Transcript fluid-1

FLUID MECHANICS
Basic Fluid Properties and Governing Equations
Density (): mass per unit volume (kg/m3 or slug/ft3)
Specific Volume (v=1/): volume per unit mass
Temperature (T): thermodynamic property that measures the molecular activity
of an object. It is used to determine whether an object has reached thermal
equilibrium.
Pressure (p):pressure can be considered as an averaged normal force exerted
on a unit surface area by impacting molecules.
( P  lim  F , N/m2 or pascal; lb/in2 or psi)
A 0
 A
Pascal law: (under static condition) pressure acts uniformly in all directions. It also
acts perpendicular to the containing surface.
If a fluid system is not in motion, then the fluid pressure is equal its thermodynamic
pressure.
Atomspheric pressure (patm): pressure measured at the earth’s surface.
1 atm = 14.696 psi = 1.01325 x 105 N/m2 (pascal)
Absolute pressure: pressure measured without reference to other pressures.
Gage pressure: pgage= p absolute - patm
Properties (cont’d)
We begin with the simplest case – FLUID STATICS – the study of fluids at rest
or in rigid-body motion (Chapter 3 – YAC).
Pressure Measurement -Atmospheric pressure can be measured using a barometer:
p=0
Vacuum
p=0
Patm=1.01x105 Pa
L
p=patm
Force balance
patm A  W  mg  ALg
Patm  gL
 is the density of the fluid, g is the gravitational constant
Note: Patm is an absolute pressure since it is referenced to vacuum (more on this later)
Pressure Measurement (cont’d) YAC: 3-1 – 3-3
(Read: infinitesimal)
Similarly, this balance can be applied to a small fluid element as shown
pA  ( p  dp) A  mg  Agdy,
dp/dy = - g
Free surface, p=p
h
p
y
dy
x
p+dp
dp
  g, integrate from fluid element to
dy
the free surface p(h)  p  gh
Rigid Body Motion – (YAC: 3-7)
Example: If a container of fluid is accelerating with an acceleration of ax to the
right as shown below, what is the shape of the free surface of the fluid?
Balancing the forces and applying
Newton’s 2nd Law:
Balancing the forces
on a fluid element
p
p+dp
ax
dy
dp
pA  ( p  dp ) A  ma x   Adxa x , 
  ax
dx
dp
a 
dy
 g ax
tan(a ) 

 , a  tan 1  x 
dx dp
g
 g 
 ax
a
dx
Buoyancy of a submerged body (YAC: 3-6)
free surface
h1
h2
p1=p+Lgh
1
Net force due to pressure difference
dF=(p2-p1)dA=Lg(h2-h1)dA
p2=p+Lgh2
Total net force (buoyancy)
z z
FB  dF   L g (h2  h1 )dA   L gVdisplaced
The Principle of Archimedes:
The buoyancy acting on a submerged object is equal to the weight of the
displaced fluid due to the presence of the object.
This law is valid for all fluid and regardless of the shape of the body. It can
also be applied to both fully and partially submerged bodies.
Buoyancy (cont’d)
Example: Titanic sank when it struck an iceberg on April 14, 1912. Five
of its 16 watertight compartments were punctuated when it collides with
the iceberg underwater. Can you estimate the percentage of the iceberg that
is actually beneath the water surface? It is known that when water freezes
at 0 C, it expands and its specific gravity changes from 1 to 0.917.
When the iceberg floats, its weight balances
the buoyancy force exerted on the iceberg by
the displaced water.
weight
W = FB
 ice gVice  berg   water gVsubmerged
Vsubmerged
buoyancy
Vice  berg

 ice
0.917

 91.7%
 water
1
Therefore, more than 90% of the iceberg
is below the water surface.
Fluid Properties - Viscosity
Due to interaction between fluid molecules, the fluid flow will resist a shearing motion.
The viscosity is a measure of this resistance.
Moving Plate
constant force F
constant speed U
H
Stationary Plate
From experimental observation, F  A(U/H)=A(dV/dy)
F dV

, where  is shear stress
A dy
dV
lb sec N sec
 
, where  is dynamic viscosity , The unit of  is
or
dy
ft 2
m2


ft 2
m2
kinematic viscosity   , has unit of
or

sec sec
Boundary Layers
Immediately adjacent to a solid surface, the fluid particles are slowed
by the strong shear force between the fluid particles and the surface.
This relatively slower moving layer of fluid is called a “boundary layer”.
Laminar
dV
 
dy
Turbulent
Question: which profile has larger wall shear stress?
In other words, which profile produces more frictional
drag against the motion of the solid surface?
Partial Differential Equations
Many physical phenomena are governed by PDE since the physical functions
involved usually depend on two or more independent variables (ex. Time, spatial
coordinates). Their variation with respect to these variables need to be described by
PDE not ODE (Ordinary Differential Equations).
Example: In dynamics, we often track the change of the position of an object
in time. Time is the only variable in this case. X=x(t), u=dx/dt, a=du/dt.
In heat transfer, temperature inside an object can vary with both time and space.
T=T(x,t). The temperature varies with time since it has not reach its thermal
equilibrium.
T
Cp
 qin  qout  0
t
The temperature can also vary in space as according to the Fourier’s law:
T
T
q   KA , if q  0, then
0
x
x
Basic equations of (motion) Fluid Mechanics
• Mass conservation (continuity equation):
The rate of mass stored = the rate of mass in - the rate of mass out
dm
 m in  m out
dt
m in
Area A
L
m  V
m out
Within a given time t, the fluid element with
a cross-sectional area of A moves a distance of
L as shown.
The mass flow rate can be represented as
m
L
m 
 A
 AV
t
t
Hence, for constant density (incompressible) flows, continuity is given by:
dm
 ( AV )in  ( AV ) out
dt
For steady state condition: mass flow in = mass flow out
( AV )in  ( AV )out
Now, looking at the left hand term:
dm d ( V )
dV
d


V
 m in  m out
dt
dt
dt
dt
Keep density constant,
volume changes with time
ex: blow up a soap bubble
Keep volume constant,
density changes with time
ex: pump up a basketball
Example: filling up an empty tank
h(t)
Water is fed into an empty tank using
a hose of cross-sectional area of 0.0005 m2.
The flow speed out of the hose is measured
to be 10 m/s. Determine the rate of increase
for the water height inside the tank dh/dt.
The cross-sectional area of the tank is 1 m2.
0, no mass out
dm
dV
dh

 Atan k
 m in  m out  AhoseV
dt
dt
dt
dh  Ahose 
 0.0005 



V 
(10)  0.005(m / s)

dt  Atan k 
 1 
Example: Incompresible Flow Through a Nozzle
Section 2, A2=1 m2
V1=20 m/s
Section 1, A1=5 m2
m in  m out
1 A1V1   2 A2V2 , for constant density
 A1 
5


V2   V1   (20)  100(m / s)
1
 A2 
V2=?
Momentum Conservation: (Newton’s second law)
Net external forces lead to the change of linear momentum
 d

 F  dt (mV )
p+dp
First, neglect viscosity

dL
dz
dx
P
This is the well-known Euler’s Equation
d
(mV )
dt
 dz 
 dV 
 Adp  A(dL) g    A(dL)

 dL 
 dt 
 dV dL 
 A(dL)
  A(VdV )
 dL dt 
dp

 gdz  VdV
pA  ( p  dp) A  W sin  

dp

 VdV  gdz  0
Simplifications/Special Cases of Euler’s Equation
dp

 VdV  gdz  0
dp
Case I, dz=0 no elevation change

VdV  0
Implication: If the flow accelerates, dV>0, then
pressure has to decrease, i.e. dP < 0 and vice versa.
 Case II: If dp=0, no external pressure gradient
VdV  gdz  0
If dz<0, fluid flows to a lower point, dV>0, its velocity increases
and vice versa
Case III: If dV=0, no flow
dp

 gdz  0
If dz<0, into the lower elevation inside the static fluid system,
dp>0, pressure increases
Example: Flow Through a Nozzle
Section 2
Section 1
Air flows through a converging duct as shown. The areas at sections 1 & 2
are 5 m3 and 1 m3, respectively. The inlet flow speed is 20 m/s and we know
the outlet speed at section 2 is 100 m/s by mass conservation. If the pressure
at section 2 is the atmospheric pressure at 1.01x105 N/m2, what is the pressure
at section 1. Neglect all viscous effects and given the density of the air
as 1.185 kg/m3. dp
 VdV  0, integrate from section 1 to section 2


2
1
dp

2
   VdV , 
1
p2  p1

V12  V22

2

p1  p2    V22  V12
2
 1.185 
5
2
2
5
 1.0110  
(100  20 )  1.0110  5688( Pa)
 2 

