Aula 1
Sandro R.
Lautenschlager
Mecânica dos Fluidos
1-Considerações
Básicas
Mecânica dos Fluidos
• Definição
– Estudo dos líquidos e gases no qual não há
movimento (estáticos) e naqueles no qual há
movimentos (dinâmica)
History
Faces of Fluid Mechanics
Archimedes
(C. 287-212 BC)
Navier
(1785-1836)
Newton
(1642-1727)
Stokes
(1819-1903)
Leibniz
(1646-1716)
Reynolds
(1842-1912)
Bernoulli
Euler
(1667-1748)
(1707-1783)
Prandtl
Taylor
(1875-1953)
(1886-1975)
Weather & Climate
Tornadoes
Thunderstorm
Global Climate
Hurricanes
Vehicles
Aircraft
High-speed rail
Surface ships
Submarines
Environment
Air pollution
River hydraulics
Physiology and Medicine
Blood pump
Ventricular assist device
Sports & Recreation
Water sports
Auto racing
Cycling
Offshore racing
Surfing
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Propriedades dos Fluidos
  g
• Peso específico
Onde:
g-gravidade local (9,8 m/s2);
-massa especifica (kg/m3);
-peso específico (kg/m2s2) ou N/m3
Propriedades dos Fluidos
• Densidade
S

 água


 água
Temperatura de referência 40C
Variação de  e  com a
temperatura
H O
2

T  4
 1000
H O
2
2
180

T  4
 9800
2
18
Massa Especifica (H2O)
3
Massa especifica (kg/m )
Massa Específica
1001.0
1000.0
999.0
998.0
997.0
996.0
995.0
994.0
993.0
992.0
H O
2

T  4
 1000
25
30
2
-15
-10
-5
0
5
10
15
20
0
Temperatura ( C)
180
35
40
45
 H 2O(00 C)   H 2O(40 C)
Peso Específico
Peso Especifico (H2O)
Massa especifica (kg/m3)
9810.0
9800.0
H O
9790.0
2
2

T  4
 9800
18
9780.0
9770.0
9760.0
9750.0
9740.0
9730.0
9720.0
-15
-10
-5
0
5
10
15
20
25
Temperatura (0C)
30
35
40
45
Nas condições Normais
Massa
Especifica
(kg/m3)
Peso
Especifico
(N/m3)
Densidade
Ar
1,23
12,1
0,00123
Água
1000
9810
1
Some Simple Flows
• Flow between a fixed and a moving
plate
Fluid in contact with the plate has the same
velocity as the plate
u = x-direction component of velocity
y
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
x
u=0
Some Simple Flows
• Flow through a long, straight pipe
Fluid in contact with the pipe wall has the
same velocity as the wall
u = x-direction component of velocity
R
r
x
  r 2 
u (r )  V 1    
  R  
V
Fluid
Fluid Deformation
• Flow between a fixed and a moving
plate
• Force causes plate to move with
velocity V and the fluid deforms
continuously.
y
Moving plate
t0
u=V
t1 t2
Fluid
Fixed plate
x
u=0
Fluid Deformation
Shear stress on the plate is proportional
to deformation rate of the fluid
da 
dL
dy
dt 
dL
dV
da dV

dt dy


da
dt
dV
dy
y
dL
t
da
dy
Moving plate
u=V+dV
t+dt
dx
Fluid
Fixed plate
x
u=V
Shear in Different Fluids
• Shear-stress relations for
different types
of fluids
• Newtonian fluids: linear
relationship
• Slope of line (coefficient of
proportionality) is “viscosity”
dV
dy
dV
 
dy

Viscosity
•
Newton’s Law of Viscosity
 
•
Viscosity
•
Units
•
N / m2 N  s

m / s / m m2
o
Water (@ 20 C)
–
•

dV / dy
V+dv
 = 1x10-3 N-s/m2
Air (@ 20oC)
–
•

dV
dy
 = 1.8x10-5 N-s/m2
Kinematic viscosity



V
Flow between 2 plates
Force is same on top
and bottom
F1  1 A1   2 A2  F2
A1  A2
1   2
y
1  
du
du

2
dy 1
dy 2
Thus, slope of velocity
profile is constant and
velocity profile is a st. line
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
Force acting
ON the plate
x
u=0
Flow between 2 plates
Shear stress anywhere
between plates
 
du
V

dy
B
  0.1 N  s / m 2 ( SAE 30 @ 38o C )   (0.1 N  s / m 2 )( 3 m / s )
0.02 m
V  3 m/s
B  0.02 m
y
 15 N / m 2
Moving plate
V
B
u( y) 
V
y
B
u=V


Fixed plate
Shear
on fluid
x
u=0
Flow between 2 plates
• 2 different coordinate systems
B
r
x
  r 2 
u (r )  V 1    
  B  
V
u ( y)  C yB  y 
y
x
Example: Journal Bearing
• Given
–
–
–
–
–
–
Rotation rate, w = 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
Shear Stress   
dV
w (d / 2)

dy
(D  d ) / 2
 2

*
1500

(0.06 / 2)
60

 (0.88*1000* 0.003) 
 124kN / m 2
(0.0002) / 2
Torque M  (2
d d
l)
2 2
 (2 *124,000 *
0.06
0.06
* 0.4)
 281 N  m
2
2
Power P  Mw  281*157.1  44,100 N  m / s  44.1 kW
Compressibilidade
Módulo de elasticidade volumétrica
p 
p

B  lim 

lim

V 0
 V / V  T  0  / 
p
 V
V
T
p


T
T
Módulo de elasticidade para água
• B=2100MPa ou
21000x atm
Mudança 1%
H O
21Mpa (210atm)
2
Líquidos podem
ser considerados
incompressíveis
Cálculo velocidade do som
c
p


T
B

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
NP1  NP2  NP3
Média 
3
Média  6,0 Aprovaçãodireta
Média  6,0 Deverá realizar avaliação final
Avaliação final Cap. 1, 2, 3, 4 e 6