Lecture 1
Contents
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Characterization of a Fluid
Continuum Hypothesis
No-Slip Condition
The Boundary Layer
Viscosity
Newton’s Law of Viscosity
Examples: Newton’s Law of Viscosity
Contact Details
Peter J. Thomas
(a.k.a. Eddy DeCay)
Office: F331
Office Hours: Presently Wednesdays 11am12noon
and Fridays 4pm-5pm
Email: Eddy.DeCay@eng.warwick.ac.uk
or: pjt1@eng.warwick.ac.uk
Tel. Ext.: 22200
All Slides From Lectures Can Be
Downloaded From …
http://www.eng.warwick.ac.uk/staff/pjt/teachdocs.html
BOOKS USED FOR COURSE
• Chapters in Tech. Sci. 2, 4th edition are from:
Douglas, J.F., Gasiorek, J.M, Swaffield, J.A., Jack,
L.B.
Fluid Mechanics
5th edition, Pearson, 2008
• If you want to buy additional book, I recommend:
White, F.M.
Fluid Mechanics
6th edition, McGraw-Hill, 2008
Next year’s course
“ES3D6: Fundamental Fluid
Mechanics for Mechanical
Engineers”
will be based largely based on the
book by White
Course Structure
• There will be
Eight Lectures plus One Example Class
• If you spot any TYPOS please let me know
so that I can correct them.
• Please
ask QUESTIONS
LECTURES and/or afterwards
DURING
• And/Or come to see me during my OFFICE
HOURS, but let me know in advance if you
are intending to turn up
GENERAL REMARK
This course will provide you with
many of the…
required terminology and the
fundamental concepts
… that will be developed in more
detail and in greater generality in next
years follow-on course.
If you understand this year’s material
then many of the key topics of next
year’s course will be easier for you!
Fluids And Their Properties
• Technical distinction between fluid and solid lies
with the reaction of the two to an applied shear stress
or tangential stress.
• Fluid is substance that deforms continuously when
subjected to shear stress; no matter how small shear
stress maybe.
• Solid can resist a shear stress by a static deformation;
a fluid cannot.
• Fluid moves and deforms continuously as long as the
shear stress is applied.
Fig. from White (1994), p.3
Fig. Deformation resulting from application of constant
shear force. Streeter et al. (1998), p.4.
CONTINUUM HYPOTHESIS
• All fluids (liquids or gases) constitute a large number of
molecules in motion ...
• Strictly there are ‘voids’ between neighboring molecules ...
• This creates a problem !!! How can we treat this
mathematically? What now!?.
For mathematical description of fluid flow
the actual molecular structure is replaced by
a hypothetical continuous medium called the
CONTINUUM.
Notes:
• Molecular theory must be used to calculate fluid properties
(e.g. viscosity) which are associated with molecular
motion.
• Continuum hypothesis not always satisfied. For instance
RAREFIED GAS DYNAMICS. Mean free path of
molecules is not small compared to characteristic length of,
say, a body in the flow. Example: Re-entry of spaceships
into the atmosphere of Earth
NO-SLIP CONDITION
• When fluid moves over a surface then the molecules in
contact with the wall stick to the wall.This is called the
NO-SLIP CONDITION.
Consequently …
Flow velocity immediately at a stationary wall is
ALWAYS zero!
This is how flow would
look if no-slip condition
would not exist.
This is how flow looks
in reality as a
consequence of no-slip
condition.
The BOUNDARY LAYER
• Flow velocity zero on wall but non-zero further away. Hence,
there must be a transition region where flow velocity increases
from zero to the free-stream value. This region is called
BOUNDARY LAYER.
• The BOUNDARY-LAYER THICKNESS δ is usually taken as that
distance from wall where flow speed is equal to 99% of
undisturbed free-stream velocity far away from wall.
• Boundary layer intoduced by LUDWIG PRANDTL, Math.
Congress, Heidelberg Germany, 1904, 8 page paper.
To visualize:
Electric wire
generates
‘bubbes’ by
electrolysis.
U0, Free-Stream Velocity
0.99 U0
FLOW
Wall
U=0 at wall
The VISCOSITY
The VISCOSITY is that property of a fluid by virtue
of which it offers resistance to shear
• Macroscopic (lump) property viscosity results from forces
governing interactions of many molecules on microscopic
level!
Consider Scenario Below:
Pull plate over a plane. Narrow gap between plate and
plane filled with fluid (e.g. oil). What one gets is ...
Upper plate moves with velocity: U
U
Gap Width: D
Velocity = zero
NEWTON’s LAW OF VISCOSITY
• Evidently there are shear forces arising in connection with
motion and viscosity!
• How can we determine them?
Shear Strain:
φ =
•
Rate of Shear Strain:
φ =
x
y
t
=
x
y
x 1 u
=
t y y
u
Assuming experimental result that SHEAR
STRESS (=Force/Area) is proportional to shear
strain yields:
du
τ =µ
dy
This is Newton’s Law of Viscosity!
Continued...
Constant of proportionality in Newton’s Law of Viscosity...
F
= τ
A
= µ
du
dy
… is the called ...
Dynamic Viscosity:
µ
Units :
kg m
s
2
Ns
kg
s
=
Pa
s
=
=
… sometimes called POISEUILLE
2
2
s
m
m
m
Kinematic Viscosity
Units :
kg
s m m2
=
kg
s
m3
µ
ν=
ρ
... and 10
−6
ρ : Density of Fluid
m2
= 1 centistokes = 1 cSt
s
Two simple examples illustrating typical applications
of Newton’s law of viscosity
Example 1: Plate Sliding Down a Ramp
d=?
Consider ...
Assumptions ...
• Area of board:
• Weight of board:
• Viscosity of oil in gap:
• Angle:
• Board slides with velocity:
A = 1m × 1m
W = 30 N
µ = 0.05 Nsm -2
α = 16o
U = 0.025 ms-1
• Edge effects negligible
• Linear velocity distribution across gap
QUESTION: What is gap width d between board
and ramp?
Continued …
Gap is narrow, hence we can assumelinear
velocity profile across gap…
Continued...
Solution:
• Board slides at constant velocity when
Fg : i.e. Weight
α
Ftangential = Fshear
Ftangential = W sin α
(1)
• Shear stress can be calculated from Newton’s law
of viscosity
du
(2)
τ =µ
dy
• Since we assumed linear velocity profile Eq. (2)
becomes
∆U
τ =µ
∆y
• The shear force is then
Fshear
∆U
∆U
µ
A
A=
= τ A= µ
d
∆y
Eq. (1) into Eq. (3)
(3)
Continued...
W sin α = µ
∆U
A
d
(4)
• Solving Eq. (4) for d yields ...
µ ∆U A
d=
W sin α
• Substituting in given values ...
d=
0.05
Ns
m
2
0
.
025
1
m
s
m2
= 0.000151 m = 0.151 mm
o
30 N sin16
Example2:
Journal Bearing
Consider the simple journal bearing sketched below ...
Assumptions ...
250 revs/min
• Shaft rotates at:
• Clearance between shaft and bearing: h = 0.1 mm
µ = 0.1 Nsm -2
• Viscosity of oil in clearance:
• Length of bearing (into paper):
l = 300 mm
• Shaft radius:
R = 60 mm
• Edge effects negligible
QUESTIONS:
(a) Draw an appropriate sketch illustrating the oil
flow in the clearance.
(b) Find the viscous shear stress resisting the
rotation of the shaft
(c) Find the torque required to rotate the shaft
(d) Find the power required to rotate the shaft
Continued...
Solution:
(a) Clearance between rotating shaft and journal
bearing is small. Hence, can assume a linear
velocity profile ...
(b) From Newton’s law of viscosity viscous shear
stress is ...
(2π f )R
du
U
ωR
τ = µ =µ =µ
=µ
dy
h
h
h
All values given and, hence, ...
250 revs ⎞
⎛
⎛ 26.18 ⎞
−3
⎜2 π
⎟ 60 mm
⎜
⎟ 60 × 10 m
Ns
Ns ⎝ s ⎠
60s ⎠
= 0.1
τ = 0.1 2 ⎝
0.1 mm
m2
0.1 × 10 −3 m
m
= 1570.8
N
m
2
= 1570.8 Pa
Continued...
(c) The torque T is given by (see data book)
T = Acting Force × R = τ (2 π R L ) × R
(
)
= 1570.8 N 2π 60 × 10−3 m 300 × 10 −3 m × 60 × 10 −3 m
= 10.659 Nm
(d) The power P is given by (see data book)
P = F ×v
= F ×ω R = T ω
⎞
⎛ 250 revs
= 10.659 Nm ⎜
2π ⎟
⎠
⎝ 60 s
= 279.1 W