Lubrication of Plain Cylindrical Bearings

8/12/2019 Lubrication of Plain Cylindrical Bearings

1/19

2014

ETSIAE

LUBRICATION OF PLAIN

CYLINDRICAL BEARINGS

Egido Fernndez Andrs

Fernndez Mndez Carlos

Snchez Hernndez Jorge


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INDEX

INTRODUCTION .......................................................................................................................................... 2

MATHEMATICAL MODEL............................................................................................................................ 3

DISCRETIZATION OF THE PROBLEMS PDE................................................................................................. 5

ANALISYS OF RESULTS ................................................................................................................................ 8

Case 1: =constant ............................................................................................................................... 8Case 2: =constant............................................................................................................................... 10

Curly behaviour .................................................................................................................................... 12

ANALITIC SOLUTIONS ............................................................................................................................... 13

1. Little loaded bearing ( )....................................................................................................... 132. Long bearing (

) .................................................................................................................. 15

3. Short bearing ( ) ................................................................................................................. 17


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INTRODUCTION

A bearing is a piece that supports and spins around a drive shaft of a machine. It reduces the friction

between the drive shaft and the machine. Basically, there are two different kinds of bearings. The

most common of them use rolling elements to work (balls, rollers, etc). On the other hand, some

bearings work with one piece spinning directly over the other one. To make this possible, it isnecessary a fluid inside the bearing in lubrication conditions.

As the fluid mechanics behaviour is quite complicated to determine analytically, a numeric resolution

is needed most of the times. That is why this is the problem we proposed for this year.

List of variables and parameters:

-

External radius of the bearing

- Internal radius of the bearing- Length of the bearing- Eccentricity of the bearing- Thickness between external and internal radius.- Fluid density- Fluid viscosity parameter- Angular velocity of the interior cylinder- Fluid velocity- Volume of flow-

Angular variable

- Length variable- Thickness variable- Pressure- Adimensional pressure- Adimensional length variable- Relative eccentricity- Elongation


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MATHEMATICAL MODEL

In order to obtain lubrication of plain cylindrical bearings general equation, hypothesis that rule this

physical model must be defined, which are:

~

e ~ Small Effective Reynolds: 1

Besides, some geometric relations are needed to proceed with the problem. The following figures

describe a plain bearings geometry:

Where = 1 c o s , being the relative eccentricity = .The induced fluid velocity comes from a constant angular velocity () of the interior cylinder.From the momentum equations of Navier-Stokes equations the following differential equations are

obtained:

1 = 0 ( 1 )

1

= 0

( 2 )

Which must verify the following boundary conditions:

= 0 = 0; = = ( 3 ) = 0 = 0; = = 0 ( 4 )Once these equations are solved, flow gets calculated:

=

=

12 2 ( 5 )


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= = 12 ( 6 )

Reynolds equation for volume:

= 0( 7 )

Terms and are geometric factors. In this case = 1and = . Substituting all these terms,the following equation is obtained:

12 2 12 = 0 ( 8 )In the end, adding

, bearings general equation is obtained:

[1 c o s ] 1 c o s = 6 sin ( 9 ), 2 = , 2 = ; , x = 2 , x; , x = 2 , x ( 10 )

In which there are periodic conditions (representing the polar behaviour of the bearing) and boundary

conditions on the limits of the bearing. This equation and their boundary conditions can be normalized

in order to simplify the analysis.

[1 c o s ] 1 c o s = s i n ; , 0,21,1 ( 11 ),1 = , 1 = 0 ; , = 2 , ; , = 2 , ( 12 )

Where = is the adimensional pressure, = L the adimensional variable of length and = LRthe elongation.Resolving the lubrication of plain cylindrical bearings problem, eccentricity (

) and elongation (

) are

considered data of the problem. In the end, the force suffered on the bearing from the fluid may be

calculated. Such force may be adimensioned too.

However, it must be taken in consideration that is normally impossible to know and, actually, it is thenumber we wish to obtain. Consequently, it might be thought that this problem is useless. On the

other hand, we must consider it is quite easy to measure empirically the force suffered on the bearing.

Had we the solution of this problem, it would be very easy to draw curves that link this force and its

eccentricity (considering a constant elongation). These curves would allow to know the bearings

eccentricity starting out its elongation and the force it suffers, and they will be represented later.


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DISCRETIZATION OF THE PROBLEMS PDE

Given the previous equation, it is intended to solve it numerically. After several transformations, the

equation and boundary conditions which describes its behaviour are ( 11 ) and ( 12 ).

To solve this problem numerically it will be used a bidimensional discretization with an equally spaced

net. To begin with, we follow this development:

3sin 1 c o s 1 c o s 1 c o s = s i n ( 13 )Defining:

=3sin 1 c o s ; = 1 c o s

= 1 c o s

; = s i n

Using Scheme 1s notation, we first consider discretization on

variable (for an arbritary ) with 1nodes on axis . = 1 1 ; = 1 , , 1 ; = 2 ; =

This way, and using numeric derivation centred with 3 nodes, expression ( 13 ) becomes:

=

+ 2

( 14 )

Discretizing boundary conditions:

= + = 0 ( 15 )Therefore, the systems matrix is:

0

0

0

0 = (

1 0

1 2 1

0 1)

+

(

0

10) ( 16 )

And considering boundary conditions are disengaged:

=

2 1 1 2 1

1 2

(

111)

( 17 )

,

, 1 1 ,

1 ,

, 1

Scheme 1.Discretizations scheme.


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Proceeding the same way than before, variable is discretized. = 1 ; = 1 , , 1 ; = 2 ; = ,

= ; = ; = ; = Applying first and second centred derivatives expression for 3 points we obtain from ( 14 ): ,+ , 2, 2 ,+ , = +, , 2, ( 18 ),+ 2 , 2 2 , 2 ,+ , = ( 19 )Leaving periodic boundary conditions as:

, = ,+ ( 20 )Thanks to ( 20 ) every node = 1is taken as = 1. To fully define the system it is also considerthat , = ,because of periodicity respect . This way ( 9 ) has been closed for = 1 , , .Besides:

= = 0 ( 21 )Considering the previous ideas, equations for = 1and = are:

, , 2 2 , +, , = 0 ( 22 ), 2 , 2 2 , 2 +, , = ( 23 )

This way, the system is fully defined with the boundary conditions defined, and variables ,, +,and ,+ are excluded (their value is already known as boundary conditions ( 15 ) and ( 20 ) ). Inorder to express it, the following matrixes are defined:

, =2 2 2 2

2 2 2 2 2

2 2 2 2 ( 24 )

, = 2 ( 25 )


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, = 2 ( 26 ), = ( 27 )( 24 ), ( 25 ), ( 26 ) y ( 27 ) are considered for

= 1 , , and

= 2, , (For ( 27 )

= 1).

In the end, the system of lineal equations that determines the discretized equations solution is

defined by the following matrix divided in boxes:

(

, , , , , , , , , )

(

,,, ,,, ,+,+, )

=

(

11

)

( 28 )

As it may be observed, the final matrix is a 1x 1size, as the vectors are 1size. This system is qualitative huge and possess a large quantity of zeros, which means that it is

needed to solve it with sparse matrix, as it is shown in the following graphic.

0 50 100 150 200 250 300 350

0

50

100

150

200

250

300

350

nz = 1860


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ANALISYS OF RESULTS

In order to obtain the numerical error from the numerical scheme, it is performed a Richardsons

extrapolation. Due to the three points derivation formulae, an order two error is supposed to be

obtained:

= 2 4.0005 4 ( 29 )Which means a parameter 2, an error of .Related to the analysis of the solutions, there are two parameters that may vary, which are and . Inorder to proceed, we shall study the solution depending on one parameter while the other remains

constant.

Case 1: =constantThroughout this case, it is intended to check how pressure varies with for a constant . Maximumpressure is reached when = and = 0. This maximum grows exponentially, making a singularsolution when 1.The following figures show the behavior for either 1and 1.

Figure 1. Richardson.


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Figure 2.

Figure 3.


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When = 0 . 1, adimensional pressure lowers as ~ 1, while when = 1 0, initial overpressure ishigher and more uniform along variable .

Case 2: =constant

In this case it is studied the normalized pressure s evolution along an incremental value of forconstant . Pressure tends asymptotically to a maximum, which means that for a high certain value of, the solution does not depend on anymore (except on the boundaries) and does not vary aswell. In order to obtain low pressure, a very low is needed.

Figure 4.


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One application of this solution consists on drawing curves that allow relating eccentricity and

adimensioned pressure for a constant , giving the chance to measure the bearings eccentricity.

The higher is, the more narrow these curves are, as it is explained before.

Figure 5.

Figure 6. Contour line map.


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Curly behaviour

Another fact to consider is a curly solution when the number of nodes is low. This is a consequence of

having a bad conditioned matrix when and 1, as functions g and p in ( 24 ) (which arelocated in the principal diagonal as

) tend to zero, while function f in ( 25 ) and ( 26 ) ( )does not. As we wish to avoid this effect, we must reduce the length of the step (increasing the

number of nodes), so g and p terms (divided by ) might have similar or superior order than f (justdivided by 2).

Figure 8. Solution with 200 nodes.Figure 7.Solution with 100 nodes.


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ANALITIC SOLUTIONS

Lubrication of plain cylindrical bearings is a problem which generally requires numeric calculus in order

to be solved. However, in certain cases (assuming some hypothesis and not considering some

variables) an analytic solution may be found, being as good as a numeric one.

As this analytic solution can only be described when and reach asymptotic limits ( 1, 1and 1) these different cases are named asymptotic solutions.

1. Little loaded bearing ( )Asymptotic limit 1 is related to small distributions of pressure. Within this limit, 1 1,which simplifies the problems equation to:

1 = ( 30 )With the usual boundary conditions. The analytic solution of this problem is:

= sin; where = 1 p+pp+p ( 31 )As it can be seen in the next figure, the numeric solution presents the same behaviour as the analytic

solution. Along the dimension it clearly follows a sine shape, whereas the functions value is zero atthe extremes of (-1 and 1), as its modulus grows as reaches the centre (0).


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Figure 10. Sine behaviour of on dimension.

Figure 9.


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2. Long bearing (

)

When 1it is a case of a very long bearing. Under this circumstance, is a very small termcompared to 1. Therefore, after some simplifications, the equation to solve is:

[1 ] = 0 ( 32 )As there are not derivatives with respect to , bounder conditions cannot be imposed in = 1, sothat the solution might not be valid near the bearing bounders. Looking at figure 5 its clear that the

numeric solution is good near those bounders, as quickly goes to zero. Only the periodicity conditioncan be applied ( = 2 ). In the end, solving the equation we obtain: = 2 2 1 ( 33 )

As it can be confirmed on the next figure (obtained from the numeric pr oblem) this solution doesnt

vary with , as with it behaves as a sine multiplied by a varying positive number (which depends on too).

Figure 11. Hyperbolic cosine behaviour of on dimension.


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Figure 13. Constant behaviour (except on thebounders) of on dimension.

Figure 12.


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3. Short bearing ( )To solve this problem with short bearings ( 1) we consider solving the equation for instead of .As a result of this, the original equation becomes:

1 c o s 1 c o s

= ( 34 )

When 1, the first term of the equation can be unconsidered. Consequently, the equation thatsolves this case is:

= 1 c o s ( 35 )The bounder conditions of this problem are = 0when = 1. They cannot be imposed on because in this equation derivatives with respect to havent been taken into consideration.With thesolution of this problem, the pressure distribution can be obtained:

= 121 c o s ( 36 )Checking the numeric solution of the problem under these conditions, its easy to see that this solution

and the analytic one are alike. On the

dimension, it becomes like a sine multiplied by a varying

positive number (which depends on and ). On the other hand, the functions value reaches zero atthe extremes of (-1 and 1), as its modulus grows as gets close to the centre (0). It is important tocheck the scale of on this figure (10), as the term multiplying lowers its value considerably(~).

Figure 14.


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Figure 15. Sine behaviour of on dimension.

Figure 16. Parabolic behaviour of

on

dimension.

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Lubrication of Plain Cylindrical Bearings