I'm trying to work out how efficient a mechanical water rocket would be, and I'm a bit confused about the theory:
Rocket 1 accelerates 1Kg water / Second of exhaust to a velocity of 20M/S Rocket 2 accelerates 0.5 Kg water / Second to an exhaust velocity of 40M/S
By my reckoning, both these rockets give the same amount of thrust (Mass x exhaust velocity) but it takes twice as much energy to power rocket 2 as it does to power rocket 1 (kinetic energy of the exhaust = 0.5MV^2).
I'm sure I must have made a mistake - can anyone point it out to me?
Energy efficiency in a rocket is a tough thing to define, because the performance of a rocket motor/engine is appropriately measured as thrust integrated over time (average thrust times time) which gives an impulse, or momentum imparted. Work (kinetic energy imparted) is measured by thrust integrated over distance (thrust times distance for a constant speed and thrust, which generally doesn't happen) Because power depends on speed, and a rocket's speed is generally constantly changing, depending on mass and time of flight, calculating power and energy is very difficult and not particularly useful.
At one point, Goddard calculated the energy efficiency of a rocket by determing the kinetic energy imparted to the exhaust from a rocket held in a fixed mount. He found that something like 2/3 of the chemical energy in a propellant could be converted to the kinetic energy of the exhaust gas.
But in general, it is better to measure the effectiveness of a rocket by determining its exhaust velocity or specific impulse. Specific impulse is the total impulse (thrust integrated over time, or average thrust times time, the momentum imparted by the rocket--total impuls is the quantity specified by the letter in an NAR designation, such as D12-3) divided by the weight of the propellant consumed. Turns out that these are proportional (except for some wacky nozzle/external air pressure effects), such that exhaust velocity = specific impulse x g.
The exhaust velocity is actually the best measure of efficiency right there.
0r if you prefer, specific impulse = 20 m/s /9.8 m/s^2 = 2.04s
Again, the exhaust velocity is the best measure of efficiency, though you could express it as Specific impulse = 4.08 s.
exhaust velocity)
Mass x exhaust velocity gives kg m/s, units of momentum or impulse, not thrust. They give the same total impulse. 20 m/s * 1 kg = 20 kg m/s = 20 Newton-s (=20 N-s) which is the impulse of a full D engine. This is the same for both rockets you describe. And total impulse is a good measure of total "kick."
Yes, you impart twice the energy into the water in rocket two. You pumped more air pressure into rocket two. You are rewarded with a higher exhaust velocity. this is good, not because you get more total kick out of the rocket (20 N-s either way) but because the weight of propellant you had to accelerate is less. If the weight of propellant is a significant part of the total weight of the rocket, rocket two will be significantly lighter, and will fly higher.
neglecting gravity and drag, rocket performance is described by Tsiolkovsky's formula. Note that burnout velocity (well, the water isn't burning, but the veloocity at the end of thrust), Vf is the measure of performance. Ve is exhaust velocity, and performance improves with that. Mf is final mass, at burnout--it is the empty or "dead" weight. Mi is the launch mass, including propellant. ln (lowercase LN) in the natural log function:
Vf = Ve ln (Mi/Mf)
This takes into account the advantages of a higher exhaust velocity, more propellant (bigger Mi) and less dead weight (smaller Mf)
Notice that energy doesn't really come into play here. Not that energy isn't exchanged, but that energy calculations aren't really all that useful in this particular aspect of rocket physics.
You've got me thinking about something that's been kicking around in the back of my head since I BAR'd.
If Work=INT(F*ds), then what am I missing in this scenario?
A constant force is applied to an object at rest for a fixed duration accelerating that object, and moving it some distance. W=F*d1
That same force is applied for the same duration, this time to an object already moving with some (non-zero) initial velocity. Again some acceleration is imparted, but more importantly, a different distance is traversed since the object was already moving. W=F*d2
IOW, in each case, the same impulse was applied, but a different amount of work was done (or was it?). I can see where the same change in momentum occurred, but using W=F*d seems to indicate more work was done to the moving object than to the object initially at rest.
What's wrong here? What am I missing? Where did I blow it? Why did I skip so many sessions of physics in college?
Actually, what I'm considering is a water-rocket powered glider that would operate at a roughly constant airspeed. I was wondering whether the energy in my compressed air would be put to better use powering a propellor or pushing water out of the back. The thrust doesn't have to last very long - just enough to get some altitude.
Perhaps I should have said this before - I figured that practicalities don't always help in understanding the theory, but perhaps I was wrong.
I gave Mass / second - which I think converts my units into thrust?
Perhaps another application where this could make a difference would be something like an ion-drive, where the power is fixed (by the size of the solar-cells / reactor...) but the propellant mass can be much lower compared to the mass of the spacecraft.
So as the specific-impulse of a rocket engine increases, its energy efficiency really does decrease? I never realised this before.
You can't realy compare work done with momentum. Rather, the work done is equal to the change in energy (kinetic plus potential), neglecting drag and other things not modelled. It is useful for things like piston launchers, and not so useful for rocket flight.
I know that's normally the case, but for me I'm not so sure - I'm considering a rocket-powered glider where the lift is generated aerodynamically.
The amount of power I have is fixed (the amount of air in a CO2 cylinder). I want to use it as efficiently as possible, and I don't need to worry as much as usual about the mass of the propellant (water).
The maximum theoretical amount of kinetic energy transfer from the propellant to the spent (single stage)rocket is 64.761% at a mass ratio of 4.921553.
Interesting but largely useless tidbit of rocket info.
What is useful is that the fraction of this maximum energy transfer increases rapidly from a mass ratio of one but decreases slowly at higher mass ratios. What this means for hobby rocketry is that almost all hobby rockets operate at very low energy efficiencies. A mass ratio of 1.2 only uses about 17% of that 64.761%---approximately 10% efficiency. And then drag losses eat up most of that. And this all doesn't even include the losses converting combustion heat energy into kinetic energy of the exhaust.
It's very easy for the energy efficiency of a hobby rocket measured as gravitational potential energy at apogee divided by energy of combustion of the the propellant to be low single digit percentages.
One advantage of water rockets is that they can operate at mass ratios that are much more energy efficient.
No mistake. But the mathematics of optimizing water rockets gets very difficult--systems of simultaneous differential equations. Most folks use simple, approximate formulas that work "good 'nuff." These are available on some water rocket web sites. I got partway through doing the math the right way before I burned out on the idea. Way too much work for something that wouldn't make me money! But it was interesting--I actually understand the basic internal interplay of energy and momentum in the things. What people forget about water rockets is that the derivative of momentum wrt time has *two* terms. For "regular" rockets you can ignore one, but not for water rockets.
Interesting. I just got this very result in water rocket calculations. The problem was to adjust the density of the working fluid (traditionally water) to maximize the cutoff velocity. In outer space, this would be
Vcutoff = Ve*ln[(Mprop + Mdead)/Mdead]
where Ve is effective exhaust velocity, Mp is propellant (working fluid) mass aMdead is dead mass (structural mass)
The kinetic energy of the exhaust gas is then
E = .5*Mp*Ve^2
Thus
Ve = Sqrt[2E/Mp]
Substituting into the top equation, and letting Mp = Fract * MDead
you discover the maximum cutoff velocity at constant energy comes at Fract = 3.92..., corresponding to an optimal mass ratio of
4.92...
(It turns out that when thrust to weight ratios are very high, and altitudes at cutoff are short, then cutoff velocities on earth are very near outer space cutoff velocities. All this holds true for water rockets.)
a different distance is traversed since the object was already
Kinetic energy is imparted to the exhaust as well as to the rocket. For the rocket starting at zero velocity, and, say, moving very slowly compared to the exhaust for the duration of the interval in question, the exhaust is moving very fast, and the rocket is moving very slowly, os the exhaust gets almost all of the kinetic energy.
When the rocket is moving at the exhaust velocity (very near the exhaust velocity for the time interval in question), the exhaust ends up effectively at a standstill, and so the rocket itself gets almost all the kinetic energy.
When the rocket is going much faster than the exhaust velocity, the exhaust is actually travelling more slowly than it was when it was in the rocket, and so the rocket actually "steals" the kinetic energy from the exhaust.
That's kind of an ad-hoc explatation, as work and energy just aren't the best tools to understand reaction propulsion. The work-energy equations work out and energy is conserved, but I think it's better for understanding for your head to be in the land of impulse and momentum.
I am blitheringly ignorant in the ways of propellers. But I can't help but notice that there are more prop-driven planes than rocket powered planes in the world. I think there must be a reason.
Kg/s is a mass flow rate, not a thrust. Thrust is a force, in units of kg m/s^2 (mass times acceleration). Mass flow rate times velocity gives units of thrust. (I'm 90% sure that mass flow rate times velocity is thrust, and 100% sure it is proportional to thrust. Pardon my rust on this issue)
Ion drive gives a much higher exhaust velocity, giving a much higher specific impulse.
I don't think I said that anywhere. You can increase specific impulse by using a propellant with more energy to begin with (liquid oxygen vs gunpowder, or in this case, high-pressure air and water vs. low-pressure air and water) or you can increase specific impulse by increasing the energy efficiency of the engine (de laval nozzle vs a simple hole).
A liquid hydrogen rocket burning with a low energy efficiency (Kinetic energy of exhaust in static test / chemical energy input) could outperform a gunpowder rocket with a higher energy efficiency.
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