*Matt Deady, Professor of Physics, Director of the Physics Program, Bard College*

How much hydropower can a stream produce? There are many limiting factors – including restrictions on water usage, seasonal variations, and efficiency of turbines – but let’s ignore all of those for a moment and calculate the absolute maximum power possible, as an upper bound estimate.

Hydropower is created from the velocity that the water gains as it flows downhill. In order to calculate the maximum potential power at a site, we translate the energy from water flowing downhill into kinetic energy using the following calculator:

### Calculating Power Output

Let’s take a look at what the calculator is doing. If a **mass (M)**, in this case water, falls a **distance (h)** under the influence of **gravity (g)**, the **Power (kW)** that it gains is given by the gravitational energy: ** kW = (M)(g)(h)**. We need to determine those three numbers to find out how much energy our falling water could generate.

The gravity force constant is **g**, which we know to be **9.8 m/s²**.

**h** is the height the water drops through, which we call head. You will need to determine this vertical drop for your site, measuring from where the water starts out to where it enters the turbine. We will use head measurements in meters, but if **h **is in feet, the conversion factor is **1 meter = 3.28 feet**.

The mass we are interested in is the mass of water that is flowing downhill. In particular, we want to know the mass per unit time that flows in the stream. This measurement is called the flow rate, which will tell you flow (**Q**) in **cubic meters per second**. If measurements are made in English units, the conversion factors are: **1 m³ = 1000 liters = 35.3 ft³ = 264 gallons**.

For example, if the Saw Kill has a flow rate of Q = 1.2 m³/s, and the drop from the lower dam is h = 2.7 m, that gives a maximum power there: P = (9.8m/s²)x(1.2m³/s)x(2.7m) = 31.6 kW. But letting the water flow all the way down to where the Sawkill enters the Hudson would give h = 18 m, for a maximum power output: P = (9.8m/s²) (1.2m³/s) (18m) = 212 kW.

Note, this is the maximum power that you could derive from the site. It is likely that you will not be able to divert all of the stream’s water into your micro-hydro system, and there will be various inefficiencies in the micro-hydro system that yield a smaller power result than the simple formula above gives you. The point of this treatment is just to give an estimate of the potential for micro-hydro at your site.