# BACK PAGE BASICS: Hydro Power Calculations

Intermediate

Hydropower is a combination of vertical drop (“head”) and flow. But how much power (W) and energy (kWh) can a particular hydro resource yield?

A basic formula, confirmed by experience, tells us that head (in feet) times flow (in gallons per minute; gpm) divided by a factor gives continuous power output (W). The factor can range from about 9 to 14, depending on the size and efficiency of the system, whether it’s battery-based or not, and other parameters. For conservative convenience, let’s use a factor of 12, which is perhaps best for a whole-house system estimate if we don’t have more information. (In reality, each resource will need a different factor, and the comparisons here are for demonstration only.)

Head (ft.) × Flow (gpm) ÷ 12 = Power (W)

It’s important to understand the relationship between watts and watt-hours. Watts are instantaneous power. Watt-hours are units of energy—what we use, generate, buy, and sell. Multiplying watts by length of time in hours gives us watt-hours. Dividing by 1,000 gives us kilowatt-hours (kWh). Since hydro systems usually run all the time, we can easily calculate daily energy generation—wattage times 24 hours.

Hydro turbine power × 24 ÷ 1,000 = kWh per day

Examples can help confirm the important hydro fact—it’s all about the resource. How much energy you want, the rating of the turbine, and any other factors are unimportant in the basic calculation of your site’s hydro potential!

A site with only 3 feet of head will need a lot of flow to generate significant energy. This is why you see this sort of site with a dam and nearly all of the water going through the turbine.

3 ft. × 2,000 gpm ÷ 12 = 500 W

× 24 hrs./day = 12,000 Wh/day

÷ 1,000 = 12 kWh/day

“High head” might be used to describe any site with more than 10 feet of head.

300 ft. × 400 gpm ÷ 12 = 10,000 W

× 24 hrs./day = 240,000 Wh/day

÷ 1,000 = 240 kWh/day

While stream heads and flows vary dramatically, finding a site with a head ranging from 40 to a few hundred feet and a flow of 20 to 100 gpm is fairly common.

120 ft. × 45 gpm ÷ 12 = 450 W

× 24 hrs./day = 10,800 Wh

÷ 1,000 = 10.8 kWh/day

Industrial Downspout

Even a large building in a rainy environment does not present a realistic hydro “collector,” so you can forget about your home’s roof. A typical hydro system has a collector area measured in square miles. An 80,000-square-foot flat roof on a warehouse in Seattle (which receives 38 inches of rain per year) will yield an average downspout flow of 3.6 gpm. Even if that warehouse roof is 60 feet tall, the production numbers are not impressive:

60 ft. × 3.6 gpm ÷ 12 = 18 W

× 24 hrs./day = 432 Wh/day

÷ 1,000 = 0.432 kWh/day

Dreaming about hydropower is fun. Really making hydropower happen must start with real measurements of the head and flow, and calculations of the power and energy available.

Thanks Ian,
The tough variable is the flow. My low head stream varies quite a lot over a year and from year to year. Still something that produces 24/7 means I can reduce my battery bank size.

Best,
Rob

Hi Rob, Variable flows can be an issue, but once you start using renewable energy, you realize that working with variability is part of the equation. Hydro is the least variable over the course of a day or week, but can be seasonably variable. Fortunately, it is often a good match with solar electricity, since in most places, it's wetter (more flow, less sun) in one season and drier (less flow, more sun) in the other. So hybrid hydro-PV systems can work quite well. Best, Ian
Hi Rob, Variable flows can be an issue, but once you start using renewable energy, you realize that working with variability is part of the equation. Hydro is the least variable over the course of a day or week, but can be seasonably variable. Fortunately, it is often a good match with solar electricity, since in most places, it's wetter (more flow, less sun) in one season and drier (less flow, more sun) in the other. So hybrid hydro-PV systems can work quite well. Best, Ian