PHY131H1S – Class 24
Today:
• Equation of Continuity
• Bernoulli’s Law
• Viscosity
• Poiseuille’s Law
Problem Set 9 on
MasteringPhysics due
Friday at 11:59PM
According to Wikipedia:
“Because the density of pure ice
is about 920 kg/m³, and that of
sea water about 1025 kg/m³,
typically only one-tenth of the
volume of an iceberg is above
water.”
If a very large iceberg were to
float to warmer waters and melt,
what, in principle, would
happen to the overall sea-level?
A. sea-level would rise very slightly.
B. sea-level would fall very slightly.
C. sea-level would stay the same.
A little pre-class reading quiz…
The aorta, leading away from the heart, has a diameter of
2.6 cm and a cross section area of about 5 cm2. It then
splits into many smaller arteries, which further divide into
a system of tiny capillaries. The flow is “parallel”,
meaning fluid flows through all the capillaries
simultaneously. The total cross-section area in all the
capillaries is
A. much smaller than the cross-section area of the aorta.
B. slightly smaller than the cross-section area of the aorta.
C. exactly equal to the cross-section area of the aorta.
D. slightly greater than the cross-section area of the aorta.
E. much greater than the cross-section area of the aorta.
Archimedes Principle
• When an object is immersed in a fluid, the fluid exerts an
upward force on the object equal to the weight of the fluid
displaced by the object.
• Iceberg not melted has weight W and volume V1. It
displaces a weight of water W with volume Vw which is
less than V1. So it floats: some of the iceberg sticks up
above the water.
• Iceberg melted has the same weight W and less volume.
But it still displaces the same amount of water. It
displaces a weight of water W.
• So melting an iceberg which is floating does not change
sea level.
• If the iceberg were not floating, but sitting on a land-mass,
and it melted and added water to the ocean, this would
increase the sea-level.
Fluid Dynamics
Comparing two points in a flow tube of cross section A1 and
A2, we may use the equation of continuity:
where v1 and v2 are the fluid speeds at the two points.
The flow is faster in narrower parts of a flow tube, slower
in wider parts.
This is because the volume flow rate I, in m3/s, is constant.
I=
ΔV
= A⋅ | v |
Δt
1
• A tube widens from a cross-sectional area
A1 to a cross sectional area A2 = 4 A1. As
a result the speed of an ideal dynamic fluid
in the tube changes from v1 to
A.
B.
C.
D.
E.
v2 = v1/16
v2 = v1/4
v2 = v1
v2 = 4v1
v2 = 16v1
Bernoulli’s Law
• We study the steady flow of water from a
water tap, e.g., in your kitchen sink. The
jet of water
A. broadens as it falls.
B. narrows as it falls.
C. does not change its
cross-sectional shape.
D. slows before hitting the
bottom of the sink.
Bernoulli’s law
• Consider an ideal fluid (laminar flow), flowing
through a tube which narrows.
• It increases its velocity. This means the kinetic
energy per volume of the fluid will increase.
• How can this be? There must be a force which
does work on the fluid to speed it up.
• The force must come from a pressure difference.
• Pressure must be lower in the region of increased
fluid velocity.
Bernoulli’s Law
• Bernoulli’s law is an expression of the conservation of
energy for a closed system.
• It states that an increase in the speed of an ideal dynamic
fluid is accompanied by a drop in its pressure.
• When combined with Pascal’s law for pressure drop with
height in a fluid, it is written:
or:
Rank in order, from highest to lowest, the
liquid heights h1 to h4 in tubes 1 to 4. The air
flow is from left to right.
A. h1 > h2 = h3 = h4
B. h2 > h4 > h3 > h1
C. h2 = h3 = h4 > h1
D. h3 > h4 > h2 > h1
E. h1 > h3 > h4 > h2
2
Bernoulli’s Law
Vascular Flutter:
• Plaque accumulates on the inner walls of an
artery.
• To maintain a constant volume flow rate, the
blood must travel faster than normal through
the constriction.
• This reduces the pressure in the artery,
relative to the stationary extracellular fluid
surrounding the artery.
• The artery collapses, stopping the flow.
• The pressure then increases until it is
enough to reopen the artery.
• The flow resumes, and pressure drops again,
creating a repeated temporary interruption of
blood flow that can be heard with a
stethoscope.
Example.
• A 5.0 cm thick coating of glycerine is
placed between two plexiglass sheets,
each with an area of 0.25 m2. The sheets
slide at a constant speed of 0.10 m/s
relative to each other. What is the force
required to move one of these sheets?
Viscosity
• Two parallel plates
are immersed in a
fluid.
• The lower plate is
at rest.
• The upper plate is
moved sideways by
an external force.
• There is a
resitance force R
which balances
Fext.
• The upper plate
moves at a
constant velocity.
• Viscosity, η, has
units of N s/m2.
In a wire, a voltage drop ΔV gives rise to an electric
current I which is related to the resistance R by
Ohm’s Law:
ΔV = RI
In a tube containing a Newtonian fluid, a pressure drop
Δp gives rise to an volume flow rate I which is related
to the resistance R by Poisseulle’s Law:
Δp = RI
where the resistance is given by:
R=
8lη
4
π ⋅ rtube
| Δv |
Δy
Fluid Flow through a
cylindrical vessel
• A pressure difference along the
tube creates a force in the
direction of motion.
• Viscosity causes a resistance
force acting opposite to the
direction of motion.
• The velocity of the fluid is highest
along the centre of the cylinder,
and decreases to zero at the
edges.
• Poiseuille’s law gives the volume
flow rate as a function of pressure
difference along the tube.
I=
Fluid Flow Ohm’s Law,
Analogy with electric circuits
R = Fext = η ⋅ A
ΔV π 4 Δp
rtube
=
l
Δt 8η
• A Newtonian fluid is forced through a tube with a
pressure difference Δp1 to obtain a volume flow rate
of I. If the same fluid is then forced through a tube of
the same cross-sectional area, but double the length,
what pressure difference Δp2 must be maintained in
order to observe the same volume flow rate I ?
A.
B.
C.
D.
E.
Δp2 = Δp1
Δp2 = 2 Δp1
Δp2 = 4 Δp1
Δp2 = 8 Δp1
Δp2 = 16 Δp1
3