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First of all, what is the correct equation to determine the thrust produced by a water rocket?

$$ F_T = \dot{m} V_e \tag{1} $$

$$ F_T = \dot{m} V_e + (p_e - p_0)A_e \tag{2} $$

Using the first equation it should be possible to increase the thrust by using a convergent nozzle because the water's exhaust velocity $V_e$ increases while the mass flow rate $ṁ$ out of the rocket is constant.

The second equation additionally incorporates the cross sectional area of the exhaust as well as the difference between the exhaust and ambient pressure.

Given the mass flow rate out of the rocket is unchanged, can a nozzle change the thrust of a water rocket?

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    $\begingroup$ What are you keeping constant? If you make the nozzle smaller and keep mass flow the same you must increase the pressure. If you keep pressure the same, mass flow must drop; which is it? $\endgroup$ – Floris Aug 3 '15 at 16:27
  • $\begingroup$ @Floris: I assume to keep the pressure the same. When mass flow must drop when the cross sectional area gets smaller, a nozzle cannot change the thrust of a water rocket. A converging nozzle will increase the exhaust velocity but decrease the mass flow rate while thrust remains the same, right? $\endgroup$ – Chris Aug 5 '15 at 12:59
  • $\begingroup$ @Chris I don't know weather i understand your question. If you are able to maintain pressure in pressure chamber that will not reduce mass flow rate even if you increase area ratio. Please refer this link nptel.ac.in/courses/112104118/lecture-40/… $\endgroup$ – Arun Govind Neelan A Aug 5 '15 at 14:12
  • $\begingroup$ If you have a closed water bottle and poke a small hole in it and squeeze it, the water jet that will come out of the hole will generate very little thrust. This is because the water that goes through the small hole experiences a lot more friction compared to a larger hole. So there is an optimum of the nozzle area which gives the most thrust. $\endgroup$ – fibonatic Aug 5 '15 at 14:56
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prerequisite

Thrust produced by a nozzle can be given by

$$F_T = \dot{m} V_e + (p_e - p_0)A_e$$
Thrust component in a nozzle can be split into two component that is pressure thrust ($(p_e - p_0)A_e$) and momentum thrust ($\dot{m} V_e$). In most of the nozzles we try to achieve exit pressure equal to ambient pressure, this phenomena is called fully expanded nozzle. But some time, this may not be possible in converging-diverging nozzle, if that is not operating in designed mach number so pressure thrust will be added or subtracted from momentum thrust. Momentum thrust is higher than pressure thrust for a given pressure, so we always try to achieve thrust by momentum thrust. Pressure thrust is because of design problem or limitations.

Kind of thrust in Converging nozzle and Converging -Diverging nozzle: Converging nozzle always try to achieve ambient pressure at exit after reaching equilibrium state. Time to reach this equilibrium state is negligible for most of the calculations. Because pressure waves travel at the speed of sound, so time taken by this pressure waves to tell the exit pressure information to the pressure chamber is very less because length of the nozzle is very small compared to speed of sound. In C-D nozzle exit pressure not necessarily equal to atmospheric pressure because characteristic waves do not tell that information to upstream of flow due to hyperbolic nature of the flow, its completely depends on area ratio after reaching sonic velocity.

Answer:

Most of the water rockets up-to my knowledge used converging nozzle, since converging nozzle expands to ambient pressure (explained earlier) there is no pressure thrust . You can use any equation 1 or 2, both of them will give same result for converging nozzle. In-fact equation 1 is sub-case of equation 2!

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  • $\begingroup$ I am not sure how applicable this is for water rockets, since water is much harder to compress then the gasses used in rocket engines. But this can be used aw an argument why a water rocket should have only an converging nozzle. $\endgroup$ – fibonatic Aug 5 '15 at 14:36
  • $\begingroup$ @fibonatic,Water rocket is not made up of water alone, it has compressed gas to push water. Please see this link en.wikipedia.org/wiki/Water_rocket $\endgroup$ – Arun Govind Neelan A Aug 5 '15 at 15:02
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    $\begingroup$ But this only passes through the nozzle after most/all water has been used up. Also the air will also offer only a very small amount of impulse, compared to the water (assuming a near optimal amount of water is used) because it has much lower density. A C-D nozzle might become interesting if you start using foam. $\endgroup$ – fibonatic Aug 5 '15 at 15:04
  • $\begingroup$ @fibonatic Ya that air increases pressure energy of water, compressibility comes in to picture, because Mach =1 in water is difficult to achieve, actually not at all possible! $\endgroup$ – Arun Govind Neelan A Aug 5 '15 at 15:06
  • $\begingroup$ @fibonatic Compressibility, $\beta =-\frac{1}{V}\frac{dV}{dp}$, this is about 5.1×10−10 Pa−1 for water, since change in volume of water for a given pressure is low, this is very low value. If we put a 10 kg piston on a cylinder contains water or air, both experience same pressure but different compression ratio. If we want to solve that problem mathematically one of boundary conditions is $p_{air}$=$p_{water}$ at interface, because nature doesn't allow discontinuity for this case. If you are still not convinced please post it, I will try if I know. $\endgroup$ – Arun Govind Neelan A Aug 5 '15 at 23:31
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No, water moving out of a bottle rocket is not nearly fast enough to generate the energy sapping shock waves and mach which the converging-diverging nozzle seeks to eliminate.

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