Today is...
Sunday, December 21, 2014
Welcome to Control.com, the global online
community of automation professionals.
Advertisement
Featured Video...
Featured Video
A quick introduction to EtherCAT for motion control and I/O...
Advertisement
Our Advertisers
Help keep our servers running...
Patronize our advertisers!
Visit our Post Archive
Generating electricity using household water pressure
how can I generate 12 volts AC or DC using only household water pressure, and what is the smallest this generator can be? Can it be smaller than 120cc?
By Earl Smith on 21 March, 2008 - 11:46 pm
0 out of 1 members thought this post was helpful...

I would like to generate 12 volts AC or DC using only household water pressure. What is the smallest this generator can be? Is there a component/device already out there on the shelf?

earlsmith3rd @ aol. com

Thanks!

Moderators note: Due to multiple requests and comments from forum members and readers, no new messages will be posted to this topic. The thread is closed.

By Phil Corso, PE on 22 March, 2008 - 7:10 pm
1 out of 1 members thought this post was helpful...

Responding to Earl Smith's 21-Mar-08 (23:46) query... unless your utility can provide water flow-rates much greater than typical, or you have a way of beating the laws of physics regarding fluid-flow, I suggest a return to Physics 101:

With an available house pressure of 35-psi a flow-rate of 10-gpm will produce about 1/8-Hp (approx 90-W!)

Please note: this calculation ignores efficiency of whatever you intend to use to convert fluid-flow energy to electric energy.

Regards, Phil Corso (cepsicon@aol.com)

By Phil Corso, PE on 23 March, 2008 - 7:10 pm
1 out of 1 members thought this post was helpful...

Earl, I aplogize for neglecting to answer the second part of your query... the term 120cc:

If you mean a flow-rate of 120 cm^3 per sec, then you will generate 30-milliwatt!

Regards, Phil Corso (cepsicon@aol.com)

By Robert Scott on 24 March, 2008 - 11:25 pm

Earl,

Following on to Phil's analysis, you should also include the cost of the water. Typical municpal water rates are 1 cent per gallon (in the US). So a 10 GPM flow is costing you $6/hour. But the electricity generated (90W) makes the rate for this electricity equal to $67 per kW-hr, as compared to a reasonable grid rate of $0.10 per kW-hr. So using household water pressure is a really expensive way to make electricity.

Robert Scott
Real-Time Specialties

By Stevan Lieberman on 31 August, 2008 - 4:47 pm

All you've said below is true, but multiple units turning water flow into electricity that run only when some one is going to use the water any way would more then offset the cost of the water itself plus feeding electricity back in to the grid. Further, the idea is to not try to get big gains, but for thousands of households to install such units obtaining a very small gain from each. Even installing such units in gutters would garner a little bit of electricity. A lot of little steps equal one huge step.

By Robert Scott on 1 September, 2008 - 11:19 am

Stevan,
Yes, a lot of little steps forward equals one huge step forward. But if all the steps are backward, they make one huge step backward. And here is why they are all backward steps:

Relying once again on Phil Corso's analysis posted earlier, a typical 10 gpm flow would generate 90 watts while it is flowing. It is generating zero watts when water is not flowing. And since you are hypothesizing this generator as something that only runs when you would be using water anyway, we must consider how much of a typical day is spent drawing water at the rate of 10 gpm.

Let's leave farming and golf course watering out of the discussion and focus only on residential homes. They consume water when you draw a bath, do the laundry, or wash dishes. I think it would be generous to say that 10 gpm flows for about 1 hour per day. 90 watts one hour per day is 32.85 kw-hr over a one year period. At $0.10 per kw-hr, that produces $3.29 worth of electricity per year. Whatever water turbine/generator you install to capture this power is certainly going to cost you more than $200. With this kind of investment, and getting $3.29 back per year, you will recoup your $200 investment in 61 years. I doubt that any $200 turbine/generator is going to last 61 years without maintenance. So you will never break even, so it was a step backward.

Another thing that extracting energy from the domestic water suppy does is lower the pressure at the point of use. In fact, to gather 90 watts, you will have to lower the pressure to almost nothing. That is going to be pretty inconvenient for taking a shower, or just about anything else. If there was excess pressure in the water supply, then the proper thing to do would not be to deploy millions of little generators to capture that energy. The proper thing to do would be to lower the municipal water pressure and realize the savings by not having to pump as hard - all without any capital investment at all.

Robert Scott
Real-Time Specialties

@ Robert

A few things.

First off, there it typically a pressure gradient in the water system depending on how far the residence is from the reservoir and the elevation difference between the home and the reservoir. So some houses (I believe most houses have one installed) have a pressure regulator installed on the house side of a water meter, so, in essence, a turbine could be just as easily used to lower the pressure from the street as a valve, while recovering some of the energy.

That being said, I doubt there would be a generator/ turbine anywhere near efficient enough to make this plausible or useable.

At least it's forward thinking, even if it results in backwards results.

Alternatively, a home could set up their own water reservoir in their attic, to gain an energy boost from gravity. I envision a large tank (appropriately supported) which has a level switch so that it turns on water from the street as water is used to maintain a constant volume of water in the house. Depending on the amount of water in the tank, a large potential energy store could be created, potentially with a larger pressure gradient than the street pressure. Slap a turbine on the outlet valve for the reservoir and generate electricity there.

Again, I don't feel like doing the fluids and thermodynamics to find out how much height/ volume would give a usable pressure, and I'm sure that there would be a significant danger in thousands of pounds of water in the attic. So it's probably not at all plausible, but it's an interesting problem to think about.

By Stuart Hiley on 13 March, 2009 - 8:31 pm

We know that you can't get something for nothing and that tapping the house water pressure will take away from the pressure at the tap. But if you utilize a water line where the pressure drop is not an inconvienence, like the washing machine, you could realize modest returns.

Modest is the key word. Using the house pressure to run a pump that pumps a low volume of rain water from a cistern up to an attic resevoir would be useful. Use it to flush the toilets. Toilets are big water wasters and do not require potable water. So using the otherwise unused potential of your house's water pressure to save water makes sense. (Not to mention it is enviromentally friendly).

Based on annual rainfall, the area of your collection surface (your roof) and your water bills, you could calculate how long it will take you to break even. In my neck of the woods (South suburbs of Chicago) it will take under three years.

Dear Friend,

you are right, its costly to generate from water from municipality. But how about using free water flow, the one from sewer or drainage flowing in city. its free and waste, do you think we can use the flow for generating electricity. the flow is not strong enough but I think its present throughout year and endless, I think we can use this resource just like solar or wind, that's useful from waste.

Please advice me on same.

Thanks and regards

By Laurin Dykstra on 2 January, 2011 - 11:30 pm

Robert,

We may assume too much. In a mobile home park, water is generally included in the rent. (I am a manager of one) That being said, if a resident left their water on continually, then we could take your number multiplied by 24. 3.29 = 78.96 then multiply it by 3 faucets =$ 236.88/ year. I am a manager, I have the park pressure set at 48psi, and there is way more gph than needed. I have a feeling the 3.29/month is quite low.

I would like to offer a better solution or better question to the answer. If you would like to augment the power generation which charges a battery system that is fed by other means of charging then this is a great idea. Other systems meaning solar, or wind. There are off peak times when solar and wind cannot produce the power needed to charge a batter. A simple shower or watering of the grass may provide that extra boost. Have a nice day.

What about people that happen to run swimming pools with pump systems that run from 4 to 8 hours per day. Could some advantage be taken of this? It is a closed system that needs to run anyway.

What if we were to tap the main the comes from the water tanks that feed the water systems? Yes a pump is used to fill the tank but the water that comes from these tanks, and feeds the whole community is gravity based why waste this energy resource?

By Robert Scott on 23 November, 2009 - 7:57 am

About people that happen to run swimming pools with pump systems, there is no advantage that could be taken, unless the pump system was seriously over-powered. If a swimming pool pump system is properly designed, then the pump is just powerful enough to circulate water through the pool. If the pump is more powerful, and water is being circulated faster than needed, then the thing to do is not to try to recapture that excess energy with a turbine. The thing to do would be to replace the pump by a smaller pump and not waste that energy in the first place.

As for tapping the gravity-based outflow of a municipal tank, that too is not the best use of resources. Those tanks are built as high as they are to get the needed pressure to run the system. If there is excess energy in the system, then the tanks should have been built lower in the first place. The only truly excess energy that has been cited in this thread is the energy that gets wasted in the pressure-reducing regulators. And even in that case the economics of the situation is such that the gain from a turbine would never outweigh the capital expense of installing the system.

Robert Scott
Ypsilanti, Michigan

I've been interested in all the information on this topic. I have a small stream running right through my property and I'd like to be able to harness the potential of being able to generate enough electricity for our residential needs. Can anyone suggest anything to help me please??

thanks

Check out http://www.ecoinnovation.co.nz/c-55-hydro-generators.aspx
Mike can customise a system for any head, any available flow.

Cheers,
Bruce.

By curt wuollet on 3 December, 2009 - 10:30 pm

Yes, see if you can conveniently enlarge the stream :^). Seriously though, people need to consider just how much power it takes to run our modern conveniences. It's a lot for anything that heats or is motorized. From a small stream, if you have enough head on your property, you might be able to provide basic LED lighting for example. For an example, look to the once common water driven grain mills. It was a lot of doing to get a few HP, but it's free once it's paid for. Thoughts of "going off the grid" have often occurred to me lately, I have the Rum River flowing through my property and I've pondered a water wheel driving a car alternator to charge batteries. The local power cooperative has been most encouraging in these schemes, not that they would want me to do it, but by the incentive their 10.4 cents per kwh rate provides. The Rum will freeze solid soon, which curbs those thoughts.

Regards

cww

If you can find it, 20-something years ago Smithsonian Magazine did a feature on the use of low-head pelton wheel turbines to power residences with streams available. There is/was a Chinese company that was making an Allis-Chalmers knock-off pelton wheel that was being used with good results.

Unfortunately, Smithsonian's online archives only go back to 2003, and this was sometime in the early 1980s or late 1970s.

Suffice it to say that you can make a small bypass dam in your stream and put a Pelton Wheel style turbine generator in it that requires relatively low head to operate.

Good luck,

Walt Boyes
Editor in Chief
Control and ControlGlobal.com
555 W. Pierce Rd Suite 301
Itasca, IL 60143

wboyes [at] putman.net
www.controlglobal.com
630-467-1300

You need to calculate the energy available based on physics before you go any further. What is the water flow? What size of head (drop) do you have available? Would you need to build a dam (and would you be able to get planning permission for it)? What is the *minimum* seasonal water flow?

1 W = 1 N*m/s = 9.8 * 1 cubic-metre * 1000 / sec

So, 1 cubic metre per second of water falling a distance of 1 metre would give you 9.8 * 1000 kg * 1 m = 9.8 kw. That's a useful amount of power, even once you subtract efficiency losses.

However, 1 cubic metre per second is also a fair amount of flow for a "small" stream. If you only have 1 litre per second of flow, then you only have 9.8 w (not kw!). That's not enough to light a light bulb, even a small CF bulb. If you have more head (drop), then you can generate more power will less flow. However, unless you have very favourable geography, you probably don't have a lot of head to work with on a small property.

Small 19th century water powered grist, flour, or saw mills typically operated on no more than a couple of kilowatts. That's why most of them were abandoned when electricity became available instead of being converted into power generation plants. Unless your location happens to be very good, generating your own hydro tends to be more of an interesting hobby than anything else.

This popped up during a related search:

http://www.ashdenawards.org/

I haven't looked at it very closely, but it seems to have some interesting projects of all kinds.

By Norman Elliott on 24 November, 2010 - 4:10 pm

A simple way to find out your water flow is to make a temporary dam in the stream with a V notch in it. Measure a known distance, say 1 yard, back upstream from it. Drop a piece of wood (or something else which floats) in at this point. Time how long it takes to cover the 1 yard. Measure how high up the V notch the water is from the bottom of the V and how wide it is at that point. The flow of water is worked out from these measurements.

We are generating 20KW from a stream/waterfall. See how at:

http://energyindependence-rob.blogspot.com/

Rob

By bhaveshchanndra on 6 March, 2011 - 10:24 pm

hi contact at keeth22@gmail.com for further details

By Namatimangan08 on 1 May, 2011 - 10:06 am

First of all you find the average of the stream flow (Q_avg). Next determine firm stream, i.e. stream flow (Q_firm) that has exceedance probability of 98%.

Then you have to find optimal pressure gradient line. Normally we should be looking for the highest pressure gradient between two points along the stream path, such as water fall.

I think I will let you to calculate the cost to install it. I will show how to calculate the return. I will show you how to determine the firm and the average return.


1. Firm return, E_f(kWh/annum)

E_f =0.80*0.90*Q_firm*H*density *g*365*24*3600 J

Note that 3.6million Joule =1kWh

2. Average return

Same as 1 above except that Q_firm shall be replaced by Q_avg

The first and second constants, 0.80 and 0.90 represent overall conversion efficiency from potential to electricity and availability factor of the proposed turbine generator. H should be in m and the others are in m-k-s

Over long run you can expect to get E_avg but you shall anticipate to get only E_firm for at most 3 consecutive years. E_avg always greater than or equal to E_firm.

I hope this post is helpful.

I would like to make private contact with:
earlsmith3rd [at] aol. com

mine Email:
daliborrmilicevicc [at] gmail.com

By David Nguyen on 9 February, 2010 - 8:21 am

I agree with the fact that using your utility water to generate electricity is not efficient but how about using the turbine at the water entrance to the house in a close loop? That means the turbine will rotate everytime water is used, the more water is used the more we will save. I saw one like this online a month ago but did not remember the website. I will post the product as soon I find it. In the mean time, please contact me at david.vapc [at]gmail.com if you find any. Best,

By curt wuollet on 9 February, 2010 - 8:30 pm

The problem, again, is not that it can't be done or won't work. It's that the power that can be generated this way is insignificant in terms of the needs of a modern household or even paying for the equipment, or the water bill for that matter. If it would, nothing would stop people from just running their dynamo from the meter to the drain and making money:^) One ten cent kWh is a lot of power.

Regards
cww

By jmpromotes on 5 April, 2011 - 7:02 pm

Here is a website that I found.... maybe this will help- http://www.nooutage.com/HydroElGen.htm#Hydroelectric Turbine Generators

Dear Readers,

I am going to generate electricity through ground water and i want the calculation material. how 4cf water discharge of water extracting turbine generate electricity. head of Tube well id 30' and which electricity generating turbine is good for generation.
i am very advance thankful to all reader how give me correct information.

Eng. Jahanzab

I would like to recharge small batteries from the water flow for our household usage. Any micro turbines out there?


REPLY...

Type in...water powered electricity/rainfall. Top of the page FOR MICRO TURBINE;PLASTIC SPOONS.

2nd reply...go to ...re-enegy.ca

We have a commercial building with four restaurant, offices etc and probably thirty toilets. The water service is a 2", and I know that the water tower is at least sixty feet above us and one mile away. Pretty impressive head. Seems like we should be able to harness some benefit from a small turbine. (knowing full well it will only generate power when water is being used).

By curt wuollet on 23 August, 2010 - 3:24 pm
1 out of 1 members thought this post was helpful...

This keeps coming up and it's worth answering. Not because generating power in this way is practical, but because it shows that people don't understand just how much energy we are using. With that amount of water usage, it's unlikely you would generate enough power to keep the coffee warm, on average. And with the cost of equipment, it would be very expensive power. That's why, even with all the very smart people who want to solve the problem, whether to save the planet, or to make millions, we still have the problem. The amount of energy we use is almost beyond human comprehension with our current lifestyle. That's why all the success stories you see about self sufficiency involve very drastic reductions in energy use, to the point that small scale green generation can provide enough. I've pondered this a lot lately, as just staying alive in Minnesota takes a vast amount of energy the way we are doing it. 9 cords of wood or 9 tons of corn or a couple _thousand_ gallons of propane or oil just to keep from freezing to death. I wish there was a simple answer, but I've got to go and pick trees for cutting when the weather cools. The key statement is probably "The way we are doing it" The whole breakdown is when you suggest to people that they change that. They are not interested.

Regards
cww

Conservation of energy - 1 m^3 of water falling through a pressure drop of 1 Pa will give 1 J of energy.

In more convenient units - 1 litre through 1 kPa = 1 J

You have 60 feet = approx 2 bar pressure drop = 200 kPa

1 litre through 200 kPa = 200 J

At a flow rate of 1 litre/second, the total power available from your water tower is 200 W.

Neglecting losses.

The water supply people use energy to put the water into the tower - it is highly unlikely that there is going to be a lot of unnecessary extra energy added so you can get it for "free". Now, if you had a water collection point at the top of a hill and had to throttle it to make use of it at a reasonable pressure at the bottom, that's a different story.

Bruce.

I believe that most of that energy represented in that head is lost as friction in the delivery pipes. One way of looking at this is if you turned the flow on full and directed it straight up, how high would it go? In most cases, not very high. You're thinking of how much excess energy you could extract from the water. The water utility looks at that as wasted energy (and extra expense). There really isn't that much excess to extract. If there were, the water utility would be installing turbines in their own water mains.

It would take a very substantial flow of water to produce a significant amount of energy. What you should be looking at is how to consume *less* energy by eliminating waste. The savings there can be substantial. That is a much more fruitful avenue than trying to generate trivial amounts of electricity.

1 out of 1 members thought this post was helpful...

David... the energy available is rather easy to test. Just note the inlet pressure to your building, with and without water flow!

Regards, Phil Corso

By philip scafordi on 8 January, 2011 - 3:10 pm

I would like to know, if a house used the water that went down its drains in a year, to turn a generator...how much electricity would it produce? enough to power the house?? thanks for your time..

It is suggested that you refer to wikipedia.org or use your favorite Internet search engine to learn the concept of head and its relationship to power generation.

Once that water exits the faucet, it has essentially zero head, unless you have a large storage area that is located hundreds or thousands of feet about the water wheel connected to the generator.

And if you were to use the pressure of the water entering the building to drive water wheel or small hydro turbine connected to a generator, you would again have essentially zero head, which would make for an interesting shower experience.

As the eloquent and esteemed contributor Phil Corso said in the first reply to this post, there isn't much to be had in this household water pressure idea. Either on the incoming side or the outgoing side.

By Namatimangan08 on 2 May, 2011 - 2:39 am

> I would like to know, if a house used the water that went down its drains in a
> year, to turn a generator...how much electricity would it produce? enough to
> power the house?? thanks for your time..

Look in bigger perspective. Probably there is an opportunity to set a power plant.

Assuming your township is located 100m above sea level. It is possible to collect used water and natural drainage and channel them to a man made reservoir. Then you can use this water to run a conventional hydro power plant.

This approach is the most likely hood to success.

hi there

i have been obsessed to get energy from mains water for ages. had a breakthrough lately :) instead of using the flow i used just the pressure in the system with a relief valve. this works on the principle of a 2 stroke engine. i managed to get a small bike generator running 24/7 without water wastage. the only physics law i am breaking is that = i get free energy yes but it is coming from the mains pump run by the water suppliers:) so it only free energy for me. well i have many gadgets at home eg connected to the tv cable and telephone line. in all i get 5 amps 12 24/ 7 free. enough to run a pc, tv and a refrigerator in winter. see ya

By Robert Scott on 21 January, 2011 - 11:02 am

ray,

You cannot extract energy from water pressure unless you let that water flow. You can use a relief valve to reduce the flow, and that will reduce the maximum energy that you could extract. If you reduce the flow to zero, then the energy you get from it is zero.

If you extract energy from water that you would use anyway, then you are reducing the effective pressure available at the point of use. Your showers would not have as much zing! Your washing machine would take longer to fill. If you can put up with all that inconvenience, then you are welcome to the little bit of energy you would be generating. And if everybody could put up with that inconvenience, then the municipal water suppliers would reduce the pressure at the source and save money for them. There is no free lunch.

Hello there,

I live in Brazil, and in here every house and building has it's own "water tower"(common houses have 1 m water towers on top of the construction, buildings i don't know :[ )

Most residential buildings have about 15 floors, going from 5 to 10 apartments per floor.

I am a economist (student), not so familiar with physics (and i know it's lame to ask for you to do the math), but could it work?

A common house would generate 60w, right? A building (with a water tower on top) have a greater water column, and more potential energy. If one home would save 3.x dollars per year, a 15 floor, times 5 apartments (75 homes) would save about 215 dollars, and could pay for the system in one year. With some batteries, it could generate power for the elevators, perhaps? Or emergency lights, power up the gate, hallways, i don't know.

Today i live in Copan (it's a 32 floor building, with 1160 apartments, one of the biggest residential complex in the world) and the pressure wastes my wash machine hose every month, so i start to think about using building pressure to generate electricity.

Sorry for my bad english.

By Robert Scott on 25 January, 2011 - 12:34 pm

No, Fabio, it will not work. If you read back through the other postings in this thread you will see that whenever the math is done, it shows that it does not work well enough to be cost-effective. Not even close. You are making up numbers (60w? 3 dollars per year). This is just wishful thinking. It is also wishful thinking to say that a system that extracts this energy would cost only $215 and not have on-going maintenance costs.

Here is some math for you. A 32-floor building like you live in, if it is all supplied by one feed of water, would have a difference of about 114 PSI between the top floor and the bottom floor. So if they are getting 30 PSI on the top floor, it would be about 144 PSI on the bottom floor. No wonder your washing machine hoses are failing. Modern construction takes special measures to equalize the pressure on difference floors. Perhaps your building does not have those provisions.

Remember, you can only extract energy from that pressure if you let the water flow. Static water pressure cannot deliver energy. And if you let the water flow even when you are not using it, then you will have to pay more for the water. (Or someone will have to pay.) If you don't have to pay extra for using the water, and if everyone in the building started flowing water to extract energy, then the building management would notice that their water bill is going up. Since you are an economics student, you know what that means. You will no longer get free water. The management will install meters and everyone will have to pay for their extra water usage.

No, if you do the math as an economist, you will see that extracting energy from household water is a losing proposition.

I know it's an old quote, but it's yours (i cut some parts so it won't be huge):

"Posted by Robert Scott on 1 September, 2008 - 11:19 am

Relying once again on Phil Corso's analysis posted earlier, a typical 10 gpm flow would generate 90 watts while it is flowing. It is generating zero watts when water is not flowing. And since you are hypothesizing this generator as something that only runs when you would be using water anyway, we must consider how much of a typical day is spent drawing water at the rate of 10 gpm.

Let's leave farming and golf course watering out of the discussion and focus only on residential homes. They consume water when you draw a bath, do the laundry, or wash dishes. I think it would be generous to say that 10 gpm flows for about 1 hour per day. 90 watts one hour per day is 32.85 kw-hr over a one year period. At $0.10 per kw-hr, that produces $3.29 worth of electricity per year. Whatever water turbine/generator you install to capture this power is certainly going to cost you more than $200. With this kind of investment, and getting $3.29 back per year, you will recoup your $200 investment in 61 years. I doubt that any $200 turbine/generator is going to last 61 years without maintenance. So you will never break even, so it was a step backward.

Robert Scott
Real-Time Specialties"

The point is activating the generator by regular use (i imagine that not having a constant flow would waste a lot of potential, but anyway...), and it could specially work on skycrapers. Commercial building usually don't have wash machines and showers, but still have a lot of toilet flushing (6 l/f in here, we don't have those with pump for water saving like US).

It's a thing to add to those barely viable "green buildings", where it would be a medium term profit (1 to 5 years refund is a great investment).

My building was build in the 60's, so i wouldn't even risk to say it has anything modern besides people's furniture. ;)

And that just reminded me, the main hall is like a mall, we have lot's of restaurants/bars and a laundry.

It's not that odd to see constructions like this, a huge building with some stores on the ground floor (though most of them won't have laundries).

As well as US's tv shows (like big bang theory) often shows residential buildings with laundries on the basement, where we could extract most of the pressure with a relatively constant flow, for a short period though, without great impact on the daily use. You know better than me if it's common or not, i went to US once for about 2 weeks (and sure want to go again, and stay a lot more).

My point is not using it to extract from one household water, but a big group of them, in buildings 100+ apartments or 1000+ workers. I know there is a maintenance cost and it's way more than 200 bucks to build the system, but in a larger scale could it work?

By Infamous Outfamous on 26 January, 2011 - 6:57 am

It's not free. The energy extracted means that the rate of flow through your 'generator' is slightly reduced, depending on the ratio of conversion. Either way, the water company must then increase the pressure to maintain flow because of your power drain on the system - granted, this may only be a small amount, but it is a drain nonetheless. Anyway - long term view is, if everyone starts doing it the Energy the water company will need to push water will be greater, they will have to invest in stronger materials and joints to cope with the backpressure now powering everyone's generators and yes you guessed it, that fee we pay for the water will grow exponentially. We don't get free 'leccy, just a higher bill. - so you're better off building a water tower to collect rainwater and create a mini hydropower-station. Then you've also got some free water for the washing machine, garden etc.

Infamous,

If you read my first post, we don't have reliable water companies in here (Brazil is the country of the future for about 200 years, and it seems that we're finally getting there, just some more decades :D ), and have to ration water (tough most people just wastes it to wash the damn walkway instead of using a freaking broom, really pisses me off seeing them pushing one leaf away with water, so lazy), we don't have water from street 24/7, so every construction has it's own water tower. The pressure don't come from the company, it's from the "stock" of water every building must keep.

I know it's better that the company plans and keep a constant service, but it's an old "habit" and many people would be without water because of the old "leeches" who have to fill their stock. As well as it could use it's floodgates to generate more energy cheaper than a building.

In Robert's outdated approximation(2 years and a half ago, damn inflation), we would save $3,29 per year per house from common use.

So it's a scenario where you must keep a stock of water on top of the building, with 100+ apartments or offices for 1000+ employees (very common scenario around here), with a very suitable laundry on the basement(residential).

In a larger scale could it work?

PS. I'm not one of those "save the planet" freaks, so I know if it's not profitable, it's useless because it's wasting resources. I'm pretty skeptic about their green solutions, unfortunately most of them are composed by a huge waste of resources.

Look, if you have water in a hose under pressure, with a nozzle on the end of the hose and that nozzle is closed, then the pressure doesn't do anything. The only way the pressure does something is if the nozzle on the end of the hose is opened and the water flows out of the nozzle.

So, the only way to derive work from pressure is for there to be flow, and if the water is flowing through a device driving a generator then the pressure on the outlet side of the device driving the generator will be less than it was going into the device driving the generator. And, the flow must be relatively continuous to produce a reliable electric current. Think of how many minutes or hours/day that water is flowing in a typical household, and that just isn't very much.

And, because there will need to be a pressure decrease on the outlet of the device driving the generator then the water "pressure" will be much less useful. (Think of a dribbling shower or a tub that takes an hour-and-a-half to fill.)

If you keep the water flowing through the device driving the generator and dump the water down the drain (which would be a huge waste of water!), then all you're doing is using the water company's pump energy, transmitted through pipes, to a generator to convert the electricity driving the pump back into electricity for some other purpose. But, remember, the water company had to use energy to pressurize the system to provide the flow that people need, or even to fill the tanks on the tops of houses and buildings in many parts of the world.

Now, I'm interested in hearing much more detail about the poster who said he's using a relief valve in some kind of two-stroke motor configuration to produce power with household water pressure. But, I'm not holding my breath, either.

By Steve Myres on 27 January, 2011 - 12:56 pm

He said he did it without any flow, IIRC, which means it's not possible. Either he's dumping some water, or he's got the "output" interconnected to some real source of power that's illuminating a light or something and he's taking that as proof that his scheme works.

Looking at all of this, it gave me an idea. I'm not saying that this is cost efficient yet, but its something to think about.

I had an old 125 gallon fish tank with a decent pump that moved A LOT of water 24/7 just recycling it. IT got me thinking about an energy efficient pump that was circulating the water and collecting electricity.

A pump that requires 5Watts of electricity can move about 80GPH. A 10Watt pump will move 160GPH.

I don't know if this helps or gives anyone any ideas, but its a thought.

Here's a link to a few pumps. There might even be more energy efficient ones available. This was just a cursory glance. http://bit.ly/hE2GAX

By Robert Scott on 1 February, 2011 - 9:06 am

Rishard, I'm not sure what you are proposing, but it sounds like you are saying that you could move the water with the electric pump and then extract energy from that moving water. If that is what you mean, then that would violate the laws of thermodynamics and would be equivalent to a perpetual motion machine. The fact is no matter what you do, you cannot get more energy out of a system than you put into it.

As for the fish tank pump, the amount of water it circulates might sound impressive, but the pump is only capable of circulating that much water because there is little or no back pressure to the flow. To extract energy from a stream of water you need to create more back pressure.

On a somewhat related topic, I had an idea for using water pressure as an energy storage medium. Imagine an off-grid residence with solar and/or wind power and a well for water. When The panels or wind turbine produce electricity in excess of what the house is using, a pump in the well turns on to pump water up into a cistern. When the load from the house exceeds the output from the wind and solar sources, a valve in the cistern opens, allowing the water to drain back down to the well, flowing past a water turbine to produce more electricity in the process.

Naturally, there is no free energy there, and you would actually lose some energy due to the inefficiencies of all of the components involved. Basically, the whole cistern / water pump system would be a replacement for a battery system. Any ideas on how big the cistern would need to be?

By Steve Myres on 7 March, 2011 - 9:45 am

"Any ideas on how big the cistern would need to be?"

BIG! One KWH is equivalent to 2.65 million ft-lbs, so you would need to raise 2.65MM lbs of water one foot, one lb 2.65MM feet, or some combination.

Lets assume your cistern is 100 feet above the well water level. You'll need to move 26500 lbs or 3200 gallons of water to store 1 KWH. And you probably use 30-100KWH at your house in a typical day.

You seem to have a much better grasp of where energy comes from and when you're storing it vs making it than the others who've posted.

Google Pumped storage....

Haven't fried my brain reading the entire thread, but power in water is just like power with electricity.

power = flow X pressure
power = amps X volts

Since most here deal with electricity I'll ask this question. Can you run a pump and move water using household voltage for free? Why can't you just use the voltage to spin the pump motor and not use any current?

By Zacharia, Tomy on 7 March, 2011 - 11:26 pm

Regarding the energy storage model. Also look at the efficiency of the motor-pump combination. In the 5 KW range, it can be 20-25%. (roughly 30% for the centrifugal single volute pump and 80+ % for the motor). Higher sizes definitely have higher efficiencies. Then cost of high efficiency systems is an issue.

Regards,

Tomy Zacharia

why couldn't the city or electrical company put micro turbines in the water pipes for each block or neighborhood, if there is one branch of the water pipe that serves 20 houses the water flow for 1 year would be 20x and instead of 60 years to make up the $200 turbine it would be 3. This would barely lower water pressure at all. also no water is wasted because all water that flows through the pipes is going to someone's house.

Because it would be more efficient, if they don't need the pressure, not to develop it in the first place.

Power in W = pressure drop in Pa x volume flow in m^3/s = .05 x drop in psi x flow in ft^3/s

Or perhaps we can run a pump to push water up to the top of a hill, stick a turbine in a line, and use the turbine output to power the pump.

Bruce

By Get A Grip People on 21 March, 2011 - 8:26 pm

I highlighted your mistake. In order to get useful energy you need either a large flow or a large pressure drop or both. Arguments about "I only want 100Watts" are spurious because that's quite a lot of energy.

This discussion refuses to go away but really it's quite simple ...

You cannot get something for nothing. Any energy you try to get from household water must be the excess energy that the water company left in there to keep the system working reliably. Even if there is plenty of excess energy, its still not very useful if you want to run more than a couple of LED's. On top of this, you typically have to pay for the water you use.

You will also notice that the only people who keep asking this question also profess their complete ignorance of the science involved. EVERYONE who understands the basic school physics tells you its possible but is a total waste of time, effort and money.

Can we please stop trolling this question every few months and go do something useful instead.

Regarding the pumped water energy storage post, this is actually already done on a grid size scale in many places. Here in Colorado, we have at least one location run by the XCEL Electric utility up in the mountains where 2 reservoirs, probably about a square mile each, are linked with a turbine/pump arrangement. During off-peak load times, energy is used from the grid to pump water up into the upper reservoir from the lower reservoir. During peak load times, water is allowed to flow down through the turbine to generate peak power. This makes economic sense because even though there are energy losses in the system, the peak power generated is financially worth enough to offset the financial cost of pumping the water up using off-peak energy.

I don't know if the same economics could apply to an individual sized installation unless your peak power costs were extremely high, or you were completely off the grid and needed energy storage, in which case it would have to somehow beat batteries in cost and performance. Batteries are pretty hard to beat, but maybe in some cases with natural geography working to reduce capital costs and no regulation problems it might be feasible.

I find this message thread very interesting. I recollect a couple of items from my past. A book was written about me and a few other people that changed the world of lighting. Whenever somebody told me something could not be done I made a point to prove them wrong.

I was confronted by some very senior scientists in the world of physics and told that what I wanted to do could not be done - I found it funny to be sitting in front of the person at the Department of Energy 6 years later that was in charge of creating a USA testing system to qualify "what could not be done". I also remember sitting in front of the man from California Energy Commission who told me that what we wanted to do could not be done - now it is on every Home Depot and Lowes shelf.

I love the challenge you all present here. I get excited about working with a bunch of smart folks to solve these type of questions. I have not failed yet.

The world was once flat folks, but not anymore...but boy did it take a few brave people to brave the train of thought.

Arsenic killed everything, but not anymore, Nasa showed us how things on earth live off it.

The gentleman from Brazil makes some good points - there is a much bigger world than just the USA folks - I lived in New Zealand this past year and while energy prices where only $0.12 KWHour - relative to a person's income it was equivalent to $0.40 KWHour in the USA. I also heard constant stories about parents teaching their kids to take 2 minute showers.

I also think the gentleman creating the 2 stroke motor idea is not a bad one - you all are just focusing on available pressure of water and on water waste.

Take a step back folks and re-read this whole post. The answer for how this can be easily done is right here in all the words.

Would be fun to show you all in the coming year how we will do this.

Last night at an event in Las Vegas we introduced our LED light tube that replaces fluorescent tubes, it pays for itself in 3 months...

In reply to Manuel Lynch: With regards to water driven piston engines (as opposed to turbines), that isn't a new idea. They were widely used in Europe for industrial power in high head applications in the 19th century but were eventually replaced by turbines (and electric power) due to the better efficiency and reliability of turbines.

As for the person who in January said he was using this, I think a bit of skepticism about his claims is in order. He said he was running this 24/7 without wasting water, but didn't provide any details as to how he did this. Where did the water to run the engine go when he wasn't using it for anything else?

He also said he as using "a small bike generator" to run "a pc, tv and a refrigerator". A typical bike generator outputs about 5 watts. As to how you run "a pc, tv and a refrigerator" on 5 watts would be something that I'm sure we would all like to know. Without any details on that, I'm a bit skeptical.

Here's some typical electric power usage figures as an example (in kw-hours): PC (1248), television (350), refrigerator (2700). The total for those would be: 4298 kwh.

A 5 w bike generator running 24/7 could produce: 5 x 24 x 365 / 1000 = 43.8 kwh. That's 2 orders of magnitude less than what the appliances consume before we even account for the losses involved in converting 5 VDC to 120 (or 240) VAC. It's possible to quibble about actual consumption patterns or appliance sizes, but I doubt those would amount to 2 orders of magnitude of difference.

So, when someone tells us he can generate useful amounts of power from his kitchen faucets, I think we are entitled to say "show me the numbers!"

I think it is possible to design a machine capable of generating useful electricity via recirculated fluid flow. But the laws of thermodynamics make it very difficult.

The three laws of thermodynamics may be summarized by saying: "You can break even but only at absolute zero".

So we know for sure we will never be able to get more energy out of a system than we put into it in the first place.

Secondly, the value of entropy in any system governs what is possible or impossible.

If entropy is zero or negative, then whatever we hoped to achieve is impossible. It is simply not permitted in this universe by the laws of physics that govern it.

An example would be trying to reverse the act of smashing an egg on the floor. Any such reversal of the egg breaking process would require a value for entropy (S) which was negative. Accordingly, we know the proposed act to be impossible.

Unfortunately, the laws of thermodynamics and the related law concerning entropy reflect reality. If we don't like it, we must go and live in another universe.

Bearing in mind the laws we are stuck with, the next step is to consider what type of system might be useful bearing in mind these significant limitations.

There are three types of system. Isolated systems, closed systems and open systems.

In isolated systems, neither mass nor energy may pass through the system boundaries (think of a perfectly insulated thermos flask).

Needless to say, we will never be able to generate useful electricity from an isolated system because for a start we will never be able to get any energy out of it in the first place.

So we can forget about isolated systems right away.

Then there are Closed systems. They allow energy to pass through the system boundaries, but prevent mass passing through the boundaries.

Again the laws of thermodynamics prevent useful electricity being generated using a closed system, because you will always get less energy out of a closed system then you put into it in the first place.

Our only hope is to use an open system.
The laws of thermodynamics will only ever allow us to generate useful electricity using systems in which both mass and energy can pass through the system boundaries.

So this has to be our starting point.

We must invent a machine (an open system) that allows mass (for example water and/or compressed air) to pass in and out of the system boundaries, and which also allows energy to pass in and out of the system boundaries.

In practical terms this means we need help from the environment.

External work must be done on the system (eg solar energy or water from a flowing stream) to provide energy from outside the system boundary)and also external mass (for example air or water) from outside the system boundary must be able to pass through the boundaries as well.

The best idea I have come up with began as a mathematical thought experiment:

There are two 25m high cylinders each of diameter 1m.

An impulse turbine (for example a Pelton turbine) has been placed 3m from the bottom of the second cylinder (allowing a space beneath it for tailgate water to accumulate).

I think I have found a way to connect these cylinders in such a way that fluid can circulate from cylinder A into cylinder B allowing it to flow down a 20m+ drop in cylinder B before striking the impulse turbine.

At a mass flow rate of one cubic meter per second (which is an enormous flow rate)and allowing for system efficiency of 0.85 (this being a unit-less fraction where 0= total inefficiency and
1= perfect efficiency), the electrical power output in watts (using water as the working fluid) would be as follows:

Pw = 1000kg/m3 x 20m x 9.81 m/s/s x 0.85
Pw = 166770 watts = 166.77kW.

So this might look promising.
But there are serious practical problems.

The water that collects at the bottom of cylinder B (the tail-gate water) will rise until the point it prevents the impulse turbine turning. It will swamp the turbine and stop it moving.

However, I think I have found a way of recirculating fluid from cylinder A back into cylinder B without having to use enormous amounts of external energy (to lift or pump fluid back into tank B).

I would much appreciate private (not to be published) dialogue hopefully with a co-inventor familiar with thermodynamics, Bernoulli, and Newtonian fluid flows.

My aim is to share with a candidate co-inventor a recirculation novelty compliant with the laws of thermodynamics for the purpose of applying for a patent relating to industrial electricity generation.

I am only interested in applying for a patent if the model works mathematically.

Thank you for reading this post.

I can be contacted at:
pmrlaw@hotmail.com

By Robert Scott on 5 May, 2011 - 7:39 pm

Paul Turbine wrote:

>I think it is possible to design a
>machine capable of generating useful
>electricity via recirculated fluid flow.
>...
>However, I think I have found a way of
>recirculating fluid from cylinder A back
>into cylinder B without having to use
>enormous amounts of external energy (to
>lift or pump fluid back into tank B).

Oh no, not another perpetual motion idea! Give it up. You seemed to have some understanding of thermodynamics in most of your posting. But then you go and completely contradict everything you said by proposing exactly what you said is not allowed.

By Namatimangan08 on 7 May, 2011 - 5:44 am

>> I think it is possible to design a machine capable of generating useful electricity via recirculated fluid flow.

>> However, I think I have found a way of recirculating fluid from cylinder A back into cylinder B without having to use enormous amounts of external energy (to lift or pump fluid back into tank B). <<

> Oh no, not another perpetual motion idea! Give it up. You seemed to have some understanding of thermodynamics in most of your posting. But then you go and completely contradict everything you said by proposing exactly what you said is not allowed. <

I think he was just joking.

By quantumtangles on 9 May, 2011 - 9:59 pm

As a matter of fact it is not a perpetual motion machine. Perpetual motion machines are not possible. But neither was I joking.

Here is a system summary followed by calculations:

System summary:

Two Cylinders A and B, each 25m high and 1m in diameter, stand side by side.

Cylinder A is 90% full of seawater of density 1020kg/m3. It also contains approximately 10% by volume of castor oil (density 961kg/m3) which floats on top of the seawater. A small air gap is left at the top of cylinder A and a pressure relief valve sits at the top of cylinder A.

A siphon of diameter 0.12m leads from the top of cylinder A down into cylinder B. An electric pump primes the siphon and begins fluid flow into cylinder B.

Cylinder B contains only air to begin with, but there is an impulse turbine connected to an alternator motor at the base of cylinder B.

A 3m space at the bottom of tank B (underneath the turbine) is needed to allow tailgate oil to accumulate without interfering with the movement of the turbine.

The siphon (the short end of which ascends from the oil on the surface of tank A) has its longer end in Cylinder B, so that the longer end of the siphon allows oil to flow into tank B and strike the turbine.

The flow rate of the oil is 1 cubic metre per second and it falls 20m (after exiting the nozzle at the long end of the siphon) before striking the turbine.

An electric pump is used to prime the siphon flowing at a flow rate of 1 cubic metre per second. The pump consumes 30kW but requires only pulsed power because once the siphon has started working, it will continue working without help from the electric pump until the level of working fluid in tank A drops below the input nozzle of the siphon.

For reasons that will become clear, the level of the working fluid in tank A cannot fall below the level of the input nozzle of the siphon.

Now we come to the critical factor. An air compressor at the top of tank B (as well as being directed to increase the velocity of the working fluid as it travels down with the help of gravity to strike the turbine) is also used to pressurise the oil that accumulates at the bottom of tank B (in the tailgate area after having struck the turbine).

This higher pressure in tank B is needed to prevent the 350kPA (absolute) pressure at the base of tank A flooding tank B through the lower connecting pipe (also of diameter 0.12m) and it is also needed to force tailgate oil back into tank A (which is full of seawater and oil).

The pressure at the base of tank A is approximately 350,000 Pascals absolute, whereas the pressure at the base of tank B (which only contains a small height of oil) would only be 130,000 Pascals absolute before the air compressor operates.

The air compressor consumes 11kW and can pressurise the volume of air inside tank B (which is hermetically sealed) to 800,000 pascals within 7 minutes.

A constant pressure of at least 350Kpa must be maintained in tank B to prevent water from tank A forcing its way into the turbine tank (into tank B) and flooding the turbine housing.

This pressure is also needed to evacuate tailgate oil from tank B. So we are using cheaply generated pressure to control the flow of fluid, and we know the fluid must flow towards the lower pressure area (in other words it must flow where we want it to flow).

There is a pressure release valve at the top of tank A to prevent P1V1 = P2V2 equilibrium in the air gap at the top of tank A.

Preventing pressure equilibrium is critical as the system will want to equalise fluid and pressure levels immediately (and would most certainly do so were it not for the pulsed power air compressor, the air gap in tank A, and the pressure relief valve at the top of tank A).

The air compressor is float activated, so that when the level of oil at the base of tank B gets too close to the rotating turbine, the air compressor is then triggered, pressurising tank B from 350Kpa to in excess of 350Kpa and forcing tailgate oil through the lower connecting pipe back into tank A where it floats to the surface of the tank.

The flow of oil onto the turbine generates 160kW.

Power (watts) = 20(m) x 961(kg/m3) x 9.81 m/s/s x 0.85 (efficiency fraction)

However the pumps used to recirculate the working fluid (the siphon pump and air compressor) together consume only 41kW if operated continuously.

Note that neither the pump nor the air compressor must work continuously. Both use pulsed power when required. A computer controlled pressure/valve regulator can prevent pressure equalisation and maximise efficiency. I would be grateful if someone would be kind enough to send me a schematic for a board to regulate pressure automatically (can we build it...yes we can).

The siphon at the top of the cylinders minimises (to zero) the work that has to be done to move oil from tank A to tank B.

The principles underlying siphons are well established and do not need expansion here.

However the positive buoyancy of the oil leading it to float back to the top of tank A does not in fact provide energy benefit.

I used the example of a less dense working fluid (castor oil) simply to illustrate that work does not have to be done to 'lift' working fluid to the top of tank A (in the sense of work performed during the lifting process up through the height of tank A).

Of course work has to be carried out to force tailgate fluid back into tank A, but that is actually the only work the system must perform in order to operate. So although gravity is a conservative force and is not path dependent, the recirculation of fluid in this system is path dependent in terms of efficiency,and pressure regulation is an efficient way of carrying it out.

Strong ionic bonds between water molecules, if expressed (allbeit inelegantly) in terms of pressure, attract one another with relative pressure of 3000kpa. In other words, the water molecules are virtually chained together. This enables processes in nature to work as well, including the phenomenon of Giant Redwood trees being able to lift water 115m into the air at a rate of 160 gallons per day.

In these trees, water evaporating from the leaves creates a partial vacuum pulling water up from the roots.

In this system, water leaving tank A in the siphon pulls more water towards the input nozzle of the siphon.

Seawater throughout tank A (and also used as a working fluid) actually performs more efficiently because it has a higher density. We know from Newton's formula (F=m.a) that a fluid of higher density will cause higher energy output from the turbine because the mass in kg will be higher and we will not have to worry about the higher viscosity of oil at low temperatures decreasing the velocity of the working fluid.

My point is that the energy ostensibly gained by having a less dense working fluid (that uses buoyancy to float up a more dense substrate) is matched and neutralised by the lower energy generated by less dense working fluid striking the turbine.

These large cylinders can form the pillars of energy pyramids. Platforms at the top of the cylinders would make excellent locations for wind turbines. Solar panels can cover the pyramids enclosing the cylinders. Geothermal energy may be generated from beneath the cylinders.

Calculations:


Cylinder A = height 25m and diameter 1m
Cylinder B = identical to cylinder A
Working fluid = seawater of density 1020kg/m3
Flow rate of working fluid: 1 cubic meter per second
Height water falls before striking turbine = 20m
Diameter of turbine = 0.9m
Pitch Circle Diameter of turbine = 0.87m

First of all, we must calculate the maximum total electrical power in watts the system is capable of generating.

To do this we multiply the density of the working fluid (seawater = 1020kg/m3) by the height the fluid falls before hitting the turbine (20m), then by acceleration due to gravity (9.81 m/s/s) by the flow rate in cubic metres per second of the seawater (here 1 m3/s) and finally by a unit-less fraction representing turbine efficiency (85% = 0.85).

Pw ([power in watts) = 1020kg/m3 x 20m x 9.81 m/s/s x 1m/3/s (flow rate) x 0.85 (efficiency of turbine)

Pw = 170,105.4 watts = 170.1054 kW

So this is the maximum output we can get out of the turbine if 'the turbine' is 85% efficient. Other inefficiencies in the system must later be taken into account, but this is a good starting point based on head and flow rate.

We can calculate the force applied to the turbine by using Newton's equation F = m.a

If we assume for the moment that the acceleration of the working fluid will be the same as acceleration due to gravity (9.81 m/s/s), then we can calculate the force in Newtons that will be applied to the turbine by the flow rate.

F (Newtons) = Mass (kg/s) x Acceleration (m/s/s)
F = 1020kg/s x 9.81 m/s/s
F = 10006 Newtons

This is a useful piece of information.

We now know that over 10,000 Newtons of force will be applied to the buckets of the turbine by the flow of fluid striking it from above. This is an enormous amount of force.

Once we have decided the diameter of the impulse turbine (which must be less than 1m in diameter to fit inside the cylinder and must be greater than 0.2m in diameter if we are to avoid serious inefficiencies), we can also use this force figure (Fjet) in Newtons to calculate the angular velocity of the turbine in radians per second (which I converted to RPM below).

So we should be able to calculate how fast the turbine will rotate just from knowing the value of Fjet (which we calculated to be 10,006 Newtons) as well as other variables we established earlier.

I have chosen to use a Pelton impulse turbine of diameter 0.9 meters. I could have chosen a smaller diameter. I decided not to. However I could not have chosen a much larger diameter because it would not have fitted inside the system cylinder.

In fact a 0.9m diameter turbine is a giant by Pelton turbine standards. Lets look at the maths.

The equation for determining the mechanical power output in watts of a turbine is as follows:

Pmech (watts) = Fjet x Njet x pi x h x w x d / 60

Explanation:

Pmech = 170,000 watts (from the first calculation of maximum power output)

Fjet = Force in Newtons of the water striking the turbine = 10,006 Newtons (calculated above)

Njet = number of water jets = 1 jet nozzle

pi = 3.141592654

h = efficiency coefficient (unit-less fraction between 0 and 1). This is always going to be a figure between 0.69 and 0.94 for Pelton turbines. The larger they are, the more efficient they become, so 0.85 efficiency for a large turbine is an acceptable estimate = 0.85)

d = pitch circle diameter of the turbine in meters (this will be a slightly smaller diameter than the outer diameter of the turbine = 0.87m)

w = rpm (here not rad/s) which is the mystery value

The mechanical power output in watts is going to be approximately the same as the electrical power output in watts (if we make allowance for heat dissipation and the inefficiency of components other than the turbine itself which we can do at any later point).

Accordingly, applying the Pmech equation:

170,000 (watts) = Fjet (10,006 Newtons) x Njet (1) x pi (3.141592654) x h (0.85) x w (mystery value in rpm) x d (0.87m) / 60

= 10,006 X 1 X Pi x 0.85 x RPM x 0.87 / 60
= 23246 x w / 60
170,000 = 23246w / 60
170,000 = 387.4333w
w = 438.78 RPM

So the RPM figure look reasonable. It we had obtained a very high RPM figure this would been worrying because high RPM values would cause a turbine under this sort of force to fail after a few months of use.

Pressure calculations:

When considering the pressure in the two cylinders, the first thing we need to calculate is the pressure at the base of cylinder A (which is 25m high and full of seawater).

We can ignore the small air gap at the top of cylinder A for the moment because the air gap would reduce base pressure rather than increase it.

The formula for calculating the pressure at the bottom of a cylinder is as follows:

P = height(m) x density(kg/m3) x gravity (9.81m/s/s)
P = 25m x 1020kg/m3 x 9.81m/s/s
P = 250155 Pascals

However this is gauge pressure. We need to add atmospheric pressure to obtain the absolute pressure value of the fluid in the base of cylinder A.

Adding 101,325 Pascals of atmospheric pressure gives us an absolute pressure value at the base of Cylinder A of 351,480 Pascals.

This means that the tailgate water in Cylinder B must somehow force its way back into Cylinder A despite there being a pressure of 351.5 Kpa in cylinder A.

The pressure in Cylinder B, which is hermetically sealed and is not subject to atmospheric pressure, is due only to the height of the tailgate water contained in it. We have not switched on the air compressor yet.

The tailgate water is only 2.5m high. So the pressure at the base of cylinder B is only 25,000 Pascals. Even if it were also subject to atmospheric pressure (which it is not) it would only have a pressure of 125,325 Pascals.

However, the air compressor at the top of cylinder B comes to the rescue. It can pressurize the volume of air in tank B to 800,000 Pascals in 10.58 minutes.

The compressor in question is the Abac Genesis 1108 air compressor which can provide a maximum pressure of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute.

The volume of tank B (h=25m d=1m)
I calculated using the formula V= pi.r2.h
V= pi x (r x r) x h (where r = radius in metres and h= height in metres)
V = 3.141592654 x (0.5 x 0.5) x 20m = 15.7m3.

The volume of air in tank B (after deducting the tailgate water taking up 10% by volume of the cylinder) is 14.13m3.

The air compressor takes just over 8 minutes to pressurize the 14.13m3 of air inside tank B to 800,000 Pa.

In fact, the air compressor does not need to create 800,000 Pascals of pressure inside tank B.

It only needs to exceed the pressure at the base of Cylinder A (351,480 Pascals).

Once 351Kpa pressure has been exceeded, the system will try to equalise pressure in both of the connected cylinders as per the formula:

P1V1 = P2V2

This formula means that the pressure multiplied by the volume in one cylinder (P1 x V1) will always equal P2 x V2 in a connected vessel (unless some force prevents equalisation).

Here the force preventing equalisation is provided by the air compressor. The pressure relief valve in Tank A breaks the equalising pressure circuit from continuing its journey into tank B.

Note that the output of the pressure relief could be used to perform work if I need to make amendments to the schematic.

In any event, once the air compressor kicks in, tailgate water must move from the area of higher pressure (at the base of tank B) into the base of tank A (which has now become the lower pressure area).

So the tailgate water is forced through the lower connecting pipe back into tank A, whereupon the siphon recirculates it back into tank B.

The pressure relief valve at the top of tank A prevents the pressure in the air gap exceeding 350Kpa. So any compressed air or excess fluid forced into tank A (which will try and cause the pressure in tank A to become the same as in tank B) will be released by the pressure relief valve, thus ensuring no equalisation of pressure in the two tanks (or dangerous pressure build up in tank A).

This air compressor consumes 11kW of electricity when operating at maximum capacity.

Maximum capacity involves generating 800,000 Pascals of pressure.

The air compressor should be able to expel tailgate water from tank B at less than 50% of its operating capacity. In other words, it will consume approximately 5.5kW of power to pressurize tank B to just over 351,480 Pascals.

Only when the pressure in tank B falls below 351,480 Pascals will the float trigger the air compressor (only when the water level in tank B nears the turbine will the air compressor be activated).

However, even if the air compressor continuously consumed 11 kW, it would still consume only a small fraction of the 170kW output of the turbine.

By Get A Grip People on 9 May, 2011 - 11:41 pm

> Complicated Stuff ....

>However, even if the air compressor
>continuously consumed 11 kW, it would
>still consume only a small fraction of
>the 170kW output of the turbine.

So what you're proposing is a completely closed system which you put 11Kw into, but get 170kW out of? That's what engineers call a violation of Thermodynamics.

Or have you got a clearer explanation for an idiot like me of where the 159kW of energy actually comes from.

Note to moderators: Peg, maybe its time this thread just got locked.

By quantumtangles on 10 May, 2011 - 8:58 am

It is not a closed system. It is an open system. Both mass and energy may enter and leave the system boundary.

What interests me is why the system gives the appearance of being able to work. Just as perpetual motion wheels give this appearance but cannot work as all forces and all torques in the system are not taken into account by inventors.

Any fool can quote the laws of thermodynamics. But only people who actually understand them they have learnt parrot fashion. But only someone who actually understands them (as opposed to parroting what they think will be well received) will be able to pinpoint the precise reasons why this system does not work.

Doing so would be educational and very welcome to the inventor.

Consider it an IQ test.

If I understand your proposal correctly, you intend to use your siphon to fill up the bottom three feet of tank B with seawater, but then you will have to pressurize tank B to force out the tailgate seawater, meaning that water will not be flowing from tank A during the pressurization cycle.

So, you'll be moving a volume of seawater 3 meters tall by 1 meter in diameter at a rate of 1 cubic meter per second over your turbine. That means you're going to be moving about 2.35 cubic meters of water in about 2.35 seconds. Your theoretical output of 170 kw from the generator will only take place over those 2.35 seconds, but then your air compressor will have to draw 5.5 kw for at least 8 minutes.

It's been a long time since my high school physics class, but if I'm doing the conversions correctly, and if I'm understanding your plan correctly, your air compressor is going to draw about 7 times the kWh in 8 minutes as your turbine produces in 2.35 seconds. Even that is assuming that your turbine operates at peak efficiency for the entire 2.35 seconds, when I suspect that a good portion of the energy from the water flowing over it will be expended just to get the turbine spinning from a state of rest.

If you think about it, that only makes sense. You are converting the kinetic energy of water dropping 20 meters into electricity, but then you use electricity to raise a 22 meter water column an additional 3 meters (more or less). I understand about the cohesive properties of water that you described (the redwoods and all that), but that doesn't help the fact that you still have to inject all of your tank B water into the bottom of a water column in tank A. Injecting the water against the pressure of that water column is pretty much the same as lifting the water column.

You could probably make your idea closer to efficient by making tank B only 4 meters tall so that you don't have to pressurize as much volume. After all, you only need the pipe to drop 20 meters, not the whole tank. You would still need a valve to close your pipe before you begin pressurization of course. You could also save a bit of energy using a cascading system of air tanks to recapture some of the pressure from Tank B at the end of the pressurization cycle, thus reducing the workload of your air compressor a bit.

Still, you will not get over, or even near, 100% efficiency. Physics just can't get around the fact that you're trying to raise a 22 - 25 meter water column with the energy from a 20 meter waterfall.

By quantumtangles on 10 May, 2011 - 9:50 am

Quote
"If I understand your proposal
>correctly, you intend to use your siphon
>to fill up the bottom three feet of tank
>B with seawater, but then you will have
>to pressurize tank B to force out the
>tailgate seawater, meaning that water
>will not be flowing from tank A during
>the pressurization cycle."

That is very interesting. In fact it would be filling the bottom 2.5m of tank B with seawater. But yes you are correct. The water has to be forced out of tank B by pressure generated in tank B by the air compressor.

However the pressure in tank B can maintain over 350Kpa throughout the cycle (spending energy only as is required to maintain higher pressure than tank A).

Flow would not have to stop at any point in the cycle because the pressure relief value at the top of tank A prevents pressure equalisation between the two tanks.

This allows tank B to remain pressurised to over 350Kpa at the same time as tank A always has a pressure of less than 350Kpa.

You make an interesting further point about the energy needed to move the water.

But if the fluid must flow from high pressure to low pressure areas (eg from tank B to tank A) at the bottom of the system, and if the pump assisted siphon keeps water flowing at the top of the system, this is the point that needs clarification.

Many thanks for your objections.

By Steve Myres on 10 May, 2011 - 2:02 pm

I suspect that at least one of two things is the case.

Either

1) You will not be able to sustain a siphon from a low-pressure vessel into a high pressure one, even with a starter pump, or

2) You've under-estimated the hp requirement to lift the flow rate at the effective lift to get it back to the full tank.

By quantumtangles on 10 May, 2011 - 4:51 pm

You have thought the system through carefully and your objections are logically sound.

First objection: (whether the siphon can work if it flows from a low pressure to a high pressure area). The fact it is a siphon (with a longer column of fluid hanging over the high pressure area) may help, but you may right in spite of the powerful pump that assists the siphon.

Your second objection (about needing power to lift water) is also interesting. As devil's advocate, I argue that a castor oil system (where one could use less dense working fluid which would float up through the seawater in tank A because of positive buoyancy) illustrates that work does not have to be done to lift working fluid (though I accept there would be no energy gain by using less fluid). So it is not about lifting working fluid a certain distance. Rather it is all about how much work must be done to pressurise the working fluid so as to force it back into tank A.

However you may have identified the dilemma. One is forcing fluid to enter the base of tank A under high pressure, the same high pressure that may prevent the upper siphon from flowing at the top of the system. You are saying in effect that you cannot have it both ways (from high to low pressure and from low to high pressure) in terms of fluid flow and this is an interesting objection.

Thank you for thinking it through.

By Steve Myres on 10 May, 2011 - 6:33 pm

"So it is not about lifting working fluid a certain distance. Rather it is all about how much work must be done to pressurise the working fluid so as to force it back into tank A."

Well, when I said "lifting" I meant in the same more general sense in which you speak. If the liquid ends up at a higher elevation by the time force is no longer being applied to it, it has been "lifted" in that sense. The energy for the lift must be added to the fluid.

By quantumtangles on 10 May, 2011 - 11:38 pm

Yes I understand. I think your earlier post pinpointed why the system cannot work.

Fluid will indeed flow from the high pressure tank B to the lower pressure tank A, but the problem is in getting this fluid from the low pressure tank it is now in, back into the high pressure of tank B.

Also my calculation of the energy required to generate enough pressure to expel tailgate water is suspect. My figures for the air compressor are in cubic metres per minute, whereas the water flow rate is 1 cubic metre per second.

Accordingly, I have to ask myself whether constant high pressure in tank B would do all the required work (in terms of shifting water in the lower part of the system) or whether a cubic metre of air per second would be needed to expel a cubic metre of water per second (though I don't think the latter would be right as the fluid must flow from the high pressure to the low pressure area regardless of the work that was needed to create the pressure differential).

It is the fact of there being a pressure differential that is critical. How that pressure differential was generated is irrelevant. How much energy was needed to create the pressure difference does not matter (in the sense the fluid must move from the high to the low pressure area).

So the system fails not because the fluid wont move back into tank A via the lower pipe, but because the siphon wont work when tank B is at higher pressure than tank A.

And this is a reason for failure quite distinct from the more pressing issue of energy expenditure.

Thanks again.

>But if the fluid must flow from high
>pressure to low pressure areas (eg from
>tank B to tank A) at the bottom of the
>system, and if the pump assisted siphon
>keeps water flowing at the top of the
>system, this is the point that needs
>clarification.

Ok. I think this is your problem then. If you're going to maintain enough pressure in Tank B to force fluid into the bottom of the water column in Tank A, then that same pressure will also force the water in your siphon back into Tank A. The pump on your siphon will have to do just as much work to put fluids into tank B as your air compressor is doing to put fluids into the bottom of Tank A.

Thought this to be a moderated forum. Rather strange, hardly a useful subject for a technical forum

Are you familiar with the term 'comic relief'?

2 out of 2 members thought this post was helpful...

>Are you familiar with the term 'comic relief'?

Your friendly local moderator here. I've gotten serveral off line comments and complaints about the direction this thread has taken. Comic relief or not, I'm going not going to be posting any more messages to this thread. It think it's time has come and gone judging by the mail in my inbox.

Happy Wednesday everyone!
Peg Ferraro