CHAPTER 9 Solids and Fluids

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CHAPTER 9Solids and FluidsFluid Mechanics
(only)pages 261-277
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Fluids and Fluid Mechanics
Density mass of substance divided by its volume
The ratio of a substances density to the density
of water at 4C (which is 1.0x103
kg/m3) The specific gravity of a substance
is 1. Dimensionless 2. Magnitude differs from
the substances density by a factor of 103
Specific Gravity (sp.gr.)
Pressure A force applied to an area
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Example Problem (Density/Sp.gr./Pressure) A lead
cube with a side of .250m rests on the floor.
Calculate the mass of the cube, the force exerted
on the floor, and the pressure on the floor
beneath the cube (? lead 11.3x103 kg/m3)
? M/V M ?v M (11.3 x 103 kg/m3)(.250m x
.250m x .250m)
F mg F (177 kg)(9.8m/s2)
F 1730 N
P F/A P (1730 N) / (.250m x .250m)
NOTE 1 atmosphere of pressure 101,300 pascals
P 27,700 N/m2 P 27,700 pascals
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Pressure In/Under Fluids
Po atmospheric pressure above the water P
pressure in the water at a point under the shaded
block of water If the shaded block of water is
not falling or rising, then FNet,y 0
?hg Po P
P Po ?gh
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Example Problem (Head Pressure) A water tower
sits on top of a 50.m platform in the middle of
the city (which is on level ground). The towers
water tank is 10.m high. Assuming the tank is
full of water, what is the pressure of the bottom
of the tank and what is the pressure in the water
pipe at grade level below the tower.
Sketch
PA Po ?gh PA 101.3kpa (1.0x103kg/m3)(9.8m/
s2)(10.m) PA 101.3kpa 98,000N/m2 1 N/m2 1
pa PA 101.3kpa 98.0kpa
PA 199.3kpa
PB PA (1.0x103kg/m3)(9.8m/s2)(50.m)
PB 689.32kpa
NOTE 101.3kpa 1.00atm 14.7psi 760mmHg
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Example Problem (Barometer) How high will liquid
Mercury rise in a barometer tube when the air
pressure is 1.00atm? note ?Hg 13.6x103kg/m3
Analysis Pressure at bottom of rising mercury
column is equal to Po 1.00atm. Po is forcing
the liquid up. When in equilibrium, the static
head pressure of the column plus the pressure
above the column (0.00atm) equals Po.
Po ?gh P 1.00atm 101,300N/m2 101,300N/m2
(13.6x103kg/m3)(9.80m/s2)h
h .760mHg h 760mmHg
h 29.9 inches Hg
As in the barometric pressure is at 29.2inches
and rising.
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Buoyant Forces
Archimedess Principle Any body completely or
partially submerged in a fluid is buoyed up by a
force whose magnitude is equal to the weight of
the displaced fluid. (What else did Archimedes
say 2200 years ago that is still repeated today?)
P1 A F1 (Downward Force) P2 A F2 (Upward
Force)
FNet,up FBuoyant F2 F1 P2A
P1A A (P2 P1) A (?gh2 ?gh1) A
?g (h2 h1) A ?g?h V?g Mg
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Example Problem (Floating Boat) A particular boat
could displace 4.50m3 of water before water would
pour over the sides of the boat. The boat itself
weighs 35,000N when filled with gasoline and
other provisions. If each fisherman wishing to go
out in the boat has a mass of 100.kg, how many
fisherman can the boat hold?
Analysis At maximum load the FBuoyant WBoat
WMen FBuoyant ?Vg (1.0x103kg/m3)(4.50m3)(9.
80m/s2) 44,100N 44,100N 35,000N
N(100.kg)(9.80m/s2) N 9.29 Fisherman
N 9 Fisherman
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Example Problem (Sunken Treasure) A box
containing .250m3 of gold bars has been located
at the bottom of the sea. How much force will it
require to lift the box upward through the water?
(Neglect the mass of the box.)
Analysis If lifting at constant velocity, Fup
Fdown FT FBuoyant Fg FT
(?water)(Vgold)(g) (?gold)(Vgold)(g)
FT 44,800 N
Notice FT (?gold ?water) Vg
FT ??Vg
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Continuity Equations for Flowing Fluids
Analysis Mass passing point 1 each second must
equal mass passing point 2 each second (at
steady-state). ?v1A1 ?v2A2
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Bernoullis Equation for Flowing Fluids
Bernoullis Equation is really a conservation of
energy equation and pressure x volume has units
of energy (N/m2 m3 N m Joule). Kinetic
and Potential Energy are also conserved.
P1 ½ (M/V) v12 (M/V) gy1 P2 ½ (M/V) v22
(M/V) gy2
P1 ½ ?v12 ?gy1 P2 ½ ?v22 ?gy2
P ½ ?v2 ?gy Constant
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Bernoullis Equation w/Closed Systems
At constant elevation P1 ½ ?v12 P2 ½ ?v22
P3 ½ ?v32 constant v2 gt v1 v3 lt v1
P2 lt P1
P3 gt P1
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Flight
v1 gt v2
?p ½ ?v12 ½ ?v22 ?p Area (wing) ?F ?F
FNet
FNet is in upward direction