Download Lec. 36 notes - High Energy Physics

Fluids, Continuity, and Bernouli
Announcements:
•  Exam Tomorrow at 7:30pm in
same rooms as before.
Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Clicker question 1
Set frequency to BA
A satellite, mass m, in uniform circular orbit around the earth, at a steady
height H above the surface of the earth. At time t=0, it is above Manhattan,
traveling northeast at 17,000 mi/hr. At this moment, what is the direction
of the acceleration of the satellite?
A) There is no “direction”, because the acceleration is zero.
B) Straight down (towards the center of the earth.)
C) Straight up (directly away from the center of the earth.)
D) In the same direction as the velocity, northeast.
E) Some other direction, significantly different from all the above answers.
Clicker question 1
Set frequency to BA
A satellite, mass m, in uniform circular orbit around the earth, at a steady
height H above the surface of the earth. At time t=0, it is above Manhattan,
traveling northeast at 17,000 mi/hr. At this moment, what is the direction
of the acceleration of the satellite?
A) There is no “direction”, because the acceleration is zero.
B) Straight down (towards the center of the earth.)
C) Straight up (directly away from the center of the earth.)
D) In the same direction as the velocity, northeast.
E) Some other direction, significantly different from all the above answers.
In uniform circular motion we always have
centripetal acceleration towards the center of the
circle.
Clicker question 2
Set frequency to BA
A satellite, mass m, in uniform circular orbit around the
earth, at a steady height H above the surface of the earth.
Clicker question 2
Set frequency to BA
A satellite, mass m, in uniform circular orbit around the
earth, at a steady height H above the surface of the earth.
Clicker question 3
Set frequency to BA
The magnitude of the buoyant force on an object which is fully
immersed in a liquid depends on…
A. The mass of the object.
B. The depth of the object.
C. The density of the object.
D. The volume of the object.
E. None of the above.
Clicker question 3
Set frequency to BA
The magnitude of the buoyant force on an object which is fully
immersed in a liquid depends on…
A. The mass of the object.
The buoyant force is always
B. The depth of the object.
equal to the weight of the liquid
C. The density of the object. displaced by the object.
D. The volume of the object.
E. None of the above.
For a fully immersed object, the weight of the displaced water is
the density of the water times the volume of the object times g.
That is, the volume of the displace water is equal to
the volume of the object:
so
.
Clicker question 4
Set frequency to BA
A solid piece of plastic of volume V, and density ρplastic is
floating in a cup of water. (The density of water is ρwater.)
What is the buoyant force on the plastic?
A. 0
B. ρplasticV
C. ρwaterV
D. ρplasticVg
E.  ρwaterVg
Clicker question 4
Set frequency to BA
A solid piece of plastic of volume V, and density ρplastic is
floating in a cup of water. (The density of water is ρwater.)
What is the buoyant force on the plastic?
A. 0
Since the plastic is floating, some of it
B. ρplasticV
is out of the water and so the volume
C. ρwaterV
of displaced water is less than the
D. ρplasticVg
volume of the plastic object.
E.  ρwaterVg
Since the object is in equilibrium, the net force is zero.
Therefore, the object weight must be exactly balanced by the
buoyant force. The object weight is W = mplasticg = ρplasticVg.
Can also use this to figure out the volume of displaced water.
Archimedes principle
A body partially or fully immersed in a fluid feels an
upward force equal to the weight of the displaced fluid.
This force is called the buoyant force:
As shown, it is due to the increase of pressure with depth in a fluid.
If the object is fully immersed then the volume of the displaced
fluid is equal to the volume of the object:
Note that volume is related to mass and density:
If an object is only partially submerged, the volume of the
displaced fluid is less than the volume of the object:
Buoyancy example
A 2 cm by 2 cm by 2 cm cube of iron (ρ=8 g/cm3) is weighed
with the iron outside, half in and fully in the water, as shown
in the diagram. What is the measured weight in each case?
Iron mass:
so
Out of the
water:
In the
water:
so
so
Archimedes used this principle to determine whether
King Hiero’s gold crown had any silver mixed in it
Mercury barometer
Flip a tube filled with mercury upside down in a bucket of mercury.
Some mercury will run out, creating a vacuum in the top.
Let’s examine the pressure at the dotted line.
Outside the tube, pressure is atmospheric:
Inside the tube, pressure is
These two are at the same height so must be equal:
Atmospheric pressure holds up the column of mercury
The height of the column measures atmospheric pressure
1 atm = 105 Pa = 760 mmHg = 760 torr = 30 inHg = 14.7 psi
Atmospheric pressure pushes! The vacuum does not pull!
Clicker question 5
Set frequency to BA
What happens when the air is evacuated from a 55 gallon drum?
A. Nothing
B. Squashed like stepping on a soda can
C. Implodes from all sides
D. Explodes
Atmospheric pressure is 14.7 psi = 14.7 lb/in2.
Clicker question 5
Set frequency to BA
What happens when the air is evacuated from a 55 gallon drum?
A. Nothing
B. Squashed like stepping on a soda can
C. Implodes from all sides
D. Explodes
I will use English units for this exercise.
Atmospheric pressure is 14.7 psi = 14.7 lb/in2.
What is the force on the top due to the atmosphere?
This is the weight of an H2 Hummer!
Usually have the same air pressure underneath the top so
the net force is 0. Applies to the sides and bottom as well.
What happens when that air pressure is removed?
Fluid flow
So far, all of our fluid studies have dealt with static situations
(the fluid was in static equilibrium).
Now we will investigate flowing fluids.
We only consider ideal fluids. Ideal fluids…
are incompressible – good assumption
for liquids, not for gases.
are nonviscous – no kinetic friction
between liquid parts
have laminar flow – steady flow with
no turbulence
Streamlines
Consider water flowing through a tube.
One can imagine paths taken
by a particle in the fluid.
These paths are streamlines.
In an ideal fluid, streamlines have two important properties:
1. Streamlines never cross.
2. Fluid speed is higher where
streamlines are closer together.
Continuity equation
For a steady flow, the amount of
water entering point 1 in time Δt
1
must be equal to the amount of
water exiting point 2 in time Δt.
How much fluid goes through points 1 and 2 in time Δt?
An area A traveling at a speed
of v will travel d = vΔt in time Δt.
For steady flow, the two volumes A1d1
& A2d2 must be equal:
Therefore:
A1
d1 = v1Δt
2
A2
d2 = v2Δt
This is a continuity equation which says whatever goes
in must come out; water is not created or lost inside.
Clicker question 6
Set frequency to BA
The end of a hose has a diameter of 4 cm. If one wants the velocity
of the water coming out to be 4 times higher, what should be the
diameter of the nozzle on the end?
A. ¼ cm
B. ½ cm
C. 1 cm
D. 2 cm
E. 4 cm
Clicker question 6
Set frequency to BA
The end of a hose has a diameter of 4 cm. If one wants the velocity
of the water coming out to be 4 times higher, what should be the
diameter of the nozzle on the end?
A. ¼ cm
Solving the continuity equation
B. ½ cm
for the unknown final area gives us:
C. 1 cm
D. 2 cm
If we want v2/v1 = 4 then we need A2 = A1/4.
E. 4 cm
Since area depends on the diameter squared, the diameter needs
to be reduced by a factor of 2 to get a factor of 4 reduction in area.
In equations:
About the continuity equation
The continuity equation says that the same amount
(volume) of liquid goes in as comes out. Note that
area times velocity (Av) has units of volume/second.
Thus, knowing this could tell us how quickly a
vessel of a given volume may be filled or emptied.
Continuity equations show up where there is a
conserved current. Here it is liquid but next
semester it will be electrical current.
Bernoulli’s equation
Our general conservation of energy equation looks like
or
We consider each term with regards to fluid flow starting with Wother.
Consider two equal volumes V of
p2
p
1
water entering and leaving a tube.
A2
A1
The water entering feels a force from
d2
the pressure of p1A1 over a distance of
d1
d1 so the work done is W=p1A1d1=p1V
The water exiting feels a force of p2A2 over a distance of d2 but
force and displacement are in opposite directions so the work
done is negative: W = –p2A2d2 = –p2V
The net work done by pressure is W = p1V – p2V
Bernoulli’s equation
We are dealing with the same volumes of water: V = A1d1 = A2d2.
Gravitational potential energy is mgy.
A2
For a volume V of liquid m = ρV
So at points 1 and 2 we have
and
Kinetic energy is ½mv2 and m = ρV
so
and
d2
A1
y2
y1
d1
Now that we have all three components, we can combine them.
Bernoulli’s equation
Conservation of energy equation:
The net work done by pressure is
Gravitational potential energy:
Kinetic energy:
and
and
Conservation of energy:
Dividing through by the volume V and rearranging gives
Bernoulli’s equation:
Bernoulli’s equation is a restatement of conservation of energy!
Problems often require both Bernoulli & continuity equations.