Download case-study-teaching-material-intro-to

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Magnetorotational instability wikipedia, lookup

Hemodynamics wikipedia, lookup

Stokes wave wikipedia, lookup

Wind-turbine aerodynamics wikipedia, lookup

Lattice Boltzmann methods wikipedia, lookup

Euler equations (fluid dynamics) wikipedia, lookup

Lift (force) wikipedia, lookup

Flow conditioning wikipedia, lookup

Coandă effect wikipedia, lookup

Compressible flow wikipedia, lookup

Flow measurement wikipedia, lookup

Airy wave theory wikipedia, lookup

Pressure wikipedia, lookup

Magnetohydrodynamics wikipedia, lookup

Aerodynamics wikipedia, lookup

Computational fluid dynamics wikipedia, lookup

Navier–Stokes equations wikipedia, lookup

Hydraulic machinery wikipedia, lookup

Fluid thread breakup wikipedia, lookup

Rheology wikipedia, lookup

Reynolds number wikipedia, lookup

Turbulence wikipedia, lookup

Derivation of the Navier–Stokes equations wikipedia, lookup

Bernoulli's principle wikipedia, lookup

Fluid dynamics wikipedia, lookup

Transcript
FLUID MECHANICS
Fluid mechanics is the study of the behaviour of fluids (liquids and gases).
These materials will look at the various aspects of fluid behaviour.
PRESSURE DUE TO DEPTH IN A LIQUID
If we consider a point below the surface of a fluid, then we can see that there
must be sufficient pressure in the fluid to support the column of fluid above
that point.
Surface of liquids
h
Column of
liquid
Area of column
The weight (W) of the column of fluid is given by the equation
W = mg
Where m is the mass of fluid and g is the acceleration due to gravity.
Also, m = V x 
Where V = Volume and  = Density
Therefore W = V x  x g
The volume of the column of fluid is given by
V=Axh
Therefore: W = A x h x  x g
This weight must be supported by the pressure of the liquid, providing the
force F
F=PxA
Where P = Pressure of fluid
Therefore if W = F
Axhx  xg=  xA
Fluid/docs/Science/DC/SL
Therefore h x  x g = P (cancelling A)
P =  gh
Or
We can now check that the units of the equation balance
N
m2
=
kg
m3
x
m
s2
x
m
Substitute 1N = 1 Kg m
s2
Kgm
S2m2
=
Kgm2
S2m2
cancelling
Kg
ms2
=
Kg
ms2
We can see that it is important that we use values of  , g and h in base units
if we are to calculate pressure in N/m2
Fluid/docs/Science/DC/SL
CONTINUITY EQUATION
1
2
Flo
w
.
If fluid flows along a pipe then the mass flow rate (m) at any point (1) is equal
to the mass flow rate at any other point (2), assuming there are no leakages.
.
.
Therefore
m, = m2
If the fluid has constant density then as
.
m=Qx 
Where
Q = Volumetric flow rate (m3/s)
 = density of fluid (Kg/m3)
Q1 = Q 2
And
A1 x C1 = A2 x C2
Where A = cross sectioned area
And C = velocity of fluid flow
.
If the fluid does not have a constant density (gas) then we have m = m 2
Q1 x  1 = Q 2 x  2
And as Q = A x C
A1 x C1 x  1 = A2 x C2 x  2
Fluid/docs/Science/DC/SL
Bernoulli’s Equation
Bernoulli’s equation states that the total energy at one point (1) of flow in a
pipe is equal to the total energy at another point (2) in the pipe providing that:
 The flow is steady
 The points lie on a streamline
 There is no friction
1
2
Pressure Energy1 + Kinetic Energy1 + Potential Energy1
= Pressure Energy2 + Potential Energy2
P1 V1 + mC12 + mgh1 = P2 V2 +mC22 + mgh2
2
2
Where V = Volumetric flow rate
Dividing through by mass gives
P1 V1 + C12 + gh1 = P2 V2 + C22 +gh2
m
2
m
2
As
m= 
V
P1 + C12 + gh1 = P2 + C22 + gh2
2 2
2
1
Multiply by 
P1 +  C12 +  1 gh1 = P2 +  C22 +  gh2
2
2
Or we can state
P +  C2 +  gh = Constant
2
Fluid/docs/Science/DC/SL
If we consider a constant density fluid flowing horizontally (i.e. neglecting
potential energy) then we have:
P1 + C12 = P2 + C22 1
2
2
Also from continuity, we have
A1 C1 = A2 C2
2
From equation 2 we can see that as the area of flow decreases, the velocity
increases.
A1 C1 = A2 C2
Also, from equation 1 we can see that the pressure decreases.
P1 + C12 = P2 + C22
2
2
Therefore, when fluid passes over a stationary body forcing the streamlines to
get closer together, the velocity must increase (as the area decreases), and
therefore the pressure decreases. This is how an airfoil works, the flow over
the top surface is faster than the flow over the bottom surface, therefore there
is low pressure on the top and an upward force is produced on the airfoil (lift)
The Bernoulli equation allows us to determine how energy transfers take
place between elevation, velocity head and pressure head. We can examine
piping systems and determine the variation of fluid properties and the balance
of energy. Sometimes the changes in the energy levels are confusing and
seem to contradict common sense; students can often find the outcomes of
the Bernoulli equation confusing. For example, if an incompressible fluid is
flowing along a horizontal pipe which expands in diameter, then, as the area
increases the velocity must decrease (continuity, Q = CA). As there is no
change in potential head (horizontal pipe), the loss in velocity head must
result in a gain in pressure head (conservation of energy or head). Therefore,
as the velocity decreases the pressure increases. This may seem counterintuitive (surely fluid cannot flow in the direction of increasing pressure),
however, we must remember that the fluid has inertia and the increase in
pressure does have the effect of decelerating the fluid (F=ma), in the same
way that a car moves forward when a brake (force in the opposite direction to
motion) is applied.
There are many fluid problems to which Bernoulli can be applied, it can be
applied to problems in which more than one flow may enter or leave the
system at the same time, or to series and parallel piping systems.
Fluid/docs/Science/DC/SL