Max Velocity of a Water Rocket

[flash]

Find the maximum velocity of a PET bottle rocket on a day with temperature 30 degrees celcius and pressure 1 atmosphere, if the pressure inside the bottle was 40 psi and the volume of water in the bottle was 300 mL.

I know of Tsiolkovsky's equation but I'm not quite sure how to use it here. Any help appreciated

EDIT: Pretty sure I need to find the exhaust velocity of the water given the above details. I have a formula that says $$\Delta p = \frac{1}{2} \rho v^2$$ where $$\rho$$ is the density of water and p is pressure that (I think) gives the exhaust velocity v but I don't know where the formula came from or if it is even correct.

 

[flash]

Feel free to ignore the first part of the question. I just need someone to explain that relationship between pressure and exhaust velocity there. Also I'm guessing that the density of water changes with temperature?

 

[kyle8921]

Thank you for your question. I am happy to help you with this problem.

Firstly, Tsiolkovsky's equation is used to calculate the velocity of a rocket based on its mass, exhaust velocity, and the change in mass during the rocket's flight. This equation is not directly applicable to your situation as it involves the mass of the rocket and the change in mass, which are not given in the problem.

To calculate the maximum velocity of a water rocket, we can use the ideal gas law, which states that pressure and volume are inversely proportional at constant temperature. This means that as the pressure inside the bottle increases, the volume of air inside decreases.

Using the ideal gas law, we can calculate the volume of air inside the bottle at 40 psi and 1 atmosphere pressure. We know that the volume of the bottle is 300 mL, so we can set up the following equation:

(P1)(V1) = (P2)(V2)

Where P1 is the initial pressure (1 atmosphere), V1 is the initial volume (300 mL), P2 is the final pressure (40 psi), and V2 is the final volume (unknown).

Solving for V2, we get V2 = (P1)(V1)/(P2) = (1 atm)(300 mL)/(40 psi) = 7.5 mL

This means that the remaining volume inside the bottle is 7.5 mL of air and 292.5 mL of water.

Now, to find the exhaust velocity, we can use the equation you mentioned, \Delta p = \frac{1}{2} \rho v^2. This equation relates the change in pressure to the density of the fluid and the exhaust velocity.

In this case, the change in pressure is 40 psi (the pressure inside the bottle), and the density of water is 1000 kg/m^3. Plugging in these values and solving for v, we get v = 6.32 m/s.

Therefore, the maximum velocity of the water rocket would be 6.32 m/s. However, this calculation assumes that all the water is expelled from the bottle in one instant, which is not realistic. In reality, the water would be expelled gradually, leading to a lower maximum velocity.

I hope this helps to clarify the process of finding the maximum velocity of a water rocket. Please let me know if you have any further questions.