Plane Hydrostatic and Aerostatic Bearings

CHAPTER 7

Plane Hydrostatic and Aerostatic Bearings Summary of Key Design Formulae Ae ¼ A$A W ¼ Ps Ae W Ps h3o qo ¼ $bB h 
Ps h3o pro þ pa qo ¼ $Kgo B$ h 2pa Ps Ae l¼ $l ho h ¼ ho $X

Effective pad area Load Hydrostatic flow Aerostatic flow Film stiffness Film thickness

7.1 Use of the Design Charts Chapter 4 gave effective area shape factors and flow shape factors for various plane pads. This chapter gives flow, load, and stiffness data with film thickness varying from the design condition from minimum to maximum. Design charts are for any shape of pad for four situations: • • • •

Single plane pads Equal opposed-pad bearings Unequal opposed-pad bearings Complex configurations.

Gauge pressures are given in upper case symbols. Absolute pressures are given in lower case.

7.2 Choice of Land Width For most hydrostatic and aerostatic applications it is recommended that the ratio of the land width a to the bearing width L should be a/L ¼ 0.25. However, for aerostatic pads, it is desirable to design virtual recesses for reduced recess volume, as described in Section 4.2. The recommended land-width ratio yields minimum pumping power to support a given load on a given total area. Hydrostatic, Aerostatic and Hybrid Bearing Design. DOI: 10.1016/B978-0-12-396994-1.00007-3 Copyright Ó 2012 Elsevier Inc. All rights reserved.

125

126

Chapter 7

At higher speeds, friction power must be taken into account. If required, land-width ratio can be reduced to increase flow for hydrostatic bearings. Reducing land-width ratio may often be preferred to increasing film thickness as a means of increasing flow. It is not recommended to reduce land-width ratio to extreme values. Usually, a land-width ratio of 0.1 is a practicable minimum. Very thin land widths can be vulnerable to damage. Small land width is not recommended for aerostatic pads with virtual recesses (Section 4.2).

7.3 Flow Variation with Film Thickness Flow q from a pad varies with film thickness. Hydrostatic flow is given by q¼

Pr h3 Ps h3o 3 $B ¼ $B$Pr $X h h

(7.1)

where B, the flow shape factor, is the same for hydrostatic and aerostatic bearings. Pr ¼ Pr =Ps is given on charts for single pads and X ¼ h=ho takes account of film thickness variation from the design condition. The design condition is defined by h ¼ ho. This yields concentric flow qo at a recess pressure Pro ¼ bPs ¼ KgoPs. The term b is the design pressure ratio for hydrostatic bearings and Kgo is for aerostatic bearings: qo ¼

Ps h3o $bB h

Hydrostatic

p þ p  Ps h3o ro a qo ¼ $Kgo B$ h 2pa

(7.2) Aerostatic

Shape factors for a range of pad shapes are provided in Chapter 4.

7.4 Load Variation with Film Thickness Bearing load W ¼ PrAe. Effective pad area is Ae ¼ A$A, where A is the area shape factor described in Chapter 4. More generally, for single- and opposed-pad bearings: W ¼ Ps Ae W

(7.3)

where W is a load factor that varies with film thickness. For single pads, load support is W ¼ Ps Ae $Pr , where Pr ¼ Pr =Ps . In this case, W ¼ Pr

Single pads

(7.4a)

For opposed pads, W ¼ Ps Ae $ðPr1  Pr2 Þ=2, so that W ¼ ðPr1  Pr2 Þ=2

Opposed pads

(7.4b)

Plane Hydrostatic and Aerostatic Bearings 127 For unequal opposed pads, the load factor takes account of the area ratio of the opposed pads. Charts are given for several methods of flow control.

7.5 Stiffness Variation with Film Thickness Bearing film stiffness l is given by l ¼ ðPs Ae =ho Þ$l, where l depends on the flow control, as discussed in Chapter 5. Values of l are given in this chapter for plane pads as film thickness h varies from the design value ho. Variations of film thickness are expressed on the charts as X ¼ h=ho varying from 0 to 1. Stiffness varies between a maximum at best to zero at extremes of load. Extreme loads should be avoided. Stiffness factors with capillary control give a better indication of average stiffness within a moderate load range.

7.6 Single-Pad Bearings Expressions for load, pressure, stiffness, and flow factors are given in Table 7.1 for hydrostatic pads and in Table 7.2 for aerostatic pads. Data charts are presented in Figures 7.1e7.9. Aerostatic bearings involve more complex expressions than hydrostatic bearings but are still straightforward when employing slot restrictors. Orifice-fed aerostatic bearings are best designed to operate with unchoked orifices to minimize the risk of pneumatic hammer. Figure 7.4b shows the regions where jet flow remains unchoked. The consequence of design for unchoked flow is a reduced range of loads that may be applied to the bearing. Orificefed aerostatic pads are better loaded with increasing loads compared with the design condition to avoid choked flow. Aerostatic pads should be designed to avoid large recess volume, as described in previous chapters. The application of virtual recesses is described in Section 4.2 but virtual recesses reduce load support, requiring slightly increased pad area in some cases. Minimizing recess volume is essential for good dynamic performance. Table 7.1: Thrust, stiffness, and gap for single hydrostatic pads (W [ Pr ) Capillary or Slot Pr ¼

1 1b 3 1þ X b

Orifice 6

X ¼

Constant Flow

1  Pr b2 2 P 1b

Pr ¼

2

l¼3

b X :   1b b 2 3 X þ 1b

l X ¼ 3Pr ð1  Pr Þ Q ¼ Pr B X

3

l¼

l X ¼ 6Pr

ð1  Pr Þ 2  Pr

Q ¼ Pr B X

3

Diaphragm

b X

3

3b X

lo ¼

4

3bð1  bÞ 3b 1  ð1  bÞ ld

l X ¼ 3Pr Q ¼ Pr B X

3

Q ¼ Pr B X

3

128

Chapter 7 Table 7.2: Thrust, stiffness, and gap for single aerostatic pads Capillary or Slot pr ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 p2s þ p2a mc X 1 þ mc X

3

X ¼ mo $

3

mc X

W ¼ ðpr  pa Þ=ðps  pa Þ

Q¼

p2r  p2a

2

ðps þ pa Þ 3 $ 2p r ½1 þ mc X 2 1  Kgo ps þ pro mc ¼ $ Kgo pro þ pa

l ¼ 3$

Orifice qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðpr Þ2=k  ðpr Þðkþ1Þ=k

p¼

p ps

p2ro  p2a mo ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðpro Þ2=k  ðpro Þðkþ1Þ=k (
k=ðk1Þ ) 2  and p ¼ max p or kþ1

p2r  p2a 3 BX 2pa ðps  pa Þ

(a)

Q¼

p2r  p2a 3 BX 2pa ðps  pa Þ

1 HYDROSTATIC Capillary or slot

0.8 = 0.8 0.7 0.6 0.5 0.4 0.3 0.2

W = Pr 0.6

0.4

W = Ps AeW

0.2

0

0

0.5

1 1.5 2 Bearing gap, X = h / ho

2.5

3

4 = 0.8 3

qη Ps Bh3o

0.7

2

0.6 0.5

1

0.3 0

0

0.5

1

1.5

2

2.5

3

Figure 7.1: (a) Load and Flow: Capillary and Slot-Controlled Hydrostatic Pads.

Plane Hydrostatic and Aerostatic Bearings 129

(b) 1

AEROSTATIC Capillary or slot 0.8

Kgo 0.8 0.7 0.6 0.5 0.4 0.3 0.2

W = Pr 0.6

0.4

W = Ps AeW Ps / pa = 6

0.2

0

0

0.5

1 1.5 2 Bearing gap, X = h / ho

2.5

3

8

Ps / pa = 6

Kgo = 0.8

6 qη PsBh3o

0.7

4 0.6 2

0.5 0.3

0

0

0.5

1

1.5

2

2.5

3

Figure 7.1: (b) Load and Flow: Capillary and Slot-Controlled Aerostatic Pads.

Selection of Design Pressure Ratio b [ Kgo [ Pro / Ps for Single Pads The general recommendation is b ¼ Kgo ¼ 0.5. This allows the maximum range of applied loads. Higher or lower values may be employed for special cases. However, for orifice-fed aerostatic bearings, pressure ratio should not be less than 0.5 to avoid choked flow and possible instability.

Adjusting Flow Rate to a Convenient Value Often, flow for a low-speed single circular pad is very small. Or, sometimes, flow is too high for the pump proposed. There are a number of ways to reduce flow, such as designing thicker lands or increasing bearing size while reducing supply pressure. Alternatively, if bearing area may be increased it may be possible to reduce pressure ratio, b, to about 0.2, which has the advantage that a higher maximum load may be applied. However, the most sensitive parameter controlling flow is gap h because it appears to the third power.

130

Chapter 7 1.4 Capillary or slot

1.2

=Kgo 0.2 0.3 0.4 0.5 0.6 0.7 0.8

1 λ

0.8 0.6

Stiffness PA λ= s eλ ho

Liquid Gas (Ps /p a=6)

0.4 0.2 0

0

0.5

1 1.5 2 Bearing gap, X = h / h o

2.5

3

Figure 7.2: Stiffness and Gap for Capillary and Slot-Controlled Pads. 1.6 Capillary or slot

= Kgo 0.2

1.4 1.2

0.3 Gas (Ps /pa = 6)

1

λ

0.4 0.5

0.8

Liquid 0.6 0.7

0.6

0.8 Ps Ae λ ho Bearing thrust W = Ps AeW

0.4

Stiffness λ = 0.2 0

0

0.2

0.4

0.6

0.8

1

Bearing thrust, W = Pr

Figure 7.3: Stiffness and Thrust for Capillary-Controlled Pads.

n

Example 7.1 A Single-Pad Hydrostatic Bearing A circular capillary-compensated pad bearing with central oil admission must support a load of 4.5 kN (1011 lbf ). The bearing gap is to be 50 mm (0.002 in) and the bearing is to operate at zero speed. Determine: 1. Suitable dimensions for the pad 2. Operating stiffness

Plane Hydrostatic and Aerostatic Bearings 131

(a) 1 HYDROSTATIC

0.8 Orifice

W = Pr

0.6

= 0.8 0.7 0.6 0.5 0.4 0.3 0.2

0.4

0.2

0

0

0.5

Thrust W = Ps AeW

1 1.5 2 Bearing gap, X = h / ho

2.5

3

2 = 0.8

1.5 qη PsBho3 1

0.6

0.5

0.4

0.7

0.2

0

0

0.5

1

1.5

2

2.5

3

Figure 7.4: (a) Load and Flow for Orifice-Controlled Hydrostatic Pads.

3. Load support and flow when X is reduced to 0.4 4. Load when h ¼ 10 mm (0.0004 in). It has been decided that b is to be 0.5 and a convenient pressure for the pump is 4 MN/m2 (580 lbf/in2). The oil to be used has a dynamic viscosity h ¼ 35 cP (5  106 reyn) at 38  C (100  F). Solution

1. At the design condition Pr ¼ 0.5Ps ¼ 0.5  4 ¼ 2 MN/m2 (290 lbf/in2) Ae ¼ A$A ¼

W ¼ 4500=ð2  106 Þ ¼ 2:25  103 m2 ð3:487 in2 Þ Pr

132

Chapter 7

(b) 1 AEROSTATIC Orifice

0.8 W = Pr

0.6

0.4

0.2

0

Thrust W = Ps AeW

Kgo 0.8 0.7 0.6 0.5 0.4 0.3 0.2

0

Ps /pa=6

Choked jets

0.5

1 1.5 2 Bearing gap, X = h / h o

2.5

3

4 Ps / pa = 6

qη Ps Bho3

Kgo = 0.8

3

Choked jets 0.7

2

0.6

1

0.4 0.2

0

0

0.5

1

1.5

2

2.5

3

Figure 7.4: (b) Load and Flow for Orifice-Controlled Aerostatic Pads.

From Figure 4.2 and choosing a/D ¼ 0.25, so that R ¼ R2 =R1 ¼ 2:0 and A ¼ 0.542, A¼

AA 2:25  103 ¼ pR22 ¼ ¼ 4:15  103 m2 ð6:435 in2 Þ 0:542 A

R2 ¼ 36:3 mm ðsay 36 mmÞ ð1:42 inÞ R1 ¼ R2 =R ¼ 36=2 ¼ 18 mm ð0:71 inÞ The design gap is achieved by setting the flow. From Figure 4.2, B ¼ 0:75, therefore qo ¼

Ps h3o 4  106  503  1018  0:5  0:75 bB ¼ h 0:035

¼ 5:36  106 m3 =s ð0:327 in3 =sÞ

Plane Hydrostatic and Aerostatic Bearings 133 2 Liquid

Gas Ps /pa= 6

1.5

Orifice

= Kgo 0.2 0.3 0.4 0.5 0.6 0.7 0.8

λ 1

Stiffness

λ=

Ps Ae λ ho

0.5

0

0

0.5

1

1.5

2

Bearing gap, X = h / ho

Figure 7.5: Stiffness and Gap for Orifice-Controlled Pads.

2 = K go 0.2

Orifice

1.5

0.3

λ

0.4 0.5 0.6 0.7 0.8

Liquid

1

Gas Ps/pa = 6

Ps Ae λ ho Bearing thrust W = Ps AeW

0.5

0

Stiffness λ =

0

0.2

0.4 0.6 Bearing thrust, W = Pr

0.8

1

Figure 7.6: Stiffness and Thrust for Orifice-Controlled Pads.

2. From Figure 7.2, for b ¼ 0.5 and X ¼ 1, the stiffness factor l ¼ 0.75. Actual stiffness is therefore l¼

Ps AA 4  106  2:25  103  0:75 l¼ ¼ 0:135 GN=m ð0:771  106 lbf =inÞ ho 50  106

134

Chapter 7 1 Constant flow

W = Pr 0.8

= 0.8 0.7 0.6 0.5 0.4 0.3 0.2

0.6

0.4

Thrust W = Ps AeW

0.2

0

0

0.5

1 1.5 2 Bearing gap, X = h / ho

2.5

3

Figure 7.7: Load and Gap for Constant-Flow Controlled Pads.

6 Constant flow

= 0.2 0.3 0.4 0.5 0.6 0.7 0.8

5

4

λ 3

Stiffness

λ =

Ps Ae λ ho

2

1

0 0

0.5

1

1.5

2

2.5

3

Bearing gap, X = h / ho

Figure 7.8: Stiffness and Gap for Constant-Flow Controlled Pads.

3. When X ¼ 0.4, h ¼ 50  0.4 ¼ 20 mm (0.0008 in). From Figure 7.1, the load factor W ¼ 0.94. Therefore, maximum applied load is W ¼ Ps AA W ¼ 4  106  2:25  103  0:94 ¼ 8:46 kN ð1900 lbf Þ

Plane Hydrostatic and Aerostatic Bearings 135 5 Constant flow 4

Stiffness λ =

3

Ps A e

ho

λ

Thrust W = Ps A eW

2

= 0.2 0.3 0.4 0.5 0.6 0.7 0.8

λ 1

0

0

0.2

0.4

0.6

0.8

1

Bearing thrust, W = Pr

Figure 7.9: Stiffness and Thrust for Constant-Flow Controlled Pads.

Maximum load at this gap compared with the design load is 8460/4500 ¼ 1.88 and the dimensionless stiffness is reduced to 0.42 compared with the design value of 0.75. From Figure 7.1 or Table 7.1, dimensionless flow is Qo ¼ bB ¼ 0.5  0.75 ¼ 0.375 at the design gap. At the reduced film thickness, dimensionless flow is reduced to 3 Q ¼ B Pr X ¼ 0.75  0.94  0.064 ¼ 0.0451 and actual flow is q¼

Ps h3o 4  106  503  1018  0:0451 $0:0451 ¼ h 0:035

¼ 0:6443  106 m3 =s ð0:0393 in3 =sÞ 4. Taking h ¼ 10 mm (0.0004 in) as the minimum allowable film thickness, X ¼ h=ho ¼ 10=50 ¼ 0:2 and W ¼ 0:99 W ¼ Ps AA$W ¼ 4  106  2:25  103  0:99 ¼ 8:91 kN ð2000 lbf Þ However, dimensionless stiffness is now reduced to 0.1. This demonstrates that minimum allowable film thickness is not a good criterion for maximum allowable load. A better approach is to decide minimum X . For b ¼ 0.5 a suitable range for X is from 0.4 to 1.6. If, however, b ¼ 0.2, a suitable range for X is from 0.2 to 1.1. n

136 n

Chapter 7

Example 7.2 A Single-Pad Aerostatic Bearing An aerostatic slideway bearing must support a load of 2 kN (450 lbf). The bearing gap is to be 25 mm (0.001 in). It has been decided that Kgo is to be 0.5 and a convenient air pressure is 0.45 MN/m2 (65.3 lbf/in2). Air has a dynamic viscosity h ¼ 18.3  106 Ns/m2 at 18  C. Ambient pressure is 0.101 MN/m2 (14.7 lbf/in2). Determine: 1. 2. 3. 4.

Suitable dimensions for the pad and entry ports Required flow at the design condition Load support when the film thickness is reduced by 50% Stiffness when the film thickness is reduced by 50% employing capillary flow control.

It is decided to choose a rectangular pad with a virtual recess. This requires a number of holes or entry ports to be arranged around the virtual recess. Solution

1.

To determine suitable dimensions. Increase the specified load by 15% using virtual recesses to allow for dispersion from discrete entry ports: Wo ¼ 2300 N. See Figure 4:4 for shape factors : B=L ¼ 2; a=L ¼ 0:3; A ¼ 0:57; B ¼ 1:2 At the design condition; Pro ¼ 0:5Ps ¼ 0:5  0:45  106 ¼ 0:225 MN=m2 ð32:7 lbf =in2 Þ: W ¼ Pr A$A so that AA ¼ 2300=ð0:225  106 Þ ¼ 10:22  103 m2 ð15:84 in2 Þ A ¼ 10:22  103 =0:57 ¼ 0:01793 m2 ð27:79 in2 Þ B $L ¼ 2L2 ¼ 0:01793 m2 L pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 0:01793Þ L¼ ¼ 0:06695 m ð2:636 inÞ 2 B ¼ 2L ¼ 0:1339 m ð5:272 inÞ a ¼ 0:3  0:06695 ¼ 0:0201 m ð0:791 inÞ

Plane Hydrostatic and Aerostatic Bearings 137 Assume the virtual recess will be surrounded at least 70% by slot-entry ports. The perimeter of the virtual recess is 2ðB þ L  4aÞ ¼ 2ð0:1339 þ 0:06695  4  0:0201Þ ¼ 0:2409 m ð9:484 inÞ For 70% recess surround: Total length of the entry slots must be 0.7  0.2409 ¼0.169 m (6.64 in). Assume 16 slot shaped entry ports. The length of the slots must be 0.0105 m (0.415 in). Suggested port dimensions are 1 mm wide  0.1 mm deep. Total port volume is 16  0.011  0.001  0.0001 ¼ 17.6  109 m3. The suitability of this port volume can be compared with the flow. See next step. 2. To determine free air flow at the design condition. Absolute recess pressure is pr ¼ 0.225 þ 0.101 ¼ 0.326 MN/m2 (47.3 lbf/in2). B ¼ 1:2. From equation (7.2): 
Ps h3o pro þ pa Kgo B qo ¼ h 2pa
 0:45  106  253  1018  0:5  1:2 0:326 þ 0:101 qo ¼ 0:0000183 2  0:101 ¼ 487  106 m3 =s ð0:487 l=sÞ

3.

This is the flow for slot-entry ports that 100% surround the virtual recess. Flow may be reduced by approximately 15% employing a virtual recess with 70% surround. The ratio of the flow to the port volume is f ¼ (487  106)/(17.6  109) ¼ 27.7  103 Hz. See Equation (5.16). This frequency is much larger than 1000 Hz, which suggests the size of the entry ports should not cause a problem, although this is not a guarantee. To determine load when film thickness is reduced by 50%. X¼

h ¼ 0:5 ho

From Figure 7.1, W ¼ 0.88. Maximum applied load for this gap is W ¼ Ps AA W ¼ 0:45  106  0:01793  0:57  0:88 ¼ 4:05 kN ð910 lbf Þ The shape of the graph suggests that this load represents an absolute maximum that should be applied under any circumstances.

138

Chapter 7

4.

l¼

From Figure 7.2, for b ¼ 0.5 and X ¼ 0.5, dimensionless stiffness l ¼ 0.62. Stiffness is Ps AA 0:45  106  0:01793  0:57  0:62 l¼ ¼ 114 MN=m ð0:65  106 lbf =inÞ ho 25  106 n

7.7 Equal Opposed-Pad Bearings The design of hydrostatic bearing systems having equal opposed pads is simpler in practice than for single-pad bearings. Data are given below in Figures 7.10e7.17. For aerostatic 0.5

= 0.2 0.3

W

Capillary

0.4

0.3

0.4

0.2

0.5

0.6

0.7

Liquid: = 0.8 Gas: Ps /pa = 6, Kgo=0.8 Thrust W = Ps AeW

0.1

Ae = Ae1 + A e2 0

0

0.2

0.4 0.6 Bearing gap, X 1 = h1 / ho

0.8

1

0.8

= 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.6 λ

0.4 Stiffness λ = 0.2

Ps A e λ ho

Ae = A e1 + A e2

0 0

0.5

1

1.5

2

Bearing gap, X 1 = h1 / ho

Figure 7.10: Load and Stiffness for Capillary-Controlled Equal Opposed Pads.

Plane Hydrostatic and Aerostatic Bearings 139

(a)

HYDROSTATIC

3.0

qη PsBho3

= 0.8 2.5

Capillary

2.0 0.7 qη PsBho3

1.5

0.6

1.0

0.5 0.4 0.3 0.2

0.5 0

0

0.2

0.4 0.6 Bearing gap, X1 = h1 / ho

0.8

1

1.6

= 0.8

1.4

Orifice

1.2

0.7

qη 1 PsBho3

0.6

0.8

0.5 0.6

0.4 0.3

0.4

0.2 0.2

0

0.2

0.4 0.6 Bearing gap, X1 = h1 / h o

0.8

1

Figure 7.11: (a) Flow: Hydrostatic Equal Opposed Pads.

bearings, the same charts can be employed as for hydrostatic bearings. The differences are relatively minor. Dashed lines are included for aerostatic bearings to show where reduced load support applies compared with hydrostatic bearings. The reduction is greatest for low design pressure ratio. The effective area of an equal opposed-pad bearing is twice the effective area of each of the single pads employed to make up the bearing. In other words, AA ¼ Ae ¼ Ae1 þ Ae2, where 1 and 2 denote the opposing pads. Load W, stiffness l, and flow qo are given by W ¼ Ps Ae W

where Ae ¼ Ae1 þ Ae2

(7.5)

140

Chapter 7

(b)

AEROSTATIC K go = 0.8

6

qη PsBho3

5 qη 4 Ps Bho3 3

0.7

2

0.5

1

0.4 0.3

0

Capillary

0.6

0.2

0

0.2

0.4

0.6

Bearing gap, X 1 = h1 / h o

6

0.8

1

5 Orifice qη PsBho3

K go = 0.8

4 3

0.7 0.6 0.5 0.4 0.3

2 1 0

0.2 0.2

0

0.4

0.6

0.8

1

Bearing gap, X 1 = h1 / h o

Figure 7.11: (b) Flow: Aerostatic Equal Opposed Pads, Ps/pa [ 6.

l¼ qo ¼

Ps Ae l ho

where Ae ¼ Ae1 þ Ae2

Ps h3o $2bB h


Ps h3o pr þ pa qo ¼ $2Kgo B h 2pa

(7.6)

Hydrostatic (7.7) Aerostatic

Values are given on the data charts that follow where film thicknesses of the two pads are X 1 ¼ h1 =ho

and X 2 ¼ 2  X 1

(7.8)

Plane Hydrostatic and Aerostatic Bearings 141 0.5

= 0.2 0.3 Orifice

0.4

0.4

W 0.3

0.5

0.6 0.7 = 0.8 Liquid: Gas: Ps /pa = 6, K go = 0.8

0.2

Thrust W = Ps AeW

0.1

A e = A e1+ A e2 0

0

0.2

0.4

0.6

0.8

1

Bearing gap, X 1 = h1 / ho

1

= 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.8 λ 0.6

0.4 Stiffness λ = 0.2 0

Ps Ae

ho

λ

Ae = Ae1 + A e2 0

0.5

1

1.5

2

Bearing gap, X 1 = h1 / ho

Figure 7.12: Thrust and Stiffness: Orifice-Controlled Equal Opposed Pads.

Selection of Design Pressure Ratio for Opposed-Pad Bearings The optimum design pressure ratio is b ¼ Kgo ¼ 0.5 and in most cases it is unnecessary to choose any other value for opposed-pad bearings. Where tolerances are given on clearance, a suitable range for pressure ratio is 0.4 < b < 0.7 or for orifice-fed aerostatic bearings a suitable reduced range is 0.5 < Kgo < 0.7. As explained in previous chapters, it is generally recommended that choked flow is avoided for orifice-fed aerostatic bearings to reduce the risk of instability. With pressure-sensing flow control devices the maximum allowable load will not be lower than that for capillary compensation and therefore the same value may be used.

142

Chapter 7 + 1.0 W = Ps AeW

Maximum positive load

+ 0.5

Operating region

W 0

Maximum negative load – 0.5

1.0

1.5

2.0

2.5

E = A e1/A e2

Operating load range for unequal opposed pads

7 6 W positive /W negative 5 4 3 2 1

1

1.5

2 E = A e1/A e2

2.5

Figure 7.13: Unequal Opposed Pads: Operating Load Range and Selection of Area Ratio.

n

Example 7.3 An Opposed-Pad Hydrostatic Bearing A capillary-compensated opposed-pad bearing is required that will not deflect more than 2.5 mm (0.0001 in) under an applied load of 450 N (101 lbf). The available supply pressure is 3.5 MN/m2 (508 lbf/in2) and in the design condition there is negligible load. The design bearing gap is to be ho1 ¼ ho2 ¼ 25 mm (0.001 in). The pressure ratio b is to be 0.5. Determine: 1. Suitable bearing dimensions for circular bearing pads 2. The maximum allowable applied load.

Plane Hydrostatic and Aerostatic Bearings 143 0.8 2.5

0.6

Ae1

2.0 1.5

0.4

Capillary

E = 1.0

Ae2

0.2 W

0

W = Ps AeW

-0.2 Ae = Ae1 + Ae2

-0.4 E = Ae1 / Ae2

-0.6

0

0.5

0.8

1 1.5 Bearing gap, X 1 = h1 / ho

2 Ae1

2.5

0.6

2.0 1.5 E = 1.0

0.4

Orifice

Ae2

0.2 W

0

W = Ps AeW

-0.2 Ae = Ae1 + Ae2

-0.4 E = Ae1 / Ae2

-0.6

0

0.5

1 1.5 Bearing gap, X 1 = h1 / ho

2

Figure 7.14: Unequal Opposed Pads: Thrust and Gap for Capillary and Orifice Control. Solution

1. Bearing dimensions. Stiffness must be at least 450/2.5  106 ¼ 180 MN/m (1.03  106 lbf/in). From Figure 7.10, stiffness factor is l ¼ 0:75. Actual stiffness is given by l ¼ 180  106 ¼

Ps Ae 3:5  106  Ae l¼  0:75 ho 25  106

Ae must be greater than 1714  106 m2, Ae1 ¼ Ae2 must be greater than 857  106 m2. From Figure 4.2, read A ¼ 0:54 at R2 =R1 ¼ 2: Ae1 857  106 ¼ ¼ 1587  106 m2 ¼ pR22 0:54 A R2 ¼ 22:5 mm ð0:886 inÞ

A1 ¼

R1 ¼ 11:25 mm ð0:443 inÞ

144

Chapter 7 0.8 Ae1

E = 2.5 Capillary

0.6

E = 1.0 Ae2

λ 0.4 λ= 0.2

Ps Ae λ ho

Ae = Ae1 + Ae2 E = Ae1 / Ae2

0

0

0.5

1 1.5 Bearing gap, X 1 = h1 / h o

2 Ae1

E = 2.5

1

Orifice

0.8

Ae2

E = 1.0

λ 0.6

λ= 0.4

Ps Ae λ ho

Ae = Ae1 + Ae2

0.2

E = Ae1/Ae2 0

0

0.5

1 1.5 Bearing gap, X 1 = h1 / h o

2

Figure 7.15: Unequal Opposed Pads: Stiffness and Gap for Capillary and Orifice Control.

2. Maximum allowable load. Examining Figure 7.10, it is seen that stiffness is greatly reduced for W greater than 0.3. A sensible maximum load would be W ¼ Ps Ae W ¼ 3:5  106  1714  106  0:3 ¼ 1:8 kN ð407 lbf Þ n n

Example 7.4 An Opposed-Pad Aerostatic Bearing An aerostatic capillary-compensated opposed-pad bearing is to have a design bearing film thickness ho1 ¼ ho2 ¼ 25 mm (0.001 in). The maximum applied load is to be 450 N (101 lbf ). It is required that the gap will not reduce by more than 50%. The available supply pressure is 0.45 MN/m2 (65 lbf/in2) and in the design condition there is negligible load. The pressure ratio is to be Kgo ¼ 0.5. Determine suitable dimensions for circular bearing pads and check flow rate.

Plane Hydrostatic and Aerostatic Bearings 145 Solution

1. Bearing dimensions. Fifty percent minimum film thickness corresponds to X ¼ 0.5. From Figure 7.10, load factor W z 0.3. Increasing required load by 15% to allow for losses due to discrete entry ports: W ¼ 517:5 N ð116:3 lbf Þ Effective bearing area is AA ¼

W 517:5 ¼ 0:003833 m2 ð5:941 in2 Þ ¼ Ps W 0:45  106  0:3

Assume opposed circular pads with a virtual recess and a/D ¼ 0.3, so that R2/R1 ¼ 2.5. From Figure 4.2, A ¼ 0.54 and B ¼ 0.57, A ¼ 0:003833=0:54 ¼ 0:007098 m2 ð11 in2 Þ total area of two opposing pads A1 ¼ A2 ¼ 0:5A ¼ 0:0003549 m2 ð5:5 in2 Þ area of each pad rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  0:003549 D¼ ¼ 0:06722 m ð2:647 in2 Þ pad outer diameter p Dr ¼ 0:06722  0:4 ¼ 0:02689 m ð1:059 inÞ virtual recess diameter Perimeter of virtual recess is p  0:002689 ¼ 0:08448 m ð3:326 inÞ Assume eight small-entry recesses. For 70% surround of the large virtual recess, the entry recess diameter or groove length: dr ¼ 0:7  0:08448=8 ¼ 0:007392 m ð0:291 inÞ hr ¼ 0:5 mm ð0:0197 inÞ depth of recess Recess volume for two pads is Vr ¼ 2  8  p  0:0073922  0:0005=4 ¼ 0:3434  106 m3 ð0:02096 in3 Þ 2. Flow. From Chapter 4, the flow shape factor for a circular pad where D/Dr ¼ 2.5 is B ¼ 0:57: Flow qo ¼ 2

Ps h3o pro þ pa for two opposed pads Kgo B h 2pa

pro ¼ 0:225 þ 0:101 ¼ 0:326 MN=m2 ; qo ¼ 2 

pa ¼ 0:101 MN=m2

0:45  106  253  1018 0:326 þ 0:101  0:5  0:57  6 18:3  10 2  0:101

¼ 463  106 m3 =s ð0:463 l=sÞ

146

Chapter 7

Flow may be reduced approximately 15% employing a virtual recess with 70% surround. Check recess volume from equation (5.16): f ¼

qo 463  106 ¼ ¼ 1348 Hz: Vr 0:3434  106

At first sight, this looks sufficiently high. n

7.8 Unequal Opposed-Pad Bearings Unequal opposed-pad bearings may be used to advantage where the principal forces are in one direction but where forces in the reverse direction may also occur. At speed there is the additional advantage of lower power losses and temperature rise than with the equivalent equal opposed-pad configuration. The basic schematic arrangement is apparent from the design charts (Figures 7.14 and 7.15), which are intended to be self-explanatory. One side of the bearing, usually the top, carries the main load and is made bigger than the other side. Thus, if the top bearing has an effective area Ae1 and the lower has an effective area Ae2, the bearing has an area ratio E ¼ Ae1/Ae2. The result of the difference between the top and bottom areas is that the bearing carries a load at the design operating gap where X ¼ 1. Another consequence is that the maximum positive bearing thrust Wpositive in the upward direction is larger than the maximum negative bearing thrust Wnegative in the downward direction. The bearing must therefore be designed so that the range of applied loads lie within the acceptable operating range. The charts in Figure 7.13 provide an approximate guide to an acceptable load range. Charts are given for bearing thrust with gap variation for both capillary and orifice control in Figure 7.14. The area ratios cover a range from E ¼ 1 to 2.5, which should be adequate for most purposes. The charts are all computed for a value of the design pressure ratio b ¼ Pro/Ps ¼ 0.5. The load and stiffness relationships are the same as previous cases for equal opposed pads: W ¼ Ps Ae W

where Ae ¼ Ae1 þ Ae2

(7.9)

Plane Hydrostatic and Aerostatic Bearings 147 and l¼

Ps Ae l ho

where Ae ¼ Ae1 þ Ae2

(7.10)

Charts of stiffness data for capillary and orifice control are given in Figure 7.15. Flow at the design condition is the sum of the flows for the two pads: qo ¼

Ps h3o ðbB1 þ bB2 Þ h

(7.11)

It is assumed that the proportions of the two pads will be the same, only the size will be different, so that pad shape factors A and B are identical for the two pads. Flow data may therefore be read from Figure 7.11, which gives flow charts for equal opposed pads. Pad size and area ratio have no effect on flow data as long as the pad shapes are identical on each side of the bearing. The shape of the characteristic curves for unequal opposed pads depends on the ratio of the areas of the two opposing pads and is termed the area ratio E: E¼

Ae1 Ae2

(7.12)

A design procedure is given below, which is intended to be self-explanatory as far as possible, and an example is given to illustrate how the procedure can be used. The operating load range of unequal opposed-pad bearings is indicated in Figure 7.13 for various area ratios. It was assumed that it would be undesirable to operate with stiffness lower than approximately half the stiffness value at the design condition. This condition ensures a small margin for overload before bearing collapse occurs. The information in Figure 7.13 is presented in a form convenient for use in the design procedure. Area ratio E is indicated for a given ratio of maximum positive load to maximum negative load. The charts are all computed for design pressure ratio b ¼ 0.5. Other values of design pressure ratio are not considered to offer significant advantage.

7.9 Complex Arrangements of Pads (Capillary Controlled) Complex bearing arrangements can be analyzed without too much difficulty for a four-pad problem such as the example in Figure 7.16. Analysis can be carried out for even more pads if there is symmetry or if some pads are arranged perpendicular to others.

148

Chapter 7

Figure 7.16: Bearing Coordinates for a Complex Arrangement of Pads.

Given a set of bearing deflections or gaps, it is a simple matter to find the bearing thrusts. In practical problems, it is more often the other way round. One has to find the deflections from a known set of applied loads. Unfortunately, there is no simple method. The designer usually overcomes this problem by using the following approximate analysis for deflections and ensuring a margin of safety. The procedure is for capillary compensation and will ensure a safe design even for other types of compensation, since load capacity will not be greatly altered and operating stiffness will be improved.

Procedure The procedure basically involves two stages: 1. Calculate pad thrusts at the design condition to ensure a suitable design pressure ratio b and effective pad area Ae for each pad. 2. Calculate approximate dimensionless deflections for the maximum operating loads using the design stiffness values. The dimensionless deflections should represent gap changes that are safe. For example, dimensionless gap should lie within an approximate range between 0.5 and 1.5. If the dimensionless deflections are excessive, the alternatives are to choose a new Ae and a new b or to rearrange the bearing layout and repeat steps 1 and 2. In Figure 7.16, the pad thrusts at the design condition are shown as W1, W2, W3, and W4, where W1 ¼ Ps Ae1 b1 W2 ¼ Ps Ae2 b2 ; etc: We assume that loads Fx, Fy, and Fz applied on the bearing are resolved into three coordinates centered at some point O and act in a two-dimensional plane. For example, a gravity load G acting at a radius RG creates a moment Fz ¼ G  RG. Of course, the applied loads will

Plane Hydrostatic and Aerostatic Bearings 149 usually act in three dimensions and involve six coordinates. The analysis can be extended to the more complex situation using the same notation. Applied loads must balance pad thrusts, so that Fx ¼ W1 cosqx1 þ W2 cosqx2 þ W3 cosqx3 þ W4 cosqx4 Fy ¼ W1 cosqy1 þ W2 cosqy2 þ W3 cosqy3 þ W4 cosqy4

(7.13)

Fz ¼ W1 R1 þ W2 R2 þ W3 R3 þ W4 R4 Equation (7.13) represents three equations with four unknown Ae values and four unknown b values. It is therefore necessary to choose or guess five of these values arbitrarily. For example, choosing Ae1, Ae2, Ae3, Ae4 and one b value, it is then possible to solve for the remaining b values. If the b values do not lie in the range from 0.2 to 0.75, some rearrangement of the areas is required. To analyze deflections: deflection ¼

change in load average stiffness

W  Wo ho  h ¼ lo

(7.14) for small deflections

This may be stated in dimensionless form as 1X ¼

W  Wo lo

(7.15)

so that W  Wo ¼ Ps Ae lo ð1  XÞ

(7.16)

Equation (7.13) can be rewritten for maximum loading to yield approximate dimensionless deflections (a prime symbolizes maximum loading): Fx0  Fx ¼ ðW10  W1 Þcosqx1 þ ðW20  W2 Þcosqx2 þ ðW30  W3 Þcosqx3 þðW40  W4 Þcosqx4 Fy0  Fy ¼ ðW10  W1 Þcosqy1 þ ðW20  W2 Þcosqy2 þ ðW30  W3 Þcosqy3 þðW40  W4 Þcosqy4 Fz0  Fz ¼ ðW10  W1 ÞR1 þ ðW20  W2 ÞR2 þ ðW30  W3 ÞR3 þ ðW40  W4 ÞR4

(7.17)

150

Chapter 7

Substituting in equations (7.17) from equation (7.16) yields three equations in four unknown values of ð1  XÞ. Reducing the number of unknowns is often facilitated by symmetry in the bearing layout, as in the following design example. However, for the general case, the number of unknowns is reduced to three by solving in terms of the three coordinate movements: 1  X1 ¼

x y zR1 þ þ ho1 cosqx1 ho1 cosqy1 ho1

x y zR2 1  X2 ¼ þ þ ho2 cosqx2 ho2 cosqy2 ho2

(7.18)

and so on. It is preferable that each bearing should preload another, so that a larger bearing area can be employed and deflections will be small. In Example 7.5, bearing pad 1 is not preloaded and the load range is limited. In many cases, for economic or other reasons, it is not possible to employ pads in opposition. Under such circumstances, the following rules tend to give confidence.

Choice of Design Pressure Ratio Pro /Ps for Complex Pad Arrangements The optimum is b ¼ Kgo ¼ 0.5 but where the greatest load-bearing capacity is required without serious loss of stiffness, the following values are chosen: 1. For bearings where the range of applied loads is small, choose b ¼ 0.5. 2. For heavily loaded pads that operate at clearances less than the design value, choose b in the range 0.2e0.4. 3. For lightly loaded pads that operate at clearances greater than the design value, choose b in the range 0.7e0.75.

n

Example 7.5 A Complex Hydrostatic Pad Arrangement Choose suitable values of the effective area and the pressure ratio for each pad. For the complex pad arrangement shown in Figure 7.17, the supply pressure is 3.5 MN/m2 (508 lbf/in2). The loading conditions are: 1. Light loading condition: Fy ¼ 5:5 kN ð1012 lbf Þ Fz ¼ 340 Nm ð3009 lbf inÞ

Plane Hydrostatic and Aerostatic Bearings 151 z

2

x

1 50 mm

O

y

4

5

3 150 mm

50 mm

Figure 7.17: Design Example: A Complex Pad Arrangement.

2. Extreme loading condition: Fx0 ¼ 5 kN ð1124 lbf Þ Fy0 ¼ 12 kN ð2698 lbf Þ Fz0 ¼ 900 Nm ð7966 lbf inÞ Solution

Select initial pad areas that appear reasonable in the light of the extreme loading condition. In this case pad 1 needs to be large enough to support half the y-loading and a substantial proportion of the z-loading. Initial consideration suggests the following pad areas: Ae1 ¼ 8000 mm2 ð12:4 in2 Þ Ae2 ¼ Ae3 ¼ 2Ae1 ¼ 16; 000 mm2 ð18:6 in2 Þ Ae4 ¼ Ae5 ¼ Ae1 ¼ 8000 mm2 ð12:4 in2 Þ Using the initial selection of areas, the light and extreme loading situations must now be analyzed and solved to find appropriate values of design pressure ratio. It is possible that the pad areas may need to be adjusted and the procedure repeated until a suitable combination is found. Analyzing for light loads as a design condition x direction:

Fx ¼ W4  W5 0 ¼ 3:5  106  8  103 ðb4  b5 Þ b4 ¼ b5 and choose a value of 0:5 y direction:

Fy ¼ W1 cos y1 þ W2 cos y2 þ W3 cos y3 ¼ W1 þ W2  W3 where Wn ¼ PsAenbn. Inserting values for the applied load Fy and the pad thrusts W1, W2, and W3 yields

152

Chapter 7 5:5  103 ¼ 3:5  106  8  103 ðb1 þ 2b2  2b3 Þ b1 þ 2b2  2b3 ¼ 0:1964

z direction:

Fz ¼ 0:05W1  0:15W2 þ 0:15W3  0:05W4 þ 0:05W5 But W4 ¼ W5 since Fx ¼ 0, therefore 340 ¼ 3:5  106  8  103 ð0:05b1  2  0:15  b2 þ 2  0:15  b3 Þ b1  6b2 þ 6b3 ¼ 0:2429 Since there are now two equations in three unknown b values, one value must be chosen arbitrarily, say b2 ¼ 0.5. Solving for the x and y directions simultaneously: b3 ¼ 0:506; say 0:51 b1 ¼ 0:208; say 0:21 Analysis of approximate deflections from equations (7.16) and (7.17) and Figure 7.2 x direction:

5  103 ¼ 3:5  106  8  103 ½0:75ð1  X 4 Þ  0:75ð1  X 5 Þ ð1  X 5 Þ ¼ ð1  X 4 Þ 1  X 4 ¼ 0:119 y direction:

6:5  103 ¼ 3:5  106  8  103 ½0:5ð1  X 1 Þ þ 2  0:75ð1  X 2 Þ  2  0:75ð1  X 3 Þ Since (1  X 3 ) ¼ (1  X 2 ), the equation simplifies to 0:232 ¼ 0:5ð1  X 1 Þ þ 3ð1  X 2 Þ z direction:

560 ¼ 0:05ðW10  W1 Þ  0:15ðW20  W2 Þ þ 0:15ðW30  W3 Þ 0:05ðW40  W4 Þ þ 0:05ðW50  W5 Þ 560 ¼ 3:5  106  8  103 ½0:05  0:5ð1  X 1 Þ  2  0:15  0:75ð1  X 2 Þ þ2  0:15  0:75ð1  X 3 Þ  0:05  0:75ð1  X 4 Þ þ 0:05  0:75ð1  X 5 Þ 0:4 ¼ 0:5ð1  X 1 Þ  6  0:75ð1  X 2 Þ þ 6  0:75ð1  X 3 Þ 0:75ð1  X 4 Þ þ 0:75ð1  X 5 Þ 1  X 4 ¼ 0:119 from above 0:5785 ¼ 0:5ð1  X 1 Þ  9ð1  X 2 Þ

Plane Hydrostatic and Aerostatic Bearings 153 Solving simultaneously from the y and z directions yields the summary of bearing gap values at the extreme load condition: ð1  X 2 Þ ¼ 0:0289 ð1  X 1 Þ ¼ 0:2907 Summarizing the resulting values of b and approximate film thickness X at extreme loading: b1 b2 b3 b4 b5

¼ 0:21 ¼ 0:5 ¼ 0:51 ¼ 0:5 ¼ 0:5

X1 X2 X3 X4 X5

¼ 0:709 ¼ 1:0289 ¼ 0:9711 ¼ 1:119 ¼ 0:881

Inspection of Figures 7.1 and 7.2 reveals that the design is completely satisfactory. For example, in the case of the first pad, the approximated gap is reduced to X 1 ¼ 0.709. The stiffness will actually be increased at this gap and so the deflection is overestimated. The gap values for the other pads are all sufficiently close to the design values to avoid any concern. n

154

Chapter 7

Appendix: Tabular Design Procedures Procedure 7.A1 Single Hydrostatic Pad

Step

Symbol

Description of Operation

Notes

1 2 3 4 5 6 7 8 9 10 11 12 13 14

W W0 Ps b

From design requirements From design requirements Suggestion: 2 MN/m2 (291 lbf/in2) Suggestion: 0.5 or see Chapter 7 Choose optimum values from Chapter 4

ho U

Specify: normal load Specify: extreme load Specify: supply pressure Specify: pressure ratio, Pro/Ps Specify: land-width ratios Specify: land-width ratios Read: area factor Read: flow factor Calculate: pad area A ¼ W =ðPs bAÞ Calculate: first pad dimension Calculate: land width Calculate: second pad dimension Specify: bearing gap Specify: sliding speed

15 16

Ar Af

Calculate: total recess area Calculate: friction area

17

h

Specify or calculate: viscosity

18 19 20 21 22 23 24 25 26 27

qo l l 0 W X h0 Hp Hf K DT

Calculate: flow Read: stiffness factor Calculate: stiffness Calculate: extreme load factor Read: extreme gap factor Calculate: extreme gap Calculate: pumping power Calculate: friction power Calculate: power ratio Calculate: max temp rise

A B A a

See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type If projected area A is too large, increase Ps From A calculate dimension From land-width ratio From A calculate dimension More than 5e10 times flatness tolerance If U ¼ 0 or is very small, go to step 17 and specify h From recess dimensions From Af ¼ A  3/4Ar sffiffiffiffiffiffi Ps h2o bB From h ¼ U Af qo ¼ Ps h3o bB=h From Figures 7.1e7.9 From Ps AA l=ho 0 From W ¼ W 0 =ðPs AAÞ 0 From Figures 7.1e7.8 using value of W From h0 ¼ ho X From Hp ¼ Psqo From Hf ¼ hAf U2/ho From K ¼ Hf/Hp SI units: DT ¼ 0.6  106Ps(1 þ K)  C

Plane Hydrostatic and Aerostatic Bearings 155

Example 7.A1 Single Circular Hydrostatic Pad, Capillary Control, Speed U [ 0.5 m/s, Normal Load 5000 N, Maximum Load 7000 N; Refer to Chapter 4 for Circular Pad Factors and This Chapter for Stiffness Factor

Step

Symbol

Example of Working

Result

1 2 3 4 5 6 7 8

W W0 Ps b a/D

Normal load Extreme load Supply pressure Pressure ratio Land-width ratio

5000 N (1124 lbf) 7000 N (1574 lbf) 2 MN/m2 (291 lbf/in2) 0.5 0.25

A B

Area factor Flow factor

0.541 0.755

9

A

10 11 12 13 14 15 16

D ¼ 2R2 a R1 ho U Ar Af

0.009259 m2 (14.35 in2) 0.1086 m (4.274 in) 0.0271 m (1.069 in) 0.0271 m (1.069 in) 50 mm (0.002 in) 0.5 m/s (19.68 in/s) 0.002307 m2 (3.576 in2) 0.007529 m2 (11.67 in2)

17

h

5000 2  106  0:5  0:54 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  0:009259=p 0.1086  0.25 0.1086/2  0.0271 Bearing gap Sliding speed p  0.02712 0.009259  0.75  0.002307 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  106  502  1012 0:5  0:755 0:5 0:007529

18 19 20 21 22 23 24 25 26 27

qo l l 0 W 0 X h0 Hp Hf K DT

2  106  503  1018  0.5  0.755/0.0708 For value X ¼ 1 2  106  0.009259  0.541  0.75/50  106 7000/(2  106  0.009259 0.541) 0 For W ¼ 0.6987 50  106  0.75 2  106  1.33  106 0.0708  0.007529  0.52/50  106 2.66/2.66 6 0.6  10  (1 þ 1)  2  106

0.0708 N s/m2 (70.8 cP) 1.33  106 m3/s (0.08 l/min) 0.75 150 MN/m (0.86  106 lbf/in) 0.6987 0.75 37.5 mm (0.0015 in) 2.66 W 2.66 W 1.0 2.4  C

156

Chapter 7

Procedure 7.A2 Single Aerostatic Pad with Virtual Recess Step

Symbol

Description of Operation

Notes

1 2 3 4 5

W W0 Ps Kgo pro

Specify: normal load  1.15 Specify: extreme load  1.15 Specify: gauge supply pressure Specify pressure ratio Pro/Ps Calculate: absolute entry pressure Specify: land-width ratio Read or calculate Read or calculate Calculate: A ¼ W =ðPs bAÞ Calculate: first pad dimension

From design requirements From design requirements Suggestion: 0.45 MN/m2 (65 lbf/in2) Suggestion: 0.5 or see Chapter 7 pro ¼ KgoPs þ pa

6 7 8 9 10

A B A

11

Calculate: virtual recess sizes

12

Cvr

13 14

n dr

15

ho

Calculate: virtual recess perimeter Sources per virtual recess Specify: entry recess or slot width Specify: bearing gap

16

h

Specify: viscosity for gas

17

qo

18 19 20 21 22 23 24

l l 0 W X h0 Hp Hf

Calculate: free air flow with 70% virtual recess surround Read: Calculate: Calculate: extreme load factor Read: displacement factor Calculate: extreme gap Calculate: pumping power Calculate: friction power

Choose optimum values from Chapter 4 See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type If projected area A is too large, increase Ps From A calculate pad dimension For example: Dr ¼ (1  2a/D)D for circular pad From step 11

Suggestion: dr  0.7Cvr/n ho should be 5e10 times flatness tolerance For air h ¼ 0.0000183 N s/m2 at 18  C 
Ps h3 Kgo B pro þ pa qo z0:85 o 2pa h From Figures 7.1e7.9 From Ps AA l=ho 0 W ¼ W 0 =ðPs AAÞ 0 Refer Figures 7.1e7.8 for W From h0 ¼ ho =X From Hp ¼ paqaln(ps/pa) From Hf ¼ hAU2/ho

Plane Hydrostatic and Aerostatic Bearings 157

Example 7.A2 Single Circular Aerostatic Pad, Orifice Control, Speed U [ 0.5 m/s, Normal Load 5000 N, Maximum Load 7000 N

Step Symbol

Example of Working

Result 5750 N (1293 lbf) 8050 N (1810 lbf) 0.45 MN/m2 (65 lbf/in2) 0.5 0.326 MN/m2 0.25 0.541 0.755

1 2 3

W W0 Ps

Normal load  1.15 Extreme load  1.15 Supply pressure

4 5 6 7 8

Kgo pro a/D A B

Pro/Ps (0.5  0.45 þ 0.101)  106 Land-width ratio Area factor Flow factor

9

A

10 11 12 13 14 15 16

D Dr Cvr n dr ho h

17

qo

18

l

19 20 21 22 23 24

l

5750 0:45  106  0:5  0:541 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  0:04724=p (1  2  0.25)  0.2453 p  0.1227

0.04724 m2 (73.22 in2) 0.2453 m (9.66 in) 0.1227 m (4.83 in) 0.3855 m (15.18 in) 8 0.03373 m (1.33 in) slot width 25 mm (0.001 in) 18.3  106 N s/m2

dr  0.7  0.3855/8 Bearing gap Viscosity of air 0:85  0:45  106  253  1018  0:5  0:755 18:3  106 0:326 þ 0:101  2  0:101 For X ¼ 1 6

0

W X h0 Hp Hf

0.0002606 m3/s (15.6 l/min)

0.65 6

0.45  10  0.04724  0.541  0.65/25  10

8050/(0.45  106  0.0411  0.541) 0 For W ¼ 0.8045 25  106  0.5 6 0.101  10  0.0002606  ln (0.551/0.101) 0.0000183  0.04724  0.52/25  106

299 MN/m (1.71  106 lbf/in) 0.8045 0.5 12.5 mm (0.0005 in) 44.7 W 0.0088 W

158

Chapter 7

Procedure 7.A3 Equal Opposed Hydrostatic Pads

Step

Symbol

Description of Operation

Notes

1 2 3

W0 Ps b

From design requirements Suggestion: 2 MN/m2 (291 lbf/in2) Suggestion: 0.5 or see Chapter 7

11 12

a

13

ho

Specify: extreme load Specify: supply pressure Specify: pressure ratio, Pro/Ps Specify: land-width ratios Specify: land-width ratios Read: area factor Read: flow factor Read: load factor Calculate: pad area A ¼ W =Ps A W Calculate: first pad dimension Calculate: land width Calculate: second pad dimension Specify: bearing gap

14

U

Specify: sliding speed

15

Ar

16

Af

Calculate: total recess area Calculate: total friction area

17

h

18

qo

19 20 21

l l Hp

22 23 24

4 5 6 7 8 9

A B W A

10

Specify or calculate: viscosity Calculate: flow

Choose optimum values from Chapter 4 See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type See Chapter 7 for acceptable value If projected area A is too large, increase Ps From A calculate dimension From land-width ratio From A calculate dimension ho should be at least 5e10 times flatness tolerance If U ¼ 0 or is very small, go to step 17 and specify h From recess dimensions From Af ¼ A  3/4Ar sffiffiffiffiffiffiffiffi Ps h2o 2bB From h ¼ U Af qo ¼ 2Ps h3o bB=h From Figures 7.1e7.9 From Ps AA l=ho From Hp ¼ Psqo for two pads

Hf K

Read: stiffness factor Calculate: stiffness Calculate: pumping power Calculate: friction power Calculate: power ratio

DT

Calculate: max temp rise

SI units: DT ¼ 0.6  106Ps(1 þ K)  C

From Hf ¼ hAfU2/ho From K ¼ Hf/Hp

Plane Hydrostatic and Aerostatic Bearings 159

Example 7.A3 Square, Opposed Plane Hydrostatic Pads, Capillary Control, Sliding Speed 0.5 m/s, Maximum Applied Load 7000 N

Step

Symbol

Example of Working

Result

1 2 3 4 5 6 7 8

W0 Ps b a/L

Extreme load Supply pressure Pressure ratio Land-width ratio

7000 N (1574 lbf) 2 MN/m2 (291 lbf/in2) 0.5 0.25

A B W

Area factor Flow factor Load factor from Figure 7.10

0.525 0.85 0.4

9

A

10 11 12 13 14 15 16

L a

7000 2  106  0:5  0:525  0:4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:03333=2 0.1291  0.25

0.03333 m2 (51.66 in2) 0.1291 m (5.083 in) 0.03225 m (1.269 in)

ho U Ar Af

Bearing gap Sliding speed 2  (0.1291  2  0.3225)2 0.03333  0.75  0.008346

25 mm (0.001 in) 0.5 m/s (19.68 in/s) 0.008346 m2 (12.94 in2) 0.02707 m2 (41.96 in2)

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  0:5  0:85 0:02707

17

h

2  106  252  1012 0:5

18

qo

2  2  106  253  1018  0:5  0:85 0:01401

19

l

For value X ¼ 1

20

l

2  106  0:01666  0:525  0:75 25  106

21 22 23 24

Hp Hf K DT

2  106  2.681  106 0.01401  0.02707  0.52/25  106 3.792/3.792 0.6  106  (1 þ 1)  2  106

0.01401 N s/m2 (14.01 cP) 1.896  106 m3/s (0.114 l/min) 0.75 525 MN/m (3  106 lbf/in) 3.792 W 3.792 W 1.0 2.4  C

160

Chapter 7

Procedure 7.A4 Equal Opposed Aerostatic Pads

Step

Symbol

Description of Operation

Notes

1 2

W0 Ps

Specify: extreme load  1.15 Specify: supply pressure

3 4 5

Kgo pro

Specify: pressure ratio, Pro/Ps Calculate: absolute entry pressure Specify: land-width ratios

From design requirements Suggestion: Ps ¼ 0.45 MN/m2 (65 lbf/in2) Suggestion: 0.5 or see Chapter 7 From KgoPs þ pa Choose optimum values from Chapter 4

6 7 8 9 10

A B W A

Specify: land-width ratios Read: area factor Read: flow factor Read: load factor Calculate: pad area A ¼ W=Ps A W

11 12 13

a Cvr

14 15

n dr

16

ho

Calculate: first pad dimension Calculate: land width Calculate: virtual recess perimeter Specify: number of sources or slots Specify: entry recess or slot width Specify: bearing gap

17

h

Specify: viscosity of gas

18

qo

19 20 21

l l Hp

Calculate: flow with virtual recess Read: stiffness factor Calculate: stiffness Calculate: pumping power

22

Hf

Calculate: friction power

See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type See charts for acceptable value If projected area A is too large, increase Ps From A calculate dimension From land-width ratio From A calculate dimension Suggestion: equi-spaced Suggestion: dr  0.7Cvr/n ho should be at least 5e10 times flatness tolerance For air h ¼ 0.0000183 N s/m2 at 18  C qo z0:85 

2Ps h3o Kgo B pro þ pa  h 2pa

From Figures 7.1e7.9 From Ps AA l=ho From Hp ¼ paqaln(ps/pa) for two pads From Hf ¼ hAfU2/ho

Plane Hydrostatic and Aerostatic Bearings 161

Example 7.A4 Square, Opposed Aerostatic Pads, Orifice Control, Sliding Speed 0.5 m/s, Maximum Applied Load 7000 N

Step

Symbol

Example of Working

Result

1 2

W0 Ps, ps

7000  1.15 Gauge and absolute supply pressure

3 4 5 6 7 8 9

Kgo pro a/L

Pressure ratio (0.5  0.45 þ 0.101)  106 Land-width ratio

8050 N (1810 lbf) Ps ¼ 0.45 MN/m2 ps ¼ 0.551 MN/m2 0.5 0.326 MN/m2 0.25

A B W

Area factor Flow factor Load factor from Figure 7.12

0.525 0.85 0.4

10

A

11 12 13 14 15

L a Cvr n dr

8050 2  106  0:5  0:525  0:4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:03833=2 0.1384  0.25 (0.1384  2  0.0346)  4 Multiple of 4 for a square pad dr  0.7  0.2768/8

16 17

ho h

Bearing gap Viscosity of air

0.03833 m2 (59.42 in2) 0.1384 m (5.45 in) 0.0346 m (1.362 in) 0.2768 m (10.9 in) 8 0.02422 m (0.9535 in) slot width 25 mm (0.001 in) 18.3  106 N s/m2

18

qo

19

l

20

l

21 22

Hp Hf

1:7  0:45  106  253  1018  0:5  0:85 0:0000183 0:326 þ 0:101 2  0:101 For value X ¼ 1 0:45

 106

 0:03833  0:525  1:0 25  106

0.101  106  587  106  ln(0.551/0101) 18.3  106  0.03833  0.52/25  106

587  106 m3/s (35.2 l/min)

1.0 362 MN/m (2.07  106 lbf/in) 101 W 0.007 W

162

Chapter 7

Procedure 7.A5 Unequal Opposed Hydrostatic Pads

Step

Symbol

Description of Operation

Notes

1 2 3 4 5 6 7 8 9 10 11

Wpos Wneg E Ps

From design requirements From design requirements Area ratio E ¼ Ae1/Ae2 Suggestion: 2 MN/m2 (291 lbf/in2) Choose optimum values from Chapter 4

A B W pos W neg Amin

12

Amin

13 14 15 16 17 18 19 20 21 22 23 24 25

Aactual A1 A2

ho U Ar Af

Specify: maximum positive load Specify: maximum negative load Read from Figure 7.13 Specify: supply pressure Specify: land-width ratios Specify: land-width ratios Read: area factor Read: flow factor Read: load factor Read: load factor Calculate: pad area A ¼ Wpos =Ps A W Calculate: pad area A ¼ Wneg =Ps A W Specify: convenient value Calculate Calculate Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Specify: bearing gap (single) Specify: sliding speed Calculate: total recess area Calculate: total friction area

26

h

Specify or calculate: viscosity

27 28 29 30 31 32 33

qo l l Hp Hf K DT

Calculate: flow Read: stiffness factor Calculate: stiffness Calculate: pumping power Calculate: friction power Calculate: power ratio Calculate: max temp rise per pass

See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type See Figure 7.14 See Figure 7.14 Amin ¼ A1 þ A2

A1 ¼ A $ E/(1 þ E) A2 ¼ A/(1 þ E)

ho should be at least 5e10 times flatness tolerance If U ¼ 0 or is very small, go to step 25 and specify h Ar ¼ Ar1 þAr2 From Af ¼ A  3/4Ar sffiffiffiffiffiffiffiffi Ps h2o 2bB From h ¼ U Af qo ¼ 2Ps h3o bB=h From Figure 7.15 From Ps AAl=ho From Hp ¼ Psqo for two pads From Hf ¼ hAfU2/ho From K ¼ Hf/Hp SI units: DT ¼ 0.6  106Ps(1 þ K)  C

Plane Hydrostatic and Aerostatic Bearings 163

Example 7.A5 Unequal Opposed Hydrostatic Pads: Capillary Control, Sliding Speed 1.27 m/s, Maximum Load 8900 N, Design Pressure Ratio 0.5, Rectangular Pads 2:1 Length/Width Step

Symbol

Example of Working

Result

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Wpos Wneg E Ps a/L

Maximum positive load Maximum negative load Approximate value of 2 Supply pressure Land-width ratio

8900 N (2000 lbf) 2450 N (550 lbf) 2 2 2 MN/m (291 lbf/in2) 0.25

A B W pos W neg Amin Amin Aactual A1 A2 L1 B1 a1 L2 B2 a2 ho U Ar Af

Area factor Flow factor Load factor from Figure 7.14 Load factor from Figure 7.14 8900/(2  106  0.63  0.5) 2450/(2  106  0.63  0.14) Convenient value 0.0144  2/3 0.0144/3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0096=2 2  0.0693 0.0693  0.25 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:0048=2 2  0.049 0.049  0.25 Bearing gap Sliding speed 0.00361 þ 0.018 0.0096 þ 0.0048  0.75  0.0054

0.63 1.45 0.5 0.14 0.0141 m2 (21.9 in2) 0.014 m2 (21.7 in2) 0.0144 m2 (22.32 in2) 0.0096 m2 (14.88 in2) 0.0048 m2 (7.44 in2) 0.0693 m (2.73 in) 0.1386 m (5.46 in) 0.0173 m (0.6732 in) 0.049 m (1.93 in) 0.098 m (3.86 in) 0.01225 m (0.482 in) 50 mm (0.002 in) 1.27 m/s (50 in/s) 0.0054 m2 (8.4 in2) 0.0103 m2 (16.04 in2)

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  0:5  1:45 0:0103

26

h

2  106  502  1012 1:27

27

qo

2  2  106  503  1018  0:5  1:45 0:047

28 29

l l

For value X ¼ 1 2  106  0.0144  0.63  0.75/50  106

30 31 32 33

Hp Hf K DT

2  106  7.71  106 0.047  0.0103  1.272/50  106 15.4/15.6 0.6 106  2  106  (1 þ 1)

0.047 N s/m2 (47 cP) 7.71  106 m3/s (0.463 l/min) 0.75 272 MN/m (1.55  106 lbf/in) 15.4 W 15.6 W 1.0 2.4  C

164

Chapter 7

Procedure 7.A6 Unequal Opposed Aerostatic Pads

Step

Symbol

Description of Operation

Notes

1 2 3 4 5 6 7

Wpos Wneg E Ps Kgo pro

Specify: maximum positive load Specify: maximum negative load Read from Figure 7.13 Specify: gauge supply pressure Specify: pressure ratio, Pro/Ps Calculate: absolute entry pressure Specify: land-width ratios

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

A B W pos W neg Amin Amin Aactual A1 A2

23 24 25

n dr ho

Read: area factor Read: flow factor Read: load factor Read: load factor Calculate: pad area A ¼ Wpos =Ps A W Calculate: pad area A ¼ Wneg =Ps A W Specify: convenient value Calculate Calculate Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: pad dimensions Calculate: virtual recess perimeter Specify: number of entry sources Specify: source width Specify: bearing gap

From design requirements From design requirements Area ratio E ¼ Ae1/Ae2 Suggestion: Ps ¼ 0.45 MN/m2 Suggestion: 0.5 From KgoPs þ pa Choose optimum values from Chapter 4 See Chapter 4 for chosen pad type See Chapter 4 for chosen pad type See Figure 7.14 See Figure 7.14 Amin ¼ A1 þ A2

26

h

Specify: viscosity of gas

27

qo

Calculate: flow for virtual recess

28

l

Read: stiffness factor

From Figures 7.1e7.9

29

l

Calculate: stiffness

From Ps AA l=ho

30 31

Hp Hf

Calculate: pumping power Calculate: friction power

From Hp ¼ paqaln(ps/pa) for two pads From Hf ¼ hAfU2/ho

Cvr

A1 ¼ A $ E/(1 þ E) A2 ¼ A/(1 þ E)

Also calculate for small virtual recess Suggestion: equi-spaced Suggestion: dr  0.7Cvr/n ho should be at least 5e10 times flatness tolerance For air h ¼ 0.0000183 N s/m2 at 18  C qo z0:85 

2Ps h3o Kgo B pro þ pa  h 2pa

Plane Hydrostatic and Aerostatic Bearings 165

Example 7.A6 Square, Unequal Opposed Aerostatic Pads, Orifice Controlled, Sliding Speed 0.5 m/s, Maximum Applied Load 4500 N, Ambient Pressure 0.101 MN/m2 Step Symbol

Example of Working

Result

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Wpos Wneg E Ps Kgo pro a/L A B W pos W neg Amin Amin Aactual A1 A2 L1 a1 L2 a2

Maximum positive load  1.15 Maximum negative load  1.15 Approximate value of 2 Gauge supply pressure Gauge pressure ratio (0.5  0.45 þ 0.101)  106 Land-width ratio Area factor Flow factor Load factor from Figure 7.14 Load factor from Figure 7.14 5175/(0.45  106  0.525  0.5) 1450/(0.45  106  0.525  0.14) Convenient value 0.044  2/3 0.044/3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:02933 0.1713  0.25 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:01467 0.1211  0.25

5175 N (1163 lbf) 1450 N (346 lbf) 2 Ps ¼ 0.45 MN/m2 0.5 0.326 MN/m2 0.25 0.525 0.85 0.5 0.14 0.0438 m2 (67.9 in2) 0.0438 m2 (67.9 in2) 0.044 m2 (68.2 in2) 0.02933 m2 (45.5 in2) 0.01467 m2 (22.73 in2) 0.1713 m (6.743in) 0.0428 m (1.686 in) 0.1211 m (4.768 in) 0.03028 m (1.192 in)

Cvr n dr ho h

(0.1713  2  0.0428)  4 Multiple of 4 for square pad dr  0.7  0.3428/8 Bearing gap Viscosity of air

0.3426 m (13.66 in) and 0.2422 m 8 0.03 m (1.18 in) and 0.0212 m (0.835 in) 25 mm (0.001 in) 18.3  106 N s/m2

27

qo

1:7  0:45  106  253  1018  0:5  0:85 0:0000183 0:326 þ 0:101 2  0:101

587  106 m3/s (35.22 l/min)

28 29

l

For value X ¼ 1 0.45  106  0.044  0.525  1.0/25  106

30 31

Hp Hf

0.101  587  ln (0.551/0.101) 18.3  106  0.044  0.52/25  106

1.0 416 MN/m (2.38  106 lbf/in) 101 W 0.0081 W