Energy From Raindrops

conlaw writes to share that according to Discovery.com scientists have found a way to extract energy from rain. A new technique could utilize piezoelectric principles of a special kind of plastic to generate power from falling water in rainstorms or even commercial air conditioners. "The method relies on a plastic called PVDF (for polyvinylidene difluoride), which is used in a range of products from pipes, films, and wire insulators to high-end paints for metal. PVDF has the unusual property of piezoelectricity, which means it can produce a charge when it's mechanically deformed."

Re:Let's think about this for a second...

[FroBugg]

We upgraded to hydroelectric dams, which provide a very significant amount of power both in the United States and worldwide. China's still working on the Three Gorges Dam, the biggest ever.

Unfortunately, the US is tapped out on hydroelectric. There really is nowhere for us to put in additional ones, and the ones we already have are often cited as concerns with regards to environmental impact and municipal water supplies.

Let's do the math on this one

[Ancient_Hacker]

Hmm, raindrops... Why didn't someone think of this before? For a good reason:

• Let's be generous and assume it's raining all the time.
• And it's a real gully-washer, say an inch an hour.
• And let's be really generous and assume this gadget captures 50% of the energy of falling water.
• And each raindrop is at maximum terminal velocity, about 10MPH.
• So that's about 700000 gallons of water per day per acre falling at 10MPH.
• Which works out about 6 million pounds per acre per day at 10MPH.
• Which is about 100 million foot-pounds per day per acre.
• But that's only 1157 foot-pounds per second, barely 2 horsepower.
• Roughly 750 watts at 50% efficiency.
• Or roughly 17 milliwatts per square foot.
• Or at ten cents a kwh, it's making almost 100 watt-hours a year, or almost a penny.

Some Back of The Envelope Calculations

[Grond]

A typical raindrop has a fall velocity of about 8 m/s [madsci.org]. Assuming a pretty healthy rainfall of 10cm (4 inches) we get 100 liters of water per square meter of land. 100 liters of water weighs 100kg, of course, and plugging that into the equation for kinetic energy gives us 6400 joules. Spread out over 2 hours, that's a whopping .89 watts per square meter.

All of that assumes 100% conversion efficiency and no losses due to standing water absorbing the impact of the drops. If the overall efficiency is, say, 50%, then you'd need something like 30 square meters to light a single compact fluorescent bulb. To generate a megawatt would require over 2 million square meters (over 500 acres).

Given that in most places it rains less often than the sun shines, this seems like an astonishingly inefficient way to generate electricity. There just isn't that much energy in rainfall.

Re:Let's think about this for a second...

[Aaron Isotton]

Annual rainfall where I live is around 1-3 metres (more slightly inland than on the coast). Let's say 2m as an average. Cumulous cloud (the kind that typically causes rain) forms at 2-16km. Picking a number somewhere in the middle, let's say 8km for the average distance rain drops fall. That means, every year, two cubic metres of rain fall 8km per square metre of ground. That's 2,000 litres, which means roughly 2,000 kg. The total energy in this is calculated as mgh, so: 2,000 x 9.8 x 8,000 = 156,800,000 J.

The calculation seems to be correct but the concepts don't hold.

The *potential* energy of the rain can indeed be calculated using m*g*h as you said. The piezoelectric panels convert the rain's *kinetic* energy to electricity. The kinetic energy on impact is *not* equal to the potential energy, because most of it is lost to air friction.

As others pointed out, the speed of a rain drop is around 8 m/s. This means that the kinetic energy of your 2t of water is E = mv^2 = 2000 * 64 = 128,000J. You're three orders of magnitude off.

Re:Let's think about this for a second...

[ultranova]

Annual rainfall where I live is around 1-3 metres (more slightly inland than on the coast). Let's say 2m as an average. Cumulous cloud (the kind that typically causes rain) forms at 2-16km. Picking a number somewhere in the middle, let's say 8km for the average distance rain drops fall. That means, every year, two cubic metres of rain fall 8km per square metre of ground. That's 2,000 litres, which means roughly 2,000 kg. The total energy in this is calculated as mgh, so: 2,000 x 9.8 x 8,000 = 156,800,000 J.

Unfortunately, this is wrong. A raindrop doesn't keep on accelerating all of these 8 kilometers; it will reach it's terminal velocity, at which point the deceleration due to air resistance exactly cancels the acceleration due to gravity. Since raindrops are small, their surface area is large compared to their mass, so I'd imagine the terminal velocity to be rather small - which is a good thing, otherwise we'd get our skulls crushed to powder by rain, but sadly means that we can't extract all that much power from a single raindrop.

Actually, I checked, and according to WonderQuest [wonderquest.com], the average speed of a raindrop is between 2 (for small ones) to 9 (for large ones) meters per second. Since kinetic energy is mv^2, this works out to between 2000kg * 2m/s * 2m/s = 8000J (= 0.002 kWh) and 2000kg * 9m/s * 9m/s = 162 000J (= 0.045 kWh) per square meter per year.

Since the price of electricity is about 0.07 euros per kWh where I live, and a square meter of this thing would need about 22 years to produce a single kWh under optimal conditions and assuming a 100% efficient conversion, I don't think that it is a good investment.

Re:Let's think about this for a second...

[ultranova]

Hups ! That should be 0.5*mv^2. So the above figures are twice as good as they should be - it's going to take 44 years to produce a single kWh per square meter :(.

Re:meh

[MadnessASAP]

I'd be willing to bet that a raindrop reaches terminal velocity in a very short distance making the difference in height between a roof top and a cloud irreleveant.

Re:meh

[FailedTheTuringTest]

Terminal velocity for raindrops is around 9 m/s [hypertextbook.com] (slower for smaller drops, like drizzle). Acceleration is 9.8 m/s/s. So big raindrops reach terminal velocity in 9/9.8 = 0.9 seconds, during which time they fall 0.5*a*t*t = 0.5*9.8*0.9*0.9 = 4 metres = 13 feet.