Norman,
The reason you’re having difficulty comprehending it (apart from the fact
it's not very intuitive) is primarily due to the way you prefer to analyse
thrust which is via pressure distribution across the active surfaces which can
be an incredibly complicated method of analysis, but provides a clear and
intuitive method of getting accurate results.
The far easier method of analysis is to calculate the approximate shift in
momentum of the exhaust mass and utilise Newton's 3rd law or the conservation
of momentum principles, but that by itself has subtle limitations, not in the
fundament principle of doing so, but in the methods typically utilised.
So, here's my crack at explaining my interpretation of Pressure Thrust and
Momentum Thrust:
Pressure thrust:
Just say we have a closed cylinder with one atmosphere pressure inside and we
(also at 1 atm surroundings) poked a hole in one end. Result: nothing. Now, the
simple explanation is Pinside=Poutside or Pc=Pa results in no thrust. However
the more fundamental explanation is that pressure on opposing area of the hole
(on the opposite internal side of the hole) = pressure on the area of the hole.
Now we repeat the same exercise with an identical closed cylinder but in a
vacuum. The result is the pressure on the inside (1 atm) is higher than the
outside, but again, fundamentally, the pressure on opposing area of the hole
(on the opposite internal side of the hole) > pressure on the area of the hole
thereby producing a thrust.
This thrust produced is 100% *pressure thrust* ie. It's 100% derived from the
imbalance of pressure between the hole area and the opposing equal area on the
opposite internal face of the cylinder.
Momentum thrust:
Taking the same cylinder (1 atm Pc) in the same vacuum environment, if we
substituted the hole with a de Laval nozzle, depending on how optimised that
nozzle was to the operating conditions, we start to convert much of that
pressure thrust into momentum thrust by physically changing the directional
vector of the gas molecules from zero (completely random) to a number greater
than one in the aft direction. This provides more use of the available energy
within the molecules to deliver thrust. The more you expand the exhaust within
the confines of the exit cone, the more pressure (random molecular motion) get
converted into directional motion (momentum) resulting in an efficiency gain up
to the point of when your exit pressure = surroundings pressure in which case
all your pressure thrust is effectively converted to momentum thrust.
We (in a non vacuum) can still keep expanding the exhaust beyond Pe=Pa (over
expansion) resulting in an even greater increase in momentum and exhaust
velocity and the equations used to calculate avg exhaust velocity will keep
providing us will higher and higher numbers as we keep expanding. But, as you
know, there is a consequence to this - the pressure exposed to the internal
surfaces of the exit cone beyond Pa will be less than the opposite side of the
nozzle wall thereby producing negative thrust on these areas. The exhaust
velocity equations don't capture this consequence, they happily tick along
telling you that less Pe = higher Ve so given that mass flow is constant,
thrust must be higher with lower Pe which is only true in a vacuum. This is
where the pressure thrust term is used to offset the ever increasing Ve with a
reduction of Pe.
Regards,
Troy
-----Original Message-----
From: arocket-bounce@xxxxxxxxxxxxx [mailto:arocket-bounce@xxxxxxxxxxxxx] On
Behalf Of Norman Yarvin
Sent: Tuesday, 1 November 2016 4:02 AM
To: arocket@xxxxxxxxxxxxx
Subject: {Spam?} [AR] Re: Rocket thrust change with altitude, RPA
Clicking on that link, I just get "Restricted Page. You have reached your
viewing
limit for this book." That's likely because the URL contains some code number
identifying who the reader is, and others clicked on it first; or maybe it's
because I'm not in Australia and that's Google's .au site. In any case,
changing
it to
https://books.google.com/books?id=V4_s91-
Vun4C&pg=PA9#v=onepage&q&f=false
did work. That's page 9 of "Astronautics: The Physics of Space Flight", by
Ulrich
Walter. Looking at it, the author states that the argument I'd been
criticizing
from Huzel and Huang is indeed false, for much the reasons I gave, but that
the
resulting equation is nevertheless true. (He doesn't cite them by name, but
states that "you will often find" that argument, so they're not alone.) His
own
argument for the equation's truth, though, is rather mysterious, involving
taking an idealized cylindrical combustion chamber+nozzle and closing over its
exit but still somehow getting thrust. And he doesn't explain how the
equation
can be true when in vacuum it violates the conservation of momentum.
The textbooks' error here seems to me to be double counting: you can calculate
the thrust in vaccum either by conservation of momentum or by integrating all
the pressure forces on the engine, and the two numbers have to be equal, but
it's a mistake to mix the two. But when you introduce an atmosphere,
conservation of momentum isn't so simple, since the atmosphere can pick up
momentum too. The vacuum equation is still the best starting point, since
nobody (*) wants to integrate forces over the inside of the combustion
chamber, but there has to be some sort of correction added in. Still, it's a
mistake to add in effects that were already taken into account in the vacuum
conservation of momentum calculation.
In situations where the textbooks start out with a basic equation like this
and
then go on to build on that basic equation and validate it either via
experiments
or via considering it from other theoretical points of view, I'd assume that I
must be wrong somehow.
(Conservation of momentum can't be wrong, but I sure can.) But here I don't
think they do make the sort of use of the equation that would really validate
it.
--
(*) except for computational fluid dynamics masochists, of course.
On Mon, Oct 31, 2016 at 04:57:19PM +1100, Troy Prideaux wrote:
Norman,Vun4C&pg=PA9&lpg=PA9&dq=mom
https://books.google.com.au/books?id=V4_s91-
entum+vs+pressure+thrust&source=bl&ots=Ry_5D8t5i1&sig=HlCeEb0aARbxTuAhX
iLtVkakGno&hl=en&sa=X&ved=0ahUKEwjMl6PAx_zPAhWJjJQKHe2uCMIQ6AEINDAE#v=o
nepage&q=momentum%20vs%20pressure%20thrust&f=falsebe:
Troy
-----Original Message-----
From: arocket-bounce@xxxxxxxxxxxxx
[mailto:arocket-bounce@xxxxxxxxxxxxx] On Behalf Of Norman Yarvin
Sent: Monday, 31 October 2016 3:11 PM
To: arocket@xxxxxxxxxxxxx
Subject: [AR] Re: Rocket thrust change with altitude, RPA
On Mon, Oct 31, 2016 at 09:20:22AM +1100, Troy Prideaux wrote:
As regards your equation, I think Pe should be removed -- that it should
exhaust velocity by a longer nozzle, but wasn't. It's a lost opportunity for
F = q * Ve - Pa * Ae.
(1) it's not *my* equation. It's the equation you'll find in just
about every propulsion text.
Well, I indeed see it in Huzel and Huang, for one. (Right up front,
page 2 in SP-
125.) I didn't realize I was challenging the authorities here, but
in for a penny, in for a pound.
(2) by removing Pe you're virtually removing any additional pressure
thrust contribution from your thrust which will show a significant
erroneous understatement with under-expansion, but it will also
erroneously understate the Pe=Pa conditions and Pe<Pa conditions.
For example, for the condition where Pe=Pa (I don't dare say optimum
expansion), your adaptation basically suggests that the entire exit
pressure is zero absolute ie a complete vacuum strictly in terms of
pressure thrust. In fact, it suggests that for all conditions which
is obviously wrong.
You don't need to be hesitant about "optimum"; that point was cleared
up in your favor quite graciously by Alexander Ponomarenko. (If you
want to be precise, it's "optimum" from the point of view of the
nozzle designer who is optimizing for a given atmospheric pressure.)
And yes, this "pressure thrust" is what I'm dubious about; see below.
That is, the effect of varying Pe is implicit in the other terms;
it shouldn't appear here as an explicit term. Certainly in vaccum
it should just be
F = q * Ve
No, not necessarily. That equation implies one thing and one thing
only
- optimum expansion (dang... there I go again with the O word).
Whether in atmosphere or vacuum. The equation still wouldn't be
correct for an under expanded nozzle operating within a vacuum. Not
without adjusting the Ve equation.
and to add Pe*Ae to that would be wrong.
Why?
Because F=q*Ve is just conservation of momentum -- about as
indubitable as an equation gets. There is no "optimum expansion" or
any other implication involved; the only assumptions are that Ve is
the exhaust velocity (well, to be precise, the average exhaust
velocity, or, well, to be really precise, the average of the
component of the exhaust velocity that points in the thrust
direction) and that q is the mass flow rate. Well, and of course
that the engine is in vacuum so there are no pesky atmospheric
pressures to complicate things. In that situation, Pe (exhaust
pressure) represents thermal energy that could have been translated into
thrust,
not something that yields thrust.
pressures.
To return to the general case (where there might be an atmosphere
involved), the reason I'm dubious about "pressure thrust" is that
pressure doesn't propagate upstream against supersonic flow. (Well,
unless it overwhelms the flow, as happens in flow separation -- and
that actually can give increased thrust since it leads to higher
pressures inside the nozzle -- but for the moment let's assume that
isn't happening; engines are usually designed not to have flow
separation, even if they are overexpanded, and things are complicated
enough without flow separation.)
Huzel and Huang's argument for adding pressure thrust into the
equation is
that:
"Equation (1-3b) states that if a mass is flowing out of a
container, the sum of all internal and external forces acting
on all surfaces of this container is equal to the total
momentum flowing out of the surface. The liquid propellant
rocket thrust chamber, with the inclusion of the exit plane,
is such a container."
The "inclusion of the exit plane" is what gives them the pressure
thrust term, yet the exit plane isn't a physical surface, so any
force that "acts on it" isn't directly pushing on the engine.
Instead the pressure there would have to work by affecting upstream
pressures -- and for that, it would have to be able to affect upstream
Since this is supersonic flow, I don't think it can. The upstream
engine has no way of knowing that the nozzle isn't much longer,
expanding into vacuum, and with negligible exhaust pressure.
In any case, they go on to explicitly state that the pressure thrust
term enters into the equation even in vacuum, and that's just wrong:
they started their analysis with the conservation of momentum, but
ended up with something that contradicts the conservation of momentum.
