Fluids in motion

Physics 201: Lecture 27, Pg 1

Fluids in motion

Goals Understand the implications of continuity for Newtonian

fluids Distinguish pressure and force for fluids in motion Employ Bernoulli’s equation

http://boojum.as.arizona.edu/~jill/NS102_2006/Lectures/Lecture12/turbulent.html


Pascal’s Principle: ExampleNow consider the set up shown on right.

Mass M is placed on right piston,

A10 > A1 = 2A1

How do dA and dB compare?

Equilibrium when pressures at P (left & right) are equal and

P1 = P2

F1 / A1 = F2 / A2

(A1dA) g/ A1 = (A2d2) g/ A2

dA = dB

A1 A10

A2 A10

M

MdB

dA

Case 1

Case 2

P2

P1


Fluids in Motion

Real flow vs. ideal flow non-steady / steady state compressible /

incompressible rotational / irrotational viscous / frictionless


Types of Fluid Flow Laminar flow

Each particle of the fluid follows a smooth path

The paths of the different particles never cross each other

The path taken by the particles is called a streamline

Turbulent flow An irregular flow

characterized by small whirlpool like regions

Turbulent flow occurs when the particles go above some critical speed


Types of Fluid Flow

Laminar flow Each particle of the fluid

follows a smooth path The paths of the different

particles never cross each other The path taken by the

particles is called a streamline Turbulent flow

An irregular flow characterized by small whirlpool like regions

Turbulent flow occurs when the particles go above some critical speed


Continuity The mass or volume per unit time of an ideal fluid moving past

point 1 equals that moving past point 2

Flow obeys continuity or mass conservation

Volume flow rate (m3/s) Q = A·v is constant along tube.

Mass flow rate is just Q (kg/s)

A1v1 = A2v2


Example problem The figure shows a water stream in steady

flow from a faucet. At the faucet the diameter of the stream is 1.00 cm. The stream fills a 1000 cm3 container in 20 s. Find the velocity of the stream 10.0 cm below the opening of the faucet.

Q = A1v1 = A2v2

Q = V / t =1000 x 10-6 / 20 m3/s

v1 = Q / A1= 5 x 10-5 / 0.25 x 10-4 m/s

v1 = Q / A1= 0.64 m/s

K2 = K1 +mgh= ½ mv12 + mgh

v2 = (v12 + gh)½ = (0.642 + 9.8 x 0.1)½ = 1.2 m/s


Ideal Fluid Model (frictionless, incompressible)

Streamlines do not meet or cross

Velocity vector is tangent to streamline

Volume of fluid follows a “tube of flow” bounded by streamlines

Streamline density is proportional to velocity

A1

A2

v1

v2


Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe?

Exercise Continuity

A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house.

v1 v1/2

(A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1


For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast.

But if the water is moving 4 times as fast then it has

16 times as much kinetic energy.

Something must be doing work on the water

Exercise Continuity

v1 v1/2

(A) 2 v1 (B) 4 v1 (C) 1/2 v1 (D) 1/4 v1


Experimentally we observe a pressure drop at the neck and

this can be recast as work (i.e., energy transfer)

P V = (F/A) (A x) = F x

Exercise Continuity

v1 v1/2

F1 and F2 maintain the pressure in the tube as the water flows

v1 v1/2

F1 F2


W = (P1– P2 ) V = K

W = ½ m v22 – ½ m v1

2

= ½ (V) v22 – ½ (V)v1

2

(P1– P2 ) = ½ v22 – ½ v1

2

P1+ ½ v12 = P2+ ½ v2

2 = const.

and with height variations (potential energy):Bernoulli Equation P1+ ½ v1

2 + g y1 = constant

Conservation of Energy for an Ideal Fluid

P1 P2

Smaller diameter implies a pressure drop


Torcelli’s Law (Bernoulli in action)

The flow velocity v = (gh)½ where h is the depth from the top surface

P + g h + ½ v2 = const

A B

P0 + g h + 0 = P0 + + ½ v2

2g h = v2

d = ½ g t2

t = (2d/g)½

x = vt = (2gh)½(2d/g)½ = (4dh)½

P0 = 1 atm

A Bd

d

d


Applications of Fluid Dynamics Streamline flow around a moving

airplane wing Lift is the upward force on the

wing from the air Drag is the resistance The lift depends on the speed of

the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal

But Bernoulli’s Principle is not applicable (open system) and air is very compressible

higher velocity lower pressure

lower velocityhigher pressure

Note: density of flow lines reflectsvelocity, not density. We are assumingan incompressible fluid.


Fluids: A tricky problem

A beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1 and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure.

What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water?

Hint: The magnitude of the buoyant force (directed upward) must exactly equal the magnitude of the gravitational force (directed downward). The buoyant force depends on d. The total buoyant force has two contributions, one from each of the two different fluids. Split this force into its two pieces and add the two buoyant forces to find the total force


Fluids: A tricky problem

A beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1

and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure.

What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water?

Soln:

Fb = W1 = ρ1 V1 g = ρ1 L3 g

= Fb2 + Fb3

= ρ2 d L2 g + ρ3 (L-d) L2 g

ρ1 L = ρ2 d + ρ3 (L-d)

(ρ1 - ρ3 ) L = (ρ2 - ρ3 ) d


For Thursday

• Read Chapter 15.1 to 15.3Read Chapter 15.1 to 15.3

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Fluids in motion