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Master of Technology1

Numerical Simulation and2Experimental Validation of3

Compliant Surface Gas Film Bearings4For high speed applications5

Balaji Sankar6

7

Academy of Scientific and Innovative Research8

Propulsion Division9

National Aerospace Laboratories10


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11

Co-ordinator

Dr.Vidyadar Mudkavi

Group HeadDr. Soumendu Jana

Guide

Mr. Sadanand Kulkarni

Date of the Graduation

26th September 201212


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Do what you can with what you have, where you are.13


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Contents14

Abstract 115

1. Introduction 3161.1. Gas Foil Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3171.2. Literature Survey . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

1.2.1. Simulation of Foil Bearings and Experimental Validation . . . 4191.2.2. Thermo-Hydrodynamic Modeling and Measurements . . . . . 620

1.2.3. Structure Model . . . . . . . . . . . . . . . . . . . . . . . . . . 8211.2.4. Encircling Foil Coatings . . . . . . . . . . . . . . . . . . . . . 922

1.3. Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023

2. Models 13242.1. Fluid Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1325

2.1.1. Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13262.1.2. Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . 16272.1.3. Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . 18282.1.4. Simplifications of Incompressible Reynolds Equation . . . . . 1929

2.1.5. Compressible Reynolds Equation . . . . . . . . . . . . . . . . 2230

2.1.6. Large Harrison Number Solution . . . . . . . . . . . . . . . . 23312.1.7. Infinitely long bearing . . . . . . . . . . . . . . . . . . . . . . 26322.1.8. Infinitely Short Bearing . . . . . . . . . . . . . . . . . . . . . 2633

2.2. Structure Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2734

3. Solution of Steady Reynolds Equation Using Multi Dimensional Newton35Raphsons Method 33363.1. Reynolds equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3337

3.1.1. Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 34383.2. Finite Difference form . . . . . . . . . . . . . . . . . . . . . . . . . . 3539

3.3. Examples of Multi Dimensional Newton Raphsons Method . . . . . . 39403.3.1. One Dimensional Example . . . . . . . . . . . . . . . . . . . . 39413.3.2. Multi Dimensional Example . . . . . . . . . . . . . . . . . . . 4142

3.4. Linearizing the Reynolds equation . . . . . . . . . . . . . . . . . . . 42433.4.1. Applying Multi-Dimensional Newtons Method . . . . . . . . . 49443.4.2. Matrix form of equations: Method 1 . . . . . . . . . . . . . . 49453.4.3. Matrix form of equations: Method 2 . . . . . . . . . . . . . . 52463.4.4. Gauss-Seidel solver . . . . . . . . . . . . . . . . . . . . . . . . 5347

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Contents Contents

3.4.5. Force computation . . . . . . . . . . . . . . . . . . . . . . . . 54483.5. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5549

4. Solution Using Iterative Non-Linear Scheme 59504.1. Iterative Non-Linear Scheme . . . . . . . . . . . . . . . . . . . . . . . 5951

4.2. Finite Difference form . . . . . . . . . . . . . . . . . . . . . . . . . . 59524.3. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6153

5. Solution Using Finite Volume Method 63545.1. Reynolds equation in Vector Form . . . . . . . . . . . . . . . . . . . 6355

5.1.1. Greens Theorem in Normal Form . . . . . . . . . . . . . . . . 63565.2. Finite Volume Procedure . . . . . . . . . . . . . . . . . . . . . . . . . 6457

5.2.1. Evaluation of line integral . . . . . . . . . . . . . . . . . . . . 64585.2.2. Flux At The East Face . . . . . . . . . . . . . . . . . . . . . . 65595.2.3. Flux At The West Face . . . . . . . . . . . . . . . . . . . . . . 6560

5.2.4. Flux At The North Face . . . . . . . . . . . . . . . . . . . . . 6661

5.2.5. Flux At The South Face . . . . . . . . . . . . . . . . . . . . . 67625.2.6. Evaluation of Surface Integral . . . . . . . . . . . . . . . . . . 67635.2.7. Assembled equation . . . . . . . . . . . . . . . . . . . . . . . . 67645.2.8. Evaluation of Fluxes . . . . . . . . . . . . . . . . . . . . . . . 69655.2.9. Evaluation of convective and diffusive flux using exact solution 71665.2.10. Handling the Non-Linearities . . . . . . . . . . . . . . . . . . 7767

5.3. Description of Program . . . . . . . . . . . . . . . . . . . . . . . . . . 77685.3.1. Outline of the program . . . . . . . . . . . . . . . . . . . . . . 77695.3.2. Description of the object model . . . . . . . . . . . . . . . . . 77705.3.3. Flow chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7971

5.4. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7972

6. Pseudo Spectral Method 81736.1. Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8274

6.1.1. Estimation of Derivatives and transformations . . . . . . . . . 83756.1.2. Fourier approximation of pressure distribution . . . . . . . . . 86766.1.3. Fourier derivative matrix . . . . . . . . . . . . . . . . . . . . . 8777

6.2. Chebyshev polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 91786.2.1. Estimation of derivatives and transformations . . . . . . . . . 95796.2.2. Chebyshev approximation of pressure distribution . . . . . . . 9680

6.2.3. Chebyshev derivative matrix . . . . . . . . . . . . . . . . . . . 97816.3. One dimensional solution . . . . . . . . . . . . . . . . . . . . . . . . . 98826.4. Two Dimensional solution . . . . . . . . . . . . . . . . . . . . . . . . 9983

6.4.1. Choice of Chebyshev polynomials in the axial direction . . . . 99846.4.2. Time Marching Scheme . . . . . . . . . . . . . . . . . . . . . . 9985

6.5. Integration of pressure to get force. . . . . . . . . . . . . . . . . . . . 101866.6. Grid dependency studies . . . . . . . . . . . . . . . . . . . . . . . . . 102876.7. Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10388

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Contents

7. Comparative study 117897.1. Validation case, Eccentricity =0.79 and Bearing Number =0.4985 . . 117907.2. Behavior at Other Eccentricity and Bearing Numbers . . . . . . . . . 11991

7.2.1. High Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . 119927.2.2. Low Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . 12093

8. Experimental Validation 131948.1. Experimental setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131958.2. Shaft center location. . . . . . . . . . . . . . . . . . . . . . . . . . . 13496

8.2.1. Using minimization approach . . . . . . . . . . . . . . . . . . 136978.2.2. Conventional method . . . . . . . . . . . . . . . . . . . . . . . 13698

8.3. Prediction of pressures . . . . . . . . . . . . . . . . . . . . . . . . . . 138998.3.1. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139100

Acknowledgments 143101

A. Appendix 145102A.1. Navier Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . 145103A.2. Stokes Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 147104A.3. Grosss approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148105A.4. Conversion to cylindrical co-ordinates using chain rule . . . . . . . . . 150106A.5. Greens Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152107

Bibliography 153108

Bibliography 153109

Index 157110

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Abstract111

The work carried out in the field of Numerical simulation and experimental valida-112tion of Gas foil bearings, in the National Test Facility for Rolling Element Bearings113in NAL-CSIR, as a part of masters course work is documented here. After a brief114presentation of essential theory, 6 fluid models are developed to solve the Reynolds115equation. This Reynolds equation governs the fluid between the shaft and encircling116foil of a foil bearing. Then each of these fluid models are coupled with the same117structural model and shaft model to generate 6 simulation models. These simulation118

models are then used to predict steady state pressure distribution around a bearing119 for which published data is available. Based on comparison with published data, the120two dimensional pseudo spectral scheme is chosen among the fluid models due to its121accuracy and run time characteristics. This simulation model is then applied to the122bearing that was fabricated and tested at NAL-CSIR at various RPMs from 20000123to 40000. Two algorithms to predict the shaft center location at various RPMs are124coupled with the simulation model. The predicted shaft center and measured shaft125center are compared at various RPMs. The shaft center in the experiments are126measured using ADRE system employing eddy current probes. Then the simulated127pressure around the shaft at these RPMs are validated by measuring the pressure128around the shaft using ESP scanners.129

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Chapter 1 Introduction

Figure 1.2.: Foil bearing, Shaft at an eccentric location

be sufficient to form a layer of air film between the journal and bearing surface and148completely separates them apart as depicted in Figure 1.1 and Figure 1.2. Though149the picture the shaft is shown to be at the center of the sleeve, its usually at an150eccentric position. The eccentric location of the shaft is shown in Figure 1.2. The151occurrence of this lift-off speed can be found by the sharp reduction in the torque152required to drive the shaft. The pressure distribution in the gas film exerts a pres-153sure force that supports the shafts weight. 1. The bump foil provides support for154the top foil and its compliance allows the top foil to deform under the action of hy-155drodynamic pressure. This deformation changes the geometry of the flow field. This156geometry change in turn affects the pressure distribution. Thus it is necessary that157we need to consider the elastic deformation of the top foil and the compressibility158of the gas to adequately model the bearing.159

1.2. Literature Survey160

1.2.1. Simulation of Foil Bearings and Experimental Validation161

On the numerical simulation front, Tadjbakshi provided a scheme to numerically162solve for pressure distribution considering the foil deflection in 1983 in [32] . The163compressibility of gas was included in the analysis. He showed that the basic equa-164tion for determination of pressure distribution becomes a third order boundary value165problem in terms of film thickness. A simple numerical scheme for solution of the166nonlinear boundary value problem was also presented.167

Prior to that in 1977, Langlois improved his own previous foil bearing model in168

[23]. Initially, his foil bearing model had an assumption of light loading. Un-169der light loading assumption, the change in lubricating film thickness arising from170

1The numerical simulation of this pressure and the consequent load carrying ability forms thebulk of this work.

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1.2 Literature Survey

the foil deflection was ignored insofar as determination of the lubrication pressure171is concerned. Ref.[23] described a series of computations carried out without the172assumption of light loading. He introduced the basic approach of working back and173forth between a foil-deflection calculator and an iterative Reynolds equation solver.174At each stage of the procedure the foil deflection calculator produces an enlight-175

ened guess for the foil shape. Next, the pressure distribution required to balance176tension forces so as to support this shape exactly is determined. This is used as the177zero-iterate in the next application of the Reynolds equation solver. This approach178is followed till date in the iterative solution schemes of Reynolds equation.179

Peng in Ref [26], developed a model of the foil bearing considering the compressibiltiy180of the fluid into consideration. Though such a model had initially been put forward181by Tadjbakshi in Ref[32] ,a novel way to estimate the load carrying capacity of182foil bearing using minimum film thickness as the determining parameter weas put183forward that agreed very well with the experimental results on the load capacity of184these bearings. A series of predictions of the load-carrying capacity based on the185

numerical solution for pressure was presented that cover a wide range of operating186speeds.187

A theoretical analysis model was presented in [10] to investigate the nonlinear dy-188namic behavior of bump-type foil bearings, i.e., the instability and unbalance re-189sponse. In their developed model, the foil structure of bump-type foil bearings was190simulated using the link-spring model, which was presented and validated in a pre-191vious study[9]. They coupled the structure and fluid model through pressure and192film thickness. An iteration solution method was applied for the shaft orbit. They193conducted parametric studies using the following design parameters:194

1. Bump number195

2. Length ratio ( ratio of segment between two bumps and a bump)196

3. Foil thickness197

4. Bump height198

5. Youngs Modulus, were presented upon the analysis of static199

They noted that more bumps or lager length ratio lead to larger load capacity.200But the bearing load capacity did not change with different bump heights. They201concluded that more the numbe rof bumps, lower is the threshold speed of instabil-202

ity,especially when the number of bumps is low. However, it almost does not affect203by other four parameters.204

NASA has developed an analysis and prediction tool for foil bearings named XL-205GFBTH, documented in Ref [17]. A test rig at GRC was used as a simulated case206study to compare rotor dynamic analysis using output from the code to actual rotor207response as measured in the test rig. The comparison shows that the rotor dynamic208coefficients calculated using XLGFBTH represented the dynamics of the system209reasonably well especially as they pertain to predicting critical speeds.210

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The Role of Radial Clearance on the Performance of Foil Air Bearings was studied211by Kevin in Ref[28]. Load capacity tests were conducted to determine how radial212clearance variations affected the load capacity coefficient of foil air bearings. In-213creases in radial clearance were accomplished by grinding the shaft. The test results214indicate that, each test bearing exhibited an optimum radial clearance that resulted215

in a maximum load capacity coefficient. Bearings operating with radial clearances216less than the optimum exhibit load capacity coefficients that are a strong function of217radial clearance and are prone to a thermal runaway failure mechanism and bearing218seizure. Conversely, a bearing operating with a radial clearance twice the optimum219suffered only a 20% decline in its maximum load capacity coefficient and did not220experience any thermal management problems.221

An experimental test program was conducted at NASA [4], to determine the highly222loaded performance of current generation gas foil bearings at alternate pressures223and temperatures. Load capacity tests were conducted at 3, 6, 9, 12, 15, 18, and22421 kRPM at temperatures from 25 to 500 C and at pressures from 0.1 to 2.5 at-225

mospheres. Results showed an increase in load capacity with increased ambient226pressure and a reduction in load capacity with increased ambient temperature. Be-227low one-half atmosphere of ambient pressure a dramatic loss of load capacity was228noted. Additional lightly loaded foil bearing performance in nitrogen at 25 C and229up to 48 atmospheres of ambient pressure was also reported. In the lightly loaded230region of operation the power loss increases for increasing pressure at a fixed load.231

In order to address the problem of misalignment in foil bearings,NASA [19] con-232ducted a preliminary experimental investigation of the misalignment tolerance of233gas foil journal bearing systems. Using a notional gas foil bearing supported rotor234and a laser-based shaft alignment system, increasing levels of misalignment were im-235parted to the bearing supports while monitoring temperature at the bearing edges.236The amount of misalignment that induced bearing failure was identified and com-237pared to other conventional bearing types such as cylindrical roller bearings and238angular contact ball bearings2. They found that due to misalignment , the foil239bearings became more edge loaded. They identified that foil bearings tolerated a240misalignment of83 ( equivalent to 8 microns per mm of misalignment).241

1.2.2. Thermo-Hydrodynamic Modeling and Measurements242

Thermo-Hydrodynamic modelling of foil bearings are equally important to load pre-243diction capability due to the ever present danger of a thermal runaway. In [22], the244authors present a model for the thermal energy transport in a Gas foil bearing sys-245tem operating at high temperature with typical inner and/or outer cooling flows.246The film temperatures they predicted agree with previous test data, demonstrat-247ing the effectiveness of an outer cooling stream to remove heat and to control the248operating temperature. They showed that the inner flow stream is not as efficient249

2These bearings can take up to 4e-3 misalignment

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in cooling the system.Theyheated the rotor electrically to 1320C and measured the250temperature at various points and measured the shaft motion. In speed-up tests251to 26 krpm, the rotor motion amplitude dropped suddenly just above the critical252speed, thus, evidencing the typical hardening of compliant bearings. At the hottest253test condition, since air is more viscous, the rotor peak motion amplitude decreased,254

not showing a jump. The coast-down tests showed that the critical speed increased255slightly as the temperature increases.256

A similar Thermo-Hydrodynamic study was also carried out by San Andres in [2].257His model also included thermal energy transport in the gas film region and with258cooling gas streams, inner or outer, as in typical rotor-GFBs systems. His analysis259also accounted for material property changes and the bearing components expan-260sion due to temperature increases and shaft centrifugal growth due to rotational261speed. He considered a detachable top foil, one in which there is no possibility of262sub ambient pressure being developed. He predicted that the film peak temperature263occurs just downstream of the maximum gas pressure. The film temperature was264

shown to be higher at the bearing middle plane than at the foil edges. The journal265speed, rather than the applied static load, influences more the increase in film tem-266perature and with a larger thermal gradient toward the bearing sides. Obviously267the gas film temperature increases rapidly due to the absence of a forced cooling268air that could carry away the recirculation gas flow and thermal energy drawn by269the spinning rotor. He compared the model considering temperature effects to the270isothermal model and showed that the THD model produced a smaller journal ec-271centricity (larger minimum film thickness) and larger drag torque. He explained this272effect by reasoning that a higher temperature causes higher gas viscosity and hence273a higher drag.274

Another theoretical model that incorporates thermal structural effects was presented275in Ref[33].This model accounted for the following factors:276

1. Bending and membrane effects in the top foil resulting from temperature277change278

2. Thermal expansion of the journal, subfoil, and bearing housing.279

3. Thermal transport through the journal, foils, and bearing housing.280

Pressure in the gas film was predicted using the Reynolds equation, and a thermal281bulk flow model was used to predict temperature. The authors demonstrated that282

models will overpredict film thickness along the side edge of a bearing if thermal283strain in the top foil is not included.284

An experimental investigation into the temperature profile of a compliant foil air285bearing was made by Kevin Radil in Ref[30]. A series of tests were performed to286determine the internal temperature profile in a compliant bump type foil journal air287bearing operating at room temperature under various speeds and load conditions.288The temperature profile was collected by instrumenting a foil bearing with nine, type289K thermocouples.The thermocouples were tack welded to the backside of the bumps290

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that were in direct contact with the top foil. Tests were conducted at speeds from29120 to 50 kRPM and loads ranging from 9 to 222 N.The results indicated that, over292the conditions tested, both journal rotational speed and radial load are responsible293for heat generation with speed playing a more significant role in the magnitude of294the temperatures. The temperature distribution was nearly symmetric about the295

bearing center at 20 and30 k RPM but became slightly skewed toward one side296at 40 and 50 kRPM. Surprisingly, the maximum temperatures did not occur at297the bearing edge where the minimum film thickness is expected but rather in the298middle of the bearing where analytical investigations have predicted the air film to299be much thicker.Thermal gradients were common during testing and were strongest300in the axial direction from the middle of the bearing to its edges, reaching 3.78301C/mm. The temperature profile indicated the circumferential thermal gradients302were negligible.303

1.2.3. Structure Model304

Regarding the structural model, in [9], the author has developed an analytical model305of bump-type foil bearings considering the effects of four factors,306

1. Elasticity of bump foil307

2. Interaction forces between bumps308

3. Friction forces at the contact surfaces309

4. Local deflection of top foil310

They simplified each bump into two rigid links and a horizontally spaced spring,311

the stiffness of which is determined from Castiglianos theorem. The interaction312forces and the friction forces are coupled with the flexibility of bumps through the313horizontal elementary spring. The local deflection of the top foil was described314using a finite-element shell model and added to the film thickness to predict the air315pressure with Reynolds equation. They showed that the bump deflections of a strip316with ten bumps calculated using their model under different load distributions were317consistent with the published results. Moreover, the predicted bearing load and film318thickness obtained from a foil bearing with a bump circumferential extend of 360319deg also agree very well with the experimental data, especially for predictions with320a proper selection of radial clearance and friction coefficients. They noted that the321

radial clearance and friction force variations in the foil bearing significantly changed322the performance of the foil bearing. They demonstrate that the radial clearance of323the foil bearing has an optimum value for the maximum load capacity.324

Structure models that do not consider the interaction between the foils tend to325over estimate the flexibility. Le Lez in Ref[24] presented a model that describes326the foil bearing structure as a multi-degree of freedom system of interacting bumps.327Each bump included three degrees of freedom linked with elementary springs. The328stiffnesss of these springs are analytically expressed. Once the stiffness matrix of329

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the whole foil bearing structure is obtained, the entire static system is solved taking330friction into account. Despite its relative simplicity, comparisons with finite elements331simulations for various static load distributions and friction coefficients showed a332good correlation. This analytical model was integrated into a foil bearing prediction333code. The load capacity of a first generation foil bearing was then calculated using334

this structure model as well as other simplified theoretical approaches. Significant335differences were observed, revealing the paramount influence of the structure on the336static and dynamic characteristics of the foil bearing.337

In [31] , the author has developed a new foil bearing with compression springs as338the support for the encircling foil. This foil gas bearing uses a series of compression339springs as a compliant structure instead of corrugated bump foils. A spring model340to estimate the stiffness of compression springs was developed and showed a good341level of agreement with the experimental results. The spring dynamics model was342combined with a non-linear orbit simulation to investigate the non-linear behavior343of foil gas bearings. His approach could also predict the structural loss factor given344

the geometry of the underlying springs.The structural stiffness of this new type of345support was highly nonlinear and showed different behavior for static loading and346the sinusoidal dynamic loading. The constructed bearing with rather soft springs347showed a load capacity of 96N at 20,000 rpm under no cooling.348

1.2.4. Encircling Foil Coatings349

Regarding the coatings used on the encircling foil to reduce the friction coefficient3,350notable work was presented in [14] by Heshmat. He tested and evaluated the per-351formance of Korolon Coatings and showed that thought eir friction coefficient was352a function of temperature, in most cases a minimum coefficient of friction less than3530.1 was observed during start up/shutdown periods.354

NASA [[7]] had also developed a coating PS304, a modified chrome oxide based355coating, for foil gas bearings.The coatings were tested at a load of 10 K Pa (I.5 psi)356and the bearings were run under start/stop cyclic conditions. The data show good357wear performance of the bearing especially at temperatures above 25 C. However358the estimated friction coefficient was 0.4, over four times the value reported by359Heshmat in [[14]].360

A comparative study of different foil bearing coatings was done by NASA in Ref[29].361

In this study, room temperature load capacity tests were performed on journal foil362air bearings operating at 14,000 rpm. Different shaft and foil coating technologies363such as plasma sprayed composites,ceramic, polymer and inorganic lubricant coat-364ings were evaluated as foil bearing lubricants. The results indicated that bearing365performance is improved through the individual use of the lubricants and treatments366tested. Further, combining several solid lubricants together yielded synergistically367better results than any material alone.368

3No such coatings were used in this study.

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In Ref[8], Dong-Hyun Lee carried out parametric studies similar to [10] while pre-369senting an algorithm that accounted for the stick slip motions, thus considering the370effect of coulomb friction. The predictions of the unbalance response show that371the foil bearing is very effective on the restriction of the vibration at the resonance372frequency and there exist optimum values of design parameters such as friction co-373

efficient, bump foil stiffness, and the number of bump foil strips with regard to374minimizing the amplitude at the resonance frequency375

1.3. Applications376

Gas foil bearings are currently used in many commercial applications, both terres-377trial and aerospace. Aircraft Air Cycle Machines (ACMs) and ground-based micro-378turbines have demonstrated histories of successful long-term operation using GFBs.379Small aircraft propulsion engines, helicopter gas turbines, and high-speed electric380

motors are potential future applications [6].381

The NASA report[18] extensively documents the gas foil bearing advances from the382point of view of application in a closed circuit Brayton cycle power plant. Since the383working fluid is typically an inert gas like Argon in a CBC, contamination of the384working fluid by the lubricant is intolerable. Gas foil bearings are ideal for such an385application.386

In Ref [34], Christof Zwyssig gives a review of recent applications of high speed387drive systems. He notes of a new topic of research called megaspeed drive sys-388tems, (MegaNdrives-1 million r/min) introduced by the Power Electronics Systems389Laboratory (PES) at Swiss Federal Institute of Technology (ETH) Zurich, in 2004.390

MegaNdrives have very high power densities and are typically limited to smaller391dimensions and/or power levels in the range of 100W due to the material strength392limitations and thermal limitations. Gas foil bearings are ideal for such high speeds,393small loads and do not complicate the system by further adding a lubrication system394to the small setup. The main reason to strive toward the megaspeed range is emerg-395ing applications in the areas of noninvasive imaging techniques in medicine, dental396technology, material processing, air compressors for high-compact fuel cells,and gas-397turbine-driven portable power systems.398

High speed grinding , milling and drilling are increasingly being used in the elec-399tronics industry were a very high precision is required. These require very high tool400

speeds, 150000 RPM in the case of grinding of silicon wafers. High speed(400000401RPM) dental drills powered by compressed air also promise to be a prominent area402of application for gas foil bearings. Gas turbine power generation is commonly used403in largescale power generation systems up to hundreds of megawatts, where the ro-404tational speed is in the order of 10 000 RPM. There are emerging applications for405portable, low-power gas-turbine based power generation systems. At power levels406around 10 W, the gas turbine system occupies a very small volume if the rota-407tional speeds are increased to over 500000 RPM. These systems typically use Gas408

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1.3 Applications

bearings , rigid/flexible. As an example, the Massachusetts Institute of Technol-409ogy (MIT) wafer-based gas turbine prototype uses rigid gas beaing to support the410disk that is both the compressor and the turbine. Additional applications for small411portable power supplies are in unmanned surveillance vehicles, autonomous robots,412and medical applications. Stanford University, in cooperation with M-DOT, has413

been developing a gas turbine with a predicted output power of 200 W and a ro-414tational speed of up to 800000 RPM. The main application is for powering Micro415Air Vehicles. Low mass-High RPM energy storing flywheels proposed to be used416in satellites by NASA are also a prominent application for gas bearings. Their low417mass and compactness would make them ideal for space based applications.418

The authors in [13] describe an oil-free, 150 Hp turbocharger that was successfully419operated with compliant foil bearings in a range of pitch and roll angles, including420vertical operation, thereby demonstrating its viability for aircraft applications.They421ran the test setup upto 120000 RPM. In addition, they also showed that the compli-422ant foil bearing-supported turbocharger successfully tolerated shock and vibration423

of 40 g. They also demonstrate that with state of the art coatings, a coefficient of424friction of less than 0.1 is possible over a wide range of temperatures.425

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2. Models426

The fluid and the structure models used in this simulation study are presented in427this chapter. Reynolds equation is used to model the fluid between the shaft and428the encircling foil. The structure model is the simplified Heshmats model which429does not account for friction between the bumps and the encircling foil. The concise430derivation given here follows Camerons approach from Ref[1]. Grosss Approach431is more involved and is detailed in the appendix along with other mathematical432preliminaries used in this report. This model is used mainly due to its simplicity433and wide spread usage.434

2.1. Fluid Model435

Instead of using the Navier Stokes (NS) equation in its entirety to simulate the fluid436between the journal and the bearing, a simplified model is used here. Solving the full437NS equation is complicated and computationally very expensive. The computational438cost becomes huge particularly when orbit simulations are carried out where the fluid439equations need to be solved at every time step and a huge number of time steps are440involved. Comparative studies have been carried out by Piekos in Ref[27] comparing441

results from full NS simulations and Reynolds equations. The assumptions that442were made to reduce the NS equations to Reynolds equation such as neglection443of fluid inertia were shown to have little effect for his bearing configuration. The444author had concluded that a less expensive tool was needed to simulate the fluid445flow and proceeded to use pseudo spectral methods towards that end. Moreover446for the bearing diameter and RPM of this study(23 mm Diameter, 30000 RPM) ,447Reynolds equation has been used extensively[10, 26, 31, 17, 21].448

2.1.1. Assumptions449

Reynolds equation is derived from Navier Stokes through a series of assumptions.450Justifications that the author could think off are also given below.451

1. Body Forces are neglected.452

a) Except in magnetohydrodynamics 1 Examples of such fluids include plas-453mas, liquid metals, and salt water or electrolytes, body forces can be454

1Magnetohydrodynamics (MHD) (magneto uid dynamics or hydromagnetics) is an academicdiscipline which studies the dynamics of electrically conducting uids.

13


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Chapter 2 Models

neglected.455

2. Pressure is constant through the thickness of the film.456

a) The thickness of the film when oil is the working medium is 25 to 125457microns.The thickness of the film when air is the working medium is 0 to45825 microns. For such small thickness, we can safely assume the pressure459is constant through the thickness of the film.460

3. Curvature of surfaces is large compared to film thickness.461

a) The statement should be that the radius of curvature is large compared to462the thickness of the film. The implication of this assumption is that the463equations representing the fluid flow in the gap can be written in carte-464sian form. This justifications for using cartesian coordinates is explained465below in points.466

b) Polar coordinates are used when the streamlines and boundaries are467

changing in direction that is similar to the change in direction of the unit468vectors of the polar coordinate system. If the surface velocities/stream469lines are straight/not changing in direction, Cartesian coordinates would470be easier. To show that surface velocities are not changing in direction,471It needs to be only shown that the normals to the surfaces are parallel.472The angle between the top and bottom surface normals after traveling a473distance of dx from the y axis is d = dx

(RinnerROuter). Since the difference474

in their radii is small, the angle is small and can be assumed 0. This475makes them parallel.476

c) Andreas S. Szeri puts the same assumption this way. The normals to the477 shaft and foil intersect at the center of the shaft. However he assumes478that the normals are parallel to each other. He says We focus on such479short distances (He means the thickness of the gas film) along the normal480vectors such that they intersect at what seems to us a a very great distance481. To show that this assumption is valid , we show that the difference482in curvature of top and bottom surfaces is small. It is shown in the next483point.484

d) In the bearing studied in this work, diameter of the shaft is 22mm.485Inside diameter of the encircling foil which wraps around the shaft is486

22.050mm(At location of maximum thickness of the film)= 0.022050m.487Thus the radii of the two surfaces are almost same. Thus their cur-488vature(Inverse of radii) differ by not more than 0.20 m1 as shown in489Figure 2.1. In terms of percentage of shaft curvature this is 0.22%. Thus490it is shown that the difference in curvature of the two surfaces is small491and the normal vectors are almost parallel.492

e) Further this scenario can be compared with the boundary layer over the493earths surface. The planetary boundary layer is 2000m thick over tropics494

14


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2.1 Fluid Model

in the trade wind zone.2. Comparing with the radius of the earth, the dif-495ference in curvature between the top and bottom of the boundary layer is4960.3722 107m1. In terms of percentage of earths curvature it is around49723%. The table Figure 2.1 below summarizes the values in this point. It498can be seen that the difference in curvature between the top and bottom499

surfaces is high compared to the difference in curvatures encountered in500the bearing. Navier Stokes equations in Cartesian coordinates are often501used in the preliminary studies of planetary boundary layers.502

4. There is no slip at the boundaries503

a) What this means is that the velocity of oil adjacent t the boundary is the504same as the boundary. The fluid sticks to the surface. There has been505much work on this and is universally accepted. This assumption also506requires the assumption of continuum flow.The assumptions that follow507are put for the simplification of the derivation.508

5. Stress is proportional to rate of shear.509

a) Non Newtonian fluid have much more complicated partial differential510equations that relate strain rate and stress in the fluid and are usually511more difficult to handle. A Newtonian model is assumed here.512

6. Laminar flow513

a) Flow in big turbine bearing is not laminar and their theory is slowly being514developed. For smaller bearings such as those used in this work (23 mm515diameter), flow can be assumed to be laminar.516

7. Fluid inertia is Neglected517

a) Several studies show that even if Reynolds number is 1000, the pressures518are only modified by 5% due to the use of this assumption.519

8. Viscosity is constant through the thickness520

a) This certainly not true but leads to great complexity if not assumed. The521actual variation of viscosity with temperature for an ideal gas is given by522

Surtherlands formula as = 0T0+CT+C

T

T0

3= 2.523

9. Not all velocity gradients are significant524

a) Compared to uz

and wz

, every thing else( meaning all other velocity525gradients) are negligible.526

2Source: Wikipedia, Planetary Boundary Layer

15


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Chapter 2 Models

Parameter Foil Bearing Planetary Boundary layer

Radius of Top Surface 11.025 mm 8400000 mRadius of Bottom Surface 11 mm 6400000 m

% difference in radii 3 0.22 .03125Curvature of Top Surface 90.70m1 1.190 107m1

Curvature of Bottom Surface 90.90m1 1.562 107m1

% difference in curvature 0.22 23.68

Table 2.1.: Curvature Comparison

Figure 2.1.: Curvature of Surfaces

2.1.2. Continuity Equation527

Consider the fluid elementFigure 2.2 as one extending from the top to the bottom528surface. Let its height be h. The roof and floor are impermeably bounded. Its cross529section is of length dx and width dy.530

Top Surface

Bottom Surface

Fuid element

x direction, u, q_x

h

dy dx

Figure 2.2.: Fluid Element

Let u be the velocity of the fluid in x direction. Then fluid flow per unit width qx is531given by

h0

(u)dy . The volume flow rate is qxdy , for the column is dy wide. Rate532of flow out - Rate of flow in is given by533

16


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2.1 Fluid Model

(qx +qx

xdx)dy (qx +

qx

xdx)dy

Similarly in the y direction534

(qy +qy

ydy)dx (qy +

qy

ydy)dx

Let wh is the velocity of the roof of the element and w0 , that of the floor. Due to535this movement, the volume of the fluid changes at a rate of536

(wh w0)dxdy

where dx*dy is the area of the base and top of the element.537

Using the above , volume balance (constant density fluid) can be written as538

qx

xdx dy +

qy

ydy dx + (wh w0)dxdy = 0 (2.1)

where The continuity equation in cartesian coordinates, for a fluid of constant den-539sity can then be written (in terms of mass) as540

( qx)

x+

( qy)

y+

( h)

t= 0 (2.2)

In terms of volume, the equation reduces to

(qx)x

+ (qy)y

+ (h)t

= 0

The fluid element to which the above equation corresponds to extends from top to541the bottom surface. Since the flux across the floor and roof are zero(they are not542porous), flux across them does not occur in the above equation. However the height543of the fluid column can change, which may be due to different velocities of the floor544and the roof.545

17


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Chapter 2 Models

2.1.3. Momentum Equation546

2.1.3.1. X direction:547

Consider only the force balance in the x direction. The forces that concern us are

pressure forces across the front and back faces and the shear forces at roof and floorfaces .

p

x=

zx

z(2.3)

We have not considered the yzterm and thewx

term due to assumption 9.548

p

x=

z

u

z

(2.4)

Since pressure is not a function of z, we can integrate the above as549

u

z=

p

xz+ C1 (2.5)

Considering the viscosity constant along rthe height of the fluid film, we can write,550

u =p

x

z2

2+ C1z+ C2 (2.6)

The boundary conditions for evaluating C1 and C2 are551

z = h, u = U1

z = 0, u = U2

552

The expression for velocity is553

u = 12

px

(z2 zh) + (U1 U2) zh

+ U2 (2.7)

Thus fluid flow per unit width in x directionh

0(u)dy

is554

qx =h3

12

p

x+ (U1 + U2)

h

2(2.8)

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2.1 Fluid Model

2.1.3.2. Y direction:555

Similarly in the Y direction,556

py

= z

V

z

(2.9)

Following a similar procedure as followed in the x direction, we obtain the fluid flow557per unit width in the y direction as558

qy =h3

12

p

y+ (V1 + V2)

h

2(2.10)

2.1.3.3. Full Reynolds Equation(Incompressible)559

Substituting expressions for qx and qy in the continuity equation we get,560

x(

h3

p

x) +

y(

h3

p

y) = 6

x(U1 + U2) h +

y(V1 + V2) h + 2(wh w0)

(2.11)

ThisEquation 2.11 is the Reynolds equation with an incompressible fluid approxi-561mation.562

2.1.4. Simpli cations of Incompressible Reynolds Equation563

The full Reynolds equation with an incompressible fluid approximation is written564as565

x h3

p

x+

y h3

p

y = 6

x[(U

1+ U

2) h] +

y[(V

1+ V

2) h] + 2(w

h w

0)

(2.12)

This equation can be simplified to yield566

x

h3

p

x

+

y

h3

p

y

= 6U

dh

dx(2.13)

19


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Chapter 2 Models

for steady and567

x

h3

p

x

+

y

h3

p

y

= 6

U

dh

dx+ 2

dh

dt

(2.14)

for unsteady conditions.568

The process of simplification is detailed below.569

2.1.4.1. Negligible axial velocity570

First consider the term571

x

(U1 + U2) h +

y

(V1 + V2) h

Considering the directions of the velocities a shown inFigure 2.3 we can see that for

Figure 2.3.: Direction of velocities

572

v to be non zero,the inner race has to be slide in and out of the outer race. This573wouldnt be practical for a journal bearing. In a journal bearing only the tangential574velocities,u are of prime importance. Thus the term considered here reduces to575

x(U1 + U2) h +

y(V1 + V2) h

x{(U1 + U2) h}

20


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2.1 Fluid Model

2.1.4.2. Rigid inner race576

Since the inner race is rigid, for a cylinder rotating at constant angular velocity ,the tangential velocity at the surface is constant every where and is not a functionof x. If the shaft is rotating and the inner surface of the bearing is fixed, U2 = 0.Hence the term further reduces to

x{(U1 + U2) h} U

h

x

The term (wh w0) is just the variation of height of the fluid film with time thatcan be written by dh

dt. Thus the rhs is now

6

x[(U1 + U2) h] +

y[(V1 + V2) h] + 2(wh w0)

6

U

dh

dx+ 2

dh

dt

2.1.4.3. Steady running bearings577

The inner race is not going up and down inside the outer race or sideways. Thismeans that the film thickness at a point is not a function of time. This effect is thesqueeze film term.Thus the rhs is now

6

x[(U1 + U2) h] +

y[(V1 + V2) h] + 2(wh w0) 6U

dh

dx

2.1.4.4. Constant viscosity578

Viscosity has already been assumed constant along z direction, along the film thick-579ness. Now assume it is constant everywhere. This reduces the lhs as shown580

x h3

p

x +

y h3

p

y 1

x h3 p

x +

y h3 p

y Thus the simplified steady Reynolds equation can now be written as581

x

h3

p

x

+

y

h3

p

y

= 6U

dh

dx

(2.15)

21


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Chapter 2 Models

2.1.5. Compressible Reynolds Equation582

However since we are concerned with compressible gas, we need to user mass conti-583nuity and not volume continuity. Using the mass conservation equation instead of584volume conservation equation we get,585

x(

h3

p

x)+

y(

h3

p

y) = 6

x( (U1 + U2) h) +

y( (V1 + V2) h) + 2 ( (wh w0))

(2.16)

For our case, (Gas bearings) a further simplifications can be made using either of586the 2 options587

1. Use gas law pv = Constant588

2. Isothermal condition = 1, or p / 589

Using the second and replacing all using p , we get the equation as590

x(ph3

p

x)+

y(ph3

p

y) = 6

x(p (U1 + U2) h) +

y(p (V1 + V2) h) + 2 (p (wh w0))

(2.17)

upon using the simplifications discussed earlier we have,591

x

ph3

p

x

+

y

ph3

p

y

= 6

U

x(ph) + 2

(ph)

t

(2.18)

2.1.5.1. Non-dimensionalization592

The Non-dimensionalization of pressure is wrt ambient pressure. x () is wrt radius,593y wrt length of the bearing. The clearance is non-dimensionalized wrt initial clear-594ance. The new variables are non dimensional terms relating to the original terms595according to the relations given below.596

pN D ( p

pah ( chN D

x ( RxN D

y ( L

2yN D

22


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2.1 Fluid Model

where,Pa is the ambient pressure, c is the radial clearance, R and L are the radius597and length of the bearing. The use of non-dimensionalizing wrt L

2and not L is to598

have the LD

term explicitly appear in the equation. The subscripts are not used after599that to make the equations clearer. Multiplying by R2 on both sides yields600

x

ph3

p

x

+

4R2

L2

y

ph3

p

y

=

6U R

c2pa

x(ph)

+

12B2

c2pa

(hp)

t

x

ph3

p

x

+

1LD

2 y

ph3p

y

=

6U R

c2pa

x(ph)

+

12B2

c2pa

(hp)

t

(2.19)

In the above equation, the term next to the normal wedge action, is the Harrison601number or the compressibility number or bearing number.602

Bearing Number =

6U R

c2pa

In a more conventional form,603

Bearing Number =6!

pa

R

C

2

And once time is also non-dimensionalized wrt inverse of angular velocity(time taken604to rotate once), the equation becomes simpler.605

x

ph3

p

x

+

1

LD2

y

ph3

p

y

=

(ph)

x+ 2

(ph)

2.1.6. Large Harrison Number Solution606

Consider a bearing without the squeeze film term (Steady running bearing). For an607infinitely long bearing, derivatives wrt y are not important compared to derivatives608wrt x. In the non dimensional form , this assumption is inserted by approximating609that R

L= 0.610

23


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Chapter 2 Models

x

ph3

p

x

=

6U B

c2pa

x(ph)

For large value of Harrison number 6U Bc2pa ! 1 , a restriction can be placed on611the product p*h . This large Harrison number case arises in cases of612

Large tangential velocities in the film613

High viscosity fluids614

Small clearances615

large bearing radii616

Hence in order to keep the LHS finite, the term x

(ph) has to small. This means617that (ph) ! constant.It has been shown by Gross that618

ph =1 + 1.52

1 + 2(2.20)

h = 1 + cos () (2.21)

p =1 + 1.52

(1 + cos ()) (1 + 2)(2.22)

The second Equation 2.21 above is geometrical and is substituted in to the first to get619the expression for pressure in terms of only . Plots of the above equation for pressure620is shown next. For the derivation ofh = 1 + cos (), refer to structural model. The621

plot of the solution is shown in Figure 2.4. The code listing for generating the plot622is also given below.623

1 c le ar a l l ; c l c ; c l f ;6242 e =2;%Non d i me ns i on a l e c c e n t r i c i t y , 0 t o 16253 the ta =0:365 ;%t h e t f ro m p o i n t o f maximum c l e a r n c e 6264 i =1;%I nd ex f o r l o t t i n g 6275 for e =0 : . 1 : . 46286 p_non_dim( i , : )=(1+1.5 e^2) ./( (1+e^2) .(1+ e . co s ( the ta pi /180 ) ) ) ;6297 i=i +1;6308 end631

9 %p l o t and l a b e l 63210 plot ( th et a , p_non_dim ) ;63311 xlabel ( Theta from poi nt of maximum cl ea ra nce , deg re es ) ;63412 ylabel ( Non dimen siona l pre ss ur e ) ;63513 t i t l e ( Pressure d i st r i bu t i on in a Gas jou rna l bear ing ) ;63614 legend ( E cc e nt r ic i t y r a t i o = 0 , 0. 1 , 0. 2 , 0. 3 , 0. 4 ) ;63715 %t o mark l o c a t i o n o f min c l ea r an ce i n p l o t 63816 x=[180 180 ] ;639

24


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2.1 Fluid Model

.

Figure 2.4.: Large Harrison number pressure distribution

17 y = [0 .6 2 ] ;64018 z =[0 0 ] ;64119 l i n e (x , y , z ) ;64220 text (1 80 , . 6 , Loca tion of minimum cl ea ra nce , 180 deg re es )6432164422 GrossPressureEqn = $ p=\f ra c {1+1.5\ ep si lo n {2}}{(1+\ ep si lo n cos \ l e f t (\ thet645

23 h = text ( s t r i n g , Gross Pressur eEqn , . . .646

24 i n t e r p r e t e r , l a t e x , . . .64725 f on ts iz e , 4 0 , . . .64826 un i t s , norm , . . .64927 pos , [ 2 00 1 ] ) ;65028 mytexstr = $p=\f ra c {1+1.5\ ep si lo n {2}}{(1+\ ep si lo n cos \ l e f t (\ theta \ ri gh t )65129 h = text ( s t r i n g , mytexstr , . . .65230 i n t e r p r e t e r , l a t e x , . . .65331 f on ts iz e , 2 6 , . . .65432 un i t s , norm , . . .65533 pos , [ .0 1 . 9 ] ) ;65634 saveas ( gcf , Pre ssu reD ist rib uti on Gro ss Eqn , eps )657

The theoretical solutions of the Reynolds equation can be obtained by omitting658either of the co-ordinate directions. In the long bearing assumption, length wise659gradients are omitted and and in the short bearing assumption, tangential gradients660are omitted.661

25


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Chapter 2 Models

2.1.7. In nitely long bearing662

The first term on the lhs ofEquation 2.19 contains the variation of pressure in tan-663gential direction . The second term contains pressure variation in axial, y direction.664Considering the bearing as infinitely long and looking at only the central part of it,665

we can put all derivatives wrt y as 0. The Reynolds equation then becomes666

x

h3

p

x

= 6U

dh

dx

(2.23)

Integrating once, we get ,667

h3dp

dx

= 6U h + C

where C is the integration constant. Letting h be the film thickness at point of zero668pressure gradient, we can evaluate the constant in terms ofh .669

The equation can be written as670

dp

dx= 6U

h h

h3(2.24)

In order to get the expression of pressure from the above, we need671

1. An expression for h in terms of x672

2. The film thickness at point of zero pressure gradient, h.673

2.1.8. In nitely Short Bearing674

This means that gradient in the x direction is small than the gradient in the y675direction, which is very small now. The Reynolds equation can now be written as676

y

h3

p

y

= 6U

dhdx

(2.25)

Film thickness is a function of x and not of y. So,677

h3

y

p

y

= 6U

dh

dx

26


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2.2 Structure Model

Figure 2.5.: Infinitely long bearing, conceptual

h3d2p

dy2= 6U

dh

dx

The above equation is known as Ocvirks equation. Upon integrating twice and678using the boundary condition that pressure is zero at both +/- l/2 , where l is the679length of the bearing, we get the following relation for pressure.680

P = 3U dxdh

h3

y2

l2

4

(2.26)

2.2. Structure Model681

The primary objective of the structural model is to provide the film thickness be-682tween the journal and the encircling foil to the fluid model. This film thickness will683be uniform when the pressure is atmospheric everywhere and the journal is at the684center of the bearing.685

When journal center is off set from the bearing center, the film thickness distributionbecomes a function of and is given by

h = H + cos ( o) (2.27)

Here, h is the film thickness at location and H is the original of film thickness when686the journal was concentric with the bearing. is the distance between the center687

27


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Chapter 2 Models

Figure 2.6.: Infinitely Thin Bearing, Conceptual

of the bearing and center of the journal. The conventions for and are shown688in Figure 2.7. is usually from the line of centers in clockwise direction from the689maximum clearance end while is from between the line of centers and the load line690. But here, Heshmat uses a different convention, where is from the vertical line in691anti clockwise direction and o is between the location of minimum clearance and692the vertical line. The origin of is the other end of the vertical line, meaning , for693the same location, the measurements of and will differ by 1800. As a check on694

the above expression, consider the location of minimum thickness of film, for which695

0 = 1800, and e gets subtracted(cos () = 1) from C . At max thickness,696

= 0in magnitude, and e gets added(cos (0) = 1) to c in full.697

When the effect of pressure deforming the encircling foil is considered, the equation698becomes699

h = H + cos ( o) + k1 (p pa) (2.28)

Here ,p

ais the ambient pressure. If the pressure at a point is above ambient at some700

location ,the foil deflects down and the film thickness at that point increases. Thus701k1 denotes the structural compliance of the foil structure and is lesser for stiffer702structure. When the foil structure is rigid, then K becomes zero. This equation can703be non-dimensionalized wrt initial clearance distribution to give704

h = 1 + e cos ( o) +k1

H(p pa) (2.29)

28


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2.2 Structure Model

Figure 2.7.: Angle conventions, Heshmats approach

and taking the pressure term out from the last term we get,705

h = 1 + e cos ( o) +k1pa

H(P 1) (2.30)

Now pressure occurs in a non-dimensional form in the equation. The force acting706per pitch is s pa [P 1]. IfKb is the stiffness per unit transverse length, then the707non dimensional deflection due to pressure alone is given by708

Deflection =s pa [P 1]

H Kb(2.31)

29


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Chapter 2 Models

The expression for Kb is given by (from Ref.[15])709

Kb =Et3

12(1 2

) l30

lois the half length of the bump, s is the pitch of the bump, E is the Youngs modulusof the material of the bumps, is the Poissons ratio and t is the thickness of the foil.This expression can also be derived as follows. Let x be the strain in x direction, ybe the strain in the transverse direction. Since the bump deforms as a whole, strainin the transverse direction is zero.

y = 0 = y x

E

x = x yE

This implies that, y = x and

x = x

E

1 2

(2.32)

x =Ex

(1 2)(2.33)

If w is the deflection of the mean line from the un-deflected position, then the radiusof curvature

1

R=

d2w

dx2

and the strain at any location z4 from the center line can be written as

x = z2w

x2

Then Equation 2.33 can be written as

x =E

z2w

x2

(1 2)

4z is the coordinate along the thickness of the bump

30


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2.2 Structure Model

This stress distribution causes a bending moment of

M =

t2

t

2

( xz) dz

M =

t2

t2

E

z

2wx2

(1 2)

z

dz

M =Et3

12(1 2)

2w

x2

Comparing with the Eulers bending equation , we get the stiffness of the plate per710unit length in the transverse direction as as G = Et

3

1 2. Please refer to Figure 2.8

Figure 2.8.: Bump Model

711

for the bump model. The non dimensional compliance to be used in the equation is712now713

=s pa

H Et3

12(1

2)l30

The structure model is now,714

h = 1 + e cos ( o) + (P 1) (2.34)

31


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41/165

3. Solution of Steady Reynolds715

Equation Using Multi Dimensional716Newton Raphsons Method717

3.1. Reynolds equation718

The compressible Reynolds equation was derived in the previous chapter as

x

ph3

p

x

+

y

ph3

p

y

= 6U

(ph)

x

(3.1)

Non dimensionally, when axial non-dimensionalisation is wrt L/2,719

x

ph3

p

x

+ (1= (L= D)2)

y

ph3

p

y

=

(ph)

x(3.2)

In the above case the limits of the bearing will be from -1 to +1 in the non dimen-720sional space.When axial non-dimensionalization wrt L,721

x

ph3

p

x

+ (R= L)2

y

ph3

p

y

=

(ph)

x(3.3)

In the above case the limits of the bearing will be from 0 to +1 in the non dimensional722space. This method of non dimensionalization is used in this method of solution.723Please note that in Equation 3.2, the variables are non-dimensional. New symbols724are not introduced for clarity. The above equation cannot be solved exactly unless725either of the 3 assumptions are made.726

1. Large Harrisons number, tending to infinity.727

2. Infinitely long bearing.728

3. Infinitely short bearing.729

All three approaches were shown in the previous chapter . Solutions using approach7301 and 3 were also plotted. It is possible to numerically solve the above equation731keeping both the co-ordinate directions in consideration.1. Of the several possible732means of discretizing the equation, finite difference method is one and is shown here.733

11 ly long and 1 ly short bearing consider only co-ordinate direction.

33


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Chapter 3Solution of Steady Reynolds Equation Using Multi Dimensional Newton

Raphsons Method

3.1.1. Boundary Conditions734

The construction of foil bearing is non rigid. So whenever pressures less than the735ambient act on the top of the encircling foil, the foil lifts from the bumps and changes736the pressure distribution such that the negative pressures(gauge) do not occur. This737

has been accounted for both at the start and the end of the encircling foil. However738no clear info is available regarding what happens when the sub ambient pressures739develop in the middle of the encircling foil . This can happen in our case as there is740a single encircling foil from start to finish and the pressures develop in the middle741of the foil. We need to know where the film starts and ends. the front and back of742the film are communicated to the ambient. The boundary conditions are given as743

Front end of the Bearing( ppa

= 1744

Back end of the Bearing( ppa

= 1745

Start of the film( ppa

= 1746

End of the film( ppa

= 1 and p

= 0747

34


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3.2. Finite Di erence form748

Consider the grid structure shown in ?? . Expressing the outer derivatives, finite differen

ph3 p i,j+ 12 ph3 p i,j

12

i,j+ i,j+12

+ (R= L)2ph3 pz i+ 12 ,j ph3 pz i

12 ,j zi,j+ zi+1,j

2

= (ph)i,j+ 12 (ph i,j+ i,j+

2

Multiplying throughout by i,j+ i,j+12

zi,j+ zi+1,j2

, we get749

ph3 p

i,j+ 12

ph3

p

i,j 12

zi,j + zi+1,j

2

+ (R= L)2

ph3 p

z

i+ 12 ,j

p

=

(ph)i,j+ 12 (ph)i,j

Getting like terms having zi,j+ zi+1,j2

together on one side , we have750

ph3 p

ph

i,j+ 12

ph3

p

ph

i,j 12

zi,j + zi+1,j

2

+

(R= L)2ph3 p

z

i+ 12 ,j

ph3

p

z

i 12 ,j

i,j + i,j+1

2

= 0

For uniform spacing in i and j direction , we get

ph3 p

ph

i,j+ 1

2

ph3

p

ph

i,j 1

2

z+ (R= L)2 ph3 pz

i+ 1

2,j

ph

Next, finite difference formulas have to be written for individual terms.75135


44/165

ph3 p

ph

i,j+1

2

z :752

Employing Central Difference Scheme (referred to as CDS hence forth) about i, j + 12

for

pi,j+ 12h3

i,j+ 12 (pi,j+1 pi,j)

z

pi,j+ 12

hi,j+ 12 z

ph3 p

ph

i,j1

2

z :753

Employing CDS about i, j 12

for the inner derivative, we get

pi,j 12 h

3

i,j

12 (pi,j pi,j1) z

pi,j 12 hi,j 12 zph3 p

z

i+ 12 ,j

:754

Employing CDS about i + 12

, j for the inner derivative wrt z, we get755

pi+ 12

,jh3i+ 1

2,j

(pi+1,j pi,j)

z

ph3 p

z

i 12 ,j

:756

Employing CDS about i 12

, j for the inner derivative wrt z, we get757

36


45/165

pi 12 ,jh3

i 12

,j (pi,j pi1,j )

z

Assembling all the finite differences together gives us,

0 =

pi,j+ 12h3

i,j+ 12

(pi,j+1 pi,j)

z

pi,j+ 12

hi,j+ 12 z

pi,j 12

h3i,j 1

2 (pi,j p

+ (R= L)2

pi+ 12 ,jh3

i+12

,j (pi+1,j pi,j)

z

(R= L)2

pi 12 ,j

h3i 1

2,j

(

where h is

h = 1 + (cos ( j 0)) + (pi,j 1).

The terms to be determined in the above equations are pi,j , pi,j+1 , pi,j1 , pi+1,j , pi1,j758we get,759

pi,j

h n

z

p

i,j+ 12

h3i,j+ 1

2

oFrom1st

+

n

z

p

i,j 12

h3i,j 1

2

oFrom2nd

+

n

z (R=L)2 p

i+ 12,j

h3i+ 1

2,j

oFrom

pi,j+1

h n+

z

p

i,j+ 12

h3i,j+ 1

2

oFrom1stTerm

ipi,j1

h n+

z

p

i,j 12

h3i,j 1

2

oFrom2ndTerm

ipi+1,j

h n+

z (R= L)2 p

i+ 12,j

h3i+ 1

2,j

oFrom3rdTerm

ipi1,j

h n + z (R= L)2 pi12 ,jh3i 12 ,j o From4thTermi pi,j+ 12 hi,j+ 12 z + pi,j 12 hi,j 12 z = 0In the above equation terms like pi,j+ 12

occur frequently. There are no grid points at thes760

values to their left and right/ top and bottom to get the pressure values at these interm76137


46/165

mans land.762

pi,j " +z

(ph3)i,j + (ph3)i,j+1

2 From1stTerm + z

(ph3)i,j + (ph3)i,

2

+

z (R= L)2

(ph3)i,j + (ph

3)i+1,j2

From3rdTerm

+

z (R= L)2

(

pi,j+1

" +

z

(ph3)i,j + (ph

3)i,j+12

From1stTerm

#

pi,j1

" +

z

(ph3)i,j + (ph

3)i,j12

From2ndTerm

#

pi+1,j

" +

z (R= L)2

(ph3)i,j + (ph

3)i+1,j2

From3rdTerm

#

pi1,j

" +

z (R= L)2

(ph3)i,j + (ph

3)i1,j2

From4thTerm

#

"z

(ph3)i,j + (ph

3)i,j+12

#

+

"z

(ph3)i,j + (ph

3)i,j12

#= 0

38


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3.3 Examples of Multi Dimensional Newton Raphsons Method

The characteristics of the above equations are given below:763

1. Number of points in j () direction = nj;764

2. Number of points in i (y) direction = ni;765

3. Before application of BC, number of nodes= ni*nj;766

4. Before application of BC, number of unknown pressures= ni*nj;767

5. After application of BC, number of unknown pressures= (ni-2)*(nj-2);768

6. After application of BC, number of equations= (ni-2)*(nj-2);769

Since we write 1 equation for each point at which pressure is unknown,Number Of770Equations=Number of Unknowns. What is important is each one of the equations771has all the unknowns in it, It is just that some of the co-efficients are zeros. Also the772above equation has the unknown quantities in the co-efficients, making it non linear.773So we linearize the equation first in order to solve it. The method is illustrated using774

the following example.775

3.3. Examples of Multi Dimensional Newton776

Raphsons Method777

A one dimensional example of Newton Raphsons method is given below which then778leads to a multi-dimensional example. This Multi Dimensional Newton Raphsons779method is actually used to solve the Reynolds equation here.780

3.3.1. One Dimensional Example781

Consider the following equation

x3 + 2x2 + 3x + 4 = 0 (3.5)

In functional form

f(xtrueSolution) = 0

The above statement is true for the value of x that satisfies the equation Equation 3.5.782We do not know the point which satisfies the equation now. So we guess some point783xo and assume that it satisfies Equation 3.5. Unless the guesser is really good, it784will not. So we move a small amount from the point x and enforce the condition785that the new point satisfies the equation. This statement can be written as786

f(xguess + h) = 0

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Raphsons Method

To know the distance we have to move, we use the truncated taylors series, leaving787out all terms of order 2 and more.788

f(xn + h) = f(xn) + h f

xn = 0f(xn + h) is the same as f(xguess + h) for the first guess. Now we can express the789distance we have to move by790

h = f(xn)

f

x

n

Note that we have to compute the partial derivative(It is ordinary derivative for791this example since we have only x) of the function to be solved at the current guess792

point. The new point to which we have moved to will be793

xn+1 = xn + h

= xn f(xn)

f

x

n

We keep movin by small distances until we reach the point where the function794evaluates to 0. For the example we have considered, the procedure is as follows:795

1. Assume x0 = 1;796

2. Evalute function at x0

as f(1) = 13 + 2 12 + 3 1 + 4 = 10797

3. Derive the partial derivative as fx

= 3x2 + 4x + 3798

4. Evaluate derivative at current point as 3 12 + 4 1 + 3 = 10799

5. Compute the next point =1 1010

= 0800

6. Keep iterating until step 2 evaluates to zero.801

The Matlab code to do the above is given below:802

1 %%8032 c l c ;8043 c le a r ;805

4 x=1;8065 fx=x^3+2x^2+3x+4;8076 while ( abs ( fx )>=1)8087 fx=x^3+2x^2+3x+4;8098 fdx=3x^2+4x+3;8109 h=fx /fdx ;811

10 x=x+h ;81211 end813

40


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3.3 Examples of Multi Dimensional Newton Raphsons Method

This code return x=-1.6544. The answers can be checked by issuing814

s ol ve ( x^3+2x^2+3x+4)815

Which returns x= -1.6506.816

3.3.2. Multi Dimensional Example817

Consider the following equation set818

f(x, y) = x3 3xy2 1

g(x, y) = 3x2y y3

The equations are in 2 variables, they have 2 unknowns x and y. We need to linearize819

the system so that it can be brought to a form that can be solved.Let us do the820

linearization about the point {xn, yn}. Let h be the small step in x direction, k821be the step in y direction. Since we need to find these steps h and k , we proceed822similar to the previous 1D case823

f(x + h, y + k) = f(x, y) + h

f

x

+ k

f

y

+ o(h2) + o

k2

(3.6)

g(x + h, y + k) = g(x, y) + h

g

x

+ k

g

y

+ o(h2) + o

k2

(3.7)

Omitting higher order terms, we get824

f(x + h, y + k) = f(x, y) + h

f

x

+ k

f

y

g(x + h, y + k) = g(x, y) + h

g

x

+ k

g

y

This is equivalent to the system

f

x

f

y

gx

gy " hk# =

f(x, y)

g(x, y) (3.8)Again finding the

"h

k

#vector and finding the solution vector as in the above 1D

example, we get

"xn+1yn+1

#=

"xnyn

#

f

x

f

yg

x

g

y

1

f(xn, yn)g(xn, yn)

41


50/165


Raphsons Method

Taking an initial guess as

"xnyn

#=

".60.6

#, and computing the Jacobian as Jn =825

3 (x2 y2) 6xy6xy 3 (x2 y2)

, its inverse isJ1n =

1Jn

3 (x2 y2) 6xy6xy 3 (x2 y2)

. The826

Matlab code for the above is given below:827

1 c l c ;8282 c le a r ;8293 x= [ .6 ;8304 0 .68315 ] ;8326 f x= [ x(1)^33x(1)x(2)^2 1;8337 3x(1)^2x(2)x(2)^38348 ] ;8359 for i =1:5836

10 fx= [ x(1)^33

x(1)

x(2)^2 1;837

11 3x(1)^2x(2)x(2)^383812 ] ;83913 J= [ 3(x(1)^2x(2)^2) 6x(1) x (2 ) ;84014 6x(1) x (2) 3(x(1)^2x(2)^2)84115 ] ;84216 x=xJ^1 fx84317 end844

The output from the code is given in table Table 3.1 : Now that the multi di-

Iteration 1(Guessed) 2 3 4 5x -0.6 -0.4 -0.504789782 -0.499885398 -0.500000004y 0.6 0.862962963 0.856464305 0.86603764 0.866025391

Table 3.1.: Solution at different Iterations

845

mensional Newton Raphson method has been illustrated, it can be applied to our846problem. An example of a finite difference grid with uniform spacing is shown in847Figure 3.1848

3.4. Linearizing the Reynolds equation849

Consider the grid for finite differencing shown in Figure 3.1. The discretization850discussed in the following content is applied to this grid. In order to construct the851partial derivative matrix( size =number of variables), we need to find the partial852derivative of all the equations wrt all the unknowns. For example, the equation853corresponding to the unknown pressure at point (2,2) will be our first equation.854

42


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3.4 Linearizing the Reynolds equation

Figure 3.1.: Finite difference grid

This equation has to be differentiated wrt all the unknown pressures. Thankfully855not all the unknown pressures enter into the equation and hence many terms in the856

first row of the Jacobian are zero. Pressures(unknowns) on which the pressure at857

(2,2) depend are (2,2), (2,3), (2,1), (1,2) and (3,2).858

43


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The repeated colors are due to non availability of more colors. Please differentiate using t859

pi,j

"+

z

(ph3)i,j + (ph

3)i,j+12

From1stTerm +

z

(ph3)i,j + (ph

3)i,j2

+ z (R= L)2 (ph3)

i,j

+ (ph3)i+1,j

2

From3rdTerm+ z (R= L)2

(

pi,j+1

" +

z

(ph3)i,j + (ph

3)i,j+12

From1stTerm

#

pi,j1

" +

z

(ph3)i,j + (ph

3)i,j12

From2ndTerm

#

pi+1,j

" +

z (R= L)2

(ph3)i,j + (ph

3)i+1,j2

From3rdTerm

#

pi

1,j " +

z (R

=L

)

2

(ph3)i,j + (ph

3)i1,j

2 From4thTerm#

"z

(ph3)i,j + (ph

3)i,j+12

#

+

"z

(ph3)i,j + (ph

3)i,j12

#= 0

44


53/165

f(pp, pe, pw, pn, ps)Compass =

pp h

3p pp

z

2

1

pe h

3e pp

z

2

1

pp h3p pp

z

2

2

pw h3w pp

z

2

2

pp h

3p pp (R= L)

2

2 z

3

pn h

3n pp (R= L

pp h

3p pp (R= L)

2

2 z

4

ps h

3s pp (R= L

+

pp h

3p pe

z

2

5

+

pe h

3e pe

z

2

5

+pp h3p pw z2 6

+ pw h3w pw z2 6

+

pp h

3p pn (R= L)

2

2 z

7

+

pn h

3n pn (R=

+

pp h

3p ps (R= L)

2

2 z

8

+

ps h

3s ps (R= L

+

hw pw

z

2

+

hp pp

z

2

he pe z

2 hp pp z

2 ;

So the partial derivatives that concern us are86045


54/165

f

pi,j=

"pp

z

h3p

#1

"pe

z

2 h3e

#1

"pp h

3p

z

#2

"pw h

3w

2

"pp h

3p (R= L)

2

z

#3

"pn h

3n (R= L)

2

2 z

#3

"pp h

3p (R= L)

2

z

#4

"ps h

3s (R= L)

2

2 z

#4

+

"pe h

3p

z

2

#5

+

"pw h

3p

z

2

#6

+ " pn h3p (R= L)2 2 z#7

+

"ps h

3p (R= L)

2

2 z

#8

+

" hp

z

2

#

" hp

z

2

#

f

pi,j+1=

"pp

z

2 h3e

#1

+

"pe

z

h3e

#5

+

"pp

z

2 h3p

#5

z

2 he

f

pi,j1= " pp

z

2

h3w# 2+ " pw z

h3w# 6 + " pp z

2

h3p# 6+ z

2

hw

f

pi1,j=

"pp (R= L)

2

2z h3s

#4

+

"ps (R= L)

2

z h3s

#8

+

"pp (R= L)

2

2z

f

pi+1,j=

"pp (R= L)

2

2z h3n

#3

+

"pn (R= L)

2

z h3n

#7

+

"pp (R= L)

2

2z

46


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The Pseudo-code for the estimation of derivatives is given below for comparison.

t1=i,j =

2 pp hp3 zb2d

pe he3 zb2d

2 pp hp3 zb2d

pw h

2 pp rbl2 hp3 tb2z pn rbl2 hn

3 tb2z

2 pp rbl2 hp3 tb2z

ps rbl2 hs3 tb2z

+pe hp3 zb2d

+pw hp3 zb2d

+pn hp3 rbl2 tb2z

+ps hp3 rbl2 tb2z

+ (gamma hp dz= 2) (gamma hp dz= 2)

t4=i,j+1 =pp he3 zb2d pp hp

3 zb2d + 2 pe he3 zb2d (gamma he

t5=i,j1 = pp hw3 zb2d

+

2 pw hw3 zb2d

+pp hp3 zb2d

+ (gamm

t2=i+1,j = (pp hn3 rbl2 tb2z) + (2 pn hn3 rbl2 tb2z) + (pp hp3 rbl2 t

t3=i1,j =pp hs3 rbl2 tb2z

+

2 ps hs3 rbl2 tb2z

+pp hp3 rbl2 tb

47


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Raphsons Method

Now all the Partial derivatives required have been evaluated. We have alreadyderived the function

f(pi,j, pi,j+1, pi,j1, pi+1,j, pi1,j)Numbered = f(pp, pe, pw, pn, ps)Compass

.With these two in hand, we may write the full-linearized Reynolds equation. The861linearized form of our Reynolds equation can be written in line with Equation 3.6862as863

f(x + h, y + k) = f(x, y) + h

f

x

+ k

f

y

f(f pi,j, pi,j+1, pi,j1, pi+1,j, pi1,j)n+1 = f(pi,j, pi,j+1, pi,j1, pi+1,j, pi1,j)n

+

f

pi,j (pn+1

ij pn

i,j) +..

This is in line with the equation presented in Heshmat Figure 3.2, so we are going864in the right track. Heshmat says this equation was obtained by linearisation using865Newton Raphson method. An alternative way to say is, the equation is linearized866using Taylors series approximation and solved using the multi dimensional Newton867Raphson method. The methodology he uses is consistent with Newton Raphsons868Multi Dimensional method, he has only named it differently . The full equation is

Figure 3.2.: Heshmats equation

869

given in Equation 3.10.870

RHS f(pi,j, pi,j+1, pi,j1, pi+1,j, pi1,j)n +

f

pi,j(pn+1ij p

ni,j) +

f

pi+1,j(pn+1i+1j p

ni,+1j)

+f

pi1,j(pn+1i1j p

ni1,j) +

f

pi,j1(pn+1ij1 p

ni,j1)+ = 0

f

pi,j+1(pn+1ij+1 p

ni,j+1)+

ThankfullyZero f

pi,j+2(pn+1ij+2 p

ni,j+2) + .... (3.10)

48


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3.4 Linearizing the Reynolds equation

3.4.1. Applying Multi-Dimensional Newtons Method871

We represent the incremental change in pressures by operators henceforth. In a872more readable form these equations can be written as873

fpi,j

n

pi,j + f

pi+1,j

n

pi,j + 1 + f

pi1,j

n

pi1,j +

f

pi,j+1

n

pi,j+1 +

f

pi,j1

n

pi,j1

= f(pi,j, pi,j+1, pi,j1, pi+1,j, pi1,j)n

In matrix form, writing this equation only for the interior unknown points, for the874small grid shown in Figure 3.3 , we have Equation 3.11 .

Figure 3.3.: Finite difference grid for demonstartion

875

3.4.2. Matrix form of equations: Method 1876

Once we have the equation to be solved Equation 3.10, it can be solved in two877ways. The two methods differ in their representation of the unknown x vector only.878Consider the first method, which is also the one commonly used in Fintie Element879Methods. The incremental pressures are the unknowns(x) . The A matrix(aka the880Jacobian) will have rows corresponding to the unknown rows in x only. The Rhs will881be of same structure. The number of columns on the A matrix will be = number882of grid points so that the boundary conditions may be applied. It is illustrated883

in figure belowFigure 3.4. Certain terms of the A matrix get multiplied with the884boundary conditions and become rhs terms as shown in Figure 3.5 The assembled885set of equations look like the one in Figure 3.6. For our Reynolds equation, the886assembly procedure is more complex as boundary pressures not only occur at the887top and bottom but also in between the pressure vector. Our Reynolds equation,888before applying the boundary conditions(corresponding to Figure 3.4) looks like889Equation 3.11.890

49


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Raphsons Method

Figure 3.4.: Assembly procedure for points

Figure 3.5.: Boundary condition Implemetation

Figure 3.6.: Assembled set

50


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0

f

p

2

0 0

f

p

5

f

p

6

f

p

7

0 0

f

p

10

0 0 0

0 0

f

p

3

0 0

f

p

6

f

p

7

f

p

8

0 0

f

p

11

0 0

0 0 0 0 0

f

p

6

0 0

f

p

9

f

p

10

f

p

11

0 0

0 0 0 0 0 0 fp 7 0 0 f

p 10 f

p 11 f

p 12 0

Please note that in the x vector, the unknowns are in red, the boundary conditions are in891

that get multiplied with the boundary pressures are in magenta . This color coding mak892characteristics. Thanks to the [5]. Note that the equations are non zero only for the unk893of all pressures, including known boundary conditions and unknown pressures.Since we are894not proceed to boundary condition application to the above set to save space.895

51


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3.4.3. Matrix form of equations: Method 2896

Since we have kept the the increments in the pressure as unknown (x), we have to take 897

in the X vector. This approach also means we have to add the increments in a separate 898alternative approach of treating the unknowns is a lot easier to program as shown belo899unknown as difference of pressures at a point rather than as p.900

0@ f

@ p

2

0 0@ f

@ p

5

@ f

@ p

6

@ f

@ p

7

0 0@ f

@ p

10

0 0 0 0 0

0 0

@ f@ p

3

0 0

@ f@ p

6

@ f@ p

7

@ f@ p

8

0 0

@ f@ p

11

0 0 0 0

0 0 0 0 0@ f@ p

6

0 0@ f@ p

9

@ f@ p

10

@ f@ p

11

0 0@ f@ p

14

0

0 0 0 0 0 0@ f@ p

7

0 0@ f@ p

10

@ f@ p

11

@ f@ p

12

0 0@ f@ p

52

thesis withlinenumbers selfcorrected compressed

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