physics 207: lecture 21, pg 1 lecture 21 goals: chapter 15 chapter 15 understand pressure in...
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Physics 207: Lecture 21, Pg 1
Lecture 21
Goals:Goals:
• Chapter 15Chapter 15 Understand pressure in liquids and gases Use Archimedes’ principle to understand buoyancy Understand the equation of continuity Use an ideal-fluid model to study fluid flow. Investigate the elastic deformation of solids and liquids
• AssignmentAssignment HW9, Due Wednesday, Apr. 8th Thursday: Read all of Chapter 16
Physics 207: Lecture 21, Pg 2
In many fluids the bulk modulus is such that we can treat the density as a constant independent of pressure:
An incompressible fluid
For an incompressible fluid, the density is the same everywhere, but the pressure is NOT!
p(y) = p0 - y g p0 + d g Gauge pressure (subtract p0,
pressure 1 atm, i.e. car tires)
Pressure vs. DepthIncompressible Fluids (liquids)
y1 y2
Ap
1
p2
F1
F2
mg
0p
F2 = F1+ m g = F1+ VgF2 /A = F1/A + Vg/Ap2 = p1 - g y
Physics 207: Lecture 21, Pg 3
Pressure vs. Depth
For a uniform fluid in an open container pressure same at a given depth independent of the container
p(y)
y
Fluid level is the same everywhere in a connected container, assuming no surface forces
Physics 207: Lecture 21, Pg 4
Pressure Measurements: Barometer Invented by Torricelli A long closed tube is filled with mercury
and inverted in a dish of mercury
The closed end is nearly a vacuum
Measures atmospheric pressure as
1 atm = 0.760 m (of Hg)
Physics 207: Lecture 21, Pg 5
Exercise Pressure
What happens with two fluids??
Consider a U tube containing liquids of density 1 and 2 as shown:
At the red arrow the pressure must be the same on either side. 1 x = 2 (d1+ y)
Compare the densities of the liquids:
(A) 1 < 2 (B) 1 = 2 (C) 1 > 2
1
2dI
y
Physics 207: Lecture 21, Pg 6
Archimedes’ Principle: A Eureka Moment
Suppose we weigh an object in air (1) and in water (2).
How do these weights compare?
W2?W1
W1 < W2 W1 = W2W1 > W2
Buoyant force is equal to the weight of the fluid displaced
Physics 207: Lecture 21, Pg 8
Archimedes’ Principle
Suppose we weigh an object in air (1) and in water (2).
How do these weights compare?
W2?W1
W1 < W2 W1 = W2W1 > W2
Why?
Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.
Physics 207: Lecture 21, Pg 9
Sink or Float?
The buoyant force is equal to the weight of the liquid that is displaced.
If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink.
F mgB
y
We can calculate how much of a floating object will be submerged in the liquid:
Object is in equilibrium mgFB
objectobjectliquidliquid VgVg
liquid
object
object
liquid
V
V
Physics 207: Lecture 21, Pg 10
Bar Trick
piece of rockon top of ice
What happens to the water level when the ice melts?
A. It rises B. It stays the same C. It drops
Expt. 1 Expt. 2
Physics 207: Lecture 21, Pg 11
Exercise
V1 = V2 = V3 = V4 = V5
m1 < m2 < m3 < m4 < m5
What is the final position of each block?
Physics 207: Lecture 21, Pg 12
Exercise
V1 = V2 = V3 = V4 = V5
m1 < m2 < m3 < m4 < m5
What is the final position of each block?
Not this But this
Physics 207: Lecture 21, Pg 18
Pascal’s Principle So far we have discovered (using Newton’s Laws):
Pressure depends on depth: p = g y Pascal’s Principle addresses how a change in pressure is
transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel.
Physics 207: Lecture 21, Pg 19
Pascal’s Principle in action: Hydraulics, a force amplifier
Consider the system shown: A downward force F1 is applied
to the piston of area A1.
This force is transmitted through the liquid to create an upward force F2.
Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid.
F F
1
2
d
2d
1
A A21
F2 > F1 with conservation of energy
Physics 207: Lecture 21, Pg 20
A1 A10
A2 A10
M
MdB
dA
Exercise
Consider the systems shown on right. In each case, a block of mass M is
placed on the piston of the large cylinder, resulting in a difference di in the liquid levels.
If A2 = 2A1, how do dA and dB compare?
V10 = V1 = V2
dA A1 = dB A2
dA A1 = dB 2A1
dA = dB 2
Physics 207: Lecture 21, Pg 22
Fluids in Motion
To describe fluid motion, we need something that describes flow: Velocity v
There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal
Physics 207: Lecture 21, Pg 23
Types of Fluid Flow Laminar flow
Each particle of the fluid follows a smooth path
The paths of the different particles never cross each other
The path taken by the particles is called a streamline
Turbulent flow An irregular flow
characterized by small whirlpool like regions
Turbulent flow occurs when the particles go above some critical speed
Physics 207: Lecture 21, Pg 24
Types of Fluid Flow Laminar flow
Each particle of the fluid follows a smooth path
The paths of the different particles never cross each other
The path taken by the particles is called a streamline
Turbulent flow An irregular flow
characterized by small whirlpool like regions
Turbulent flow occurs when the particles go above some critical speed
Physics 207: Lecture 21, Pg 25
Flow obeys continuity equation
Volume flow rate (m3/s) Q = A·v is constant along flow tube.
Follows from mass conservation if flow is incompressible.
Mass flow rate is just Q (kg/s)
A1v1 = A2v2
Ideal Fluids Streamlines do not meet or cross
Velocity vector is tangent to streamline
Volume of fluid follows a tube of flow bounded by streamlines
Streamline density is proportional to velocity
A1
A2
v1
v2
Physics 207: Lecture 21, Pg 26
Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?
Exercise Continuity
A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house.
v1 v1/2
(A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1
Physics 207: Lecture 21, Pg 27
For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast.
But if the water is moving 4 times as fast then it has
16 times as much kinetic energy.
Something must be doing work on the water
(the pressure drops at the neck) and we recast the work as
P V = (F/A) (A x) = F x
Exercise Continuity
v1 v1/2
(A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1
Physics 207: Lecture 21, Pg 29
W = (P1– P2 ) V and
W = ½ m v22 – ½ m v1
2
= ½ (V) v22 – ½ (V)v1
2
(P1– P2 ) = ½ v22 – ½ v1
2
P1+ ½ v12 = P2+ ½ v2
2 = const.
and with height variations:Bernoulli Equation P1+ ½ v1
2 + g y1 = constant
Conservation of Energy for Ideal Fluid
P1 P2
Physics 207: Lecture 21, Pg 30
Lecture 21
• Question to ponder: Does heavy water (DQuestion to ponder: Does heavy water (D22O) ice sink O) ice sink or float?or float?
• AssignmentAssignment HW9, due Wednesday, Apr. 8th Thursday: Read all of Chapter 16