Flowing Water

I relieze there are many variables and only an approximation is possible, but giving the following situation, what formulae are required and what are they:

I have a stream of water flowing through a nozel rotated in two axis (theta, and ???( up and down rotation) by motors. I need to calculate (based on the angle of launch) the approximate distance traveled by the stream assuming no wind. The pressure constant needs to be determined by trial and error(ie, it would fire a stream at a specified angle and i would give it the distance traveled.

Is this possible at all? and if so, are there also equations for calculating water pressure changes as the diameter of the pipe changes?

Thanks

Hello gwdash.

I know very little (read: nothing) about fluid dynamics etc etc but from your first paragraph it sounds like the water could be treated as a projectile? If you don't know the equations for these post back and I'll help some more.

Of course, it's entirely possible that falling water doesn't ressemble a projectile but then I'm sure another member will point out my mistakes...

I believe that is a valid assumption and how i was planing to approach the problem. What are these equations(relating launch velocity and angle) to distance traveled?




Thanks alot

The hotizontal and vertical motions are independent of each other so i guess the s u v a t straight line equations fit:



s = .5(u+v)/t
v = u + at
v[sup]2 = u² + 2as
s = ut + .5at²


horiz: a=0

s = ut
s = ut cos(@) were @ is inclination from horiz


vert: a =g=9.8
s = ut - .5gt²
s = ut *sin(@) - .5gt² were @ is as above


but these are for balistic shots/projectile motion, i think water flowing in air acts different because and tend to form a non parabolic path the faster it goes try using a garden hose to deminstrate this

Could you provide a drawing?

You could use Bernoulli's equation.

P + pgh + 1/2 pv^2 = constant

Where P is the external pressure, p is the density of the fluid, h is the height difference, and v is the velocity of the fluid. This is energy conservation for fluids. You could use this to determine the pressure. Keep in mind that this is for ideal fluids (no viscous losses). What exactly are you trying to do? You could always stick a pressure gauge where you want to measure the pressure. That is the best way.


Do you have a constant pressure system? In that case, I would just measure the velocity of the water coming out. You can do that by measuring the time it takes to fill a container of given volume, say a liter. The flow rate is the volume of the container that you filled divided by the time it took to fill it. The velocity is then given by the flow rate divided by the area of your outlet.

Q = Vcontainter/fill time (Volumetric flow rate)

v = Q/Aoutlet (speed of your fluid)


After doing all that I think you could use the general equations for the motion of projectiles, but I'm not sure.

Fonzdude has the correct equations.

Thanks for all the help, i will attempt with projectile formulas and the filling the liter technique. As for what i'm doing, i will attatch photographs once i finsh this programing.

In Fonduze's equations, what is "u"

and am i correct in assuming the horizontal distance is given by:

d_horiz= ut cos(@) where @=theta from horizontal

It's a robotic controlled squirt gun basicly

I forgot about your last question: How pressure changes with pipe diameter... For most fluids (like water) moving through a pipe you can use the Hagen-Poiseuille equation (see attachment).

I'm not sure how valid the projectile motion equations are with water. The best way is to check and see whether or not they are accurate enough.

By the way,


if all you want is the horizontal distance travelled you can use the following expression:

d = 2*V^2*sin(x)*cos(x)/g


where V is the total speed in the direction of x, x is the angle from the horizontal, and g is the acceleration due to gravity.

thanks alot, when i solved that for x, i got sqrt(cos^-1(sin^-1(9.8d/(2V²)))).

i know this is wrong, but where?

Well, that' s just not how you solve those types of equations. Maybe I misunderstood. Are you trying to choose the pressure and angle given a certain distance you want the water stream to go?

Anyway, if you want to compute the angle, x, made between the nozzle and the horizontal if you're given the distance and initial launch speed you'll need to convert the equation I gave before.

Use the following trig relationship to help you out:

sin(x) cos(x) = 1/2 * sin(2x)

Now you can solve directly for x.


x = 1/2 * arcsin(g*d/V^2)

thanks. i'm calculating the angle for lauch to go a distances in a certain range(determined by the avalible water presure which i can regulate manually). thanks, i will post when i impliment it all

by the way in my equations

a = acceleration (m/s^2)
v = final velocity (m/s)
u = initial velocity (m/s)
s = displacment (m)
t = time( seconds)

m = meters
s = seconds

if you use the set si units you get si units as a result, by the way if your from a country that uses feet or foot or miles or rods i THINK you just change the meters in the into foots.

why is that america in its infinte wisdom still use foot/pound/acre ie imperial system of mesuring things?

i agree, Metric SO much better, but is only taught in science :confused:

Ignoring wind resistance for a solid object is not a terrible approximation. For water spurting out of a nozzle, I suspect that there would be huge effects due to interaction between the air and the water.

a fluid flowing through air (as a projectile) i believe will act diferently than a solid, i think its vicosity (and temputure, atmospheric pressure all related to viscosity of a liqiud) its diameter (how thick it is) the fact that it is one beam and not a single object of mass being projected would play a significant role in its path, which i believe would not be parabolic in shape.


also when water (i dont know of anything else) creates a electric current when flowing through a hose bcause of the the dipole effect of the hydrogen-oxygen bonding


i could be wrong......please speculate

a computer-controlled squirt gun that shoots electric water. I hope to see these in Walmart while I'm young enough to duck and still stand back up :)

the electric current is not noticible
1. it is earthed
2. the resistance of human skin is to great even when wet


after thinking about it probably have no effect on trajectory what so ever