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Unformatted text preview: What is a fluid? Fluids are “substances that flow”…. “substances that take
the shape of the container” Lecture 1
Fluids: density, pressure,
Pascal’s principle. Density, pressure
Density: ρ= m
V Pure water: 1000 kg/m3 Atoms and molecules must be free to move .. No long range
correlation between positions (e.g., not a crystal).
Gas or liquid… or granular materials (like sand) Atmospheric pressure
The atmosphere of Earth is a fluid, so every object in air is
subject to some pressure.
At
At the surface of the Earth, the pressure is Pressure: Units: p= F⊥
A patm ~ 1.013 x 105 Pa = 1 atm
10 Pascal (Pa) = 1 N/m2
psi (pounds per square inch)
atmosphere
bar 1 atm = 1.013 × 105 Pa Area of a hand ~ 200 cm2 = 0.02 m2 F = patmA ~ 2000 N on your hand due to air! DEMO: Piston
and weight 1 bar = 105 Pa 1 DEMO:
Plastic tube
with cover Pressure vs. depth Imaginary box of fluid with
density ρ with bases of area A
and height h Ftop Fbottom = Ftop + mg
m = ρAh Example: How deep under water is p = 2 atm? h= pbottom − ptop
ρg = (10 1.01 × 105 Pa
3 kg/m 3 )(9.81 m/s )
2 (i .e ., 1 atm is produced by a 10.3 m high column of water) Two liquids Y and G separated by a thin,
light piston (so they cannot mix) are placed
in a Ushaped container. What can you say
about their densities? A.
A. ρG < ρY If
If liquid height
was higher above
A than above B pA > pB Net force
→ Net flow
→ •
B
This is not
equilibrium! Water towers
Water towers are a common sight in the Midwest… because it’s so flat! h1 Y h3 G h2
•
A B.
B . ρG = ρY •
A = 10.3 m ACT: U tube •
B C. ρG > ρY
Pressure at A and B must be the same: Since h1 < h3 − h2 h phouse = patm + ρwaterhg ρ Y gh1 + ρG gh2 + patm = ρG gh3 + patm ρYh1 = ρG (h3 − h2 ) p(y) Fluid level is the same everywhere in
a connected container (assuming no
surface forces) A DEMO:
Utube with
water and
kerosene y Fbottom Fbottom/top pbottom = ptop + ρ gh Fluid in an open container Pressure is the same at a
given depth, independently of
the container. h mg Net force must be zero! Pbottom/top = DEMO:
Pascal’s
vases ⇒ ρ Y > ρG 2 So physics sucks, but how much?
Your physics professor sucks on a long tube that rises
out of a bucket of water. She can get the liquid to rise
5.5 m (vertically). What is the pressure in her mouth at
this moment? Pascal’s principle
xB
h Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls
of the containing vessel. A. 1 atm
B. 0.67 atm xA C. 0.57 atm
D. 0.46 atm Pascal’s Principle is most often applied to incompressible
fluids (liquids): pmouth + ρwater gh = patm E. 0 atm pmouth = patm − ρwater gh = 10 Pa − (10 kg/m
5 DEMO:
Sucking
through a
hose 3 3 )( 9.8 m/s ) (5.5 m)
2 Increasing p at any depth (including the surface) gives
the same increase in p at any other depth = 46100 Pa = 0.46 atm Hydraulic chamber
F1 F1 F2
=
A1 A2
A
F2 = 2 F1
A1 F2 can be very large… ACT: Hydraulic chambers
F2 d2
d1 In each case, a block of mass M is placed on
the piston of the large cylinder, resulting in
a difference di between the liquid levels. If
A2 = 2A1, then: A.
A. dA < dB dA A1 A1 ⎛ W = F1d1 = ⎜ F2
⎜
⎝ A1 ⎞⎛ A2 ⎞
⎟⎜ d
⎟=Fd
A2 ⎟⎜ 2 A1 ⎟ 2 2
⎠⎝
⎠ A10 B. dA = dB
A2 C. dA > dB dB No energy is lost:
Pressure depends only on the height of the
water column above it. A2 M M A10 Incompressible fluid: A1d1 = A2d2 3 Measuring pressure with fluids
Barometer
Measures absolute pressure
Top of tube evacuated (p = 0)
Bottom of tube submerged into pool of mercury
open to sample (p)
p
Pressure dependence on depth: h =
g ρHg vacuum
Vacuum
p p=0
=0 Barometer
atmosphere
Sample
p=p
at p h
h 0 Manometer
Measures gauge pressure: pressure relative to
atmospheric pressure.
p − p0
Δh =
Pressure dependence on depth:
g ρHg p
p patm
p0
∆h
Δ A unit for pressure
760 mm Hg = 1 torr = 1 atm 4 ...
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This note was uploaded on 09/21/2011 for the course PHYS 222 taught by Professor Johnson during the Spring '07 term at Iowa State.
 Spring '07
 Johnson
 Physics

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