3/19/2014
Floating
Apparent weight in a fluid
When a body floats:
•
The gravity force Fg balances the buoyancy force Fb!
The gravity force is equal to the weight of the
displaced fluid
The mass M of the floating body is equal to the mass of
the displaced fluid mf
Fb
•
•
•
Apparent weight = Actual weight - Buoyancy force
Fg = Fb
Fb
Fg = m f g
Fg
M = mf
Wapparent
Wactual
Example problem: Iceberg
What percentage of the volume of an iceberg can
be seen above the surface of the sea?
•
Fb = Fg
V
•
Vabove 1.024 − 0.917
=
= 0.10 = 10%
V
1.024
The equation of continuity
•
For an ideal fluid the volume flow rate Rv
is constant along a tube of flow.
Rv =
•
∆V Aυ ∆t
=
= Aυ = constant
∆t
∆t
For a tube with variable cross section A:
Rv = A1υ1 = A2υ 2
•
v
no friction
Irrotational flow:
•
V
ρ −ρ
Vabove
= 1− f = f
V
V
ρf
the density is constant.
Nonviscous flow:
•
ρ
ρf
the velocity does not change with time.
Incompressible flow:
•
•
ρ f V f g = ρVg
=
Steady flow:
•
•
m f g = Mg
Vf
Ideal Fluids in Motion
•
There is no rotation
Streamline: the path of a fluid element.
•
•
•
The velocity of a fluid is tangent to the streamline.
Two streamlines do not intersect.
A surface fallowing streamlines acts like a tube.
Example (text problem 9.41): A garden hose of inner radius
1.0 cm carries water at 2.0 m/s. The nozzle at the end has
radius 0.20 cm. How fast does the water move through the
constriction?
A1v1 = A2 v2
A 
 πr 2 
v2 =  1 v1 =  12 v1
 A2 
 πr2 
2
 1.0 cm 
=
 (2.0 m/s ) = 50 m/s
 0.20 cm 
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3/19/2014
Bernoulli’s Equation
•
Example: parallel ships
Conservation of energy for fluids:
•
the work done by the pressure force
is equal to the change of the kinetic
and the potential energy of a fluid
element.
•
p1
v
1
1
p1 + ρυ12 + ρgy1 = p2 + ρυ 22 + ρgy 2 = constant
2
2
•
A1 p1
A2 p2
If the motion is horizontal:
p1
1
1
p1 + ρυ12 = p2 + ρυ 22 = constant
2
2
•
Two ships move parallel with a velocity v.
A1υ1 = A2υ 2
Applications of Bernoulli’s Principle:
Venturi Tube
•
•
•
A1
A2
1
1
p1 + ρυ12 = p2 + ρυ 22
2
2
1
p1 = p2 + ρ(υ 22 − υ12 ) ⇒
2
If the velocity is increasing, the
pressure is decreasing!
•
⇒ υ 2 = υ1
Shows fluid flowing
through a horizontal
constricted pipe
Speed changes as
diameter changes
Can be used to measure
the speed of the fluid
flow
Swiftly moving fluids
exert less pressure than
do slowly moving fluids
⇒ υ 2 > υ1
A net force pushes the ships
towards each other
p1 > p2
Example (text problem 9.49): A nozzle is connected to a
horizontal hose. The nozzle shoots out water moving at 25
m/s. What is the gauge pressure of the water in the hose?
Neglect viscosity and assume that the diameter of the nozzle
is much smaller than the inner diameter of the hose.
Let point 1 be inside the hose and point 2 be outside
the nozzle.
1
1
P1 + ρgy1 + ρv12 = P2 + ρgy2 + ρv22
2
2
The hose is horizontal so
y1 = y2. Also P2 = Patm.
1
ρv 2 + P = constant
2
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Example continued:
Substituting:
Example continued:
1
1
P1 + ρv12 = Patm + ρv22
2
2
1 2 1 2
P1 − Patm = ρv2 − ρv1
2
2
1 2 1 2
ρv2 − ρv1
2
2
1
1
2
= ρ v2 − v12 ≈ ρv22
2
2
1
2
3
= 1000 kg/m (25 m/s )
2
= 3.1× 105 Pa
P1 − Patm =
(
v2 = 25 m/s and v1 is unknown. Use the continuity equation.
  d2 2 
π  
2
 A2 
d 

2 
v1 =  v2 =   2 v2 =  2  v2
A
d
 1
 1
 π  d1  


 2 
(
)
)
Since d2<<d1 it is true that v1<<v2.
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3/19/2014
Bernoulli’s Equation: Applications
Bernoulli’s Equation: Application
• Siphon:
• Example (velocity of efflux):
The figure shows a siphon, which is a device for
removing liquid from a container. Tube ABC must
initially be filled, but once this is done, liquid will
flow until the liquid surface of the container is
level with the tube opening A. With what speed
does the liquid emerge from the tube at C? What
is the greatest possible height h1 that a siphon can
lift water?
We can use Bernoulli’s equation to calculate the speed of
efflux, v2, from a horizontal orifice (and area A2) located a
depth h below the water level of a large talk (with area A1).
P1 + 21 ρv12 + ρgy1 = P2 + 21 ρv22 + ρgy2
P1 = P2 = Patm
v1 = v2A2/A1
(1↔2)
v22 = v12 + 2 gh = ( A2 / A1 ) 2 v22 + 2 gh
Patm + 12 ρV 2 = PA + 12 ρv 2 − ρgd
Area A1
y-axis
v1
PA + 12 ρv 2 − ρgd = Patm + 12 ρv 2 − ρg ( d + h2 )
h
(Torricelli’s Law)
v2
(2)
PHY 2053
y=0
Area A2
ρ
University of Florida
Page 13
1
2
ρv 2 = ρg ( d + h2 )
v = 2 g (d + h2 )
University of Florida
An Object Moving Through a Fluid
•
Many common phenomena can be explained
by Bernoulli’s equation
•
•
At least partially
In general, an object moving through a
fluid is acted upon by a net upward force
as the result of any effect that causes
the fluid to change its direction as it
flows past the object
V = va / A
 a 
Patm = PA + 12 ρv 2 1 − 2  − ρgd a
→ PA + 12 ρv 2 − ρgd
<< A
 A 
(1)
2 gh
v2 =
→ 2 gh
(1 − ( A2 / A1 ) 2 ) A2 << A1
VA = va
(S↔A)
2
A = Area of container
a = area of tube
S
y=0
V
(A↔C)
v
PB + 12 ρv 2 + ρgh1 = Patm + 12 ρv 2 − ρg ( d + h2 )
h1 =
( Patm − PB )
− (d + h2 )
ρg
(B↔A)
(h1 ) max =
Patm
− (d + h2 )
ρg
PHY 2053
v
Page 14
Application – Golf Ball
•
•
•
•
The dimples in the golf
ball help move air along
its surface
The ball pushes the air
down
Newton’s Third Law tells
us the air must push up on
the ball
The spinning ball travels
farther than if it were
not spinning
3