How Much Power is in that Stream?Everyone that has a stream running through their property has thought of turning some of that flow into electrical energy. The idea of free power from a natural fast flowing stream is an enticing one, but is it practical? Well, let's see.
We have two things to figure out. One, just how much power is there in falling water. This is a simple "physics" problem. The other challenge is to estimate the amount of water we have in our stream. This is a tedious process of estimation and guessing. That's the way things usually work, the physics is simple and the field work is tedious.
We'll do the physics first.
The power we get from the stream results from water falling. The more water that falls, the more power we get, and the further the water falls, the more power we get. The equation is this:
P = (Q * H * E) / 11.8
where P is power in kilowatts Q is water flow rate in cubic feet per second H is the height the water falls in feet E is our conversion efficiency and 11.8 makes all the units work out
Here's an example.
Suppose we have an overall efficiency of .6, a working height of 10 feet and a flow of 1 cubic foot per second (cfs). Our total generated power would be:
P = ( 1 * 10 * .6 ) / 11.8 = .508 Kw = 508 watts
Nothing to write home about, but it is free!
I should note that the .6 efficiency is figured on a reaction turbine with .8 efficiency multiplied by .75 to account for gearing and alternator losses. Higher and lower values are possible, but .6 is a reasonable value.
The power we get is directly proportional to both flow rate and height, so doubling either one will double resulting power. If we had a 100 foot drop and 5 cfs of water flow our power output would be 25 kilowatts. Wow! A very respectable amount. But wait a minute, just how much water is 5 cubic feet per second?
There are 6.22 gallons to a cubic foot and 60 seconds to a minute so 5 cfs is 1866 gallons per minute. Is that a lot? It certainly is a big number. But is it a big stream?
Let's go look at a stream. To figure water flow rate we will need to measure how fast the water is flowing and what the cross sectional area of the stream is. The cross section in square feet multiplied by the stream flow rate in feet per second will give us the stream volume flow in cubic feet per second.
Streams are not usually simple box shapes, so we're going to have to measure, average, and estimate.
We need two points along the stream. We will drop a float at the upstream point and record the time it takes to float to the downstream point. The distance between the two points divided by the time is our stream speed.
At both the upstream and downstream points we will measure the width of the water surface and the depth of the water in several places across the width. The depths we average and multiply by the width to arrive at a (rough!) estimate of the cross sectional area of the stream at that point. Figure the area of the stream at both the upstream and downstream point, and average the two averages to get the stream cross section in square feet.
Stay with me, we're almost done.
Multiply the stream cross section by the stream speed to get total volume flow in cubic feet per second.
Whew! Let's do an example. We pick two points in the stream, one 30 feet downstream from the other. We drop a ping pong ball in the water at our upstream point, and exactly 10 seconds later it passes our point downstream. Dividing 30 by 10 we get a stream speed of 3 feet per second.
The upstream location we measure to be three feet wide, two feet deep in the middle, and six inches deep halfway from middle to each edge. These depths average to one foot, so the cross section is 3 square feet.
Measuring our downstream location we find that it is 2 feet wide and fully 2 feet deep from edge to edge (our make believe stream passes between some make believe boulders) so the area is 4 square feet.
We can now compute the total flow of our stream. The average stream cross section is (3+4)/2 = 3.5 sq. ft., and the stream speed is 3 ft/s so our total flow is 10.5 cfs.
If our stream drops just 5 feet in 30 then we have:
P = ( 10.5 * 5 * .6 ) / 11.8 = 2.67 Kilowatts
This is from a stream 2 - 3 feet wide that flows about as fast as a person walks and is sloped more gently than a standard metal roof (3 in 12). And sure enough, 5 cfs is not that much water for a stream. Our example is by most definitions a small stream, and it flows 10.5 cfs. This is more than 3900 gallons per minute.
There are a number of practical challenges to be solved in making water power work, but nearly any stream that makes noise will make usable power with no fuel cost.
This article first appeared in the November 11, 1999 issue of OffgridOnlie, our free weekly newsletter. |
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