Calculating Water Rockets
Calculating Water Rockets (no calculus)
Water Rockets are a great way of getting outside and doing some experiments. When we observe how a water rocket performs, we can make some deductions about the effect of changing the air pressure and water volume. If we keep the air pressure constant and change the water volume, we quickly find there’s a point where there is too much water volume, and the rocket doesn’t perform as well. Likewise, keeping the water volume constant and varying the air pressure will also alter the performance. After some experimentation, an optimum amount of water for each pressure can be determined.
To understand the physics of what is going on, we can also predict how the rocket will perform by modelling the physics occurring in the water rocket. This page goes through how to calculate various aspects of the water rocket’s performance that can be used later for modelling the flight profile.
It is important to note that the calculations done in this analysis are an approximation. It gives a good idea of how you can calculate the rocket’s performance from the starting conditions.
Pressure
The difference in air pressure between the container and the outside air drives the water rocket. To pressurise the water bottle some effort is required to pump it up, that energy is stored as compressed air which is then released when the mechanism activates to release the rocket (upsidedown water bottle).
When the mechanism is released, water starts to flow out of the nozzle. This is caused by gravity and the compressed air pushing the water out. The volume of water in the bottle decreases over time, and this makes the space where the air is increased. There are a couple of ways to consider pressure in the water rocket. We can imagine that the pressure in the bottle reduces by a certain amount with every instant in time, or we can take a rough approach and assume that the pressure reduces linearly and calculate the average of the initial pressure and the final pressure.
Boyle’s Law
Boyle’s Law states that for a constant temperature pressure multiplied by volume is constant. So if you increase the volume you decrease the pressure. In the water bottle as the volume of air in the bottle increases, as the water flows out, then the pressure will decrease.
If we know the pressure and the gravity acting on the water then we can calculate how fast the water is going to leave the bottle.
Calculation
We know the volume of the bottle, in this case 1 litre and we know the amount of water, we’ll go with 0.15 L. For this example we will use a pressure of 25 psi, we need to convert the pressure to pascals (or Newtons per metres squared). 1 psi is 6894.76 pascals so 25 psi is 171,869 pascals.
25 psi on the gauge on the water rocket launcher is called gauge pressure, the actual pressure is the 25 psi plus the ambient pressure. We will use a standard atmospheric pressure of 101325 pascals for the ambient pressure and this added to what we worked out above gives 273194 pascals inside the bottle. We call this the absolute pressure.
Boyle’s Law can be represented with this equation:
PiVi=PfVf (1)
we make this in terms of Pf :
Pf=PiVi/Vf (2)
plugging in the numbers, our volumes have to converted to cubic metres and are all about the volume of air, so Vi=0.00085m3 and Vf=0.001m3:
Pf=(273194 x 0.00085)/0.001
Pf= 232215 pascals
We want to find the average pressure acting on the water so that is found by:
Paverage=(Pi+Pf)/2
Paverage=(273194 + 232215)/2
Paverage=252705 pascals
Mass Flow Rate
The next import thing we need to work out now that we have the pressure is the mass flow rate, this is the amount of mass flowing out of the bottle per second. The formula for that is derived from Bernoulli’s equation and the continuity equation and is:
m˙ = AnozzleCd√(2ρwater(Paverage-Patmosphere))
It looks complex but it’s not as we have already calculated Paverage and the density of water (ρwater) we can us 1000 kgm3 and we use a discharge coefficient (Cd) of 0.98.
The nozzle attached to the bottle has an area of 6.36 x 10-5m3.
Putting all of those numbers in the equation:
m˙ = 6.36 x 10-5 x 0.98 x √(2 x 1000 x (252705-101325))
m˙ = 1.085 kgs-1
Water Velocity at Nozzle
The next key thing to work out is the water velocity at the nozzle and this is simply given by the formula:
vnozzle = m˙/(ρwater x Anozzle)
and we have all of those values so we just plug them in:
vnozzle = 1.085 / (1000 x 6.36 x 10-5)
vnozzle = 16.98ms-1
Force
We’re getting closer to deriving some parameters we can use later on. Thrust is defined at the mass flow rate multiplied by the water velocity at the nozzle:
Fthrust = m˙ x vwater
and if we plug the number in that we just calculated:
Fthrust = 1.085 x 16.98
Fthrust = 18.42N
we do need to subtract the force due to gravity though, as that is working in the opposite direction to thrust. This is simply the average of the mass of the rocket as it’s expelling the water multiplied by the acceleration due to gravity of 9.81ms2:
Fgravity = ((Minitial – Mfinal)/2) x 9.81
Fgravity = ((0.3 + 0.15)/2) x 9.81
Fgravity = 2.21N
therefore the total force is:
Ftotal = Fthrust – Fgravity
Ftotal = 16.21N
The Final Calculations
The last few things to work out is the acceleration of the water rocket, the time it takes for the water to be expelled and then, finally, the velocity of the rocket once the water has all gone:
To get the acceleration we calculate it from:
F=mass x acceleration, we have Ftotal = 16.21N and the average mass of the rocket, which is the starting mass of 0.3kg plus the mass with the water gone of 0.15Kg divided by 2, so 0.225kg:
acceleration = Force/mass
acceleration = 16.21/0.225
acceleration = 72 ms-2
The time taken to expel the water is simply the mass of the water (0.15kg) divided by the mass flow rate (1.085kgs<sup>-1</sup>) giving:
t = 0.14s
finally we can calculate the velocity of the rocket once the water is expelled using:
v = a x t
v = 72 x 0.14
v = 10.08ms-1
This means that when all the water has been expelled from the rocket, the rocket travels at about 10 metres per second, or about 36 km/h.
