chapter_8_dynamic analysis of fydrodynamic bearings

8/10/2019 Chapter_8_Dynamic Analysis of Fydrodynamic Bearings

1/27

CHAPTER 8 DYNAMIC ANALYSIS OF HYDRODYNAMIC BEARING

In this chapter the analyses of the hydrodynamic bearings such as plane slider bearing

and journal bearing are discussed. Briefly different types of lubrications are describedand the mechanism of pressure development in the oil film is studied. The Petroffs

equation for a lightly loaded journal bearing is derived. The derivation of Reynolds

equation is carried out and it is applied to idealized plane slider bearing with fixed and

pivoted shoe and journal bearings.

Lubrication

Lubrication is the science of reducing friction by application of a suitable substance

called lubricant, between the rubbing surfaces of bodies having relative motion. The mainmotive of using a lubricant is to reduce friction, to reduce or prevent wear and tear, to

carry away heat generated in friction and to protect against corrosion. The basic modes of

lubrication are thick and thin film lubrication.

Thick Film Lubrication:

Thick film lubrication describes a condition of lubrication, where two surfaces of bearing

in relative motion are completely separated by a film of fluid. Since there is no contact

between the surfaces, the properties of surface have little or no influence on theperformance of the bearing. The resistance to the relative motion arises from the viscous

resistance of the fluid. Therefore, the performance of the bearing is only affected by the

viscosity of the lubricant. Thick film lubrication is further divided into two groups:

hydrodynamic and hydrostatic lubrication.

Hydrodynamic Bearing:Hydrodynamic lubrication is defined as a system of lubrication

in which the supporting fluid film is created by the shape and relative motion of the

sliding surfaces.

The principal of hydrodynamic bearing is shown in fig.1. Initially the shaft is at rest (a)

and it sinks to the bottom of the clearance space under the action of load W. As the

journal starts to rotate, it will climb the bearing surface (b) and as the speed is further

increased, it will force the fluid into the wedge-shaped region (c).

142


2/27

(a) (b) (c)Figure 1 Formation of Continuous Film in a Journal Bearing

Figure 2. Hydrodynamic Lubrication (Oil Wedge Region)

Since more and more fluid is forced into the wedge-shaped clearance space, pressure is

generated within the system. Fig.3 shows the pressure distribution around the periphery

of a journal.

Since, the pressure is created within the system due to rotation of the shaft, this type of

bearing is known as self acting bearing. The pressure generated supports the external load

W. This mode of lubrication is seen in bearings mounted on engines and centrifugal

pumps.

143


3/27

Figure 3 Pressure Distribution in Hydrodynamic Bearing

Hydrostatic Lubrication:Hydrostatic lubrication is defined as a system of lubrication in

which the load supporting fluid film, separating the two surfaces, is created by an

external source, like a pump, supplying sufficient fluid under pressure. Since the

lubricant is supplied under pressure, this type of bearing is called externally pressurized

bearing. Hydrostatic bearings are used on vertical turbo-generators, centrifuges and ball

mills.

Thin Film Lubrication:

Thin fluid lubrication, also known as boundary lubrication, is defined as a condition of

lubrication, where the lubricant film is relatively thin and there is partial metal to metal

contact. This mode of lubrication is seen in door hinges and machine tool slides. The

conditions of boundary lubrication are excessive load, insufficient surface area or oil

supply, low speed and misalignment.

Figure 4 Boundary Lubrication

144


4/27

The hydrodynamic bearing also operates under the boundary lubrication condition when

the speed is very low or when the load is excessive.

Under the extreme conditions of load and temperature, the fluid film gets completelyruptured, direct contact between the two metallic surfaces takes place and thus, extreme

boundary lubrication exists.

Figure 5 Contacts at High Points (Extreme Boundary Lubrication)

The phenomenon of extreme boundary lubrication is based on the theory of hot spots.

These hot spots, also known as high spots are the spots on the metallic surfaces where the

welding of the two surfaces takes place, owing to extreme temperature conditions, which

is a consequence of the shearing action of the high points. However, due to the relative

motion between the two surfaces, the welding too gets ruptured.As a consequence of the

phenomenon of the high spots, occurring at extreme conditions of load and temperature,the physical properties get severely damaged.

LIGHTLY LOADED JOURNAL BEARINGS:

The following assumptions are made while deriving the characteristic equations for the

lightly loaded journal bearings:

1. The radial load is almost zero.

2.

Viscosity of the lubricant is very high.

3. Journal speed approaches very large values.

4. Film thickness is very small as compared to radius of the journal i.e.h


5/27

Figure 6 Journal Bearing

Figure 7: Unwrapped Film

Fig.7 shows the unwrapped film. The length is 2rand the width isL into the plane of the

paper. Also, the film thickness is equal to the clearance i.e. h = C.

Now, we have

2U N'= andF A= (1)

where N =journal speed

= shear stress acting on the fluid

A = 2rL, area of the journal surface.

Assuming constant coefficient of viscosity of the fluid and from Newtons law, we have

U

h = (2)

or2 rN

h

'

= (2a)

Hence,

2 24 'N L rF

C

= (3)

Further, the frictional torque may be obtained as

146


6/27

2 34 '.f

N L rT F r

C

= = (4)

This equation is known as thePetroffs equation, for lightly loaded journal bearings.

The coefficient of friction may be obtained asF

fW

= (5)

We define unit bearing loadPas the radial load per unit projected area.

2

WP

rL= (6)

Hence, the coefficient of frictional is

2 '2N r

fP C

= (7)

PRESSURE DEVELOPMENT IN THE OIL FILM:

Consider two parallel surfaces, one stationary and the other moving with uniform

velocity U, as shown in fig.8.

Figure 8 Two Parallel Surfaces in Motion

Here, we assume that the two surfaces are very large in a direction perpendicular to the

plane of motion and therefore, their velocity in this direction is zero. Since, the velocity

of the oil film varies uniformly from zero at the stationary surface ST to Uat the moving

surfaceMN, therefore, the pressure developed in the oil film is zero. That is, the moving

surface cannot take any vertical load and even a small load is applied, the oil film will

squeeze out.

147


7/27

Consider another case similar to the previous case, the only difference being that here the

direction of motion of the moving surface is vertical and not horizontal. Due to the

motion of the surfaceMN, oil film is squeezed out and the velocity increases from zero at

the central section CC1to a maximum at the outlet sectionsMS and NT. The distribution

of velocity is shown below.

Figure 9 Two Parallel Surfaces, One Stationary and the Other in Vertical Motion

We observe from the figure that the maximum velocities occur at the midpoints for each

cross-section. This type of velocity distribution occurs only if the maximum pressure is at

the central cross-section CC1, falling out to zero value at the outlet cross-sections MC and

NT. Such a kind of flow is known aspressure induced flow.

Lastly, consider another case similar to the first case, the only difference being that the

stationary surface here is inclined at an angle to the line of motion.

Figure10 Stationary Surface Inclined at an angle to the Line of Motion

148


8/27

The velocity distribution of the oil film is shown in fig.11(a).

Figure11 (a) Velocity Distribution of the Oil Film

Considering only unit thickness perpendicular into the plane of paper. Volume of fluid

entering the space is given by SMO and that leaving is given by NPT, with MO and NP

representing the velocities at the moving surface. Since some vertical load is applied,

therefore some amount of fluid is squeezed out of the space between the two plates. Thevelocity distribution due to this pressure induced flow is shown in fig.11(b).

Figure11 (b) Velocity Distribution of the Oil Film

Fig.11(c) shows the resultant velocity distribution, thereby balancing the volume of fluid

entering and leaving the space between the two surfaces. Also, owing to pressure induced

flow, pressure is developed in the oil film with a maximum value at the cross-section

CC1, where such a flow is zero, as at the outer sections MS and NT. The pressure

distribution is also shown in fig.11(c).

149


9/27

Figure11 (c) Velocity Distribution of the Oil Film

DERIVATION OF REYNOLDS EQUATION:

The theory of hydrodynamic bearing is based on a differential equation derived by

Osborne Reynold. Reynolds equation is based on the following assumptions:

1. The lubricant obeys Newtons law of viscosity.

2.

The lubricant is incompressible.3. The inertia forces of the oil film are negligible.

4. The viscosity of the lubricant is constant.

5. The effect of curvature of the film with respect to film thickness is neglected. It is

assumed that the film is so thin that the pressure is constant across the film

thickness.

6. The shaft and bearing are rigid.

7. There is a continuous supply of lubricant.

An infinitesimally small element having dimensions dx, dy and dz is considered in theanalysis. uand vare the velocities in xandydirection. xis the shear stress along the x

direction whilepis the fluid film pressure.

150


10/27

Figure 12 Converging Oil Film

Figure13 Infinitesimal Element in Equilibrium

On balancing the force acting in thex-direction, we get

0xxddp

pdydz p dx dydz dxdz dy dxdzdx dy

+ + + =

(9)

xddp

dx dy

= (10)

From Newtons viscous flow, we have

x

du

dy = (11)

where uis the velocity in thex-direction.

Hence,

2

2

dp d u d u

dx dy y dy

= =

(12)

151


11/27

or2

2

1d u dp

dy dx= (13)

On integrating we get,

1

1du dpy C

dy dx

= + (14)

2

1

1

2

dpu y C y

dx= + 2C+ (15)

The boundary conditions are:

Aty= 0, u= U (16.1)

Aty= h, u= 0 (16.2)

On applying the boundary conditions, we obtain the constants as

C1= U (17)

and 21

2

dp U

C hdx h= (18)

Hence,

( )21

2

dp h yu y hy U

dx h

= + (19)

Now, considering the flow between the two surfaces STandMN, where the distribution

of velocity for a sectionABis represented.

Volume of fluid entering the element = udydz + vdxdz (20)

Volume of fluid discharging =u u

u dx dydz v dy dxdz x y

+ + +

(21)

On applying the conservation of mass, we get

v

y x

u=

(22)

Substituting the value of uand on rearranging, we get

( ) ( )21 U h ydp

dv y hy dyx dx h

= +

(23)

( ) ( )2

0 0

10

y h y h

y y

U h ydpdv y hy dy

x dx h

= =

= =

= + =

(24)

On simplifying we get,

3

012 2

d dp h d Uh

dx dx dx

=

(25)

152


12/27

The above equation can be rearranged as

3 6d dp d

h Udx dx dx

=

h (26)

The equation represents the Reynolds equation in two dimensions, expressing thepressure gradient in a converging oil film as a function of film thickness, viscosity of the

lubricant and the relative velocity of the moving surface.

IDEALIZED PLANE SLIDER BEARING (Fixed Shoe):

Consider a plane slider bearing with a fixed shoe.

Figure 14. Plane Slider Bearing With Fixed Shoe of Length L

Figure 15. Film Thickness and Inclination Angle

153


13/27

Let

Length of the shoe = L

Surface velocity (uniform) = U

Force acting = F

External load acting vertically = WWidth of the moving surface = w

Thickness of the film (at entrance) = h1

Thickness of the film (at exit) = h2

Angle between the fixed shoe and the x-axis =

The thickness of the oil film at any distance can be expressed as

1 21

h hh h x

L

= (27)

Defining some non-dimensional terns1 2h h

L

= , 2

ha

L= and X

L= (28)

Hence, the expression for the thickness of the oil film at any cross-section can be re-

expressed as

1h LX h= + (29)

But 1h La L= (30)

Hence, the oil film thickness can be expressed as

( )h L X a = + (31)

From Reynolds equation, we have

3 6d dp d

h Udx dx dx

=

h (26)

On integration, we get

3

16dp

h Uhdx

C= + (32)

or 2 3

1

6

dp B

Udx h h

= +

(33)

where 1

6

BB

U=

From equations (28), (31) and (33), we get

154


14/27

( ) ( )

2

6 1dp U B

dx L a X L a X

= +

+ + 2

(34)

or

( ) ( )

2

6 U dX BdX dp

L a X L a X

= +

+ +

2 (35)

On integrating equation (35), we get

( ) ( )2

6

2

U dX BdX 2

p CL a X L a X

= + +

+ + (36)

The boundary conditions are:

AtX =0,p=0 (37.1)

AtX =1,p=0 (37.2)

On substituting equation (29) in equation, we get

( )6

2

UC

L a

=

(38)

22

aB La

a

=

(39)

Hence, the pressure distribution along the idealized plane slider bearing can be expressed

as

( )

( ) ( )2

6 1

2

X XUp

L a a X

=

+

(40)

The load carrying capacity can be expressed as

0

L

W wpdx= (41)

1

0

W wLpdX = (42)

On substituting the value ofp from equation (40), we get

( )

1 2

2

0

62

X XW wU dX a a X

= + (43)

On integrating, the load carrying capacity can be expressed as

2

26 ln

2

w a aW U

a a

=

+

(44)

155


15/27

For calculating the total frictional force acting on the moving surface, the shear forces

acting on the elemental areas need to be determined.

1

n

i i

i

F A=

= or F dA=

Now, from Newtons law, we have

x

du

dy = (11)

where uis the velocity, which can be expressed as

( )21

2

dp h yu y hy U

dx h

= + (19)

Differentiating equation (19) w.r.t.y, we get

1 2

2

du dp y h U

dy dx h

=

(45)

On substitutingLdX = xand equation (45) in equation (11), we get

1 2

2x

dp h y uU

L dx h

=

+ (46)

On differentiating equation (40) w.r.t.Xand simplifying, the pressure gradient obtained is

( )

( )( )3

26

2

a X aX dp U

dX L a a X

+ =

+ (47)

Hence, from equations (45), (47) and (11), the shear force acting at any point is

( )( ) ( )

( ) ( )33 2 2 1

2x a X aX a X yU

L La a X

+ + = + + +

a X(48)

On the moving surface y = 0, and hence

[ ] ( )

( ) ( ) ( )0 20

3 2 1

2x y

a X aX U

L aa a X

= X

+ = = + + +

(49)

Hence, the total frictional forceF0acting on the moving surface is1

0 0 0

0 0

L

F w dx dx = = (50)

On substituting equation (49) in equation (50) and integrating, we get

0

4ln

2

aF Lw

a a

6

= +

(51)

The coefficient of friction,f is

156


16/27


17/27

Figure16 Location of the Pivot Point of the Shoe

Consider the following equation

1

2

1h

rh a

= = (58)

2h r

L = (59)

Using equation (59), the equations for the performance of the plane slider bearing with

pivoted shoe can be obtained as

( )( )

2

2 2

2

6 1 2ln 1

2

UwLW r

h r r r

= +

+ (44a.1)

or ( )2

2

2

6w

UwLW g

h

= r (44a.2)

158


18/27

where ( ) ( )( )2

1 2ln 1

2w

g r rr r r

= +

+

( )02

4ln 1

2

UwLF r

h r r

= + 6

+ (51a.1)

or ( )00

2

F

UwLF g r

h

= (51a.2)

where ( ) ( )0

4 6ln 1

2F

g r rr r

= + +

( ) ( ) ( )

( ) ( ) 2

51 3 ln 1 3

22 ln 1 2

r r r r r

l Lr r r r

+ + + +

=+ +

(57a)

( )

( )02 1

6

F

w

g rhf

L g r

=

(60)

( )2 fh

f g r

L

= (60a)

The shoe has tangential as wall as radial degrees of freedom. The friction force at the

pivot junction , balances with fluid friction of the shoe. Moreover, the friction force

and the normal reaction together balance the vertical thrust. Thus, the value of rchanges,

automatically, to meet the equilibrium conditions.

pF

PRESSURE DISTRIBUTION IN JOURNAL BEARING:

Consider a full journal bearing as shown in figure below.

159


19/27

Figure 17: Idealized Full Journal Bearing

Figure.18 Unwrapped Film

Let

Radius of the journal =r

Radius of the bearing = r + c

Clearance = c

Eccentricity = e

Vertical Load = W

Angle of Inclination of the vertical load =

h = CA CB = r + c CB (61)

From and , we haveCBE 'C BE

sin

sin sin

BE rCB

= = (62)

Since, =

160


20/27

(sin cos cos sinsin

rCB )

= (62a)

From and , we have'C DC 'C BD

sin sine r = (63)

22

2cos 1 sin

e

r = (63a)

On substituting the values of sin and cos from equations (63) and (63a) in equation

(62a), we get

2 2 2sin cosCB r e e = (64)

Substituting equation (64) in equation (61), the final expression for film thickness as a

function of is obtained as

2 2 2sin cosh r c r e e = + + (65)

On neglecting 2 2sine in comparison with r2, and further using

en

c= (66)

where n is called the eccentricity ratio, the equation (65) can be simplified as

( )1 cosh c n = + (67)

On substituting x r= in Reynolds equation, we get

3 6d dp d

h Ud d d

h

=

(26a)

On integrating, we get

3 6 6dp dh

h Ur d Urh K d d

= = + (68)

or( ) ( )

1

22

6 1

1 cos 1 cos

Kdp Ur

d c n c n

3

= +

+ + (69)

or( ) ( )

1

22

6

1 cos 1 cos

K dUr dp

c n c n

3

= +

+ + (70)

Further, we have

0 2 2 0 0p p p p = = = == = (71)

which gives

( ) ( )

2

1

2 32

0

6 10

1 cos 1 cos

KUrp d

c n c n

=

=

= +

+ + = (72)

161


21/27

or( )

( )

2

2

01

2

3

0

1 cos

1 cos

d

nK

c d

n

=

=

=

=

+=

+

(73)

On simplifying, we get

( )21 2

2 1

2

c nK

n

=

+ (74)

From equation (69), it can be easily observed that 0dp

d= when ( )1 1 cosK h c n = = +

and the minimum film thickness hmat minimum and maximum pressure is obtained as

( ) ( ) ( )2

1 2max min

2 1

2m p p

c nh h h K

n= =

= = = =

+ (75)

Correspondingly, the value of where the maximum and minimum pressured occur isgiven by

2

3cos

2

n

n

=

+ (76)

Substituting the value ofK1in equation (72), the pressure at any angle from the radial

line is given by

( ) ( )0 22

0

6 1

1 cos 1 cos

mhUr3

p p dc n c n

= +

+ + (77)

or( ) ( )

2

0 2 32

0

6 1 1

cos cos

mh AUrAp p dc cA A

=

+ + (78)

On solving, we get

( )

( )( )0 22 2

2 cos sin6

2 1 cos

n nUrp p

c n n

+ = + +

(79)

162


22/27

Figure 19 Pressure Distribution in the Film of a Journal

CHARACTERISTIC OF JOURNAL BEARING:

Considering the equilibrium of forces iny-direction, we get

( )2

0

sin sin 0Lrp d W

= (80)

Since, for an idealized bearing, = 900, the load carrying capacity is given by

2

0

sinW Lr p d

= (81)

Substituting the value for p and solving, we get

( )

2

2 2 2

12

2 1

r ULnW

c n n

=

+ (82)

Choosing '

2

UN

r= and

2

WP

Lr= , we get

( )2 22 '2

2 1

12

n nr N

c P n

+ =

(83)

The quantity

2 'r N

c P

is called the Sommerfeld number, which is a function of the

eccentricity ratio (figure 20).

163


23/27


24/27

Substituting the values of the film thickness h and the pressure gradientdp

d from

equations (75) and (78), the frictional force on the Journal surface J is obtained as

( )( )

( )( )

2

22

6 14

1 cos 2 1 cosJ

nU

c n n n

= + + +

(86)

The total frictional force may be obtained as

( )

( )( )( )

22

220

6 14

1 cos 2 1 cosJ J

nUF d Lr

c n n n

d

= =

+ + + (87)

On integrating and simplifying, we get

( )

( )

2

2 2

4 1 2

2 1J

nULrF

c n n

+=

+ (88)

The non-dimensional form of the equation can be expressed as

( )( )

2 2'

2

2 1

4 1 2f

n nr N

c P n

+ =

+ (89)

where '2U N= and2

Jf

FP

rL= .

165


25/27

Sommerfeld Number

Figure 21'r N

c P

v/s Sommerfeld Number

(Horizontal Line represents'

2

10.0506

2f

r N

c P

= , for lightly loaded bearings)

From figure 21, it can be observed that the value for'

f

r N

c P

remains constant for

Sommerfeld number greater than 0.15. In other words, the value of the Frictional Force

JF is independent of nfor Sommerfeld values greater than 0.15. Hence,

2 24J

Lr NF

c

=

'

for

(90)

0.15S>

Equation (90) is similar to the Petroffs equation (while Petroffs equation, the effect of n

is neglected), applicable for lightly loaded bearings.

Substituting n= 0 in the above equation, we get

166


26/27

'

2

10.0506

2f

r N

c P

= =

For an idealized journal bearing, the coefficient of friction can be found as

( )

( )

( )

2

2 2 2

2

2 2 2

4 1 2

2 1 1 2

12 3

2 1

J

nULr

c n nF c nf

r ULnW r

c n n

+

+

n

+= = =

+

(91)

Rearranging the above equation,

21 2

3

r

fc n

n +

= (92)

The equation (92) shows thatr

fc

is a function of nonly.

Now, multiplying both sides of equation (7), the relationship ofr

fc

with Sommerfeld

number of a lightly loaded bearing can be obtained as

2

2 '2r r

fc c

N

P

=

(93)

Exercise problems

1. Obtain the Pressure distribution (p v/s x) plot and determine the maximum pressuredeveloped for a plane slider bearing with the following data:

Length of the Bearing 10cm

Width of the Bearing = 6 cmVelocity =4 m/s

Viscosity of the lubricant = 100 cp

Minimum Fluid Film Thickness = 0.002 cmMaximum Fluid Film Thickness = 0.006 cm

2. In a journal bearing, diameter of the bearing = 3 cm, length of the bearing = 6 cm, speed =

2000 rpm, radial clearance = 0.002 cm, inlet pressure 0.3 Mpa. Location of the inlet hole = 300 ,

viscosity = 25 cp, eccentricity ratio = 0.1. Radial load = 500 N and Sommerfeld number is

calculated to be 0.1688. Find friction torque on the journal, coefficient of friction and power loss

and load carrying capacity.

0

167


27/27

chapter_8_dynamic analysis of fydrodynamic bearings

Documents