Water Rockets

[Christie]

Why does a rocket fly when water and pressure from a pump is added??? I know this is probably very easy, but I'm not good at science and I'm new to this forum so I don't know how things works yet.

[timo]

The basic idea is the following:

- Due to the pressure, the water wants to escape the container. It can do so only in one direction (at the rear end of the rocket).

- The water leaves the rocket with a certain velocity and hence carries a certain momentum p (the actual value is not relevant here, just note that it´s not zero).

- Momentum is a conserved property which means that if you give the water a momentum p at the same time somewhere else a momentum -p must be obtained. In this case, it is the rocket whose momentum changes by -p.

- Often, a change in momentum means a change in velocity. In this case, from p = m*v is follows that v = p/m. So if the momentum of the rocket changes by -p, its velocity changes by -p/m.

 

Well, that´s about it. The momentum of the outstreaming water is compensated by the rocket by gaining momentum and hence increasing velocity. Please note that above is only an attempt to explain what is happening. Several terms like the mass where not explained properly and can become a bit tricky as soon as you really start doing calculations.

 

 

That leads me to some request to the other sfn members: On sfn, there is the tendency to greet every new member who presents his theory of everything with the words "show us the math" whereas very little (well, basically "no") math or real physics is used in the "real physics sections", either. I think some work on this subject (rocket equation) should be challenging but feasible and most importantly interesting for many people in here. So it would be nice to see some explanations/discussions on the rocket equation in this thread, even if it´s just failed attempts to solve the problem (oh, and since I see that coming already: With that I explicitely don´t mean the link to the corresponding wikipedia article(s)).

 

EDIT: Oh, and btw: Welcome to SFN, Christie.

[YT2095]

another way to look at it is this; Imagine a Gun or a Cannon, when the pressure from the gunpowder builds up beween the cannon barrel and the cannon ball inside it it pushes the ball out of the barrel, but it also pushes the barrel backwards too, like a Kick/recoil.

the only difference in this case is thet the cannon ball is water and the barrel is the rocket body or bottle, and instead of hot gasses from burning gunpowder your using pressurised air instead.

[D H]

Very good summary, Atheist. The water rocket (or any rocket, for that matter) accelerates because it ejects mass at some non-zero velocity relative to the rocket. Conservation of momentum dictates a corresponding change in the velocity of the rocket. It's pretty simple, conceptually.

 

I think some work on this subject (rocket equation) should be challenging but feasible and most importantly interesting for many people in here. So it would be nice to see some explanations/discussions on the rocket equation in this thread, even if it´s just failed attempts to solve the problem (oh, and since I see that coming already: With that I explicitely don´t mean the link to the corresponding wikipedia article(s)).

 

EDIT: Oh, and btw: Welcome to SFN, Christie.

 

Welcome to SFN, Christie.

 

Do you want the answer, Atheist (this problem is one of my jobs), or do you want me to help others get to the answer?

[timo]

Do you want the answer, Atheist (this problem is one of my jobs), or do you want me to help others get to the answer?

 

I was hoping that some people who don´t already know the whole physics behind it get interested in the topic and try to find a/the/some solution for themselves or even together. Your input is very welcome, of course. But I think we should wait for some more replies by others or some feedback from Christie, first.

[Ragib]

Am I missing something here? Which rocket equation are you referring to? I had a feeling there was more than one...

[timo]

I´m not even sure there is any. I made up the term. What I was meaning basically was any investigation of the matter beyond hollow phrases. Be it a complete solution of the movement equation, be it an investigation of the amount of fuel necessary to achieve a specific change in velocity, be it a quantitative statement of the momentum change that results from a small amount of fuel spent or be it whatever interests you.

[Ragib]

I´m not even sure there is any. I made up the term. be it an investigation of the amount of fuel necessary to achieve a specific change in velocity

 

I Think i have the perfect thing, perhaps you could put it into latex for me later though. Velocity= Escape Velocity x Natural Log of (Mass/Orignal Mass)

 

This shows the velocity through different times of mass, changing by say, expenditure of fuel, in this case water.

[CPL.Luke]

well there is a rocket equation, namely

 

dp/dt=m dv/dt +dm/dt v

 

however it gets really interesting when you try to figure out how the quantities of mass being ejected out the back create the force that the rocket feels.

 

Or how a person in the rocket feels a constant force but with respect to any given frame he is not acceleerating linearly.

 

I've always wondered how you would formulate a proper coordinate transformation to take the force percieved by the man in the rocket too the force observed by a person in anouther frame.

[D H]

well there is a rocket equation, namely

 

dp/dt=m dv/dt +dm/dt v

 

That is not the "rocket equation". That is simply the chain rule.

 

I Think i have the perfect thing, perhaps you could put it into latex for me later though. Velocity= Escape Velocity x Natural Log of (Mass/Orignal Mass)

 

Not quite, but closer.

 

Google is your friend ... Google rocket equation

 

BTW, the "rocket equation" is erroneous in the sense that it ignores the momentum of the fuel flow inside the rocket.

[CPL.Luke]

the fuel flow inside of the rocket can be ignored as the momentum exchange will simply cancel out, the only thing that matters is the momentum being carried away by the fuel exiting the rocket, and the rocket.

 

also I've seen that application of the chain rule reffered to as the rocket equation for reasons that it allows for very intricate applicattions for thrust functions. also if you note the rocket equation in the wikipedia article appears to merely be the solution of the differential equation I posted.

[D H]

the fuel flow inside of the rocket can be ignored as the momentum exchange will simply cancel out.

 

It does not "simply cancel out". Here's a simple example.

 

Drill holes in the center of the top of a covered hopper car (one of these). Assume the hopper is quite deep so that the center of mass of any fluid poured into the hopper will always be at the center of the hopper, lengthwise. Place a channel along the length of the top of the hopper car, sloped so that any fluid poured onto the track will run into the hole in the top of the hopper. Put a water tank over the channel and located some distance from the hole. Put the hopper car at rest on a frictionless railroad track. Open the spigot to the water tank so the water drains into the hopper.

 

What happens to the hopper car? Answer: it moves. To an observer fixed on the hopper car, the horizontal location of the system (water+water tank+hopper car) center of mass changes as water flows from the off-center water tank to the hopper. However, no horizontal external forces act on the entire system, so to an observer fixed on the ground, the horizontal location of the system center of mass does not move. The hopper has to move to make the system center of mass stationary.

 

Now repeat this experiment, but with a hole drilled in the bottom of the hopper as well. Now what happens to the hopper? It still moves, but with a different velocity than the original problem. Imagine that a frictionless channel is placed between the frictionless tracks so that any water that spills from the hopper maintains its momentum. The center of mass of the entire system (hopper car + water tank + water flowing on top of the hop + water under the hopper) is stationary.

 

Think of this latter case as a zero-thrust rocket. The fuel flow on top of the hopper has to be included in the equations of motion.

 

also if you note the rocket equation in the wikipedia article appears to merely be the solution of the differential equation I posted.

 

A falling rock, a rocket, an intricate set of pulleys and weights, any physical system for that matter, obeys [math]d\vec p/dt = m\,d\vec v/dt + dm/dt \,\vec v[/math]. Why? Because the equation is a tautology.

[timo]

I Think i have the perfect thing, perhaps you could put it into latex for me later though.Velocity= Escape Velocity x Natural Log of (Mass/Orignal Mass)

[math] \vec v = \vec v_e \, \ln \frac{m}{m_0} [/math]. Hover your mouse over the equation to see the latex code. To write TeX in this forum you must enclose the code like "[ math] here goes the code [ /math]" (without the spaces, of course). I highly recommend getting used to TeX as soon as possible. It´s easy to learn and very powerful in writing articles/posts containing mathematical formulas.

 

The equation is what I originally thought of, although I was rather thinking in terms of trying to derive it than simply stating it (since simply stating it is a no-brainer, it´s written in the wikipedia article).

 

Also if you note the rocket equation in the wikipedia article appears to merely be the solution of the differential equation I posted [dp/dt=m dv/dt +dm/dt v].

You can derive the equation on wikipedia from conservation of momentum. Since total momentum is conserverd you can set dp/dt=0. If you plug in a suitable term for dp/dt, you can solve something for something (change in velocity for amount of fuel spent solved by integration, in this case). What I want to say: dp/dt = d(mv)/dt = dm/dt v + m dv/dt or something similar can be an important step in deriving the equation on wikipedia and the result might resemble an integrated version of it. But that is not because it was anything rocket-specific but because of a more fundamental principle of physics: Conservation of momentum.

[D H]

[math] \vec v = \vec v_e \, \ln \frac{m}{m_0} [/math].

 

I assume you meant exhaust velocity when you wrote the above. This is indeed THE "rocket equation". Ragib said "escape velocity", which is quite a different concept from exhaust velocity.

 

I highly recommend getting used to TeX as soon as possible. It´s easy to learn and very powerful in writing articles/posts containing mathematical formulas.

 

I couldn't agree more. While math and physics are beautiful in and of themselves, LaTeX makes papers written about math and physics beautiful to behold.

 

The equation is what I originally thought of, although I was rather thinking in terms of trying to derive it than simply stating it (since simply stating it is a no-brainer, it´s written in the wikipedia article).

 

Quoting (or misquoting) some equation written elsewhere doesn't take much ability at all. Deriving the rocket equation requires some simplifications. The ability to simplifying things (but avoiding oversimplification) is an important skill in applied physics.

[CPL.Luke]

who says that the center of mass of the system has to be stationary?

 

if I have a train car and I drop a 5 ton wait on one end of it, the center of mass moved, but the train car most certainly did not.

 

However the main thing in your problem is that there is a delay between the time that the liquid hits the top of your hopper car, and he time it hits he inside of the car and comes to rest. once the liquid has impacted the inside of the car and come to rest, the momentum cancels out, so shortly after you turn the water off the hopper will come to rest.

 

same thing happens in your second model with the hole in the bottom of the hopper. because the hole leaves the liquid falling straight down, the total momentum still goes to zero after the fuel stopped moving.

 

In your third model where the fuel is directed as it leaves the hopper, that one actually does have a net effect, as your effectively taking the gravitational potential energy that the fluid has and using it to move the hopper.

 

but if you look carefully at your third setup the net force that the hopper feels over the course of its "burn" is due only to the water being directed out the bottom of the hopper.

 

 

A similar thing will happen in the rocket, when you first start the pumps inside of the rocket you may get some odd motion, but all of the momentum that was carried by the fuel internally will total out to zero, and any temporary velocity changes will also go to zero by the time the rocket has finished its burn.

 

also if you disagree with my interpretation of the hopper situation, can you please post a diagram of your setup.

[D H]

CPL.Luke,

 

In the first setup, the car will come to a rest as soon as the water flow is turned off. The main points of this setup were (1) to show that the vehicle does indeed move as a result of fuel flow and (2) to set things up for the variable mass situation.

 

In the second setup, the car does not stop moving when the fuel flow terminates. The fuel flow inside the vehicle forces a residual velocity that balances the flow outside the vehicle.

 

There is no third setup --- yet. The third setup follows. This setup does indeed involves rockets. Two of them on one vehicle.

 

The vehicle is symmetric except for the location of the fuel tank. Two rockets are placed on the vehicle, one on each end, each thrusting outward. The ascii diagram below depicts this simple vehicle. The [] represents the off-center fuel tank location.

 

>=====[]==<

 

The two thrusters have identical specific impulses and fuel consumption rates. The net force on the vehicle is therefore zero. The asymmetric fuel flow dictates that the vehicle itself gain some momentum while the thrusters are firing. This momentum does not magically disappear when the fuel flow terminates.

[CPL.Luke]

actually if you note there are three setups

 

the second setup has the water hitting the side of the hopper, and then trickling down a whole in the bottom. this means that once the water flow stops all the "borrowed" momentum will be returned to the hopper, and the hopper will come to rest.

 

then in a final setup you have added an additional ramp to the hole, thus directing the water out towards the "back" of the hopper and then there is nothing to "catch" the momentum again and bring the system to rest.

 

in the final setup the energy comes from the gravitational potential energy in the water.

 

In your first two setups your essentially trying to put a boat engine in a bathtub and make the bathtub move, outside of wiley cyote this doesn't work.

 

 

now for your new setup

 

where is this assymetric fuel flow that you speak of? you already said that the thrusters produce the same amount of thrust, and use the same amount of fuel to do so, if this is the case then there is no assymetric fuel flow to create your momentum.

 

however if you mean that because of the difference in the length of pipe to the thrusters one of the thrusters will fire first compared to the other, then of course you will have a net gain in momentum by the hopper.

 

 

also the rocket equation only involves the final velocity of the rocket for any amount of expended mass. So you because in your first two setups the final velocity was zero, you can see how the internal fuel movement doesn't matter. Also in your third setup (the one I call your third) the hopper did eject fuel and is thus able to gain a net velocity.

 

actually your experiment does make for an interesting idea, a spaceship could potentially produce "impulse" without expending any mass as long as it had a very large amount of energy at its disposal

[D H]

then in a final setup you have added an additional ramp to the hole, thus directing the water out towards the "back" of the hopper and then there is nothing to "catch" the momentum again and bring the system to rest.

 

If you read that in my post, I apologize for the bad write-up. There is no third setup in my original post on this concept. There is no additional ramp.

 

Now for your new setup: where is this assymetric fuel flow that you speak of?

 

The fluid flow momentum from the tank to the i^th thruster is [math]\dot m_{t_i}(\vec r_{s\rightarrow t_i} - \vec r_{s\rightarrow f})[/math] where [math]\vec r_{s\rightarrow t_i}[/math] and [math]\vec r_{s\rightarrow f}[/math] are vectors from some fixed point on the rocket to the i^th thruster and the fuel tank. The flow momentum to the thruster further from the fuel tank is larger (in magnitude) than the flow momentum to the nearer thruster. It is this flow momentum that the rocket equation ignores. Nothing wrong with that per se; a lot of physics is knowing what effects to ignore and what effects need to be modeled.

 

More later; my wife and kids are waiting on me.

[CPL.Luke]

out of curiosity why are we generalizing to i thrusters?

 

but only in your latest setup do you get a net gain of momentum in the hopper, and as far as I can tell that is mainly because the thruster closer to the tank ignites before the other one does.

 

I don't see how this shows how the rocket equation is invalid, unless your tryin to say that it doesn't take into account opposing forces (which its not supposed to)

[D H]

I don't see how this shows how the rocket equation is invalid.

 

Of course it is "invalid", in the sense that most simple models are "invalid". (More precisely, they are only approximately correct.) A lot of simplifying assumptions are made in deriving the rocket equation. Quoting from the Wikipedia article,

From http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Although an extreme simplification, the rocket equation captures the essentials of rocket flight physics in a single short equation.

 

Ignore all the details of timing. The rocket equation ignores the fluid flow momentum (and a host of other items, such as gravity, thrust variation, and thrust vectoring).

[Ragib]

O sorry my bad it was exhaust velocity, not escape. I just saw the small e and i automatically assumed escape. I don't normally delve into Rocket Science too often.

 

As to deriving it, it says on the wikipedia article:

 

The equation is obtained by integrating the conservation of momentum equation

 

mdv = (ve)dm

 

where ve is exhaust velocity. Fair Enough.

[CPL.Luke]

well I'll give you that it ignores thrust vectoring, and thrust variation

 

 

I don't think anyone in their right mind would want to use it in a gravitational field.

 

but you have yet to show me how internal fluid movements matter in this instance. you have shown me that a rocket can temporarily gain some velocity without ejecting any mass, but I have shown you that that rocket will imediatly come to rest once the fuel shuts off.

[D H]

but I have shown you that that rocket will imediatly come to rest once the fuel shuts off.

 

No, you have not.

 

First, I'll examine what happens to the hopper car that leaks water to the ground.

 

Nomenclature:

 

[math]\dot r_{v_{flow}}[/math] Hopper car velocity while water is flowing

[math]\dot r_{v_{final}}[/math] Hopper car velocity after water flow terminates

[math]r_{v\rightarrow f}[/math] Location of the water tank wrt the center of the hopper car

[math]m_v[/math] Dry mass of the hopper car

[math]m_f[/math] Mass of the water in the tank prior to opening the valve

[math]\dot m_f[/math] Constant rate at which water tank mass changes

 

Scenario:

A hopper car on a frictionless track starts at rest with respect to the ground with the valve to the water tank on top of the hopper car closed. The valve is opened at some time [math]t_0[/math], at which time water immediate begins to flow at a constant rate [math]\dot m_f[/math] from the tank toward the center of the hopper. The water drains from the hopper to a frictionless channel underneath ground. The velocity of the drain water with respect to the hopper car is zero. Water drains from the tank to depletion at some time [math]t_f = t_0 + m_f/\dot m_f[/math].

 

Analysis:

 

Ignoring end effects (which I can make arbitrarily small by increasing the amount of water in the tank), the hopper car moves with a constant velocity while the water is flowing. Call this velocity [math]\dot r_{v_{flow}}[/math].

 

To conserve momentum, the hopper, the water in the tank, and the water under the track must have a velocity that opposes the flow inside the hopper:

 

[math](m_v + m_f) \dot r_{v_{flow}} = - \dot m_f r_{v \rightarrow f}[/math]

 

Note that [math]\dot m_f[/math] is negative. Thus the vehicle moves in the direction of the water tank while water is flowing.

 

After tank depletion, the water under the track continues to flow with the velocity [math]\dot r_{v_{flow}}[/math]. To conserve momentum, the hopper must have a final velocity

 

[math]m_v \dot r_{v_{final}} = -m_f \dot r_{v_{flow}}[/math]

 

or

 

[math]\dot r_{v_{final}} = \frac {\dot m_f\, m_f}{m_v (m_v+m_f)} r_{v \rightarrow f}[/math]

 

In other words, the hopper does not stop when the flow stops. Instead, it changes direction.

 

The exact same situation occurs with the dual-thruster rocket. The rocket moves one way while the thrusters are firing, then reverses direction when the flow stops.

 

When applied to a realistic rocket, the result is that the velocity is always a tad smaller than the result acheived by integrating the momentum transfered to the exhaust gas. The effect on the velocity is small. The effect on position is not small -- at least to me. I have to certify that the propagation techniques used to simulate vehicle behavior are accurate to 1 part in 10^-12 or so.

[CPL.Luke]

in the setup you first used in your last post did you mean an inclined channel at the bottom? or a level channel?

 

if it was a level channel and water impacted it while moving directly down, than the fuid will flow with equal velocity in both directions, and no momentum will be gained in the hopper car.

 

if the channel is inclined, than the water will flow in one direction, and depart some amount of momentum into the earth causeing the earths rotation to speed up by a very small amount.

 

you seem to be over thinking this problem, it doesn't matter what the water does after its left the hopper, there's no way for that information to be returned to the hoppper for it to matter, so the hoppers velocity is only dependant on the velociy with respect to the hopper that the water left the hopper at. (downward velocity doesn't matter as that momentum is conserved against the earth.

 

 

as for your second setup with the dual thrusters, like you said the effects of the internal fuel movements cancel out in the end. if your simulating the behavior of the rocket, then it matters, not if your just analyzing end behavior like the rocket equation does.

 

as for your simulations, out of curiosity what are you doing that requires that precission?

[D H]

in the setup you first used in your last post did you mean an inclined channel at the bottom? or a level channel?

 

The water falls from the hopper with the same horizontal velocity as the hopper. Ignoring vertical motion, this is akin to a rocket for which [math]v_e=0[/math].

 

The intent of this contrived setup was to create the same equations of motion as those of the vehicle in space with a pair of opposing thrusters, one on each end of the vehicle.

 

it doesn't matter what the water does after its left the hopper, there's no way for that information to be returned to the hoppper for it to matter, so the hoppers velocity is only dependant on the velociy with respect to the hopper that the water left the hopper at.

 

One very good reason for employing conservation laws in analyzing some system is that doing so enables one to avoid investigating the mechanisms that change behavior. I know that in the initial configuration (full tank, closed valve), the total system (vehicle+water) is at rest with respect to some inertial observer. In the final configuration (empty tank, water flowing under the tracks), the water flowing under the tracks has some momentum. To conserve total momentum the vehicle must have exactly the opposite momentum as that of the water. In this case, the mechanism that results in the change in vehicle momentum is opening and later closing the valve. The change in fluid momentum flow inside the vehicle changes the vehicle's momentum.

 

 

as for your simulations, out of curiosity what are you doing that requires that precission?

 

Simulating real rockets. We want to ensure that the propagation techniques themselves are very accurate so that any discrepancies between the simulation and reality result from stochastic effects and modeling errors.