About Midterm Exam 3 l When and where çThurs April 21 th, 5:45-7:00 pm TODAY! çRooms: Same as Exam I and II, See course webpage. çYour TA will give a.

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).

104/18/23 Phys 201, Spring 2011

04/18/23 Phys 201, Spring 2011 2

Physics 201: Lecture 25Fluids (cont’d)

Today’s lecture will coverPascal’s Principle, Archimedes’ Principle revisitFluids in motion: Continuity & Bernoulli’s equation

04/18/23 Phys 201, Spring 2011 3

Archimede’s and Pascal’s Principles

Archimede’s Principle: Buoyant Forceweight 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

A

1

F1

F2

A

1

A

2

04/18/23 Phys 201, Spring 2011 6

Fluid Flow

• 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

Fluid flow without friction

04/18/23 Phys 201, Spring 2011 7

Faucet

A stream of water gets narrower as it falls from a faucet (try it & see).

Explanation: the equation of continuity

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

A1

A2

V1

V2

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.

04/18/23 Phys 201, Spring 2011 8

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. 0.1 v

2. 0.2 v

3. 0.5 v

4. v

5. 2 vF1=F2 ⇒ v1A1=v2A2

A1=1 cm2

A2=20 ×0.5 cm2=10 cm2

v2=v1

A1

A2

=0.1v

04/18/23 Phys 201, Spring 2011 9

Bernoulli’s Equation

Bernoulli’s Equation

04/18/23 Phys 201, Spring 2011 10

Bernoulli’s Equation Pressure drops in a rapidly moving fluid

whether or not the fluid is confined to a tube For incompressible, frictionless fluid:

€

P +1

2ρv 2 + ρgh = constant

1

2ρv 2 =

1

2mv 2 1

V=KE

V

ρgh =mgh

V=PE

VW = Fx = PAx = PV

W = ΔE = ΔKE + ΔPE

Bernoulli equation states conservation of energy

04/18/23 Phys 201, Spring 2011 11

(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.

Applications of Bernoulli’s PrincipleWings 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

04/18/23 Phys 201, Spring 2011 12

(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.

€

v1 =F1A1

=40 ×10−3m3/s

π (0.045)2= 6.29 m/s

v2 =F2A2

=40 ×10−3m3/s

π (0.015)2= 56.6 m/s

P1 − P2 =1

2ρ(v2

2 − v12) =1.58 ×106 N/m2

h =v 2

2g=

(56.6)2

2 × 9.8m =163 m

Pressure drop

04/18/23 Phys 201, Spring 2011 13

Velocity Measurement: Pitot tube

Dead spot, v1= 0

∴P1= P2+ 12ρv2

2 , but, P1- P2 ∝ h

h∝12ρv2

2 , or wind velocity, v2 ∝ h

Two openings at 1 and 2:

04/18/23 Phys 201, Spring 2011 14

Torricelli’s Theorem

Bernoulli's equation at constant pressure (P1 =P2 )

P1 +12ρv1

2 +ρgh1 =P2 +12ρv2

2 +ρgh2

12ρv1

2 +ρgh1 =12ρv2

2 +ρgh2

v22 =v1

2 + 2g(h1 −h2 )Same as kinematics equation for any object falling h=h1 −h2 with negligible friction.

h

P2=P1 , v2 , h2

P1, v1, h1