Physics of the Water Rocket

rallain — Fri, 30 Jan 2009 22:38:19 +0000

I said I would come back to this, and I am. I am a man of my word. Yesterday, I posted a link to a video of this really cool water rocket thingy.

How does this work? What is the physics going on here? I think this can be best explained with the momentum principle.

Let me start by pretending like I have some object that shoots out a piece of water (or really it could be anything). Also, let me pretend like this is in space ore something where there are no external forces.

Above is a before and after picture. Initially, there is something (the box) with water inside. Through some process, the water is shot out of the box. The initial total momentum (water plus box) is zero (vector), so the final total momentum is zero (since there are no external forces). Note that the water will be going much faster than the box if the water has a smaller mass.

What if I want to just look at the box? In that case, I DO need to consider the force between the box and the water. Who knows what makes that water shoot out - maybe it is a little rubber band. Anyway, while the water is being shot out, it exerts a force on the box for a certain amount of time. I can write the following for the water:

So this gives me a way to get to the force exerted on the rocket (box, I am calling it a box). Clearly, if the box just shot one little piece of water out, I wouldn't need to do much more. However, suppose I want the box to keep shooting water out so that it can hover over the ground on Earth. If this is the case, I can write the force on the box as:

I wrote delta m because it is not the total mass, but the little piece of water being shot out in the time there is an interaction between the water and the box. However, the change in mass over the change in time can be represented as the "flow rate" of the water. Note that this is not really the normal rocket thrust derivation because I assumed that the mass of the box doesn't change (although this wouldn't really change the force on the box anyway). Let me call the flow rate, k in units of kg/sec. Then the force (thrust) on the box is:

How much thrust would it take to keep this guy up in the air? The thrust, I can estimate. However, I will not know both the flow rate AND the water velocity. I will, instead, completely off-the-wall estimate the water velocity and then calculate the flow rate.

Actually, I am going to look at the video more closely to see if there is something there that can help me estimate the water speed. In the past I had used Download Helper (a firefox add on) to get streaming movies, but for some reason this was not working. Instead, I used KeepVid - an online thingy for getting videos (free). KeepVid seemed to only work when it gave me the file in .flv format. I used MPEG Streamclip (also free) to convert to mp4 or something.

I looked at the movie, and the only thing I could think of was the the water seemed to be faster than a fire hose. If you pointed the thing straight up, I would be it would go 20 meters (complete guess). There would be some air resistance, but if there weren't, I could use work-energy to find out the speed. I am going to call h the height the water goes when pointed straight up and v₀ the initial speed. If I take the water+Earth as the system, then the work-energy says:

If I put in 20 meters for h, then v₀ = 20 m/s (about). Let me draw a free-body diagram for the rocket-dude.

The thrust from the two water-rockets must be equal in magnitude to the weight of the rocket-dude + backpack + big hose of water. I will call all the mass M. Putting in my expression for the thrust from above:

In case you didn't notice, the two above equations are the y-components of force and velocity. What is the mass? Well, if the guy is 5 meters above the water, and the feeder hose is 0.15 meters in radius, then the mass of just the water would be:

The estimated mass of the person plus back pack could be 80 kg (another complete guess). This will give a flow rate of 1930 kg/sec. That is the total flow rate (each water rocket would have half that). Is that too large of a value? I could calculate the flow rate inside the feeder tube, but that really is just based on some other estimates. How about I calculate the power of engine needed to pump that water? This shouldn't be too difficult. The motor has to do two things. It must lift the water and speed it up. If I assume a constant hose diameter, then the water is always going at the same speed. In one second, 2000 kg of water has to be increased in speed to 20 m/s and lifted 10 meter (the distance traveled in one second). So, the power would be:

800 hp seems rather high, but not impossible. Of course, the problem could be with my estimations. Needless, I don't think this is fake. Dangerous, yes. Expensive, yes. Cool, yes. But not fake.

rallain Fri, 01/30/2009 - 17:38