMounting: trunnion mountedSpeed: 0-2400 RPMPower: 25 hp at 2400Control: tachometer feedback speed regulationShaft dia: 1.600”Motor ControllerBrand name: RandtronicsType: regenerative DC motor controllerModel: TB 750 seriesRated Power: 25 hpLine Voltage: 240 V, 3 phaseOutput: 0-240 V DC, 0-55 AmpsField: 100 VMode: tachometer feedback speed regulationSynchronous BeltBrand Name: BrowningType: gearbeltModel: HPT 8M191Drive Pulley: B348M50SH34 teeth3.409 pitch diameterLow SpeedDriven Pulley: B1928M50E192 teeth19.195 pitch diameter5.56 speed ratio20.03” center distance with 2000 mm long belt (2000 8M50)17.46 hp rating at 205 RPM driven pulley speedHigh SpeedDriven Pulley: B 1 128M50E112 teeth11.175 pitch diameter3.29 speed ratio19.62” center distance with 1600 mm long belt (1600 6M50)22.91 hp rating at 530 RPM driven pulley speedFlywheelOutside dia: 16”Width: 4”Weight: 45 kg192APPENDIX FInstrumentation Specifications and CalibrationThe purpose of this Appendix is to describe the transducers used in the measurement ofthe intensifier performance. Details of the calibration of these instruments and the calibration fileused by the data acquisition software are also included.The following transducers are examined:• Loadcell• Thermocouple• Strain gage pressure transducer• Mass flow rate meter• TachometerLoadcellModel: Interface SM-250 (serial # C26470)Type: Strain gage force transducerCapacity: ± 250 lbfOutput: 3.218 V/VExcitation: 10 VDCSignal Conditioning: 5B38 isolated strain gage input193The calibration of the load cell was done with the load cell installed in its usual position(figure 4.4). A load arm, which measured 16 in, was attached to the motor shaft. The rotor andcasing of the electric motor were linked so that a torque applied to the shaft would be transmittedto the casing (and thus measured by the load cell). The following calibration curve wasgenerated:Figure F.1Loadcell CalibrationIn this graph, as well as all the calibration curves to follow, the square points represent thevalues read and the line is a fitted curve which was used in the calibration file. Also the outputvoltage was read after all signal conditioning.Loadcell CalibrationVoltage(mV)0 4 8 12 16 20Applied Torque (Nm)24 28194ThermocouplesModel:Type:Capacity:Output:Excitation:Signal Conditioning:Omega ThermocoupleJ type (iron÷ constantan-)0-760 °C0-40 mV10 VDC5B37-J-0 1 isolated thermocouple inputThere were four thermocouples used in the test rig to measure inlet, interstage, outlet andwall temperature. The calibration was done by placing the thermocouples in boiling water whichwas allowed to cool. The voltage after signal conditioning was read after each 10 °C drop in thewater temperature. The calibration curves are found in figures F.2-F.5.Figure F.2Inlet Temperature CalibrationInlet Temp. Transducer Calibration1.05I0.950.90.85Voltage0.750.70.650.60.550.510 30 50 70 90Temperature (C)195Figure F.3Interstage Temperature CalibrationFigure F.4Outlet Temperature CalibrationInterstage10.95Temp. Transducer Calibration0.90.85Voltage(V)0.80.75U0.70.650.60.550.510 30 50 70Temperature (C)90Voltage(V)Outlet Temp. Transducer Calibration1. 30 50 70 90Temperature (C)196Figure F.5Wall Temperature CalibrationPressure TransducersModel:Type:Capacity:Output:Excitation:Signal Conditioning:Data Instruments model SAHigh gain strain gage0-206 bar1-6 VDC10 VDC5B38 isolated strain gage inputThere are three pressure transducers used to measure inlet, interstage and outlet pressure.The calibration curves can be found in figures F.6-F.8. The calibration was done with a deadweight gage tester. Different hydraulic pressures covering the operation range were applied tothe transducer while the voltage output was read after the signal conditioning.Wall Temp. Transducer Calibration1.110.9Voltage(\/) 30 50 70 90Temperature (C)197Figure F.6Inlet Pressure CalibrationFigure F.7Interstage Pressure CalibrationInlet Pressure Transducer Calibration765Voltage(V) 43210 0 40 80 120 160 200Applied Pressure (bar)Interstage Pres. Transducer Calibration654Voltage(V)210-1 0 40 80 120 160 200Applied Pressure (bar)198TachometerModel:Type:Output:Signal Conditioning:SignerGenerator7V/1000 RPM5B38 isolated strain gage inputThe tachometer was calibrated with a hand held optical tachometer trained on the shaft ofthe motor. The same method was used to verify the speed ratio of the gearbelt transmission. Thecalibration curve generated is:Outlet Pres. Transducer Calibration76Voltage(V)432I0-1 0 40 80 120 160 200 240Applied Pressure (bar)Figure F.8Outlet Pressure Calibration199Mass Flow Rate MeterModel:Transmitter Model:Type:Capacity:Pressure Rating:Temperature Rating:Signal Conditioning:Micro Motion DHO12S100Micro Motion RFT9712Coriolis meter0-300 kg/hr393 bar-240 to 204 °C5B32Speed Transducer Calibration43Voltage(V)2101200Speed (RPM)Figure F.9Tachometer CalibrationThe calibration of the mass flow rate meter is done with a supplied calibration computer.200CALIBRATION FILE FOR INTENSIFIER DATA ACQUISITIONLast modified 02-18-1994Intensifier Version 2.00, 2-Stage Rotary-Reciprocating, Mechanical Drive, E-MoiorTorque0Nm-.284,-.023662, .284, 0, 0, 0, 0, 0, 0, 0-29.3625, 0, 29.3625, 0, 0, 0, 0, 0, 0, 0Wall Temperature6C.54 172, .995 73, 0, 0, 0, 0, 0, 0, 0, 010, 90.556, 0, 0, 0, 0, 0, 0, 0, 0Motor Speed5RPM.0013, .185 1, 3.86651, 0, 0, 0, 0, 0, 0, 00, 104, 2171, 0, 0, 0, 0, 0, 0, 0Interstage Press0bar-5,-. 199, .05291, 4.706 1, 0, 0, 0, 0, 0, 00, 0, 6.8948, 189.6058, 0, 0, 0, 0, 0, 0Outlet Pressure8baro‘o‘0‘0‘0‘0‘0‘0‘t6‘L1717c9du1jsiui0‘0‘0‘0‘0‘0‘1790Z‘8176W9‘0‘0o‘o‘0‘0‘0‘0‘186ç‘8cFT‘8L6‘0n?q•17ojnssaijtui0‘0‘0‘0‘0‘0‘0‘oc‘0‘oco’o’o’o’o’o’o’c’o’ciq/j0P/mP9N3o‘o‘0‘0‘0‘0‘0‘0‘906‘010‘0‘0‘0‘0‘0‘0‘0‘L186‘17o1717c9no0‘0‘0‘0‘0‘0‘0‘0‘9cc06‘Ot0‘0‘0‘0‘0‘0‘0‘0‘ZT86‘L1717c3Lininidmjiui0‘0‘0‘0‘0‘0‘c9w1171‘8176W9‘0‘00‘0‘0‘0‘0‘0‘lz86Lc‘l0tcu-’c981-’ot-I0110zo‘o‘o‘o‘o‘o‘o‘o‘oooc‘oo’o’o’o’o’o’o’o’c’oci(1p!Ju30‘0‘0‘0‘0‘0‘0‘0‘9cc06‘OT203APPENDIX GData Acquisition Hardware SpecificationsThe data for the following components was taken from C. Aichinger’s thesis [2]. Since nomodifications were made to the data acquisition hardware, this data is still valid.Signal Processing ModulesTachometer: 5B31 Isolated Voltage inputMass flow: 5B32 Isolated Current InputThennocouple: 5B37-J-0l Isolated Thermocouple Input - Type JPressure Transducer: 5B38 Isolated Strain Gage InputTorque Load cell: 5B38 Isolated Strain Gage InputA/D BoardModel: Advantech, PCL 818, PC-LabCard SeriesType: high speed, high performance I/O card for IBM A/D conversionAnalog Input:channels: 16 single ended or 8 differentialResolution: 12 bitsConverter: ADC-774Conversion Time: 8 jsInput Range Bipolar: ± 1OV, 5V, 2.5V, 1V, 0.5VInput Range Unipolar: 0-1OV, 0-5V, 0-2V, 0-1VInput Range Selection: ProgrammableTrigger Mode: by software204Digital Output:Channels: 16A/D Pacer and Counter:Device: Intel 8254Pacer: 32 bit with 10 MHz or 1 MHz lime basePacer max. rate: 2.5 MHzCounter: 16 bit counter with 100 kHz time basePersonal ComputerBrand Name: ANOType: IBM compatibleCPU: Intel 80286205APPENDIX HPerformance Data Calculation SubroutineSUB PerfCalc* * Subroutine to calculate performance values *I **************************************************‘Set constant values:rho.air = 1.205’ kg/mA3 Density of air @ 2OøC, 1.0133 barR.GEN = 8314.34 IJ/Kmol*KI GENERAL GAS CONSTANT‘STANDARD VALUES FOR NATURAL GAS:‘density: ON NATURAL GASrho.STD = .713 142 ‘kg/m”3 PROPERTIES SEE:C:\123\CNGTABL‘Molecular weight:M.STD = 17.1456999# ‘kg/Kmol‘Gasconstant:R.STD = R.GEN / M.STD IJ/kg*K‘B.C. HYDRO VALUES FOR NATURAL GAS:‘density:rho.BCH = .690351 Ikg/mA3‘Molecular weight:M.BCH = 16.597737# ‘kg/Kmol‘Gasconstant:R.BCH = R.GEN / M.BCH IJ/kg*KKelvin = 273.15k.isentr = 1.3FOR MORE INFORMATION‘Set variables:Torque = EngData(VAL(chan$(1)))’ N-rnRPM = EngData(VAL(chan$(2)))’ Motor Speed in RPM206IF speed .rat = 0 THENspeed.rat = 1END IFintens.rpm = RPM / speed.ratin.temp = EngData(VAL(chan$(3)))’ CNG inlet temperature in øCout.temp = EngData(VAL(chan$(6)))’ CNG end temperature (after 2nd stage) in ø = EngData(VAL(chan$(4))y CNG inlet pressure in = EngData(VAL(chan$(7)))’ CNG end pressure (after 2nd stage) in barwall.temp = EngData(VAL(chan$(8)))’ wall temperature of intensifier cylinder in øCCNG.mass EngData(VAL(chan$(5)))’ CNG massflow in kg/hrinter.temp = EngData(VAL(chan$(9)))’ interstage temperature after 1st stage in ø = EngData(VAL(chan$(1D)))’ interstage pressure in barIF LCASE$(chan$(11)) = “none” THENstroke = 2 * VAL(cnst$(11)) * 2.54’ Stroke in cm (2* eccentricity)ELSEstroke = 2 * EngData(VAL(chan$(1 1))) * 2.54’ Stroke in cmEND IF‘Calculate Intensifier Displacement:‘FIRST STAGE:disp.1 = (bore.1 ‘ 2) * pi / 4 * stroke * ncylinders.1 ‘in cm3‘SECOND STAGE:disp.2 = (bore.2 ‘ 2) * pj / 4 * stroke * ncylinders.2’ in cm3‘Conversion of temperatures to Kelvinin.tempK = in.temp + Kelvin’ inlet temperature (K)out.tempK = out.temp + Kelvin’ outlet temperature (K)inter.tempK = inter.temp + Kelvin’ interstage temperature (K)wall .tempK = wall.temp + Kelvin’ wall temperature (K)temp.diff. 1 = inter.temp - in.temp’ first stage in øCtemp.diff.2 = out.temp - inter.temp’ second stage in = out.temp - in.temp’ total temperature difference in ØCIF in.tempK < 250 OR inter.tempK < 250 OR ouLtempK < 250 THENtemp.ratio.1 = 999temp.ratio.2 = 999ELSEtemp.ratio.1 = inter.tempK / in.tempKtemp.ratio.2 = out.tempK / inter.tempKEND IF‘Pressure ratios:IF < .1 THENp.ratio.1 = = 999ELSEp.ratio.1 = / ‘first = / ‘overall pressure ratioEND IFIF < .1 THENp.ratio.1 = 999ELSEp.ratio.2 = / ‘second stageEND IF‘Theoretical temperaturestheoret.out.tempK.1 = in.tempK * p.ratio.1 “((k.isentr - 1)1 k.isentr)theoret.out.tempK.2 = inter.tempK * p.ratio2 “ ((k.isentr - 1) / k.isentr)theoret.out.temp.1 = theoret.out.tempK.1 - Kelvintheoret.out.temp.2 = theoret.out.tempK.2 - Kelvin‘Theoretical mass flow:‘FIRST STAGE:IF intens.rpm <5 THENtheoret.mass.1 = 999ELSErhol = / (R.BCH * in.tempK) ‘density of intake gas (i0*kg/crnA3)theoret.mass.1 = rhol * disp.l / 10 * (intens.rpm * 60)’ kg/hrEND IF208‘Volumetric Efficiency:‘OVERALL:vol.eff.tot = CNG.mass / theoret.mass.1 * 100’ in %‘Power:‘FIRST STAGE: [kW theoretical (isentropic) power consumption]A = (k.isentr / (k.isentr - 1)) * CNG.mass * R.BCH * in.tempK / 3600 / 1000isentr.power.1 = A * ((p.ratio.1 A ((k.isentr - 1)! k.isentr)) - 1)‘SECOND STAGE: [kW theoretical (isentropic) power consumption]A = (k.isentr / (k.isentr - 1)) * CNG.mass R.BCH * inter.tempK / 3600 / 1000isentr.power.2 = A * ((p.ratio.2 A ((k.isentr - 1) / k.isentr)) - 1)‘OVERALL:isentr.power.tot = isentr.power.1 ÷ isentr.power.2act.power = Torque * RPM * 2 * pi / 60 / 1000’ kW actual power consumptionIF CNG.mass <= .001 THENpower.per.mass = 999ELSEpower.per.mass = act.power / CNG.mass ‘ power per mass (kW/kg!hr)END IF‘Isentropic Efficiency:IF act.power <= .01 THENisentr.eff = 999ELSEisentr.eff = isentr.power.tot / act.power * 100 ‘ isentropic efficiency in %END IFcontinue:‘Put calculated values in an array for subsequent use‘See list of Calcname$() in SUB SetCalcNarnesCalcData(1) = p.ratio.1CalcData(2) = p.ratio.2209CalcData(3) = p.ratio.totalCalcData(4) = power.per.massCalcData(5) = act.powerCalcData(6) = isentr.power. 1CalcData(7) = isentr.power.2CalcData(8) = isentr.power.totCalcData(9) = vol.eff.totCalcData(10) = isentr.effCalcData(11) = intens rpmCalcData(12) = disp.1CalcData(13) = disp.2CalcData(14) = temp.ratio.1CalcData(15) = temp.ratio.2CalcData(16) = theoret.out.temp.1CalcData(17) = theoret.out.temp.2CalcData(18) = theoret.mass.1END SUB ‘Last edited: Dec. 6, 1993210APPENDIX ISimulation Program**ssimu1ation program for rotary intensifier***1* * *dimensioning variables* * *‘crank angle variablesDIM xl#(360), Voll#(360), pl#(36O), tl#(360), M1#(360)DIM x2#(360), Vo12#(360), p2#(36O), t2#(360), M2#(360)DIM Fradl#(360), Ftanl#(360), Ftotl#(360)DIM Frad2#(360), Ftan2#(360), Ftot2#(360)DIM torq1#(360), torq2#(360), torqtwo#(360), torqfour#(360)DIM Mres#(360), pres#(360), tres#(360), alpha#(360)‘revolutions variablesDIM MFlowl#(200), MFlow2#(200), MFlowbyps#(200)‘intake pressure variablesDIM MFlow#(20), Pmax#(20), prl#(20), pr2#(20), prt#(20), power#(20)* *getting compressor dimensions* * *stroke = (1) * 25.4eccmax = stroke / 2 ‘maximum eccentricityRinnner = (.6875) * 25.4 ‘distance from top of cyl to centerRouter = (6.8125) * 25.4 ‘outer ring radiusborel = (1.625) * 25.4bore2 = (1) * 25.4rodlen = Router - Rinnner ÷ eccmax ‘connecting rod length‘* * *5jflg constants* * *pi# = 3.141592665359#R = 500000 ‘gas constantk = 1.3 ‘isentropic coefficient -ncom = 1.25 ‘polytropic coefficientnexp = 1.4 ‘polytropic coefficientVres# = 375000 ‘intercooler volumeclvoll# = pi# * 2 * (borel “2) / 4 + 4950 ‘clearance volume of 1st stageclvol2# = pi# * 2 * (bore2” 2)/4 + 1500 ‘clearance volume of 2nd stageal# = pi# * (borel “2)! 4 ‘1st stage piston areaa2# = pi# * (bore2” 2) / 4 ‘2nd stage piston area211tblk# = 320 ‘bulk temperatureI***default values***Pintmin# = 2Pintmax# = 20nopoints# = 1Pexh# = 20Tint# = 290ecc# = eccmaxrpm# = 200drive$ =filename$ = “dataoutsimgraph$ =I***inputs***CLSPRINT “Intake tempreature (K) “, Tint#;INPUT temp$IF temp$ <> “ THEN Tint# = VAL(tempS)PRINT “Minimum intake pressure (MPa)”, Pintmin#;INPUT temp$IF temp$ <> ““ THEN Pintmin# = VALQemp$)PRINT “Maximum intake pressure (MPa)”, Pintmax#;INPUT temp$IF temp$ <> “ THEN Pintmax# = VAL(ternp$)PRINT “Exhaust pressure (MPa) “, Pexh#;INPUT temp$IF temp$ <> THEN Pexh# = VAL(temp$)PRINT “Number of points “, nopoints#;INPUT tempSIF temp$ <> THEN nopoints# = VAL(Lemp$)PRINT “Eccentricity (inches) “, ecc# / 25.4;INPUT temp$IF temp$ <> “i’ THEN ecc# = VAL(temp$) * 25.4PRINT “RPM “, rpm#;INPUT temp$IF temp$ <> THEN rpm# = VAL(temp$)212PRINT “data output drive “, drive$;INPUT temp$IF temp$ <> “ THEN drive$ = temp$PRINT “data ouput filename ‘, filename$;INPUT temp$IF temp$ <> “ THEN filenameS = temp$PRINT “do you want graphical output of P and T (yin)”, graph$;INPUT temp$IF temp$ <> ““ THEN graph$ = temp$•***variable intake pressure loop (counter is m)***m=1DOPint# = (m - 1) * ((Pintmax# - Pintmin# / nopoints#) + Pintmin#I***first iteration guesses***pl#(O) = Pint#p2#(O) = Pint#pres#(0) = (bore I / bore2) A 2 * Pint#tl#(0) = Tint#t2#(0) = Tint#tres#(0) = tblk#I***deterIning values of- x (distance from BDC) for every crank angle***- volume- contact angle (alpha)i=0xl#(0) = 0DOi=i+2theta# = i * pi#/ 180alpha#(i) = ecc# * SIN(theta#) i Routerasn# = alpha#(i) + (1/ 6) * alpha#(i) A 3 + (3 / 40) * alpha#(i) A 5 + (15 / 336) * alpha#(i) A7 + (105 / 3456) * alpha#(i) “ 9 + (945 I 42240) * alpha#(i) A + (10395 / 599040) * alpha#(i) A13 + (135135 / 9676800) * alpha#(i) A 15 + (2027025 / 175472640) * alpha#(i) A 17xl#(i) = ((-Router * SIN(pi# - theta# - asn#)) / SIN(theta#)) + Rinnner + rodlenIF i = 180 THEN xI#(180) = (stroke / 2) + ecc#IF i = 360 THEN xl#(360) = (stroke / 2) - ecc#Vol 1#(i) = (pi# * (stroke - xl#(i)) * (borel A 2)! 4) + clvoll#LOOP UNTIL i = 360i=0213DOi=i+2ii = i - 90IF ii 0 THEN ii=i+ 270x2#(i) = xl#(ii)Vo12#(i) = (pi# * (stroke - x2#(i)) * (hore2 A 2)! 4) + clvol2#LOOP UNTIL i = 360Voll#(0) = Voll#(360)Vo12#(0) = Vol2#(360)Pmax#(m) = 0M1#(0) = pl#(O) Vol 1#(0) / (R * tl#(0))M2#(0) = p2#(O) * Vol2#(0) / (R * 12#(0))Mres#(0) = pres#(0) * Vres# / (R * tres#(0))***revolutions loop (counter is j)***j=0DOj=j+1***crank angle ioop (counter is i)***IF graph$ = ‘y” THENSCREEN 12VIEW (1, 1)-(638, 420), 0, 1WINDOW (-1, 0)-(360, 50)LINE (0, -1)-(0, 50), 7LINE (-1, 0)-(360, 0), 7LINE (90, -1)-(90, 1), 7LINE (180, -1)-(180, 1), 7LINE (270, -1)-(270, 1), 7LINE (-1, 2)-(1, 2), 7LINE (-1, 4)-(1, 4), 7LINE (-1, 6)-(1, 6), 7LINE (-1, 8)-(1, 8), 7LINE (-1, 10)-(1, 10), 7LINE (-1, 12)-(1, 12), 7LINE (-1, 14)-(1, 14), 7LINE (-1, 16)-(1, 16), 7LINE (-1, 18)-(1, 18), 7LINE (-1, 20)-(1, 20), 7LINE (-1, tblk# / 10)-(360, tblk# /10), 5LINE (-1, 30)-(360, 30), 5LINE (-1, 21)-(360, 21), 1LOCATE 2, 5: PRINT ‘Temperature”LOCATE 17.5, 5: PRINT “Pressure”214END IFi=ODOi=i+2***first stage***IF xl#(i)> xl#(i - 2) THEN‘compressionpl#(i) = pl#(i - 2) * (Voll#(i - 2) / Voll#(i))” ncomIF pl#(i)> pres#(i - 2) THENpl#(i) = pres#(i - 2)valve 1 = 1ELSEvalvel = 0END IFtl#(i) = tl#(i - 2) * ((pl#(i) / pl#(i - 2)) A ((neom - 1)! ncom))ELSE‘expansionpi#(i) = pl#(i - 2) * (Voll#(i - 2) / Voll#(i)) A (nexp)IF pl#(i) <Pint# THENpl#(i) = Pint#END IFtl#(i) = tl#(i - 2) * ((pl#(i) / pl#(i - 2)) A ((nexp - 1)! nexp))valvel = 0END IFM1#(i) = pl#(i) * Voll#(i) / (R * tl#(i))IF valve 1 = 1 THENMres#(i) Mres#(i - 2) + (M1#(i - 2) - M1#(i))tres#(i) = ((tl#(i) * (M1#(i - 2) - M1#(i))) + (tres#(i - 2) * Mres#(i))) /(Mres#(i) + M1#(i - 2) - M1#(i))pres#(i) = Mres#(i) * R * tres#(i)/ Vres#mtoTl# = mtoTl# - M1#(i) + Mi#(i - 2)ELSEMres#(i) Mres#(i - 2)pres#(i) = pres#(i - 2)tres#(i) = tres#(i - 2)END IFIF (-M1#(i - 2) + M1#(i))> .0000001# AND xl#(i) <xl#(i - 2) THENtl#(i) = ((tl#(i - 2) * M1#(i - 2)) + Tint# * (M1#(i) - M1#(i - 2)))! M1#(i)END IFs***first stage forces and torque***Fradl#(i) = pl#(i) * al#Ftanl#(i) = Fradl#(i) * TAN(alpha#(i))215Ftotl#(i) = SQR((Fradl#(i) A 2) + (Ftanl#(i)) A 2)torql#(i) = Ftanl#(i) * ((Router + ecc# - xl#(i)) / 1000)* *reservoir bypass* * *IF pres#(i) > Pexh# THENpres#(i) = Pexh#mb# = pres#(i) * Vres# / (R * tres#(i))mtotb# = mtotb# + Mres#(i) - rnb#Mres#(i) = mb#END IFI***second stage***IF x2#(i)> x2#(i - 2) THEN‘compressionp2#(i) = p2#(i - 2) * (Vo12#(i - 2) / Vo12#(i)) ‘ ricomIF p2#(i) > Pexh# THENp2#(i) = Pexh#END IFt2#(i) = t2#(i - 2) ((p2#(i) I p2#(i - 2)) A ((ncom - 1)! ncom))valve2 = 0ELSE‘expansionp2#(i) = p2#(i - 2) * ((Vo12#(i - 2)1 Vo12#(i)) nexp)IF p2#(i) < pres#(i) THENt2#(i) = t2#(i - 2) * ((p2#(i) / p2#(i - 2)) ((nexp - 1) I nexp))p2#(i) = pres#(i)valve2 = 1ELSEIF (tres#(0) * (Vo12#(90) / Vo12#(270)) A (ncom - 1))> t2#(270)THEN= t2#(270)ELSE t# = tres#(0) * (Vo12#(90) / Vo12#(270)) A (ncom - 1)END IFt2#(i) = t2#(i - 2) * (Vo12#(i - 2) / Vo12#(i)) A (nexp - 1)valve2 = 0END IFEND IFM2#(i) = p2#(i) * Vo12#(i) / (R * t2#(i))IF valve2 = 1 AND x2#(i) < x2#(i - 2) THENMres#(i) = Mres#(i) - (M2#(i) - M2#(i - 2))pres#(i) = Mres#(i) * R * tres#(i) / Vres#mtoT2# = mtoT2# + M2#(i) - M2#(i - 2)t2#(i) = ((t2#(i - 2) * M2#(i - 2)) + (tres#(i) * (M2#(i) - M2#(i - 2)))) /M2#(i)216END IFI***second stage forces and torque***ii i - 90IF ii <= 0 THEN ii = i + 270Frad2#(i) = p2#(i) * a2#Ftan2#(i) = Frad2#(i) * TAN(alpha#(ii))Ftot2#(i) = SQR((Frad2#(i) “2) + (Ftan2#(i))” 2)torq2#(i) = Ftan2#(i) * ((Router ÷ ecc# - x2#(i)) / 1000)torqtwo#(i) = torql#(i) + torq2#(i)iii = i - 180IF iii <= 0 THEN iii = i + 180torqfour#(i) = torqtwo#(i) ÷ torqtwo#(iii)torqavg# = torqavg# + torqfour#(i)* *graphing* * *IF graph$ = “y” THENPSET (i, (tl#(i) / 10)), 11PSET (i, (t2#(i) / 10)), 12PSET (i, (tres#(i) / 10)), 10PSET (i, pl#(i)), 11PSET (i, p2#(i)), 12PSET (i, pres#(i)), 10END IF***heat transfer from bulk***IF tl#(i) < tblk# THENtl#(i) = t1#(i) + .005 * (tblk# - tl#(i))pl#(i) = (M1#(i) * R * tl#(i)) / Vol 1#(i)END IFIF t2#(i) < tblk# THENt2#(i) = t2#(i) + .005 * (tblk# - t2#(i))p2#(i) = (M2#(i) * R * t2#(i)) / Vo12#(i)END IFIF tres#(i) <> tblk# THENtres#(i) = tres#(i) + .005 * (tblk# - tres#(i))pres#(i) = (Mres#(i) * R * tres#(i)) / Vres#END IF•***end off crank angle loop’LOOP UNTIL i = 360I***mass flow rate calculation***217MFIowl#(j) = mtoTl# * 2 rpm# * 60MFlow2#j) = mtoT2# * 2 * rpm# * 60MFlowbyps#(j) = mtotb# * 2 * rpm# * 60MFlow#(m) = (MFlowl#(j) ÷ MFlow2#(j) + MFlowbyps#(j)) /2trqavg# = torqavg# / 180power#(m) = trqavg# * rpm# 2 ‘ pi# / 60 / 1000torqavg# = 0mtoTl# = 0mtoT2# = 0mtotb# = 0I***continuity***tl#(0) = tl#(360)pl#(O) = pl#(36O)M1#(0) = M1#(360)t2#(0) = t2#(360)p2#(O) = p2#(36O)M2#(0) = M2#(360)Mres#(0) = Mres#(360)pres#(0) = pres#(360)tres#(0) = tres#(360)I***verifying equilibrium criterion and max flow rate reqt***IF ABS(MFlowl#(j) - MFIow2#(j) - MFlowbyps#(j)) < .2 AND j> 10 THENj = 120END IF***end of revolution ioopLOOP UNTILj = 120***calcu1ation of output data***pintersta# = 0i=0DOi=i+2pintersta# = pintersta# + pres#(i)LOOP UNTIL i = 360pintersta# = pintersta# / 180prt#(m) = Pexh# / Pint#prl#(m) = pintersta# / Pint#pr2#(m) = Pexh# / pintersta#I***end of intake pressure 1oop’m = m + 1LOOP UNTIL m = (nopoints# + 1)I***data output to disk***OPEN drive$ + filename$ FOR OUTPUT AS #2m=1DOPRINT #2, prt#(20), prl#(rn), pr2#(m), MFlow#(m), power#(m)m = m + 1LOOP UNTIL m = (nopoints# + 1)CLOSE #2END218


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