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Pico-Hydro in house renewable energy idea  RSS feed

 
Jonathan Hopkins
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Disclaimer: First and foremost Id like to start off by saying I am by no means an expert in science, math, or hydro electric, just a college undergrad interested in renewable energy. Therefore this idea could be totally useless and physics wise impossible.

Alright, with that being said, I've been formulating an idea for the past few weeks about an in house pico-hydro small electric generator power plant. The idea starts with a 55 gallon drum filled with water. At the bottom of the drum is a 3/4" hole with a valve. After the valve a hose would run to a water pump inlet. A hose would then be ran from the outlet of the pump to a 1/2" nozzle, or two, which would be directed at a pelton wheel. The pelton wheel would be attached to a pma generator directly at the shaft, or through a pulley system. The water would then fall to the bottom of the pelton wheel housing and out a tube that would be redirected into the 55 gallon drum, thus recycling the water and completing the system. I have done some research on pond pumps and have discovered a few that produce a decent gph and draw relatively low watts. One I have been thinking about using is a pondmaster pm-9.5 which can pump 950 gph at zero ft and draws 93w. Assuming I would have to raise the water 6 ft to get to the pelton wheel housing, it would be pumping 650 gph, or 10.8 gpm. Now, I have researched pma's and have come across some from Mikes windmill shop. This website claims that their 24v pma will create 35.8v and 4.1 amps at 430rpm, therefore creating 146.78w. Now if the system was ran all day at 430rpm that would be about 3.5Kwh of power generated per day. Now subtract the 2.3Kwh that would be drawn from the pump, and you get a gain of 1.2Kwh. In theory to me this sounds viable, but like I said, I am no expert or even close to one. So it makes me wonder, if it is viable, Im sure someone much smarter and educated that me would have already come up with this idea. As of now Im not sure if I could even produce 430 rpms, but with a pulley system it seems like it could be possible. Another concern I had was energy loss during the conversion to electricity. Critiques and info as to why this wouldn't work are welcome. Please let me know what you think.
 
Tom OHern
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Nope.

You can not get more energy out of a system than you put into it. And due to the losses from friction, and conversion from mechanical energy to electrical energy, you would get significantly less out of the system than you put in. The amount of energy required to pump the water up, is going to be greater than the energy you receive on it's way back down. Every first year physics student ever has run through this thought experiment, and every single one of us has come up against the law of conservation of energy. This just doesn't work.
 
Jonathan Hopkins
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I am familiar with the law of conservation of energy, but I would not be using gravity to turn the wheel, It would be a pressurized stream due to the diameter reduction at the nozzle. Decreased flow with increased velocity.
 
Tom OHern
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To determine the energy in the system, we need to measure the forces in the system. Diameter reduction does not increase the forces. It does increase the pressure but it reduces the volume so you will still have constant force and therefore the energy output is the same. The only force behind that water is that due to gravity.
 
Marcos Buenijo
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Jonathan Hopkins wrote: Alright, with that being said, I've been formulating an idea for the past few weeks about an in house pico-hydro small electric generator power plant. The idea starts with a 55 gallon drum filled with water. At the bottom of the drum is a 3/4" hole with a valve. After the valve a hose would run to a water pump inlet. A hose would then be ran from the outlet of the pump to a 1/2" nozzle, or two, which would be directed at a pelton wheel. The pelton wheel would be attached to a pma generator directly at the shaft, or through a pulley system. The water would then fall to the bottom of the pelton wheel housing and out a tube that would be redirected into the 55 gallon drum, thus recycling the water and completing the system.


I concur with Tom. It absolutely will not work. However, I think it's important for you to understand how it doesn't work.

All of the energy you have to work with here is gravitational. Specifically, it is gravity acting on the mass of water contained in the drum. This is otherwise known as the "weight" of the water. However, weight (a force) is not energy. Applying a force over a distance IS energy. Actually, it's more useful to consider that a force acting over a distance TRANSFERS energy at a rate that can be quantified by the product of the applied force and the displacement over which the force is applied in that same direction. With water contained in a drum, it doesn't matter whether or not we suspend the drum with a rope connected to a pulley, connect the pulley to an alternator, then allow the drum to fall ten feet all the while driving the alternator to make electricity.... OR, if we allow the water to drain out of the drum down and through a line ten feet below to a pelton wheel that is connected to an alternator. The principle is the same. The weight of the water (a force) falls a distance of ten feet in the same direction at which that force (the weight) is directed. Now, you're proposing that the system essentially regenerate itself. This is just like connecting one drum filled with water to an identical drum filled with water through a pulley system that also drives an alternator. Once the two drums are at the same level, then there is no net force on the pulley and the system stops. No net energy. In the context of your system, you drain an inch of water from the drop of the drum through the pelton wheel (no different than dropping the drum by an inch) which delivers energy to the ouput shaft of the pelton wheel. Now, to return the water to the drum you have to raise the water level by one inch to maintain equilibrium (this is just like raising the drum of water by one inch). Of course, this means ALL the energy delivered by the pelton wheel shaft is used to drive the pump use to return the water. In reality, due to friction and other losses, the pelton wheel cannot return all the water to the drum let alone provide any NET energy out.

Think of it this way. You have a two pulleys. One is 100 feet above the other and they are connected by a long belt such that when one turns the other also turns. A shaft is connected to the lower pulley such that it drives an alternator. Let's say you have heavy weight that can be attached to the rope, and this weight is located at the top pulley. A person clamps the weight onto the rope and the weight falls. This pulls on the pulleys to cause a torque that turns the alternator. Now, once the weight falls to the lower pulley, then the motion stops. There is no more energy available for the system. The only way to return the system back to the initial state is to raise the weight back up to the top pulley. However, this requires one to expend at least as much energy that the falling weight delivered. The same principle applies with multiple weights, or with a mass of water (which consists of trillions of little molecular weights).

Returning the water from the pelton wheel housing and back to the drum requires the delivery of at least the same amount of energy that is delivered to the pelton wheel when that same mass of water flowed through the pelton wheel. There can be no net energy delivered if one stipulates that the system remain in equilibrium (i.e. that the water level in the drum not fall, assuming the water is not replenished from an outside source).
 
Marcos Buenijo
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Jonathan Hopkins wrote:I am familiar with the law of conservation of energy, but I would not be using gravity to turn the wheel, It would be a pressurized stream due to the diameter reduction at the nozzle. Decreased flow with increased velocity.


All of the energy that turns the wheel is derived from gravitational potential energy. To make this clear, consider that if you flow all the water in the drum through the pelton wheel when it is 20 feet below the drum, then you will get about twice the output from the wheel as compared to the case when the pelton wheel is only 10 feet below the drum. Of course, in both cases if the system must return the water to the drum at the same rate that it flows through the pelton wheel, then all the energy delivered to the pelton wheel output shaft will be required to drive the pump that returns the water. In practice, this requires MORE energy due to friction and other losses.
 
Jonathan Hopkins
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Right I understand what your saying. My original idea was to pressurize air in the same container as the water with a manual air pump. which i have done and experimented with. It works to an extent, but I feel i would have to have a very large water container and air container connected to it to produce any real power. which would require a lot of manual labor. Not sure if it would be worth it or not. Any thoughts?
 
Lyvia Dequincey
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To me, the way to improve this is to allow rain water to replenish it somehow.
 
John Devitt
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Location: Belfair WA
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The basic problem of all alternate energy systems is storage. Storage is the means of having energy available when it is needed. That's why most alternate energy systems have battery banks.


I have though of the same system, where the water tank is your storage system. With my concept I would use solar power to pump the water back to the tank


The road block I came up with was the amount of water storage required to make it work as well as the amount of head required/available.
 
r john
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John Devitt wrote:The basic problem of all alternate energy systems is storage. Storage is the means of having energy available when it is needed. That's why most alternate energy systems have battery banks.


I have though of the same system, where the water tank is your storage system. With my concept I would use solar power to pump the water back to the tank


The road block I came up with was the amount of water storage required to make it work as well as the amount of head required/available.


Your road block was solved by Armstrong in the UK in the 1850's see the wicki

http://en.wikipedia.org/wiki/Hydraulic_accumulator

We still have a few weighted accumulators in use in the UK mainly for swing bridges.
 
Marcos Buenijo
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r john wrote:

Your road block was solved by Armstrong in the UK in the 1850's see the wicki

http://en.wikipedia.org/wiki/Hydraulic_accumulator

We still have a few weighted accumulators in use in the UK mainly for swing bridges.


Unfortunately, these do not store appreciable energy (low energy density). However, they are useful in some settings because they can deliver energy efficiently at an extremely high rate (high power density). I maintained several very large hydraulic accumulators pressurized to 3000 psig. The purpose of these was to raise an elevator platform weighing more than 250,000 pounds by more than 40 feet in under 10 seconds (aircraft elevators on U.S. Navy aircraft carrier vessels). The system was recharged after each cycle by running four large piston hydraulic pumps powered by electric motors rated at roughly 300 hp. All four pumps working together could recharge the system in roughly 90 seconds. While such a massive storage system at high pressure appears to supply an enormous amount of energy (10 million foot pounds sure seems like a lot), the system actually shows very poor energy density (the figure is about 13000 btu - about one pound of coal). This same amount of energy in the form of electricity (about 3.8 KWh) can be delivered by a fully charged lead acid battery weighing about 250 lbs.

I don't see this as a practical system for appreciable energy storage.
 
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