both nozzle is 2cm diameter
one "rocket" has air pump which hold constant 5 bar
second "rocket" it has water inside and also air pump which hold cosntant 5 bar,(water column is only 30cm haight,so hydrostatic pressure is neglibile,dont take into account)
rocket diameter 25cm
How much is thrust in air and water case?(load cell measure thrust)
NOTE: FLUID IN TANK HAS SOME VELOCITY,SO PRESSURE IN TANK IS NOT 5BAR IN EVERY POINT!
One can calculate this from first principles. Consider a water packet with volume $V$ going across a pressure differential $p$. Kinetic energy of this particle is $$E=pV.$$ Water is incompressible so the mass of the packet $m=\rho V$. Since $E=\dfrac{1}{2}mv^2$ we can calculate $$v=\sqrt{\dfrac{2E}{m}}.$$ The momentum of our water packet $q=mv$. Filling in all the equations so far, we get $$q=V\sqrt{2p\rho}.$$ Thrust is momentum per unit of time, so we get $$F=Q\sqrt{2p\rho}$$ with Q the water flow.
This leads to the logical conclusion we can double the thrust by adding twice as many nozzles (but we'll run out of water twice as fast). For the air case, simply replace the density of water $\rho=1000\mathrm{kg/m^3}$ with that of air. For a given volumetric flow, thrust will be a lot less, as expected.
A simpler way to look at things may be a force balance. The internal pressure $p$ is balanced everywhere except at the nozzle. If the nozzle has a cross section $A$, we get $$F=pA.$$ This is independent of the fluid used, but looking at our previous equation we see that for a less dense fluid we will need a higher flow rate.
Note that all of the energy from your setup comes from your fictional ideal air pump, so while in your question your can just increase the pressure to get better thrust, the reality is that you will increase the energy requirements faster than your thrust will grow (because simply the linear vs quadratic relation for momentum and trust).
Addendum: I did a quick check to see what the exit velocity of the air should be and it's over two times the speed of sound at sea level. So unless you have a high tech convergent-divergent nozzle you're better off using water
Consider control volume around the whole apparatus, and apply the momentum balance using the steady-state Euler's equations in the integral form:
$$\vec{F}_{body} = \oint_S(\rho\vec{V} \cdot d\vec{S})\vec{V} + \oint_Spd\vec{S}$$
If the nozzle is aimed parallel to the ground, then it simplifies to the 1D thrust equation, noting that the exit pressure is equal to the ambient pressure ($p_{atm}$):
$$T = \rho V^2 A_e \tag{1}$$
where $T$ is thrust, $\rho$ is the density of the flow right before the nozzle exit, $V$ is its exit speed, and $A_e$ is the nozzle frontal area.
We are going to assume there is zero loss in this whole endeavour, so the flow is isentropic everywhere, which means we can apply Bernoulli's to derive the exit speed (noting that inside the tank, the fluid velocity is zero, assuming the nozzle is small compared to the total volume):
$$V=\sqrt{\frac{2(p-p_{atm})}{\rho}} \tag{2}$$
where $p$ is the pressure of the fluid inside the tank.
Now substitute (2) into (1), simplify, and we have 1:
$$T=2(p-p_{atm})A_e=Q\sqrt{2\rho(p-p_{atm})} \tag{3}$$
where $Q=VA_e$ is the volumetric flow rate.
So the thrust is independent of what kind of fluid you use! However, it should be cautioned that this does not mean that a compressed air rocket will receive the same $\Delta V$ as the water one when their tanks are emptied. More factors, such as empty weight and tank volume, must be taken into account. For instance, the water rocket will have much lower volumetric flow rate than the air one, giving it a longer "burn" time.
1 Note the conclusion reached is identical to @sanchises' answer, despite the slightly different approaches.
