Fluids in motion
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Transcript of 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
Physics 201: Lecture 27, Pg 2
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
Physics 201: Lecture 27, Pg 4
Fluids in Motion
Real flow vs. ideal flow non-steady / steady state compressible /
incompressible rotational / irrotational viscous / frictionless
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
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 7
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
Physics 201: Lecture 27, Pg 8
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
Physics 201: Lecture 27, Pg 9
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
Physics 201: Lecture 27, Pg 10
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
Physics 201: Lecture 27, Pg 11
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
Physics 201: Lecture 27, Pg 12
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
Physics 201: Lecture 27, Pg 13
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
Physics 201: Lecture 27, Pg 14
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
Physics 201: Lecture 27, Pg 15
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.
Physics 201: Lecture 27, Pg 16
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
Physics 201: Lecture 27, Pg 17
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
Physics 201: Lecture 27, Pg 18
For Thursday
• Read Chapter 15.1 to 15.3Read Chapter 15.1 to 15.3