Unformatted text preview: t includes the force contribution from
the weight of all the ﬂuid above it. Well known to
divers: pressure increases as you dive deeper
underwater. The pressure goes up by 1atm for
every 10.3m of water depth.
y
p( y )
Assuming constant
dy extra weight of water
in the range dy p(y + dy ) = p(y ) + ρgdy
Monday, September 30, 2013 density (incompressible)
good approx for water, not
air. Will do air later. p = p 0 + ρg ∆ y Pressure and Buoyancy
E.g. try to submerge a water polo ball underwater:
Water p = p 0 + ρg ∆ y Ball
pressure
a little bigger
here than on top
net force: Side pressures cancel, bigger
pressure below than above, so
net buoyant force up. Fb = − Around 250 BC
Monday, September 30, 2013 pda = 0 Buoyant force from
bigger force on bottom Buoyancy, cont.
King to Archimedes: how pure is the gold in my crown?
Archimedes mulled, and came to the idea in his bath.
Buoyant force: upward, equal to weight of the ﬂuid
displaced by the body.
E.g.
mship g = ρH2 O g Vdisplaced
Vdisplaced = mship /ρH2 O Determines how much volume is hidden
below the water surface in the picture.
Monday, September 30, 2013 Icebergs
W = gVtotal ρice What fraction is above
the surface?
ρice = 917 kg/m3
ρsea Fb = gVsub ρsea Monday, September 30, 2013 Fb = W water (Vtotal − Vsub )/Vtotal = 1 − ( water = 1024 kg/m3 ρice
ρsea water ) ≈ 10% Fluid Flow
If a ﬂuid is (approximately) incompressible (like water),
dm = ρA1 v1 dt = ρA2 v2 dt
A1 v 1 = A2 v 2 compressible case: Monday, September 30, 2013 ρ1 A1 v1 = ρ2 A2 v2 Bernoulli Equation
Expresses conservation of energy
Work done on ﬂuid element: dW = p1 A1 dx1 − p2 A2 dx2
dW = (p1 − p2 )dV (suppose incompressible) 1
2
2
Change of its kinetic energy: dK =
ρdV (v2 − v1 )
2
Change of its potential energy: dU = ρdV g (y2 − y1 )
dW = dK + dU Monday, September 30, 2013 12
p + ρgy + ρv = const.
2 Bernoulli, physically 12
p + ρgy + ρv = const.
2 Monday, September 30, 2013 Large tank with small
hole example
hole 12
p + ρgy + ρv = const.
2
(Approx.) Since small hole, and big tank. Top: p = patm , y = h, v = 0. On side away
from the hole Bottom: p = patm + ρgh, y = 0, v = 0.
Hole:
p = patm , y = 0, v = 2gh Equilibrium
with outside
at the hole.
Monday, September 30, 2013 Same velocity as an
object dropped from
a height h....
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 Fall '09
 Hicks
 Thermodynamics

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