The physics of the hydrodynamic rocket

How many times have you watched to the shuttle takeoff on TV? And every time you've admired it as if it was a miracle… actually there are many specific studies and experiments behind it.
We also got curious about these events and we tried to better understand them by studying the laws of physics they are based on.

We also got curious about these events and we tried to better understand them by studying the laws of physics they are based on.
Firstly let's analyse the motion our rocket is following after leaving the launching pad, neglecting, as initial hypothesis, the air aerodynamic drag.
As soon as the rocket is released, it accelerates uniformly (a motion is said to be uniformly accelerated when a constant force acts on a constant mass) from an initial velocity equal to zero on a rectilinear trajectory (this is true in an ideal model which has a constant thrust even if mass is decreasing). Once reached the maximum speed, the rocket, in accordance with the second law of dynamics, keeps going with a rectilinear but decelerating motion, and the speed decreases till the maximum height is reached. From now on, the bottle starts its descent, with acceleration equal to 9.81m/s2 (gravitational acceleration).

When the bottle, our rocket, is still anchored to the launching pad, the resultant of all the acting forces is 0. Inside the rocket there is a quantity of water, which is going to act as propellant, and air at a pressure of about 2 atm. When the rocket is released, water gets instantaneously pushed outside by the air (the air expands along an isotherm and the water obeys Bernoulli law) and provides the bottle with the takeoff thrust.
The thrust keeps then acting on the rocket until the water hasn't completely been discharged. From now on, the rocket continues its motion just because of the force of inertia.
As we know, the rocket isn't conceived to utilize the wings aerodynamic lift, in fact, once the motor power comes to vanish, it enters in a free fall motion instead of gliding.
Actually it's really the motor thrust to make the rocket win the gravitational field, taking off.
Once departed, the rocket structure could rotate around its barycentre as a virtual pin. Air flowing around the surfaces preceding the barycentre tends to destabilize the flight, while air acting on the surfaces following the barycentre acts in exactly the opposite way. That's the reason for which flaps are nearly always situated on the lowest part of rockets.
In order to simplify the phenomenon, we can consider the resultant of all the forces acting on the rocket as applied in a single point called centre of pressure. In other researches we have seen there's a relation between the centre of pressure and the barycentre, but we have decided to face this topic in the future .Let's suppose to follow the rocket motion through many little time steps, for example with a time interval equal to 0.01s. At the launch, a little mass of water discharged by the compressed air with velocity acquires a momentum equal to . An identical momentum will be transferred to the rocket, which, from null speed will take off with velocity . In questo intevallo di tempo sufficientmente breve, possiamo approssimare il moto del razzo ad un moto uniforme. We can consider, with little approximation for such a brief time interval, the motion of the rocket as a uniform motion.
During the next 0.01s long time interval, the internal pressure of the air will be slightly decreased (it can be computed through the relations of an isothermal expansion), but will be still sufficient to discharge a new quantity of water with velocity . The rocket will therefore newly acquire a momentum equal to * and will increase its velocity to the value . We'll then have a new little time step again approximatively defined by a uniform motion, this time with velocity . The phenomenon goes on like this for all the following time steps, with always decreasing air pressures and, consequently, with water masses discharged progressively with lower velocities, this until the pressure of the air inside the bottle equals the atmospheric pressure. Now the thrust phase finishes and the rocket proceeds with a decelerating motion. Once reached the maximum height, the rocket, as previously mentioned, falls down with a uniformly accelerated motion. To be precise, taking into account the friction due to the aerodynamic drag, the rocket, during the free fall, quickly approaches the critical velocity and then falls with a uniform motion.
By the end we can summarize the phenomenon as follows hereafter:

1. In order to move a rocket it's necessary to win all the antagonist forces (inertia, aerodynamic drag, etc..)

2. Thrust exceeds the opponent forces;

3. Thrust is generated by the acceleration transmitted to the water, pushed by the internal air pressure (prevailing over the atmospheric pressure), while exiting from the rocket nozzle

4. Air (as well as water) contained in the bottle, in order to be discharged outside, must prevail over the ambient pressure

5. A compressor has to be used in order to inflate air inside the bottle.

A bit of theory

The following symbols will be used:

We'll follow the rocket motion through constant time intervals with . After the inflation of the bottle (filled with ½ l of water), let's use the perfect gas law for modelling the behaviour of the air inside the bottle, which, while pushing out the water, is undergoing an isothermal expansion:

From which it can be derived the following equation:

Therefore, knowing P and V at the initial moment and calculating the mass of the discharged water in a , we can as first step obtain the , and then, through the use of the last equation above reported, the which is the pressure acting on the water for the next . We'll repeat the same procedure for all the time intervals till the water won't have been completely discharged from the bottle.
We can apply the Bernoulli's formula in order to derive the velocity of the water passing through the discharge nozzle:

Since the relative velocity, with reference to the mass centre of the internal air and its pressure can be expressed as

we can obtain the following equation:

from which:

Now the velocity of the water exiting through the discharge nozzle can be easily derived:

By means of the volumetric flow equation:

we can then calculate the quantity of discharged water per second. If you want to know the quantity of discharged water for a given time interval , it will be sufficient just to multiply the flow Q by tha t amplitude:

If on the other hand you want to determine the corresponding discharged water mass in this it will be necessary to insert the water density (we assumed it equal to 1 g/cm^3) in the calculation:

Applying the momentum conservation law, it can be derived the increase of the speed the rocket is experiencing as a function of the quantity of discharged water.
In the reference frame of the rocket mass centre, the velocity of the water still present inside the bottle (not yet discharged) is 0. When discharged, this little quantity of water acquires a momentum, which, in accordance to the conservation law, must be equal to the momentum of the rocket. Through the equation Pressure * Section = Force, we can calculate the force acting on the discharged water quantity. This force times gives the force impulse. This impulse will be applied to the rocket as well. Referring to the velocity acquired by the rocket with the symbol vr we can write the following formula:

from which it is possible to derive the speed variation the rocket has undergone during the time interval :