Is weight a better way to derive energy than flow?

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CherryB

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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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russ_watters

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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

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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?
 
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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.

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%
 

berkeman

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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?
 
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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... :woot:
 

russ_watters

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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".
 

russ_watters

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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.
 

cjl

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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.
 
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My google gets plenty of image hits for "I-X-L steel overshot water wheel":
1573070283254.png


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. :oldshy:
 

russ_watters

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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. :oldshy:
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.
 

russ_watters

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(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%.
 

cjl

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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?).
 
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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?
 

cjl

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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.
 
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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.

1573071741999.png


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.

1573071945151.png
 
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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

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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?
 

CherryB

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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.
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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CherryB

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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.

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?
 

Paul Colby

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

russ_watters

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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.
 
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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?"
 

jack action

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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.

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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