Transcript Power Point File (PPF)

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
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 gravitatio nal constant
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
Fluids in RIGID BODY MOTION (FM, Ch. 3)
Example: Consider a container with a fluid in Rigid Body Motion. If
the 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
This says that the angle (or shape) of
the free-surface is determined by the
relative magnitudes of the accelerations
in the two directions.
Does that seem physically plausible ?
Buoyancy of a submerged body
free surface
h1
h2
p1=p+Lgh
1
p2=p+Lgh2
Perfroming a force balance:
Net force due to the pressure
difference
dF=(p2-p1)dA=Lg(h2-h1)dA
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
fluid displaced by the object.
This law is valid for all fluids and regardless of the shape of the body.
It can also be applied to both fully and partially submerged bodies.
Buoyancy Example
The Titanic sank when it struck an iceberg on April 14, 1912. Five
of its 16 watertight compartments were punctured when it collided with
the partially submerged iceberg. Can you estimate the percentage of the iceberg that
was 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.
Solution:
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 (cont.)
Viscosity: Due to the 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 (PDE) and their uses in Thermal Sciences:
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 PDEs not ODEs
(OrdinaryDifferential 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.
Energy
= Energy
Generation
Net
T=T(x,t).
TheStorage
temperature
varies with
time since it+has
notHeat
reach Transfer
its thermal
equilibrium.
T

 c pT  dxdydz
Cp qdxdydz
 qin qqxout qx 0dx

t
t
qx
T
The
temperature
can
also
vary
in
space
given
by
Fourier’s
cp
dxdydz  qdxdydz  qx  (qx 
dx)law:
t
xT
T
q   KA , if q  0, then
0

T x
x
 qdxdydz  (k
)dxdydz
x
x
E.g. The Heat Diffusion
Equation
in Cartesian Coordiates:
T

T

T
 T
cp
 q  (k
)  (k
)  (k
)
t
x x
y y
z z
Will be discussed at the
end of semseter
Basic equations of Fluid Mechanics
• Mass conservation (commonly called the continuity eqn.)
The rate of mass stored = the rate of mass in - the rate of mass out
dm
 m in  m out
dt
Consider fluid moving in & out of a tank as shown below.
Define the region of interest or system by the dashed lines.
(Question: Is it an open or a closed system?)
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, continutiy is given by:
dm
 ( AV )in  ( AV ) out
dt
For a steady state condition (compressible or incompressible):
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
Conservation of Mass Example
Example 1: Filling 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 2: Flow through a nozzle
V1=20 m/s
Section 2, A2=1 m2
Section 1, A1=5 m2
V2=?
m in  m out
1 A1V1   2 A2V2 , for constant density
 A1 
5
V2   V1   (20)  100(m / s)
1
 A2 
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 a well-known relationship, called
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
Case I, dz = 0, i.e. no elevation change
dp

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, i.e. as fluid flows to a lower point => dV>0, i.e. the velocity increases
and vice versa
Case III: If dV=0, no flow
dp

 gdz  0
This says that as elevation decreases, i.e. dz<0, in a static fluid, the pressure
increases, i.e. dp>0,
Example : Flow through a nozzle
Section 1
Section 2
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 assume the density of the air
as 1.185 kg/m3.
dp


2
1
 VdV  0, integrate from section 1 to section 2
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 

