Sd 3

Design of externally pressurised gas-!ubricated porous journal bear=ngs B. C. Majumdar*

Design charts for the evaluation of the performance characteristics of externally oressurized, porous gas bearings under static loading are Dresented for a wide range of parameters for direct use in practice. The theoretical results are obtained from a rigorous analysis by simultaneous solution of the continuity equation (derived from Darcy's law) and the modified Reynolds' equation satisfying the boundary conditions of sealed end bearings. The effect of the feeding parameter on static behaviour is discussed. Finally, a design procedure of one such bearing is described. Nomenclature C D e

G,G H k x, ky, k z

L P p' Pa Ps , -fis R S,S T W,W X, y, Z 7? 6

0 A

radial clearance diameter of journal eccentricity mass rate of flow and flow parameter thickness of porous bushing permeability coefficients of bushing material in circumferential, axial, and radial directions respectively bearing length pressure (absolute) in the bearing clearance pressure (absolute) in the porous media atmospheric pressure (absolute) supply pressure (absolute), Ps = Ps/Pa journal radius gas constant stiffness and stiffness parameter absolute temperature load capacity and load parameter coordinates absolute viscosity of gas eccentricity ratio, e = e/C (dimensionless) angular coordinate, 0 = x / R feeding parameter A = 12kzR2/C3H (dimensionless)

Note: A dimensionless quantity has a bar on the top.

A porous material can be used as a bearing surface, with the porousity of the material controlling the gas supply by serving as a restrictor between the supply manifold and the bearing. More even distribution of gas is possible compared with a discrete hole-admission bearing 1. Moreover, a bearing with a porous surface has an inherent damping capability. Hence a porous bearing may be preferred to a conventional hole-admission bearing in many industrial applications. Probably the first experimental work on gas porous bearings is that of Montgomery and Sterry 2 who showed that these * Department of Mechanical Engineering, Indian Institute of Technology, Kharagpur, India

bearings, when supported on rubber 'O' rings, could be rotated up to a speed of nearly 4200 Hz with a shaft diameter of about 6 ram. The most complete theoretical analysis, using the perturbation method and considering only radial gas flow in the porous wall, is due to Sneck and Yen 3. Later, Mori et al 4,s, gave an incompressible solution assuming an equivalent clearance model. Mori et al 6 in another paper indicated a theoretical method for the solution for a porous bearings as a boundary value problem. Although this work 6 considered three dimensional flow in the porous media, it failed to give sufficient information for the design of a journal bearing. The limitations of the work of Sneck and Yen 3 are that the results are applicable only to a lightly loaded bearing and are only valid for a thin porous wall. On the other hand, Mori e t al 4,s used incompressible flow theory as a basis of their solution. Hence the results are only true for low gas supply pressures. The present work considers the problem for conditions between these two extremes.

Theoretical considerations For an externally pressurized journal bearing with sealed ends as shown in Fig 1, gas is fed at constant pressure from behind the porous bushing throughout the length of the bearing. The flow of gas takes place in the axial, circumferential and radial directions within the bush and then exhausts through the bearing clearance to the atmosphere. Assuming the flow of gas through the porous media is characterized by Darcy's law and no slip occurs at the porous surface, the flow equations in the porous wall and in the bearing clearance are respectively6: ~2p'2

c32p,2

~2p,2

~x ~xx2-+ky -0~- +k z ~zY = 0

(I)

and O ~x

h3 ~)p2t + h3 -~x] -~y2 = 12 k z ~ 3z ]z=H

(2)

With the following substitutions: 0 = x / R , ~ = y/(L/2), z = z/H, h = h/C (= 1 + e cos 0),P- ' = (p'/pa) 2 and ff = (p/pa) 2,

TRIBOLOGY international April 1976

71

x,8

~

The static stiffness of this type of bearing is given by:

L

dW

S= -

(9)

de

and in dimensionless lbrm is:

dW Porous beorin 9

de

#%0 A gas-lubricated porous journal bearing

Fig 1

where

equations (1) and (2) for the case of uniform permeability coefficients (ie k x = ky = kz) will be respectively: 02p'

02p'

002

~y2-

---- + (D/L) 2

+ (R/H) 2

02p' T~-

=0

(3)

and

SC

S-=

(10)

LD(Ps - Pa) Hence the slope of W versus e at any particular value of e gives the stiffness. However, a simplified formula in terms of calculated W for small values ofe (e ~< 0.5) can be computed using a linear perturbation theory with respect to e. In these cases the stiffness

02]6

3 3P dh

002

h 30 dO

--- +

+ (D/L) 2

32P 3y 2

-

A (OP'] h 3 ~32-]z=l

(4)

S-

where

A-

12 kzR 2

L/2 2n w=-2f

f o

(s)

p'RcosOdOdy o

and 2.tr

2C 3

(1 +ecosO)3RdO (6)

127 ~ T

f 0

p, 3p' y=L/2

The load and flow in equations (5) and (6) are computed numerically and expressed in dimensionless form as: W

(7)

and _

72

Ce

(ll)

LD(p, pa)~

24 • ~T

c 3% 2 -

G

TRIBOLOGY international April 1976

Ce

(12)

Therefore, the stiffness at any radial clearance C can be calculated from the dimensionless load W using equation (7). The theoretically predicted results of bearings having LID = 0.5, 1.0, and 2.0 for various design parameters are shown in Figs 2 - 4 . The load capacity in all cases increases with increase in the feeding parameter, reaches a maximum and then decreases with further increase in A. Hence an appropriate value of A can be selected for maximum load capacity. With an increase in the bushing thickness the load falls considerably and therefore, a one dimensional solution 3 which does not have bush thickness as a parameter is inadequate to predict the above behaviour. The flow rate increases continuously with increases in the feeding parameter particularly at high values of A. It may also be observed that the flow does not change appreciably with e. In this connection it may be mentioned that there is no significant decrease in flow rate with increase in H. For this reason the flow is shown only for HID = 0.1 (Figs 2 - 4 ) . From the results it appears that the dimensionless load W reaches a maximum value when LID = 1.0 for most feeding parameters. The absolute value is, of course, higher at higher LID ratios. Thus, an increase in the LID ratio does not result in a proportional increase in load capacity.

Design procedure

LD(Ps - Pa)

G-

W

-

S=

C3H

Equations (3) and (4) are solved by iteration 7 in a finite difference form with a digital computer to determine the pressure distribution in the bearing clearance. Thus, with the pressure distribution known, the load W and the mass rate of flow G at the bearing ends can be calculated from:

W=

de

-

where

As pressure p is given by pressure p' in tile porous media at the bearing surface, which is the boundary between the porous media and the bearing clearance, ~ in equation (4) can be replaced by ~' and as the pressure gradient across the bearing clearance is zero, ~' at ~-= 1 is the film pressure.

G_

dW

(8)

The bearing can be designed either for maximum load or for maximum stiffness. The values of A which yield these criteria are different. Therefore, to fulfil a particular design criterion the dimensions of the bearing must be found separately, the following worked example illustrating the method. A journal bearing is operating at e = 0.2 with the following specifications.

80

0.8

ko

/15o , = o a----/// ,=o 5-~f/

70

0.7 , = 0.8.~,,.//

,~= 0 . 5 - - - - ~

0.6

=

0.5

,E=0.27.40

08

~=0.2///-~/'" 60

,

//

50 o.s

I~ 0.4

_

• =o.8

40 IO

/ 03

.:o5~./

.

02

C--"

.:o5~..':'$.,3;"-~-,....

~ - - ~

\

~.

,,// - :5.0 /,~-,oo 8 ~/.d'---, =0 5 ,o

- 30

/

• -T2.a__

f~

~-- ~,=o.z

o,

20

-

J "%

.

1.0

I0

06.5

,io

5c,0

,b

06

A

,.'o

6 A

~0

80

0.8

,° 0.8-~///50 , =0s--y//

,=0.8~

07

,=0.2

70

E = 0.5-~--~

• =0.2 . - ~ 06

60

05

50

08

40

0.6

30

40 if,.9

1~04

Io

:50

0:5

0.2

//"

r~,=o.2

/.~/"/

- " - ...

"-,,~. -

I.O

0.4

~

~

,

:

0

.

5

~-~,

i

02

10

~'---

=0.2 I C)

I .lO

5, 0

.3

b 0.8

80

Q7

• =0.8-~ • = 0.5-~/ 70

0.6

60

0.5

50

~

.'50

02

20

0.1

I0

--14o

. • =0.8

~

c

L5

I PO

I I0

~

~

~

- 30 "

"=0.8

-,=0.5

• = 0.8 ..-.-.~ - - ~

.~..~.

--,

20 ""

~----~"~/

Io

= 0.2

"~,E =0.5

0.2 "-----,=0.2

0

Z"

0.4

oc

5

o.SF

I~=

2 ,0

I0

A

IOi

40 Io

0.5

I

IJO

0.6-

~~ - - o & ~ ~

I~= o 4

20

_...~-..

20 I0

00.5

• =o.5

"'--

, _ _----------~ " " ' = 0.2 " "

IO

5,) 0

A

Fig2 Load capacity for L / D = 0.5. (a) Ps = 3.0, (b} ~s= 6.0, (c) ~ = 8.0* * Figs 2 4: solid line represents w for HID = O.1, the dotted line w for H/D = 0.2, and the broken line

c Fig 3

°(~3

I})

A

I()

2~°

Load capacity fi~r L / D = ]. O. (a} Ps = 3. O, (b) Ps =

6.0, (c) -Ps = 8.0*

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international April 1976

73

L=5xl0-2m,

D=5x

10-2m,

H=5x

10-3m

T a b l e 1: Stiffness of a bearing having

L/D=

1.0,

H/D=

0.1,

c = 0.2 and P-s = 6.0

and C x 10-6, m

S x 106, N/m

8.88 7.94 6.30

106.5 113.0 119.0

9.0

5.84

114.0

12.0

5.73

101.0

A k z = 1 x 10 - 1 5 m 2. 2.1 3.0 6.0

Using air as a lubricant and assuming Ps = 6 bar and Pa = 1 bar, and an air temperature of 293 K, design the bearing for both maximum load and maximum stiffness. From the above dimensions, L I D = 1.0, H I D = 0.1 and Ps = 6.0. The actual design will find the required bearing clearance C for the above criteria. Assume ~ 293 K.

are: W = 150 N , G = 7.19 x 10 - 5 kg/s and S = 119.0 x 106 N/m (since W = 0.120 and G-= 28.783 for A = 6.0).

= 288 Nm/kg °C and r~ = 18.3 NS/m 2 at

Designing for maximum

From the above example it may be noted that a bearing designed for maximum stifiness will have smaller radial clearance and less flow rate than a bearing designed for maximum load.

load

Referring to Fig 3(b), the maximum load will occur when A ~- 1.70 and the corresponding values of W = 0.152 and G = 1 1.638. The required radial clearance is J12 3

kzR2

References

1 Majumdar, B. C. On the general solution of externally pressurized gas journal bearings', J Lub Tech, Trans. ASME, Series F, vol 94, no 4, 1972, pp 291-296 2 Montgomery, A. G. and Sterry, F. 'A simple air bearing motor for very high rotational speed', Atomic Energy Research Establishment. Harwell, Berkshire, England, AERE, ED/R 196 71, 1955. 3 Sneek, H. J. and Yen, K. T. 'The externally pressurized, porouswall gas-lubricated journal bearings - I', ASLE Trans. vol 7, no 3, 1964, pp 288-298 4 Mori, H., Yabe, H., Yamakage, H. and Furukwa, J. 'Theoretical analysis of externally pressurized, porous gas journal bearings (lst Report)',Bulletin of JSME, vo111, no 45, 1968, pp 527535 5 Mori,H., Yabe, H., and Yamakage, H. 'Theoretical analysis of externally pressurized, porous gas journal bearings (2nd Report)', Bulletin of JSME, vo112, no 54, 1969, pp 1512-1518 6 Mori, H., Yabe, H. and Shibayama, T. 'Theoretical solution as boundary value problem for externally pressurized porous gas bearings,' J Basic Eng Trans. ASME, Series D, vo187, no 1, 1965, pp 622-- 630 7 Majumdar, B. C. 'Analysis of externally pressurized porous gas journal bearings - 1,' Wear, vol 33, no 1, 1975, pp 25 35

- 9.618 x 10 . 6 m.

HA The load and flow rate are 190 N and 10.3 x 10 - 5 kg/s. Using the simplified equation (12), the stiffness S = 98.5 x

106 N/m. Designing for maximum

stiffness

As the bearing carries only a light load ( i e e = 0.2), equation (12) can be used to find the stiffness.__Taking various values of A corresponding to high values of W, at e = 0.2, from Fig 3(b), the necessary values of the clearance C are calculated to give the maximum values of stiffness S. The results are given in Table 1. Referring to Table 1 the stiffness is seen to be a maximum when A = 6.0, the corresponding radial clearance being C = 6.30 x 10 - 6 m. The load, flow and stiffness in this case 0.6

60

0.6

60

0.6

60

05

50 ,=0.8 40

04

. . . . . . .

s ~ ~,:o

• = O.8-~-~

_

501~

_

20

0.I

0.I -

=---'°C).l a

Fig 4

74

A

b

~

II0

"---,:o.z I.O A

L o a d capacity f o r L / D = 2.0. (a) Ps = 3.0, (b) Ps = 6.0, (c) Ps = 8 . 0 *

T R I B O L O G Y international April 1976

0.I

~o c

~ . . ~ " ~ - - • =0 5 -~.-- ,~=0.2 I O6= I.O A

Io