The principle of propulsion of any rocket is based on the momemtum conservation law, also known as action and reaction law. To illustrate this law, you can do the following experience : Take a ball and get on a roller skate, if you launch the ball behind you, you go backward.
The momentum of an element is the product of its massm by its speed v, the one of a system is the sum of the momentum of each parts of its system.
The momemtum conservation law says :
"The momemtum of a system is null or stays constant if the resultant of the external forces, on this system, is null".
In the previous example, if we suppose that there is no friction force beetween the roller skate and the ground, and you are balanced and motionless. The system is composed of you, with your ball, on your skate board. The resultant of the external forces are null (the weight of your system is offset by the ground's reaction). As the speed is null, the momemtum is null : m*v = 0. If you launch the ball, the two parts of your system have a momemtum equal to the product of their respective mass by their speed. As there is system, we have : momemtum conservation of the
m_{1}*v_{1} + m_{2}*v_{2} = m*v = 0 et donc v_{1} = v_{2}*(m_{2}/m_{1}).
This means that if the ball with a mass m_{2} is launched with a speed v_{2}, you, on your roller skate, will go in the opposite direction of the ball with a speed v_{1} equal to the product of v_{2 }by the mass ratio m_{2}/m_{1}. If m_{1} = 50 kg and m_{2} = 1kg, v_{1} will be 50 times lower than v_{2}.
Thus, for a rocket, the rule of the game is to eject a mass of material (liquid or gas) with a speed as high as possible to propels the rocket in the opposite direction.
In a water rocket, the material is water ejected by high pressure air.
Suppose our water rocket in the outer space and far from the earth, thus without external forces ( no weight and no drag forces), at the time t after the launch, the mass of the rocket is m, its speed is v and the eject speed of the water relative to the rocket is v' (thus relative to the earth, the eject speed of the water is vv').
Consider the time infinetely close t + dt, during the small amound of time dt, a mass dm of water will have ejected, thus the mass of the rocket will have lost dm, but its speed will have gained the small amound of speed dv.
At this time the momemtum of the rocket will be (mdm)*(v+dv), while the momentum of the ejected water is dm(vv').
Since the momentum is conserved, we have :
(mdm)*(v+dv) + dm*(vv') = mv (Momentum at t+dt = Momentum at t)
if we neglect dm*dv very small relative to the others, we arrive to :
m*dv = dm*v'
If we call thethe mass flow of water q = dm/dt, we have dm=q*dt, thus :
m*dv=q*dt*v' or m*(dv/dt)=q*v'
but dv/dt is the acceleration and as the Nedwton's second law of motion says m*(dv/dt) is representative to a force, generated by the ejection of the water, this is the thrust P. So, we can see that the thrust is proportional to the mass of water ejected per second and its speed of ejection :
P=q*v'.
The mass flow of water ejected per second is equal to the density of water time the volume flow ejected per second. This last one is the product of the eject speed by the area of the nozzle. so the new equation of the thrust is :
P = r*s*v'^{2}
Now we have to calculate the eject speed of water and for that we have to use the theorem of Bernoulli. The Bernoulli's family had generated many genius of mathematics and physics, it's good to precise that this theorem is due to Daniel Bernoulli (17001782).
This theorem says that for an incompressible and non viscous fluid, the mechanical energy is conserved in steady fluid flow. This mechanical energy is the sum of three components : The kinetic energy, the potential energy and the pressure energy. Between two points 1 and 2 of a steadyflowing, nonviscous and incompressible fluid, we have the equation :
(v_{1}^{2}/2*g) + p_{1}/(r*g)+h_{1} = (v_{2}^{2}/2*g) + p_{2}/(r*g)+h_{2}
At the area 1 between the water and the air, we have a pressure p_{1}, an area s_{1} (those of the body of the bottle) and a speed v_{1}. At the end of the neck of the bottle, we have a pressure p_{2} which is the atmospheric pressure Patm, an area s_{2} (those of the neck) and a speed v_{2}. Between these two areas we have a height of water h which the difference of altitude between the two areas 1 and 2. The liquid being incompressible, the flow rate in 1 is the same as the flow rate in 2, thus s_{1}*v_{1} = s_{2}*v_{2} and v_{1} = v_{2} * (s_{2}/s_{1}) Then in the Bernoulli's equation, we will have : ((v_{2} * (s_{2}/s_{1}))^{2}/2*g) + p_{1}/(r*g)+h_{1}h_{2} = (v_{2}^{2}/2*g) + P_{atm}/(r*g) We can simplify, so : v_{2}^{2} = (2/(1(s_{2}/s_{1})^{2}))*(((p_{1}P_{atm})/r)+g*h) We can simplify this result because (s_{2}/s_{1})^{2} is very smaller than 1, and then g*h is very smaller than (p_{1}P_{atm})/r. Thus the equation becomes : v_{2}^{2} = 2*((p_{1}P_{atm})/r) Now if we use this result in the expression of the thrust that we find previously, we obtain : P = 2 * (p_{1}P_{atm}) * s_{2} Where p_{1} is the absolute pressure in the bottle, that is the pressure we can read on the gauge plus the atmospheric pressure. Thus p_{1}  P_{atm is the pressure gauge.} Numeric example : Let a normal bottle of 1,5 l, with a diametre neck of 22mm, be pressurized at 6 bars (607950 Pa) at gauge. The thurst at the beginnig will be : P = 2*607950*3,14*0,022_{2}/4 = 462 N 
Now we know how to calculate the thrust, it's time to calculate how high does this rocket go.
For that, we can distinguish 2 phases of flight :
The boost phase itself is divided in 3 steps :
But in these 3 steps, the rocket is submitted to 3 forces :
a = F/M
Of course this step has sens only if you use a launcher with a rod inside the bottle, like those I describe in this site.
Here the thrust is given by the pressure acting on the launch rod. No mass of water or air is ejected (even if there is a leak we neglect it), thus the mass of the rocket don't change. The only one thing which change is the volume allocated to the air, actually, when the rocket go up on the rod, this one release space. If the volume of the air increase, that means that pressure decrease. This phenomenon being very fast, we can consider that we are in an adiabatic process (without heat exchange with outside) so we can use the relationship pV^{gamma} = constante (gamma is the adiabatic constante , that is 1,4 for air). If p_{1} and V_{1} are respectively the pressure and the volume at beginning then p_{2} and V_{2} the pressure and the volume at the output of the launcher. We have p_{2} = p_{1}*(V_{1}/V_{2})^{gamma}.
The volume V_{2} at the output of the launcher is equal to the bottle's volume, and V_{1} = V_{2}  V_{t} where V_{t} is the volume of the rod inside the rocket at the beginning. If this rod has a diameter d and a length L, V_{t} = p*d^{2}*L/4 thus :
p_{2}=p_{1}*(1(p*d^{2}*L/4*V_{2}))^{gamma}.
To simplify the calculations, we will consider that the pressure is constant and equal to the average of p_{1} and p_{2}, while running along the launch rod. We will neglect also the air drag, so with these conditions the acceleration is : a = F/M, with F = ((p_{1}+p_{2})/2)*s  M*g and, if we replace p_{2} by its value found previously :
a= (p_{1}(1+(1(p*d^{2}*L/4*V_{2}))^{gamma})*(p*d^{2}/4)2*M*g)/2*M
Be careful, here, p_{1} is the absolute pressure, that is the pressure gauge plus the atmospheric pressure.
As we have the acceleration, we can calculate the speed v= a * t. However, we don't know the time t, but we know that an increase of altitude for a time t and with a constant acceleration a is given by dh = 1/2 * a * t^{2}. Here, this increase of altitude is the length of the launch rod L. Thus t = sqaure(2*L/a) and then :
v = square(L *(p_{1}(1+(1(p*d^{2}*L/4*V_{2}))^{gamma})*(p*d^{2}/4)2*M*g)/M)
Numeric example : Take our 1.5 liter bottle, suppose its empty weigth of 100g . Fill in at 33% thus with 0,5 L of water. Tshe launch rod has a diameter of 20mm and a length of 20 cm.
The total mass of the rocket will be M = Mf + Me + Ma where Mf = the empty mass of the rocket, Me = Mass of water and Ma = mass of air. This last one seems to be negligible, but we will see below that it is not so negligible. We calculate this mass with the formula : P*V = m*R*T where P is the pressure, V the volume, m the mass, R the mass constant of the air (286,91 J/kg°K) and T the ambiant temperature in ° Kelvin. In our example and if we suppose the temperature at 27°C thus 300 °K, so the mass of the air will be Ma = p_{1}*V_{1}/R*T that is
Ma = 709275 *(10,063)/(1000*286,9*300) = 7,7 g so the tatal mass will be
M = 0,1 + 0,5 + 0,0077 = 0,6077 kg
Now we can calculate the speed v = square(0,2* ((709275*(1+(1(6,3*10^{2}))^{1,4})*(3,14*10^{4})2*0,6077*9,81)/0,6077) = 11,7 m/s that is a little bit more than 42 Km/h. Notice that we have an average thrustof 213 N and an acceleration of 340 m/s^{2} (that is 34 times the earth gravity). The drag at the end of the launch rod is R = K*S*v^{2}, thus with a coefficient K of 0,3 and an area of 3,14*0,088^{2}/4 = 0,006 m^{2} : R = 0,26 N so we can neglect it when faced with
the thrust (213 N) and the weight M*g = 0,6077 * 9,81 = 5,96 N
This is the (very short) period during which the water is expelled.
During this phase, the forces applied to the rocket are :
The Thrust P = 2*(p_{1}  P_{atm}) * s_{2}, which we have demonstrated above.
The weight of the rocket which stand in the way of the thrust : P_{f} = M*g (M is the mass in kilos and g the earth gravity = 9,81 m/s^{2})
The drag : R = K*S_{f}*v^{2} where K is a coefficient, Sf the front area of the rocket and v the speed. The coefficient K is a function of the density, the viscosity and the pressure of the air, the shape of the rocket (Coefficient of drag : Cd). This coefficient must be determined experimentally. In a first time, we can consider that K = 0,3 (a good fork seems to be between 0,2 and 0,6 ) then we have to correct it in accordance to the results.
The second law of motion a = F/M is always applied :
a = dv/dt = (2*(p_{1}P_{atm})*s_{2}  M*g  K*S_{f}*v^{2}) /M
Here, all the terms changed at any time. As the water is expelled, the mass M of the rocket decrease (thus the acceleration and the speed increase) and the volume of the air increase thus the pressure p_{1} decrease and if the speed increase, then the drag increase also.
The resolution of this equation in not very simple, and I admit that I don't know any more how to resolve this kind of equation. But, fortunately, the power of the computers allows us to find numerically the solution by computing step by step the value of each terms, considering that they don't change during the step. Two steps are a small amount of time apart. These calculations give good results and many wellinformed Water rocketeers made good simulators .
When all the water is expelled, it remains compressed gas in the bottle. Of course the pressure is lower, but this makes a thrust not negligable.
The calculation of the thrust in this case is more difficult because the gas is compressible, so we have to use the generalised Daniel Bernoulli's equation. But here I surrender, so I trust to the results given by Bruce Berggren, in November 97 on the maillist. Bruce says that this "gas boost" can be considere as an instantaneous impulse (momentum) applied to the rocket and which produces an increase of the speed of impulse/mass. The mass is the average between the beginnig and the end of this phase. The impulse is given by the formula :
Imp = V*4050/racine(T_{1})*(p_{1}/P_{atm}1)^{1,24}
V is full chamber volume in m^{3}, T_{1} and p_{1} are respectively the temperature in °K and the pressure in Pa of the gaz in the bottleat the beginning of the air burst.
To compute T_{1}, then p_{1} , we assume that we are in an adiabatic process (without heat exchange with outside, which is probably true due to the very short time). In these conditions, we can use the ideal gas equations :
T*V^{(gamma1)} = constante
p*V^{(gamma)} = constante
Thus T_{1}*V_{1}^{(gamma1)} = T_{2}*V_{2}^{(gamma1)}, then T_{1} = T_{2}*(V_{2}/V_{1})^{(gamma1)}. If we consider V_{2} and T_{2} as the volume and the temperature of the air just before the launch and V_{1} and T_{1} the volume and the temperature of the air at the beginning of the air burst. But V_{1} is the full chamber volume and V_{2} = V_{1} minus the volume of water. If Cr is the fill ratio then V_{2} = V_{1} * (1  Cr), thus
T_{1} = T_{2} * (1Cr)^{(gamma  1)}
In the same manner, we have :
p_{1} = p_{2} * (1Cr)^{gamma}
Numeric example : With the values of our previous example, where T_{2} = 300°K, Cr = 33%, p_{2} = 7 bars thus 709275 Pa (This is the absolute pressure) and gamma = 1,4 for the air. So, we have : T_{1} = 300*(10,33)^{0,4} = 255,6 °K (17,4°C) and p_{1} = 709275 * (10,33)^{1,4} = 404873 Pa
the impulse = 0,0015 * 4050 / square (255,6)*(404873/101325  1)^{1,24} = 1,48 N*s
Now we can compute the increase of speed = impulse/mass. We know the impulse, we have to compute the average mass, This is the rocket mass plus the half of the air mass expelled during this phase. We have previously computed the gaz mass just before the water boost phase , this mass doesn't change during the water boost phase because, only the water is expelled out of the rocket (in any case, we do this hypothesis). Now we compute the mass of daz at the end of gaz boost phase with the same principle as above : Ma = p*V/R*T with p = Patm and V = volume of the bottle, R = 286,91 J/kg°K, but we have to compute T. During the phase of adiabatic process, the volume stays constant, so we can use the relation :
T * P^{((1gamma)/gamma)} = constante .
and thus T = T_{1} * (p_{1}/P_{atm})^{((1gamma)/gamma)}
Numeric example : T = 255,6 * (404873/101325)^{((11,4)/1,4)} = 172 °K ( 101 ° C ) thus the mass is Ma = 101325 * 0,0015/286,91 * 172 = 0,0031 kg.
We have seen previously that the gaz mass at the lift off and thus at the end of water boost phase was 7,7 g (0,0077 kg). So the average mass of air is donc la masse moyenne d'air is : 5,4 g and the one of the rocket is therefore 105,4 g, as a result, the increase of speed du to the ejection of the residual air is v = 1,48 / 0,1054 = 14 m/s what is not negligible (more than 50 km/h).
In this phase, there is no more propulsion force, so the rocket, always submit to its weight and the air drag, continue on its way until its speed becomes null ( apogee), then it fall down. During the ascent, The weight and the drag will combine to decelerate the rocket while in descent, the drag is opposed to the weight of the rocket.
The relation a = F/M always apply, of course, therefore :
a = dv/dt = ( M*g  (v/v)*K*S_{f}*v^{2}) /M
where (v/v) is the sign of the speed (if the speed is positive, the rocket goes up, the air drag is added to the weight , if the speed is negative, these forces must be substrasted) thus the speed at the instant t is equal to the speed at the instant tdt plus the variation due to the acceleration, that is : a*dt. As for the variation of altitude, it is the product of the average speed during this period by the interval of time dt :
dh = Vmoy*dt where Vmoy = (Vt1 + Vt)/2. Vt1 is the speed at t1 et Vt is the speed at t.
There, again, the numeric computation allow us to know the altitude reached, then the time of fall down et the speed of touch down, if there is no device to slow down the descent. We can introduce the influence of the openning of a parachute given that the coefficient K increase (this is now those of the parachute that is about 1) and the area S_{f} becomes also those of the parachute.
Numerous rocketeers had realised some flight simulators. I just cite those I consider as the mains :
