Download Physics 207: Lecture 2 Notes

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.
Case 1
 Mass M is placed on right piston,
A10 > A1 = 2A1
 How do dA and dB compare?
M
dA
P1
A1
 Equilibrium when pressures at P
(left & right) are equal and
A10
Case 2
M
dB
P2
P1 = P2
F1 / A1 = F2 / A2
r (A1dA) g/ A1 = r(A2d2) g/ A2
dA = d B
A2
A10
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 (m3/s)
Q = A·v
is constant along tube.
A1v1 = A2v2
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 cm3 container in 20 s.
Find the velocity of the stream 10.0 cm
below the opening of the faucet.
Q = A1v1 = A2v2
Q = DV / Dt =1000 x 10-6 / 20 m3/s
v1 = Q / A1= 5 x 10-5 / 0.25p x 10-4 m/s
v1 = Q / A1= 0.64 m/s
K2 = K1 +Dmgh= ½ Dmv12 + Dmgh
v2 = (v12 + gh)½ = (0.642 + 9.8 x 0.1)½ = 1.2 m/s
Physics 201: Lecture 27, Pg 8
Ideal Fluid Model (frictionless, incompressible)
A2
 Streamlines do not meet or cross
A
1
 Velocity vector is tangent to
streamline
v1
 Volume of fluid follows a “tube of
v2
flow” bounded by streamlines
 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.
v1
v1/2
 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 v1
(B) 4 v1
(C) 1/2 v1
(D) 1/4 v1
Physics 201: Lecture 27, Pg 10
Exercise
Continuity
v1
(A) 2 v1
(B) 4 v1
v1/2
(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
Physics 201: Lecture 27, Pg 11
Exercise
Continuity
v1
v1/2
 Experimentally we observe a pressure drop at the neck and
this can be recast as work (i.e., energy transfer)
P DV = (F/A) (A Dx) = F Dx
v1
F1
v1/2
F2
F1 and F2 maintain the pressure in the tube as the water flows
Physics 201: Lecture 27, Pg 12
Conservation of Energy for an Ideal Fluid
W
= (P1– P2 ) DV = DK
W
= ½ Dm v2 – ½ Dm v1
2
2
P1
P2
= ½ (r DV) v22 – ½ (r DV) v12
(P1– P2 ) = ½ r v22 – ½ r v12
P1+ ½ r v12 = P2+ ½ r v22 = const.
and with height variations (potential energy):
Bernoulli Equation  P1+ ½ r v12 + r g y1 = constant
Smaller diameter implies a pressure drop
Physics 201: Lecture 27, Pg 13
P0 = 1 atm
Torcelli’s Law (Bernoulli in action)

d
The flow velocity v = (gh)½ where
h is the depth from the top surface
P + r g h + ½ r v2 = const
d
d
A
B
A
B
P0 + r g h + 0 = P0 + 0 + ½ r v2
2g h = v2
d = ½ g t2
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
higher velocity
lower pressure
the airplane, the area of the wing,
its curvature, and the angle
between the wing and the
lower velocity
horizontal
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) 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 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?
 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 17
For Thursday
• Read Chapter 15.1 to 15.3
Physics 201: Lecture 27, Pg 18