Journal Bearing

(1)

KELOMPOK 2

Darma

Darma Adhi

Adhi W.

W. (11210009)

(11210009)

Galuh

Galuh Intan

Intan P.

P. (11210012)

(11210012)

Bhatara

Bhatara Putra

Putra M.

M. (11210014)

(11210014)

Mulyani

(11210020)

Mulyani

(11210020)

Muliyani

(11210023)

Muliyani

(11210023)

Ali

Ali Akbar

Akbar

(11210024)

Tia

Tia Utari

Utari

(11210028)

Eko

(2)

Bhatara Putra Mediriyanto

Konsep Dasar

Apa itu Bearing?

(3)

Tipe beban pada bearing

1. Bantalan radial

1. Bantalan radial (Journal Bearing) (Journal Bearing)

Arah beban yang di tumpu bantalan ini Arah beban yang di tumpu bantalan ini adalah tegak lurus dengan sumbu poros adalah tegak lurus dengan sumbu poros

2. Bantalan aksial

2. Bantalan aksial (Thrust Bearing) (Thrust Bearing)

Arah beban bantalan ini adalah sejajar Arah beban bantalan ini adalah sejajar

dengan sumbu poros dengan sumbu poros

3. Bantalan kombinasi

3. Bantalan kombinasi (combination bearing) (combination bearing)

Bantalan ini menumpu beban yang Bantalan ini menumpu beban yang arahnya sejajar dan tegak lurus dengan arahnya sejajar dan tegak lurus dengan sumbu poros

(4)

Introduction

•• _{Journal bearing termasuk salahsatu sliding}_{Journal bearing termasuk salahsatu sliding bearing dan keterbalikkan}_{bearing dan keterbalikkan}

dari ball bearing dari ball bearing

•• _{Journal bearing secara umum digunakan pada mesin}_{Journal bearing secara umum digunakan pada mesin piston kendaraan}_{piston kendaraan}

bermotor berbahan bakar bensin atau diesel bermotor berbahan bakar bensin atau diesel

•• _{Kelebihan :}_{Kelebihan :}

•• _{Bearing type ini mampu menopang shaft yang berat.}_{Bearing type ini mampu menopang shaft yang berat.} •• _{Awet dan tahan lama}_{Awet dan tahan lama}

•• _{Efek redaman dari film minyak membantu membuat mesin}_{Efek redaman dari film minyak membantu membuat mesin}

beroperasi dengan tenang dan halus. beroperasi dengan tenang dan halus.

•• _{Kekurangan :}_{Kekurangan :}

•• _{Membutuhkan suplai minyak pelumas yang besar}_{Membutuhkan suplai minyak pelumas yang besar}

•• _{Hanya cocok untuk temperatur dan kecepatan rendah}_{Hanya cocok untuk temperatur dan kecepatan rendah} •• _{Pembentukan lapisan minyak pelumas lambat}_{Pembentukan lapisan minyak pelumas lambat}

(5)

Bearing Diagram

Journal bearing

Journal bearing – –_{Berfungsi sebagai bantalan poros engkol yang berputar}_{Berfungsi sebagai bantalan poros engkol yang berputar}

Oil inlet

Oil inlet – –_T_{Tempat masuknya minyak}_{empat masuknya minyak pelumas}_pelumas

Ketika oli pelumas masuk ke dalam bearing, oli akan memenuhi Ketika oli pelumas masuk ke dalam bearing, oli akan memenuhi clearance/ gap antara shaft dan bearing sehinggga mengakibatkan clearance/ gap antara shaft dan bearing sehinggga mengakibatkan tekanan fuida meningkat dan daya angkat hidrodinamis terhadap shaft tekanan fuida meningkat dan daya angkat hidrodinamis terhadap shaft

(6)

(7)

Type

Type Typical Typical Loading Loading ApplicationApplication

(

(aa)) PPaarrttiiaal al arrcc Unidirectional Unidirectional load load Shaft Shaft guides, guides, dampersdampers

(

(aa)) CCiirrccuum fm feerreennttiiaall

groove, Axial groove

types

Variable

Variable load load direction direction Internal Internal combustion combustion enginesengines

(

(aa)) CCyylliinnddrriiccaall Medium to heavyMedium to heavy Unidirectional load Unidirectional load

General machinery General machinery

(

(aa)) PPrreessssuurre e ddaamm Light Light loads, loads, unidirectional unidirectional High High speed speed turbines, turbines, compressorcompressor

(

(aa)) OOvveerrsshhoott Light Light loads, loads, unidirectional unidirectional Steam Steam turbinesturbines

(

(aa)) MMuullttiilloobbee Light Light loads, loads, unidirectional unidirectional Gearing, Gearing, compressorcompressor

(

(aa)) PPrreellooaaddeedd Light Light loads, loads, unidirectional unidirectional Minimize Minimize vibrationvibration

(

(8)

Movement of the bearing

(9)

Video

(10)

Infinitely Long Approximation (ILA)

Menentukan jari-jari shaft dan

(11)

ILA

Menentuk

(12)

(13)

Boun

(14)

8.3 BOUNDARY CONDITIONS

Assumsi :

ṔṔ

= 0

θ θ

= 0

Ṕ Ṕ  −

− 





Dimana :

Ps = tekanan suplai

C

C =

= radial

radial clearence

clearence

R

R =

= radius

radius bearing

bearing



=

= viskos

viskositas

itas pelumas

pelumas



= kecepatan

putaran poros

(15)

8.4 FULL SOMMERFELD BOUNDARY

CONDITIONS

Asumsi :

ṔṔ

= 0

θθ

= 2

ππ

(360)

(16)

cos

cos +θ

_{11 ++ ɛ ɛ θθ}

+θ

Substitusi Sommerfeld :

8.4 FULL SOMMERFELD BOUNDARY

CONDITIONS

Tekanan puncak terjadi ketika

cos

cos

 −3

_(2+

−3

_

₎₎

Dimana :

cos

cos

=

= sudut

sudut angular

angular pada

pada

tekanan maksimum



=

= rasio

rasio eksentrisita

eksentrisitass



= =







= eksentrisitas

C

(17)

Dari persamaan 8.9 asumsi P = 0 pada

Ɵ  π

,,

besarnya tekanan

puncak tak berdimensi adalah :

8.4 FULL SOMMERFELD BOUNDARY

CONDITIONS

Ṕ Ṕ   3(4−

3(4−

_2(1−



)(4−5

_

₎₎

_

_(2+



++ 

_

₎₎



))

.





−3

22 ++ 



Yang terjadi pada :

Besarnya tekanan puncak tak berdimensi pada distribusi tekanan

adalah :

(18)

Load Carrying Based

on Full

(19)

Load Carrying Based

on Full

Sommerfe

ld Condition

ѾѾ   

        

_



Arah Radial

Ѿ

Ѿ     

_



   

Arah

Arah Tangensia

Tangensiall

(8.10)

(8.11)

Dari

Dari substitusi

substitusi tekanan

tekanan tak

tak berdimensi

berdimensi pada

pada persamaan

persamaan 8.9

8.9 dengan

dengan

persamaan 8.10 dan 8.11 maka

persamaan 8.10 dan 8.11 maka didapatkan :

didapatkan :

0

_Ѿ

12

(1−

(20)

Load Carrying Based

on Full

Sommerfe

ld Condition

Dimana Beban tak berdimensi:

(8.13)

Resultan dari

Ѿ

dan

Ѿ

(8.14)

ѾѾ  











Dengan



_











= beban yang diproyeksikan

Ns = kecepatan poros dalam rev/s

Ѿ Ѿ



_++Ѿ

_Ѿ



_



(

(21)

Load Carrying Based

on Full

Sommerfe

ld Condition

Attitude Angle

(8.15)

  ɸɸ  −Ѿ

−Ѿ

_Ѿ

_{Ѿ   ∞∞}

ɸɸ 



_

..

Dalam berbagai kasus, jika

Ɛ

_{= 0}

ѾѾ 

₀

Ɛ

(22)

8.5

8.5 DEFINITION OF THE

DEFINITION OF THE

SOMMERFELD NUMBER

(23)

Bilangan Sommerfeld (S) merupakan bilangan tak berdimensi yang

merupakan

parameter

parameter karakterisasi

karakterisasi

performansi

sebuah

sebuah bearing.

bearing.

Bilangan ini menunjukkan karakteristik gesekan total dari bantalan.

8.5

8.5 DEFINITION OF THE

DEFINITION OF THE

SOMMERFELD NUMBER

  



_



_



substitusikan ke dalam persamaan

Sommerfeld Number (8.14) maka :

ѾѾ  11

Dan penyelesaian S menjadi

  (1−

(1−



_12

))



(2+



))

(8.16)

(8.17)

(24)

8.6 HALF SOMMERFELD BOUNDARY

CONDITION

Ѿ

6

11 −− 



₎₎



_(2+



Ѿ

_{11 −− }

12

_

_)(2+



_

(8.18)

(8.19)

Total kapasitas beban dukung dan attitude angle adalah:

Ѿ

_{11 −− }

_

66

_)(2+

_





_{−− }



_((



_{−− 44))}

.

(8.20)

∅∅   



 



(25)

Eko Rusdiyanto

(26)

Contoh Soal 8.1

(27)

(28)

(29)

(30)

(31)

(32)

Fenomena Kavitasi

Kavitasi

Gaseous Cavitation Gaseous Cavitation Vapor Cavitation Vapor Cavitation

(33)

Gaseous cavitation merupakan kavitasi yang

disebabkan oleh adanya bagian dari minyak

pelumas yang terlarut dengan udara pada

kondisi jenuh (sekitar 10%), dan ketika tekanan

sekitar menjadi turun bagian yang terlarut ini

akan membentuk suatu kavitasi tetapi dibagian

yang berbeda dari fluid film, hal ini yang

menyebabkan kavitasi jenis gaseous tidak terlalu

berbahaya.

Gaseous

(34)

Vapor cavitation disebabkan oleh tingginya

fluktuasi tekanan yang ada diantara film dari

pelumas dan bearingnya itu sendiri, kavitasi

jenis

ini

cukup

berbahaya

berbahay

a

karena

karena bisa

bisa

menyebabkan kerusakan pada bearing (fatigue

damage)

V

(35)

SWIFT-STEIBER (REYNOLD)

SWIFT-STEIBER (REYNOLD) BOUNDAR

BOUNDARY

Y

CONDITION

Perhitung

Perhitungan beban

an beban bearing dengan

bearing dengan

memperhatikan kavit

memperhatikan kavitasi

asi didalam

didalam

perhitungannya





  0 

0  

   



(36)

(37)

(38)

Ali Akbar

INFINI

INFINITEL

TELY SHORT JOURNAL BEARI

Y SHORT JOURNAL BEARING

NG

APPROXIMATION (ISA)

(39)

A. Infinitely Short Journal Bearing

Approximation (ISA)

Integral 2 kali

Integral 2 kali Length-to-Diameter ratios up toLength-to-Diameter ratios up to

L/D = ½ dengan trends rata-rata L/D = ½ dengan trends rata-rata L/D = 1

(40)

Table Infinitely Long Journal Bearing Solutions with the Reynolds

Boundary Condition

(41)

(42)

B. Full and Half

(43)

Figure Short Bearing

(44)

Figure Short Bearing Attitude Angle vs Sommerfeld Number

(45)

(46)

Darma

(47)

FINITE BEARING DESIGN &

ANALYSIS

This section focused on design and

performance analysis based on the full

solution of Reynold equation.

(48)

FINITE BEARING DESIGN & ANALYSIS

(49)

FINITE BEARING DESIGN & ANALYSIS

•• _{Minimum film}_{Minimum film}

thickness thickness

h

h_min_min = C ( 1 = C ( 1– – Ɛ )Ɛ )

•• _{Friction force}_{Friction force}

F = f W F = f W

•• _{Power loss}_{Power loss}

E

E _p_p= F 2π R N= F 2π R N_ss

•• _{Temperature rise}_{Temperature rise}

ΔT ΔT==

(50)

FINITE BEARING DESIGN & ANALYSIS

(51)

FINITE BEARING DESIGN & ANALYSIS

••

_Example:

A large pump has a horizontal rotor weighing 3200 lb supported

on two

plain 360

oo

_{journal bearings}

_{, one on either side of the pump impeller. The}

specifications of the bearings are as

specifications of the bearings are as follo

follows:

ws:

R

R =

= 2

2 in

in

L

L =

= 4

4 in

in

C

C =

= 0.002

0.002 in

in

N

N =

= 1800

1800 rpm

rpm

the lubricant viscosity

µ = 1.3 x 10

-6-6

_{reyns (SAE 10 at an inlet temperature}

of 166

oo

_F)

Determine:

a)

Eq

Equil

uilibr

ibrium p

ium posi

ositio

tion of the s

n of the shaf

haft cen

t cente

ter and l

r and loca

ocatio

tion of fil

n of film rup

m ruptur

ture

e

b)

M

Min

inim

imum

um fi

fillm

m th

thic

ickn

knes

esss

c)

Lo

Loca

cati

tion

on an

and m

d mag

agni

nitu

tude

de of

of ma

maxi

ximu

mum p

m pre

ress

ssur

ure

e

d

d))

P

Po

ow

we

er

r llo

ossss

e

(52)

FINITE BEARING DESIGN & ANALYSIS

(53)

FINITE BEARING DESIGN & ANALYSIS

(54)

FINITE BEARING DESIGN & ANALYSIS

(55)

FINITE BEARING DESIGN & ANALYSIS

(56)

Muliyani

(57)

ATTITUDE ANGLE FOR OTHER BEARING

CONFIGURATION

Where:

(58)

LUBRICANT SUPPLY ARRANGEMENT

Supply Hole

Axial Groove

Circumferential Groove

(59)

SUPPLY HOLE

A common supply methode with small bearing and

A common supply methode with small bearing and bushing is to place bushing is to place an inlet portan inlet port

at the bearing midplane opposite to the load line

(60)

AXIAL GROOVE

(61)

VARIOUS GROOVE POSITIONS AND

INLET ARRANGEMENT

(62)

CIRCUMFERENTIALS

CIRCUMFERENTIALS GROO

GROOVE

VE

Alur yang melingkar

(63)

FLOW CONSIDERATION

Where fL is a corretion factor for the film position as given below:

1.

1. Oil holOil hole or axiae or axial grol groove poove positiositioned in thned in the unloade unloaded sectied section of thon of the bearie bearing oppong opposite tsite too

the load line:

2.

2. Oil hole or axial grOil hole or axial groove positionoove positioned at the maximum ed at the maximum film thicknessfilm thickness 

(64)

3. For double axial grooves runn

3. For double axial grooves running parallel at ±90º angles to the ing parallel at ±90º angles to the load line:load line:

4. For a full film starting from the maximum film thickness position

(65)

FLOW CONSIDERATION



Pressure Induced

Pressur

e Induced Flow

Flow

Inlet hole of diameter DH:

••

_{Qp : Pressure induced}

flow

••

_{Ps : Supply pressure}

(66)

FLOW CONSIDERATION

1.

1. The film thickness parameter for an oil hole ar an axial groove

The film thickness parameter for an oil hole ar an axial groove

positioned in the unloaded section of the bearing opposite the load

line is

2. Positioned at the maximum film thickness

3. For double axial groove

(67)

Total leakage flow rate

1.

1. For For an oil an oil hole hole or or an axan axial grial groovoove posie positiontioned in ed in the unthe unloadeloaded sectd section oion of the f the bearibearingng

opposite to the load line

2.

2. For aFor an axian axial grol groove oove of lengf length Lg (Lg/L= 0.th Lg (Lg/L= 0.3 to 0.83 to 0.8) posit) positioned ioned at the mat the maximaximum filum filmm

thickness or two

(68)

Mulyani

(69)

Example 8.4

•• _{Consider a journal bearing with the}_{Consider a}_{journal bearing with the follo}_{following specification that correspond to}_{wing specification that correspond to}

an actual bearing tested by Dowson et al , (1966):

an actual bearing tested by Dowson et al , (1966): L/DL/D = 0.75;= 0.75; R/CR/C = 800; D == 800; D = 0.102 m;

0.102 m; W W = 11000 = 11000 N, and operating speed isN, and operating speed is NN_ss= 25 rev/s.= 25 rev/s.

•• _{An axial groove was cut into the bearing surface in the unload portion of the}_{An axial groove was cut into the bearing surface in the unload portion of the}

bearing, opposite the load line. the groove width is

bearing, opposite the load line. the groove width is ωω_g_g = 4.76 x 10- = 4.76 x 10-33_{m, and it is}_{m, and it is} LL_g_g = = 0.067 m long. 0.067 m long. LubricanLubricant is t is supplied to the bearing at supplied to the bearing at tempertemperature Tature T_ii = 36.8 = 36.8 o

o_{C at a supply pressure of P}_{C at a supply pressure of P}

ss = 0.276x10 = 0.276x1066 Pa (40 Psi) . The lubricant viscosity is a Pa (40 Psi) . The lubricant viscosity is a function of temperature and varies according to µ =

function of temperature and varies according to µ = µie µie--ββ(T-Ti)(T-Ti)_with_with_µi_µi _{= 0.03 Pa.s,}_{= 0.03 Pa.s,}

and the temperature viscosity coefficient is estimated to be

β

= 0.0414. = 0.0414.

lubricant thermal cond

lubricant thermal conductivity k = uctivity k = 0.13 W/mK, and 0.13 W/mK, and thermal diffusivitythermal diffusivity ɑɑ_tt = 0.756 = 0.756 x 10-7 m

x 10-7 m22_{/s. Determine the flow rates, power loss, attitude angle, and maximum}_{/s. Determine the flow rates, power loss, attitude angle, and maximum} pressure.

pressure.

Paramete

Parameter r Nilai Nilai ParametParameter er NilaiNilai

L/D L/D 0.75 0.75 TT_ii 36.836.8oo_C_C R/C R/C 800 800 PP_ss 0.276x100.276x1066_{Pa (40}_{Pa (40} Psi) Psi) D

D 0.102 0.102 m m µi µi 0.03 Pa.s0.03 Pa.s

W W 11000 N11000 N ββ 0.04140.0414 N N_s_s 25 25 rev/s rev/s k k 0.13 0.13 W/mKW/mK ω ω_g_g 4.76 x 10-4.76 x 10-33_m_m _ɑ_ɑ tt 0.756 0.756 x x 10-710-7 m m22_/s_/s L L 0.067 m0.067 m

(70)

••

_{Menggunakan rumus Sommerfeld number :}

  



_



_{ <<}

  



_



_{ << }

0.03  25.0 

0.03  25.0  0.0762  0.102

₁₁₀₀₀

0.0762  0.102

800

800 

_0.338

Dari table 8.6 Dari table 8.6 dengan L/D = 0.75 dengan L/D = 0.75 didapat didapatɛ =͠ɛ =͠ 0.450.45

(71)

••

_{Maka didapat}

_Ǭ

LL

= 0.7821 ; (R/C) f = 7.4017; ɸ =

59.19

oo

••

_Sehingga

_{Leakage flow rate : Ǫ}

LL

=

Ǭ

LL





N

ss

DLC

Ǫ

_LL

=

0.7821





25 x 0.102 x 0.0762 x 6.35.10

-5-5

= 1,51.10

-3-3

_m

33

_{/s = 15.1 c}

_{/s = 15.1 cm}

_m

33

_/s

Cara 2 dengan curve fit equation, table 8.8

 

= [ 1- 0.22 (0.75)

1.91.9

_(0.45)

0.020.02

_{= 0.875}

 

ƒ

_L_L

= ɛ. ƒ

₁₁

=

= 0.45

0.45 (0.875)

(0.875) =

= 0.394

0.394

 

=

π (0.762) (0.102)(25)(

6,35.10

-5-5

_)(0.394)

= 1,52.10

-5-5

_m

33

_/s

ƒ

₁₁

= 1- 0.22





1.91.9

ɛ

0.020.02 Ǫ Ǫ_LL= π= π N N _ssDLC DLC ƒ ƒ _LL

(72)

••

_{Attitude angle}

Asumsi h

_max

, δ = 0.18,

dengan menggunakan

persamaan 8.45 dan 8.46 maka didapat :

ɀ = 4

11 ++





((11 −− 11..225 5 εεᵟᵟ))

(8.46)

= 4

11 ++ 00..7755 ((11 −−11..225 5   00..4455

00..18

18 ))

= 3.75

ɸ = tan

-1





1−<

(8.45

(8.45 ))

= tan

-1





1−0.45<

= 59

o

(73)

••

_{Next, the pressure include flow must be}

_{Next, the pressure include flow must be determined. Fro}

_{determined. From table 8.8 determine the}

_{m table 8.8 determine the}

groov

groove function and

e function and the related film thickness :

the related film thickness :



 =(1+ɸ)=

=(1+ɸ)=

 ((11 ++ 00..4455ccoos s 5599))==  11..8877

ƒ

ƒ  1.1.2255 −− 00..2255 0.0.0067

_{33 ((00..00776622/ / 00..006677))}



67/0/0..007762

_{−− 11 ++}

62 33 0.0.0076

762/2/00..110022 ((11 −− 0.0.0067

4.76 x 10ˉ=/0.102

67/0/0.0.077662)2)

= 0.838

ƒ

ƒ  11..2255 −− 00..2255 //

_{33 ((/ / ))}



_{−− 11 ++}

33 // ((11 −− //))

ωω

gg//DD

Ǫ

_p_p

=

ƒ

_g_g

ℎ



=

P

ss

=

ᵢ

=

..

 .

 . ((..





)(.ˉ?)=

.

=

3.69 x 10

-6-6

_m

33

_{/s = 3.69 cm}

33

_/s

(74)

S’ = 0.75 S’ = 0.75





0.70.7 + 0.4 + 0.4 = 0.75 = 0.75

.

.

0.70.7 + 0.4 + 0.4 = 1.085 = 1.085

Total leakage is determinated as follows. First, the datum flow rate

Ǫ

_m_m

is

estabili

estabilished

shed ::

Ǫ

_m_m

=

Ǫ

_LL

+

Ǫ

_p_p

–

0.3

0.3 ǪǪ

_{LL . .ǪpǪp}

= 15.1 + 3.69

–

0.3

0.3 (15.1)(3.69)

(15.1)(3.69)

=

= 16.6

16.6 cm³

cm³/s

/s (16.

(16.6x10

6x10

-6-6

_m

33

_/s)

From table 8.8 for an axial groove

positioned in the unloaded section of

the bearing opposite to the load line,

we have :

Therefore , the total leakage

flow rate becomes :

Ǫ

_{L total}_{L total}

=

Ǫ

_m_mSS’’

_Ǫ

p p 1-S1-S’’

= (16.6)

1.0851.085

_(3.69)

-0.085-0.085

= 18.81 cm

33

_/s

(18.81 x 10

-6-6

_m

33

_/s)

(75)



Power loss :

Ep = FU =

ƒW

(2π R Ns)

= 7.4017 (1/800) (11000) (π x 0.102 x 25)

= 812 W (1.09 hp)

Dari therma

Dari thermal diffusivity,

l diffusivity,

ρCp

= k/α

_t_t

= 1.72 x 10

66

_{W.s/ (m}

33

_K)

∆

∆ 

_(Ǫ



(76)

••

_{Effectiive temperature for evaluating next iteration is}

(section 8.17)

T

_cc

= T

_ii

+ ΔT = 36.8 + 25.1 = 61.9

oo

_C

••

_{The corresponding viscocity is :}

µ =

(77)

••

_{Dengan diketahui viskositasnya, rasio eksentrisitas}

dapat dihitung dengan

dapat dihitung dengan menggunak

menggunakan data dari tabel

an data dari tabel

8.9 dengan

8.9 dengan prediksi effective temperaturn

prediksi effective temperaturnya 50

ya 50

oo

C

••

_{Dengan interpolasi didapat}

..

.

(78)

••

θmax−;



 <

=

cos

-1-1

  .

(.)<

= 138.13

oo

_C

••

_Prediksi

_θ

_max_max

₌

_138.13

oo

_{C, which if measured from the load}

_{C, which if measured from the load line}

_line

would be =

138.13

138.13 +

+

ɸ

=

(79)

Dengan menggunakan

Dengan menggunakan tabel 8.6

tabel 8.6 maka

maka maximum pressurenya

maximum pressurenya

:

••

..

..



..

Ṕ.

.

.



Ṕ.

.

••

Ṕmax13.24≅13

••







<<

ṔṔmax max + P + Pss = (0.016) (25) (800) = (0.016) (25) (800)22_{(13) + 2.67.105}_{(13) + 2.67.105} = 3.60 Mpa (520 Psi) = 3.60 Mpa (520 Psi)

(80)

(81)

Circumferential Groove

Ǫ

_cc









′

((11 ++ 11..5 5 



))

(8.61)

Ǫ

= Ǫ

_LL

+ 2

Ǫ

_cc

(8.62)

(82)

Example 8.5

Consider a plain

Consider a plain journal bearing with

journal bearing with the following specificati

the following specifications:

ons:

D

= 8 in; L = 4.00;

C

= 6x10

-3-3

_{in; operating speed}

_N

_{= 3600rpm. The}

load imposed on the bearing is W= 4800lbf. A narrow

circumferential oil feed groove is cut into the bearing at is

midlength, abd lubricant (

μ

= 10 cp at

T

= 120

oo

_{F) is supplied at 10}

psi. determine the temperature rise.

With the full length

LL

= 4 divided in half,

l’/D

= 0.25.

Load of each two bearing segments is

Load of each two bearing segments is W W _l_l_’_’ = =







24

00

00 





Projected pressure on each bearing

Projected pressure on each bearing PP_l’_l’==



(  )

 15

150 

0 

Operati

Operating viscocity ng viscocity isis

μ

= 10cp ( = 10cp ( 1.45x10-7) reyns/cp1.45x10-7) reyns/cp

= 1.45 x 10-6 reyns

(83)

Sommerfe

Sommerfeld n

ld numbers :

umbers :

 

 (/)<

′

0.258

Operating eccentricity Ɛ= 0.8

Dimensionaless

leakage

leakage flow

flow rate

rate Ǭ

Ǭ

_LL

= 1.5753

Friction coefficient

(84)

••

_{In dimensional form, the leakage flow rate due to}

shaft rotation is :

Ǫ

_LL

_{= Ɛ}





NsDl’C

= 0.8





60. 8. 2.

610⁻



= 14.25 in

33

/s (14.25x60/231 = 3.70 gpm)

••

_Pressur

_{Pressure-induce fl}

_{e-induce flow is :}

_{ow is :}

Ǫ

_cc









′

((11 ++ 11..5 5 



))

=

  (⁻



)

  ..⁻@  

((11 ++11..5 5 ((00..88))



))

= 3.06 in

33

_/s

_{/s (0.79}

_{(0.79 gpm)}

_gpm)

(85)

••

_{Total leakage flow becomes}

Ǫ

= Ǫ

_LL

+ 2

Ǫ

_cc

= 14.25 in

33

_{/s + (2 x 3.06 in}

33

_/s)

_{/s) =}

_{= 20.37}

_{20.37 in}

_in

33

_{/s (5.29}

_/s

_{(5.29 gpm)}

_gpm)

••

_Po

_{Power loss is for each half}

_{wer loss is for each half-length bearing segment is :}

_{-length bearing segment is :}

Ep =

Ep = ƒW

ƒW

_l l

_’ ’

π

D Ns

= 0.01329 x 2400 x

π x 8 x 60

= 4.81 x 10

44

in.lbf/s (0.83hp)

••

_Prediction

_{Prediction temperatur}

_temperature

_{e rise}

_{rise is :}

_{is :}

∆

∆  2  

_(Ǫ

2  

,,

))



77778 8 1122 00..4488 00..00331155 ((2200..3377)) 33.4:

2  4.81.10>

33.4:

Mean outlet temperature is

T

(86)

BEARING STIFFNESS, ROTOR VIBRATION,

AND O

(87)

••

_{Spring mass syst}

_{Spring ma}

_{ss system}

_em

(88)

Tia Utari

(89)

Determine whirl stability for a horizontal

rotor and its bearings with

rotor and its bearings with the following

the following

characteris

characteristics

tics ::

D = 2R = 5 in D = 2R = 5 in L = 2.5 in L = 2.5 in C = 0.005 in C = 0.005 in N = 90 N = 90 rad/rad/s (5400 s (5400 rpm)rpm) Ks = 5 x

Ks = 5 x 10^6 lb/in rotor stiffness10^6 lb/in rotor stiffness W = 5000 lb rotor weight (m = W/

W = 5000 lb rotor weight (m = W/g = 5000/386 = g = 5000/386 = 13 lb13 lb





/in rotor mass/in rotor mass

μ

μ = 2 x 1 = 2 x 1

00 



lbs/ lbs/





(reyns) viscosity (reyns) viscosity

Example Example

8.6 8.6

(90)

Example Example 8.6 8.6Con’tCon’t

Analysis Using the

Analysis Using the graphgraph

Unit bearing load Unit bearing load P = P = _(DL)_(DL)



= =

 /

  ∗. 

= 200 psi = 200 psi Sommerfeld number Sommerfeld number S = S = μμNN

(/)





= 2*1 = 2*1

00 



/



∗90/

(. ∶. )



 

= 0.225 = 0.225

Characteristic bearing number Characteristic bearing number = S(L/D = S(L/D

))



= 0.225 (2.5 in/ 5 in = 0.225 (2.5 in/ 5 in

))



= 0.05625 = 0.05625

Stability on rotor stiffness Stability on rotor stiffness (C/W)

(C/W)



_

= = (0.005 (0.005 in/ in/ 5000 5000 lb)*(5*1lb)*(5*1

00 

lb/in) lb/in) = 10

= 10

next next

(91)

Example Example 8.6 8.6Con’tCon’t Stability on case Stability on case (C/W)m

(C/W)m

ωω

= (0.005 in/ 5000 lb)*(13 lb

= (0.005 in/ 5000 lb)*(13 lb





/in*(2/in*(2ππ*90*90

/)

))

= 8.28

Rotor will be free of Rotor will be free of

oil whip

(92)

General

Design

Guides

Effective Effective Temperature Temperature Maximum Maximum Bearing Bearing Temperature Temperature Turbulent Turbulent and Parasitic and Parasitic Loss Effect Loss Effect Flooded Flooded versus versus Starved Starved Condition Condition Bearing Load Bearing Load Dimensions Dimensions Eccentricity Eccentricity and and Minimum Minimum Film Film Thickness Thickness Operating Operating Clearance Clearance Misalignment Misalignment and Shaft and Shaft Deflection Deflection

(93)

Eff

ective

T

emperature

Temperature rata-rata pada viskositas tertentu

Global effective temperature Global effective temperature

Dimana : Dimana :

JJ  panas mekanik panas mekanik

ρ

ρ densitas oildensitas oil





 leakage leakage flowrateflowrate





 temperatur awal temperatur awal





 kapasitas panas kapasitas panas

ΦΦ



 conduction & radiation conduction & radiation





 power loss power loss For small bearing

(94)

Maximum Bea

ring T

emperature

Temperature Temperature Minimum Minimum film thickness film thickness

(95)

Turbulent and Parasitic Loss Effect

Turbulent : Turbulent :

•• _{Bearing diameter}_{Bearing diameter}

•• _{Large film thicknesses}_{Large film thicknesses (clearance)}_(clearance) •• _{High surface speed}_{High surface speed}

•• _{Low fluid viscosities}_{Low fluid viscosities} •• _{High Reynolds numbers}_{High Reynolds numbers}

Par

Parasitic loss asitic loss ::

•• _Putar_{Putaran dan}_{an dan turbulensi pada oil}_{turbulensi pada oil grooves dan clearence}_{grooves dan clearence}

•• _{Losses pada percepatan feed oil terhadap surface speed yang tinggi}_{Losses pada percepatan feed oil terhadap surface speed yang tinggi} •• _{Vortex pada feed dan oil grooves}_{Vortex pada feed dan oil grooves}