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Quote:
Momentum is not pressure, and therefor the equalization of pressure in the pipe has nothing to do with where that momentum goes.


think of all the little air molecule thingies as being
little rockets flying around inside the big old pipe thingie.

some of the rockets fly north , some fly south , some fly east , some fly west , some fly down, the others fly up.

the little rockets that fly up , dont have far to go.
but the little rockets that fly down have a much further trip to take.
most of the litle rockets begin their trip by flying down but they have a harder and harder time finding a place to stop because of all the other rockets getting in the way.

but a few of them get to go all the way down.

and because the little rockets all move inside the pipe the pipe moves , up.
------------------------------------------------
here we go again , the equalization of pressure inside the pipe

has everything to do with the momentum

the pressure is the air , the air mass is what moves , and
the momentum of the moving mass of air

is the momentum.


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Originally Posted By: paul
there really is nothing that your twisting of words and math can accomplish with me that is , I know Im right

And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL

Originally Posted By: paul

even though you have repeatedly re-stated the same old stuff over and over , never once admitting that the mass of air comming out of the nozzle has velocity , simply because the nozzle isnt really a nozzle.


And now you're lying - I've repetitively stated the mass of air has velocity upon exiting the tank. If you re-read my last few posts I've even given the exact speed it will move at - the speed of sound.

Odd, that you have to lie about my claims, in order to make your "point".

Originally Posted By: paul
a nozzle only focuses the thrust.


Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how nozzles help provide additional thrust. Apparently you don't even read your own sources.

Originally Posted By: paul

in this case a nozzle is not the reason there is a thrust
the thrust comes from the m*Ve.


You're wrong here, as has been pointed out time and time again. Ignoring reality doesn't make it go away.

how m*Ve is calculated is broadly published. You ignore those formula, that doesn't make you right.

Originally Posted By: paul
so why do they bother putting fuel in jet aircraft all they really need is a nozzle.


Wow, the ignorance grows:
1) Jet engines work on a completely different concept. Ergo, not comparable to simple rocket engines

2) The combustion produces pressure. A nozzle is useless without pressure.

Originally Posted By: paul
paul is just using a tube therefore no velocity , right!!!


No increase in exhaust velocity, therefore m*Ve is zero. There will be the velocity intrinsic to [Pt-Pe]Ae.

Originally Posted By: paul
if I have a rifle and I cut the barrel off just beyond the bullet , according to your genius the bullet would just stay there when I fired the riffle.


Nope, and if that is what you think than you completely missed the point. The fact you think guns and rockets work on the same principals is particularity telling...

So there ya'll have it - Pauls argument in a nut shell:
a) ignore basic physics,
b) re-write formulas to suit his purposes,
c) lie about the claims of others, and
d) come up with irrelevant examples

Bryan

Last edited by ImagingGeek; 06/08/10 08:44 PM.

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Quote:
And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL


I see where kallog also agrees that m*VE cannot be ZERO.
and given that us three are the only participants in this
thread , your wrong.

also who have you convinced?


Quote:
And now you're lying - I've repetitively stated the mass of air has velocity upon exiting the tank. If you re-read my last few posts I've even given the exact speed it will move at - the speed of sound.


the speed of sound is a constant it is the speed at which SOUND travels in air , not the speed limit that air can travel at.

not once have you even tried to calculate or give the exact speed or velocity.

and if you did you just simply stated that it was ZERO.

I calculated the velocity or speed at 313600.466382069 m/s
which is 194.86 miles per second.

Quote:
Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how
nozzles help provide additional thrust. Apparently you don't even read your own sources.


provide additional thrust , as in the .37N

but you want to neglect the 45N comming from m*Ve

Quote:
You're wrong here, as has been pointed out time and time again. Ignoring reality doesn't make it go away.

how m*Ve is calculated is broadly published. You ignore those formula, that doesn't make you right.


Im not the one that ignores formulas and rearanges them to suit.

Quote:
Wow, the ignorance grows:
1) Jet engines work on a completely different concept. Ergo, not comparable to simple rocket engines

2) The combustion produces pressure. A nozzle is useless without pressure.


so its ignorance that typed the above , you just dont believe
how much you discredited yourself.

you have constantly been saying exactly the opposite of the above.

the combustion produces pressure , in the tank the pressure is
already produced... LOL

so the nozzle is useless without the pressure
did it ever cross your mind? that m*Ve is the result
of the stored pressure.

Quote:

No increase in exhaust velocity, therefore m*Ve is zero. There will be the velocity intrinsic to [Pt-Pe]Ae.


there it is again , the brainwashing must be complete
earlier he attempts to admit that he has given the velocity of air comming out of the tube at the much slower speed , the speed of sound.

So there ya'll have it - Bryan's argument in a nut shell:
a) ignore basic physics,
b) re-write formulas to suit his purposes,
c) lie about the claims of others, and
d) come up with irrelevant examples

BTW , Bryan its a,b,c,d not a,b,b,c

you be the judge.

---------------------------
I have a great idea , can you find an article that tells
how the nozzle itself produces more thrust than the
pressure that is either stored as in the air tank or the pressure that is produced by a jet engine or rocket engine.

meanwhile heres a bottle rocket that just uses air pressure
I noticed how the air came out of the bottle even though it
shouldnt because it doesnt have a space ship type nozzle , just a straight tube.

it just comes out of the bottles neck.

not one shaped like this.


but it sure takes off fast .. ZOOOOOOOM

pressurized air bottle rocket



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I was going to wait a while longer but decided to go ahead and
let you know that the side of the equation you use is used
in determining the amount of thrust and duration of flight
in a combustion based rocket engine.

the picture above that I linked to gave it away on this page.

http://www.aerospaceweb.org/design/aerospike/nozzles.shtml

so in reality in my example that does not have a nozzle
your side of the equation is unecessary as it only allows
a area to focus the thrust in a direction and with adjustable
nozzles can be used to adjust thrust amounts.

ie.. less thrust when less thrust is needed.
more thrust when more thrust is needed.

and of course the m*Ve side of the equation is the same as
the momentum equation and is the only side of the equation that we should use , given we have no nozzle.

I was thinking you would eventually figure that out , but
this has dragged on too long and you never did.

F=m*v
or
P=m*v

http://en.wikipedia.org/wiki/Momentum

I was just waiting to see if you could / would ever comprehend it Bryan.






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when Hydrogen is burned in a rocket engine.

the combustion produces a thrust.

then the recombination produces water.

you could burn a Hydrogen rocket engine inside a pipe
and still have all the water you started with.

given you have a big enought pipe.

and a centrifuge to capture and hold the water.
the thrust can spin the centrifuge , and the centrifuge can
seperate the water into H2 and O using membrane technology.

you now have a propulsion system for space travel.





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Originally Posted By: paul
Quote:
And yet are completely unable to provide one iota of proof. Strange, isn't it - you're right and yet you've been able to convince no one here...LOL


I see where kallog also agrees that m*VE cannot be ZERO.


he said no such thing - he said it would appear to be non-zero and would wait for my return to explain why I was saying it was zero. And I did just that, in my first post today.

Originally Posted By: paul
and given that us three are the only participants in this
thread , your wrong.

also who have you convinced?


kallog agrees with me that momentum of a closed system is conserved, and thus your engine is impossible. the closest he's been to agreeing with you is on the m*Ve, where his exact statement was "But we'll see better when he returns."

kirbygillis stopped by to make fun of you, which I assume means (s)he disagrees with you

So all three who've bothered to post vis-a-vis your claims agree that basic physics makes your engine impossible. None were not willing to take your claims vis-a-vis at face value...

Originally Posted By: paul
the speed of sound is a constant it is the speed at which SOUND travels in air , not the speed limit that air can travel at.


Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature. Different mixtures of gasses will have different speeds of sound - for example, the speed of sound in air is highly dependent on the amount of water vapor present. its also dependent on temperature.

The speed of sound is also the maximum speed that a gas can flow at under its own pressure. The speed of sound is determined by a gasses elastic modulus. That same elastic modulus also defines the back pressure of a flowing gas, and thus the degree to which a gas can accelerate itself due to its own pressure.

In simple terms, as a gas accelerates itself it also creates backpressure - resistance to flow generated by the elastic qualities of the gas itself. Gas can only accelerate itself, by its own pressure, to a point where the backpressure equals the gas pressure. Ignoring friction, backpressure will equal gas pressure when the speed of that gas is equal to its speed of sound.

You can get a gas to flow faster than its speed of sound, if you provide it with a either a source of energy other than its own pressure, or with reduced back-pressure. In the case of a nozzle you get the latter; a lowering of back pressure due to expansion of the gas stream.

Originally Posted By: paul
not once have you even tried to calculate or give the exact speed or velocity.


Oh, but I have, several times:

Speed of gasses leaving the tube: speed of sound (assuming room temp air, no humidity, approx 343m/s)

Additional speed of gasses due to acceleration by nozzle (Ve): 0m/s

Originally Posted By: paul
and if you did you just simply stated that it was ZERO.


Another lie - the only thing I stated was zero was Ve. And I provided both the formula and math to show that was the case.

Originally Posted By: paul
I calculated the velocity or speed at 313600.466382069 m/s which is 194.86 miles per second.


Which is wrong - the gas is choked at ~343m/s. Even your own sources described that - maybe you should read your own links...

Originally Posted By: paul
Quote:
Nope. And once again the irony - both the page I linked to, as well as the one you linked to, clearly state how nozzles help provide additional thrust. Apparently you don't even read your own sources.


provide additional thrust , as in the .37N

but you want to neglect the 45N comming from m*Ve


I ignore nothing; you're the one ignoring how those numbers are calculated.

m*Ve produces ZERO newtons of thrust when there is no divergence in the nozzle, as outlined numerous times before. [Pt-Pe]Ae produces the same amount of thrust, regardless of the nozzle design.

Originally Posted By: paul

I have a great idea , can you find an article that tells
how the nozzle itself produces more thrust than the
pressure that is either stored as in the air tank or the pressure that is produced by a jet engine or rocket engine.


It was provided to you two or three pages ago:
http://www.nakka-rocketry.net/th_nozz.html
http://www.nakka-rocketry.net/th_thrst.html

That is written by a fellow named Richard Nakka, a proverbial god in the EX rocketry community for the better part of two decades. On his webpage you'll find everything from basic theory, to engine design, to propellant formulation, to ready-to-go engine designs. He has calculators, applets, and various examples to illustrate every aspect of rocket engines you can imagine.

The relevant quote, in regards to the speed of sound, is:
"The critical point where the flow is at sonic velocity (M=1 at A/A*=1) is seen to exist at the throat of the nozzle. This shows the importance of the nozzle having a diverging section -- without it, the flow could never be greater than sonic velocity!"


Note: underlining provided by me, rest is original.

As for the amount of thrust added by the nozzle:
"The Thrust Coefficient determines the amplification of thrust due to gas expansion in the nozzle as compared to the thrust that would be exerted if the chamber pressure acted over the throat area only . . . The slope of the curve is very steep initially, then begins to flatten out beyond Po/Pe = 5. This is significant, as it indicates that even a nozzle provided with a minimal expansion will be of significant benefit. With such a pressure ratio of 5, the resulting thrust is about 60% of maximum theoretical"[/u]


And here is an applet, and more theory, describing these various factors:
http://www.engapplets.vt.edu/fluids/CDnozzle/cdinfo.html


Originally Posted By: paul
meanwhile heres a bottle rocket that just uses air pressure...


And? the amount of force derived in the case of the bottle rocket filled only with air will described by [Pe-Po]Ae; so that doesn't exactly help your case any. Water complicates things, due to two-phase flow, which I described earlier.

And you may want to look at the source you got the rocket engine picture from:
http://www.aerospaceweb.org/design/aerospike/nozzles.shtml

Even in this page - as in YOUR OWN SOURCE they counter your claim:
"the ultimate purpose of the nozzle is to expand the gases as efficiently as possible so as to maximize the exit velocity (v exit)"

LOL, disproven by your own sources - must hurt.

Bryan


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Originally Posted By: paul
I was going to wait a while longer but decided to go ahead and let you know that the side of the equation you use is used in determining the amount of thrust and duration of flight in a combustion based rocket engine.

Sorry, that dog doesn't hunt - the properties of gas flow doesn't magically change due to the source of pressure. It doesn't matter if the air is compressed by a fan, piston, chemically (i.e. combustion) or through magic - once compressed, the release of that gas will obey the exact same physical laws.

Originally Posted By: paul
so in reality in my example that does not have a nozzle your side of the equation is unecessary as it only allows a area to focus the thrust in a direction and with adjustable nozzles can be used to adjust thrust amounts.

Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure. The fact you have compressed air, while rockets have fire to compress their air, doesn't change the fact that both will produce pressurized streams of air that can act against that nozzle - or in your example, will be wasted due to the lack of a nozzle.

Originally Posted By: paul
and of course the m*Ve side of the equation is the same as the momentum equation and is the only side of the equation that we should use , given we have no nozzle.

No, it is not, as described by both YOUR and MY links. m is mass flow, not mass. Ergo m*Ve calculates a FORCE NOT MOMENTUM.

Originally Posted By: paul
I was thinking you would eventually figure that out , but this has dragged on too long and you never did.

Figured out what - that you don't read your own sources, or that you don't understand the difference between a force and momentum?

Bryan


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Quote:
he said it would appear to be non-zero


that doesnt sound like he is agreeing with you , in fact it
sounds alot more like he can read formulas better that you.

Quote:
kirbygillis stopped by to make fun of you, which I assume means (s)he disagrees with you

So all three who've bothered to post vis-a-vis your claims agree that basic physics makes your engine impossible. None were not willing to take your claims vis-a-vis at face value...


kirbygillis was correcting the thing I missed when I used
seconds istead of hours in light speed.

ok , I made 1 mistake , but I corrected it in the next post below it , so anyway that makes it about
50 to 1 in my favor.

Quote:
kallog agrees with me that momentum of a closed system is conserved


I agree that momentum is conserved also , but its not like Im agreeing with you , as that
is something that makes sence.

yes the momentum is conserved in the moving pipe itself.
nothing left the pipe.

Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.


so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.

if you will notice everything you are refering to is
about rocket engines , im not using a rocket engine.

so all your links to rocket engines are null.

a rocket engine uses a nozzle to direct the thrust


AFTER IT PASSES INTO THE NOZZLE...

so just like the bottle rocket does not have a nozzle
after the thrust passes the lip of the bottle , my pipe
and tank anology also does not have a nozzle it
just has a 1 inch sq area tube , why do you keep insisting on using nozzles when there isnt a nozzle?

think of it as a smaller pipe that has a hole in it.

Quote:
Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure.


keep reading further down it also says that it directs the gasses in a direction.

Quote:
Wrong again - nozzles harness the expansion of the compressed gas once they leave the tank, regardless of the source of that pressure. The fact you have compressed air, while rockets have fire to compress their air, doesn't change the fact that both will produce pressurized streams of air that can act against that nozzle - or in your example, will be wasted due to the lack of a nozzle.


I never said it had a nozzle , you did.
in fact at first you acknowledged it was just a tube.
so ...WHAT HAPPENED TO THE TUBE?

then you acted as if it had a nozzle from that point
on , but theres no nozzle.

you included a formula to compute the nozzle thrust
and disreguarded the only thing that actually was there
the tube.

theres is no nozzle.

ok , I want to allow the air to expand as it wants...
I dont want the air to be directed towards the rear as you seem to want it to.

I dont want a nozzle , nope no nozzle , none.

OK...no nozzle.

now what , we didnt install a nozzle.

will the pipe move?

is like saying will the bottle rocket go anywhere.

I say it will , what do you say it will do .

go towards the ground instead.

that video was photoshopped wasnt it?
dang conspirators.

they shouldnt promote such realism on the internet.




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Quote:

I agree that momentum is conserved also , but its not like Im agreeing with you , as that is something that makes sence.

so let me get this straight - you agree with the fundamental physical principal which clearly states your "engine" is a physical impossibility, and yet you insist your "engine" will impart momentum...

A little schizoid, are we wink

Quote:
yes the momentum is conserved in the moving pipe itself. nothing left the pipe.

Ergo, the pipe does not move.

Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.

so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.

The point is the very part you dishonestly cropped - the speed of sound sets the maximum velocity a gas can accelerate itself to, using its own pressure as the source of that acceleration.

Nor is the speed of sound anywhere near as constant as the gravitational acceleration of earth - the speed of sound changes with the constituents of the gas and temperature.

Quote:
if you will notice everything you are refering to is about rocket engines , im not using a rocket engine.

so all your links to rocket engines are null.

The physical principals are the same, regardless of the source of the pressure. The fact you expect otherwise defies not only the laws of physics,but basic logic as well.

Quote:
a rocket engine uses a nozzle to direct the thrust

Nope, even your own sources refute that claim - the nozzle accelerates the gas, providing thrust beyond that of the pressure differential alone.

Quote:
so just like the bottle rocket does not have a nozzle after the thrust passes the lip of the bottle , my pipe and tank anology also does not have a nozzle it
just has a 1 inch sq area tube , why do you keep insisting on using nozzles when there isnt a nozzle?

But I'm not insisting its a conical nozzle!!!! In fact, I've repetitively stated tat your nozzle is linear - as in a tube, exactly as you describe.

That is, after all, the entirety of the point I've been trying to make the past 10 or so posts - for Ve to have a non-zero value, the nozzle must have a divergent section. No divergent section, no change in the velocity of the gas escaping the "tube", thus Ve = 0.

Quote:
I never said it had a nozzle , you did.
in fact at first you acknowledged it was just a tube.
so ...WHAT HAPPENED TO THE TUBE?

Nothing, I simply used the word "nozzle" to differentiate that part of your contraption from the pipe and the tank. Too many tube-like objects.

As I've stated time and time again, your nozzle is tube-shaped; as in it has no divergent angle.

No divergent angle = no acceleration of the gas within the nozzle

No acceleration of the gas in the nozzle = no Ve

Quote:
you included a formula to compute the nozzle thrust
and disreguarded the only thing that actually was there
the tube.

Wrong again. If you go back and check my calcs you'll see I was using the exact conditions you set in the formula - no divergence in the nozzle (i.e. it is a tube).

So basically you're complaining that I did the math right for the conditions you explained, but chose to use what I thought was a less confusing term for the particular part of your contraption.

Quote:
ok , I want to allow the air to expand as it wants...I dont want the air to be directed towards the rear as you seem to want it to.

Wants and desires have no role in physics - physics obeys set, unalterable laws. In the case of the force generated when a gas moves from a region of higher pressure there are two factors involved - the pressure differential over the exit area; determined mathematically by [Pt-Pe]Ae, and the force generated by the gas as it expands in a divergent nozzle (if there is a divergent nozzle); determined mathematically by m*Ve.

As you said, and as I've agreed since the beginning, you do not have a divergent nozzle; simply a straight pipe. Under that condition, Ve is zero, leaving the sole source of force as the pressure differential - [Pt-Pe]Ae.

As for whether your pipe moves or not, that's a matter of the momentum in the system not the total force generated by the expulsion of air from the tube/nozzle.

Your contraption, nozzleless or nozzle containing, imparts momentum to the tank, which then transfers that momentum to the pipe. However, the same amount of momentum will be transferred to the air coming out of the pipe/nozzle, in the opposite direction. That momentum will be transferred to the opposite end of the pipe, where it will neutralize the momentum induced by the tank.

The nozzle doesn't change the momentum factor - all it does is more effectively transfer the pressure in the tank into momentum. But regardless, that momentum will be equal, but opposite, in the tank verses the air, and thus when the tank and air impart their momentum onto the pipe, the momentum cancels out. The only thing the nozzle does is determine how much momentum you "extract" from the pressurized gas. But regardless of your "conversion efficiency", the momentum imparted onto the tank will always be equal and opposite to the momentum imparted into the air.

Quote:

will the pipe move?

is like saying will the bottle rocket go anywhere.

Completely different situations - the bottle rocket is part of an open system - its exhaust transferes its momentum into the environment, not back into the bottle rocket. If you were to seal that rocket into a pipe, the rocket would move - the pipe would not.

And the reason is simple - the water coming out of the back of the rocket will propel it forward in the pipe. The expelled water will be propelled backwards in the pipe. Both will have the same momentum. When the rocket hits the front of the pipe, it transfers that momentum to the front of the pipe. When the water hits the back of the pipe, it transfers its momentum to the back of the pipe. Since both the rocket, and the water, have the same momentum, the net momentum imparted on the pipe is zero.

Ergo, the pipe will not move.

And since you're stuck on the nozzle issue, bottle rockets are propelled by a liquid - water - not a gas. Because water is incompressible at water-bottle pressures, the water will not expand in volume when it leaves the "tube". Ergo, having a divergent nozzle will not improve performance, as there is no expanding gas to take advantage of.

http://en.wikipedia.org/wiki/Water_rocket#Nozzles

But fill that bottle with compressed air alone, and a nozzle will give you great dividends:
http://en.wikipedia.org/wiki/De_Laval_nozzle

But, and I emphasize this again, regardless of how efficiently or inefficiently you harness the expansion of that gas, the amount of momentum imparted onto the tank will always be equal, but opposite, to the amount of momentum imparted on the gas leaving that tank. And since your system is closed, the net effect will be no net change in momentum of your pipe.

Bryan

Last edited by ImagingGeek; 06/09/10 01:14 AM.

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Quote:
In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.


momentum , thermodynamics , n such

so you were saying something?

Quote:
although it may be hard to believe, a human sneeze travels at about 120 mph when it exits. the air moves faster, sometimes maybe 200 mph. some people have been able to prove that their sneeze traveled supersonic (breaking the sound barrier). most children sneeze at about 90 mph. adults from 40-50 have the fastest sneezes at sometimes over 800 mph.


the fastest sneezers 800 mph 40-50 yr old people

I guess that what I should get is a bunch of 40-50 year old people because they dont seem to have as hard a time at breaking the sound barrier when they sneeze. LOL

that way we wouldnt need a air tank that is pressurized to 100 psi , they could just sit up front facing backwards and
I could sprinkle pepper over them -- deep space here we come.

----------------------------------
nozzel stuff


Rocket Nozzle Design: Optimizing Expansion for Maximum Thrust
A rocket engine is a device in which propellants are burned in a combustion chamber and the resulting high pressure gases are expanded through a specially shaped nozzle to produce thrust. The function of the nozzle is to convert the chemical-thermal energy generated in the combustion chamber into kinetic energy. The nozzle converts the slow moving, high pressure, high temperature gas in the combustion chamber into high velocity gas of lower pressure and temperature.
------------------ STOP JUST A MINUTE HERE--------------
Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles.
.......................................................
note : thats slightly faster than the speed of sound
wouldnt you say?
speed of sound 340 meters per second.

hmmm .... 4500 m/s - 340 m/s = 4160 meters per second faster than the slower speed of sound...
------------------------- continue --------------------
The nozzles which perform this feat are called DeLaval nozzles (after the inventor) and consist of a convergent and divergent section. The minimum flow area between the convergent and divergent section is called the nozzle throat. The flow area at the end of the divergent section is called the nozzle exit area.

Hot exhaust gases expand in the diverging section of the nozzle. The pressure of these gases will decrease as energy is used to accelerate the gas to high velocity. The nozzle is usually made long enough (or the exit area great enough) such that the pressure in the combustion chamber is reduced at the nozzle exit to the pressure existing outside the nozzle. It is under this condition that thrust is maximum and the nozzle is said to be adapted, also called optimum or correct expansion. To understand this we must examine the basic thrust equation:

F = q × Ve + (Pe - Pa) × Ae

where F = Thrust
q = Propellant mass flow rate
Ve = Velocity of exhaust gases
Pe = Pressure at nozzle exit
Pa = Ambient pressure
Ae = Area of nozzle exit

The product qVe is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust.

Let us now consider an example. Assume we have a rocket engine equipped with an extendible nozzle. The engine is test fired in an environment with a constant ambient pressure. During the burn, the nozzle is extended from its fully retracted position to its fully extended position. At some point between fully retracted and fully extended Pe=Pa (see figure below).



Finally, we calculate the thrust,

F = q × Ve + (Pe - Pa) × Ae
F = 100 × 2,832 + (0.05x106 - 0.05x106) × 0.409
F = 283,200 N

notice the above , take into account the
100 * 2832 !!!! because it equals 283,200
you add that amount to the ZERO from the Pe-Pa side and
you get the same number...

amazing

I might add that air can be compressed , combustible gasses
can not be compressed as high as air.

theres a big difference in compressed air and combustible gasses expanding , much higher velocities can be obtained
using compressed air than can be obtained by combustible gasses expanding.

if Uncle Al were here he would put you across his knee and give you a good lashing.





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Quote:
Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature.

so does g 9.8m/s^2
but we use it as a constant --- ACTUALY---
whats your point other than nic picking.


As I mentioned earlier, my point was dishonestly cropped by you. But this point is worth revisiting, and as such here is a separate post...

The speed of sound in a given gas is determined by:
S = sqrt(C/d), where
S = speed of sound
C = elastic/bulk modulus of the gas/gas mixture,
1.42×10^5 Pa for air at STP
d = density

The density of a gas is dependent on pressure:
d = p/RT, where:
d = density
p = absolute pressure
R = specific gas constant, for dry air this is
287.058J kg−1 K−1
T = temperature, in kelvins

So at 100PSI, at room temp, the density of your gas is:
d100 = p/RT
d100 = 689,475Pa/(287.058J kg−1 K−1 * 293K)
d100 = 8.2 kg/m^2

and the speed of sound is:
S100 = sqrt(C/d)
S100 = sqrt(1.42×10^5 Pa/8.2 kg/m^2)
S100 = 131.61 m/s

At atmospheric pressure (14.7PSI):
d14 = 101,352Pa/(287.058J kg−1 K−1 * 293K)
d14 = 1.2 kg/m^2

thus the speed of sound is:
S14 = sqrt(1.42×10^5 Pa/1.2 kg/m^2)
S14 = 343.28 m/s
---------------------------------
At this point you're probably asking "WTF is the point?"

The point is simple, and two-fold:
1) Contrary to your claims, the speed of sound is not a constant. In fact, it will be roughly three times SLOWER in your pressurized tank than it will be in your pipe.

2) You've mistakenly tried to calculate the force generated by your system, using m*v where m is the mass of the air, and v is the speed. You're wrong two ways here - m*v in using your measures gives you momentum, not force. And on top of that not-so-insignificant foopah, your v is wrong - its restricted to the speed of sound of the gas, due to back pressure. In the case of your 100PSI tank, that's 131.61 m/s.

So two basic claims of yours - the second of which is te foundation of your "proof" my math is wrong - are in and of themselves 100% wrong.

Kinda like your whole argument, beginning to end...

Bryan

Last edited by ImagingGeek; 06/09/10 02:07 AM.

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Originally Posted By: paul
Quote:
In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.


momentum , thermodynamics , n such

so you were saying something?


Yep, that momentum in a closed system cannot change - a point which wikipedia does not contest either in the page you cite, or in its page on momentum:

"Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot change"

http://en.wikipedia.org/wiki/Momentum

Quote:
...some people have been able to prove that their sneeze traveled supersonic (breaking the sound barrier)


Two point here:

1) I can find no evidence to substantiate your claim. The maximum speeds I've found in various sources - medical and layman is 90-150mph. The fastest speed I've seen claimed by a pseudo-medical source (JFK Health World Museum) is 650mph; about 85% the speed of sound. According to wikianswers, the highest confirmed speed is 166.7km/h; about 103mph.

2) Even if your claim is true, it hardly disproves what I said. Your nose is a divergent structure; the nasal cavity expands in cross section nearly 15X relative to the nasopharinx (where the air enters the nose when you exhale/sneeze). Ergo, your nasal cavity acts exactly like a rocket nozzle would - it allows for acceleration of expired gas, due to the expansion of pressurized gas within the nasal cavity.

In fact, airflow through the nose is a serious area of research, and is very well understood. A few examples:
http://www.springerlink.com/content/x48u517p27517169/
http://www.springerlink.com/content/7869gchy3g05fhut/

Once again, the very physical phenomena I've been describing, in action...well understood, and completely within the laws of physics I have described.

Bryan


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Quote:

------------------ STOP JUST A MINUTE HERE--------------
Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles.
.......................................................
note : thats slightly faster than the speed of sound
wouldnt you say?
speed of sound 340 meters per second.


You've added this since my last reply - trying to sneak stuff through, are we?

You need to go back and read what I've written before - it is the nozzle of the rocket that allows for the gas to be accelerated faster than sound. I said that clearly, on several occasions now.

It is that acceleration that creates Ve.

So bascially, you just supported my argument - you need a divergent nozzle to get faster-than-sound traveling air.

Quote:

as you can see the Pe-Pa is just the calculatable part of the moveable nozzle.


And once again paul shows he doesn't even read his own sources. Even your own source confirms Pe-Pa is the pressure force, not the force due to nozzle acceleration.

LOL, your own sources showing you to be wrong. Funny!

Bryan


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I think your loosing it , the speed of sound is not relevant.

it is the speed that air will travel that is relevant.

Quote:
d100 = 689,475Pa/(287.058J kg−1 K−1 * 293K)


could you go back to the page where you copied the above
work that someone else did , and insert the correct characters.

so it will be ledgible.




Originally Posted By: ImageingGeek

Actually, the speed of sound is neither a constant, nor is it just the "speed that sound can travel at". The speed of sound varies with the physical makeup of the medium and its temperature. Different mixtures of gasses will have different speeds of sound - for example, the speed of sound in air is highly dependent on the amount of water vapor present. its also dependent on temperature.

The speed of sound is also the maximum speed that a gas can flow at under its own pressure. The speed of sound is determined by a gasses elastic modulus. That same elastic modulus also defines the back pressure of a flowing gas, and thus the degree to which a gas can accelerate itself due to its own pressure.



as in the rocket nozzle the gasses are NOT under their own pressure ,,,,, they are COMPRESSED GASSES.

just like in the air tank the air is COMPRESSED not under its own pressure.

so either your right and we have never launched a rocket
or
Im right and we have...

havent you ever seen an aircraft pass the sound barrier?
how do you think it reached that speed , with slower than the speed of sound gasses flowing out of the nozzles?

Quote:
And once again paul shows he doesn't even read his own sources. Even your own source confirms Pe-Pa is the pressure force, not the force due to nozzle acceleration.


and it also states that m*Ve is the momentum and velocity thrust.

and the calculations below show the the m*Ve thrust is much larger than the Pe-Pa thrust.

did you happen to see that part , I have alreasdy covered it for you in a earlier post.




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Hi Paul, sorry to disturb you two's mad fight, but if you find time, could you please respond to my post #34791.

I'm not asking you check the working if you don't want, but at least to identify if that analogy is good enough, because it's a hell of a lot easier than this rocket business!!

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Originally Posted By: ImagingGeek

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae


Hi Bryan. Yea I pretty much got the gist of that, cheers. But it still doesn't explain an equation Paul put up, off NASA's website:

Exit Velocity: Ve = Me sqrt(gamma R Te)

According to both you and that same site, for a straight nozzle, Me = 1
gamma is probably on the order of 1
R > 0
Te > 0
So Ve > 0 for a straight nozzle.


You have a different equation for Ve, which gives zero if Pe=Pt. But Pe <> Pt because you need a difference to get any pressure thrust.


Quote:

Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.


Wikipedia: "Ve = Exhaust velocity at nozzle exit"
No mention of being relative to the pre-nozzle velocity. NASA's page equally doesn't say anything about it being an "extra" velocity, they just call it velocity.


So I'm stongly leaning in Paul's direction.

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Originally Posted By: kallog
Originally Posted By: ImagingGeek

Ftotal = Fn + Fp = [m*Ve] + [Pt-Pe]Ae


Hi Bryan. Yea I pretty much got the gist of that, cheers. But it still doesn't explain an equation Paul put up, off NASA's website:

Exit Velocity: Ve = Me sqrt(gamma R Te)

According to both you and that same site, for a straight nozzle, Me = 1
gamma is probably on the order of 1
R > 0
Te > 0
So Ve > 0 for a straight nozzle.


I assume you mean this page:
http://www.grc.nasa.gov/WWW/K-12/airplane/rktthsum.html

The difference is one of ideal verse non-ideal gases. NASA has the classical (i.e. ideal gas-based) formula on their page, while I provided the one derived for non-ideal gases (the page that my image comes from goes through that entire derivation, if you're interested). basically, paul has the version you'd teach in school to get an approximation of force; I used the one that is used by engineers to actually design and build rocket engines.

The way Mdot is calculated is also different for non-ideal gases, so you cannot really mix. Here is what the non-ideal gas formula looks like, when Mdot (for non-ideal gases) is added into the Ve formula provided earlier:



Everything before the (Pe-Pa)Ae is Mdot*Ve, for non-ideal gases.

I guess I owe paul half an appology - if he were using an ideal gas (the only one we know of is helium, below ~90K) he would have been correct.

Unfortunately, NASA's example math doesn't work (accurately) in the real world.

Originally Posted By: kallog

You have a different equation for Ve, which gives zero if Pe=Pt. But Pe <> Pt because you need a difference to get any pressure thrust.


No, that is not correct. But its my fault - I see I've been mixing pauls and my terms. To be clear:

Pe = pressure of the environment
Pt = pressure of the tank (in pauls case) or throat of an
actual rocket engine
Po = pressure at the end of the nozzle, which is almost
always higher than Pe

Pressure force is [Pt-Pe]Ae - as in the pressure difference between the tank (or wherever pressure is highest in the engine) and the environment.

Ve uses the ratio of Pt/Po - as in the ratio of the pressure difference between the front of the nozzle and the end of the nozzle. In a straight tube Pt will equal Po, ergo [1-(Pt/Po)] will be zero


Quote:

Ve = the exit velocity of the gas from the nozzle, relative to the velocity of the gas when it enters the nozzle.


Quote:
Wikipedia: "Ve = Exhaust velocity at nozzle exit"
No mention of being relative to the pre-nozzle velocity. NASA's page equally doesn't say anything about it being an "extra" velocity, they just call it velocity.


Once again, we have the issue of ideal verses non-ideal gases. Ve, as given in Nasa's formula, is the ideal exit velocity - what you would get if you had a gas that essentially doesn't interact with itself - i.e. no in-flow friction. However, in the real world we don't have ideal gases, so we need to calculate the effective exit velocity:

http://en.wikipedia.org/wiki/Specific_impulse
http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation

For a non-divergent nozzle, with a real gas passing through it, effective Ve is 0.

Bryan

EDIT: Just to add some more sources in regards to theoretical verses real-world calculations of ideal verses effective Ve, thrust calcs, etc in real-world situations:

NASA SP-8039 - derivation of all the formulas I used above, starting from ideal-gas, going to real-world

NASA SP-8115 - formulas for the design of rocket nozzles; covers the nozzle theory and derivation of the equation I provided in this post

NASA TN D-467 - comparisons between the performance of real-world engines and their ideal-gas calculated performance.

Some Considerations on the use of a Pressurized Tank System for a Rocket Engine. R.Sandri, Canadian Aeronautics and Space Journal, Oct.1967 - basically a description of the physics of using pressurized air

All but the latter are available on NASA's webpage.

Last edited by ImagingGeek; 06/09/10 02:04 PM.

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its not a rocket engine Bryan.

its just a tube


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Originally Posted By: paul
its not a rocket engine Bryan.

its just a tube


The physics are the same, paul. A rocket engine is, afterall, just a lumpy tube.

Bryan


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Quote:
The physics are the same, paul. A rocket engine is, afterall, just a lumpy tube.


then what is the velocity of the air that will come out of the
streamlined tube.


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