# A proposal for a hypersonic craft.

1. Jun 20, 2006

### RGClark

In this post to sci.astro I suggested a method for air-breathing propulsion to orbit:

From: Robert Clark
Date: Sat, Jun 17 2006 2:43 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.space.policy, sci.physics, sci.mech.fluids, sci.engr.mech
Subject: Proposals for air breathing hypersonic craft. II
http://groups.google.com/group/sci.astro/msg/b9c19abd8f97a5d2

What I want to ask about is the portion of the post copied below where I derive a version of the rocket equation. I wanted some feedback on the validity of the derivation.

- Bob Clark

=====================================================
I shall argue that the method of not slowing the incoming air at all
but accelerating the fuel up to the relative air speed will result in a
marked improvement in fuel efficiency. Specifically, the exponential
increases in fuel according to velocity given by the rocket equation
will no longer be needed.
Let X be the mass of the rocket with fuel, v the rocket velocity, r
the ratio of air to fuel in mass, and e the nominal exhaust velocity of
combusting still air with still fuel.
For this method to work I will assume that the force produced by the
combustion of the air with the fuel can be fully communicated to the
craft. To derive the thrust equation take the craft including the fuel
to be a closed system and the air to be outside the system and take the
rest frame to be the Earth, or likewise the still air.
Now if we did not combust the ejected fuel with the air then by
momentum conservation we would have:

(X + dX)(v + dv) + 0(-dX) = Xv

In the first term on the left we add dX to X because dX is negative
since the mass is decreasing as fuel is consumed. So the first term
represents the mass of the rocket less the ejected fuel times the
increased velocity of the rocket. In the second term we are multiplying
the velocity with respect to ground of the ejected fuel times the mass
of the fuel ejected. Since we are ejecting the fuel at a speed to stay
at zero relative velocity to air, i.e., to the ground, this velocity
here is 0. The negative sign in front of dX again is because dX is
negative so -dX is the positive mass of the fuel.
This equation expanded out is Xv + Xdv+ vdX + dXdv = Xv. So the change
in momentum is Xdv + vdX + dXdv = 0, and the rate of change of momentum
is:

0 = Xdv/dt + vdX/dt + (dXdv)/dt = Xdv/dt + vdX/dt , because the term
with two differentials dXdv vanishes as dt ---> 0.

Now when we do combust the fuel with the air, then the rate of change
in momentum of the system is the force on the craft due to the
combustion of the air and fuel. This is the thrust produced by this
combustion which equals mass flow rate, air + fuel, times the nominal
exhaust velocity of the combustion of still air and still fuel:

Xdv/dt + vdX/dt = -ed(rX+X)/dt = -e(r+1)dX/dt , where the minus sign
comes from dX being negative.

Let c = e(r+1). Then the equation becomes Xdv/dt + (c + v)dX/dt = 0,
which is equivalent to:

d[(c +v)X]/dt = 0

This has solution (c + v)X = constant. Let X0 be the initial mass and
v0 the initial speed of the rocket. Then the solution is (c +v)X = (c +
v0)X0.
Therefore X0/X = (c + v)/(c + v0), i.e., the mass ratio of the fully
fueled rocket to the empty rocket is just a linear function of ending
velocity.
==============================================

2. Jun 20, 2006

### Staff: Mentor

That doesn't describe anything about how it would actually work and I think if you read up on jet propulsion a little, you'd find that the slowing of the air (to compress it) is a critical part of the jet propulsion process. That energy isn't lost (well, a little is lost due to friction/heat inside the engine), you get it back when you expand the air out the exhaust. Plus, the pressure needs to be higher upstream than downstream, otherwise the flow would be backwards. Think of a pulse-jet engine. It is basically just a stovepipe with a swinging damper in the front. When the jet fires, pressure would send air out both ends if the damper didn't shut to stop it from going out the front.

Jets in general:
http://en.wikipedia.org/wiki/Jet_engine
Here is more info on a scramjet:
http://en.wikipedia.org/wiki/Scramjet

Last edited: Jun 20, 2006
3. Jun 20, 2006

### rcgldr

Scramjets operate with supersonic internal air flow, but need to be running at hypersonic speed to be in full scramjet mode. So far this stuff is cutting edge and/or experimental. A scram jet doesn't need a turbine compressor stage, relying instead on the compression of hypersonic air and shockwaves (view the links for detailed info)

Edit / update: One of the issues is that scram jets won't operate until you get them to hypersonic speeds. So some other means of propulsion is required to get up to speed. The most common method currently used in experiements is rocket power.

NASA succeeded in this test:
http://www.nasa.gov/missions/research/x43-main.html

Other links:

http://www.aip.org/tip/INPHFA/vol-10/iss-4/p24.html [Broken]

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

Do a web search for scramjet and you'll find a lot of hits.

Last edited by a moderator: May 2, 2017
4. Jun 24, 2006

### RGClark

There are problems with getting scramjets to work at the high hypersonic speeds required to reach orbit. One comes from getting the fuel and air to combine chemically for combustion when the air is moving at such high speed with respect to the fuel, another is maintaining combustion at the high temperatures produced when slowing the air from hypersonic speeds.
That is why I was suggesting not slowing down the air at all and in fact accelerating the fuel rearward to stay at zero relative velocity to the air.
The proposal was to use both the thrust produced by accelerating the fuel rearward and the thrust produced by combustion of the air and fuel.
A problem now is how is this combustion thrust going to be transmitted to the rocket when is moving away from this fuel/air at up to 8000 m/s.
Question: if the air is made to flow in a circle or for example in a helical path, does this air now become part of the rocket system? Does momentum imparted to this air also become transmitted to the rocket? If so, this may provide a means to impart most or all of the thrust produced from combustion to the rocket.
A key question though is would this create the same type of heating problems as when the air is decelerated from hypersonic speed? Though the air is not decelerated, it is being accelerated in moving in a curved path.

Bob Clark

5. Jun 25, 2006

### Staff: Mentor

Did you read my post? Jets require compression of the air.

6. Jun 25, 2006

### FredGarvin

EDIT: Sigh...nevermind.

Last edited: Jun 25, 2006
7. Jun 25, 2006

### RainmanAero

Mr. Clark:

I didn't bother trying to follow your derivation, because when I got to the bottom-line result I saw a problem:
It is not linear, but rather exponential. You have attempted (but commited a significant error) to derive the basic mass ratio form of the rocket equation. The actual mass ratio equation is:

(Mass-initial/Mass-final) = EXP(Burnout_Velocity/g0/Specific_Impulse)

The exponential is a result of deriving the thrust equation as a function of mass flow rate in differential form. When you do so you will end up with a (dM/M) term which integrates to the natural log function of initial-to-final mass. Moving the "ln" to the other side of the equation is what gives you the "e^y" form for mass ratio.

Like I say, I didn't spend much time trying to follow your derivation, but it seems to me that you are trying to merge equations for airbreathing propulsion thrust production with closed (rocket) equations. They are two different physical situations where the mass flow rate of air entering an airbreathing engine cannot be ignored.

Rainman

8. Jun 26, 2006

### RGClark

Thanks for the response. You're actually *assuming* the equation will turn out to be the same as for the usual rocket case, which is exponential.
The key difference is that the exhaust velocity is *changing* in this proposal. If you follow the derivations of the usual rocket equation an important factor in making the answer be exponential is that the exhaust velocity is *constant*.
Now if you repeat the *derivation* used in the rocket equation but this time use the fact the exhaust velocity is matching the speed v of the rocket, you see you do indeed get a linear dependence of fuel used on final velocity.
As for the combining of rocket and jet aspects, the impetus of this proposal was to avoid stopping the air flow. I want it to flow freely unimpeded. This is different from turbojets which impede the flow in order to send it through turbines or ramjets/scramjets which slow down the air from supersonic or hypersonic speeds.
Because of this you don't have a ram drag term that you subtract off from the thrust as with usual jets: the net thrust *is* the gross thrust. This makes the proposal more analogous to the rocket case.
However, I do have some questions about the proposal. In order to derive that equation for fuel vs. velocity, the key physics fact I was using was that the rate of change of total momentum of a system is the external force applied to it. However, the way this proposal would work in practice is that there would be some time delay between when the fuel is ejected rearwards and when this fuel is combusted and when the force due to that combustion is applied to the rocket.
I'm a little uncertain about the applicability of that physics law when there is that time delay.

Bob Clark

9. Jun 26, 2006

### RGClark

Certainly for the best efficiency for *jets* you want the highest compression, but anything that heats the air flow so that it exits at higher speed than it enters will create a jet.
However, an important difference with this proposal is that the fuel and air are still with respect to each other. In this respect it is more similar to a rocket. Indeed you may want to use turbopumps for both the air and the fuel to inject them into a rocket engine which is moving rearward along with the fuel/air.

Bob Clark

10. Jun 26, 2006

### RainmanAero

Mr. Clark,

You need to pay attention to an important point that TWO other respondents have tried to point out to you about your idea. We'll get to that in a moment
The only thing I am "assuming" is that a CONTROL VOLUME approach must be used, and this approach will look different and result in different equations depending on whether your engine is airbreathing (open) or rocket (closed).
No, it isn't. I think you should study the following reference to see that the derivation of the rocket equation does not even require a consideration of the exhaust velocity (provided you know the engine's specific impulse). I told you above why the exponential result occurs. Read: Introduction To Flight, John D. Anderson Jr., Chapter 9.10 "Rocket Equation".
This is your mistake: You are using the rocket equation (which does not need to consider any inlet mass flow of the oxidizer(air)) but applying it to an airbreathing engine (SCRamjet). That is just wrong. As I mentioned above, they are two DIFFERENT control volume problems.
And this is where you have completely avoided understanding the IMPORTANT aspect of why the flow is slowed down that the gentlemen above have pointed out to you. COMPRESSION. More air molecules per unit volume...equals more stuff to burn...equals more energy released... equals higher exit velocity at the nozzle. In order for local pressure to go UP (compression) the local velocity needs to go DOWN (review Bernoulli's principle while you are at it).
No. You are in error, and again you are mixing apples and oranges. Jets are NOT equal to rockets. Ergo you cannot mix the analysis techniques any way you wish.
I don't mean to be nasty, but honestly you should just abandon the approach. It is analytically flawed.
That is actually a bit backwards to the conventional way to derive the jet equation (i.e. the net force exhibited on a control volume is equal to the total momentum change).
There is no reason or need to model any sort of "time delay" that happens within the control volume. Both the jet equation and rocket equation are based on simple, continuum mechanics.

Once again, you need to understand that the basis of why the SCRamjet is so efficient at high Mach numbers is a result of using the internal shock wave (to slow the flow) in place of a mechanical compressor. But rest assured, without employing compression in your "proposal" you are going to have one terribly poor jet engine that will not be worth the money it costs to operate it.

Rainman

Last edited: Jun 26, 2006
11. Jun 26, 2006

### RainmanAero

More analytical errors:
No, no, NO it is NOT! :uhh: It is still an open system, whereas a rocket is NOT. IOW, you are not carrying your oxidizer in a jet engine... you are "borrowing" it from the atmosphere. In a rocket you are carrying your oxidizer with you (and that means an exponential weight penalty).

When/where did you receive your degree?
Rainman

12. Jun 26, 2006

### Staff: Mentor

Well sure, but you can't just tell the air what to do - it doesn't obey you, it obeys the laws of physics and the laws of physics say it is going to compress and slow down whether you want it to or not. If nothing else, your engine has an intake and shock waves will form at the intake (indeed, that's the main source of the compression in engines designed for high supersonic/hypersonic speed).

By the way, you are using pumps to pressurize/accelerate the fuel for mixing with the airstream. In a typical jet, the air accelerates the fuel. How efficient do you think that pump is compared to how efficient that nozzle is at accelerating the fuel? Hint: the energy that goes into the airstream stays in the airstream while the energy for the pump is bled-off at the pump. Basically, you're using a 65% efficient pump to do something that you could be using the nozzle to do at 100% efficiency, while simultaneously attempting to destroy what makes a jet a good idea (and being only partially successful, since a jet is going to be a jet whether you want it to be or not).

Maybe another way would help: in a jet, you use the energy of the airstream to keep the jet running. In your proposal, you are throwing that energy away and replacing it with energy generated at a pump. What you describe isn't physically possible, but if you tried to get it to work, all you'd end up with is a less efficient jet engine (perhaps even one that produced no net energy output).

Last edited: Jun 26, 2006
13. Jun 27, 2006

### RGClark

Thanks for the analysis. Here's one page among many on the net that derive the rocket equation:

Rocket Motion.
http://www.physics.upenn.edu/courses/gladney/mathphys/subsubsection3_1_3_3.html [Broken]

At one point it comes to this equation:

M*dv/dt = -u_ex*dM/dt, M the mass of the rocket including the fuel, v the rocket's velocity and u_ex the exhaust velocity.
Here's one way often presented for solving this:

dv/dt = -u_ex*(1/M)*dM/dt = -u_ex*d(lnM)/dt.
Now if u_ex the exhaust velocity is *constant* we can integrate both sides to get:

v = -u_ex*lnM + C, where C is a constant determined by the initial conditions. You see this will result in an exponential dependence of the total rocket mass with fuel according to final velocity.
Now if u_ex the exhaust velocity is not constant but in fact is kept equal to v, then the equation becomes:

dv/dt = -v*(1/M)*dM/dt,
(1/v)*dv/dt = - (1/M)*dM/dt,
d(lnv)/dt = -d(lnM)/dt.
Integrating, the equation becomes:

lnv = -lnM + C,
lnv + lnM = ln(v*M) = C,
v*M = e^C, a constant.
Since vM must stay constant it must stay the same as it was initially so
v*M = v0*M0, and M0 = v*M/V0. Remember M0 is the initial mass when the rocket is fully loaded with fuel and M is the ending velocity after the fuel is expended, so the equation says the total mass of rocket plus fuel is a linear function of the ending velocity.
Note though the equation also says the ending velocity will be 0 if it starts at 0. So you need to apply it either when you already have some initial velocity or as I was initially proposing have some additional momentum source as by burning the fuel.
Your reference by Anderson might not explicitly give the exhaust velocity in the rocket equation but it is there because the ISP is just the exhaust velocity divided by g (where ISP is given in seconds, when ISP is given in meters/second ISP *is* exhaust velocity.)
The proposal is somewhat of a hybrid of the rocket and jet systems. Key to keep in mind is the thrust from expelling the fuel rearward at velocity v does not need the external air. In this regard this thrust is like the rocket situation. However, as I said this alone if started at 0 would not give you any added velocity. That is why I need the other portion of the scenario where the fuel is then burned with air which it is still with respect to.
Keep in mind the method presupposes there is some method for expelling pure *fuel*, not combustion products, rearward at the steadily increasing speed v. This is why you can still combust it with air.

Bob Clark

Last edited by a moderator: May 2, 2017
14. Jun 27, 2006

### RGClark

The proposal is just to use a thin walled pipe with no ramps, diffusers, turbines or the like inside to impede the air flow. Imagine this pipe say 10 meters across with a wall 1 cm thick. As the front edge of the wall hits the air flow certainly there will be disturbance of the air flow. There will also be disturbance of the air flow due to boundary layer effects. This will have an effect very close to the wall. But the overwhelming majority of the air will still flow at its entering speed.
Remember with ramjets/scramjets your have ramps, diffusers specifically designed to slow the air flow. I want to dispense with those.
Jets can work without compression. They just aren't efficient fuel wise. That in fact is what happens with afterburners. They are just a pipe in which fuel is injected into the oxygen rich exhaust. This exhaust is not compressed before the afterburner fuel is injected. In fact to achieve the highest thrust it is known it is best to have this thrust exit at the same pressure as the surrounding air.

Afterburners though are not fuel efficient. And the part of the proposal requiring the fuel to be burned probably also would not be fuel efficient. However, key is that the overall result is that the fuel required would only be a linear function of ending velocity.
This would result in total fuel used less than of rockets or scramjets.

Bob Clark

15. Jun 27, 2006

### RainmanAero

I guess I am just going to give up, because you refuse to see how and why your analysis is incorrect.
This is the fundamental thing that you do not understand is invalid. You simply cannot have such a hybrid. If you have an open system (with an intake) it is not valid to assume you can use the rocket equation, which has no intake. Understand that if you can succeed (don't know how) in making the internal velocity to be equal to the vehicle velocity (despite the shock wave you will have that WILL decelerate the flow and compress it), you would still have to assume that the fuel-air mixture, when it combusts, will expand isentropically in all directions. This will cause increase in velocity in BOTH directions within the control volume. That means you will get reverse flow out of the inlet (again, assuming you can have zero relative velocity inside the control volume, which you cannot).

But like I say: I give up. You think you are right, and more people than just myself have told you why you are not. Your OP wanted feedback on your analysis, and you have gotten it. Yet you refuse to believe those of us who do this for a living.

Go ahead and build one if you are so sure of your analysis. If you live through it you will learn some valuable lessons.

Rainman

16. Jun 28, 2006

### Staff: Mentor

No. You cannot make air flow straight just by wishing it. When air hits the edge of the intake, it will make a large shock wave and there isn't anything you can do to prevent that.
You need to start accepting the facts/reality of how airflow works:
http://www.airandspacemagazine.com/ASM/Mag/Supp/FM99/oxcart.html [Broken]

The intake on the SR-71 engine is its most critical feature. Those things that you want to do away with are the entire reason why jet engines are a good idea. What you are doing is taking what you think are good ideas - based on your logic applied to your complete igorance of the reality of how airflow works - and combining them into a mathematical model that has nothing at all to do with real life.
You were claiming above that your design would increase fuel efficiency.... so perhaps we are making progress...
The exhaust has already been compressed and has started expanding again. You're still not understanding what makes the air flow through the engine. And you clearly don't understand how/why the exiting air needs to be at ambient pressure in an ideal nozzle.
And it doesnt matter how many times you say it - what you propose is not physically possible and you're not listening or trying to learn why.

Pick up a book on jet propulsion or take a class in high speed flow (that's 3rd year aerospace engineering) - or failing that, start listening to people who have.

Thread closed.

Last edited by a moderator: May 2, 2017
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