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#34837 - 06/09/10 04:46 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
Originally Posted By: paul
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.



I've answered this many times - whatever the speed of sound is, given the gas mixture in the tube and the temperature/pressure that gas is at.

The effective Ve however is zero, as outlined in my previous posts, as air is not an ideal gas.

Bryan
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#34838 - 06/09/10 05:54 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
whatever the speed of sound is


well Bryan , since the air is limited to the
"speed of sound"

340.29 m/s

the tube size is not important as the tube can be any size
according to your reasoning it can never travel faster than
the speed of sound.

!!! air is not a ideal gas !!!

fine lets just use the slower speed of sound.

so lets open it up fully to 58 inch diameter.
removing the need for a tube or nozzle.
after all Im only looking for momentum as I have stated.

if I put 1000 cu ft of 1 atm air into the tank that
already has 1000 cu ft of 1 atm of air inside

the pressure doubles.

I now have 2000 cu ft of air inside the tank.
and the pressure is 14.7 * 2 = 29.4 psi

if I put in another 2000 cu ft of 1 atm air inside the
tank the pressure doubles again.

I now have 4000 cu ft of air inside the tank at 58.8 psi

if I put 4000 more cu ft of 1 atm air inside the tank
I now have 8000 cu ft of 1 atm air that has been squeezed into the 1000 cu ft tank.
and the air pressure is doubled again to 117.6 psi
air weights 0.0807 lb / cu ft
8000 cu ft * 0.0807 lb / cu ft = 645 lbs
645 pounds = 292.567079 kilograms

F = (m × V) for force
or
P = (m x V) for momentum

P = 292.567 kg * 340.29 m/s = 99557.62443 kg-m/s

so the momentum is 99,557 kg-m/s

the tank is 54.5 ft long
its diameter is 58 inch

it takes apx 0.16 seconds for the mass of air to leave the tank...


if I use a large or small tube doesnt matter that much
according to your reasoning.

the air will only have a certain speed at that pressure.
and that speed is the speed of sound.

I can increase the time that I allow the air to escape from the tank by using a small orifice or tube.

and this will decrease the momentum that is felt by the pipe each second.

and the ONLY WAY you can counteract this momentum is to supply a force of 99,557 kg-m/s in the opposite direction of the momentum force to make the pipe not move at all.


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#34840 - 06/09/10 07:33 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Kallog

Quote:
I'll simplify it to be a gun instead of a rocket, and it's fixed to the front of the pipe, pointing backwards.

You fire the gun, and it gives momentum -p1 to the bullet. At the same time it gives momentum p1 to the gun and pipe. Exactly the same as recoil in a normal gun.

Clearly the pipe's now moving because it has momentum p1.

Eventually the back end of the pipe and the bullet collide, and the bullet becomes embedded in the back of the pipe. They were travelling in opposite directions, with zero total momentum (-p1 + p1 = 0). Now they're a single solid object with the same total momentum - 0.

The pipe has 0 momentum and has therefore stopped moving.


the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.

the air moves throughout the pipe , some of the air moves into every cubic inch of area inside the pipe.

because the air pressure itself is equalizing.

because the tank is at one end of the pipe most of the
air in the tank will move toward the other end of the pipe.

and since the pipe is 60 inch diameter and the tank is 58 inch diameter there will basically be just air equalizing at the end where the tank is.

all the other air moves away from the tank.

and it is this movement that moves the pipe.

in my last post I calculated the momentum to be
99,557 kg-m/s

and this movement of mass inside the pipe causes the
pipe to move.

so if you move this mass fast or slow doesnt matter
as the total momentum is the result of P = m*v

in your bullet anology you would still have the reduction
in velocity due to air resistance as the bullet travels the 500 ft distance to the other end.

then F = m* (less velocity)than you started with.

the pipe would still be moving.


_________________________
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#34841 - 06/09/10 07:34 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
Originally Posted By: paul
Quote:
whatever the speed of sound is


well Bryan , since the air is limited to the
"speed of sound"

340.29 m/s

Paul, your own source says exactly that. From your NASA page:
http://www.grc.nasa.gov/WWW/K-12/airplane/rktthsum.html

The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot is determined by the throat area

But in your case - 100PSI, the speed of sound is 1/3rd of the value you stated - 131.6m/s (see post #34826 for the calc).

Originally Posted By: paul
the tube size is not important as the tube can be any size according to your reasoning it can never travel faster than the speed of sound.

Right and wrong, depending on what you mean by size.

The radius of your tube will determine the *rate* at which air is released from the tank. And as such, it'll also dictate the force developed. Keep in mind that [Pt-Pe]Ae determines the force derived from pressure; if you double Ae you'll double the force you'll produce.

However, since you have a tank of air the total amount of thrust produced (ISP, measured in newton-seconds) will be the same regardless of the tube's radius - the only difference is a larger diameter tube will produce that thrust in a shorter period of time, while a small-diameter tube will produce that amount of thrust over a long period of time.

Originally Posted By: paul
if I use a large or small tube doesnt matter that much according to your reasoning.

As described above, it depends on what you mean by size. The speed of the air in a tube will be restricted to the speed of sound, regardless of length or diameter. The *mass flow* however will be greater in a larger diameter pipe, but will not be affected by length. Double the cross-sectional area of your pipe, and twice as much mass will flow through it every second. It'll flow at the speed of sound, but there will be more moving

Originally Posted By: paul

I can increase the time that I allow the air to escape from the tank by using a small orifice or tube.

and this will decrease the momentum that is felt by the pipe each second.


Here is where you are missing the implication of choked flow (i.e. flow limited to the speed of sound) - and I think where our entire "conflict" (in regards to your tube) lays.

Forget air for a second, and think of a metal ball in your tube. The 100PSI tank you described earlier produced ~6000N of force through its tube. Pretend for a second that we applied that 6000N of force to the metal ball, but in a way where the source of the force was unconstrained (i.e. not using air confined to the speed of sound). This would accelerate the ball as per the formula F=ma (rearranged to a = F/m to calc the acceleration). The longer the pipe, the more acceleration that ball would experience. Hypothetically speaking (and ignoring relativity) an infinitely long pipe would lead to an infinitely accelerated steel ball.

I believe you will agree with the above description, as that's essentially what you've been saying all along (with the steel ball replaced by a mass of air).

Here is the part you are missing - your flow of air is choked; restricted to a set speed. This is occurring despite the fact that you have ~6000N of force acting on that air. So unlike that steel ball, your air is not continually accelerated - something is holding it back. And, unlike your steel ball, a longer tube won't further accelerate the air - one it gets to the speed of sound it stays there.

The factor that restricts the flow of gas to the speed of sound is the air itself - "real" (as in non-ideal) gases resist their own flow. The faster you push a gas, the more it resists being pushed. When this pushing is pressure-driven, the resistance to flow will equal the force produced by the pressure once the gas reaches the speed of sound. What this means is the momentum of the gas traveling at the speed of sound is perfectly balanced by an opposing force - in this case the backpressure of a non-ideal gas.

So, in your tube you have the situation where your tank initially accelerates the air to the speed of sound - this initial acceleration of the air is generated by the force determined by [Pt-Pe]Ae. At this point you now have a stream of air traveling at the speed of sound, despite the fact the tank's 6000N of force is "trying" to accelerate it further. This lack of acceleration is due to the force provided by the airs resistance to its own flow - and that force is equal, but opposite, to the force provided by the tank.

In other words, once flow is choked your tank is pushing it with 6000N of force, but that force is being countered by 6000N of backpressure.

And keep in mind, just as the tank imparts momentum onto the air due to the 6000N of force available to it, the choked air is in turn pushing back on the tube/tank with 6000N of force. It is this equal, but opposite, forces that prevent the further acceleration of the gas, and it is that opposing force that causes m(dot)Ve to be zero in a tube.

What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.

All of that could be derived by looking at the formulas provided, in a lot fewer words, but the above is the plain-ol-english version.

Bryan
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#34842 - 06/09/10 08:24 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.


still the mass will come out of the tank given its tube is
58 inch diameter.

and when the mass comes out , and it will come out , the momentum will be mass x velocity.

P=m*V

I dont care how you word it , the pipe will move.

because the mass moves inside.

lets check this...

the pressure differential is

114.7 psi - 14.7 psi = 100 psi

the force is 100 lb/sq inch.

the cross sectional area of the 58 inch tube is
2642.079 sq inch
the air inside the tank weights 0.645 lbs / cu ft
2642.079 cu ft / 1728 cu in = 1.528 cu ft in the cross section

suppose the cross section is 1 inch long.

there is 1.528 cu ft in the cross section.

1.528 cu ft * 0.645 lbs = 0.955 lbs in the cross section

there are 2642.079 sq inches in the cross section

.955 / 2642.079 = 0.0003614577 lb per cu inch of area.

a=f/m

a = 100 psi / 0.0003614577 lbs = 276,657.545 fps

276,657.545 fps / 5280 ft = 52.3972 miles per second.

276 657.545 fps = 84,325.219 meters / second


this incorporates the resistive force and friction is pretty much null in a tube this large especially with air as the fluid.

there is no choking force involved in this situation
there is only fluid flow.

and that restriction is minimal if not non existant
in a pipe this large.

the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s








_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34843 - 06/09/10 08:34 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
Originally Posted By: paul
the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.


Paul, here again you are making a simple mistake. Forces and momentums are vectorial - meaning they have both a magnitude and direction. Pressure is non-vectorial - it has a magnitude, but occurs equally in all directions.

It's not 100% clear what you're trying to claim, but I see two ways to interpret it.

--------------------------------------------

The first interpretation is that you are claiming the pressure-mediated expantion of the air stream will dissipate its momentum, thus leaving only the momentum of the tank, and thus causing your pipe to move. This is wrong - think of the three spatial axis - x, y, z. Imagine your tank is expelling air in the -x direction - this would push the tank in the +x direction. Lets say 100kg*m/s of momentum is given to the tank. This means the tank has 100kg*m/s momentum in the x-axis, but 0 in the y/z axis. Likewise, the air expelled from the tank has -100kg*m/s momentum in the x-axis, but 0 in the y/x axis.

If I understand you correctly, you're saying the pressure of that air stream will counter the momentum of the air stream - as in the expansion of the air stream will dissipate the momentum. This is wrong. Keep in mind pressure is non-directional. So if there is 100PSI of pressure in the air stream, that means there is 100PSI pushing in the +x direction, -x directions, as well as the +/-y and +/-z directions.

So what happens? First, the obvious - the pressure expands the air stream in the +/-y and +/-z directions. In fact, if you were too look at any one air molecule it would now have more momentum, as it would have the initial -100kg*m/s of momentum along the x-axis, plus whatever momentum its now added on the y/z axis. Over all the stream of air has no net gain in momentum, as there will be equal expansion in the +/- directions on the y and z axis. Keep in mind, the expansion on the y/z axis has NO EFFECT on the momentum of the air stream, as this pressure-driven expansion along the y & z axis does not impact momentum on the x-axis.

The big Q is what happens on the x-axis, in regards to pressure. If you pick any one point along the x-axis of the air stream, and look immediately the the + and - of it, there will be essentially no pressure difference. Ergo, the pressure in the stream will equally push in the + and - x directions. Ergo, there is no net pressure-mediated force in the +/- x direction (i.e. expansion will be equal in the + and - x-directions), and thus no net change in momentum along the x-axis will occur.

Ergo, don't expect the pressure mediated expansion of the air stream to magically dissipate the momentum of the air stream - it won't. The net effect of the pressure of the air stream is it will expand it equally in all directions. What this means is any vectorial properties of the air stream (i.e its momentum or forces it applies) will not be affected by its pressure-mediated expansion, as that expansion has no vectorial value (i.e. its even in all directions). When that column of moving air hits the back of the pipe it'll still have a momentum of 100kg*m/s - the only difference is that momentum will be spread over a larger area.

--------------------------------------------

The other way to interpret your claim is that the momentum of the air and tank will equal out (which they do), but the equalization of pressure will provide a net momentum. This interpretation too is wrong - because pressure is non-vectorial, it is incapable of inducing a vectorial force on your pipe. Equalization of pressure occurs equally in all directions; there is no net direction to the force (as air molecules move in a Brownian fashion). Since you have equal momentum of air in all directions the net force - and thus the net momentum - is zero.

Even taking into account simple mass-action, your net is still zero. You start by throwing a mass of air in the -x direction - this will give your pipe an equal momentum in the +x direction. However, the gas will want to reach equilibrium, meaning it'll expand towards the front of the pipe. The force of mass-action will equal (by definition) the amount of force needed to create the initial state. The movement of the air mass forward will be of equal momentum to the initial momentum of the air mass rear-wards. Net effect - pipe doesn't move.

Bryan
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#34844 - 06/09/10 08:38 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
Originally Posted By: paul
Quote:
What this means is that your tube-based system experiences a force, and thus gains momentum, from that initial acceleration of the air from rest to the speed of sound - a force determined by [Pt-Pe]Ae. However, the force generated by mass flow - and thus that momentum - is countered by an equal but opposite "choking" force. Ergo m(dot)Ve is zero with a tube.


still the mass will come out of the tank given its tube is
58 inch diameter.

and when the mass comes out , and it will come out , the momentum will be mass x velocity.

P=m*V

I dont care how you word it , the pipe will move.

because the mass moves inside.


Now you're mixing issues.

The air that leaves the nozzle/tube will have momentum - a momentum given to it by the force [Pt-Pe]Ae.

However, the momentum of the air leaving the nozzle/tube will impart no net momentum on your pipe, for the simple reason it'll transfer that exact same amount of momentum in the opposite direction once it strikes the end of the pipe.

Equal and opposite momentums = your pipe does not move.

Bryan
_________________________
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#34845 - 06/09/10 08:40 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
whatever you wrote is no longer important to me


I no longer consider you as someone who should be listened to.

_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34846 - 06/09/10 08:43 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
whatever you wrote is no longer important to me


I no longer consider you as someone who should be listened to.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34847 - 06/09/10 08:44 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34850 - 06/09/10 09:10 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
Originally Posted By: paul
the larger area did not reduce the velocity below the
340.29 m/s^2
the same mass is involved.

so the momentum I stated earlier is the same
the momentum is 99,557 kg-m/s


Absolutely, but this comes back to another common issue with you - you consistently confuse momentum and force.

Lets use simpler quantities to make the math more obvious - your tank is at 1000Pa (1kPa), and contains 1kg of air. For the sake of simplicity, the environment is a vacuum - 0Pa.

With your straight pipe that 1kg of air will be accelerated to the speed of sound - 340m/s. This means that after leaving the tank the air has a momentum of 1kg*340m/s = 340kg*m/2 - regardless of the size of tube you use.

However, the force produced will vary depending on the size of the tube. Take the case where you have a tube with a cross-section of 1m^2 verses 2m^2:

F = [Pt-Pe]Ae
If Ae = 1m^2, then F = [1000Pa-0Pa]1m^s = 1000N
If Ae = 2m^2, then F = [1000Pa-0Pa]2m^s = 2000N

---------------------------------------------
Now what if we add a nozzle that has a divergence angle which doubles the speed of the air (i.e. accelerates it to 2X the speed of sound - 680m/s)?

Now your momentum doubles - 1kg*680m/s = 680kg*m/2 - regardless of the size of tube you use (so long as the nozzle is scaled appropriately).

Your force will also be larger:


The [Pt-Pe]Ae part will remain the same as above - i.e. a 1m^2 will impart 1000N of force. But we've also got the additional acceleration of the air to 2X the speed it would have if there was no nozzle. I've not included enough information here to calculate that, based on the above formula.

But I kinda cheated - the only time you would have a situation where a nozzle would double the speed (and thus double the momentum) of the air passing through it is when you have a nozzle designed such that it produces an amount of force equal to that of [Pt-Pe]Ae.

Ergo, the force produced by our nozzle is:
If Ae = 1m^2, then F = 2000N
If Ae = 2m^2, then F = 4000N

The above nozzle, BTW, is fairly inefficient. Most nozzles accelerate to 3-5X the speed of sound. There is a theoretical maximum you can extract, which depends mostly on the pressure and temperature of the gas exiting the nozzle. For a rocket this usually comes in around mach 6.4. It would be lower in your case, due to the relatively low temperatures involved.

Bryan


Edited by ImagingGeek (06/09/10 09:11 PM)
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#34851 - 06/09/10 09:16 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136


Edited by paul (06/09/10 09:17 PM)
Edit Reason: FOR CLARITY
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3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34853 - 06/09/10 09:22 PM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
wat je schrijft is niet meer belangrijk voor mij

ik je niet meer beschouwen als iemand die moet worden geluisterd naar

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3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34854 - 06/09/10 09:26 PM Re: Orion, Mission to Alpha Centauri [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Kallog
Denkt u dat dit Bryan is vervelend of wat?
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3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34857 - 06/09/10 09:45 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34859 - 06/09/10 11:36 PM Re: Orion, Mission to Alpha Centauri [Re: paul]
ImagingGeek Offline
Senior Member

Registered: 05/19/10
Posts: 410
Loc: Canada
So I see your arguments are now reduced to childish word games.

Guess we've established the validity of your ideas - they're childish!

LOL
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#34862 - 06/10/10 01:00 AM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
quels que soit vous avez écrit n'est plus important pour moi


Je ne vous considère plus comme quelqu'un que l'on devrait écouter.

avec vous il n'y a aucun argument, c'est strickly de cette façon la discussion.

et cette voie est toujours dans la faveur de tyour.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34865 - 06/10/10 04:38 AM Re: Orion, Mission to Alpha Centauri [Re: ImagingGeek]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: ImagingGeek

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


Hang on, for a straight nozzle:
With ideal gasses the thrust is pressure thrust + an extra, nonzero term.
With non-ideal gasses the thrust is only pressure thrust, and the other term becomes exactly zero?
What if the gas is nearly ideal but not quite? There's a discontinuity which is really rare in the real world!



Your equation's site says Ve is exit velocity too, just like every other site. None of this "relative to some other speed" or "effective velocity". Actual velocity relative to the rocket.

I think there's a big mix-up with P symbols too. Your equation's site says:
Pe is exit pressure
Po is chamber pressure
Pa is ambient atmospheric pressure

But still we don't get zero exit velocity. I think any ratio of those pressures <> 1 for a tube shaped nozzle.

Why? Because without a pressure difference between the ends of the tube, there's nothing pushing the gas through it.




Quote:

force; I used the one that is used by engineers to actually design and build rocket engines.

Why? This is all just a thought experiment to see if a closed tube can fly isn't it? There's no need for engineering. While we're at it why don't we just use that 90K helium and forget all this non-ideal business? Remember Paul never said it's burning fuel, it's just compressed gas coming out of a tank.

Imagine if Einstein had used equations for the speed of a steam engine in his train thought experiments. He'd have said "In the real world trains are powered by steam engines, and they have friction, etc. These are the (really complicated) equations actually used by engineers." It would be totally pointless. It worked fine with engineless, frictionless, imaginary trains.

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#34866 - 06/10/10 04:53 AM Re: Orion, Mission to Alpha Centauri [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul

the bullet is not itself pressurized , its molecules will not break apart and expand throughout the pipe.

Air's momentum doesn't magically dissappear because it disperses. It gets transferred to other air molecules and eventually to the walls of the pipe.

[/quote]
then F = m* (less velocity)than you started with.

the pipe would still be moving.
[/quote]

If the force is reduced by friction, then it's being applied over a longer period of time. Either you have high acceleration for a short time, or low acceleration for a long time. Same velocity in the end.

It doesn't matter if it hits the wall suddenly or if the pipe is filled with ballistics jelly, it doesn't matter if the bullet shatters into a million little pieces on its way through. Either way, when the bullet finally becomes stationary relative to the pipe, they becomes like a single object, and that object has the same total momentum that its constituent parts had, which is zero.





Edited by kallog (06/10/10 05:07 AM)

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#34867 - 06/10/10 04:55 AM Re: Orion, Mission to Alpha Centauri [Re: kallog]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Paul, if the tube can accelerate itself then do you agree it violate's Newton's 1st law?

"Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it."

Notice it must be an _external_ force. Internal stresses, or thermal motion, or mechanisms, or anything purely internal doesn't count.

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