Naked Science Forum
On the Lighter Side => New Theories => Topic started by: YourUncleBob on 07/05/2008 05:53:43
-
I've been working with my material science students on creating an artificial tree encased in a glass-like sphere to lift water over 300 meters.
Our goal is to take energy from the sun in the form of heat to create electricity. Our basic design involves several of these 'trees' releasing the water they collect at the bottom of their glasslike domes into a central reservoir.Once the reservoir reaches a predetermined level, the water is released, passes through a turbine before being deposited in another reservoir under their 'roots'. This is a complete closed system.
Obviously there has been a lot of interesting problems to overcome!
But one of our biggest problems right now is getting the balance right between evaporation and solute concentration. The evaporation rate from our artificial leaves has to be such that the concentration levels of salt&sugar in our respiratory system don't get too dense and clog up the works!
I posted a question on this in the plant section but with no replies.
We've yet to test the fluid dynamics of a solute heavy liquid travelling down a 50 meter pipe (nevermind 300meter) into a pure water reservoir (where the pipe's 'skin' is semipermeable) then back up 50 meters.
Would the flow offset the 'osmotic push'?
Would we have fresh water pushing against the downward flow of the solutes?
Or would the two forces combine to provide some additional upward lift?
Your thoughts would be much appreciated.
-
Here's a diagram of one of our artificial trees, the latice of artificial leaves is not shown.
A few things to note.
A natural Tree 'tries' not to lose it's water and yet releases anywhere from 500 liters to a 1000 liters a day (less so at night). Obviously we want our artificial tree to release as much water as possible and as long as the temperature difference inside our spheres compared to the outside remains adequate we'll get evaporation 24 hours a day.
We've been trying to increase the evaporation rate of our artificial leaves 100 times over. Giving us up to 100,000 liters for each tree per day.
Any thoughts would be appreciated.
-
I am not sure I entirely understand, are you essentially generating electricity by evaporating water from a very concentrated solution, condensing it somewhere with no solute and then generating the power by letting it fall to the bottom of the tree, only to be lifted back up again by osmotic pressure?
If so my first reaction is that is is going to be hideously inefficient, the latent heat of vaporisation for water is 2.2MJ per kg... if you can lift the water up 300m that means you are going to get out at the most 300J/kg which gives you a thermodynamic efficiency of about .015% so if you have on average about 250W/m2 of sunlight hitting a piece of ground,you are going to generate about .04W/m2...
If I have misunderstood what you are doing I apologise...
-
Thanks For your comments Dave,
I am not sure I entirely understand, are you essentially generating electricity by evaporating water from a very concentrated solution, condensing it somewhere with no solute and then generating the power by letting it fall to the bottom of the tree, only to be lifted back up again by osmotic pressure?
The evaporation is from a light solute, leaving a heavier solute to continue flowing down around and with an added osmotic push back up. The design goal of our 'leaves' is to release as much water vapor as possible into our 'dome'. The hydrophilic beads help attract the water vapor to the inside of the dome's surface, the beads themselves help create a 'dewpoint' by conducting the lower air temperature outside the dome. As the droplets fall they collect in the hydrophobic chutes and spiral down and out at the bottom of the dome, creating inner air currents helping to keep the humidity at a level where evaporation is still possible.
The water from 8 of these 'domes' collates in a central reservoir, where once the level reaches 800,000 liters a ballcock valve is triggered and the water falls 250 meters before passing through a turbine creating electricity. Finally this water is fed back into the central 'root' reservoir connecting all the 'trees'
If so my first reaction is that is is going to be hideously inefficient, the latent heat of vaporisation for water is 2.2MJ per kg... if you can lift the water up 300m that means you are going to get out at the most 300J/kg which gives you a thermodynamic efficiency of about .015% so if you have on average about 250W/m2 of sunlight hitting a piece of ground,you are going to generate about .04W/m2...
The evaporation from our artificial leaves require only a low heat as they separate the water in very small droplets much the same way natural leaves do. If the internal temperature of the dome stays above 3 or 4 degrees Celsius the system still flows at night without any sunlight.
Dave if we have 800,000 liters of water falling 250 meters through a pipe 20 cm in diameter hitting a turbine how much electricity would we generate? And what is the optimum diameter of the pipe?
yours sincerely Blaine
-
Looks like the flow and return system from a central heating unit that has no pump and relies on density changes generated by heat and cooling generated by the water tank, usually in the loft.
Something I have been saying goes on in trees but using evaporation to change density and also the cooling at the leaves to change density giving us a powerful flow and return system capable of turning a small rotor blade.
I would like to take a look at this when its running.
-
Dave if we have 800,000 liters of water falling 250 meters through a pipe 20 cm in diameter hitting a turbine how much electricity would we generate? And what is the optimum diameter of the pipe?
I realized after rereading this that I might have seemed a bit lazy here, but I've gotten into the habbit of asking questions of my students when I've already an idea of the answer just to keep their interest up!
So here goes!
To get an idea of how much power in kilowatt-hours can be derived from hydropower we need to know the height, flow and efficiency of the system.
Flow is the tricky one as anyone who has studied fluid dynamics in pipes will attest!
To get an estimate on how long it takes for something to fall from a given height to the ground we can use;
Time = SquareRoot ((Height/(0.5*Gravity))
= SquareRoot ((250meters/(0.5*9.8))
7.142 seconds
Now obviously we need to discount air resistance/drag and the friction from the pipe's
walls!
If the pipes inner walls were coated with a hydrophobic material a lot of this friction could be offset.
A single droplet of water naturally evolves into the most aerodynamic shape possible as it moves through the air, it is very difficult to estimate how aerodynamic water falling down a pipe would be, however.....
let's estimate a drag coefficient of Cd=0.18
How fast is the water moving at the bottom of the pipe?
Well let's see how far the water has fallen at time 6.142 seconds, 1 second before exiting.
Height = 0.5*Gravity*(time Squared)
= 184.848 meters
So the water's velocity if we just averaged the acceleration rate after 6.142 seconds at the end of the pipe would be (250-184.848)= 65.15 meters per second.
Now just to be clear let's look at Time 7.142 seconds and Time 8.142 seconds
@7.142 seconds = 250 meters
@8.142 seconds = 324.83 meters
So the water's velocity if we just averaged the acceleration rate after 7.142 seconds the water's velocity after falling 324.83 meters would be (324.83-250)= 74.83 meters per second.
Now we have a velocity between 65.15 meters per second and 74.83 meters per second to base our estimates on so let's take an average of 70 meters per second.
Okay now these figures don't take into account air resistance, air pressure, humidity and temperature!
Mass doesn't come into the equation if we don't think about an object falling through air, but here we have a column of water 250 meters high with a diameter of 20 cm.
Cylinder pi * radius2 * height
Volume = (pi * radius squared)*height = (3.1416 * (diameter/2)squared)* height
= (3.1416 * 0.10Squared)* 250 meters
= 7.8540 cubic meters or 7,854,000 cubic centimeters
As 1 liter equals 1000 cubic centimeters this gives us 7,854 liters
As 1 liter of water has a mass of 1 kilogram this gives us a mass of 7,854 kg
An object that is falling through the atmosphere is subjected to two external forces. The first force is the gravitational force, expressed as the weight of the object, and the second force is the aerodynamic drag of the object. The weight equation defines the weight W to be equal to the mass m of the object times the gravitational acceleration g:
W = m * g
W = 7854kg * 9.8
Weight = 76,969.2
The drag equation tells us that drag D is equal to a drag coefficient Cd times one half the air density (estimated at 1.2kg/cu m) r times the velocity V squared times a reference area A (pi * 0.10 meters Squared) on which the drag coefficient is based:
D = Cd * 0.5 * r(kg/cu m) * V Squared * A
Drag = 0.18 * 0.5 * 1.2 * 70^2 * 0.0314 cubic meters
Drag = 16.625308
The motion of any moving object can be described by Newton's second law of motion, force F equals mass m times acceleration a:
F = m * a
We can do a little algebra and solve for the acceleration of the object in terms of the net external force and the mass of the object:
a = F / m
Weight and drag are forces which are vector quantities. The net external force is then equal to the difference of the weight and the drag forces:
F = W - D
The acceleration of the object then becomes:
a = (W - D) / m
a = (76,969.2 - 16.625) / 7854
a = 9.79
The drag force depends on the square of the velocity. So as the body accelerates it’s velocity and the drag increase. It quickly reaches a point where the drag is exactly equal to the weight. When drag is equal to weight, there is no net external force on the object, and the acceleration becomes zero.
So in our case our estimates are going to be fairly accurate even if we take air resistance into account.
Height = 0.5*Acceleration*(time Squared)
= 0.5*9.79*(6.142^2)
= 184.66 meters as opposed to 184.848 meters
Taking into account pipe friction, Temperature and Humidity factors let's lower our estimated velocity for our water from 70 meters per second to 62 meters per second
Now 1 meter of a 20cm diameter pipe holds 31.4 liters of water.
and 62 meters of a 20cm diameter pipe holds 1946.8 liters.
Therefore we can estimate that at the bottom of the pipe we have a flow of
1946.8 liters per second
If we have a reservoir of 800,000 liters then dividing by 1946.8 we find that
after 411 seconds or 6 minutes and 51 seconds - all the water has gone!
The only formula I've found that estimates power from falling water is from an American website (http://www.wvic.com/hydro-works.htm) and it's in imperial.
Power = (Height of Dam) x (River Flow) x (Efficiency) / 11.8
Power = The electric power in kilowatts per Hour (one kilowatt equals 1,000 watts).
Height of Dam = The distance the water falls measured in feet.
River Flow = The amount of water flowing in the river measured in cubic feet per second.
Efficiency = How well the turbine and generator convert the power of falling water into electric power. For older, poorly maintained hydroplants this might be 60% (0.60) while for newer, well operated plants this might be as high as 90% (0.90).
11.8 Converts units of feet and seconds into kilowatts.
Converting and inputting our variables
Height = 250 meters = 820.2 feet
Flow = 1946.8 liters per second = 68.75 cubic feet per second
Efficiency = 90%
Power = ( 820.2 * 68.75 * 90% )/11.8
= 4300.8 kilowatts per hour or 71.68 kilowatts per minute
BUT our system only runs for 6 minutes and 51 seconds!
So our system (As it Stands) should generate about 491 kilowatts a day, which would provide enough electricity for 24 homes all year round!
I say as it stands as next week I'd like to address the efficiency issues, but for a quick taster the amount of energy it takes to evaporate water is not a constant for all conditions.
Most of the figures quoted would make your average leaf curl up and die!
Blaine
-
An outer water filled sleeve around your tubes would provide your tree with stability, much the same as a natural tree has a bark sleeve.
The horizontally joined tubes that increase in diameter as they go higher should cause circulation at each junction rather than providing an overall circulation. Bamboo provides a similar circuit, but has a reservoir of water at each junction. The monkey-puzzle tree and palm trees have a simpler tubular structure that is self-evident.
Cut the top off the monkey-puzzle tree and the whole tree dies and does not recover. Cut the taproot off a palm tree and it also dies. Your model has added a semi-permeable membrane at the top and bottom. I am wondering whether this would disrupt the whole system as you build the model above 24 metres?
I suspect also that the upward flowing side of your model will experience cavitations and fail after a short period, hence my suggestion to use multiple tubes linked so that cavitations can be compressed when pressures change inside the tube. For example, when one tube completely cavitates the pressure inside will change from negative to positive and will push an increased head of water in other adjoined conduits out at the top and into your sphere, caused by the water vapour expanding, presumably where it can re-enter the tube that now has vapour bubbles inside it, as one would expect from water that boils, and as was shown in the rattan experiment which caused water to boil in the enclosed ground based sealed water filled container. (an experiment which I must see for myself on IVY growing in the UK to the tops of some pretty impressive trees.
I am also confused by your reservoir of water, which is shown at the same height as the tree and flows into the same pipe as the tree draws its water from. Releasing the valve so that water flows through the turbine may increase pressure at the tree but I suspect that water will not flow through the turbine but will remain at the same water level as in a U tube filled with water. Forgive me if I am missing a part of the whole picture but I feel the diagram may need some modifications for this to work.
For example, the reason your semi-permeable membrane would allow water molecules to flow through it at the top is not because of osmosis but because of the increased head of water caused by the downward flow in the juxtapose side of your model.
Hope this makes sense.
Kind regards
Andrew
-
Power = ( 820.2 * 68.75 * 90% )/11.8
= 4300.8 kilowatts per hour or 71.68 kilowatts per minute
Power is energy transfer per second (1W is 1J/s) - you can't have a power of kW per minute, the units are not consistent.
It is important to get this absolutely right if you want a meaningful conclusion to your study. I think your sums need cleaning up because the answer you present is 'unlikely'. You must not confuse power with energy - if you want to preserve the family fortune which you plan to invest in the scheme.
Reality check:
Remember, if your original energy resource is solar (heating from the Sun) you won't get any more than 1kW per square metre, the absolute maximum for power received from the Sun. You can't get anything for nothing. You will need a huge area for power gathering - as with all 'green' systems. This is not necessarily a reason to reject the idea but it is a serious factor which needs to be considered.
When trees do this sort of thing they are only interested in shifting water for their use - not in generating significant amounts of power.
-
One square metre of timber when burned will give more than 1KW and we get this for nothing. Blaine's proposal does not follow the conventional energy from sun ratio. The enrgy from the sun provides the evaporation, the resulting density chages provide the pump to raise the water and the release of the desalinated water through the turbine provides the electrical return. So we are not looking to heat the water to provide energy, just enough heat to evaporate it effectively when it is exposed to the air over the beeds inside the sphere.
My problem is that from my own experiments the semipermiable membrane will cause the column to draw in air rather than draw water up, and the downward flow appeas to be going into the feed tank where the water is drawn from and this would cancel out much of the free falling waters energy.
-
One square metre of timber when burned will give more than 1KW
?
kW is power; what counts with a piece of wood is the Energy that it can produce. Whilst the wood was growing it was sitting there for years and years with 1kW /msquared (nearly) falling on it - which corresponds to many GigaJoules of energy, probably. The energy is no more 'for nothing' than any other resource.
Mr Blain can't get away with the fact that you can only get out what energy you put in. However you try to do it, you are deluding yourself if you think otherwise. The fact is that a tree needs very little energy to provide itself with water - a few tens of kg per day, lifted for a few tens of metres (maximum). This would correspond to a few thousand Joules of mechanical energy - enough to heat 1kg of water by, perhaps, a few degrees C - or move a motor car by a few metres. Not a serious source of actual energy but enough to do the job for the tree. Nothing magical - just simple Physics.
-
I mentioned the energy stored in timber as fuel to show you there are other ways of looking at the way energy from the sun converts. Another way to look from a different prospective is how many KW are used by the tree raising it's massive girth branches, foliage and fluids to over 100 metres cell by cell? Does this too come into your equation for energy? I am not trying to start a discussion just trying to show that there may be other ways of looking at a problem that lead to a different set of maths.
The point with the tree is that the energy used to raise it to 100 metres is free energy, it does not cost a dime. And if Blaine is using free energy to lift the water, it does not matter how many of these trees he is using to lift it because there is no cost to the process other than the construction of the artificial tree.
But one could then argue why build a tree in the first place, why not put a turbine on a free falling naturally elevated water supply?
Well the principles are exactly the same, you use the water cycle to raise the supply above ground level and you use gravity to drive the generator.
-
The energy stored in timber is from Photosynthesis. The energy for self - irrigation is thermal.
For plants which grow out of doors the energy can be looked upon as free, of course.
My point was that the energy associated with flow of sap is not significant when discussing 'creating electricity', which is what the thread is titled.
-
Sophie
Thank you for your input, I hope you can stick around for the next few weeks or so as I'll be talking about power and efficiency more.
Power is energy transfer per second (1W is 1J/s) - you can't have a power of kW per minute, the units are not consistent.
It is important to get this absolutely right if you want a meaningful conclusion to your study. I think your sums need cleaning up because the answer you present is 'unlikely'. You must not confuse power with energy - if you want to preserve the family fortune which you plan to invest in the scheme.
But for now I feel you may have misread the equations I used.
Power = ( 820.2 * 68.75 * 90% )/11.8
This is in imperial units and the the '11.8' is used to convert units of feet and seconds into kilowatts per hour.
If we can have kilowatts per hour why not per minute?
Now as I said earlier this formula was the only one I could find that estimates power from falling water and it is from an American website (http://www.wvic.com/hydro-works.htm).
If you know of another formula please tell me.
kind regards
Blaine
-
Andrew,
Thanks for keeping this thread 'alive' while i attend to some pressing 'work'.
I am also confused by your reservoir of water, which is shown at the same height as the tree and flows into the same pipe as the tree draws its water from. Releasing the valve so that water flows through the turbine may increase pressure at the tree but I suspect that water will not flow through the turbine but will remain at the same water level as in a U tube filled with water. Forgive me if I am missing a part of the whole picture but I feel the diagram may need some modifications for this to work.
Please keep in mind that the 'root' reservoir's water level will drop as the water collects in the upper reservoir, leaving space for the same water to be replaced. Obviously the artificial tree's root must always remain submerged during this process.
The horizontally joined tubes that increase in diameter as they go higher should cause circulation at each junction rather than providing an overall circulation. Bamboo provides a similar circuit, but has a reservoir of water at each junction. The monkey-puzzle tree and palm trees have a simpler tubular structure that is self-evident.
The idea is that these horizontal tubes provide additional little osmotic lifts.
Don't forget we're planning to use lots of these connected up and down tubes and that their diameters will be very small. (Recently we've been informed that one kind of optic fiber made in the UK is actually hollow in the middle and is relatively inexpensive to obtain, something to look into).
I'll put up another diagram here, which I've shown you before, try to imagine the tops of these individual circular tubes connected by huge 'vein filled' circular sheets of material, with a hole cut out in the middle. The water from the 'up tubes' enters these sheets goes around them and then exits them through the 'down tubes'. Obviously prior to exiting we want as much evaporation as possible leaving a denser solute to fall back to the roots.
Kind Regards Blaine
-
Sophie,
A quick point, there is already a lot of energy in water itself.
When the sun goes down in most countries of the world the temperature doesn't drop as far as it does in the UK.
Trees would continue to lose water through evaporation at night, if their guard cells didn't squeeze their stomata shut.
Blaine
-
If you are drawing water from the reservoir, then this reservoir is obviously open to allow air in?
If this is the case then you will have problems with dissolved gas in the water. Dissolved gas will initiate cavitations over the 10 metres.
You can use 6mm bore tubing with out any problems for the flow and return. This fact alone might tell you something about the validity of osmosis in this model.
Hollow fibre optic tubes could allow some capillary action when cavitations take place. Not sure how this will go.
Also the downward pull of the solutes induces vacuum at the sphere end of your experiment this would undoubtedly accelerate evaporation.
And of course we know the flow and return works in the tubular structure of a dead tree thanks to Strasburger's famous experiments with trees when he cot off the roots while the tree was suspended and remained submerged in a bath of picric acid or copper sulphate, killing all of the living processes in the tree, yet the circulation continued for weeks after the tree had died.
-
Wow you're quick!
Yes this 'degassing' of the water has been a bane for a while, but we may have a solution.
The system should be initiated with a full degassed root reservoir and a closed valve still below the water, just below the turbine enclosure.
At the bottom of the top reservoir the valve is closed with a low level of water covering it.
Now we extract as much air as we possibly can from the connecting tubes and turbine blades enclosure.
When the the top reservoir is at our predetermined level both valves are opened.
After almost all of the water has passed down this almost 'airless' tube, the top reservoir's valve is closed again leaving it still covered with water left in the top reservoir.
Not perfect I know, but it may help in reducing the amount of gases impregnating our water supply and help in reducing cavitations, although the salt levels may actually help in reducing any 'nanoscopic' bubbles joining forces to create bigger ones.
Blaine
-
Now as I said earlier this formula was the only one I could find that estimates power from falling water and it is from an American website (http://www.wvic.com/hydro-works.htm).
If you know of another formula please tell me.
The amount of graviatational energy in joules = m g h
where m is the mass in kg, g is the gravitational field (9.81 N/kg) and h is the height in metres
power is energy released per second
So if you want to work out how much energy your scheme can produce in an ideal world you need to work out the mass of water it can lift every second.
My understanding is that you are evaporating all the water that is then going to go down your turbine.
It takes 2.2MJ /kg to evaporate water
Energy has to come from somewhere and in this case from the sun. On the equator the sun will average about 250W per square metre
so using the heat from the sun you will be able to evaporate about 0.00011kg or 0.11g of water per second.
Assuming you have lifted it 300m, this will release at most
0.00011g * 9.81 *300m = 0.33J/s = 0.33W
You have a machine so the efficiency is about 0.1% and to produce an average of a megawatt you would have to cover 300 hectares.
As a comparison cheap solar cells which would probably cost a similar amount to build are about 10% efficient - 30 times better!!
-
Dave,
I hope that you're just not reading carefully, as you're not addressing the issue at all.
I spent some time trying to estimate how much power would be created by falling water
800,000 liters falling 250 meters through a 20 cm diameter pipe.
If you can find a better formula than the one I used please tell me.
-
Dave,
So if you want to work out how much energy your scheme can produce in an ideal world you need to work out the mass of water it can lift every second.
We already have a good idea how much water we can raise in 24 hours, our minimum figure using 8 'trees' to deposit the water in our central reservoir is 800,000 liters
Just to give you a reality check a natural tree 'trying' NOT to lose water evaporates between 500 liters and 1000 liters of water a day.
We estimated than we can increase this by a factor of a hundred, recently we've begun to think than we can increase this natural evaporation rate by closer to three hundred times.
It takes 2.2MJ /kg to evaporate water
Energy has to come from somewhere and in this case from the sun. On the equator the sun will average about 250W per square metre
so using the heat from the sun you will be able to evaporate about 0.00011kg or 0.11g of water per second.
Now it seems that you're suggesting that a tree is doing something impossible!
According to your figs 0.11g per second * 60 * 60 * 24 = 9504g or 9.5 liters a day!
9.5 liters a day doesn't come close to what a tree tries NOT to do!
As I said earlier I'll go through most of the math with regards to estimating how much water we can raise over the next few weeks.
Blaine
-
Now it seems that you're suggesting that a tree is doing something impossible!
According to your figs 0.11g per second * 60 * 60 * 24 = 9504g or 9.5 liters a day!
9.5 liters a day doesn't come close to what a tree tries NOT to do!
That is 9.5 litres per square metre, for a large tree 10m across so about 100 square metres that is 950 litres - a bloody good estimate if I do say so myself
As I said earlier I'll go though most of the math with regards to estimating how much water we can raise.
we estimated earlier than we can increase this by a factor of a hundred, recently we've begun to think than we can increase this natural evaporation rate by closer to three hundred.
I would very much like to see your maths...
-
excellent point Dave,
Yes, per square meter, silly me, now if only a tree looked like a 100 square meter pool of water, the math would pan out.
As I said a tree doesn't 'want' to release this water, and at night would continue to release this water if it didn't close its stomata.
Physics has to explain nature.
Have you ever wondered what experimental apparatus was used to get the figure of 2.2mj of energy to evaporate 1 kg of water.
Energy is required to break the hydrogen bonds between the individual water molecules so that they may eventually leave their liquid siblings and take flight.
Why do those H2O molecules on the surface leave first, because obviously they have half the bonds to break.
Does it take the same amount of energy to evaporate 1 kg of water when it is in a cube shape as it does when it is spread out a few molecules in depth with a huge surface area?
Blaine
-
Does it take the same amount of energy to evaporate 1 kg of water when it is in a cube shape as it does when it is spread out a few molecules in depth with a huge surface area?
The water molecules always evaporate from the surface so to a first approximation they will be eqiuvalent
If you are getting more subtle:
If you spread out the water on a hydrophobic surface you are doing work against surface tension to stretch the water out so you are giving the water some energy before you try and evaporate it, so the energy required would be slightly less...
However
Water won't spread out over a hydrophobic surface, the reason that water will spread out over such a large surface on a tree is that the tree's surface is very hydrophilic - the water molecules bond to the tree probably better than they do to other water molecules - so you are going to need at least as much energy if not more to evaporate a kilogram of water from your tree, than from a bucket.
In fact because energy must be conserved in your design the difference in energy to evaporate a kilogram of water from the leaves of the tree 300m up and from a bucket on the ground. must be at least the change in potential energy of the water as it rises the 300m. So it will definitely require more energy to evaporate water from your tree than from a bucket.
Although if the energy is available the larger surface area of the tree will mean a higher rate of evaporation. But if the energy isn't being added by the sun as fast as it is being lost though, your tree will cool down until the rate of energy loss equals the rate of energy input.
-
This is in imperial units and the the '11.8' is used to convert units of feet and seconds into kilowatts per hour.
If we can have kilowatts per hour why not per minute?
Kilowatt Hours (Power times time) is the unit of energy - not Kilowatts PER Hour (Power divided by time). There is a huge difference between the two and you can't just gloss over such an error.
Basics basics basics!
-
The power available to the tree to do this work will come from the thermal energy 'absorbed' by the leaves at the top. This can be from the Sun or from the surrounding air (clothes dry in the shade as well as in direct sun, although not so quickly).
The details of the actual evaporation process, although important, are not as important as the Energy consideration. With bigger or smaller leaf area and a different surface chemistry, the same Energy budget would apply.
-
In any case, where is this thread going?
Are we discussing
1.how a tree gets enough energy to supply what is needed for self irrigation (it clearly gets enough energy)
or
2. whether this is a system which could be looked upon as a possible Energy Resource for electricity generation (not in a million years - there just isn't enough power input)?
It isn't clear.
-
Sophie Dave,
I wish I could devote more time on here, hopefully soon, in the meantime let me clarify where this thread is hopefully going.
Two questions need to be addressed.
1) How much power can be generated from the falling water 800,000 liters falling 250 meters through a 20 cm diameter pipe?
2) Can our Artificial Trees lift that much water?
We have a lot of confidence in our Artificial tree's ability to lift the water. Which as I alluded to earlier has to do with the apparatus used in the experiments to measure the energy required to evaporate 1 kg of water. Did any of you check it out? (remember water is not a good conductor of heat)
Question 1 has been very confusing, have you guys checked the link I gave for the formula I used? If not please check their example, I'll copy and paste it below.
If you could kindly point out the errors I made when working through the formula with our numbers I would be very grateful.
Below is the example reprinted from this website http://www.wvic.com/hydro-works.htm
Power = (10 feet) x (500 cubic feet per second) x (0.80) / 11.8 = 339 kilowatts
To get an idea what 339 kilowatts means, let's see how much electric energy we can make in a year.
Since electric energy is normally measured in kilowatt-hours, we multiply the power from our dam by the number of hours in a year.
Electric Energy = (339 kilowatts) x (24 hours per day) x (365 days per year) = 2,969,000 kilowatt hours.
The average annual residential energy use in the U.S. is about 3,000 kilowatt-hours for each person. So we can figure out how many people our dam could serve by dividing the annual energy production by 3,000.
People Served = 2,969,000 kilowatts-hours / 3,000 kilowatt-hours per person) = 990 people
Kind Regards
Blaine
-
Question 1 is not confusing, it is trivial to put an upper bound on.
800 000kg of water raised by 250m by mgh has about
800 000kg * 10kgm/s2 * 250m
2GJ of energy
which is 555kWhr of energy.
assuming this is what your system can lift in a day (which will mean it has an area of about 8 hectares)
you will be producing an average of 23kW
about 50ish people assuming a 100% efficient conversion from potential to electrical energy, this is probably nearer 80% so you are talking about 18kW average.
-
Yes, Y U B, question 1 is triv.
This system works on evaporation of some of the water to provide energy for lifting all of it.
I have a suspicion that all this energy transfer could produce a significant modification to the microclimate. The air would be full of evaporated water and it could mean an increase in rainfall (admittedly, some of this rainfall would find itself back in the system again.)
This process is not unlike the normal water cycle, which uses the GPE of water which has been lifted there by solar energy via the clouds.
I would need some more information before accepting that an area of artificial tree covering 8 hectares would produce the quoted level of power output. It would depend, greatly, upon the actual location on the Earth and the existing climate. A cold Northerly wind would put the mockers on it, I feel.
-
The 8.5ha is the minimum possible area at the equator if all the sun's energy falling onto those 8ha was used to evaporate water and then all the water was condensed again (with some mysterious immense heat sink) 250m up.
As a comparison 8ha of 12.5% efficient solar cells on the equator would produce nearer 80 000*30W/m2 2.4MW - I think rather better than 20kw, and probably also far cheaper to build as they don't involve thousands of 205m tall 'trees', with giant glass domes at the top, and huge pumps to move cold seawater in order to cool your condensing surfaces which would probably use more energy than your turbines produce.
-
My original point about the system that trees use is that it does a particular job excellently. The available power just suits the requirement and doesn't allow for TV, heating or lighting.