cxxLjevans
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Posts posted by cxxLjevans
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Larry, I didn't notice the density appearing in my equation, so how could it affect it?
Please bear with me I haven't labelled my variables etc, as I hope that all will become clear in the next post(s).
But your post has encouraged me that the interest is still there to complete this discussion.
Good point. That equation does just describe the work on the airfoil
body. On the other hand, the energy of airfoil and fluid would require
some density terms. After all:
[math]m={\rho Vol_foil}
where Vol_foil is the volume of the airfoil. Anyway, the point
is as the airfoil rises, it gravity potential increased but
the fluid must fall and equal amount, and since the densities
are the same, there's no change in the sum of gravitational
potential energies of airfoil and fluid.
OOPS. I now realize that the two points at which the
fluid velocities and pressures height were measured
take that into account because at the airfoil is in between
these two points and aren't displacing the airfoil volume
at either of these points. Now I realize why someone
(maybe you) said density doesn't matter). Sorry for noise
Also, density:
[math]\rho
was removed from the previous equation:
[math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2} = H = {\rm{A constant}}
from some reason, and that equation describes energy in the fluid.
Also, the above equation should have some velocity squared terms
instead of just velocity terms according to:
http://en.wikipedia.org/wiki/Hydraulic_head
where it's called the velocity head.
However, you did have velocity squared in an earlier equation:
Hence, it was probably just a typo(using math markup is difficult).
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[snip]
The work equation for any object lefted in a gravitational field is
[math]W = \left( {\frac{1}{2}mV_2^2 - \frac{1}{2}mV_1^2} \right) + \left( {mg{z_2} - mg{z_1}} \right)[/math]
[snip]
But if the fluid has the same density as the airfoil, then as the airfoil rises the fluid falls
the same amount, and wouldn't that just cancel the 2nd term:
[math] \left( {mg{z_2} - mg{z_1}} \right)[/math]
in the W equation, if it were included in the W equation?
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And, for energy to be conserved, then in the rising case (where there is an increase in gravitational potential energy in the form of the risen body) there must be an equal decrease in another form of energy. And the only
Since the body has the same density as the fluid (one of the assumptions)
I'm not sure there is any increase in gravitational potential energy due
to rise of the airfoil. Am I missing something?
-Larry
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Larry, how would you start the motion?
Start with the fluid height in the left vertical channel
X meter higher than that in the right vertical channel.
This height difference could be enforced by a door of
some sort. The experiment starts when that door
is instantly removed.
I'd guess that whatever energy is used to raise the
airfoil would be reflected in the difference in height
in the opposite channel after the opposite channel
reached its maximum height during an oscillation.
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J.C.MacSwell,
You seem to know a thing or two about fluids so I don't want to try to "teach my grandfather to suck eggs".
The analysis of fluid action can be approached from at least two viewpoints.
The Eulerian approach of following a parcel of fluid and calculating its path (position), change of shape, orientation and other properties.
The control volume approach of fixing a specific volume in space so that it does not vary in shape or position and calculating the flows of energy, momentum and mass across its boundaries.
The point I have been trying to make is how to select the most appropriate for the issue at hand and in particular that these two approaches should not be mixed, as Zet is trying to do.
The examples in post83 are best handled by the control volume approach.
The entire L shaped duct is made into the control volume.
Attempting to use this control volume approach to the open air flow will fail because the volume must extend to infinity, meaning that the energy supply is infinite so the prime statement/assumption 'energy is conserved' is violated.
Would the it be simpler to answer Zet's question if the shape of the container were a U instead of an L?
In this case, the fluid would simply oscillate between the left and right vertical columns of the U
until something stopped it. I, and I think Zet, would assume, with frictionless flow, it would
oscillate forever if the arifoil were constrained vertically; however, if it were allowed to rise,
then, again I'm guessing, the energy used to raise the airfoil would dampen the ozcillation
until ist stopped or the arifoil reached its vertical limit.
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There would be no drag, and no lift, if that was the case.
The Euler equation:
http://www.grc.nasa.gov/WWW/k-12/airplane/eulereqs.html
assume viscosity can be neglected:
The Euler equations neglect the effects of the viscosity of the fluid
So are you saying the Euler equatons would not predict lift
for an airfoil. But then why does a further quote from same page:
For some problems, like the lift of a thin airfoil at low angle of attack, a solution of the Euler equations provides a good model of reality.
indicates that it would? Hmm, maybe your saying that with Euler
turbulence and vortices would be present and they would produce lift?
Sorry, I don't understand. Could you explain more?
-regards,
Larry
Larry,
The link to SU2 was fascinating (post #108). I was a c (… not a c++ …) programmer back in the mid-nineties. And … I’m guessing some things have probably changed by now. (If you ever need a mean DOS batch file programmer, I’m your guy.)
I couldn’t get the NASA link to work (post #112). I reloaded Java and downloaded the Java Control Panel and followed their instructions under “security settings” for Windows 8 and greater but I couldn’t get the Java Control Panel to let me type or paste in the web site URL in the excepted list box. I guess I’m getting old. It looks cool, and perhaps very promising for perhaps resolving this thread. I’ll keep trying to get it to work on my (new but very low end) laptop.
Could this piece of software (even if only in beta and even if only approximately) answer/address whether or not there would be the same decrease in the swirl for differently shaped airfoils of equal mass and volume all rising the same distance as opposed to when they don’t rise?
That would be exciting.
[snip]
Hi Zet,
I also, initially, could not get the java applet to work. I had to make some adjustments
to my FileFox browser prefserences (or something having to do with enableing
java applets, but I forget what exaclty) but eventually I got it to work.
I've looked at the code (which you can if you download) briefly, and I found
code that mentioned viscosity, but viscosity of different forms. I've yet to
figure out how to allow the airfoil to rise; hence, I don't know whether it can answer
the question.
I tried emailing the author listed on the web page (I think it was Thomas Benson)
;however, my mailer said there was no longer such an email address
I've also looked at the SU2 docs, but, agian, it willl talk some time to figure out
how to use it and graph the results to make the more understandable.
I did ask for help here:
http://www.cfd-online.com/Forums/cfd-freelancers/148415-effect-rising-hydrofoil-fluid.html
and got a "private" reply where some guy offered help for a "reasonable price".
I'm not sure I want to pay, but paying might be less work than going on forever here
I probably try my hand using su2 (since I've been a c++ programmer for some time)
and if it gets too hard, I might pay the money.
-regards,
Larry
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Blind misapplication of Bernoulli can lead you wildly astray.
Here are some more conundrums to ponder.
With reference to the diagram.
An airflow is proceeding from left to right with steady velocity V and pressure P until it encounters an obstacle.
Consider two parcels of air at A and B with such velocity V and pressure P.
The 'Bernoulli Explanation' is supposed to run as follows:
Parcel A passes over the top of the obstruction, gaining velocity over the stream velocity, Which I have shown as V+ at C, D, and E then finally returning to match the stream at F at velocity V.
As a result it is claimed that Bernoulli's theorem says the static pressure must fall below stream pressure at C, D, and E (shown as P-) returning to p at F.
In mirror fashion parcel B passes under the obstruction and looses velocity, denoted V- but gains pressure at G, H and I (denoted P+) , returning to V and P at J.
So the pressure difference exerts a net upward force on the obstruction, causing it to rise and extract energy from the stream.
Question 1)
If parcels A and B between them loose some energy to the obstruction, where does the energy come from to rejoin the stream at stream pressure and velocity?
Question 2)
When parcel A has velocity V+ and pressure P-, then its pressure is less than the pressure of the air immediately above it in the undisturbed stream.
So why does it not rise.
Question3)
Similarly when parcel B has velocity V- and pressure P+ its pressure is greater than the air immediately below it so why does it not fall?
If the angle of attach is positive, I can see why the velocity below would decrease and the pressure
would increase. However, if the angle of attach was 0, and the airfoil was symmetrical,
I would think the pressure above would exactly mirror the pressure below.
At:
http://www.grc.nasa.gov/WWW/k-12/airplane/foil3.html
when a symmetric airfoll (an ellipse) is specified with 0 angle of attack,
the pressures above an below look the same.
The velocities both look same too.
Could you explain more the reasoning behind your
pressures? Is your airfoil not symmetric or is
the angle of attack positive?
TIA.
-Larry
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Why does it not occur anyway? (it occurs more so...but baby steps)
The airfoil is slowing the wind. Drag is being created. What makes you think it slows the wind more by allowing it to rise?
If I do a brake stand in my car, revving the engine at full power, and finally take my foot off the brake, are you going to ask "I wonder where the energy came from to accelerate the car?"
Would there be drag if there was no viscosity and no turbulence and no vortices created?
I think in this thought experiment, those were the assumptions.
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[snip]
Now I explained, back in posts# 93, 98 & 100, that Zet has not provided enough information to distribute energy added to or taken from the fluid between these three, using Bernoulli.
A possible key missing variable is the volumetric flowrate.
[snip]
Couldn't the missing variables simply be given a name, such as VolFlowRate,
and the a solution found using in terms of that name? If not, then pick
some value for the missing varable and solve the problem.
Maybe you mean the volumetric flow rate at the start of the
channel before encountering the airfoil. I guess you could
just set that to 1 in some aribitrary units and set the
density to 1 (of both airfoil and fluid), and viscosity to 0.
IOW, for every term in the Bernoulli equatoin, pick a value
and solve the problem. Why wouldn't that work?
OOPS. Bernoulli s for steady state. Hmm. So
maybe that wouldn't work. Maybe just using some
CFD code, such as:
could answer the problem; however I suspect
it would require a lot of time to learn how to do
that
; however, since this thread seems tobe going on for so long, maybe resorting to
using a CFD to answer the question would
be desirable.
-regards,
Larry
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[snip]
> To do this look at my diagram. I have drawn two
> streamlines BC and AD and two more (B'C' and A'D') exactly
> 1 metre alongside so they form a square box section stream
> tube. The streamlines are above and below a 1 metre
> section of airfoil. So everything in the third dimension
> is measured per metre.
[snip]
> Both sections have the same area so
>
> U1x = U2x
>
> This is the obstruction does not slow the fluid
> horizontally, in accordance with D'Alembert. What does
> happen is a vertical downward velocity Uy is imparted to
> the fluid.
How? Maybe this:
http://en.wikipedia.org/wiki/Lift_%28force%29#Flow_deflection_and_Newton.27s_laws
provides an explanation for the downward velocity; however,
I haven't read it closely. I'd guess a downward flow is
caused by a positive angle of attack, like that experienced
by your flattened hand angled up and held outside of a
speeding car window.
In addition, this page:
http://www.grc.nasa.gov/WWW/K-12/airplane/shed.html
shows a "downwash" which looks like it *may* correspond to
your Uy at the exit to your stream tube.
However, my question (and I suspect Zet's ) is what happens
when there's 0 angle of attack and the only lift is provided
by the difference in the static presssure between the upper
and lower surfaces of the airfoil. Bernoulli said there'd
be a change in pressure caused by the fluid moving faster
over the top and that produces the lift. And (like Zet),
I'm wondering where the energy comes from causing that rise?
Does the fluid flowing under the wing slow down or decrease
in pressure or both in order to compensate for the energy
required to raise the foil?
>
> The mass flowrate equals the fluid density times the
> volumetric flowrate,
>
> and the momentum flowrate equals the mass flowrate times
> the horizontal or vertical velocity.
>
> At section1 there is zero vertical momentum in the fluid
> but at section 2 there is a downwards momentum flow of
>
> (pUxA2)Uy
>
Now, if we assume the flow conditions used to derive
Bernoulli's equation:
* Velocity head
* Elevation head
* Pressure head
remain constant (where these terms are as defined here:
http://en.wikipedia.org/wiki/Hydraulic_head
)
then, since the Velocity head has increased (since Ux
remains constant but Uy increases from 0) the pressure head
must decrease to compensate. Hence, the airfoil is sucked
backward (I guess you could say that is part of the drag).
> But a momentum flow is the rate of change of momentum and
> this is the definition of a force.
>
> The force exerted on the airfoil is the lift force and
> this is equal to in magnitude but opposite in direction to
> this force on the fluid.
>
> So the lift force = Fy = -(pUxA2)Uy
>
What about the difference in pressure between the upper
surface of the air foil and the lower surface of the
airfoil? I thought this was the reason given by the
Bernoulli equation to explain lift. Indeed, the page:
http://en.wikipedia.org/wiki/Lift_%28force%29#Increased_flow_speed_and_Bernoulli.27s_principle
claims:
For any airfoil generating lift, there must be a pressure
imbalance, i.e. lower average air pressure on the top than
on the bottom. Bernoulli's principle states that this
pressure difference must be accompanied by a speed
difference.
-regards,
Larry
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arc, your comments would be fine except for two things.
1)Yes I said earlier that a glider will eventually return to Earth (thereby loosing its potential energy) and that is true here as well.
However Zet's (correct) point was that this energy must have come from the fluid and therefore the fluid must have lost some energy.
But for the second thing. Zet is also adamant that he has specified zero fluid friction and here his understanding of fluid mechanics falls down.
I thought friction was caused by viscosity and 0 viscosity (and 0 friction, IIUC) is used to derive Euler's equation:
http://en.wikipedia.org/wiki/Inviscid_flow#Reynolds_number
Since Euler assumed frictionless flow was OK, I can see why Zet assumed it was.
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Wouldn't that (lifting) energy be subtracted from the 'thrust' ...or whatever horizontal force is adding energy?
That's what I would guess; however, if the energy comes from the horizontal thrust, then
that begs the question of how a horizontal force causes a change in vertical potential energy.
IOW, how can something pushing horizontally cause movement of something vertically?
OTOH, if the energy comes from the fluid, then how is that energy substracted from the
fluid. Does the fluid drop, decreasing is potential energy? Does its speed decrease?
Does its static pressure decrease? If so, then how?
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First bolded...that's correct.
Second bolded...I am not showing energy is conserved, I am assuming the law of conservation of energy holds, and know there are generally frictional type losses, inefficiencies, in any real system. No thought experiment will prove it wrong.
Can't the problem be simplified by:
1) making the fluid incompressible (as in a hydrofoil instead of an airfoil)
2) making the hdrofoil density the same as fluid density
3) making the fluid viscosity = 0 (no energy loss due to friction)
IIUC, 1) would avoid any change in fluid energy due to
compression (in air, change in internal energy is P dV).
Also, 3) would avoid change in fluid energy due to frictional heating.
2) would make the gravitational potential energy constant.
1. Generally speaking, if you restrain an airfoil both horizontally and vertically you will have higher drag than the same set up where the airfoil rises.
2. When the airfoil rises there will be slightly less angle of attack from the reference frame of the wing, so induced drag should be reduced. Less energy will be removed from the free stream, so the overall velocity will remain higher than in the first case
Energy is of course conserved in each case. The gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy, along with other losses associated with drag, which together will be less than the drag associated losses in the first case.
How is the loss of the free stream energy, mentioned here:
gravitational/potential energy gain in the second case is accounted for as a loss of the free stream energy,
accomplished. Does the velocity of the free stream decrease with respect to its
value before encountering the foil. If so, then this is, essentially, what Zet said in post#46 which
says:
when the airfoil rises there must be a decrease in the horizontal velocity of the wind
-regards,
Larry
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You will have to ask MigL exactly what he meant.
A body may, on occasion, extract temporary energy from the air, but all non powered 'heavier than air' bodies eventually fall back to the ground.
The path of a glider is basically a longer or shorter fall to ground. The fall may be extended by temporary 'borrowing' energy from some fortuituous local pressure difference and/or or air current.
A leaf may be blown up in the wind, but it returns the energy to the fluid by displacement when it falls back to earth.
For sustained, heavier than air, flight the power comes from the engines which develop thrust in some form.
You should be careful and not think in horizontal and vertical terms because the forces may well not be horizontal/vertical.
There are three main forces involved in a heavier than air craft.
Weight, which always acts vertically.
Thrust which always acts in the direction of motion (the craft may be climbing or descending)
The force exerted by the fluid on the craft.
This force is usually resolved into two components
The drag which acts in the same line as the thrust but in the opposite direction
The lift which is perpendicular to the drag, in the opposite sense to the weight.
In level flight only the lift is vertical and the drag is horizontal.
After resolution into lift and drag there four forces acting. This the usual description.
However the four forces do not, in general, meet at a point.
This means that there is also a residual moment to consider.
The moment is countered by the air force on the subsidiary planes, particularly the tailplane.
So there are five forces to consider.
Finally the weight is actually distributed and measures to pump fule and other fluids about are used to maintain trim.
As to your calculation, I have no idea from your sketchy figures and this site is not a do your calculations for you site.
Could you please provide a very simple example (hopefully by adding to my
sketchy example ) which is complete enough to calculate an answer?
Also, maybe you could at least outline the calculation steps needed in this
example to get an answer. I'd be happy to do the calculations myself then.
TIA.
-regards,
Larry
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Does it?
When the aircraft has passed by is the fluid any different from before the aircraft arrived?
In what way does it have less energy?
Did I misinterpret MigL's statement? If so, could you clarify what MigL meant by:
The fluid has a certain amount of energy and it imparts some of it to the airfoils ( heating, lift, etc. ).
Also, if the energy does not come from the air, then does it only come from the thrust?
To put concrete numbers on this, assume:
1) force of gravity, g, is 1.0 meter/(second^2)
2) thrust, T, is 2.0 (kilogram*meter)/(second^2)
3) wt. of airfoil is 1.0 kilogram
4) the airfoil and rocket engine have no drag
5) the fluid is incompressible
6) density of fluid = 1
(in "appropriate" units
[ this needed to calculate the change in p
in the bernoulli equation A on:
http://en.wikipedia.org/wiki/Bernoulli%27s_principle
]
).
7) speed of airfoil at start time is 1.0 meter/sec
8) the airfoil cross section is an isosceles triangle
with width=1 meter and height of .1 meter.
9) the airfoil is within a wind tunnel
stetching 1 meter above and below the base
of the triangle (this data is needed to calculate the
speed up in fluid flow over top of triangle, AFAICT).
10) density of airfoil is 1.1 * density of fluid.
( this is needed in order for energy to be used
in lifting the airfoil. If the airfoil density = fluid density
then there would be no change in gravity potential
no matter what the height of airfoil.
)
What is the difference in height of airfoil after 1 sec.?
Since there's no drag, and since no energy comes from
the fluid, I guess part of the thrust is used to accellerate
the airfoil horizontally and part is used to lift the
airfoil vertically. But how much is used horizontally and
how much is used vertically?
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Ok lets simplify.
We consider two stationary airfoils with the fluid moving past both, except that one is tied down and one isn't.
The fluid has a certain amount of energy and it imparts some of it to the airfoils ( heating, lift, etc. ).
In the case where the airfoil is allowed to rise, an even smaller amount is imparted to the airfoil as a rise in gravitational potential.
In the case where the airfoil is constrained, the fluid is 'deflected' more strongly. In effect the fluid keeps more of its energy
This is more easily seen with a flat plate. If you introduce a flat plate orthogonally to a fluid flow, it will tend to fly away in the direction of the fluid flow. The plate gains some energy from the fluid which loses some. The only way to keep it stationary is to hold it, in effect, by giving it energy in the way of thrust.
From:
The fluid has a certain amount of energy and it imparts some of it to the airfoils ( heating, lift, etc. ).
I gather that the energy used to lift the airfoil a certain distance, d, comes from a decrease
in the fluids energy (via decrease in pressure or slowing down). OK, so how much comes from
slowing down the air speed and how much from decrease in pressure?
BTW, I've had the same questions as Zet and I applaud him in being persistent in getting
an answer he (and hopefully I) can understand.
Thanks Zet.
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airfoil rising
in Classical Physics
Posted
In the above W equation, I assume {V,z}_1 is the airfoil {velocity,height} at the start of some
time period, and {V,z}_1 is the airfoil {velocity,height} at the end of some time period.
Is that right? But I thought for the fluid, V_1 was fluid velocity at entrance to the wind tunnel
and V_2 was the fluid velocity at exit from the wind tunnel. So V for the airfoil depends
on time, but V for the fluid depends on distance along the wind tunnel. So, how can that
be?