Sometimes you come across crazy stuff on the internet. Just check out the Qattara Depression Project; the basic idea is to create a channel to let water from the Mediterranean Sea flow into the Qattara Depression, a giant low-lying area in Egypt. This would create an enormous artificial lake that would change the local climate. It’s a prime example of a massive geoengineering project.
The project never got underway, probably because it would be too expensive. But I don’t want to talk about the financial or environmental impacts of this project. I want to focus on another aspect of the plan: to use this lake as a method for generating electrical power. Here’s how it works. The water from the lake will evaporate and cause a continuous flow of new water from the sea. The moving water in the channel can then be used to turn a turbine and generate electricity. Although someone has already estimated the amount of energy, I want to do this myself. You know … for fun.
Let’s start with the important part—why would a lake lose water? Actually, you can try this yourself. Get some water and put it in a shallow pan. Leave the pan with water on your counter for a couple of days and then check it out. Maybe all the water will be gone, or maybe there will still be some left. But likely there will be less water in the pan than at the start. This is either due to evaporation or you have a cat that came by and drank it. So let’s assume that it was evaporation.
Water can exist in three phases: solid, liquid, or gas. During evaporation, some of the liquid water turns into gas phase water (we call this water vapor). But it takes energy to change from the liquid to the gas phase, and the average water particle energy is not enough to move into the vapor phase. However, that’s the average energy. Water particles in the liquid phase have a range of different energies. Some water particles are moving slow with low energy and some are moving faster. It’s these faster-moving water particles near the liquid’s surface that have enough energy to “escape” the liquid phase and evaporate. Of course, if the highest-energy water particles leave, what remains are the lower-energy particles. That means that the average energy (and thus the temperature) of the remaining water decreases. Evaporation is indeed a method to cool stuff.
But who cares about that? We only care about the rate that the water goes from the liquid to the vapor phase. The greater the rate that water evaporates, the more water would need to pour in from the sea. So, how do you calculate the evaporation rate? It turns out to be not so simple. Actually, this is the net evaporation rate—as water evaporates from the liquid to gas phase, some water in the gas phase condenses into liquid water. You have to consider this condensation as part of the big picture. That means that the overall evaporation rate will depend on the following:
- Temperature of the water
- Temperature of the air
- Humidity of the air
- Altitude of the water
- Wind speed of the air
- Ground temperature
It’s a tough problem. If you want to calculate the evaporation rate for a lake, you could use a model like the Penman or Shuttleworth equation—but that’s too complicated for me. Instead, I will use this estimation of the evaporation rate in Lake Nasser, in Egypt, which averages perhaps 7 millimeters per day.
Now that I have a value for the evaporation rate, I need some more stuff. I am going to assume that in a given second, the mass of water evaporating is going to be equal to the water mass entering the lake from the sea. The estimated evaporation rate tells me how much the water level drops—but not the volume (or mass) of water. To calculate the mass per second, I need the surface area of the lake. According to Wikipedia, the Qattara Depression has an area of 19,000 square kilometers.
With the lake area and the evaporation rate, I can get the volume of water that needs to be for each second of time. Let’s do it.
The first step is to convert the evaporation rate into units of meters per second (instead of millimeters per day). That’s just a plain unit conversion problem, but it would look like this:
Remember that’s just a unit conversion. In case you need a refresh on unit conversions, here’s something I wrote awhile ago. Now I just need to multiply the evaporation rate by the area of the lake (in square meters). This will give me the volume flow rate of water.
I now need to find the speed of the water coming from the sea. With a flow rate of 1539 cubic meters per second, that could be water flowing at 1539 m/s going through a square-meter cross section or water going at 1 m/s going through 1539 square meters. It might seem like the speed doesn’t matter, but it does. This means I will need to estimate the size of the channel (the cross-sectional area). Let’s say the channel is 100 meters wide and 5 meters deep. This would give a cross-sectional area of 500 m2 with a water speed of 3.08 m/s.
You can picture it now: Every second a giant block of water moves into the lake at a speed of 3.08 m/s. If I could stop all this water, I would get the kinetic energy from it to use in other electrical things. But of course you can’t stop all the water—that would be crazy. Also, the hydroelectric turbine won’t be 100 percent efficient. So, I will just approximate the total efficiency of this system at about 25 percent. The power output will then be the efficiency multiplied by the kinetic energy and divided by the time of 1 second (this is why it’s nice to use a 1 second time interval).
Oh, remember that kinetic energy is calculated as:
But wait! What about the mass of the water? I guess I forgot about that part. If I know the volume of water (I do) and the density of water (I do), I can calculate the mass. The density of water is around 1000 kilograms per cubic meter. That means the mass of this water would be 1.539 million kilograms. I think we are ready to calculate the power. Since I’m a huge fan of using Python as my calculator, I am just going to put my code here.
That puts the power at 1.8 Megawatts. That seems sort of low. A nuclear power plant can produce more than 500 MW, and I would have thought this would be in that range. OK, let’s try something. What if you use a smaller channel to let the water in? If I put the channel width at 10 meters, the water is going to have to move much faster. Since the kinetic energy is proportional to the velocity squared, this will give more power. Actually, it’s really easy to recalculate since I used python. If I just change the channel width, I get a power of 180 MW. That seems better.
Wait! One more homework question for you. Where does this energy really come from? How do you get energy that is essentially from evaporation? In the big picture of energy, this is ultimately a form of solar energy. The sun heats up the water in the lake so that it can keep evaporating. This water eventually rains back into the ocean so that it could run through the hydroelectric thingy. So, it’s essentially solar.
What if you approached this as a solar energy problem? If you can get about 1000 Watts per square meter from the sun, how much power could you get from the artificial lake? What kind of efficiency for a solar panel would give the same power calculation as above?
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