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

Physics 201: Lecture 27, Pg 1

Pascal’s Principle: Example

 Now consider the set up shown on right.

 Mass M is placed on right piston, A 10 > A 1 = 2A 1  How do d A and d B compare?

Case 1 d A A 1  Equilibrium when pressures at P (left & right) are equal and Case 2 P 1

M

A 10 d B P 2

M

P

1

= P

2 r

F

1

/ A

1

(A

1

d

A

) g/ A

1

= F

2

/ A

2

=

r

(A

2

d

2

) g/ A

2

d

A

= d

B A 2 A 10 Physics 201: Lecture 27, Pg 2

Fluids in Motion

 Real flow vs. ideal flow     non-steady / compressible / rotational viscous / / steady state incompressible irrotational frictionless Physics 201: Lecture 27, Pg 4

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 Physics 201: Lecture 27, Pg 5

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 Physics 201: Lecture 27, Pg 6

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 (m 3 /s) Q = A·v is constant along tube.

A 1 v 1 = A 2 v 2

Mass flow rate is just r Q (kg/s) Physics 201: Lecture 27, Pg 7

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 cm 3 container in 20 s. Find the velocity of the stream 10.0 cm below the opening of the faucet.

Q = A

1

v

1

= A

2

v

2

Q =

D

V /

D

t =

1000 x 10

-6

/ 20

m

3

/s

v

1

= Q / A

1

=

5 x 10

-5

/

0.25

p

x 10

-4

m/s

v

1

= Q / A

1

= 0.64

m/s K

2

= K

1

+

D

mgh=

½

D

mv

1 2

+

D

mgh v

2

= (

v

1 2

+ gh

)

½

= (0.64

2

+ 9.8 x 0.1)

½

= 1.2 m/s

Physics 201: Lecture 27, Pg 8

Ideal Fluid Model (frictionless, incompressible)

 Streamlines do not meet or cross A 1  Velocity vector is

tangent

streamline to v 1  Volume of fluid follows a

“tube of flow”

bounded by streamlines A 2 v 2  Streamline density is proportional to velocity Physics 201: Lecture 27, Pg 9

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.

v 1 v 1/2

 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? (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1 Physics 201: Lecture 27, Pg 10

Exercise

Continuity v 1 v 1/2

(A) 2 v 1

(B) 4 v 1

(C) 1/2 v 1 (D) 1/4 v 1  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 Physics 201: Lecture 27, Pg 11

Exercise

Continuity v 1 v 1/2

 Experimentally we observe a pressure drop at the neck and this can be recast as work (i.e., energy transfer) P D V = (F/A) (A D x) = F D x

v 1 v 1/2

F 1 F 2 F 1 and F 2 maintain the pressure in the tube as the water flows Physics 201: Lecture 27, Pg 12

Conservation of Energy for an Ideal Fluid

W = (P 1 – P 2 ) D V = D K

P 1

W = ½ D m v 2 2 – ½ D m v 1 2 = ½ ( r D V) v 2 2 – ½ ( r D V) v 1 2

P 2

(P 1 – P 2 ) = ½ r v 2 2 – ½ r v 1 2 P 1 + ½ r v 1 2 = P 2 + ½ r v 2 2 = const .

and with height variations (potential energy):

Bernoulli Equation

 P 1 + ½ r v 1 2 + r g y 1 = constant Smaller diameter implies a pressure drop Physics 201: Lecture 27, Pg 13

d d d P 0 = 1 atm A

B

Torcelli’s Law (Bernoulli in action)  h The flow velocity v = (gh) ½ where is the depth from the top surface P + r g h + ½ r v 2 = const P 0 + r A g h + 0 = P 0 B + 0 + ½ r v 2

2g h = v

2

d = ½ g t

2

t = (2d/g)

½

x = vt = (2gh)

½

(2d/g)

½

= (4dh)

½ Physics 201: Lecture 27, Pg 14

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, higher velocity lower pressure its curvature, and the angle between the wing and the horizontal lower velocity higher pressure But Bernoulli’s Principle is not applicable (open system) and air is very compressible Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

Physics 201: Lecture 27, Pg 15

Fluids: A tricky problem

 A beaker contains a layer of oil (green) on H 2 O (blue), which has density ρ 3 with density ρ 2 floating . A cube wood of density ρ 1 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 Physics 201: Lecture 27, Pg 16

Fluids: A tricky problem

 A beaker contains a layer of oil (green) on H2O (blue), with density ρ 2 floating which has density ρ 3 . A cube wood of density ρ 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 1 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: F b = W 1 = ρ 1 V 1 g = ρ 1 L 3 g = F b2 + F b3 = ρ 2 d L 2 g + ρ 3 (L-d) L 2 g ρ 1 L = ρ 2 d + ρ 3 (L-d) (ρ 1 ρ 3 ) L = ( ρ 2 ρ 3 ) d Physics 201: Lecture 27, Pg 17

For Thursday • Read Chapter 15.1 to 15.3

Physics 201: Lecture 27, Pg 18