Fluid Mechanics
description
Transcript of Fluid Mechanics
Chapter 10
DensityRecall that the density
of an object is its mass per unit volume (SI unit is kg/m3)
The specific gravity of a substance is its density expressed in g/cm3
V
m
Pressure in FluidsFluids exert a pressure in all directions
A fluid at rest exerts pressure perpendicular to any surface it contacts
The pressure at equal depths within a uniform fluid is the same
Area
Force
A
FP
))()(( depthgdensityghP
SI Unit for Pressure is Pa1 Pa= 1 N/m2
1 atm= 101.3 kPa=760 mm-Hg
Pressure in FluidsGauge Pressure is a measure of the pressure
over and above the atmospheric pressurei.e. the pressure measured by a tire gauge is
gauge pressure. If the tire gauge registers 220 kPa then the absolute pressure is 321 kPa because you have to add the atmosphere pressure (101 kPa)
If you want the absolute pressure at some depth in a fluid then you have to add atmosphere pressure ghPP o
Pressure in FluidsPascal’s Principle: Pressure applied to a fluid
in a closed container is transmitted equally to every point of the fluid and to the walls of the container
2
2
1
1
A
F
A
F
BuoyancyBuoyant force is the force
acting on an object that is immersed in a fluid
Archimedes Principle: The buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the objectSince the buoyant force acts
opposite of gravity, an object seems to weigh less in a fluid
Apparent Weight= Fg-FB
Fb = Buoyant Force
Fg = Gravity
Sinking vs FloatingThink back to free body diagrams
If the net external force acting on an object is zero then it will be in equilibrium
If Fb=Fg then the object will be in equilibrium and will FLOAT!
Fg
FB
Cubes floating in a fluid
70% Submerged 100% Submerged20% Submerged
Density determines depth of submersionThis equation gives the
percent of the object’s volume that is submergedVf is the volume of fluid
displacedVo is the total volume of
the objectρo is the density of the
objectρf is the density of the
fluid
f
o
o
f
V
V
Summary of FloatingFor an object to float
FB= Fgρo ≤ ρf
If ρo = ρf then the Object will be
Completely submergedBut not sinking.
If ρo is less than ρf Then the amount
submerged can be found with
f
o
o
f
V
V
Continuity Equation•Continuity tells us that whatever the volume of fluid in a pipe passing a particular point per second, the same volume must pass every other point in a second.
•If the cross-sectional area decreases, then velocity increases
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Equation Continuity
vAvA
The quantity Av is the volume rate of flow
Bernoulli’s PrincipleThe pressure in a fluid decreases as the
fluid’s velocity increases.
Fluids in motion have kinetic energy, potential energy and pressure
How do planes fly?
Bernoulli’s EquationThe kinetic energy of a fluid element is:
The potential energy of a fluid element is:
ghVmghPE )(
Bernoulli’s EquationThis equation is essentially a statement of
conservation of energy in a fluid. Notice that volume is missing. This is because this equation is for energy per unit volume.
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211 2
1
2
1
Equation sBernoulli'
ghvPghvP
Sample Problem p. 306 #40What is the lift (in newtons) due to
Bernoulli’s principle on a wing of area 80 m2. If the air passes over the top and bottom surfaces at speeds of 350 m/s and 290 m/s, respectively.Let’s make point 1 the top of the wing and
point 2 the bottom of the wingThe height difference between the top of the
wing and the bottom is negligible
22221
211 2
1
2
1ghvPghvP
Sample Problem p.306 #40The net force on the wing is a result of the
difference in pressure between the top and the bottom. P1 is exerted downward, P2 is exerted upward
If we know the difference in pressure we can use that to find the force 2
22211 2
1
2
1vPvP
upward Pa 20318)290340)(/29.1(2
1
2
1
2
1 22322
2112 mkgvvPP
Sample Problem p.306 #40P2-P1=20318 Pa
upward 1063.1)80)(20318()( 6212 NxmPaAPPF
P.306 #43Water at a pressure of 3.8 atm at street level
flows into an office building at a speed of 0.60 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6 cm in diameter by the top floor, 20 m above street level. Calculate the flow velocity and the pressure in such a pipe on the top floor. Ignore viscosity. Pressures are gauge pressures.
Find the flow velocity at the top
A1 is area of first pipe= πr2 = 1.96x10-3 m2
A2 is area of second pipe= πr2 = 5.31x10-4 m2 V1= 0.6 m/s
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Equation Continuity
vAvA
s
m
mx
smmx
A
vAv 2.2
1031.5
)/6.0)(1096.1(24
23
2
112
Find pressure at the top
22221
211 2
1
2
1ghvPghvP
222
2112 2
1
2
1ghvvPP
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2 19620024201801084.3m
N
m
N
m
NPaxP
P2= 1.86 x 105 Pa= 1.8 atm
Sample Problem p.305 #37What gauge pressure in the water mains is
necessary if a fire hose is to spray water to a height of 12.0 m?Let’s make point 1 as a place in the water main
where the water is not moving and the height is 0
Point 2 is the top of the spray, so v=0 , P= atmospheric pressure, height = 12m
22221
211 2
1
2
1ghvPghvP
Sample Problem p.305 #37
)12)(m/s81.9)((1000kg/m P 23atm21 mghPP
Pax 5atm1 102.1PP Pressure Gauge
Remember that Gauge Pressure is the pressure above atmosphericpressure. So to get gauge pressure, we need to subtract atmosphericPressure from absolute pressure.