This Part One discusses how to assess the viability of a site for hydroelectric power, and simultaneously provides a basic understanding of the physics and math behind hydroelectric power generation. Future articles are planned to address other aspects, such as turbine and generator design, but it may be a while. This here is the fun part.
The first thing to understand is that virtually all of the electricity generated in the world comes from using some physical process to create spin. If you can spin a turbine, that turbine can turn a generator, and that creates electricity. Spin the turbine against greater resistance, and you get more electricity. Hydroelectric power is about using water to spin the turbine. Pretty easy; just drop the water through the turbine. If you have a lot of water or if the water is under pressure – for example, if it has a lot of water above it pushing downward – then it will spin the turbine against greater resistance and generate more power. So hydroelectric power is fundamentally about harvesting the energy difference between water held at different elevations under the force of gravity. If you have a source of water at some elevation and you can control the water’s ability to flow to a lower elevation, then you can steal that energy and make it do work for you.
How much energy can you get? That depends on how much water you have and how much of an elevation change you can produce. Let’s walk through the equations:
- The equation we’re looking for defines the relationship between water, elevation, and energy production in familiar units we can use to understand the relationship. As Americans, we understand energy in terms of kilowatt-hours because that’s how our energy utility companies bill us for them, and we understand water in terms of gallons because that’s what we use when we cook, shower, and flush. So let’s try to build an equation that relates Watts to gallons.
- A kilowatt-hour (kWh) is a kilowatt (kW) (1,000 watts) of power exerted for one hour. Or half a kW (500 watts) exerted for two hours. Or two kW (2,000 watts) exerted for half an hour. So let’s say you have a one-kW microwave oven; if you run it for five minutes, you’ll use one-twelfth of a kilowatt-hour. A typical house uses about 900 kWh per month (cite).
- Your bathtub probably holds about 40 gallons of water during a typical bath. There are 7.48 gallons of water in a cubic foot of water. A keg of beer is a little over two cubic feet of beer. A gallon of water weighs about eight pounds near its boiling point, or about 8.3 pounds at room temperature.
- Since we all understand kilowatt-hours and gallons, the goal is to relate kilowatt-hours to gallons of water under force of gravity. So, step one, let’s break down the kilowatt-hour into its component units until we find something that makes sense in terms of water and gravity.
- A Watt is a unit of power (i.e., how hard you’re pushing), and a Watt-hour is a unit of energy (how hard you’re pushing for how long). The Joule is also a unit of energy, equal to Watt-second. There are 3,600 seconds in an hour, so there are 3,600 Joules in a Watt-hour. Then, because there are 1,000 Watt-hours in a kilowatt-hour, there are 3,600,000 Joules in a kilowatt-hour. (Keeping all this relatable: a Calorie is 4.187 Joules, and is the heat energy required to heat one gram of water one degree Celsius – so a kilowatt-hour would be enough energy to heat about 3 gallons of water from room temperature (20 C) to boiling (100 C) – do bear in mind, however, that the “calories” that appear on your food’s nutrition labels are actually kilocalories – each one is 1,000 of the SI unit, the Calorie… therefore in order to burn off a 250-calorie Krispy Kreme doughnut, you’ll need to do about 0.30 kilowatt-hours of work).
- A Joule is a Newton-meter, which is starting to get us somewhere because Newtons are a measure of force, and the action of gravity on water is a force. So the only piece of the puzzle still missing is: how many Newtons of force does gravity exert on a gallon of water?
- A Newton is a kilogram-meter per second per second. Gravity pulls on objects at Earth’s surface with a force equal to their mass times 9.8 meters per second per second. That means a kilogram of water weighs 9.8 Newtons… and because a kilogram weighs 2.2 pounds, that means a gallon of water at room temperature weighs about 37 Newtons.
- So then in order to get to Joules, we have to factor in meters. If we have a gallon of water with two meters of drop, that’s 74 Joules, which is… only about 0.00002 of a kilowatt-hour. So in order to get any meaningful power out of hydroelectric generation, it takes either a lot of water or a lot of height.
- Solving the equation we have set up, it turns out that:
Wh = 0.01028 * gallons of H2O * meters of drop
… or, since there are 7.48 gallons in a cubic foot of water (and it’s useful to know cubic feet since the flow rates of most bodies of water are measured in cubic feet per second):
Wh = 0.07688 * feet3 of H2O * meters of drop
… or, to calculate wattage expressed in terms of flow rate (gallons or cubic feet per minute):
W = 0.62 * (gallons of H2O / minute) * meters of drop
W = 4.6 * (feet3 of H2O / minute) * meters of drop
Now let’s play with those equations some…
First off, let’s say you have a license to dam the Yadkin River for purposes of hydroelectric power generation. Let’s say you dam the Yadkin, giving you the Yadkin’s entire flow rate of about 2,000 cubic feet per second to work with. And let’s say you build your dam to create a 60 foot elevation drop (about 19 meters). That means every second, you can generate about three kWh… which means that every hour, you can generate about eleven megawatt-hours. However, if you really wanted to maximize your profits, you’d probably build in about 40 MW worth of generating capacity, so that you could let the water level rise during non-peak times of day, and then generate like crazy during peak usage hours when energy is selling for a high price. How much money could you make? Well, if you could sell power to the grid for $0.05/kWh, you could make $4.8 million annually.
As another scenario, let’s do a basic analysis to see if it would be worthwhile to install a home hydro system. Say you’ve measured your flow rate and found that a stream on your property has a flow rate of 60 gallons per minute, and it undergoes a 20′ change in elevation on your property. Equation says this can produce somewhere around 67 Watts. Because this is a home setup, figure some losses to inefficiency, so maybe it will actually produce about 57 Watt-hours per hour. That’s about 41 kWh per month… which comes to about four bucks a month if you pay $0.10/kWh for energy. So you probably have better things to do than try to produce energy off that stream.
How much water or drop do you need to make it worthwhile? Well, I’d be pretty happy to have a system that could both erase my power bill and supply me with energy in the event of a major blackout. For such a system, I’d want to generate about 1000 kWh per month. There are about 720 hours in a month, so the system needs to generate about 1.4 kW. Applying an 85% efficiency factor, that means I need about 1.65 kW worth of water and drop. So, at a drop of 6 meters, I’d need a flow rate of 444 gallons per minute. Or, at a drop of 12 meters, I’d need a flow rate of 222 gallons per minute. Either way, that’s a pretty good-sized waterfall.
Measuring your flow rate
To measure your flow rate, grab a five-gallon plastic bucket, divert your stream’s water into it, and see how long it takes to fill. You could also measure the cross-sectional area of your stream and measure the water’s speed and then run those numbers, but that’s probably more work than the plastic bucket. You’re going to have to build some sort of diverter to get all the water into the hydro system anyway. Remember that flow rates can vary a lot with weather and time of year, so it may be advisable to take a few measurements.
Common Questions
Q: Do I need a drop?
A: Yes. Now, I know what your next question is: what about water wheels on old mills? True, those work. The thing is, to get any kind of meaningful electricity generation out of one, it would have to be huge.
Q: Can I keep my waterfall and also use the water and drop to generate electricity?
A: Not really, no. You have to channel the water from the higher elevation to your turbine, which sits at a lower elevation. It can’t just drop; it has to be channeled so that the water at the turbine is under pressure. Of course, you could just channel part of the water from the waterfall, and let the rest cascade over. That works.
References:
- energy.gov – http://energy.gov/energysaver/articles/planning-microhydropower-system
- Micro Hydro: A New Spin on Hydro Power – http://www.motherearthnews.com/renewable-energy/micro-hydro-momentum-zwfz1210zrob.aspx#axzz3NvK0vQTU
- Alcoa – https://www.alcoa.com/yadkin/en/info_page/high_rock.asp