# Vertical flow rate of water stream under gravitation

1. Jun 6, 2007

### Positronized

Hi,

I'm trying to get the rig for my physics project set up which requires some constant water flow through the system. Unfortunately it's requiring a bit more water inflow than the tap can provide. I measured the tap's flow rate to be somewhere around 0.1L/sec to 0.3L/sec

Now because I don't want to invest in a water pump I'm planning to create some quick and dirty device to provide a bit more flow. I'm thinking of a reservoir of water (probably just a bucket) with a hole in the bottom. Now if I raise the bucket to sufficient height above my expermental device and let the water flow through the hole vertically into my rig, I might be able to get enough flow rate.

But I have no idea if this is even practical. I meant I don't know how high I have to raise the bucket to and if it's going to even provide higher flow rate than the tap. So let's say if the hole at the bottom of the bucket is half an inch diameter, how can I calculate the flow rate of water when I raise the bucket to height h above the system? (Presumably the flow rate will change along the stream due to gravitation acceleration so I would appreciate the mathematical explanation for it too if possible) I meant I otherwise understand things in physics but fluid dynamics stuff just confuse me a bit.

Oh and also, is there any direct way of calculating the force/pressure/momentum of a flowing stream of water? Say if I have a water stream flowing, assuming completely uniformly with no surface friction or viscous drag, at a some rate Phi L/sec so I can quite easily convert that into some flow velocity v ms-1 given the surface it's flowing through using the flux integral. Now if some random object is placed at that surface and obstructs the flow, what force would the water flow exert on the object?

Thanks!

Last edited: Jun 6, 2007
2. Jun 6, 2007

### cesiumfrog

Think this through.

3. Jun 6, 2007

### Positronized

Well I did do some quick kinematic calculation on it but, while I can find out the velocity of falling "droplets" of water when it reaches the rig, the continuum of water flow stream just confuses me.

*EDIT: So can I just go: v of water = sqrt(2*h*g) at the device, but water is let flowing through a circular hole of diameter 1.5cm so assuming it reaches the rig through in circular flat surface area = (5.625*10^-5)Pi cm^2 hence flow rate = v*A m^3/s whatever that comes down to?

Last edited: Jun 6, 2007
4. Jun 6, 2007

### cesiumfrog

If the bucket looses 1L/min, whatever is below will certainly receive.. 1L/min. Otherwise, well, where would the water come from? If the liquid accelerates, expect the cross-sectional area to vary (and the steady stream will break up into droplets..).

Last edited: Jun 6, 2007
5. Jun 6, 2007

### Positronized

What if, say to prevent that, I fix a pipe from the hole right down to the rig below it? So now the flow rate would depend on the *depth* of the water (which now includes the pipe length too)? And so then what calculation would this turn into?

*And I editted the original post, inserted one more paragraph concerning something completely irrelevant to the topic but I thought I shouldn't post another thread so it'd be nice if someone could also address that too :tongue:

Last edited: Jun 6, 2007
6. Jun 6, 2007

### Danger

This might be over-simplifying things, but have you considered just running hoses from a couple of different taps and combining the flows?

7. Jun 6, 2007

### ank_gl

cheh.. apply bernoulli equations and you ll get the solution to your numerical part. and you can easily calculate the momentum of water jet(by simple newtonian mechanics).
but i have got to say, the method DANGER suggested is far more easier and i guess efficient to. you wont like to run with a bucket in b/w your project presentation

8. Jun 6, 2007

### vanesch

Staff Emeritus
In fact, if you add a pipe to the bucket, you will be able to LOWER the pressure at the hole in the bucket, hence increasing the flow. Of course, once the pipe is about 10 meters long, this won't help anymore as you'd get cavitation.