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http://www.physics.usyd.edu.au/teach_res/jp/fluids09

web notes: lect6.ppt

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Ideal fluid

Real fluid

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(1) Surface of liquid

(2) Just outside hole

v2 = ? m.s-1

y1

y2

Draw flow tubes

v1 ~ 0 m.s-1 p1 = patm

p2 = patm

h = (y1 - y2)

What is the speed with which liquid flows from a hole at the bottom of a tank?

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Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied

p1 + ½ v12 + g y1 = p2 + ½ v2

2 + g y2

A small hole is at level (2) and the water level at (1) drops slowly v1 = 0

p1 = patm p2 = patm

g y1 = ½ v22 + g y2

v22 = 2 g (y1 – y2) = 2 g h h = (y1 - y2)

v2 = (2 g h) Torricelli formula (1608 – 1647)

This is the same velocity as a particle falling freely through a height h

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(1)

(2)

F

m

h

v1 = ?

What is the speed of flow in section 1 of the system?

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Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied for the flow along a streamline

p1 + ½ v12 + g y1 = p2 + ½ v2

2 + g y2

y1 = y2

p1 – p2 = ½ F (v22 - v1

2)

p1 - p2 = m g h

A1 v1 = A2 v2 v2 = v1 (A1 / A2)

m g h = ½ F { v12 (A1 / A2)

2- v12 } = ½ F v1

2 {(A1 / A2)

2 - 1}

m

1 21

F2

2

1

g hv

A

A

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How does a siphon work?

Q: What do we know?

Continuous flow

Pressure in top section > 0 otherwise there will be a vacuum pC 0

Focus on falling water not rising water patm - pC 0 patm g yC

yC

pC

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C

B

A

D

yA

yB

yC

Assume that the liquid behaves as an ideal fluid, the equation of continuity and Bernoulli's equation can be used.

yD = 0

pA = patm = pD

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pC + ½ vC2 + g yC = pD + ½ vD

2 + g yD

From equation of continuity vC = vD

pC = pD + g (yD - yC) = patm + g (yD - yC) The pressure at point C can not be negativepC 0 and yD = 0pC = patm - g yC 0 yC patm / ( g)

For a water siphonpatm ~ 105 Pa g ~ 10 m.s-1 ~ 103 kg.m-3

yC 105 / {(10)(103)} m yC 10 m

Consider points C and D and apply Bernoulli's principle.

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pA + ½ vA2 + g yA = pD + ½ vD

2 + g yD

vD2 = 2 (pA – pD) / + vA

2 + 2 g (yA - yD)

pA – pD = 0 yD = 0 assume vA2 << vD

2

vD = (2 g yA )

How fast does the liquid come out?

Consider a point A on the surface of the liquid in the container and the outlet point D.Apply Bernoulli's principle

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FLUID FLOW MOTION OF OBJECTS IN FLUIDS

How can a plane fly?

Why does a cricket ball swing or a baseball curve?

Why does a golf ball have dimples?

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Lift FL

drag FD

Resultant FR

Motion of object through fluid

Fluid moving around stationary object

FORCES ACTING ON OBJECT MOVING THROUGH FLUID

Forward thrust by engine

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C

D

BA

Uniform motion of an object through an ideal fluid ( = 0)

The pattern is symmetrical

FR = 0

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Drag force

frictional drag (viscosity)

pressure drag (eddies – lower pressure)

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low pressure region

high pressure region

rotational KE of eddies heating effect increase in internal energy temperature increases

Drag force dueto pressure difference

NO CURVEDrag force is opposite to the direction of motion

motion of air

motion of object

16low pressure region

high pressure region

Drag force dueto pressure difference

v

v

flow speed (high) vair + v reduced pressure

flow speed (low) vair - v increased pressure

vair (vball)

Boundary layer – air sticks to ball (viscosity) – air dragged around with ball

MAGNUS EFFECT

motion of air

motion of object

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Professional golf drive

Initial speed v0 ~ 70 m.s-1

Angle ~ 6°

Spin ~ 3500 rpm

Range ~ 100 m (no Magnus effect)

Range ~ 300 m (Magnus effect)

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Golf ball with backspin (rotating CW) with air stream going fromleft to right. Note that the air stream is deflected downward with a downward force. The reaction force on the ball is upward. This gives the longer hang time and hence distance carried.

The trajectory of a golf ball is not parabolic

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lift

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Direction plane is moving w.r.t. the air

Direction air is moving w.r.t. plane

low pressure drag

attack angle

lift

downwashhuge vortices

momentum transfer

low pressure

high pressure

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Semester 1, 2004 Exam question

A large artery in a dog has an inner radius of 4.0010-3 m. Blood flows through the artery at the rate of 1.0010-6 m3.s-1. The blood has a viscosity of 2.08410-3 Pa.s and a density of 1.06103 kg.m-3.

Calculate:(i) The average blood velocity in the artery.(ii) The pressure drop in a 0.100 m segment of the artery.(iii) The Reynolds number for the blood flow.

Briefly discuss each of the following:(iv) The velocity profile across the artery (diagram may be helpful).(v) The pressure drop along the segment of the artery. (vi) The significance of the value of the Reynolds number calculated in part (iii).

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Solution

radius R = 4.0010-3 m

volume flow rate Q = 1.0010-6 m3.s-1

viscosity of blood = 2.08410-3 Pa.s

density of blood = 1.06010-3 kg.m-3

(i) v = ? m.s-1 (ii) p = ? Pa(iii) Re = ?

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(i) Equation of continuity: Q = A v

A = R2 = (4.0010-3)2 = 5.0310-5 m2

v = Q / A = 1.0010-6 / 5.0310-5 m.s-1 = 1.9910-2 m.s-1

(ii) Poiseuille’s EquationQ = P R4 / (8 L) L = 0.100 m

P = 8 L Q / ( R4)

P = (8)(2.08410-3)(0.1)(1.0010-6) / {()(4.0010-3)4} Pa

P = 2.07 Pa

(iii) Reynolds NumberRe = v L / where L = 2 R (diameter of artery)Re = (1.060103)(1.9910-2)(2)(4.0010-3) /

(2.08410-3) Re = 81

use diameter not length

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Flow of a viscous newtonain fluid through a pipeVelocity Profile

Adhesive forces between fluid and surface fluid stationary at surface

Parabolic velocityprofile

Cohesive forces between molecules layers of fluid slide past each other generating frictional forces energy dissipated (like rubbing hands together)

(iv) Parabolic velocity profile: velocity of blood zero at sides of artery

(v) Viscosity internal friction energy dissipated as thermal energy pressure drop along artery

(vi) Re very small laminar flow (Re < 2000)