Hey there!
Right now, we are in the beginning of development and have to test the basic concepts first, so take this following stuff with a grain of salt. As for the mechanics, this means things like tank and valve fundamentals. Because we want to have the capability to control the thrust of the rocket, we need a solenoid valve (a controlled magnetic valve) controlled by a microcontroller.
Now, the question is how the tank-valve-etc layout should look like.
The basic idea of a water rocket is one big tank, with circa one third of the volume being water and two thirds pressurized air. To lift off, the valve at the bottom of the tank opens and so the pressure drives the water out through the valve and nozzle. In the following, we’re going to see if this concept is feasible or not.
Thrust and Valves
Thrust: The math
As covered in the Water Rocket 101 article, the thrust of a water rocket is dependent on how much and how fast water can be propelled out of the rocket – or more specifically the change of mass per time of the propellant and its exit velocity. It can be written as the change of impulse per time, or like the following if the exit velocity stays constant:
![\[ F_{thrust} \:=\: \frac{dp_{propellant, exit.}}{dt} \:=\: v_{exit} \cdot \frac{dm_{prop.}}{dt} \:=\: v_{exit} \cdot \rho_{prop} \cdot \frac{dV_{prop.}}{dt} \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-7e3335707d79334a430a1e88ec77fa7a_l3.png "Rendered by QuickLaTeX.com")
Because the density of the propellant stays the same, we can use the change of volume instead. The change of volume per time is called flowrate Q:
![\[ Q_{prop} \:=\: \frac{dV_{prop}}{dt} \:=\: \left[ \frac{m^3}{s} \right] \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-6987cc1b87b4499d2172cf715326e104_l3.png "Rendered by QuickLaTeX.com")
So the equation for the thrust looks like this:
![\[ F_{thrust} \:=\: v_{exit} \cdot \rho_{prop} \cdot Q_{prop} \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-d0563046e4b6f4f5fdf580bdc25656ad_l3.png "Rendered by QuickLaTeX.com")
Valves 101
For incompressible fluids, the flowrate Q of a valve is determined by the flow coefficient Kv, the relative pressure difference and the density of the fluid. This flow coefficient Kv is inherent to each valve and is specified in its datasheet.
For water, the equation looks like this:
![\[ Q \: := \: K_V \cdot \sqrt{\Delta p}} \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-246ff06789c7b38031a90a54c7cfdf37_l3.png "Rendered by QuickLaTeX.com")
Most solenoid valves – which are small and light enough to possibly use for a water rocket – have Kv values of around 1 to 1,5 liter/minute. This is relatively little, but small solenoid valves are constructed that way to allow for higher operating pressures. Our specific valve model has a Kv value of 1,7 l/min. This equates to a flowrate of 4,8l/min at 8bar, and 5,4l/min at 10bar.
Determening the thrust
But there is another factor determining the thrust: the exit-velocity v of the water.
![\[ Q \:=\: \frac{dV}{dt} \:=\: A \cdot \frac{ds}{dt} \:=\: A \cdot v \quad \Rightarrow \:\: v \:=\: \frac{Q}{A} \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-9795fb626cfeff7c56cbe1fdd6d3bf2c_l3.png "Rendered by QuickLaTeX.com")
The velocity of the water can be expressed by the flowrate and the cross-sectional area of the valve. If we now substitute the equation for Q in the thrust equation from before, we get:
![\[ F_{thrust} \:=\: \rho_{prop} \cdot \frac{Q^2}{A_{nozzle}} \:=\: \rho_{prop} \cdot \frac{K_V^2}{A_{nozzle}} \cdot \Delta p \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-545adc1de0b8eb78fb6e9d48011b7acd_l3.png "Rendered by QuickLaTeX.com")
As a result, the thrust of the rocket is dependent on the inverse of the nominal area of the nozzle, the square of the flow coefficient and the relative pressure difference (and the density of the propellant).
The thrust problem
However, here lies the problem. As you can imagine, a flow of about 83 milliliter per minute is not enough – certainly not for a rocket. When calculated, our valve (and to an extend all small solenoids) produces only about 0,253 Newton per bar of tank pressure.
![\[ F_{thrust} \: =\: 0,253N \cdot \Delta p \]](https://poseidon.tcrobotics.eu/wp-content/ql-cache/quicklatex.com-c50f04000622eabf4cc3cf6f2c31e68c_l3.png "Rendered by QuickLaTeX.com")
Our planned rocket could get quite heavy with all the electronics (LiPo etc.) and pneumatics (often made out of brass or steel) and so forth, and a thrust of 2 Newton at 8 bar wouldn’t be sufficient to lift the rocket.
Possible Solution
However, there might be a solution. As the flowrate for compressible fluids like air is much higher than for water, the valve could supply pressurized air to a tank filled with water, while this water tank has a much bigger nozzle opening.
If we split the fuselage into two tanks – using one for the water and the other one for the pressurized air – and use the main valve between the tanks to regulate the airflow into the water tank, we can control the exhaust without minimizing the thrust. The air from the top tank pressurizes the water tank below when the main valve opens.
To stop the thrust the main valve is closed and a secondary valve depressurizes the water tank. Because this depressurization prozess is quite slow and the water continues flowing for a while due to its inertia, the nozzle needs to be closed temporarily. A small seal or plug moved in front of the nozzle by a mini servo motor prevents the water from flowing out.
We are currently testing this concept with valves and two tanks. If you want to learn more about our protoype rockets, the testing and results, stay tuned for more updates coming soon!
Finn
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