FLUID FLOW IDEAL FLUID BERNOULLI'S PRINCIPLE

How can a plane fly?How does a perfume spray work?

What is the venturi effect?Why does a cricket ball swing or a baseball curve?

web notes: lect6.ppt flow3.pdf

Daniel Bernoulli (1700 – 1782)

Floating ball

A1

A2

v1

v2

A1

v1

Low speedLow KEHigh pressure

high speedhigh KElow pressure

Low speedLow KEHigh pressure

v small v smallv large

p large p large

p small

In a serve storm how does a house loose its roof?Air flow is disturbed by the house. The "streamlines" crowd around the top of the roof faster flow above house reduced pressure above roof than inside the house room lifted off because of pressure difference.

Why do rabbits not suffocate in the burrows?Air must circulate. The burrows must have two entrances. Air flows across the two holes is usually slightly different slight pressure difference forces flow of air through burrow. One hole is usually higher than the other and the a small mound is built around the holes to increase the pressure difference.

Why do racing cars wear skirts?

velocity increasedpressure decreased

low pressurehigh pressure

(patm)

VENTURI EFFECT

high speedlow pressure

force

force

What happens when two ships or trucks pass alongside each other?Have you noticed this effect in driving across the Sydney Harbour Bridge?

artery

External forces causes artery to collapse

Flow speeds up at constrictionPressure is lowerInternal force acting on artery wall is reduced

Arteriosclerosis and vascular flutter

y1

y2

x1

x2 p2

A2

A1v1

v2

p1

X

Y

time 1

time 2

m

m

Bernoulli’s Equation

for any point along a flow tube or streamline

p + ½ v2 + g y = constantDimensionsp [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]

½ v2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3]

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.

½ v 2 KE of bulk motion of fluid

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)

y1

y2

x1

x2 p2

A2

A1v1

v2

p1

X

Y

time 1

time 2

m

m

Mass element m moves from (1) to (2)

m = A1 x1 = A2 x2 = V where V = A1 x1 = A2 x2

Equation of continuity A V = constant

A1 v1 = A2 v2 A1 > A2 v1 < v2

Since v1 < v2 the mass element has been accelerated by the net force

F1 – F2 = p1 A1 – p2 A2

Conservation of energy

A pressurized fluid must contain energy by the virtue that work must be done to establish the pressure.

A fluid that undergoes a pressure change undergoes an energy change.

Derivation of Bernoulli's equation

K = ½ m v22 - ½ m v1

2 = ½ V v22 - ½ V v1

2

U = m g y2 – m g y1 = V g y2 = V g y1

Wnet = F1 x1 – F2 x2 = p1 A1 x1 – p2 A2 x2

Wnet = p1 V – p2 V = K + U

p1 V – p2 V = ½ V v2

2 - ½ V v12 + V g y2 - V g y1

Rearranging

p1 + ½ v12 + g y1 = p2 + ½ v2

2 + g y2

Applies only to an ideal fluid (zero viscosity)

Ideal fluid

Real fluid

(1) Point on surface of liquid

(2) Point just outside hole

v2 = ? m.s-1

y1

y2

Flow of a liquid from a hole at the bottom of a tank

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

(1)

(2)

F

m

h

v1 = ?

How do you measure the speed of flow for a fluid?

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}

m1 2

1F

2

2

1

g hv

A

A

C

B

A

D

yA

yB

yC

How does a siphon work?

How fast does the liquid come out?

Assume that the liquid behaves as an ideal fluid and that both the equation of continuity and Bernoulli's equation can be used.Heights: yD = 0 yB yA yC

Pressures: pA = patm = pD Consider a point A on the surface of the liquid in the container and the outlet point D.Apply Bernoulli's principle to these points

Now consider the points C and D and apply Bernoulli's principle to these points

From equation of continuity vC = vD

The pressure at point C can not be negative

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 )

pC + ½ vC2 + g yC = pD + ½ vD

2 + g yD

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

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).

Semester 1, 2004 Exam question

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) 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 Equation

Q = 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} PaP = 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

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)