# Converting potential energy to electricity **URL:** \ **Category:** Factual Questions\ **Created:** [July 23, 2021, 1:59pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140 "2021-07-23T13:59:46Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [July 23, 2021, 1:59pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/1 "2021-07-23T13:59:46Z")
This is not homework, I am just curious and don’t know the math. If I had a tower 100 feet tall how much weight would I have to raise to the top of the tower to produce 1 KW of electricity? Just say 50%.efficiency. ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [July 23, 2021, 2:08pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/2 "2021-07-23T14:08:15Z")
A weight at the top of a tower has a certain quantity of potential energy. 1 kilowatt is a measure of power, which is energy delivered per unity of time. So any weight, if allowed (braked) to fall at a high enough constant speed, could deliver 1 kilowatt of mechanical energy; a small weight falling very fast would do it, but so would a heavy weight falling slowly. I think you’re going to need to restate the scenario a bit. Maybe you meant 1 kilowatt-hour (a unit of energy, e.g. a 1-kilowatt system operated for one hour)? ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [July 23, 2021, 2:13pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/3 "2021-07-23T14:13:11Z")
Yes, I meant 1 KWH!, I am thinking about 20,000# but that doesn’t seem like enough for some reason. ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [July 23, 2021, 2:58pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/4 "2021-07-23T14:58:51Z")
OK, 1 kilowatt-hour = 2,655,000 foot-lbf. Divide by 100 feet, and your weight then needs to be 26,550 pounds. That assumes 100% efficiency in converting gravitational potential energy into electrical energy. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 23, 2021, 2:58pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/5 "2021-07-23T14:58:54Z")
I cannot convert your weird units in my head, but there is a very easy approximation you can use: let _g_ = 9.8 m/s2, the “standard” gravitational acceleration near the surface of the earth. Then the difference in potential after lifting it to a height _h_ is about _mgh_, where _m_ is the mass of your brick. ---
**Author:** ![Pleonast](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/pleonast/32/1183_2.png) [@Pleonast](https://boards.straightdope.com/u/Pleonast)\ **Post date:** [July 23, 2021, 2:58pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/6 "2021-07-23T14:58:57Z")
Ground-level gravitational potential energy E in J is E=mgh with m mass in kg, g as 9.81 m/s^2, and h height in m. Your 1 kWh is 3.6 MJ. Some math yields 12 Mg as the required mass at 30 m (ignoring efficiency, so divide by 50% if you like). That’s about 12 tons at 100 ft. If you were to use water, that’s 12 m^3 or about 3200 gal. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 23, 2021, 4:41pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/7 "2021-07-23T16:41:00Z")
And if you insist on using American units, note that pounds are a measure of weight (or other force), not mass. So if you want an answer in pounds, then you’re solving for mg, not m. ---
**Author:** ![am77494](https://avatars.discourse-cdn.com/v4/letter/a/d2c977/32.png) [@am77494](https://boards.straightdope.com/u/am77494)\ **Post date:** [July 23, 2021, 4:44pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/8 "2021-07-23T16:44:58Z")
Hey @HoneyBadgerDC , here is a hydro-electric dam power online calculator. > **[Hydroelectric Power Calculator](https://www.omnicalculator.com/ecology/hydroelectric-power)** > > Our hydroelectric power calculator finds the power produced by three different types of turbines: a dam, a "run-of-river" installation, and a tidal power turbine. You may play around with some numbers there to get what you want For example : A 100ft tall tower (gives a pressure head of 90ft accounting for pressure losses in the pipe), with 2 gallons per second of water flowing out and a turbine efficiency of 50% (i think this is much higher in real life) will give you a kW of power. If you want 1 kWh - then you need 2 gallons x 60 secs x 60 mins = 7200 gallons of water. This is the same as 60,000 lbs or 30 US tons. ---
**Author:** ![komolono](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/komolono/32/532_2.png) [@komolono](https://boards.straightdope.com/u/komolono)\ **Post date:** [July 23, 2021, 5:37pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/9 "2021-07-23T17:37:31Z")
Have you heard of this idea - well business today? > **[136: Gravity-based energy storage system - 366solutions](https://366solutions.com/14-/01-/2021/136-gravity-based-energy-storage-system/)** > > The energy provided by sun and wind is intermittent and often needs a back-up system; ---
**Author:** ![Acierocolotl](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/acierocolotl/32/6241_2.png) [@Acierocolotl](https://boards.straightdope.com/u/Acierocolotl)\ **Post date:** [July 23, 2021, 6:06pm UTC](https://boards.straightdope.com/t/converting-potential-energy-to-electricity/947140/10 "2021-07-23T18:06:32Z")
I’m doing this in metric units, and leave the commentary between SI to Imperial to others. I’ve paid that tax a long time ago and have my excemption stamp. For ease of calculation, I am making some basic assumptions: - The tower is now 100 meters (~330 feet). - There is no air resistance or other parasitic losses. - Because I’m extra-lazy I simplified the gravitational acceleration constant to 10 meters per second squared. - All “braking” (converting the kinetic energy of the water to electrical energy) is done in one second. - These simplifications allow us to use the basic potential energy equation: E=10mh, where m is the mass in kilograms, g is now 10 m/s^2 (and already written in), h is the height in meters. To do this in US units, E=70.2mh\*Ef, where m is your mass in pounds, h is your height in feet, and Ef is your target efficiency ratio. It’s accurate enough for back of the napkin math. The apparatus is a tower, from which we’re dropping one kilogram shots of water. There’s a turbine at the bottom that will get spun up by the water which we’ll use to generate electricity. A one kilogram shot of water will generate 100 joules of energy in a perfectly efficient system. I’m assuming, of course, that the water transfers its energy to the turbine over exactly one second. Your target of one kilowatt at 50% efficiency requires 20 kilograms dropped from 100 meters.