1 http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm

web notes: Fluidslect5web.pdf flow4.pdf flight.pdf

Lecture 5 Dr Julia Bryant

Fluid dynamics - Applications of Bernoulli’s principle

Fluid statics

• What is a fluid? Density !Pressure!• Fluid pressure and depth Pascal’s principle • Buoyancy Archimedes’ principle

Fluid dynamics!• Reynolds number • Equation of continuity!• Bernoulli’s principle • Viscosity and turbulent flow • Poiseuille’s equation

2

Bernoulli’s Equation for any point along a flow tube or streamline

p + 1 " v2 + " g y = constant 2 Between any two points along a flow tube or streamline p1 + 1 " v1

2 + " g y1 = p2 + 1 " v22 + " g y2

2 2 Dimensions p [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]

1 " v2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3] 2 "  g h [kg.m-3 m.s-2. m] = [kg.m.s-2.m.m-3] = [N.m.m-3] = [J.m-3]

Each term has the dimensions of energy / volume or energy density.

1 " v 2 KE of bulk motion of fluid 2 "  g h GPE for location of fluid

p pressure energy density arising from internal forces within moving fluid (similar to energy stored in a spring)

3

Solving Bernoulli’s Equation

1. Establish points 1 and 2 along the flow. 2. Define the coordinate system - where is

y=0? 3. List your known and unknown variables. 4. Solve for the unknown variables, possibly

using the continuity equation.

4

(1) Surface of liquid!

(2) Just outside hole!

v2 = ? m.s-1"

What is the speed with which liquid flows from a hole at the bottom of a tank? Where does it hit the ground?

R?

Fluid is flowing from the surface ==> Bernoulli’s applies

5

(1) Surface of liquid!

(2) Just outside hole!

v2 = ? m.s-1"

y2 = 0"

v1 ~ 0 m.s-1 (large tank)"y = h"

p1 = patm

p2 = patm

Firstly, what is the speed with which liquid flows from a hole at the bottom of a tank?

y1"

y2"

h = (y1 - y2)

<-- y = 0"

6

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

p1 + 1 " v12 + " g y1 = p2 + 1 " v2

2 + " g y2 2 2 A small hole is at level (2) and the water level at (1) drops slowly (if tank is large) # v1 = 0

p1 = patm p2 = patm

"  g y1 = 1 "v22 + " g y2

2 v2

2 = 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

y1"

y2"

h

y = 0"

p1"

p2"

7

Recall d=1 a t2" 2"t= √(2d/a)"

So, time taken to fall d= y1 - h is "t= √{2(y1 - h)/g}"

Distance, "R = v2 t" = $(2 g h) . √{2(y1 - h)/g}" =2√{h(y1 - h)}"

v2 = $(2 g h) Secondly, where does it hit the ground?

y1"

y2"

h

R?

8

0

5

10

15

20

25

0 2 4 6 8 10 12 14 16 18 20h

R

R =2√{h(y1 - h)}"Secondly, where does it hit the ground?

h=4

R?

DEMO

h=8.5

h=14

y1=20"

9

R =2√{h(y1 - h)}"Secondly, where does it hit the ground?

Y1=16"

R?

DEMO

4.5

10

0

5

10

15

20

0 2 4 6 8 10 12 14 16h

R

10

Venturi Meter A Venturi meter is used to measure the speed of the flow

in a pipe. What is the speed v1 of flow in section 1 of the system?

DEMO

11

Venturi Meter Bernoulli’s equation only applies along the same

streamline. Therefore we must choose points 1 and 2 along the pipe, but not in the vertical tubes.

DEMO

12

Venturi Meter

Q: What do we know? y1 = y2 no height difference"We will need to use: "• Bernoulliʼs equation (Assume liquid behaves as an ideal fluid) !p1 + 1 " v1

2 + " g y1 = p2 + 1 " v22 + " g y2

2 2 • Continuity equation A1 v1 = A2 v2 !• hydrostatic equation for the vertical pipes p1 - p2 = "m g h!

13

From Bernoulli:! p1 + 1 " v1

2 + " g y1 = p2 + 1 " v22 + " g y2"

2 2 ! y1 = y2"

p1 – p2 = 1 " (v22 - v1

2) From the continuity equation v2 = v1 (A1 / A2)" 2 "

p1 p2"

d"

v1 = 2 gh " {(A1 / A2)2 - 1}"

So, p1 – p2 = 1 " ((A1 / A2)2 v12 - v1

2)" 2" p1 – p2 = 1 " v1

2 ((A1 / A2)2 - 1) A1 > A2 so p1 – p2>0 p1 > p2" 2"From hydrostatics p1 = p0 + " g (d+h) ! p2 = p0 + " g d"p1 - p2 = " g h!

"g h = 1 " v12 {(A1 / A2)2 - 1}"

2 "

14

DEMO

or

Bernoulli’s Bar

15

How to impress your mother in a pub! From Bernoulli: p1 + 1 " v1

2 + " g y1 = p2 + 1 " v22 + " g y2

2 2 y1 = y2 = 0

p1 – p2 = 1 " (v22 - v1

2) 2 Continuity equation A2 v2 = A1 v1 A2 < A1 v2> v1 p2< p1 < patm"p1!

p2!

Ffriction!

patm!

patm!

Fatm = pA = 1.013 x 105 x 0.065 x 0.1 = 658N p2 – patm = (F2-Fatm)/A "If Fatm-F2 > Ff cans will accelerate towards each other."

F!

F!

d=0.065 "h=0.1"

The coefficient of friction between the cans and the bench is ~0.1 so, Ff =0.1 mg =0.013N (13g can) "Friction force is tiny compared to Fatm so small drop in pressure required to move cans."

Ffriction!

16

How does a siphon work?

17

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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C"

B"

A"

D"

yA"

yB"

yC"

yD = 0

What do we know? pA = patm = pD vA =0 approximately

What do we need to find? vD = ? yC = ?

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

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pC + 1 " vC2 + " g yC = pD + 1 " vD

2 + " g yD 2 2 From equation of continuity vC = vD pC = pD + " g (yD - yC) = patm + " g (yD - yC)

The pressure at point C can not be negative pC % 0 and yD = 0

pC = patm - " g yC % 0 yC & patm / (" g)

For a water siphon patm ~ 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.

C"

B"

A"

D"

yA"

yB"

yC"

yD = 0

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C"

B"

A"

D"

yA"

yB"

yC"

How fast does the liquid come out?

21

pA + 1 " vA2 + " g yA = pD + 1 " vD

2 + " g yD 2 2 vD

2 = 2 (pA – pD) / " + vA2 + 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

C"

B"

A"

D"

yA"

yB"

yC"

yD = 0

22

FLUID FLOW MOTION OF OBJECTS IN FLUIDS

How can a plane fly?

Why does a cricket ball swing or a baseball curve?

web notes: flow4.pdf flight.pdf

23

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"

B"A"

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

• Consider a cylinder. • The fluid will slide freely over the surface.

• Pressure near A and B are equal and greater than undisturbed flow (streamlines further apart). • Pressure near C and D are equal and lower then undisturbed flow

# Resultant force = zero

• No drag or lift. • The pattern is symmetrical

25

When real fluids flow they have a certain internal friction called viscosity. It exists in both liquids and gases and is essentially a frictional force between different layers of

fluid as they move past one another.

In liquids the viscosity is due to the cohesive forces between the molecules whilst in gases the viscosity is due to collisions

between the molecules.

“VISCOSITY IS DIFFERENT TO DENSITY” 26

Drag force

In a viscous fluid, a thin layer of fluid sticks to the surface of an object and the resulting friction leads to a drag force on the object.

The flow is no longer complete around the object, and the flow lines break away from the surface resulting in eddies behind the object. The pressure in the eddies is lowered and the pressure difference gives pressure drag force.

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low pressure region"

high pressure region"

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

Drag force due"to pressure difference"

Drag force is opposite to the direction of motion"

motion of air motion of object

28 low pressure region"

high pressure region"

Drag force due"to 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

What happens if the object is spinning?

29 30

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

31

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)

32

Stagnation line

Higher v, lower pressure

Lift increases with angle of attack, until stall.

0°

5°

10°

How can a plane fly?

33

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

DEMO 34