Backyard Fluid Dynamics

Here’s a picture of the ornamental pond we have in our back yard, showing the fountain that we run to keep the water circulating so it’s not just a mosquito ranch:

You can see the brick that we have sitting on top of the pump housing to keep it submerged (it tends to tip over otherwise– the filter box is a little top-heavy), and also the spray of water shooting up into the air.

You can’t tell without some reference, but the pond level is actually a little lower than I like it at the moment. The lining leaks slowly, and every so often I need to put more water in. How do I judge the level, you ask? By the height of the fountain. As the level drops, the water shoots up higher. Here’s the pond after I re-filled it this morning:

These were taken using a tripod, and cropped to the same size and position in GIMP, so as to enable an accurate comparison. So, what physics can we get from this?

Well, from looking at a picture with the pump off, we can estimate the scale:

Using a tape measure, I get the height of the brick above the water to be about half an inch (it’s an American tape measure– that’s 1.27 cm in furrin units). Looking at the picture, the exposed brick extends about 10 pixels vertically, which gives us a scale of 20 pixels to the inch.

Counting pixels in GIMP, the height of the before-filling water spray is about 210 pixels, or 10.5 inches (27 cm). We can then ask what the speed of the water leaving the surface is to reach that height. If we assume that the water is launched straight up (not a great assumption here, but no worse than the estimate of the vertical scale), and use a little introductory mechanics, we find that the speed of launch in terms of the height h is:

where g=9.8m/s² is the acceleration due to gravity. Using 27cm as the height above the surface of the water, we get a speed of 2.3 m/s, about the speed of a brisk walk.

What about the after picture? Using the same vertical scaling, we get the speed to be 1.9 m/s, a reduction of about 17%.

How do we explain this? Well, the most naive model would be to attribute the change in height to the fact that the pump needs to move a larger amount of water in the “after” picture than in the “before” picture– there’s now all this extra water sitting on top of the opening that has to be moved. So how much extra water do we need to cause this effect?

Well, we can think about this in terms of energy. The energy to lift a given mass of water m into the air has to come from work done by the pump. If we make the tremendously simpleminded assumption that the pump does the same amount of work in both cases, just acting on more mass in the “after” case, we can write down two related equations for the energy involved in the two cases:

The first equation relates the “before” height h₁ reached by a mass m of water to the work done by the pump; the second adds an extra mass m_extra and reaches a new height h₂. We don’t know what the work done is, but we also don’t care– we’re assuming this is the same in both cases. As a result, we can set the left-hand sides of these two equations equal to one another (since they’re both equal to the same right-hand side), and find the relationship between m_extra and the original mass m:

The measured heights are roughly 27 cm and 18 cm, making the right side of this a convenient 9/18 = 1/2 = 0.5. So, the extra mass is half the initial mass of water launched.

So, if we knew the initial mass of water launched into the air, we’d be golden. But, of course, we don’t know that, either. We can make another incredibly naive assumption, though, and say that the mass of water launched into the air is just the mass of the water that happens to be sitting on top of the pump opening, which is just the area of the nozzle multiplied by the height of the water. We don’t change the nozzle size in going from one case to the other, so it all comes down to the height of the water above the nozzle.

If we increase the mass moved by the pump by 50% (adding a m_extra that is half what we started with), then, we would expect the depth of the water above the pump to increase by 50%. And that’s something we can measure.

So how did we do? Well, the brick is a 2-inch brick, and half an inch was originally exposed, so we started with a depth of 1.5 inches above the nozzle of the pump. After filling the pond, the brick was covered by something less than half an inch of water (it’s really difficult to estimate. A 50% increase in depth would imply an additional 0.75 in of water, namely the half inch that was exposed, plus another quarter-inch. Which is right in the range of what we see.

That’s… shockingly good, given that the assumptions I made in the process were all pretty naive. I suspect the surprisingly good agreement may be wholly or partly accidental, with some of my silly simplifying assumptions canceling each other out. That, or I’m better at this physics stuff than I realized, but I think “accidental cancellation” is the way to bet.

(A real test would require me to set up a better distance calibration, and make better measurements of the heights involved. Also, I could let the pond drain down farther, so as to get more data points. But there is a limit to how much effort I’m willing to put into silly physics models for the blog.)

Anyway, there’s your latest peek into the world of a physics professor. If you look for it, you can find physics in all sorts of simple situations, even the splashing of a backyard fountain.

For those whose heads are hurting because of the math above, I’ll close with a sequence of pictures from before, during, and after the topping-off of the pond. Because splashing water is pretty cool even if you don’t think about the physics involved.

(I realize I’m kind of trespassing on Rhett’s territory, here, and I half expect him to show up with better image analysis tools and a better model. But that’s what blogs are for…)