kelompok 2 - journal bearing (2).pdf

KELOMPOK 2

Darma Adhi W. (11210009)Galuh Intan P. (11210012)

Bhatara Putra M. (11210014)

Mulyani (11210020)

Muliyani (11210023)

Ali Akbar (11210024)

Tia Utari (11210028)

Eko Rusdiyanto (11210030)

Bhatara Putra Mediriyanto

Konsep Dasar

Apa itu Bearing?

Tipe beban pada bearing

1. Bantalan radial (Journal Bearing)

Arah beban yang di tumpu bantalan ini

adalah tegak lurus dengan sumbu poros

2. Bantalan aksial (Thrust Bearing)

Arah beban bantalan ini adalah sejajar

dengan sumbu poros

3. Bantalan kombinasi (combination bearing)

Bantalan ini menumpu beban yang

arahnya sejajar dan tegak lurus dengan

sumbu poros

Introduction

Journal bearing termasuk salahsatu sliding bearing dan keterbalikkan dari ball bearing

Journal bearing secara umum digunakan pada mesin piston kendaraan bermotor berbahan bakar bensin atau diesel

Kelebihan : Bearing type ini mampu menopang shaft yang berat. Awet dan tahan lama Efek redaman dari film minyak membantu membuat mesin

beroperasi dengan tenang dan halus.

Kekurangan : Membutuhkan suplai minyak pelumas yang besar Hanya cocok untuk temperatur dan kecepatan rendah Pembentukan lapisan minyak pelumas lambat

Bearing Diagram

Journal bearing Berfungsi sebagai bantalan poros engkol yang berputar

Oil inlet Tempat masuknya minyak pelumas

Ketika oli pelumas masuk ke dalam bearing, oli akan memenuhi

clearance/ gap antara shaft dan bearing sehinggga mengakibatkan

tekanan fuida meningkat dan daya angkat hidrodinamis terhadap shaft

Type Typical Loading Application

(a) Partial arc Unidirectional load Shaft guides, dampers

(a) Circumferential

groove, Axial groove

types

Variable load direction Internal combustion engines

(a) Cylindrical Medium to heavy

Unidirectional load

General machinery

(a) Pressure dam Light loads, unidirectional High speed turbines, compressor

(a) Overshot Light loads, unidirectional Steam turbines

(a) Multilobe Light loads, unidirectional Gearing, compressor

(a) Preloaded Light loads, unidirectional Minimize vibration

(a) Tilting pad Moderatic Variable loads Minimize vibration

Movement of the bearing

Infinitely Long Approximation (ILA)

Menentukan jari-jari shaft dan clearance

ILA

Menentukan Dimensionless pressure

Galuh intan prawesti

Boundary Condition

8.3 BOUNDARY CONDITIONS

Assumsi :

= 0 = 0

=

2

Dimana :

Ps = tekanan suplai

C = radial clearence

R = radius bearing

= viskositas pelumas = kecepatanputaran poros

8.4 FULL SOMMERFELD BOUNDARY CONDITIONS

Asumsi :

= 0 = 2 (360)

cos = +

1 +

Substitusi Sommerfeld :


Tekanan puncak terjadi ketika

cos =3

(2 + 2)

Dimana :

cos = sudut angular padatekanan maksimum

= rasio eksentrisitas

=

= eksentrisitasC = radial clearence

Dari persamaan 8.9 asumsi P = 0 pada = , besarnya tekanan

puncak tak berdimensi adalah :


=3 (4 2)(4 52 + 4)0.5

2(1 2)2(2 + 2)

= 13

2 + 2

Yang terjadi pada :

Besarnya tekanan puncak tak berdimensi pada distribusi tekanan

adalah :

=6 sin (2 + )

(2 + 2)(1 + )2 (8.9)

Load Carrying Based

on Full Sommerfeld Condition

Load Carrying Based


= 0

2

Arah Radial

=

Arah Tangensial

(8.10)

(8.11)

Dari substitusi tekanan tak berdimensi pada persamaan 8.9 dengan

persamaan 8.10 dan 8.11 maka didapatkan :

= 0 =12

(1 2)12 2 + 2

(8.12)

Load Carrying Based


Dimana Beban tak berdimensi:

(8.13)

Resultan dari dan

(8.14)

=

2

Dengan =

2

= beban yang diproyeksikan

Ns = kecepatan poros dalam rev/s

= + =

()(+)

Load Carrying Based


Attitude Angle

(8.15)

=

=

=

.

Dalam berbagai kasus, jika

= 0 = 0

= 1 =

8.5 DEFINITION OF THE SOMMERFELD NUMBER

Bilangan Sommerfeld (S) merupakan bilangan tak berdimensi yang

merupakan parameter karakterisasi performansi sebuah bearing.

Bilangan ini menunjukkan karakteristik gesekan total dari bantalan.

8.5 DEFINITION OF THE SOMMERFELD NUMBER

=

2

substitusikan ke dalam persamaan

Sommerfeld Number (8.14) maka :

=1

Dan penyelesaian S menjadi

=(1 2)

12(2 + 2)

12

(8.16)

(8.17)

8.6 HALF SOMMERFELD BOUNDARY CONDITION

=6

1 2)12(2 + 2

=122

1 2)(2 + 2

(8.18)

(8.19)

Total kapasitas beban dukung dan attitude angle adalah:

=6

1 2)(2 + 2)2 2(2 4 0.5 (8.20)

=

( ). (8.21)

Eko Rusdiyanto

Contoh Soal 8.1

Fenomena Kavitasi

Kavitasi

Gaseous Cavitation

Vapor Cavitation

Gaseous cavitation merupakan kavitasi yangdisebabkan oleh adanya bagian dari minyakpelumas yang terlarut dengan udara padakondisi jenuh (sekitar 10%), dan ketika tekanansekitar menjadi turun bagian yang terlarut iniakan membentuk suatu kavitasi tetapi dibagianyang berbeda dari fluid film, hal ini yangmenyebabkan kavitasi jenis gaseous tidak terlaluberbahaya.

Gaseous Cavitation

Vapor cavitation disebabkan oleh tingginyafluktuasi tekanan yang ada diantara film daripelumas dan bearingnya itu sendiri, kavitasijenis ini cukup berbahaya karena bisamenyebabkan kerusakan pada bearing (fatiguedamage)

Vapor Cavitation

SWIFT-STEIBER (REYNOLD) BOUNDARY CONDITION

Perhitungan beban bearing dengan memperhatikan kavitasi didalam perhitungannya

= 0 =

Ali Akbar

INFINITELY SHORT JOURNAL BEARING

APPROXIMATION (ISA)

A. Infinitely Short Journal Bearing Approximation (ISA)

Integral 2 kaliLength-to-Diameter ratios up to L/D = dengan trends rata-rata L/D = 1

Table Infinitely Long Journal Bearing Solutions with the Reynolds Boundary Condition

B. Full and Half Sommerfeld Solutions for Short Bearings (ISA)

Figure Short Bearing Eccentricity Ratio vs Sommerfeld Number

Figure Short Bearing Attitude Angle vs Sommerfeld Number

Darma Adhi Wardhana

FINITE BEARING DESIGN & ANALYSIS

This section focused on design and performance analysis based on the full solution of Reynold equation.


Minimum film thickness

hmin = C ( 1 )

Friction force

F = f W

Power loss

Ep = F 2 R Ns

Temperature rise

T =

FINITE BEARING DESIGN & ANALYSIS Example:

A large pump has a horizontal rotor weighing 3200 lb supported on two plain 360o journal bearings, one on either side of the pump impeller. The specifications of the bearings are as follows:

R = 2 in L = 4 in C = 0.002 in N = 1800 rpm

the lubricant viscosity = 1.3 x 10-6 reyns (SAE 10 at an inlet temperature of 166o F)

Determine:

a) Equilibrium position of the shaft center and location of film rupture

b) Minimum film thickness

c) Location and magnitude of maximum pressure

d) Power loss

e) Temperature rise

Muliyani

ATTITUDE ANGLE FOR OTHER BEARING CONFIGURATION

Where:

LUBRICANT SUPPLY ARRANGEMENT

Supply Hole

Axial Groove

Circumferential Groove

SUPPLY HOLE

A common supply methode with small bearing and bushing is to place an inlet port at the bearing midplane opposite to the load line

AXIAL GROOVE

VARIOUS GROOVE POSITIONS AND INLET ARRANGEMENT

CIRCUMFERENTIALS GROOVE

Alur yang melingkar ditempatkan pada centerline

FLOW CONSIDERATION

Where fL is a corretion factor for the film position as given below:1. Oil hole or axial groove positioned in the unloaded section of the bearing opposite to

the load line:

2. Oil hole or axial groove positioned at the maximum film thickness

Axial flow due rotation

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

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

FLOW CONSIDERATION

Pressure Induced Flow

Inlet hole of diameter DH:

Qp : Pressure induced flow

Ps : Supply pressure

i : Lubricant Viscosity

FLOW CONSIDERATION

1. 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 grooves running parallel 90 angles to the load line

Total leakage flow rate

1. For an oil hole or an axial groove positioned in the unloaded section of the bearingopposite to the load line

2. For an axial groove of length Lg (Lg/L= 0.3 to 0.8) positioned at the maximum filmthickness or two axial grooves running parallel at 90 angles to the load line

Mulyani

Example 8.4

Consider a journal bearing with the following specification that correspond to an actual bearing tested by Dowson et al , (1966): L/D = 0.75; R/C = 800; D = 0.102 m; W = 11000 N, and operating speed is Ns = 25 rev/s.

An axial groove was cut into the bearing surface in the unload portion of the bearing, opposite the load line. the groove width is g = 4.76 x 10-

3 m, and it is Lg = 0.067 m long. Lubricant is supplied to the bearing at temperature Ti = 36.8 oC at a supply pressure of Ps = 0.276x10

6 Pa (40 Psi) . The lubricant viscosity is a function of temperature and varies according to = ie-(T-Ti) with i = 0.03 Pa.s, and the temperature viscosity coefficient is estimated to be = 0.0414. lubricant thermal conductivity k = 0.13 W/mK, and thermal diffusivity t = 0.756 x 10-7 m2/s. Determine the flow rates, power loss, attitude angle, and maximum pressure.

Parameter Nilai Parameter Nilai

L/D 0.75 Ti 36.8 oC

R/C 800 Ps 0.276x106 Pa (40

Psi)

D 0.102 m i 0.03 Pa.s

W 11000 N 0.0414

Ns 25 rev/s k 0.13 W/mK

g 4.76 x 10-3 m t 0.756 x 10-7

m2/s

Lg 0.067 m

Menggunakan rumus Sommerfeld number : =

=

=

0.03 25.0 0.0762 0.102

11000800 2 = 0.338

Dari table 8.6 dengan L/D = 0.75 didapat = 0.45

Maka didapat L= 0.7821 ; (R/C) f = 7.4017; = 59.19o

Sehingga Leakage flow rate : L = L

2NsDLC

L = 0.7821

225 x 0.102 x 0.0762 x 6.35.10-5

= 1,51.10-3 m3/s = 15.1 cm3/s

Cara 2 dengan curve fit equation, table 8.8

= [ 1- 0.22 (0.75) 1.9(0.45) 0.02 = 0.875

L = . 1 = 0.45 (0.875) = 0.394

= (0.762) (0.102)(25)(6,35.10-5)(0.394) = 1,52.10-5 m3/s

1 = 1- 0.22

1.9 0.02

L = NsDLC L

Attitude angle

Asumsi hmax, = 0.18, dengan menggunakanpersamaan 8.45 dan 8.46 maka didapat :

= 4 1 +

(1 1.25 ) (8.46)

= 4 1 + 0.75 (1 1.25 0.45 0.18)

= 3.75

= tan-1

21 (8.45 )

= tan-1

21 0.45

= 59o

Next, the pressure include flow must be determined. From table 8.8 determine the groove function and the related film thickness :

= (1 + )

= (1 + 0.45 cos 59) = 1.87

=1.25 0.25 0.067/0.0762

33 (0.0762/ 0. 067) 1+

4.76 x 10/0.102

3 0.0762/0.102 (1 0.067/0.0762)

= 0.838

=1.25 0.25 /

33 (/ ) 1+

g/D3 / (1 /)

p = g

Ps

= 0.838 1.87 (2.76105)(6.3510)

0.03

= 3.69 x 10-6 m3/s = 3.69 cm3/s

S = 0.75

0.7 + 0.4

= 0.75 0.067

0.07620.7 + 0.4

= 1.085

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

m = L + p 0.3 L . p= 15.1 + 3.69 0.3 (15.1)(3.69)

= 16.6 cm/s (16.6x10-6 m3/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 :

L total = mS p

1-S

= (16.6) 1.085 (3.69) -0.085

= 18.81 cm3/s

(18.81 x 10-6 m3/s)

Power loss :

Ep = FU = W (2 R Ns)

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

= 812 W (1.09 hp)

Dari thermal diffusivity, Cp = k/t = 1.72 x 106 W.s/ (m3K)

=

(, )=

812

1.72 10 18.81 10

= 25.1

Effectiive temperature for evaluating next iteration is (section 8.17)

Tc = Ti + T = 36.8 + 25.1 = 61.9oC

The corresponding viscocity is :

= ie-(T-Ti) = 0.3 e-0.00414(61.9-36.8) = 0.011 pa.s

Dengan diketahui viskositasnya, rasio eksentrisitasdapat dihitung dengan menggunakan data dari tabel8.9 dengan prediksi effective temperaturnya 50oC

Dengan interpolasi didapat

51.645.7

5044.7=0.60.55

0.55

5.9

4.3=

0.05

0.55 = 0.58

max = 3

2+

= cos-13 0.58

2+(0.58)

= 138.13 oC

Prediksi max = 138.13 oC, which if measured from the load line

would be = 138.13 +

= 138.13 + 59 = 197.13o

Dengan menggunakan tabel 8.6 maka maximum pressurenya :

0.60.55

0.580.55=14.613911.1975

11.1975

0.55

0.03=

3.4164

11.1975

max = 13.24 13

=

+ max + Ps

= (0.016) (25) (800)2 (13) + 2.67.105= 3.60 Mpa (520 Psi)

Circumferential Groove

c=3

6(1 + 1.5 2) (8.61)

L total = L + 2 c (8.62)

Example 8.5

Consider a plain journal bearing with the following specifications: D= 8 in; L = 4.00; C = 6x10-3 in; operating speed N = 3600rpm. Theload imposed on the bearing is W= 4800lbf. A narrowcircumferential oil feed groove is cut into the bearing at ismidlength, abd lubricant ( = 10 cp at T= 120oF) is supplied at 10psi. determine the temperature rise.

With the full length L= 4 divided in half, l/D = 0.25.

Load of each two bearing segments is Wl = 4800

2= 2400

Projected pressure on each bearing Pl = 2400

(2 8)= 150

Operating viscocity is = 10cp ( 1.45x10-7) reyns/cp= 1.45 x 10-6 reyns

Sommerfeld numbers : = (/)

= 0.258

Operating eccentricity = 0.8

Dimensionaless leakage flow rate L= 1.5753

Friction coefficient

= 0.8657

In dimensional form, the leakage flow rate due to shaft rotation is :

L =

2NsDlC

= 0.8

260. 8. 2. 6103

= 14.25 in3/s (14.25x60/231 = 3.70 gpm)

Pressure-induce flow is :

c=3

6(1 + 1.5 2)

= 4 (6103)

6 1.45.10 2(1 + 1.5 (0.8)2)

= 3.06 in3/s (0.79 gpm)

Total leakage flow becomes

L total = L + 2 c= 14.25 in3/s + (2 x 3.06 in3/s) = 20.37 in3/s (5.29 gpm)

Power loss is for each half-length bearing segment is :Ep = Wl D Ns

= 0.01329 x 2400 x x 8 x 60= 4.81 x 104 in.lbf/s (0.83hp)

Prediction temperature rise is :

=2

(, )=

2 4.81.10

778 12 0.48 0.0315 (20.37)= 33.4

Mean outlet temperature isT0 = 120 + 33.4 = 153

oF

BEARING STIFFNESS, ROTOR VIBRATION,

AND OIL WHIRL INSTABILITY

Spring mass system

The bearing horizontal natural frequency

Tia Utari

Determine whirl stability for a horizontal rotor and its bearings with the following

characteristics :

D = 2R = 5 in

L = 2.5 in

C = 0.005 in

N = 90 rad/s (5400 rpm)

Ks = 5 x 10^6 lb/in rotor stiffness

W = 5000 lb rotor weight (m = W/g = 5000/386 = 13 lb2/in rotor mass

= 2 x 106 lbs/2 (reyns) viscosity

Example 8.6

Example 8.6 Cont

Analysis Using the graph

Unit bearing load

P =

(DL)

= 5000 /2

5 2.5

= 200 psi

Sommerfeld number

S = N(/)2

= 2*106/2 90/(2.5 0.005 )2

200

= 0.225

Characteristic bearing number

= S(L/D)2

= 0.225 (2.5 in/ 5 in)2

= 0.05625

Stability on rotor stiffness

(C/W) = (0.005 in/ 5000 lb)*(5*106 lb/in)

= 10

next

Example 8.6 Cont

Stability on case(C/W)m2= (0.005 in/ 5000 lb)*(13 lb2/in*(2*90/)2)

= 8.28

Rotor will be free of oil whip instability

General Design Guides

Effective Temperature

Maximum Bearing

Temperature

Turbulent and Parasitic Loss Effect

Flooded versus

Starved Condition

Bearing Load Dimensions

Eccentricity and

Minimum Film

Thickness

Operating Clearance

Misalignment and Shaft Deflection

Effective Temperature

Temperature rata-rata pada viskositas tertentu

Global effective temperature

Dimana :J panas mekanik densitas oil leakage flowrate temperatur awal kapasitas panas

conduction & radiation power loss

For small bearing

Maximum Bearing Temperature

Temperature

Minimum film thickness

Turbulent and Parasitic Loss Effect

Turbulent :

Bearing diameter

Large film thicknesses (clearance)

High surface speed

Low fluid viscosities

High Reynolds numbers

Parasitic loss :

Putaran dan turbulensi pada oil grooves dan clearence

Losses pada percepatan feed oil terhadap surface speed yang tinggi

Vortex pada feed dan oil grooves

Surface drag yang terjadi antara oil dengan high surface speed

Flooded versus Starved Condotion

Flooded

Condition

< 1Starved Condition

Leakage flow rate

Kerja bearing jelek

Good Cooling

Bearing Load and Dimensions

Projected Loading

PL= W/(L*D)

Rentan vibrasi

Power loss tinggi

High oil flow

overheating

Eccentriciry and Minimum Film Thickness

= C e= C(1 )

Excessive bearing temperature

Susceptibility to wear

too small

Poorer vibration

Higher power loss

too large

Operating Clearance

Clearance is 0.002 per inch of diameter

kelompok 2 - journal bearing (2).pdf

Documents