Download Fluids Wrap up - Ms. Gamm

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Lift (force) wikipedia , lookup

Classical central-force problem wikipedia , lookup

Biofluid dynamics wikipedia , lookup

Reynolds number wikipedia , lookup

Mass versus weight wikipedia , lookup

Pressure wikipedia , lookup

Fluid dynamics wikipedia , lookup

Bernoulli's principle wikipedia , lookup

Buoyancy wikipedia , lookup

Transcript
AP Physics – Fluids Wrap Up
Here are the equations that you get to play around with:
p  p0   gh
This is the equation for the pressure of something as a function of depth in a fluid.
You would use it to figure out the pressure acting at a depth of 25.0 m in a lake for
example.
FBuoy  Vg
This is the equation for the buoyant force. It is also an equation that calculates the
weight of an object as a function of its density, volume, and the acceleration of
gravity.
A1v1  A2v2
This equation represents flow rate, which is the cross sectional area, A, multiplied by
the velocity of the fluid, v. This is set up for two locations in a flow system. The
flow rate for a fluid that is incompressible must stay constant, so this equation allows
you to calculate the linear speed of the fluid as a function of the cross sectional area
of the system.
1
p   gy   v 2  const.
2
This is a Bernoulli’s equation. This allows you to calculate pressure, linear speed,
&tc. For a system at different places within the system.
Here is the stuff you need to be able to do.
A. Fluid Mechanics
1. Hydrostatic pressure
a) You should understand that a fluid exerts pressure in all directions.
This is basic. For example, atmospheric pressure goes in all directions about an object –
under it, over it, on the sides, &tc. Good old Pascal’s Principle.
b) You should understand that a fluid at rest exerts pressure perpendicular to any surface
that it contacts.
This is also an application of Pascal’s principle. The pressure is everywhere throughout
the liquid. The direction of the force acting on a surface is always perpendicular to the
surface.
680
c) You should understand and be able to use the relationship between pressure and depth
in a liquid, p   g h .
p  p0   gh
where p0 would be some
initial pressure. We did a bunch of these problems. Guage pressure is based on the idea that the
atmospheric pressure is zero pressure. Absolute pressure uses a perfect vacuum – zero pascals –
as its zero pressure. So guage pressure differs from absolute pressure by one atmosphere. The
pressure at a certain depth would be give by p  p0   gh . For an absolute pressure you
The actual equation that is provided you is
would set
term.
p0 equal to the atmospheric pressure.
For a gage pressure you would drop the
p0
2. Buoyancy
a) You should understand that the difference in the pressure on the upper and lower
surfaces of an object immersed in a liquid results in an upward force on the object.
We went through this when the Physics Kahuna derived the buoyancy equation for you.
Because the pressure depends on depth, the pressure increases with the depth. So if the
top of a regular object is 10 m below the surface and the bottom of it is 15 m below – five
meters deeper, the force, which is pressure times area, must be greater. Thus there is a
larger force pushing up on the bottom of the body than the pressure pushing down on the
top of the body. The net force is upward and is given the name of ‘buoyant’ force.
b) You should understand and be able to apply Archimedes’ principle; the buoyant force on
a submersed object is equal to the weight of the liquid it displaces.
Well, the statement gives you Archimedes’ principle and tells you to understand it. So do
that.
3. Fluid flow continuity
a) You should understand that for laminar flow, the flow rate of a liquid through its cross
section is the same at any point along its path.
So okay, do that too.
b) You understand and be able to apply the equation of continuity,
1 A1v1  2 A2v2 .
Actually the equation that you are given is: A1v1  A2v2 the density part isn’t in the
equation. This is because in the type of problem that you’ll be doing, the density won’t change
and will remain constant. Because of that, it cancels out of the equation. The Physics Kahuna is
not at all sure why statement b) above had a different form of the equation. Probably some
miscommunication at the College Board. Anyway, we did a bunch of problems where you
got to use the equation. It is all pie.
681
4. Bernoulli’s equation
a) You should understand that the pressure of a flowing liquid is low where the velocity is
high, and vice versa.
Simple principle, simple stuff. Hey you can do it!
b) You should understand and be able to apply Bernoulli’s equation,
1
p   gy   v 2  const.
2
The Physics Kahuna is not sure what sort of questions you can expect. He provided you
with several of them, but is only guessing. So it goes.
Well. That’s it for what you need to know.
In Flanders Fields
In Flanders field the poppies grow
Between the crosses, row on row,
That mark our place; and in the sky
The larks, still bravely singing fly
Scarce heard amid the guns below.
We are the Dead. Short days ago
We lived, felt dawn, saw sunset glow,
Loved and were loved, and now we lie
In Flanders fields.
Take up our quarrel with the foe;
To you from failing hands we throw
The torch; be yours to hold it high.
If ye break faith with us who die
We shall not sleep, though poppies grow
In Flanders fields.
--- LtCol John McCrae
John McCrae, a noted poet, was a member of the first Canadian
Contingent and served on the front lines for four years in France
during the Great War (as it was called until WW II). He was killed in
combat on 28 January, 1918.
682
From 2002:

In the laboratory, your are given a cylindrical beaker containing a fluid and you are asked to
determine the density  of the fluid. You are to use a spring of negligible mass and unknown
spring constant k attached to a stand. An irregularly shaped object of known mass m and
density D (D >>  ) hangs from the spring. You may also choose from among the following
items to complete the task.



A metric ruler
A stopwatch
String
(a) Explain how you could experimentally determine the spring constant k.
The mass of the weight is known, suspend the mass from the spring in air, measure the
displacement of the spring and calculate k from the equation Fs   kx where Fs is the
mg, the weight of the thing.
(b) The spring-object system is now arranged so that the object (but not the spring) is immersed
in the unknown fluid, as shown above. Describe any changes that are observed in the
spring-object system and explain why they occur.
The mass will have less weight in the fluid because of the buoyant force. It will decrease
by the amount of the force which is Fbuoy  Vg
(c) Explain how you could experimentally determine the density of the fluid.
Knowing k we can calculate the apparent weight of the object in the fluid. The difference
between its weight in the air and in the fluid will equal the buoyant force.
The volume of the object could be calculated using the equation for density:

m
V
The volume of fluid displaced will be the same as the volume of the object.
683
Knowing the buoyant force, we can use the buoyant force to calculate the density of the
fluid.
Fbuoy  Vg

FBuoy
Vg
(d) Show explicitly, using equations, how you will use your measurements to calculate the fluid
density  . Start by identifying any symbols you use in your equations.
Symbol
Physical quantity
Fs
This is the force that stretches the spring. In air, it will be the
weight of the object.
k
The spring constant
x
The spring displacement
Fg
m
The weight of the object
g
The acceleration of gravity
Fbuoy
The buoyant force

The density of the fluid
V
Volume of fluid displaced by the object
g
Acceleration of gravity
The mass of the object
Measure the displacement of the spring by the object in air. Calculate the weight of the object
using Fg  mg . Using this weight, calculate the spring constant from Fs   kx . Calculate
the volume of the object using the equation for density. This will be the same as the volume of
the fluid displaced.
Note the spring displacement. From this calculate the weight of the object in the fluid using the
buoyant force equation. The difference in the two forces is the buoyant force. Using the volume
displaced, the buoyant force, and the acceleration of gravity, calculate the density of the fluid.
(Using the buoyant force equation.)
So there it is. Your’re all set fluid mechanicswise.
684
FLUID CONCEPTS






















Matter is made up of atoms. Every atom of an element is alike.
We can model atoms as tiny, hard “billiard balls.”
In a substance, atoms are combined to form molecules.
When atoms or molecules collide, they collide elastically (no loss of kinetic energy).
There are three forms of matter that we encounter in this class: solids, liquids, and gases. In all
cases, the atoms or molecules are in constant motion in a material.
In a solid, the average position of the atoms or molecules remains constant. That is why solids
tend to maintain their shape, in general.
In a liquid, the atoms or molecules are free to move around, but the distance between atoms or
molecules remains relatively small. For this reason, a liquid takes the shape of its container, but
does not expand to fill it.
The molecules of a liquid feel a mutual attraction. This creates surface tension, which acts
something like a “skin” on the surface of the liquid.
In a gas, the molecules are widely separated and feel little mutual attraction. This is why a gas
tends to expand to take the shape of its container.
We will treat liquids as incompressible. It is true that the compressibility for liquids in the
situations we consider is negligible.
Pressure is force per unit measure of area. The SI unit of pressure is the Pascal (Pa) or N/m2.
Since a fluid is not rigid, it can only exert a force perpendicular to a surface. Picture the
particles as little balls colliding with the surface, as in when you bounce a ball off a wall. Can
you see that the direction of the velocity change is always perpendicular to the wall? (Try
drawing it).
A fluid has weight. This is the reason why it exerts a pressure.
The pressure of a fluid is proportional to the density of the fluid, the gravitation constant (for
earth, average g=9.8 N/kg), and the depth of the fluid. Thus the pressure exerted by a fluid only
varies with the depth, not with the volume, the shape of the container, or any other factors.
The atmosphere exerts considerable pressure on objects at the surface of the earth.
Many simple devices depend on atmospheric pressure: drinking straws, suction cups, mercury
barometers. Be sure you can explain how these and others work using the concepts in this
section.
Gauges measure atmospheric pressure as 0 Pa. Atmospheric pressure must be added to gauge
pressure to give the actual pressure (called “absolute pressure”).
Since a liquid is incompressible, a pressure exerted on it is transmitted throughout the liquid
(Pascal’s Principle).
An object floats b/c it displaces a volume of water whose weight equals the object’s weight
(Archimedes’ Principle). The sum of the vertical forces is zero. Alternate version: an object
floats b/c its average density is less than the surrounding fluid. Second alternate version: an
object floats b/c the pressure at the bottom of the object is higher than the pressure at the top (the
resulting force must be equal to the weight of the object).
We will treat only fluids that flow in thin sheets, i.e. laminar flow. The thin sheets can be
modelled in two dimensions as streamlines.
Fluids flow from high pressure to low pressure.
Because a fluid is incompressible, the volume flow rate is constant, even if the diameter of the
pipe changes (Continuity Principle). The product of Area and velocity of a fluid is a constant for
685



a particular pipe. This means the fluid speeds up in a constriction and slows down in a wider
part of the pipe.
Bernoulli made a conservation of energy statement for a fluid in laminar flow: the energy per
volume is constant in a pipe (Bernoulli’s Principle). This is true even if the diameter of the pipe
and therefore the velocity change.
Fast-moving fluids are at lower pressure than slow-moving fluids (Venturi Effect). This is a
consequence of Bernoulli’s Principle (i.e., conservation of energy).
Airplanes fly b/c of Bernoulli’s Principle and the momentum change of the air (due to the
“attack angle” of the wing). Can you explain this?
"On Les Aura!" -- Soldat Jacques Bonhomme loquitur:
See you that stretch of shell-torn mud spotted with pools of mire,
Crossed by a burst abandoned trench and tortured strands of wire,
Where splintered pickets reel and sag and leprous trench-rats play,
That scour the Devil's hunting-ground to seek their carrion prey?
That is the field my father loved, the field that once was mine,
The land I nursed for my child's child as my fathers did long syne.
See there a mound of powdered stones, all flattened, smashed, and torn,
Gone black with damp and green with slime? -- Ere you and I were born
My father's father built a house, a little house and bare,
And there I brought my woman home -- that heap of rubble there!
The soil of France! Fat fields and green that bred my blood and bone!
Each wound that scars my bosom's pride burns deeper than my own.
But yet there is one thing to say -- one thing that pays for all,
Whatever lot our bodies know, whatever fate befall,
We hold the line! We hold it still! My fields are No Man's Land,
But the good God is debonair and holds us by the hand.
On les aura!" See there! and there! soaked heaps of huddled grey!
My fields shall laugh -- enriched by those who sought them for a prey.
James H. Knight-Adkin
686