Transcript Fluids_ch13 - bba-npreiser

Fluids
Chapter 13
P
F
A
Pressure
• Pressure:
, where F is the force acting perpendicular to
surface and A is the area of surface
• SI unit for pressure: [N/m2= Pa]
• Atmospheric pressure: 101325 Pa or 1.01 x 105 Pa
•
Hold a thumb tack between your index finger and thumb with a force of
10 N. The needle has a point that is 0.1mm in radius whereas the flat
end has a radius of 5 mm.
(a) What is the force experience by your finger; what is the force
experienced by your thumb.
(b) Your thumb holds the pointy end. What is the pressure on the thumb;
what is the pressure on your finger?
Pressure and Depth
• A fluid or gas exerts pressure on a surface
• A barometer is a way to measure
atmospheric pressure
• The height of the column of liquid
depends on the density of the liquid ( =
m/V) and atmospheric pressure.
P1 =0
h
P2 = P1 + gh
Patm = gh
Measure h, determine Patm
P2 = Patm
Example--Mercury
 = 13,600 kg/m3
Patm = 1.05 x 105 Pa
 h = 0.757 m = 757 mm = 29.80 in. (for 1 atm)
Suppose you have a barometer with mercury and a barometer with water.
How does the height hwater compare with the height hmercury?
Pressure and Depth
• The pressure P at a depth h below
the surface of a liquid open to the
atmosphere is greater then the
atmospheric pressure by an amount
gh
P  P0    g  h
• The added pressure (gh)
corresponds to weight of fluid
column of height h and P0 is
atmospheric pressure.
• In a fluid, all points at the same depth
must be at the same pressure.
Sample Problem
• Blood (density = 1060 Kg/m3) in the arteries is flowing, but as
a first approximation, the effects of this can be ignored and
the blood treated as a static fluid. Estimate the amount by
which the blood pressure P2 in the anterior tibial artery at the
foot exceeds the blood pressure P1 in the aorta at the heart
when a person is (a) reclining horizontally, and (b) standing
(assume the distance between the heart and feet is 1.35 m)?
(a) P2 = P1 + gh, so P2 – P1 = gh, since h = 0 m, gh = 0 and
P2 – P1 = 0
(b) P2 – P1 = gh = (1060 Kg/m3)(9.8 m/s2)(1.35 m) = 1.4 x 104
Pa
Absolute Pressure vs. Gauge Pressure
- Absolute pressure P: absolute pressure,
including atmospheric pressure
- Gauge pressure PG: difference between
absolute pressure and atmospheric pressure
 pressure above atmospheric pressure
 pressure measured with a gauge for
which the atmospheric pressure is calibrated
to be zero.
Pascal’s Principle
• Pascal’s Principle: A change in the pressure
applied to a fluid is transmitted undiminished
to every point of the fluid and to the walls of
the container.
Application of Pascal’s Principle -Hydraulic press
•
In a hydraulic press Force F1 is applied to area A1
•
Pressure P in columns 1 is P1 = F1/A1 and the pressure in column 2 is P2 =
F2/A2 Pascal’s Principle tells P1 = P2 so
F1/A1 = F2/A2
• Force F2 on area A2 is greater than F1 by a factor A2/A1
Sample Problem
• The piston of a hydraulic lift has a cross sectional area of 3.00
cm2, and its large piston has a cross-sectional area of 200 cm2.
(a) What force must be applied to the small piston for it to raise
a 15 kN car?
(b) Could your body weight (600 N) provide the force?
Buoyant forces and Archimedes's Principle
• Archimedes’s principle: The magnitude of the
buoyant force is equals the weight of the fluid
displaced by the object.
B  m f  g   f V f  g
Equation of Continuity
Mass is conserved as the fluid flows
A1 1 v1
A2 2 v2
 2 A2v1   2 A2v2
For fluids flowing in a “pipe”, the product
of area and velocity (flow rate, Q = Av)
is constant (big area  small velocity).
-
if 2 = 1 then
A1v1  A2v2
we assume: the flow of fluids is laminar (not turbulent)
–  There are now vortices, eddies, turbulences.
Water layers flow smoothly over each other.
A1
V1
A2
V2
-
we assume: the fluid has no viscosity (no friction).
–  (Honey has high viscosity, water has low
viscosity)
Bernoulli’s equation
• Bernoulli’s equation is a consequence of the
law of conservation of energy
1 2
P  v  gy  constant
2
1 2
1
2
P1  v1  gy1  P2  v2  gy2
2
2
Bernoulli’s equation relates the steady flow
of a nonviscous, incompressible fluid
of density, , pressure P, fluid
speed v, and elevation y at any
two points (1 and 2)
Venturi Tube
• A 2 < A 1 ; V 2 > V1
• According to Bernoulli’s Law, pressure at A2 is lower.
• Choked flow: Because pressure cannot be negative,
total flow rate will be limited. This is useful in
controlling fluid velocity.
P2 + 1/2ρ v12 = P1 + 1/2ρ v12
;
ΔP = ρ/2*(v22 – v12)
Sample Problem
•
Water flows through a horizontal pipe, and then out into the
atmosphere at a speed of 15 m/s. The diameters of the left
and right sections of the pipe are 5.0 cm and 3.0 cm,
respectively.
(a) What volume of water flows into the atmosphere during a 10
min period?
(b) What is the flow speed of the water in the left section of the
pipe?
(c) What is the gauge pressure in the left section of the pipe?
Torricelli
• In 1843, Evangelista
Torricelli proved that the
flow of liquid through an
opening is proportional to
the square root of the
height of the opening.
• Q = A*√(2g(h1-h2)) where Q
is flow rate, A is area, h is
height or v = √(2g(h1-h2)),
where v is velocity.
Derivation of Torricelli’s Equation
• We use the Bernoulli Equation:
P2 + 1/2ρ v12 + ρ gh1 = P1 + 1/2ρ v12 + ρ gh1
• In the original diagram A1 [top] is much larger than A2
[the opening]. Since A1v1 = A2v2 and A1 >> A2, v1 ≈ 0
• Since both the top and the opening are open to
atmospheric pressure,
P1 = P2 = 0 (in gauge pressure).
The equation simplifies down to: ρgh1 = 1/2 ρv22 + ρgh2
1/ ρv 2 = pg(h -h )
2
2
1 2
v22 = 2g(h1-h2)
∴ v2 = √(2g(h1-h2))
Q = Av2 = A √(2g(h1-h2))
Bernoulli examples
• Bernoulli has been used to explain curve balls, how
planes fly (aerodynamic lift), and why your roof can
blow off during high winds and more.
Aerodynamic Lift
Aerodynamic Lift – Bernoulli??
•
•
•
•
•
Bernoulli effect assumptions: 1. Smooth, laminar flow and 2. Incompressible
fluid. Neither of these assumptions applies to an airplane wing.
In Bernoulli-an view, lift is produced by the different of pressure (faster velocity
on the top, slower velocity in the bottom)
In Newtonian view, lift is the reaction force that results from the downward
deflection of the air.
Both views are correct, but current arguments arises from the misapplication of
either view.
The most accurate explanation would take into account the simultaneous
conservation of mass, momentum, and energy of a fluid, but that involves
multivariable calculus.
Curve Ball
• The commonly accepted explanation is that a spinning object
creates a sort of whirlpool of rotating air about itself. On the
side where the motion of the whirlpool is in the same direction
as that of the windstream to which the object is exposed, the
velocity will be enhanced. On the opposite side, where the
motions are opposed, the velocity will be decreased. According
to Bernoulli's principle, the pressure is lower on the side where
the velocity is greater, and consequently there is an unbalanced
force at right angles to the wind. This is the magnus force.
Due to viscosity, adjacent air
molecules are swept and result in
lower pressure.
Sources
http://canteach.candu.org/library/20040602.pdf
http://geocities.com/k_achutarao/MAGNUS/ne
w_magnus.html
www.wfu.edu/~gutholdm/Physics113/Physics_1
13_Fall_2007/Chapter_14.ppt
http://www.slideshare.net/guestfda040/bernoul
lis-principle
www.physics.umd.edu/lecdem/powerpoint/Sum
mer2006AAPT.ppt