# Is weight a better way to derive energy than flow?

• CherryB
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In summary, the conversation discusses the potential of using a system of falling buckets to produce energy as a replacement or enhancement for turbines in a hydroelectric dam. The calculation of total energy is relatively simple, but the efficiency of the system is difficult to determine. The inefficiencies include friction, water spillage, and limitations on flow rate. The maximum power is achieved when all the water is caught, but running slower or faster can cause inefficiencies. The efficiency of different types of water wheels is also discussed, with modern turbines being the most efficient at up to 83%. There is also mention of the theoretical limit of 100% efficiency for water turbines and the potential for even higher efficiency with implementation details.
CherryB
Gold Member
TL;DR Summary
figure out the total energy that a system of buckets, suspended on a chain system around pulleys, assuming the pulleys are 10 meters in radius, and that each bucket is 1000 cubic meters, falling for 30 meters. I used the simple mgh formula, but not sure how to factor in the radius. Also, how fast does the system have to run to produce maximum power and whether a gearbox might be used to take off power at slower revolutions
I am trying to figure out the total energy a system of falling buckets can produce, and whether it would be a system that can replace or enhance a system of turbines in a hydroelectric dam situation. I need to figure out the total energy that a system of buckets, suspended on a chain system around pulleys, assuming the pulleys are 10 meters in radius, and that each bucket is 1000 cubic meters, falling for 30 meters. I used the simple mgh formula, but not sure how to factor in the radius. Also, how fast does the system have to run to produce maximum power and whether a gearbox might be used to take off power at slower revolutions. Thanks

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The simplistic energy calculation is easy. The difficulty in this is calculating the efficiency. The water is falling at a certain velocity when moving with the bucket, and that energy is lost when it empties at the bottom (in addition to any height it falls). That's a source of inefficiency and is hard to calculate.

My understanding is modern turbines can be something like 90% efficient and that is going to be tough to beat.

CherryB
Turbines are very efficient as I understand, but they all require very large flows to work, along with high head. What would be the rough calculation to use to figure out the energy flows from turbines vs the buckets for the same amount of flow? would you use the mgh adjust for pulley radius and bucket speed. Also, why would you lose the energy during the emptying process? Is it because the water is flowing backwards out of the bucket?

Some inefficiencies are due to friction. Friction in that setup would depend on implementation details.

Another inefficiency would be caused if the buckets become full and some water spills out over the top of the bucket. To avoid that you need to move the buckets fast enough to catch 100% of the water and never overflow.

Maximum power happens when you catch 100% of the water. Running slower makes water spill. Running faster makes more friction.

Your idea is a variation of a water wheel. Here is a site that discusses efficiency and gives some estimates.

http://www.angelfire.com/journal/millbuilder/efficiency.html
Efficiency of Different Water Wheel Types
Flutter wheel....................20% or less
Undershot water wheel..............15% to 25% or less
Low breast shot water wheel................30% to 35%
Middle breast shot water wheel..............35% to 45%
High breast shot water wheel...............45% to 65%
Pitch-back water wheel.................55% to 65%
Overshot water wheel..................55% to 70%
Modern water turbine..................45% to 83%
Modern Fitz I-X-L steel overshoot water wheel
(depending upon the type of bearings)............73% to 93%

CherryB and russ_watters
Interesting. I did some searching and didn't find any images of the "Modern Fitz I-X-L steel overshoot water wheel". Does anybody know what it looks like, and why its efficiency is so high?

CherryB
berkeman said:
Interesting. I did some searching and didn't find any images of the "Modern Fitz I-X-L steel overshoot water wheel".
As I see, due the nature of that 'Angelfire' page (collection of old resources) actually that table about the efficiency might be even a century old: you should adjust the 'modern' part accordingly...

CherryB and anorlunda
CherryB said:
Turbines are very efficient as I understand, but they all require very large flows to work, along with high head.
As demanded by conservation of energy, yes.
What would be the rough calculation to use to figure out the energy flows from turbines vs the buckets for the same amount of flow? would you use the mgh adjust for pulley radius and bucket speed.
Yes, mgh... but I don't see why pulley radius matters. bucket speed insofar as it tells you mass flow rate, yes.
Also, why would you lose the energy during the emptying process? Is it because the water is flowing backwards out of the bucket?
It depends where the water level is at the bottom. If the buckets are submerged they act like paddles. If not, they drop the water and you aren't using all the available "h".

CherryB
berkeman said:
Interesting. I did some searching and didn't find any images of the "Modern Fitz I-X-L steel overshoot water wheel". Does anybody know what it looks like, and why its efficiency is so high?
As far as I can tell, the theoretical limit for a water turbine should be 100%, so it doesn't surprise me they can achieve above 90%. This is unlike a wind turbine where the limit is 59% or 61% depending on the theory applied.

anorlunda
berkeman said:
Modern Fitz I-X-L steel overshoot water wheel
It's possible that web page was written by Fitz I-X-L

russ_watters said:
As far as I can tell, the theoretical limit for a water turbine should be 100%, so it doesn't surprise me they can achieve above 90%. This is unlike a wind turbine where the limit is 59% or 61% depending on the theory applied.

It's worth noting that this is just because you have constrained flow rather than free flow - you can get an air turbine to much higher than 59.3% (where are you getting 61 out of curiosity? I haven't seen that value before) if the air is constrained and thus unable to flow around the turbine. I agree that you should be able to achieve almost arbitrarily high efficiency though.

My google gets plenty of image hits for "I-X-L steel overshot water wheel":

Since turbines and stuff like that is well out of my wheel house, is a water wheel a reaction or impulse device?!

Must admit had a bit of a chuckle at the "theoretical upper limit of water turbine is 100%", isn't this the theoretical upper limit for everything? I think someone even wrote it into law.

CherryB and anorlunda
essenmein said:
Must admit had a bit of a chuckle at the "theoretical upper limit of water turbine is 100%", isn't this the theoretical upper limit for everything? I think someone even wrote it into law.
Depends on the theory/law being applied. If you consider only conservation of energy, sure, but here we generally go the next step and apply a theory/law specific to the process. The wind turbine is one such example. Carnot's theorem is another. Usually this second level still includes assumptions like zero friction/electrical resistance, no heat loss, etc.

cjl said:
(where are you getting 61 out of curiosity? I haven't seen that value before)
Turns out I misread, and I was in a hurry so I didn't look hard enough (I remembered 59% too). It's in the wiki:
In 2001, Gorban, Gorlov and Silantyev introduced an exactly solvable model (GGS), that considers non-uniform pressure distribution and curvilinear flow across the turbine plane (issues not included in the Betz approach).[9] They utilized and modified the Kirchhoff model,[10] which describes the turbulent wake behind the actuator as the "degenerated" flow and uses the Euler equation outside the degenerate area. The GGS model predicts that peak efficiency is achieved when the flow through the turbine is approximately 61% of the total flow which is very similar to the Betz result of 2/3 for a flow resulting in peak efficiency, but the GGS predicted that the peak efficiency itself is much smaller: 30.1%.

Interesting. I wonder what went wrong with the GGS model, because we have operational wind turbines running much higher than 30.1% (unless they're defining efficiency differently somehow?).

russ_watters
cjl said:
Interesting. I wonder what went wrong with the GGS model, because we have operational wind turbines running much higher than 30.1% (unless they're defining efficiency differently somehow?).

How do they determine effy? Is it the cylinder of moving air, with diameter of the turbine, or something else? Seems that the wind turbines I've seen around have large windows between the blades where air can move through without being impeded by the blades, compared say turbine in a jet engine where there is no line of sight through the blades?

Usually, it's defined by looking at the incoming kinetic energy of the air that would pass through the total rotor area (if the turbine were not there) and determining what percentage of that kinetic energy gets transformed to electricity, and by that measure, modern turbines are around 40-50% efficient. Also, the gaps between the blades are not as much of a problem as you might think - the turbine actually does interact with nearly all of the air that passes through it. The way it does this is because the blade speed is much higher than the wind speed, so air passing "between" the blades will still have a blade pass by relatively close in front of or behind it at some point.

russ_watters and essenmein
essenmein said:
Since turbines and stuff like that is well out of my wheel house, is a water wheel a reaction or impulse device?!
It could be either. A pelton wheel (below) is also a kind of water wheel, or turbine.

In fact, as I think about it, the dividing line between water wheels and water turbines is fuzzy. There are many variations and much ingenuity has been applied over the centuries.

You may think that a turbine is always an axial flow device resembling a propeller. But the radial flow Frances turbine is very common. A year or two ago, we had a thread on PF about radial flow water wheels.

russ_watters
In the technical sense a water wheel is a turbine.

I had to look this up lol. Define turbine:
"A turbine (from the Latin turbo, a vortex, related to the Greek τύρβη, tyrbē, meaning "turbulence") is a rotary mechanical device that extracts energy from a fluid flow and converts it into useful work"

CherryB, russ_watters and anorlunda
Yes. I mean as in a vortex driven device, and there is turbulent.com that made such a device, and I think it’s fantastic. Only problem is that I don’t have the water flow that it requires, and for that reason I’m forced to explore the use of a water wheel, or, rather, a vertical conveyor System, with buckets attached to the chain, as shown. Ok. I guess my refined question would run as such. If you had to pass through the gates as much water as falls through the Hoover dam, and setting aside construction and deployment hurdles for now. Which system would bring in more usable kWatts. And, also, am I asking the right question?

russ_watters said:
The simplistic energy calculation is easy. The difficulty in this is calculating the efficiency. The water is falling at a certain velocity when moving with the bucket, and that energy is lost when it empties at the bottom (in addition to any height it falls). That's a source of inefficiency and is hard to calculate.

My understanding is modern turbines can be something like 90% efficient and that is going to be tough to beat.

I agree with that Russ, but i’m asking 90% of what? Is that most we can do with that system?

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anorlunda said:
Some inefficiencies are due to friction. Friction in that setup would depend on implementation details.

Another inefficiency would be caused if the buckets become full and some water spills out over the top of the bucket. To avoid that you need to move the buckets fast enough to catch 100% of the water and never overflow.

Maximum power happens when you catch 100% of the water. Running slower makes water spill. Running faster makes more friction.

Your idea is a variation of a water wheel. Here is a site that discusses efficiency and gives some estimates.

http://www.angelfire.com/journal/millbuilder/efficiency.html

What if you were able to control the flow to reduce spillage to no more than 10% of flow volume, let the water exit the bucket always above the tail waters, and the two sprockets at the top and bottom were >=10 meters in radius. I think that the greater the radius the greater the torque acting on the shaft..., and say the buckets were falling at a constant speed. How would you calculate that?

I would think an increase cavitation and turbulence in the water would sap some energy that would otherwise be available.

CherryB said:
I agree with that Russ, but i’m asking 90% of what? Is that most we can do with that system?
90% of e=mgh -- you had the principle correct.

CherryB said:
If you had to pass through the gates as much water as falls through the Hoover dam, and setting aside construction and deployment hurdles for now. Which system would bring in more usable kWatts.
That make it less clear what you are asking.

I would say the answer to this new question is "The system that the Hoover Dam engineers chose, Francis turbines. Why would they choose anything less than the best possible?"

CherryB said:
Only problem is that I don’t have the water flow that it requires

You have to first identify the potential of your water flow. You do that by comparing power (as oppose to torque or energy).

If you have a river with a given mass flow rate (##\dot{m}##, in kg/s) and the river flows at velocity ##v_0##, then the power ##P## you can harvest is ##P = \frac{1}{2}\dot{m}\left(v_0^2 - v_f^2\right)## (in Watts), where ##v_f## is the final velocity of the river after you extracted the power. Obviously, to extract the maximum power, ##v_f## should be zero.

That is based on the kinetic energy (##\frac{1}{2}mv^2##). But potential energy (##mgh##) can be converted into kinetic energy, or ##\frac{1}{2}v^2 = gh##. Therefore, the previous power equation can be rewritten like so:

$$P = \dot{m}g\left(h_0 - h_f\right)$$

Where ##h_0## and ##h_f## are the initial and final heights (in meters) of the water fall.

You cannot do better than this, no matter how you rearrange your machine to extract the power from the flow.

CherryB said:
I think that the greater the radius the greater the torque acting on the shaft

The problem with increasing the radius is that, yes, you will increase the torque ##T##, but you will also proportionally reduce the rpm ##\omega##. The power output of the machine will still be ##P = T\omega## which - assuming 100% efficiency - will equal the power extracted from your river flow. The torque and rpm output of your machine only depends on gearing, but one thing is for sure, you will always have the same power.

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dlgoff, CherryB, russ_watters and 1 other person
Ok, so here's what I got so far. The one water filled bucket, falling from the top of the upper Sprocket, "uS," will exert a certain maximum downward force upon the chain it is attached to, which will pull on the uS, and force it to follow the chain that is being acted upon. The inner shaft of the uS, which is connected to a generator, will spin the shaft of that generator, and electrical current will be produced, which can then be used.

The bucket on the right side of uS, and is at 30m, and falling at a constant speed (determined by the braking power of the generator - similar to the braking power of a regen system in electric vehicles). It is 1000 cubic meters in volume, which translates to 1,000,000 liters, or 1 million kg, has a maximal potential of (1,000,000kg x 9.81m/s^2 x 30m), or 294,300,000, or 294.3 megawatts.

That figure of 294.3MW is the maximum power that could be generated by this system.
Is that correct?
Would the speed of the bucket be a factor?
Would the distance of the bucket from the shaft be a factor?

And the factors that will inhibit that maximal output include
-keeping the water in the bucket for as long as possible
-reducing the friction, heat loss, etc. of the chain to the sprocket
-overcoming the cogging effect, and electrical resistance of the generator

Thanks for all the help.

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CherryB said:
The bucket on the right side of uS, and is at 30m, and falling at a constant speed (determined by the braking power of the generator - similar to the braking power of a regen system in electric vehicles). It is 1000 cubic meters in volume, which translates to 1,000,000 liters, or 1 million kg, has a maximal potential of (1,000,000kg x 9.81m/s^2 x 30m), or 294,300,000, or 294.3 megawatts.

That figure of 294.3MW is the maximum power that could be generated by this system.
Is that correct?
Would the speed of the bucket be a factor?
Yes, that math skips a step to get to megawatts: diving energy by time. The number without units isn't megawatts, it is joules. Megawatts is joules per second, so the number in megawatts is the power if the bucket travels down the 30m in one second.
Would the distance of the bucket from the shaft be a factor?
No.

And the factors that will inhibit that maximal output include
-keeping the water in the bucket for as long as possible
-reducing the friction, heat loss, etc. of the chain to the sprocket
-overcoming the cogging effect, and electrical resistance of the generator

CherryB
Ah. So there’s the tie-in to momentum. So if it traveled the 30m in 1 minute then it would be 294.3 mw divided by 60?

CherryB said:
So if it traveled the 30m in 1 minute then it would be 294.3 mw divided by 60?
Yes.

CherryB
Well, I guess you can adapt some of those calculations to your case, but I have to agree with @anorlunda that even if the text is not directly from the inventor himself, it is likely from somebody closely involved - since it has some strong smell of advertisement

russ_watters
CherryB said:
Ah. So there’s the tie-in to momentum. So if it traveled the 30m in 1 minute then it would be 294.3 mw divided by 60?
Yep. It's worth noting that this is per bucket as well, so if you had 10 full buckets descending at once, it would be 294.3/60 per bucket * 10 buckets for a total of around 50 MW.

CherryB
russ_watters said:
As demanded by conservation of energy, yes.

Yes, mgh... but I don't see why pulley radius matters. bucket speed insofar as it tells you mass flow rate, yes.

It depends where the water level is at the bottom. If the buckets are submerged they act like paddles. If not, they drop the water and you aren't using all the available "h".
I guess I'm confusing moment with force, and translating that conflation into energy, because I can definitely hold a 10 lb barbel in my hand indefinitely if it's close to my body, but I can't do the same if my arm is outstretched. Seems to make some sense for me viscerally

russ_watters
cjl said:
Yep. It's worth noting that this is per bucket as well, so if you had 10 full buckets descending at once, it would be 294.3/60 per bucket * 10 buckets for a total of around 50 MW.
Right. So that's quite a bit of power, but it just seems like an awfully fast rotation to keep up with mechanically. I'm guessing that I would have to utilize a gearbox like the ones they use for wind turbines. I saw something about a 600 shaft rpm turbine gearbox yielding 8mw, but I'm not sure how to calculate the shaft input power needed to turn the gearbox

A full minute of descent time to cover 30m seems fast to you? Sure, you can gear them however you want, but if anything, I'd actually expect you'd want to gear it up, since generators are smaller, cheaper, and more efficient if spun faster.

EDIT: Also, the power you get out of a gearbox is the same as the power you put in (minus losses). If the gearbox is outputting 8MW at 600 RPM, the input will also be 8MW (at some other RPM - probably around 9RPM for a turbine that size).

russ_watters
<h2>1. Is weight a more accurate measure of energy than flow?</h2><p>No, weight is not a more accurate measure of energy than flow. While weight is a measure of the force of gravity on an object, energy is the ability to do work or cause change. Flow, on the other hand, is a measure of the rate at which a substance moves through a system. Both weight and flow are important factors in determining energy, but they are not interchangeable.</p><h2>2. Can weight be converted directly into energy?</h2><p>No, weight cannot be directly converted into energy. Weight is a measure of mass, while energy is a measure of the ability to do work. In order for weight to be converted into energy, it must go through a process such as combustion or nuclear reactions.</p><h2>3. How does weight affect energy production?</h2><p>Weight can affect energy production in various ways. For example, in a hydroelectric power plant, the weight of water is used to turn turbines and generate electricity. In a nuclear power plant, the weight of uranium atoms is used to produce heat and generate electricity. However, weight alone is not enough to produce energy - it must be combined with other factors such as flow or chemical reactions.</p><h2>4. Is weight a more important factor than flow in energy production?</h2><p>It depends on the specific type of energy production. In some cases, weight may be a more important factor, such as in the examples mentioned above. In other cases, flow may be a more important factor, such as in wind or solar energy production. Ultimately, both weight and flow play important roles in energy production and cannot be compared as one being more important than the other.</p><h2>5. How does the weight of an object affect its potential energy?</h2><p>The weight of an object does not directly affect its potential energy. Potential energy is determined by the position of an object in a system and its ability to do work. However, the weight of an object can indirectly affect its potential energy by influencing its position and the forces acting on it within a system. For example, a heavier object may have more potential energy at the top of a hill than a lighter object due to the force of gravity acting on it.</p>

## 1. Is weight a more accurate measure of energy than flow?

No, weight is not a more accurate measure of energy than flow. While weight is a measure of the force of gravity on an object, energy is the ability to do work or cause change. Flow, on the other hand, is a measure of the rate at which a substance moves through a system. Both weight and flow are important factors in determining energy, but they are not interchangeable.

## 2. Can weight be converted directly into energy?

No, weight cannot be directly converted into energy. Weight is a measure of mass, while energy is a measure of the ability to do work. In order for weight to be converted into energy, it must go through a process such as combustion or nuclear reactions.

## 3. How does weight affect energy production?

Weight can affect energy production in various ways. For example, in a hydroelectric power plant, the weight of water is used to turn turbines and generate electricity. In a nuclear power plant, the weight of uranium atoms is used to produce heat and generate electricity. However, weight alone is not enough to produce energy - it must be combined with other factors such as flow or chemical reactions.

## 4. Is weight a more important factor than flow in energy production?

It depends on the specific type of energy production. In some cases, weight may be a more important factor, such as in the examples mentioned above. In other cases, flow may be a more important factor, such as in wind or solar energy production. Ultimately, both weight and flow play important roles in energy production and cannot be compared as one being more important than the other.

## 5. How does the weight of an object affect its potential energy?

The weight of an object does not directly affect its potential energy. Potential energy is determined by the position of an object in a system and its ability to do work. However, the weight of an object can indirectly affect its potential energy by influencing its position and the forces acting on it within a system. For example, a heavier object may have more potential energy at the top of a hill than a lighter object due to the force of gravity acting on it.

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