LUBRICATION Process by which the friction in a moving contact is reduced. Five distinct form of lubrication are:
Hydrodynamic Hydrostatic Elastohydrodynamic Mixed Boundary Solid film 335
Quantification of LUBRICATION using dimensionless film parameter Λ (“Specific film thickness) Λ=
hmin 2 2 Rrms + R ,a rms ,b
Boundary lubrication, Λ<1 Mixed lubrication, 1<Λ<3 Hydrodynamic lubrication, Λ>5 Elastohydrodynamic , 3<Λ<5
Solid-film
Hydrostatic
336
¾ Root-mean-square deviation
1 l 2 Rq = z ( x ) dx ∫ l 0
337
To understand functioning of lubricants one needs to understand Dry FRICTION Leonardo da vinci(1452-1519): Fα W;F ≠ A ¾ “Friction made by same weight will be of equal resistance at the beginning of movement, although contact may be of different breadths or length” ¾ “Friction produces the double the amount of effort if weight be doubled”
G.Amontons, 1699: Fα Fn; F ≠ A
338
¾C.A.Coulomb 1781 (1736-1806): 1)Clearly distinguished between static & kinetic friction 2)Contact at discrete points. 3)Friction due to interlocking of rough surfaces
4)No adhesion 5)f ≠ func(v)
339
TOMLINSON’s Theory of Molecular attraction: 1929
Relation between friction coefficient & elastic properties of material involved.
f = 1 . 07 * [θ
θ =
I
+θ
II
3 . E + 4 .G G (3 .* E + G )
Clean Steel Aluminum Titanium
]2 / 3
E is young
modulus,
G is modulus
E=30 Mpsi, E=10 Mpsi, E=15.5 Mpsi
Mpsi
in shear,
G=12 Mpsi G=3.6 Mpsi G=6.5 Mpsi
Mpsi 0.6558
0.5039
340
Bowden & Tabor Model Two friction sources Generally load on bearing surface is carried on just a few points. These are subjected to heavy unit pressure, and so probably weld together. Adhesion force developed at real area of contact. Deformation force needed to plough asperities of harder surface through softer. Resulting friction force is sum of two contributing terms
Adhesion = Attractive force across an interface Interface = Contact boundary between two surfaces. 341
ADHESION Theory
342
ADHESION Theory
• Two surfaces are pressed together under load W. • Material deforms until area of contact (A) is sufficient to support load W. A = W/H. • To move the surface sideway, must overcome shear strength of junctions with force F F=As 344
Shear stress of softer of contacting materials
W = Areal H
F = Areal s
μ=
s H
For most of materials H = 3σy & s = σy /1.7321 Expected value of μ =.2 Friction of metals arises from strong adhesion or welding at the regions of real contact. Similar & dissimilar materials????? 345
PLOUGHING Effect Ploughing occurs when two bodies in contact have different hardness. The asperities on the harder surface may penetrate into the softer surface and produce grooves on it, if there is relative motion. Assume n conical asperities of hard metal in contact with flat soft metal, vertically project area of contact:
W = n(0.5 * πr ) H 2
μ=
2
π
(
A = n 0.5 * πr 2
F = (nrh) H
)
cot θ 346
Two basic reasons of ploughing are: Ploughing by surface asperities Ploughing by hard wear particles present in contact zone For θ = 45° μ = 0.6366 For θ = 60° μ = 0.3676 For θ = 80° μ = 0.1123
Slope of real surfaces are nearly always less than 10° (i.e. θ> 80°), therefore μ < 0.1. Conclusion: Total μ , representing contribution for both ploughing and adhesion terms, should not exceed 0.3. For same material, μ = 0.2.
Junction Growth
⎛ δW ⎞
δA.τ y = ⎜
2
2 ⎟ + δF ⎝ 2 ⎠
Constant
F =f (A) ???? 348
Limiting Junction Growth Presence of weak interfacial films. Assume shear stress, τi.
Fmax = τ i Amax Fmax τ i Amax μ= = 2 W (σ y2 − 4τ i2 ) Amax
μ=
2
⎛ δW ⎞ 2 δA.τ y = ⎜ ⎟ + δF ⎝ 2 ⎠
τi
2 (τ y2 − τ i2 )
Understanding this mechanism motivate to apply thin film of low shear strength materials to the surfaces. 349
μ for mixed lubrication
= 0.01 – 0.1
μ for elastohydrodynamic lubrication = 0.001 – 0.01 Coefficient of friction for hydrodynamic lubrication ??? 350
FLUID FILM BEARINGS Machine elements designed to produce smooth (low friction) motion between solid surfaces in relative motion and to generate a load support for mechanical components. Fluid between surfaces may be a gas, liquid or solid. Word film implies that fluid thickness (clearance) separating the surfaces is several orders of magnitude smaller than other dimensions of bearing (width & length). Successful design requires film thickness to be larger than the micro asperities on the surfaces, operation without contact of surfaces. Operation principles of liquid film bearings are hydrodynamic, hydrostatic or combination. 351
Hydrodynamic
Axial coordinate
Hydrostatic
Axial coordinate
Hydrodynamic
Hydrostatic
Relative motion between two mechanical surfaces is utilize the generate pressure and levitate one surface relative to other surface…. Self-acting
External source of pressurized fluid is required to levitate the shaft surface and separate it from bearing surface.. Costly
Load support is a function of lubricant viscosity.
Load support is a weak function of lubricant viscosity.
HDL provided an infinite bearing life
Infinite life if supporting ancillary equipments function well
Able to damp the external vibrations.
Able to control the effect of external vibration.. Active control
Requires lubricant flow to compensate the leakage from bearing ends.
Require pressurized lubricant.
Significant difference in static & kinetic friction coefficients
Almost same value of coefficient of frictions.
High relative speed generates much higher load capacity & destabilize the shaft-system.
Very good control on the shaft position.
VISCOSITY
Absolute viscosity Dynamic viscosity
To define viscosity consider surface 1 is moving with a velocity V on a film of thickness h. Imagine film is composed of series of horizontal layers and force F causing these layers to deform/slide on one another just like a deck of card.
du τ =η ; dy
Constant of proportionality
η ν= ρ 2
m η in Pa.s ; ν in s
354
Viscosity Physical property-resistance to flow. Due to internal friction and molecular phenomena . Dynamic Viscosity
o 1 cP=10-3 Pa.s Kinematic Viscosity
o 1 cSt = 1 mm2/s
Grades of Oils: • SAE– (Society of Automotive Engineers) ISO 355
Variation of viscosity with temp: Increase in temp decreases intermolecular forces. SAE grade
ISO grade
Viscosity In mPa.s 1000c 400c
10W
32
32.6
5.57
3.20
107
20W
68
62.3
8.81
5.01
118
SAE 30
100
100
11.9
6.25
110
SAE 40
150
140
14.7
8.0
102
5W-20
46
38
6.92
4.17
140
10W-30
68
66.4
10.2
5.7
135
10W-40
100
77.1
14.4
8.4
193
10W-50
------
117
20.5
10.53
194
VI 1300c
356
Petroff’s Equation Friction = Shear Stress * Area F = (Viscosity* V/h)*Area V = 2πRN ; A = 2πRL Friction force, F =
η * 2πRN * 2πRL
C F Coefficient of friction, μ = W η * 2πRN * 2πRL / C ⇒μ= 2 RLP Conclusion: Coefficient of 2 ηN R → μ = 2π friction is a function of P C speed, load and viscosity
C is radial clearance
357
Temperature Rise Friction, due to shear of lubricant film, generates heat (F×V)) in lubricant oil and increases the temperature of lubricant. Assuming that total generated heat is carried by the oil flowing through bearing Heat generated = Heat convected by oil flow η (2 π R N )(2 π R L ) (2 π R N ) = m CP Δt C or
Δt =
η (2 π R )3 N 2 C m CP
ρ = 860 kg / m3
(
L
C P = 1760 J / kg oC
η (2 π R )3 N 2 or Δt = L ⎛ 2πRN ⎞ C ⎜ρ C L ⎟ CP ⎝
( )
2
η 8π 2 N ⎛ R ⎞ or Δt = ⎜ ⎟ (ρ ) C P ⎝ C ⎠
⎠
2
R = 1000 C
Δt = 52.2η N
)
Δt = 52.2η N
Assume rotational speed = 900 rpm
Δt = 783η SAE grade
Viscosity in Viscosity in mPa.s 400c mPa.s 1000c
25.5258
10W
32.6
5.57
4.3660
48.7809
20W
62.3
8.81
6.8982
78.3000
SAE 30
100
11.9
9.3177
109.6200
SAE 40
140
14.7
11.5101
29.7540
5W-20
38
6.92
5.4184
51.9912
10W-30 66.4
10.2
7.9866
60.3693
10W-40 77.1
14.4
11.2752
91.6110
10W-50 117
20.5
16.0515
Logarithmic scale
VISCOSITY INDEX: Qualitative comparison index • VI relates viscosity change at 37.8 0c and 98.90c. • Pennsylvanian oil~VI=100 • gulf coast oil ~ VI=0 VI =
L-U *100 L-H
An oil with VI = 240 has less rate of viscosity change with temperature compared to oil with VI 100. 361
η = ηin e
− β (t −tin )
η log e ⎛⎜ in ⎞⎟ η⎠ ⎝ ⇒β = (t − tin )
Walther' s Equation log log( cS + 0.6) = constant − c log T β
SAE grade
Viscosity In mPa.s 1000c 400c
10W
32.6
5.57
3.20
0.0294
20W
62.3
8.81
5.01
0.0326
SAE 30
100
11.9
6.25
0.0355
SAE 40
140
14.7
8.0
0.0376
5W-20
38
6.92
4.17
0.0284
10W-30
66.4
10.2
5.7
0.0312
10W-40
77.1
14.4
8.4
0.0284
10W-50
117
20.5
10.53
0.0401
1300c
• Lubricant viscosity increases with pressure. • Barus relationship
η = η0 exp(α p )
• α for petroleum oil~2*10-8/pa Pressure
Multiplier
1.e5
1.0010
1.e6
1.0101
1.e7
1.1052
1.e8
2.7183
1.e9
22.026 363
Reynolds Equation A basic pressure distribution equation for “Fluid Film Lub.” In 1886, Reynolds derived for estimation of pressure distribution in the narrow, converging gap between two surfaces. Reynolds equation helps to predict hydrodynamic, squeeze, and hydrostatic film mechanisms. Reynolds' equation ∂ ⎛ h 3 ∂ P ⎞ ∂ ⎛ h 3 ∂P ⎞ ∂ ⎧∂ ⎫ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 6 ⎨ (U 1 + U 2 )h + (W1 + W2 )h + 2(Vh − V0 )⎬ ∂x ⎝ η ∂x ⎠ ∂z ⎝ η ∂z ⎠ ∂z ⎩ ∂x ⎭ 364
No pressure development within the parallel surfaces.
365
Pressure driven flow
366
Small element of Fluid with sides dx, dy, and dz ⎛ ∂τ ⎞ ∂p ⎞ ⎛ Force balance : pdy.dz + ⎜⎜τ + dy ⎟⎟dx.dz = ⎜ p + dx ⎟dy.dz + τdx.dz ∂y ⎠ ∂x ⎠ ⎝ ⎝
∂u For Newtonian flow τ = η ∂y ∂P ∂ ⎛ ∂u ⎞ ∂x
=
⎜⎜η ⎟⎟ ∂y ⎝ ∂y ⎠
367
∂u ∂P On integration :η y + C1 = ∂y ∂x
∂P ∂ ⎛ ∂u ⎞ = ⎜⎜η ⎟⎟ ∂x ∂y ⎝ ∂y ⎠
∂P y 2 + C1 y + C2 → ηu = ∂x 2 Using boundary conditions : y = 0, u = U 2 ; y = h, u = U 1
ηU 2 = C 2 ,
η (U1 − U 2 ) ∂P h −
= C1
h ∂x 2 ⎛ y 2 − yh ⎞ ∂P y ⎜ ⎟ ⇒u =⎜ + (U1 − U 2 ) + U 2 ⎟ h ⎝ 2η ⎠ ∂x
Check !!!
368
Flow rate in x - direction per unit width : h
q x = ∫ u.dy 0
h 3 ∂P h qx = − + (U1 + U 2 ) 12η ∂x 2
Check !!!
Similarly flow rate in z - direction h
q z = ∫ w.dy 0
h 3 ∂P h qz = − + (W1 + W2 ) 12η ∂z 2
Reynolds equation is derived using mass continuity equation ∂q x ∂q z + (Vh − V0 ) + =0 ∂x ∂z 369
Reynolds' equation ∂ ⎛ h 3 ∂P ⎞ ∂ ⎛ h 3 ∂P ⎞ ∂ ⎧∂ ⎫ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 6⎨ (U1 + U 2 )h + (W1 + W2 )h + 2(Vh − V0 )⎬ ∂x ⎝ η ∂x ⎠ ∂z ⎝ η ∂z ⎠ ∂z ⎩ ∂x ⎭
Stretching action
Simplification I : h 3
Wedge action (inclined surfaces
∂ ⎛ 1 ∂P ⎞ ⎧∂ ⎫ ⎜⎜ ⎟⎟ = 6(U1 + U 2 )⎨ h ⎬ ∂z ⎝ η ∂z ⎠ ⎩ ∂x ⎭
Can carry High loads for short duration
∂ ⎛ h 3 ∂P ⎞ ⎧∂ ⎫ ⎟⎟ = 6(U1 + U 2 )⎨ h ⎬ Simplification II : ⎜⎜ ∂x ⎝ η ∂x ⎠ ⎩ ∂x ⎭
Squeeze action (bearing surfaces move perpendicular to each other)
Comparison among pressure profiles at z = 0 7000000
Simplification II
6000000
Pressure
5000000 plong in green color
4000000 3000000 2000000
Simplification I
1000000 0 0
20
40
60
80
100 θ
Length / diameter = 0.25
120
140
160
180
Comparison among pressure profiles at z = 0 1.4E8
Simplification I
1.2E8 pshort in red
Pressure
1E8 8E7 6E7 4E7 2E7
Simplification II
0 0
20
40
60
80
100 θ
Length/ diameter = 2.5
120
140
160
180
Comparison among pressure profiles at z = 0 22000000
Simplification I
pshort in red
2E7 18000000 16000000 Pressure
14000000 12000000 1E7 8000000
Simplification II
6000000 4000000 2000000 0 0
20
40
60
80
100 θ
Length/diameter = 1.0
120
140
160
180
Short Static Bearing ∂ ⎛ 1 ∂P ⎞ ⎧∂ ⎫ ⎜⎜ ⎟⎟ = 6(U1 + U 2 )⎨ h ⎬ h ∂z ⎝ η ∂z ⎠ ⎩ ∂x ⎭ 3
∂ 2 P 6Uη dh = 3 2 ∂z h dx dp 6Uη dh = 3 z using symmetry condition dp/dz = 0 at z = 0 dz h dx 3Uη dh ⎛ 2 L2 ⎞ ⎜⎜ z − ⎟⎟ p= 3 4⎠ h dx ⎝
using p = 0 at z = ± L/2 374
Film thickness, h, depends on geometry of tribo-pair. For example, in journal bearing h = Cr + e cosθ Using expression of h in following equation 3Uη dh ⎛ 2 L2 ⎞ ⎜⎜ z − ⎟⎟ p= 3 4⎠ h dx ⎝
3Uη ⎛ ε sin θ p= 2 − 3 ⎜ R Cr (1 + ε cos θ ) ⎝
θ 2 ⎞⎛ 2 L ⎞ ⎟⎜⎜ z − ⎟⎟ 4⎠ ⎠⎝
375
Load capacity of short journal bearing Load component in direction of loc π
L
UηL3 2ε 2 Wθ = ∫ ∫ p.( Rdθ .dz ). cos θ ⇒ Wθ = − 2 2 C 2 1 − ε L r 0− 2 2
(
)
2
Load component perpendicular to loc π
L
UηL3π ε Wr = ∫ ∫ p.( Rdθ .dz ). sin θ ⇒ Wr = 2 2 C 4 r 0 −L 1 − ε 2 2
(
ε UηL π ⇒W = 4Cr2 1 − ε 2
2
1/ 2
3
(
)
3
)
2
⎧⎛ 16 ⎞ 2 ⎫ ⎨⎜ 2 − 1⎟ε + 1⎬ ⎠ ⎩⎝ π ⎭
Wr π 1− ε 2 tan φ = − ⇒ tan φ = Wθ 4 ε
Lesser the attitude angle, better the stability of bearing. eccentricity ratio vs. attitude angle 90 80 70 Attitude angle
60 50 40 30 20 10 0 0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1
Eccentricity ratio 377
Friction force in Journal Bearing Petroff equation --- inaccurate U dp h τ =η + h dx 2 L/2 π ⎛ U dp h ⎞ F = ∫ ∫ ⎜η + ⎟Rd θ .dz h dx 2 ⎠ −L / 2 0 ⎝
F=
2ηULRπ Cr 1 − ε 2
+
Cr εWr 2R
⎛ y 2 − yh ⎞ ∂P y ⎜ ⎟ u=⎜ + (U 1 − U 2 ) + U 2 ⎟ h ⎝ 2η ⎠ ∂x
ε UηL3π Wr = 4Cr2 1 − ε 2
(
)
3
2
If εÆ 0, F Æ Petroff solution 378
Temperature Rise Friction, due to shear of lubricant film, generates heat (F×V)) in lubricant oil and increases the temperature of lubricant. Assuming that total generated heat is carried by the oil flowing through bearing Heat generated = Heat convected by oil flow F (2 π R N ) = m C P Δt or
or or
F (2 π RN ) ⎛ 2πRN ⎞ Cr L ⎟ C P ⎜ρ 2 ⎝ ⎠ 2F Δt = (ρ Cr L ) C P Δt =
Δt =
1 F 756800 (Cr L )
ρ = 860 kg / m3
(
C P = 1760 J / kg oC
)
Design of Hydrodynamic Journal Bearing
1. Guess eccentricity ratio 2. Calculate load capacity, friction force, temperature rise. 3. Modify lubricant viscosity. 4. Repeat steps 1-3 so that average viscosity and load converge. 380
Ex: Determine the minimum film thickness, maximum pressure, coefficient of friction for a hydrodynamic journal bearing, which supports a 600 N load at rotational speed of 2000 rpm. The shaft dia is 40 mm. Assume bearing length = 10 mm, oil viscosity at room temp (30°C) = 15 mPa.s, β=0.029, and radial clearance 20 μm. Given: U = 4.19 m/s. Factor1 = U*L3*π*0.25/(Cr2)=8227 m2 /s, Factor2 = 2*U*L*R*pi()/Cr , STEP 1: Assume ε = 0.5 Æ W = 118 N Assume ε = 0.75 Æ W = 562 N STEP 2: Assume ε = 0.8 Æ W = 900 N, F = 6.76 N, Δt=44.68°C
W = (Factor1)η F=
η (Factor 2) 1− ε
2
1/ 2
ε
(1 − ε )
2 2
⎧⎛ 16 ⎞ 2 ⎫ ⎨⎜ 2 − 1⎟ε + 1⎬ ⎠ ⎩⎝ π ⎭
( Factor1)Cr η + 2R
ε2
(1 − ε )
2 1 .5
Δt =
1 F 756800 (Cr L )
381
STEP 3: Modify viscosity using
η = ηin e
− β (t −tin )
at Δt = 44.68o C η = 0.0041 STEP 4: For ε = 0.8 Æ W = 246 N, F = 1.85 N, Δt=12.22°C, η=0.0105Æ W = 630.2 N, F = 4.74 N, Δt=31.3°C. Now it is preferable to increase ε. Let us assume ε = 0.82 and Δt=30 °C Æ W = 471.4 N, F = 2.99 N, Δt=19.75°C, η=0.0085Æ W = 636 N, F = 4.0 N, Δt=26.6°C. η=0.0069Æ W = 516.3 N, F = 3.3 N, Δt=21.63°C. Now it is preferable to increase ε. Let us assume ε = 0.83 and Δt=25 °C Æ W = 615.3 N, F = 3.56 N, Δt=23.5°C.
Groove arrangement to feed to HB under pressure – Hybrid bearing 1.75 0.675 hg3 Psupply ⎛ d h ⎞ QP = ⎜ + 0 .4 ⎟ η ⎠ ⎝ L hg = Local film thickness , m
d h = Diameter of feed hole , m Psupply = Feed pressure , Pa
383
Ex: Determine the minimum film thickness, maximum pressure, coefficient of friction for a hydrodynamic journal bearing, which supports a 600 N load at rotational speed of 2000 rpm. The shaft dia is 40 mm. Assume bearing length = 10 mm, oil viscosity at room temp (30°C) = 15 mPa.s, β=0.029, and radial clearance 20 μm. Assume dia of feed oil is 2.5mm and supply pressure is 1.5 bar Given: U = 4.19 m/s. Factor1 = U*L3*π*0.25/(Cr2)=8227 m2 /s, Factor2 = 2*U*L*R*pi()/Cr , STEP 1: Assume ε = 0.5 Æ W = 118 N Assume ε = 0.75 Æ W = 562 N STEP 2: Assume ε = 0.8 Æ W = 900 N, F = 6.76 N
W = (Factor1)η F=
η (Factor 2) 1− ε 2
1/ 2
ε
(1 − ε )
2 2
⎧⎛ 16 ⎞ 2 ⎫ ⎨⎜ 2 − 1⎟ε + 1⎬ ⎠ ⎩⎝ π ⎭
( Factor1)Cr η + 2R
ε2
(1 − ε )
2 1 .5 384
ρ = 860 kg / m3
Heat generated = Heat convected by oil flow F (2 π R N ) = m C P Δt or
or
F (2 π RN ) ⎛ 2πRN ⎞ Cr L + QP ⎟ ρ C P ⎜ ⎝ 2 ⎠ 27.64 F Δt = 4.189 + 0.254 × (1 + ε )3 Δt =
(
)
≅ Δt = 33o C
(
C P = 1760 J / kg oC
)
1.75 0.675 hg3 Psupply ⎛ d h ⎞ QP = ⎜ + 0 .4 ⎟ η ⎝ L ⎠
hg = 20 ×10 −6 × (1 + ε ), m d h = 0.0025, m Psupply = 1.5 × 105 , Pa
⇒ QP = 2.54 × 10 −8 (1 + ε ) m3 / s 3
STEP 3: Modify viscosity using
η = ηin e − β (t −tin ) at Δt = 33o C η = 0.0058
STEP 4: For ε = 0.8 Æ W = 348 N, F = 2.615 N, Δt=12.75°C, η=0.0104Æ W = 624.2 N, F = 4.7 N, Δt=22.9°C.
Now it is preferable to increase ε. Let us assume ε = 0.82 and Δt=23 °C Æ W = 576.13 N, F = 3.65 N, Δt=17.66°C, η=0.009Æ W = 673.4 N, F = 4.27 N, Δt=20.6°C. η=0.0083Æ W = 621 N, F = 3.94 N, Δt=19°C.
