Little hydropower question.

Thread Starter

praondevou

Joined Jul 9, 2011
2,942
I have a friend who insists in an idea and I want to prove to him that it cannot work because it would violate basic physical laws.

But somehow I don't find the right words.

He says if he drilled a well of 2km into the earth and at the bottom there was a pump that pumped water through a pipe back into to the upper end of the well he could create more electric energy than the pump would be using.

He would install a few, not just one, turbines into the well, where the sum of energy generated by all turbines is greater than the energy needed to pump all the water up.

Now according to this:

A simple formula for approximating electric power production at a hydroelectric plant is:
where P is Power in kilowatts, h is height in meters, r is flow rate in cubic meters per second, g is acceleration due to gravity of 9.8 m/s2, and k is a coefficient of efficiency ranging from 0 to 1.

the only parameter that is different for all turbines is the height. Unless the flow rate is changing.

It looks like very easy to explain why it cannot work, yet...:confused:
 

KJ6EAD

Joined Apr 30, 2011
1,581
Invite your friend to bicycle up a small hill then coast down it repeatedly to increase his energy...or his understanding of practical physics, whichever comes first. :)
 

steveb

Joined Jul 3, 2008
2,436
It looks like very easy to explain why it cannot work, yet...:confused:
The simplest explanation comes from the concept of conservation of energy. He needs to tell you what the energy source is. What is so special about 2 km depth? Is he using thermal energy of the earth or tapping another energy source? If not, how is 2 km any different than 10 ft depth? Gravity is a conservative force. You can use existing gravitational potential energy, but you can't recycle it to generate over unity power utilization. You'd have to suck the energy out of something and that something will eventually be depleted. However, if he's thinking of tapping thermal energy, he could have something valid in principle because the earths internal energy is a huge energy reservoir that won't be depleted for billions of years.
 

THE_RB

Joined Feb 11, 2008
5,438
I think he's making the point that gravity is less when you are at 2km depth.

However the difference is gravity is equalised at every point along the way so there is no convenient way to move mass from the higher gravity zone to the lower gravity zone without it being equalised along the way or by cyclic movement.
 

strantor

Joined Oct 3, 2010
6,798
send him to the wikipedia page for perpetual motion machines. if he reads that and still insists his principle is sound then he's an overunity nut. If he's an overunity nut then there's no convincing him. Your only option would be to ask him to demonstrate in small scale; if he does, then it won't work, and he will blame the scale. He will insist that only if you drill down to 2km would it work, and since that will never happen, you will have to settle for a "stalemate". Just have him educate himself and see where it goes from there.
 

steveb

Joined Jul 3, 2008
2,436
Thanks guys.

But I would like to use formulas to prove it to him. Sometimes words are not enough.
Before you can make the formulas, you have to know the principle of operation. It's still not clear what the principle of operation is. Is it only gravitation potential energy being tapped, or is there more involved? If it is only gravitational energy, then one simply integrates mass times gravitational acceleration (g) over the height change. If the principle is what THE RB suggested above, then g is a function of height. Including friction in the equations is not all that easy, but the easiest thing to do is to show that even assuming no losses, the work needed to pump the water up equals the work you can do letting the water flow back down. The energy depends only on the height and not on the exact path travelled. This is the definition of a conservative force and gravity is a conservative force.

Really, the statement that gravity is a conservative force is more powerful than any equations. This statement tells anyone that the best you can do is break even.
 

steveb

Joined Jul 3, 2008
2,436
By the way, why dig a hole of 2 km when you can use a mountain? This is only 2000 m and there are mountains of greater height. I still wonder if his principle is to tap thermal energy.
 
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Thread Starter

praondevou

Joined Jul 9, 2011
2,942
but the easiest thing to do is to show that even assuming no losses, the work needed to pump the water up equals the work you can do letting the water flow back down.
He cannot be convinced by "words". I have to put formulas on the table.
The principle is based on height and gravitational force, that's what he says.

He claims to have heard that a group of engineers from our electric utility invented such a device and when they saw that it can generate more energy than was put in they shut the whole project down and bla bla bla. This is of course BS, as we all know.

I have yet to find the exact formulas to prove it to him. It's been a long time that I used them (in school) ;)
 

evilclem

Joined Dec 20, 2011
118
I'd look into the energy required to pump water 2km high (20 MPa pressure).

Then look at how much energy can be derived from the turbines.

"A simple formula for approximating electric power production at a hydroelectric plant is:
where P is Power in kilowatts, h is height in meters, r is flow rate in cubic meters per second, g is acceleration due to gravity of 9.8 m/s2, and k is a coefficient of efficiency ranging from 0 to 1. "

h*g here refers to the weight of water on the turbine (Pressure). The flow rate needs careful consideration as each turbine (assuming multiple turbines) will have a resistance to the flow of the water. This flow rate however cannot be higher than that of the 20 MPa pump mentioned earlier.

The other problem is that gravity will not carry the water 2km down the hole, I don't know what it's like in your part of the world but here in Australia I'm not aware of any places where the groundwater level is that deep.
 

steveb

Joined Jul 3, 2008
2,436
He cannot be convinced by "words". I have to put formulas on the table.
The principle is based on height and gravitational force, that's what he says.

He claims to have heard that a group of engineers from our electric utility invented such a device and when they saw that it can generate more energy than was put in they shut the whole project down and bla bla bla. This is of course BS, as we all know.

I have yet to find the exact formulas to prove it to him. It's been a long time that I used them (in school) ;)
OK, if it's gravitational force and based on the idea that the height change is sufficient to change gravitational acceleration, then the Newtonian Gravitational force formula \( F={{k m_1 m_2}\over{r^2}}\), with \( W=\int F \cdot dl \) can be used to calculate work or energy. Then power is dW/dt, the rate of change of work or energy.

If the height change is small enough to consider gravitational acceleration to be constant, then the simple F=mg (W=mgh) is sufficient.

If the height change is only 2 km out of an earth radius of over 6000 km, I would consider g to be constant. Hence, the work needed to pump the water Wp is greater than mgh and the work obtained from the water going back down Wd is less than mgh. The greater and less than signs are the result of friction. Hence Wp > Wd.

What many people don't seem to grasp is that equations are just a language, just as words and grammar are. If words don't convince him, then equations won't convince him. The words and the equations are consistent with a basic empirical fact. That is, all experiments show gravity to be a conservative force, hence you can't recycle it to obtain over-unity.
 

steveb

Joined Jul 3, 2008
2,436
I can see the T-shirt slogan now "dig a hole, recycled gravity, the ultimate green energy".
And if it were true, human nature would result in a lot of digging, hence causing the next global threat ... "Too many holes in the ground !!!", to be featured on future T-shirts. :p
 
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Thread Starter

praondevou

Joined Jul 9, 2011
2,942
Mmh. Let's assume acceleration and water density are fixed parameters and efficiency of all components is 1.

The formulas for pumps or turbines are similiar. Leaving aside water density and acceleration and losses it is height difference x flow rate.

I'm having difficulties to determine the flow rate through multiple turbines in a vertical pipe...

It doesn't need to be 2km. I know it doesn't matter because it doesn't work anyway.
 

steveb

Joined Jul 3, 2008
2,436
I'm having difficulties to determine the flow rate through multiple turbines in a vertical pipe...
How are the turbines configured? Series or parallel? Do you split the flow into separate pipes, or do you let the same flow through all turbines.

Note that your original formula has r as volume flow rate, but it should be mass flow rate. They differ by a factor of the density.

If you are making a cyclic system, then the same total mass flow rate exists at all stages of the system. If the flow is split in parallel, then the separate paths must add up to the total flow. If the turbines are in series, then the mass flow rate is the same through them all.

As you mentioned, pumps and turbines are somewhat similar and each has an efficiency factor less than one, which makes it clear that overunity is not possible. However, be careful with your calculations. Turbines operate on kinetic energy, but your original formula is a potential energy formula. There is an equality between them if efficiency is factored in, but just be careful about how you interpret potential energy and kinetic energy.

The power formula for a turbine operating on kinetic energy is

\( P=0.5 \alpha \rho A v^3\), where alpha is the efficiency factor, rho is the water density, A is the flow cross sectional area and v is the flow speed. Volumetric flow rate through the turbine is \( A v\) and mass flow rate is \( \rho A v\).

Also, keep in mind the Betz limit on a simple turbine is \(\alpha=0.59\). Hence, there is still kinetic energy in the flow after the turbine. Perhaps this is the reason for multiple turbines? I'm not sure, but if it is, there is still a limit to how much you can recover as turbines get less efficient when the flow velocity is small. Basically, a lot of kinetic energy will be wasted at the bottom. This is one of many loss mechanisms in such a system.
 

Thread Starter

praondevou

Joined Jul 9, 2011
2,942
How are the turbines configured? Series or parallel? Do you split the flow into separate pipes, or do you let the same flow through all turbines.

Note that your original formula has r as volume flow rate, but it should be mass flow rate. They differ by a factor of the density.

If you are making a cyclic system, then the same total mass flow rate exists at all stages of the system. If the flow is split in parallel, then the separate paths must add up to the total flow. If the turbines are in series, then the mass flow rate is the same through them all.

As you mentioned, pumps and turbines are somewhat similar and each has an efficiency factor less than one, which makes it clear that overunity is not possible. However, be careful with your calculations. Turbines operate on kinetic energy, but your original formula is a potential energy formula. There is an equality between them if efficiency is factored in, but just be careful about how you interpret potential energy and kinetic energy.

The power formula for a turbine operating on kinetic energy is

\( P=0.5 \alpha \rho A v^3\), where alpha is the efficiency factor, rho is the water density, A is the flow cross sectional area and v is the flow speed. Volumetric flow rate through the turbine is \( A v\) and mass flow rate is \( \rho A v\).

Also, keep in mind the Betz limit on a simple turbine is \(\alpha=0.59\). Hence, there is still kinetic energy in the flow after the turbine. Perhaps this is the reason for multiple turbines? I'm not sure, but if it is, there is still a limit to how much you can recover as turbines get less efficient when the flow velocity is small. Basically, a lot of kinetic energy will be wasted at the bottom. This is one of many loss mechanisms in such a system.
Mmh, I really appreciate your help. It's been really long that I had to work with mechanical parameters.

I think he meant the turbines are in series, in the same pipe but just in different heights.

Potential energy is transformend into kinetic energy whenthe water flows downwards due tue gravity, determined by the acceleration.

I have to read up all this stuff again.. :-/

I tried to ask for a formula on a physics forum but they immediatly shut it down. Overunity not allowed. Maybe they should distinguish between a discussion and cases like mine.

I want to help someone else not to get ripped off by some magical free energy ebook salesman...Instead of prohibiting these issues we should do all to teach people why it cannot work. this is not a waste of time.

So I guess I have to read up on basic physics again. :-/

Energy needed to pump up a mass of water vs energy created by turbines the same mass of water flows through...
 

Thread Starter

praondevou

Joined Jul 9, 2011
2,942
I think I know more or less how to explain it. Not with numbers though.
If we go back to the formula here: http://en.wikipedia.org/wiki/Hydropower#Calculating_the_amount_of_available_power

I guess the best way to logically explain it to my friend is that the height change available for energy retrieved at the first turbine is not the total height, it's only the height from the head of the well to the turbine.
The water then gets slowed down, because cinetic energy is transformed into electric energy.

That means for each step, from one turbine to the other there is only the energy available given by the distance between them... water accelerates, water slows down.

Something like this. In other words the total height/potential energy is divided into smaller individual heights each of them being responsible for only part of the total energy to be delivered.

Thanks all
 

Wendy

Joined Mar 24, 2008
23,421
The other thing that hasn't been mentioned is conversion efficiencies. Nothing is 100% efficient, and if you are doing 90% you are doing well.
 
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