0
$\begingroup$

I need to calculate the range that a fluid (water in this case) will travel after exiting a nozzle. The water will be sprayed through the air from a height of 2 m at a horizontal angle. The exit velocity is known to be 15 m/s. The nozzle will spray a full cone type spray, with a cone angle of 15°. The range is defined as the distance between the nozzle and the point where the water hits the ground (the furthest water droplet).

I have tried to use Bernoulli's equation on the uppermost streamline however only the final velocity can be solved for. I'm not sure how to get the range. I also realize that the air resistance on the fluid must be taken into account but I don't know how.

$\endgroup$
1
  • 1
    $\begingroup$ If you are building something, you might try measuring it. That is sometimes easier than calculating. $\endgroup$
    – mmesser314
    Commented May 15, 2021 at 15:48

2 Answers 2

0
$\begingroup$

A key parameter is missing, the diameter of droplets produced by the nozzle.

If you were firing out a water jet that hit the ground before breaking up then perhaps you could use Bernoulli as part of your solution, but once the jet has broken up it's not valid any more.

The larger the droplets produced by the nozzle, the more parabolic a trajectory they will take, and you could do a ballistic calculation for the droplet fired up at 15 degrees to get the maximum range.

However, if the droplets produced are quite small, they can hang in the air for a long time, carried by local air currents. The droplets will also be evaporating as they move through the air and so only a certain fraction of the water fired from the nozzle may reach the floor, the rest will be carried away.

I'd think that the easiest way to find out would be to build the set up and try it out.

$\endgroup$
3
  • $\begingroup$ Thank you! I was able to find an estimate of the droplet size so I have been to do a ballistic calculation. $\endgroup$
    – DrM
    Commented May 20, 2021 at 13:14
  • $\begingroup$ No problem. If you're happy with my answer could you mark it as accepted? $\endgroup$
    – Nick
    Commented May 20, 2021 at 15:18
  • $\begingroup$ Yes! Sorry still learning how all this works $\endgroup$
    – DrM
    Commented May 24, 2021 at 7:06
0
$\begingroup$

I studied a similar problem during my PhD. You can estimate the range using equations I developed in my dissertation, though they can get messy. Unfortunately, the dissertation is presently embargoed due to some chapters being published in a journal, but the trajectory chapter is not bound by such rules and I have uploaded it here.

The theory in my dissertation considers the breakup of the jet and the droplet size. It does not consider evaporation, which I don't believe to be a significant factor in most instances. The theory requires the firing angle to be less than about 40 degrees or so, which is fine in most instances.

The equation for range (equation 7.75, p. 193) in my dissertation follows:

$$R = d_0 \left[\frac{\langle x \rangle_\text{b}}{d_0} \cos \theta_0 - \frac{a}{b C_\text{d}^*} - \frac{1}{C_\text{d}^*} \mathrm{W}_{-1}\left(-\frac{\exp(-a/b)}{b}\right)\right],$$ where $R$ is the range, $d_0$ is the nozzle diameter, $\langle x \rangle_\text{b}$ is the jet breakup length, and $\mathrm{W}_{-1}$ is the -1 branch of the Lambert W function (available in Python, Matlab, etc., though not Excel). The remaining terms are more complicated: $$C_\text{d}^* = \frac{3}{2} \frac{C_\text{d}}{\rho_\text{l} / \rho_\text{g}} \frac{(1 - \alpha)^2}{D_\text{max} / d_0},$$ $$a = 1 + {(C_\text{d}^* \cos \theta_0)}^2 \left[\left(\frac{\langle x \rangle_\text{b}}{d_0} \sin \theta_0 + \frac{h_0}{d_0}\right) \mathrm{Fr}_0 - \frac{1}{2} {\left(\frac{\langle x \rangle_\text{b}}{d_0}\right)}^2\right],$$ $$b = 1 + C_\text{d}^* \cos \theta_0 \left(\mathrm{Fr}_0 \sin \theta_0 - \frac{\langle x \rangle_\text{b}}{d_0}\right),$$ where $\mathrm{Fr}_0 = \overline{U}_0 / (g d_0)$ is the Froude number, $\overline{U}_0$ is the jet velocity at the nozzle exit, $h_0$ is the firing height and $\theta_0$ is the firing angle.

If you want to avoid the Lambert W function, it's not hard to implement a numerical simulation along the lines of the model described in my dissertation.

For the particular case described, I'd expect the breakup length to be very short, and it might be reasonably approximated as zero. You'll need values of $D_\text{max}/d_0$ and $\alpha$. These numbers will change from case to case. I used $D_\text{max}/d_0 = 0.7$ (p. 198 says 0.8, which is in error though not by much) and $\alpha = 0.05$ for fire hose jets. Those numbers are not likely useful for the particular case described by DrM. I'd expect more air entrainment (higher $\alpha$) and smaller droplet sizes in that case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.