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1-Considerações Básicas
Sandro R. Lautenschlager
Mecânica dos Fluidos
Aula 1
Mecânica dos Fluidos
• Definição– Estudo dos líquidos e gases no qual não há
movimento (estáticos) e naqueles no qual há movimentos (dinâmica)
HistoryFaces of Fluid Mechanics
Archimedes(C. 287-212
BC)
Newton(1642-1727)
Leibniz(1646-1716)
Euler(1707-1783)
Navier(1785-1836)
Stokes(1819-1903)
Reynolds(1842-1912)
Prandtl(1875-1953)
Bernoulli(1667-1748)
Taylor(1886-1975)
Weather & ClimateTornadoes
HurricanesGlobal Climate
Thunderstorm
VehiclesAircraft
Submarines
High-speed rail
Surface ships
Environment
Air pollution River hydraulics
Physiology and Medicine
Blood pump Ventricular assist device
Sports & Recreation
Water sports
Auto racing
Offshore racingCycling
Surfing
Aplicação em engenharia
Computer Simulations Save US $150,000 in Commuter Jet Design by Reducing Wind Tunnel Testing
Aplicação em engenharia
Aplicação em engenharia
Aplicação em engenharia
Propriedades dos Fluidos
• Peso específico
g Onde:
g-gravidade local (9,8 m/s2);
-massa especifica (kg/m3);
-peso específico (kg/m2s2) ou N/m3
Propriedades dos Fluidos
• Densidade
águaágua
S
Temperatura de referência 40C
Variação de com a temperatura
e
180
41000
2
2
TOH
18
49800
2
2
TOH
Massa Específica
992.0993.0994.0995.0996.0997.0998.0999.0
1000.01001.0
-15 -10 -5 0 5 10 15 20 25 30 35 40 45
Temperatura (0C)
Mas
sa e
spec
ifica
(kg/
m3) Massa Especifica (H2O)
180
41000
2
2
TOH
)4( )0( 02
02 COHCOH
9720.0
9730.0
9740.0
9750.0
9760.0
9770.0
9780.0
9790.0
9800.0
9810.0
-15 -10 -5 0 5 10 15 20 25 30 35 40 45
Temperatura (0C)
Mas
sa e
spec
ific
a (k
g/m
3 )
Peso Especifico (H2O)
Peso Específico
18
49800
2
2
TOH
Massa Especifica
(kg/m3)
Peso Especifico
(N/m3)
Densidade
Ar 1,23 12,1 0,00123
Água 1000 9810 1
Nas condições Normais
Some Simple Flows• Flow between a fixed and a moving
plateFluid in contact with the plate has the same velocity as the plate
u = x-direction component of velocity
u=VMoving plate
Fixed plate
y
x
V
u=0
B yB
Vyu )( Fluid
Some Simple Flows
• Flow through a long, straight pipeFluid in contact with the pipe wall has the same velocity as the wall
u = x-direction component of velocity
rx
R
21)(
R
rVru
VFluid
Fluid Deformation• Flow between a fixed and a moving
plate• Force causes plate to move with
velocity V and the fluid deforms continuously.
u=VMoving plate
Fixed plate
y
xu=0
Fluid
t0 t1 t2
u=V+VMoving plate
Fixed plate
y
x
u=V
Fluid
Fluid Deformation
t t+t
x
y
L
t Shear stress on the plate is proportional
to deformation rate of the fluid
V
Lt
y
L
y
V
t
y
V
Shear in Different Fluids
• Shear-stress relations for different typesof fluids
• Newtonian fluids: linear relationship
• Slope of line (coefficient of proportionality) is “viscosity”
dy
dV
dy
dV
Viscosity• Newton’s Law of Viscosity
• Viscosity
• Units
• Water (@ 20oC)– = 1x10-3 N-s/m2
• Air (@ 20oC)– = 1.8x10-5 N-s/m2
• Kinematic viscosity
dydV /
2
2
//
/
m
sN
msm
mN
V
V+dv
dy
dV
Flow between 2 plates
u=VMoving plate
Fixed plate
y
x
V
u=0
B yB
Vyu )( Fluid Force acting
ON the plate
21
21
222111
AA
FAAF
221
1 dy
du
dy
du
Thus, slope of velocity profile is constant and velocity profile is a st. line
Force is same on topand bottom
Flow between 2 plates
u=VMoving plate
Fixed plate
y
x
V
u=0
B yB
Vyu )(
B
V
dy
du Shear stress anywherebetween plates
Shearon fluid
mB
smV
CSAEmsN o
02.0
/3
)38@30(/1.0 2
2
2
/15
)02.0
/3)(/1.0(
mN
m
smmsN
Flow between 2 plates
• 2 different coordinate systems
rx
B
21)(
B
rVru
Vy
x yByCyu )(
Example: Journal Bearing
• Given– Rotation rate, = 1500 rpm– d = 6 cm– l = 40 cm– D = 6.02 cm
– SGoil = 0.88
– oil = 0.003 m2/s
• Find: Torque and Power required to turn the bearing at the indicated speed.
Example: cont.
• Assume: Linear velocity profile in oil film
2/1242/)0002.0(
)2/06.0(1500*602
)003.0*1000*88.0(
2/)(
)2/( StressShear
mkN
dD
d
dy
dV
mN
dl
dM
2812
06.0)4.0*
2
06.0*000,124*2(
2)
22(Torque
kWsmNMP 1.44/100,441.157*281Power
Compressibilidade
TT
TTV
p
V
pV
p
VV
pB
/
lim/
lim00
Módulo de elasticidade volumétrica
Módulo de elasticidade para água
• B=2100MPa ou 21000x atm
Mudança 1% OH2
21Mpa (210atm)
Líquidos podem ser considerados incompressíveis
Cálculo velocidade do som
Bp
cT
Exemplo para água c = 1450m/s (condições normais )
Data da Provas
• P1 17/08/11 Cap. 1 e 2;
• P2 05/10/11 Cap. 3 e 4;
• P3 21/11/11 Cap 4 e 6.
• Exame 5/12/11
final avaliação realizar Deverá 0,6Média
direta Aprovação 0,6Média3
NPNPNPMédia 321
Avaliação final Cap. 1, 2, 3, 4 e 6
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