Kelompok 2 - Journal Bearing (2).pdf
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Transcript of Kelompok 2 - Journal Bearing (2).pdf
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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)
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Bhatara Putra Mediriyanto
Konsep Dasar
Apa itu Bearing?
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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
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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
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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
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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
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Movement of the bearing
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Video
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Infinitely Long Approximation (ILA)
Menentukan jari-jari shaft dan clearance
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ILA
Menentukan Dimensionless pressure
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Galuh intan prawesti
Boundary Condition
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8.3 BOUNDARY CONDITIONS
Assumsi :
= 0 = 0
=
2
Dimana :
Ps = tekanan suplai
C = radial clearence
R = radius bearing
= viskositas pelumas = kecepatanputaran poros
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8.4 FULL SOMMERFELD BOUNDARY CONDITIONS
Asumsi :
= 0 = 2 (360)
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cos = +
1 +
Substitusi Sommerfeld :
8.4 FULL SOMMERFELD BOUNDARY CONDITIONS
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 :
8.4 FULL SOMMERFELD BOUNDARY CONDITIONS
=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)
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Load Carrying Based
on Full Sommerfeld Condition
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Load Carrying Based
on Full Sommerfeld Condition
= 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)
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Load Carrying Based
on Full Sommerfeld Condition
Dimana Beban tak berdimensi:
(8.13)
Resultan dari dan
(8.14)
=
2
Dengan =
2
= beban yang diproyeksikan
Ns = kecepatan poros dalam rev/s
= + =
()(+)
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Load Carrying Based
on Full Sommerfeld Condition
Attitude Angle
(8.15)
=
=
=
.
Dalam berbagai kasus, jika
= 0 = 0
= 1 =
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8.5 DEFINITION OF THE SOMMERFELD NUMBER
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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)
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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)
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Eko Rusdiyanto
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Contoh Soal 8.1
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Fenomena Kavitasi
Kavitasi
Gaseous Cavitation
Vapor Cavitation
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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
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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
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SWIFT-STEIBER (REYNOLD) BOUNDARY CONDITION
Perhitungan beban bearing dengan memperhatikan kavitasi didalam perhitungannya
= 0 =
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Ali Akbar
INFINITELY SHORT JOURNAL BEARING
APPROXIMATION (ISA)
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A. Infinitely Short Journal Bearing Approximation (ISA)
Integral 2 kaliLength-to-Diameter ratios up to L/D = dengan trends rata-rata L/D = 1
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Table Infinitely Long Journal Bearing Solutions with the Reynolds Boundary Condition
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B. Full and Half Sommerfeld Solutions for Short Bearings (ISA)
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Figure Short Bearing Eccentricity Ratio vs Sommerfeld Number
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Figure Short Bearing Attitude Angle vs Sommerfeld Number
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Darma Adhi Wardhana
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FINITE BEARING DESIGN & ANALYSIS
This section focused on design and performance analysis based on the full solution of Reynold equation.
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FINITE BEARING DESIGN & ANALYSIS
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FINITE BEARING DESIGN & ANALYSIS
Minimum film thickness
hmin = C ( 1 )
Friction force
F = f W
Power loss
Ep = F 2 R Ns
Temperature rise
T =
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FINITE BEARING DESIGN & ANALYSIS
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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
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FINITE BEARING DESIGN & ANALYSIS
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FINITE BEARING DESIGN & ANALYSIS
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FINITE BEARING DESIGN & ANALYSIS
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FINITE BEARING DESIGN & ANALYSIS
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Muliyani
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ATTITUDE ANGLE FOR OTHER BEARING CONFIGURATION
Where:
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LUBRICANT SUPPLY ARRANGEMENT
Supply Hole
Axial Groove
Circumferential Groove
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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
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AXIAL GROOVE
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VARIOUS GROOVE POSITIONS AND INLET ARRANGEMENT
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CIRCUMFERENTIALS GROOVE
Alur yang melingkar ditempatkan pada centerline
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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
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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
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FLOW CONSIDERATION
Pressure Induced Flow
Inlet hole of diameter DH:
Qp : Pressure induced flow
Ps : Supply pressure
i : Lubricant Viscosity
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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
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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
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Mulyani
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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
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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
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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