Exit velocity ?

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HattoriHanzo
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Exit velocity ?

Post by HattoriHanzo » Tue Nov 08, 2005 1:36 pm

More questions...now about exit velocity.

We have two rocket engines. Both have exactly the same thrust, mass flow rate and burntime, but one have higher exit velocity. In theory the rocket engine with higher exit velocity will go to a higher altitude.
But would it be the same in practice? If thats the case, why?
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Anders Troberg
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re: Exit velocity ?

Post by Anders Troberg » Tue Nov 08, 2005 3:07 pm

Without knowing much about rocketry, I can still venture a guess.

Everything else being equal will in this case mean that you have a narrow, fast air column or a wide, slow one.

This is something airplane constructors have known for a long time:

* Narrow, fast air column is good at high speed
* Slow, wide is good for raw power at low speed

This is clearly seen in the differences in how much power a helicopter needs to hover compared to a Harrier. On model airplanes, the fast ones have a small, fast prop, while the ones that will do slow 3D aerobatics (flying well below stall speed, basically just dragging along the plane on raw power) use a slower and larger prop.

Since rockets are quite fast, a fastmoving narrow column of air will work better, giving it better speed and also higher altitude.

At least it sounds plausible.

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re: Exit velocity ?

Post by Ray » Tue Nov 08, 2005 6:05 pm

I am going to make an assumption here first that they also have the same thrust profile.

I don't see how there could be any difference in the altitude achieved, with one exception.

I also don't see how they could have the same thrust etc, without having a different propellent in the motors. In a theoretical world I guess this would be possible.

There are two different types of thrust you get from a rocket motor, pressure thrust, and momentum thrust. Pressure thrust is a theoretical number that can be calculated using the nozzle throat area and the chamber pressure. Momentum thrust is measured thrust and includes the added thrust of the expansion portion of the nozzle.

Exit velocity of the gasses is based on the expansion that happens in the divergent portion of the nozzle and the ISP of the propellent, it hits mach at the nozzle throat.

Exit velocity will determine your ultimate velocity. Higher exit velocity=higher rocket velocity.

SO I guess if your rocket hits "terminal velocity," based on the exhaust velocity, during the burn of the motor, then a motor with higher exhaust velocity will take your rocket higher...because "terminal velocity" would be higher.

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re: Exit velocity ?

Post by leo » Tue Nov 08, 2005 7:08 pm

When you have the same trust and the same mass flow rate, you must have the same exit velocity.
That’s how its works, Action =Reaction.
You can’t compare a rocket with a prop, because with a rocket your fuel is the stuff you accelerate and that’s a fixed mass. With different sizes off props you move different amounts off air mass.

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Re: re: Exit velocity ?

Post by Ray » Tue Nov 08, 2005 7:26 pm

leo wrote:When you have the same trust and the same mass flow rate, you must have the same exit velocity.
The divergent portion of the nozzle will have an effect on the exit velocity.

As an example, a nozzle with no divergent section will achieve local mach as the exit velocity. Same MASS flow rate, and only pressure thrust. Add a divergent section and exit velocity will increase, so will thrust, as momentum thrust.

Of course, if you have the same thrust, from the same mass flow, then the nozzle MUST be the same, so you are correct. Just wanted to point out that the divergent portion of a nozzle is very important to a rocket motor. If designed properly it can add more than 50% to the total thrust of the motor and increase ISP by the same amount.

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re: Exit velocity ?

Post by Anders Troberg » Wed Nov 09, 2005 6:45 am

You can’t compare a rocket with a prop, because with a rocket your fuel is the stuff you accelerate and that’s a fixed mass. With different sizes off props you move different amounts off air mass.
Not really. A small, fast prop will move the same amount of air in a fast moving column as a slower and bigger will do in a wide, but slower column, but the results are quite different.

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re: Exit velocity ?

Post by leo » Wed Nov 09, 2005 7:40 pm

Not really. A small, fast prop will move the same amount of air in a fast moving column as a slower and bigger will do in a wide, but slower column, but the results are quite different.
But the use different amounts off horsepower.
When you use the same engine you get different amounts off air with different props.

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re: Exit velocity ?

Post by Anders Troberg » Wed Nov 09, 2005 9:17 pm

Not really, they are geared differently, smaller props are spinning faster. There is also the angle of attack of the blades to consider.

Choosing the right prop for a certain application is not an easy task (which, by the way, is what I thing the Mythbusters effed up in the jet pack build).

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re: Exit velocity ?

Post by HattoriHanzo » Fri Nov 11, 2005 4:26 pm

Great response!


Of course it is Newtons second law, as stated above! Why didnt I think about that earlier :roll:
So it is: F=ma -> T=massflow(fuel+oxidizer) x exit velocity
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Re: re: Exit velocity ?

Post by Ray » Fri Nov 11, 2005 7:13 pm

HattoriHanzo wrote:So it is: F=ma -> T=massflow(fuel+oxidizer) x exit velocity
Well, not strictly. A in the equation is acceleration...therefore exit velocity isn't what you'd use...exit velocity is a speed, not an acceleration. You'd have to figure out the acceleration to plug the numbers into the equation. But you got the right idea.

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Re: re: Exit velocity ?

Post by larry cottrill » Fri Nov 11, 2005 7:29 pm

Ray wrote:
HattoriHanzo wrote:So it is: F=ma -> T=massflow(fuel+oxidizer) x exit velocity
Well, not strictly. A in the equation is acceleration...therefore exit velocity isn't what you'd use...exit velocity is a speed, not an acceleration. You'd have to figure out the acceleration to plug the numbers into the equation. But you got the right idea.
Ray -

Technically, you're correct, of course, and yet the old Zucrow equation of Thrust = Massflow x ExitVelocity is still a legitimate value to use. The reason it works that way is that in a rocket, the fuel mass is accelerated from rest (relative to the engine) to the exit velocity. Since the velocity and massflow are in terms of 1 second units, the velocity attained is the same value as the acceleration(!) and the massflow is the same as the slug of mass moved(!) during that same 1-second interval. Therefore, the two representations are exactly equivalent for each 1 second of burn time.

L Cottrill

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re: Exit velocity ?

Post by leo » Sat Nov 12, 2005 12:56 am

Thrust = Mass flow x Exit Velocity

And horsepower = Mass flow x( Exit Velocity squared )

This means 2 times faster Exit Velocity = 2 times the Thrust, but 4 times the horsepower.

That’s why a helicopter has to move a lot of mass to keep the needed horsepower low.
But when you fly at greater speed you need a greater Exit Velocity because you have to subtract the Intake Velocity.
Of course only the used air part of the combustion, not the fuel part.
This is something to think about when you interpret TFC numbers.
You have to take into account on what speed you want to fly.
Of course with rockets you always want the greatest Exit Velocity achievable.

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