EXERCISES 1 2 5 4 3 LEVEL PREREQUISTES LET’S HAVE FUN VIDEO n. 1 VIDEO n. 2 APPLICATIONS OF THE PRINCIPLE: THEORICAL EXPLANATION 132456 9101178 VENTURI.

EXERCISES

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LEVELPREREQUISTES

LET’S HAVE FUN• VIDEO n. 1• VIDEO n. 2

APPLICATIONS OF THE PRINCIPLE:

THEORICAL EXPLANATION

1 32 4 5 69 10 117 8

• VENTURI EFFECT• BIRDS FLIGHT

VIDEO

KEY LEARNING OBJETIVES

BERNOULLI’S PRINCIPLE

LEVELThis lesson is suitable for students attending

the fourth year of scientific high school.

• Work-energy principle • Density of a fluid

• Definition of ideal fluid• Definition of descharge of a fluid

• Continuity equation

PREREQUISITES

KEY LEARNING OBJECTIVES• To explain Bernoulli’s principle correctly and

organically.• To know its most important applications.

A LITTLE INTRODUCTION

Bernoulli’s principle is probably one of the most important principles of fluidodynamic. It

was published for the first time in Daniel Bernoulli’s book “Hydrodynamica” in 1738.

This principle has got different applications in a lot of fields of science.

PART 1

We consider a duct with a variable section. When the time is t, the fluid is present between S1 and S1‘ . When the time is t + Δt, the fluid is present between S2 and S2‘. The mass

of the liquid can be expressed this way: Δm = ρΔV

PART 2

At first the mass of the fluid is present at the height h1, in the second moment that we consider it is

present at the height h2 from the ground. We indicate with Δx1 the first distance, with Δx2 the second one.

PART 3

If Δt is very small, we can suppose that v is costant. Then, we can apply the Work-energy principle. It says that

W = ΔEc

W is the sum of the different works done by the different forces.

PART 4

The first forces are the pressures that act in the

two opposed directions.p1 = F1/S1 F1 = p1S1

p2= F2/S2 F2 = P2s2

PART 5

The work associated to the first pressure is: W1 = F1Δx1 W1 = p1S1ΔX1 because F1 = p1S1

We know that S1ΔX1 = ΔV

So our relation can be considered this way:W1 = p1ΔV

PART 6

We can do the same considerations about the second

pressure. But pay attention! This pressure acts in a direction that is opposed to the movement of the fluid.

For this reason we have to add the sign “-” (minus). W2 = - F2Δx2 W2 = - p2S2ΔX2 W2 = - p2ΔV

PART 7

The other force that we have to consider is the acceleration of gravity. The work done by this force is

indicated with the expression: W3 = Δmg(h1- h2)

PART 8

Because of the “work-energy principle”, the sum of the different works is equal to the cinetic energy difference (ΔEc). In this

case: W1+ W2+ W3 = ΔEc

PART 9 We know that:

ΔEc = ΔmV22/2 – ΔmV1

2/2

So, we can say that:p1ΔV - p2ΔV + Δmgh1- Δmgh2 = ΔmV2

2/2 – ΔmV12/2

We know that: Δm = ρΔV ΔV = Δm/ρ

We can substitute this expression to ΔV:p1Δm/ρ - p2Δm/ ρ + Δmgh1- Δmgh2 = ΔmV2

2/2 – ΔmV12/2

Finally, we divide everything for the mass Δm:p1/ρ - p2/ρ + gh1- gh2 = V2

2/2 – V12/2

PART 10 We can multiply everything for ρ:

p1 - p2 + ρgh1- ρgh2 = ρV22/2 – ρV1

2/2

And, finally:p1 + ρgh1+ ρV1

2/2 = p2 + ρgh2+ ρV22/2

Now we can give an accurate definition of Bernoulli’s principle.

PART 11If we indicate with:

• v = speed of the fluid in a certain point• p = pressure of the fluid in the same point• ρ = density of the fluid• h = height of the point we consider from the ground• g = acceleration of gravity

p + ρgh+ ρV2/2 = COSTANT

APPLICATIONS OF THE PRINCIPLE

We know that Benoulli’s principle has got different applications in several

fields of science. One of them is called Venturi effect and it can

explain al lot of phenomenons such as birds flight. Let’s examine it.

VENTURI EFFECT

If we consider an ideal fluid that flows through a perfect duct, we know that its descharge is costant:

Sv = COSTANTIt follows that if S increases, v decreases and vicecersa.If we consider Bernoulli’s equation of an ideal fluid in a

horizontal duct, it can be semplified in: p + (ρv2 )/2 = COSTANT

Because h = 0

Considering this form of the equation: p + (ρv2 )/2 = COSTANT

We can notice that if v increases, p dicreases and viceversa.

Considering again the observations that we did about the continuity equation, we can definitely adfirm that if

S Increases, also p increases!This fundamental event is called “VENTURI EFFECT”

BIRDS FLIGHT EXPLAINED THROUGH VENTURI EFFECT

The wing is flying in the air. We consider the wing motionless and the air investes it. The current lines above the wing are

denser than those under the wing. We can consider them as two sections of two flow pipes. The section “above the wing” is bigger than the one “under the wing”. Because of the “Venturi

effect”, if S Increases, also p increases. For this reason the pressure above the wing is minor than the pressure under the

wing. For this reason the wing, and the birds, can fly!

LET’S HAVE FUN

Maybe it can appear quite strange, but also phisics can be funny! Let’s watch some videos

about having fun with Bernoulli’s principle!

VIDEO N. 1

VIDEO N. 2

EXERCISES Now that you have learnt the Bernoulli’s principle and its most important application (Venturi effect), try to use them in these following exercises. The first two exercises will give you the solution. As for the others, you’ll have to do on your

own!

EXERCISE n.1 An open container contains a liquid and presents, at depth “h” from the surface of the liquid, a little hole. If the speed with which the liquid comes out is “v” on the Earth, what will be the speed on the Moon, where the

acceleration of gravity is 1/6 of Earth’s one?

a) (1/6)vb) √6v c) (√6/6)vd) 0 (zero), because there’s no atmosphere on the Moon

EXERCISE n.2 In a container that contains an

ideal liquid of density “ρ” is practiced a hole at a height “h” from the bottom. We can say that the speed of

outflow of the liquid: a) is equal to √ρghb) is equal to the speed that the liquid would get, falling in va

cuum from height “h”c) is equal to 2ghd) the other answers are all wrong.

EXERCISE n.3The pressure of a liquid flowing in a horizontal tube is “p1”,

in a part where the tube has section “S1”. What can you say about pressure “p2” of the liquid in correspondence

of a bottleneck where the section is “S2=S1/2” ?

a) p2 = 2p1

b) p2 = p1/2

c) p2 - p1 < 0

d) p2 = p1

EXERCISE n.4 An ideal liquid flows with v = 40 cm/s through a duct,

which has got a section S = 3,0 cm2. Calculate the speed that the liquid has gets if in a certain point the

section of the duct is S1= 2,0 cm2.

a) v1= 6,0 cm/s

b) v1= 60 cm/s

c) v1= 5,0 cm/s

d) v1= 50 m/s

EXERCISE n.5 In a horizontal pipe water flows with v = 4,00 m/s and its

pressure is p = 100 kPa. Because of a bottleneck of the pipe, the pressure becomes p1 = 20,0 kPa. Calculate the speed of the water in the new part of the pipe and the

ratio between the two sections.

a) v1= 13,3 m/s; 3,32

b) v1= 133 m/s; 4,32

c) v1= 67 cm/s; 5,6

d) v1= 6,7 m/s; 3,78

RIGHT!!!

Ex. 1 Ex. 2 Ex. 3 Ex. 5Ex. 4

WRONG!!!

Ex. 1 Ex. 2 Ex. 3 Ex. 5Ex. 4

SOURCES

• FISICA 1 – Antonio Caforio / Aldo Ferilli• www.youtube.com