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.


Edited by ImagingGeek (06/09/10 02:04 PM)
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