This Dam Acts Like a Water Cannon. Let’s Do Some Physics!

0
7

When people build dams—giant walls that hold back entire lakes and rivers—they have to build an overflow channel called a spillway, a mitigation against flooding.

A spillway could be something as simple as a path for water to flow over the top of the dam, or more complicated, like a side channel. Sometimes, there is just a big hole at the bottom of the dam (on the dry side) so that water can just shoot out like a massive water cannon. This is how it works at the Funil Hydropower Plant in Brazil. There’s a nice video showing the water coming out—it looks like a river in the air, because it basically is a river in the air.

But the really cool physics of this spillway is that the speed of the water coming out of the hole mostly just depends on the depth of the water behind the dam. Once the water leaves the tube, it essentially acts like a ball thrown at that same speed. Yes, you know what I’m going to do: I’m going to use the trajectory of the water leaving the spillway to estimate the depth of the water in the reservoir.

There’s actually a name for the relationship between water flow and depth—it’s called Torricelli’s law. Imagine you have a bucket full of water and you poke a hole in the side near the bottom. We can use physics to find the speed of the water as it flows out.

Let’s start by considering the change in water level during a very short time interval as the water drains. Here is a diagram:

Illustration: Rhett Allain

Looking at the top of the bucket, the water level drops—even if just a little bit. It doesn’t really matter how much the water level decreases; what we’re interested in is the mass of this water, which I label as dm. In physics, we use “d” to represent a differential amount of stuff, so this could just be a tiny amount of water. This decrease in water level at the top means that the water has to go somewhere. In this case, it’s leaving through the hole. The mass of the exiting water must also be dm. (You have to keep track of all the water.)

Now let’s think of this from an energy perspective. The water is a closed system, so the total energy must be constant. There are two kinds of energy to think about in this case. First, there is gravitational potential energy (Ug = mgy). This is the energy associated with the height of an object above the surface of the Earth, and it depends on the height, the mass, and the gravitational field (g = 9.8 N/kg). The second type of energy is kinetic energy (K = (1/2)mv2). This is an energy that depends on the mass and the speed (v) of an object.

Since the total energy must be constant, the change in kinetic energy plus the change in gravitational potential energy must be equal to zero. The water at the top of the bucket (which I will call position 1) is stationary, and the water at the bottom (position 2) has some exit velocity, v. Putting this together, I get the following:

Illustration: Rhett Allain

Notice that this is just the magnitude of the velocity. It actually doesn’t matter if this hole points straight down from the bottom of the bucket or horizontally out through its side—the water’s speed will be the same. But let’s say that the hole is on the side, so that the water shoots out parallel to the ground. If the distance from the hole to the ground is y0, how far from the hole will the water stream land when it hits the ground?

Even though this would be a stream of water, we can treat each molecule like an individual particle with just the downward-pulling gravitational force acting on it. Yes, this is your classic projectile motion problem.

The key idea here is that we can separate the horizontal and vertical motion into two separate problems. This means that in the vertical direction, the shooting water is the same as if it were just a single drop falling straight down, with an initial vertical velocity of 0 meters per second (m/s) since the water was shot horizontally. We can use this vertical motion to determine the time it takes to fall down to the ground. For the horizontal direction, it’s just a drop of water moving with a constant velocity—the same velocity that it was shot from the hole. Using the time from the vertical motion, I can calculate how far the water travels.

Illustration: Rhett Allain

Here, x is the distance along the ground from where the water came out of the container. I can use my expression for the velocity of the water, and I get the following relationship between the height of the water in the container and the distance the water shoots.

Illustration: Rhett Allain

OK, this is kind of awesome. First, notice that the gravitational field (g) isn’t in this expression—it cancels. That means that if you did this same water-leaking experiment on Mars, you would be the first human astronaut to make it to another planet. Oh, also the water would go the same distance as it does on Earth, even though the gravitational field on Mars is lower. (That’s assuming the water doesn’t freeze first.)

Another cool thing is that the distance the water travels changes with the amount of water in the container—but it’s not a linear relationship. (Remember y0 is a constant distance from the ground to the hole.)

Now it’s time to just try this thing out. I’m going to get a vertical tube and add some water and then let it shoot out the side. This is what it looks like:

Photograph: Rhett Allain

Since I was thinking ahead, I put a ruler right there. Now I can measure both the height of the water in the tube and the distance the ejected water travels. Then I can see if that equation actually works. Here’s the data I get from that image:

  • Height of water in tube (h = 0.477 meters).
  • Height of water hole above the ground (y0 = 0.334 m).
  • Horizontal distance water travels (x = 0.421 m).

Using the values for h and y0 I get a theoretical distance of 0.798 meters. This is clearly not the same value as the measured distance of 0.421 meters. But don’t worry—Torricelli’s law actually deals with this discrepancy. The actual water velocity leaving an opening is going to be the theoretical velocity (the one I calculated) multiplied by a “coefficient of discharge” (μ). This coefficient takes into account the properties of the hole that slow down the flow of water. A nice circular hole would have a coefficient close to 1.0, but an outlet that has a bent shape will slow down the exiting water. Using my measured horizontal distance, I get an actual water velocity of 1.61 m/s and a theoretical velocity (based on the water height) of 3.06 m/s. This gives a coefficient value of 0.53. OK, that’s fine.

Now, what about the massive amount of water shooting out of the spillway at the Funil Hydropower Plant? This water doesn’t come out horizontally, but with a rough measurement I get a “launch” angle of about 27 degrees above the horizontal. It’s difficult to get an exact measurement of how far it travels, but I feel like it’s maybe 50 feet horizontally, or around 15 meters. With this distance and the angle, I can use the following equation for the range of a projectile:

Illustration: Rhett Allain

Using this and solving for v, I get a water speed value of 13.5 m/s. (That’s 30 miles per hour, in case you were curious.) If I assume a coefficient of discharge similar to my experiment (I’m going with 0.5), then the theoretical water speed would be 27 m/s. With this speed, I can now calculate the depth of the water behind the dam. This gives a water depth of 37 meters (121 feet).

Oh, look at that. Wikipedia lists the dam height at 39 meters. That’s pretty close to my calculated value. But why is it different? There are a couple of reasons. First, I just estimated the coefficient of discharge—it could be some other value. Second, the lake behind the dam most likely doesn’t go all the way to the top of the wall. And third, the spillway hole is probably slightly above the bottom of the lake floor. Either way, I’m pretty sure physics still works.


More Great WIRED Stories

LEAVE A REPLY

Please enter your comment!
Please enter your name here