A reliable, earth quake and tsunami proof, power plant could be built over the deep water of the Japan trench.
The fuel I would suggest using would be, a slurry of crushed volcanic ash. Japan has several volcanoes that produce large amounts of unconsolidated ash.
How to burn ash? I haven’t got a clue!. But it can be used like this. continuously add an amount of ash slurry to the top of the water in a long wide vertical tube. This tube is open at the bottom, and sticks up above the waves at the top. The slurry disperses through the column of water, increasing its density Has the pressure caused at the bottom of the tube by the weight of the sea water and slurry is grater than the pressure at the same depth out side of the column, the fluid inside the tube will flow out until the pressures are equal This will cause the surface of the water inside the tube to be lower than the sea level outside of the tube..having created this difference in hight of the water between the sea and the inside of the tube, it is now possible to run sea water through a turbine and into the tube.[ EDIT ]I came up with this image whilst thinking about some of the objections people where raising. It makes the basic idea so intuitively understandable, that I wish I had thought of it before. 1) Make a salt water solution, with a little food colouring in it.2) Get a bit of stiff, clear plastic tube. A drinking straw or the shell of a cheep pen.3) Fill this tube with the coloured water from (1) and put a finger over both ends.4) Hold this tube vertical under normal clear water and take the finger away from the bottom hole. 5) When the finger is taken away from the top of the tube. The coloured liquid will run out of the bottom of the tube, and clear water will run in at the top.[END EDIT]
Those of you who know your alternative energy devices will recognize this as; a power tower, turned upside down, stuck in the sea and ran backwards, with sea water as the working fluid and crushed ash as the source of density difference.
The power that can be generated is proportional to the head ( the difference in hight between the water surfaces ), and the mass of water passing through the turbines in a given time (the flow rate ).
The head that can be created is proportional l to the length of the tube and the density difference.
The best flow rate will is proportional to the pressure difference at the bottom of the tube, and the area of the tube.-- j paul, Jul 06 2011 Geothermal power http://en.wikipedia...ki/Geothermal_powerWell established [8th of 7, Jul 06 2011] The energy intensity of transport http://www.inferenc...r/c15/page_95.shtml['Sustainable Energy - without the hot air', David MacKay FRS] [hippo, Jul 07 2011] OK but....
Instinctively, the energy you get out will be no greater than the potential energy of the ash you put in.
The density of volcanic ash is going to be something like 3 (it's basically small rock pieces; assuming they are fine enough not to have trapped any gases), and you're doing this in the sea, so the effective density is reduced to 2. In other words, you only get 2/3rds of the gravitational potential energy you would if it were all in air.
So, let's suppose you dump 3000kg (as measured in air, which is 2000kg in water) of ash into this thing every second, and suppose the tower is 5000 metres tall (roughly the depth of the ocean near Japan). You're going to get 2000 x 5000 x 10=100MJ of energy per second. That's 100MW.
This means that, to supply Japan's current electricity needs, and assuming all your generators are 100% efficient, you will need about 110 of these 5km tall towers, each receiving 3 tons of volcanic ash per second.
Yep. That sounds good.-- MaxwellBuchanan, Jul 06 2011 Dumping ash to lower water then running a water wheel seems circuitous. Maybe just dump the ash on the wheel, since you are set up to dump, and have ash. I will note that volcanoes, being mountains, sometimes have supplies of old ash at some altitude and this could be done on dry land, so long as the ash descended and on doing so reliquished some potential energy.
This process could be repeated, inserting ash, rocks and everything else which is high into the wheel and generating energy as it becomes low. When nothing on land remains that is higher than anything else, the ocean trench option for lower lowness becomes attractive.
The created level surface would also lend itself to a large bumper cars rink.-- bungston, Jul 06 2011 If you scrape off the top 1m of the whole of Japan and use it in a 5km-deep oceanic system as described, it will provide all of Japan's electricity needs for just over 100 years.
If you go for the land-based approach, and if you reduce all of Japan to a plateau by dropping stuff from high points to low points, you will generate enough electricity to run the country for approximately 3 months.
Personally, I think you'd be better off running the country by means of an extension lead from China.-- MaxwellBuchanan, Jul 06 2011 Geothermal steam generation is a much more mature and viable technology. The Icelanders have made excellent use of their copious geothermal resources.
<link>-- 8th of 7, Jul 06 2011 Can't you design this as a perpetual motion machine, so that once you have created the difference in water level it keeps running automatically without needing any further input? Perhaps you could shake or agitate the ash in a certain way?-- pocmloc, Jul 06 2011 // The Icelanders have made excellent use of their copious geothermal resources.//
Yes, but they are also the only nation I know of that cannot spell "th" properly.-- MaxwellBuchanan, Jul 06 2011 It's probably more efficient to find someone who needs volcanic ash, sell them the ash and use the money to buy energy from someone else.-- hippo, Jul 07 2011 //extension lead from China// Singapore has exactly that relationship with Malaysia.-- spidermother, Jul 07 2011 Logically, then, somewhere on your planet there's a totally overloaded single outlet with many, many multiway adaptors plugged into it, a la "National Lampoon's Christmas Vacation" ...-- 8th of 7, Jul 07 2011 1) The pacific coast of southern Japan has an ongoing problem with loos volcanic ash. That has to be removed before it can collapse buildings, or pose a danger from lahars and landslides. 2) Yes, volcanic ash will have a specific gravity of about 3. that means it as 3 times the mass of the same volume of water. Why have you assumed that the mass would go down if it is put in water? If you put a rock from earth on the moon it would still have the same mass! So why should its mass go down when it is in water? 3) Even if Maxwell’s numbers are right, I personally can not see 100 MW of power generated for each cubic metre of ash that as to be disposed of, as anything other than fantastic! And more than I had expected, but then I had been thinking in terms of a 3 km tube and a 100 m working head.-- j paul, Jul 07 2011 // Why have you assumed that the mass would go down if it is put in water?//
The sg is 3, (ie, 3 grams per cubic centimetre); that of seawater is about 1. Therefore, the gravitational potential energy which you can extract by dropping your ash in seawater is 2.
(Look at this way - if you dropped into a liquid with an sg of 3, it wouldn't sink at all and you wouldn't be able to extract any energy from it).
//personally can not see 100 MW of power generated for each cubic metre of ash that as to be disposed of, as anything other than fantastic!//
That's one cubic metre **per second**, or more precisely 3000kg of ash **per second**.-- MaxwellBuchanan, Jul 07 2011 ...and [Max-B] was assuming 100% efficiency, and ignoring the energy cost of gathering up all the ash and transporting it to the sea - and he was assuming that the drop would be 5000m which it wouldn't be, because the sea level isn't going to drop right to the bottom of the shaft. If we assume the actual drop is 1000m (which is a generous assumption) then you get 20MJ, or 4MJ a second if the thing is 20% efficient (again, generous). Then the energy intensity of road transport in the UK is about 1 kWh per t-km (see link), so if you're transporting 3 tonnes a second from 10 km away (i.e. a nice, close volcano - and let's not take into account the trucks' return journeys) you're using 30kWh of energy a second in transport. 30 kWh is 108MJ, so the whole thing is costing you about 104MJ a second - quite a lot really...-- hippo, Jul 07 2011 Mass is mass!
The only way to change an objects mass is to either knock a lump off, or stick a lump on!
Thing float upon liquids of grater density because of buoyancy. The object will still have the same gravitational potential energy, as measured from some arbitrary point, if it is floating or not. As an extreme example Imagine an apple falling passed the edge of a table, on witch there is another apple If both apples are the same, and are the same distance from the floor, they will have he same gravitational potential energy. Just that one of them is supported, and the other one is not.
The efficiency of modern hydro electric power stations, given a reasonable a head of water, is over 95%.
I had not thought about getting the ash to the device. But there are two obvious solutions.1) dry ash, on gravity power conveyor belts down the mountain followed by powered convey belts over the water.2) make the slurry at the quarry and pipe it under gravity the whole way.-- j paul, Jul 09 2011 But an object sinking in water displaces its own volume of water. And work is _force_ times distance, not mass times distance. If the force is less (due to buoyancy) the work is less in exactly the same proportion.
Equivalently, the difference is accounted for by the work done on the water displaced (raised) by the mass as it sinks.-- spidermother, Jul 09 2011 // The object will still have the same gravitational potential energy, as measured from some arbitrary point, if it is floating or not.//
No. Sadly not, in terms of the recoverable energy. If you are dropping something in water, you only get the potential energy available from its mass _minus_ that of the water it displaces.
If it makes it easier for you to imagine, bear in mind that if the lump of rock (or suspension of ash) drops a certain distance in seawater, the same volume of water rises in exchange (ie, its displacement). Hence, the net potential energy you recover is that of the falling mass, minus the energy needed to raise the seawater which it displaces.
Let me ask you this: if I take a gallon of seawater (which weighs about 4kg), and release it into the top of the ocean, how much potential energy can I harvest by its falling to the ocean floor? None.-- MaxwellBuchanan, Jul 09 2011 I do not like doing the detailed maths. Mainly because I can get it wrong. I am not going to attempt to prove that this could provide all of a nations power, because that is an unrealistic expectation that is some times placed upon alternative energy that is not placed upon conventional energy sources.
In doing this maths I will use the formulas; P=gQH, ke=0.5Mv^2,pe=MgH, and there derivative and will be taking the M in these formula to mean mass, not weight.
As I imagine this, it would be three kilometres deep, and have an area of fifty squire metres. And would have ten kilograms of slurry for every cubic metre of water.
Starting with that last figure, that equals a one percent increase in the density of the water inside of the tube, and that means that at equilibrium the surface of the water in the tube will be thirty metres below that of the sea.
If instead of letting it reach equilibrium, sea water is run through turbines to keep the level at ten metres below sea level. The maximum electrical out put in kW is given by P=gHQ. H is the head of water, in this case 10. and g is acceleration due to gravity, approximate 10. So kW=100* Q. Q is the volume of water flowing through the turbines per second.
The force moving the water has two components;The force caused by the mass of ash and water, above the equilibrium position And the mass of rock below the equilibrium position and the bottom of the tube.
Above the equilibrium point there will be 20 metres of water and ash. At an average head of 10 .metre Using the formula velocity equals the square root of twice the head times the acceleration due to gravity, gives a flow of 14.1 metres per second. So that’s 1.4 MW. small compared to the 460 MW of the Fukushima 1 reactor, but respectable for a piece of alternative generating technology.
The simplified formula can not be used below the equilibrium point, because it is the mass of ash that has the potential energy and the much larger mass of water that has to be accelerated. instead the potential energy will need to be calculated and then put through the kinetic energy formula to give a velocity.
There 10* 2970 = 29700 kilograms of powdered ash in the tube if it is one square metre in area and each contains ten kilograms of extra mass. With an average hight from the bottom of the tube of 1485 metres, the potential energy of the ash in the tube is 29700*10*1485 = 441045000 j. however this energy is acting on 2970*1000 kilograms of water. This gives a velocity of 17.2 metres per second.
Given a tube of 3000 metres length, one square metre in area, containing water at 10% higher density will generate 3.1 MW of electricity for each 310 kilograms of ash.-- j paul, Jul 12 2011 At the risk of being excessively sensible, I would suggest wave, wind and geothermal. There are already windfarms on the coast, and the potential for wave power on the Pacific coast is pretty shonky, geothermal is also plausible.
Seeing as I actually do live in the place, can you please use the brush and dustpan provided after you're done with the fooling around with volcanic dust stuff....-- not_morrison_rm, Jul 12 2011 Sorry NOT Morrison, but I feel that there is just to much potential in this idea to give up just yet. I am not picking upon Japan particularly, its just that you have the deep water, the ash, and the engineering skills to do this. As my thinking on his has developed I suspect most units will be over deep water near raw materials. The ore being shipped in, processed with electricity, and the wast mass then being used to generate more electricity.
I started my last submission with an improbable set up, a long tube open t both ends containing a stationary liquid that is heaver than the water around it.
I then calculated that, because there was a lot of extra mass in that tube, (297 metric tonnes ) there was there for a lot of potential energy. But that all that energy would only accelerate the mass of water in the tube to 17 metres per second.
But provided ash is still being added at the rate of 10 kilograms per cubic metre, then it will be accelerated by a further 17 metres per second, in the next second.
That is the acceleration of water, in 3 kilometre tube open to water at both ends, that has been given 1% extra mass is 17 metres per second per second!
Just how fast the final flow will be, is hard to calculate because it depends on friction and turbulence But a maximum value can not be faster than the fastest speed to which the ash in the column would travel at if there was nothing to slow it down. 172 metres per second.-- j paul, Jul 14 2011 SM / MB We seem to be talking at cross purposes. Possibly because I may not have explained myself as well as I could have.
I am taking my measurements of the energy in the system ; from the mass already in the water at the top of the tube, to the bottom of the tube. I Started with the mass in the water like that not because its ovoids the complication of the displacement of water as the mass moves the short distance from just above the water to just below its surface, but simply because it measures what I wanted to measure.-- j paul, Jul 14 2011 Hold on.
1) Let me just confirm: you are getting all this energy by allowing volcanic ash to fall, right or wrong?
2) The ash starts out in a ship or something, roughly level with the surface of the ocean, right or wrong?
3) The ash ends up at the bottom of the ocean, right or wrong?
If all the above are true, then the energy you get is the gravitational potential energy of the ash, in falling through the system, MINUS the gravitational potential energy acquired by an equal volume of seawater being lifted through the same distance. (The seawater has to go somewhere when it's displaced by the ash.)
If you still don't believe me, try this thought experiment. I create some "artificial ash" which has a density of 1 gram per cubic centimetre, the same as water. I take one ton of this ash, and shovel it off the boat and into the ocean. What will it do? It'll just drift around and disperse slowly, but it won't sink.
How much energy can I get from doing this? Zero, zilch, nada, nothing, not one squittojoule of energy. Why? Because my stuff has the same density as seawater. In other words, it displaces a mass of water which is equal to its own mass.
(In fact, seawater is a bit denser than 1g/cm3, which makes things even worse, but that's a minor point.)
If your real volcanic ash has a density of, say, 3 grams per cubic centimetre, then it is 2 grams per cubic centimetre heavier than the water it's displacing. So, you will only get 2/3rds as much energy out as you would if you dropped it the same distance in air.
Is it only me who doesn't see how you don't see this? Think about another example: I make a generator that works by dropping logs on it to make a big wheel go around. Now I take that generator out to sea and submerge it. I drop the logs on it, but they don't fall and make the wheel go around.
Anyway, let it go. OR, maybe I've got (1), (2) or (3) wrong, in which case you have my apologies and a request for an explanation.-- MaxwellBuchanan, Jul 14 2011 Run for the hills! It's ARTRR all over again!!!-- hippo, Jul 14 2011 I'm with you, [MaxwellBuchanan]. It's one of those things that seems so obvious and simple that I can't see how else to explain it.
A sinking object continues to displace its own volume, all the way down - maybe that's the point that's being missed. You could think of it as repeatedly changing places with the bit of water that was previously just underneath; for each 1m that 1m³ of substance sinks, 1m³ of water rises by 1m.-- spidermother, Jul 15 2011 when Maxwell started with the formula M*g*h. I thought it was brileant as it would give an answer without much maths. I see now why He used a different value to do the calculation, I get it.
The formula that the following I based upon is v=sqrt (2*g*H). This is the standed formula for the speed of water entering a pelton wheel. Note it dose not include Mass.
First to state an implicate assumption, explicitly. If the particles of mass that are added to the water are very small, then the force of gravity acting on them will be small compared to forces acting upon it from the water. Buoyancy, will resist the downward motion of the particles, and drag will resist motion in any direction. So a finely divided mass added to water will act as if the density of the water is increased. Important. Although the particles can not move very fast relative to the water around them, the entire volume of dense liquid can, and will, flow quickly.
From doing the maths I discovered something quite odd about this idea / device. Although the equilibrium point the point at witch the weight of denser fluid is balanced by the weight of a longer column of sea water outside of the device, is dependant upon the increase in he density of the liquid within the tube. The speed of the water flowing through the device dose not(not in my original idea) . Provided that the device is run for long enough to settle at a steady speed, then it will run at that speed if it as 1 kg of mass per ton of water or 100 kg. That is the sqrt of (10 times the length of liquid in the tube ). that is the maximum value, in practise there will be loss due to friction with the tube wall, from turbulence within the liquid, and back pressure as the fast flowing liquid enters the sea.
If this seams wrong to you it dose to me as well. But its where the maths leads-- j paul, Jul 21 2011 [jp] The whole point of doing a simple energy calculation is that it obviates the need for worrying about the mechanism. All your energy comes from the falling mass; you can't get any more out than that, however you do it (although there are plenty of ways to get less).
As to the speed of the falling water-column, it will only be density-independent if you are not trying to extract energy from it (and if, as you note, there is no friction).-- MaxwellBuchanan, Jul 23 2011 Not even then. Mass is proportional to the density of the column, but net force is proportional to the density of the column minus the density of sea water. Thus force/mass is not constant; a denser column will move faster.-- spidermother, Jul 23 2011 random, halfbakery