Lecture 2
Final Version
Contents
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Newtonian vs. Non-Newtonian Fluid
Real Fluid vs. Ideal Fluid
Pressure and Head
Variation of Pressure Vertically in a Fluid
Equality of Pressure at the Same Level in a
Static Fluid.
Hydrostatic Paradox
Pressure and Head
Pressure Measurement by Manometer
U-Tube Manometer
Manometer with Inclined Leg to Increase
Sensitivity
Inverted U-tube manometer
What Did We Do In Last Lecture?
• Characterization of a Fluid
• Continuum Hypothesis
• No-Slip Condition
• The Boundary Layer
• Viscosity
• Newton’s Law of Viscosity
Newton’s Law of Viscosity
du
τ =µ
dy
What Did We Do In Last Lecture?
Newton’s Law of Viscosity
du
τ =µ
dy
Example 1:
Example 2:
Continued …
CONSTANT
Newton’s Law of Viscosity
du
τ =µ
dy
NEWTONIAN AND NON-NEWTONIAN
FLUID
Fluids behaving according to Newton’s law of
viscosity are called NEWTONIAN FLUIDS (e.g.
water, oil, gases)
• For these fluids dynamic viscosity is constant…
• … or, equivalently, shear stress is linearly related to the velocity
gradient.
• If viscosity not constant then NON-NEWTONIAN
FLUID (blood, paints, drilling mud, polymer melts,
elastomers, many emulsions)
Shear
Stress
Rate of Shear, du/dy
RHEOLOGY
• The science that investigates deformation and flow of
matter and the properties and molecular
characteristics of fluids.
• Plastic: Shear stress must reach a certain minimum value before
flow commences. Thereafter, shear stress increases with rate of
n
shear according to
⎛ du ⎞
τ = A + B⎜
⎟
⎝ dy ⎠
where A, B and n are constants. If n=1, the material is known as a
Bingham plastic (e.g. sewage, sludge).
• Pseudo Plastic: Dynamic viscosity decreases as the rate of shear
increases (e.g. colloidal solutions, clay, milk, cement).
The harder I stir the softer material gets.
• Dilatant Substance: Dynamic viscosity increases as the rate of
shear increases (quicksand).
The harder I stir the harder material gets.
• Thixotropic Substance: Dynamic viscosity decreases with the
time for which shearing forces are applied (thixotropic jelly paints)
The longer I stir the softer material gets.
• Rheopectic Materials: Dynamic viscosity increases with the
time for which shearing is applied
The longer I stir the stiffer material gets.
• Viscoelastic Materials: Similar to Newtonian fluids under
time-invariant conditions, but if shear changes suddenly behave as
if plastic.
Like Newtonian but gets stiffer when I hit it..
IDEAL vs. REAL FLUID
• Fluids which have a viscosity (i.e. ALL fluids) are
often very difficult to treat mathematically.
THEREFORE, we approximate the real world...
The
IDEAL FLUID
is a HYPOTHETICAL
FRICTIONLESS
and
INCOMPRESSIBLE
Fluid.
• Ideal fluid fluid does not exist in reality, but predictions
based on the concept often yield results which are useful as
they are not too different from the flow of a real fluid
But!!! ... Be CAREFUL …
• Sometimes predictions based on an INVISCID MODEL
yield results which have nothing in common with what one
observes when one conducts experiment with real fluid. We
will begin to discuss details later as we progress in
course….
Pressure and Head
•Fluid will exert a force normal to a solid boundary or
any plane drawn through fluid. Pressure defined as:
FORCE
Pressure =
AREA
(Normal) Force Exerted
Area of Boundary
or
p=
F
A
• Note the subtle difference between definition of STRESS and
PRESSURE! Stress is force ‘parallel; to area divided by the area.
• Both quantities have the same dimensions, i.e. N m-2
•Usually pressure changes from point to point, hence,
use ...
Mean Pressure, p =
δF
δA
•In limit, as δA → 0
δF dF
=
δA→0 δA
dA
Pressure at a Point , p = lim
Variation of Pressure Vertically in a Fluid
• Consider vertical column of fluid ...
•Pressure force at top:
F2 = p2 A
•Weight of fluid in column:
FW = m g
= ρ g A ( z2 − z1 )
•Pressure force at bottom:
F1 = p1 A
In Equilibrium...
p1 A − p2 A − ρ g A ( z2 − z1 ) = 0
p2 − p1 = − ρ g ( z2 − z1 )
1442443
p2 < p1
<0
Thus, in any fluid under gravitational
attraction, pressure decreases with
increase of height z.
Example:
• Diver descends
from surface of sea to depth of 30m.What
would be pressure under which diver would be working
above that at surface? ( Density water: ρ w = 1025 kg m −3 )
Solution:
•Take sea level as datum:
z1 = 0
•Since z2 is lower than z1 the value of z2 is -30m.
Increase of Pressure = p2 − p1
kg
= − 1025
m
3
m
9.81
s
2
(− 30 m − 0 )
N
= 301.7 ×10
m2
3
• Note, atmospheric pressure at sea level is approximately
101× 103
N
m2
Hence, pressure on diver increases by about:
3 x atmospheric pressure !
Something to remember:
10m water column = 1x atmospheric pressure
Equality of Pressure at the Same level in
a STATIC Fluid
•By
performing a force balance similar to the
previous one but now on end planes of a horizontal
cylinder …
… one can show that:
The pressure at any two points at the
same level in a body of fluid at rest will
be the same.
Hydrostatic Paradox
• From our example with vertical column of fluid ...
z2
p2 − p1 = − ρ g ( z2 − z1 )
z1
•If
free surface of a liquid in a container datum where
pressure is atmospheric (i.e. reference zero) and writing:
h = z 2 − z1
p1 = ρ g h
Hence, pressure at bottom of a liquid
filled container depends only on filling
height of container!
Continued...
•Hence,
the pressure at the bottom of the four
containers is in each case:
p=ρ gh
h
Force on Bottom = Pressure × Area
= p A= ρ g h A
•Thus,
although the weight of the fluid is obviously
different in the four cases, the force on the base of the
vessel is the same, depending on the depth, h, and the
base area A.
Continued...
Which of the two dams in sketches
below has to be stronger to
withstand the hydrostatic
pressure?
Lake, 500 km
Quiescent
h=20m
Water
Dam
1m
h=20m
Dam
Pressure and Head
•Pressure
measured
pressure is called ...
relative
to
atmospheric
GAUGE or GAGE PRESSURE
… as distinct from the ...
ABSOLUTE PRESSURE
Abs. Press. = Gauge Press. + Atmosph. Press
p = ρ g h + patm
• If
one sets patm to zero and assumes gravitational
acceleration, g, as constant gauge pressure at point X in
sketch can be defined by stating vertical height h, called
head, of a column of a given fluid which would be
necessary to produce this pressure.
Pressure Measurement by Manometer
•Simplest
form of manometer is pressure tube or
piezometer. This is a single vertical tube, open at top,
inserted into pipe or vessel containing liquid under
pressure which rises in tube to a height depending on
pressure.
•Press. at A = Press. due to of
liquid of height h1
p A = ρ g h1
h1
h2
•Press. at B = Press. due to of
A
B
liquid of height h2
p B = ρ g h2
•If liquid moving inside vessel (e.g. pipe) then bottom
of tube must be flush with inside wall of vessel,
otherwise reading will be affected by velocity of fluid.
U-Tube Manometer
•Sketch
below shows what is known as a U-tube
manometer. Can be used to measure pressure of gases
or liquids.
D
A
h1
B
h2
ρ < ρ man
C
!!!
(I)
(II)
(III)
Press. pB at B = Press. pc at C
Left Limb
p B = p A + ρ g h1
Right Limb pC = patm + ρ man g h2
Set to zero, i.e. zero gauge
Since p B = pc
p A + ρ g h1 = ρ man g h2
p A = ρ man g h2 − ρ g h1
Example
• U-tube manometer used to measure gauge pressure of fluid
−3
P of density ρ = 800 kg m . If density of liquid Q is
ρQ = 13,000 kg m −3 what will be the gauge pressure at A if
(a) h1=0.5m and D is 0.9m above BC,
(b) h1=0.1m and D is 0.2m below BC
D
A
h1
(a)
p A = 13.6 × 103
− 0.8 × 103
(b)
kg
m
kg
kg
m
kg
p A = ρ man g h2 − ρ g h1
9.81
9.81
m3
p A = 13.6 × 103
− 0.8 × 103
3
3
ρ < ρ man
C
B
For solution use:
h2
m
s
m
s2
9.81
9.81
0 .9 m
2
0.5 m = 116.15 × 103 N m −2
m
s
m
2
( − 0 .2 m )
0.1 m = − 27.45 × 103 N m −2
m3
s2
Negative sign indicates that pA is below atmospheric
pressure.
Example
• U-tube
gauge is arranged to measure pressure difference
between points A and B in a pipeline. Calculate the pressure
difference.
B
A
b
a
h
C
D
Solution:
• Again, principle involved in calculating difference pressure
is that pressure at C and D must be same.
Left Limb :
Right Limb :
pC = p A + ρ g a
p D = p B + ρ g (b − h ) + ρ man g h
pC = p D
p A + ρ g a = p B + ρ g (b − h ) + ρ man g h
Pressure Diff.= p A − pB = ρ g (b − a ) + h g ( ρ man − ρ )
Manometer with Inclined Leg to
Increase Sensitivity
x
z
θ
•If leg was vertical (i.e.θ = 90o
) then meniscus inside
leg rises by z = x .
•If leg inclined (i.e.θ < 90o ) then meniscus rises inside
leg rises by x > z where ...
sin (θ ) =
z
x
•Since sin (θ ) → 0
x=
when θ → 0
z
sin (θ )
x→∞
Hence, when pressure small manometer can be made
as sensitive as required by adjusting angle of
inclination of leg and choosing manometer liquid with
suitable density value.
Inverted U-tube manometer
•Used for measuring pressure differences ….
x
x
h
b
a
B
A
•Same pressure at x, … and then as usual ...
Left Limb :
Right Limb :
p xx = p A − ρ g a − ρ man g h
p xx = p B − ρ g (b + h )
pB − p A = ρ g (b − a ) + g h ( ρ − ρ man )
If A and B at same level: pB − p A = g h ( ρ − ρ man )