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I Want to build a water gun that throws water 100 meter far under 1 second. is it possible? how can i build it? (please send the formulas if you can)

IF A water jet has 4mm nozzle and the pressure on the water is 55000 psi. I want to know the water coming out of the water jet will reach 100meter or not and how long will it take for the water to reach 100 meter? (use 10 g of mass if needed)

Thank you in advance for your answer.

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  • $\begingroup$ you do know that if you were able to attain anywhere near such a pressure, it would be extremely dangerous. $\endgroup$ Oct 11, 2019 at 5:23
  • $\begingroup$ It isn't just the pressure. 100 m/s is 224 mph. If you hit someone with water at that speed, you would kill them. But yes, a water cannon like that is possible. They were used in the 1800's for gold mining. They would wash away a hillside, which was much easier than digging it away. $\endgroup$
    – mmesser314
    Oct 11, 2019 at 5:30

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Throwing water 100 meters in 1 second means the water has to have a speed of 100m/s. So the question becomes "how much pressure do I need to achieve those speeds?"

What you're looking for is Bernoulli equation. They state that for any given incompressible flow (such as water out of a nozzle):

$$\frac{v^2}{2}+gz+\frac{p}{\rho}=C$$

In this equation $v$ is the velocity, $g$ is the acceleration of gravity (9.8m/s^2), $z$ is the altitude at that point in the flow, $p$ is the pressure and $\rho$ (the Greek letter "rho", despite looking confusingly like a "p") is the density of the fluid.

The neat thing about this equation is that the constant $C$ is the same for all points in the flow. So we can compare different points just by remembing that the left hand side of this equation never changes value.

In the whole system you describe, we never change altitude, so we'll ignore the $gz$ term.

While the water is inside your presurized tubes, the velocity can be very slow, as long as your pipes are big enough. So inside the tubes, the only term we have is $\frac p \rho$.

When the water is free, in the air, its pressure must be 0. That's basically the definition of a free stream. So the only term we have outside the tubes is $\frac {v^2}{2}$

This means that $\frac{v_{free}^2}{2} = \frac{p_{internal}}{\rho}$. The velocity of the free stream in the air, $v_{free}$, squared and divided by 2 is equal to the pressure inside the water gun, $p_{internal}$ divided by the density of water. With a little rearranging we get $p_{internal} = \frac{\rho}{2}\cdot v_{free}^2$

Since this is a water gun, $\rho = 1000 \frac{kg}{m^3}$. we can plug this in $$p_{internal} = \frac{1000 \frac{kg}{m^3}}{2}\cdot (100 \frac{m}{s})^2$$ $$p_{internal} = 5000000\frac{kg}{m\cdot s^2} = 5000000 Pa$$ $$p_{internal} \approx 725 psi$$

Thus, with the right nozzle design, you need roughly 725 psi.

Now the hard part is going to be keeping the stream intact. In a water-jet cutter, the water is only coherent for a very short distance. After that, the aerodynamic forces break up the water column into a bunch of droplets. This is very desirable for water cutters, which don't want to cut the table beneath the material being worked on. Its a bit of a problem for you.

We can make it stay coherent longer by making the stream bigger. However, this water gun could get brutal fast. A fire-hose is pressurized to between 150psi and 300psi, depending on whether the fire is on the ground or a high-rise building. You're talking about twice that pressure. This will not be a trivial thing.

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