All big rockets are burning either gas or fluid to create thrust. While this is so, I have filled up a plastic bottle with air at high pressure, and it can go long distances by blowing the pressurised air at reverse direction. If my bottle can do this without using any fire, why don't rockets just use air? How is the effect of combustion in thrust?
Rocket Thrust – How Fire Creates Thrust in Rockets
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It was already discussed in the comments that a water rocket needs to push "something" out. It is instructive to do the calculation in a little more detail to see where the "energy" goes. For this I will consider the relative share of energy going to the rocket and the "expelled matter" (gas, or water) as a function of the expelled mass. To simplify things, we will assume that all matter is expelled as a single entity with a certain velocity; in reality you might need to integrate, but any inequality that holds for a small amount of expelled matter will hold for the integral over many such amounts.
I will use upper case symbols for quantities relating to the "rest of the" rocket (mass M, velocity V, momentum P - without the expelled mass) and lower case for the expelled matter(m, v, p). From conservation of momentum, $P = -p$ so $M\cdot V = - m\cdot v$. The energy of the rocket $E_r$ and expelled mass $E_m$ will be respectively:
$$E_{r} = \frac12 M V^2 = \frac{P^2}{2M}\\ E_m = \frac12 m v^2 = \frac{p^2}{2m} = \frac{P^2}{2m}$$
It follows that the ratio of (energy in rocket)/(energy in expelled matter) is
$$\frac{E_r}{E_m} = \frac{m}{M}$$
In other words - the lower the mass of the expelled matter, the greater the relative amount of energy it contains. In the limit of "no water", the little bit of air mass contains virtually all the energy.
The first method is correct. In the second you have assumed that the pressure at the nozzle is still $P$ despite the water exiting with some velocity. i.e. You have neglected the dynamic pressure.
You need to use Bernoulli's principle $$ P + \frac{\rho v^2}{2} + \rho h g = {\rm constant}$$
Your first method assumes that the top surface of the water is hardly moving (because it's surface area is much bigger than the nozzle area). Applying the same idea to the second method, then we can calculate the constant both in the water and immediately below the nozzle as $$P = P_A + \frac{\rho v^2}{2},$$ where $P_A$ is atmospheric pressure and we neglect the small $\rho h g$ term which increases pressure due to to the column of liquid above the nozzle on the LHS. If we further assume that $P \gg P_A$ then $P = \rho v^2/2$ and the rate of change of momentum of liquid from the nozzle is $$ F = \rho A v^2 = 2PA$$
Best Answer
What the rocket does is holds the propellant in a lower-pressure, easier to store form, with higher energy density. Basically you have a lot of that kinetic energy you'll need in the form of chemical energy. Then you can pipe that fuel to the combustion chamber and ignite it there. At that point all that chemical energy is released in the form of expanding gases and heat.
So, instead of having a very heavy container holding all that hot pressurized gas, you just have a smaller container holding your fuel.