Download Fluids (Chapter 3,13) The term `fluids` is used to describe both gases

Fluids (Chapter 3,13)
The term 'fluids' is used to describe both gases
and liquids. What they have in common is that
they both spontaneously deform to fit their containers.
However there is a difference:
Gases fill their entire three dimensional container,
while liquids in the presence of gravity deform to
fill the bottom part of the container and form a sharp
, usually flat, surface at the top.
Another difference is that gases are compressible,
as we saw last week. If you put pressure on them,
they shrink. Liquids, by contrast, shrink very
little under pressure.
As an idealization, we mainly consider liquids
which don't shrink at all under pressure (incompressible).
Mass Density
In order to discuss pressure and flow in fluids
we need the concept of mass density ρ of the
fluid, which is defined as
ρ = M/V
For incompressible liquids, the density does not
change as the pressure changes, but for
gases, it does, because V changes, as we have
discussed.
The pressure in my bicycle tire is measured to
be 80 psi (pounds per square inch) . Atmospheric
pressure is 14 .7 psi. The tire diameter is .66 meters
and the crossectional diameter of the tire is .03m.
The density of air at atmospheric pressure and
at the time when the tire pressure was measured
was 1.2 kg/m3 . What is the mass of the air in the tire?
a. 9.58 grams
b. 11.3 grams
c. 4.69 grams
d. 5.65 grams
The densities of the gases are much less than those
of fluids and do not change very much with pressure.
Now we consider pressure in a liquid which is not moving
and is in a container at the surface of the earth
so that gravity is acting on it.
We consider a cylinder drawn
around part of the liquid as
shown.
From the top, the atmosphere
pushes down on the liquid with
pressure po. From the bottom the
rest of the liquid pushes up with
pressure p. Finally, gravity is
pulling the mass of the liquid in
the cylinder down with force
mass x g = Adρg. Force balance
(no acceleration) gives
pA=po A +Ad ρg or p-po=ρgd
Conclusion. The pressure in an incompressible
liquid at depth d at atmospheric
pressure po is po +ρgd where ρ is the mass
density of the liquid.
Notice that the pressure only depends on the
depth. Because we are assuming that the
liquid is not moving, that means that the
pressure is po +ρgd even if there is not a
column of liquid of cylindrical shape right
above some points in the liquid.
That's because the pressure exerts forces IN
ALL DIRECTIONS perpendicular to an area
A of liquid. If the pressure at depth d were
different anywhere then liquid would be
accelerated from the higher pressure to the
lower pressure and the liquid would be flowing,
not stationary.
Bouyancy
Consider a solid object submerged in a liquid.
Assuming that the object does not float, we
measure the apparent weight while the object
is submerged and observe that the weight is
reduced.
The reduction in apparent weight is called the
bouyant force. We can understand how large the
bouyant force is by the following argument originally
due to Archimedes, more than 2000 years ago.
When the solid object
is in place in the liquid
the forces pushing in
on the solid are the
same as they were when
the object was not there.
Therefore they push the
solid object up by a force
equal to the weight of the
displaced liquid.
The dashed line outlines the
place in the liquid where the
object was. The liquid inside
the dashed line is not moving.
Therefore the pressures inward
toward the region outlined add
up to the weight of the liquid
inside the boundary.
Summary:
Archimedes' principle:
The bouyant upward force on an object submerged
or partially submerged in a liquid experiences
an upward bouyant force equal in magnitude
to the weight of the displaced liquid.
The apparent weight of the cylinder which I submerged
was 13N when weighed in air and 9 newtons when
submerged in water. The density of water is
approximately 1gm/cm3 . What is the volume of
the cylindrical weight?
a. .005 m3
b. .510 m3
c. .000510 m3
d. 5 m3
The volume of the cylinder was 5.1 x 10-4 m3 and
the weights were 14 N in air and 9N in water.
What is the mass density ρ of the material in
the cylinder?
a. 1000 gm/cm3
b. 2.8 x 103 gm/cm3
c. 1.00 gm/cm3
d. 2.8 gm/cm3
A disc shaped piece of wood 7.5 cm in diameter and
3 centimeters thick is observed to float in water
with 2cm submerged and 1 cm in the air.
What is the mass density of the wood?
a. 500 kg /m3
b. 667 kg/m3
c. 333 kg/m3
d. 250 kg/m3
Pressure in the atmosphere.
The atmosphere is a gas, not a liquid. It is
compressible. The pressure at a given height
h is determined by the weight of all the gas
above that point (as in a liquid) but the density gets
higher at lower heights because the gas gets more
compressed due to the higher presssure.
h+Δh
h
p(h+Δh)=p(h)-ρgΔh
ρ=mmolep(h)/RT
so p(h+Δh)=p(h)(1-mmolegΔh/RT)
Calculation for an isothermal atmosphere then
gives
120000
100000
pressure (Pa)
80000
60000
P(Pa)
40000
20000
0
0
2000 4000 6000 8000 10000 12000 14000 16000 18000 20000
height (meters)
experimental
data
Note:
The real atmosphere
is not isothermal
Moving liquids:
Equation of continuity (conservation of mass).
Many of the same principles apply to gases but
the discussion will suppose that the fluid is
incompressible (ie an ideal liquid). Suppose a liquid
is flowing steadily through a pipe with 2 diameters
as illustrated:
Think of an imaginary piston, tied to the liquid
and moving with it at 1. It is moving to the
right with the liquid and a time Δt later it
has moved a distance v1Δt and liquid mass
ρ v1Δt A1 has moved in front of it. All along the
pipe, the same amount of mass must move to
the right in Δt because the liquid is incompressi ble.
That means that at point 2, in the narrower part
of the pipe, the same amount of mass must have
moved to the right. So
ρ v1Δt A1 = ρ v2Δt A2 or v1 A1 = v2A2
Summary: when an incompressible liquid flows
through pipes of varying crosssectional area , the
relation
v1 A1 = v2A2
holds as a consequence of the conservation of mass.
This is called the equation of continuity.
Pressure in a flowing liquid:
time t
kinetic energy of this part=
A1v1Δt ρv12/2
potential energy of same part
= A1v1Δt ρ gy1
kinetic energy of this part=
A2v2Δt ρv22/2
potential energy of same part
= A2v2Δt ρ gy2
time t +Δt
work done on the liquid in time Δt =
p1A1 v1Δt- p1A1 v1Δt
ΔK +ΔU=W so
ρv22/2- ρv12/2 +ρ gy2-ρ gy1=p1-p2
Conclusion:
The velocity, elevation and pressure at various
points in the flow of an incompressible
fluid are related by the fact that the quantity
ρv2/2 + ρ gy+ p
remains the same everywhere in the flowing
liquid. (Bernoullis equation).
You are using this in the analysis of a siphon
in Homework 3.