Transcript Lect26

About Midterm Exam 3




When and where
Thurs April 21th , 5:45-7:00 pm  TODAY!
Rooms: Same as Exam I and II, See course webpage.
Your TA will give a brief review during the discussion session.
Coverage: Chapts 9 – 12 (4 chapters)
Format
Closed book, 20 multiple-choices questions (format as in practice exams)
1 page 8x11 formula sheet allowed, must be self prepared, no photo
copying/download-printing of solutions, lecture slides, etc.
Bring a calculator (but no lap-top computer). Only basic calculation
functionality
can be used. Bring a 2B pencil for Scantron.
Fill in your ID and section # !
Special requests:
If different from Exam II, email me at [email protected]
One alternative exam: 3:30pm – 4:45pm, Thurs April 21, Cham 5280
(as before).
4/13/2015
Phys 201, Spring 2011
1
Physics 201: Lecture 25
Fluids (cont’d)

4/13/2015
Today’s lecture will cover
Pascal’s Principle, Archimedes’ Principle revisit
Fluids in motion: Continuity & Bernoulli’s equation
Phys 201, Spring 2011
2
Archimede’s and Pascal’s Principles
A change in pressure
F1
applied to an enclosed fluid is transmitted
undiminished to all portions of the fluid and to
the walls of its container.
A
A
 This principle is used in hydraulic system
11
P1 = P2
(F1 / A1) = (F2 / A2)
Archimede’s Principle: Buoyant Force
weight of fluid displaced
» B = ρfluid g Vdisplaced, W = ρobject g Vobject
» object sinks if ρobject > ρfluid
» object floats if ρobject < ρfluid
If object stays floating: B=W
» Therefore ρfluid g Vdisplaced = ρobject g Vobject
» Therefore Vdisplaced/Vobject = ρobject / ρfluid
4/13/2015
Phys 201, Spring 2011
F2
A
2
3
Fluid Flow
Fluid flow without friction
• Volume flow rate: ΔV/Δt = A Δd/Δt = Av (m3/s)
• No source, no sink
Continuity: A1 v1 = A2 v2 = Flow rate
i.e., flow rate the same everywhere
e.g., flow of river: more slowly in wider area
Water through a narrow hose moves faster
4/13/2015
Phys 201, Spring 2011
6
Faucet
A stream of water gets narrower as it falls
from a faucet (try it & see).
Explanation: the equation of continuity
A1
V1
V2
A2
The velocity of the liquid increases as the water falls
due to gravity. If the volume flow rate is conserved,
them the cross-sectional area must decrease in order
to compensate
The density of the water is the same no matter where
it is in space and time, so as it falls down and
accelerates because of gravity,the water is in a sense
stretched, so it thins out at the end.
4/13/2015
Phys 201, Spring 2011
7
An artery with cross sectional area of 1 cm2 branches into 20
smaller arteries each with 0.5 cm2 cross sectional area. If the
velocity of blood in thicker artery is v, what is the velocity of
the blood in the thinner arteries?
1.
2.
3.
4.
5.
0.1 v
0.2 v
0.5 v
v
2 v
F1 =F2  v1A1 =v2 A 2
A1 =1 cm 2
A 2 =20  0.5 cm 2 =10 cm 2
A1
v2 =v1
=0.1v
A2
4/13/2015
Phys 201, Spring 2011
8
Bernoulli’s Equation
Bernoulli’s Equation
4/13/2015
Phys 201, Spring 2011
9
Bernoulli’s Equation


Pressure drops in a rapidly moving fluid
whether or not the fluid is confined to a tube
For incompressible, frictionless fluid:
1 2
P  v  gh  constant
2
1 2 1 2 1 KE
v  m v 
2
2
V
V
m gh PE
gh 

V
V
W  Fx  PAx  PV
W  E  KE  PE
Bernoulli equation states conservation of energy
4/13/2015
Phys 201, Spring 2011
10
Applications of Bernoulli’s Principle
Wings and sails
Higher velocity on one side of sail or wing
versus the other results in a pressure
difference that can even allow the boat to
sail into the wind
(a)
Calculate the approximate force on a square meter
of sail, given the horizontal velocity of the wind is
6 m/s parallel to its front surface and 3.5 m/s along
its back surface. Take the density of air to be
1.29 kg/m3. (b) Discuss whether this force is great
enough to be effective for propelling a sail boat.
1
(v22  v12 )A  15.3 N
2
The force is small. However, when the sails are large,
the force can be high enough to propel a sail boat.
For larger boats, one can add more than one sail to
Force, F  (P1  P2 )A 
increase the surface area.
One can even sail into the wind, where (P1  P2 ) is small.
4/13/2015
Phys 201, Spring 2011
11
Pressure drop
(a) What is the pressure drop due to Bernoulli effect as water goes
into a 3 cm diameter nozzle from a 9 cm diameter fire hose while
carrying a flow of 40 L/s? (b) To what maximum height above the
nozzle can this water rise neglecting air resistance.
F1 40 103 m3 /s
v1 

 6.29 m/s
2
A1
 (0.045)
F2 40 103 m3 /s
v2 

 56.6 m/s
A2
 (0.015) 2
1
P1  P2  (v 22  v12 )  1.58 106 N/m2
2
v 2 (56.6) 2
h

m = 163 m
2g 2  9.8
4/13/2015
Phys 201, Spring 2011
12
Velocity Measurement: Pitot tube
Two openings at 1 and 2:
Dead spot, v1 = 0
1 2
 P1 = P2 + v2 , but, P1 - P2  h
2
1 2
h  v2 , or wind velocity, v2  h
2
4/13/2015
Phys 201, Spring 2011
13
Torricelli’s Theorem
P1, v1, h1
h
P2=P1 , v2 , h2
Bernoulli's equation at constant pressure (P1  P2 )
1
1
P1  v12  gh1  P2  v22  gh 2
2
2
1 2
1 2
v1  gh1  v2  gh 2
2
2
v22  v12  2g(h1  h 2 )
Same as kinematics equation for any object falling
h  h1  h 2 with negligible friction.
4/13/2015
Phys 201, Spring 2011
14