Generating electricity using household water pressure

P

paul turbine

I think it is possible to design a machine capable of generating useful electricity via recirculated fluid flow. But the laws of thermodynamics make it very difficult.

The three laws of thermodynamics may be summarized by saying: "You can break even but only at absolute zero".

So we know for sure we will never be able to get more energy out of a system than we put into it in the first place.

Secondly, the value of entropy in any system governs what is possible or impossible.

If entropy is zero or negative, then whatever we hoped to achieve is impossible. It is simply not permitted in this universe by the laws of physics that govern it.

An example would be trying to reverse the act of smashing an egg on the floor. Any such reversal of the egg breaking process would require a value for entropy (S) which was negative. Accordingly, we know the proposed act to be impossible.

Unfortunately, the laws of thermodynamics and the related law concerning entropy reflect reality. If we don't like it, we must go and live in another universe.

Bearing in mind the laws we are stuck with, the next step is to consider what type of system might be useful bearing in mind these significant limitations.

There are three types of system. Isolated systems, closed systems and open systems.

In isolated systems, neither mass nor energy may pass through the system boundaries (think of a perfectly insulated thermos flask).

Needless to say, we will never be able to generate useful electricity from an isolated system because for a start we will never be able to get any energy out of it in the first place.

So we can forget about isolated systems right away.

Then there are Closed systems. They allow energy to pass through the system boundaries, but prevent mass passing through the boundaries.

Again the laws of thermodynamics prevent useful electricity being generated using a closed system, because you will always get less energy out of a closed system then you put into it in the first place.

Our only hope is to use an open system.
The laws of thermodynamics will only ever allow us to generate useful electricity using systems in which both mass and energy can pass through the system boundaries.

So this has to be our starting point.

We must invent a machine (an open system) that allows mass (for example water and/or compressed air) to pass in and out of the system boundaries, and which also allows energy to pass in and out of the system boundaries.

In practical terms this means we need help from the environment.

External work must be done on the system (eg solar energy or water from a flowing stream) to provide energy from outside the system boundary)and also external mass (for example air or water) from outside the system boundary must be able to pass through the boundaries as well.

The best idea I have come up with began as a mathematical thought experiment:

There are two 25m high cylinders each of diameter 1m.

An impulse turbine (for example a Pelton turbine) has been placed 3m from the bottom of the second cylinder (allowing a space beneath it for tailgate water to accumulate).

I think I have found a way to connect these cylinders in such a way that fluid can circulate from cylinder A into cylinder B allowing it to flow down a 20m+ drop in cylinder B before striking the impulse turbine.

At a mass flow rate of one cubic meter per second (which is an enormous flow rate)and allowing for system efficiency of 0.85 (this being a unit-less fraction where 0= total inefficiency and
1= perfect efficiency), the electrical power output in watts (using water as the working fluid) would be as follows:

Pw = 1000kg/m3 x 20m x 9.81 m/s/s x 0.85
Pw = 166770 watts = 166.77kW.

So this might look promising.
But there are serious practical problems.

The water that collects at the bottom of cylinder B (the tail-gate water) will rise until the point it prevents the impulse turbine turning. It will swamp the turbine and stop it moving.

However, I think I have found a way of recirculating fluid from cylinder A back into cylinder B without having to use enormous amounts of external energy (to lift or pump fluid back into tank B).

I would much appreciate private (not to be published) dialogue hopefully with a co-inventor familiar with thermodynamics, Bernoulli, and Newtonian fluid flows.

My aim is to share with a candidate co-inventor a recirculation novelty compliant with the laws of thermodynamics for the purpose of applying for a patent relating to industrial electricity generation.

I am only interested in applying for a patent if the model works mathematically.

Thank you for reading this post.

I can be contacted at:
[email protected]
 
R

Robert Scott

Paul Turbine wrote:

>I think it is possible to design a
>machine capable of generating useful
>electricity via recirculated fluid flow.
>...
>However, I think I have found a way of
>recirculating fluid from cylinder A back
>into cylinder B without having to use
>enormous amounts of external energy (to
>lift or pump fluid back into tank B).

Oh no, not another perpetual motion idea! Give it up. You seemed to have some understanding of thermodynamics in most of your posting. But then you go and completely contradict everything you said by proposing exactly what you said is not allowed.
 
N

Namatimangan08

>> I think it is possible to design a machine capable of generating useful electricity via recirculated fluid flow.

>> However, I think I have found a way of recirculating fluid from cylinder A back into cylinder B without having to use enormous amounts of external energy (to lift or pump fluid back into tank B). <<

> Oh no, not another perpetual motion idea! Give it up. You seemed to have some understanding of thermodynamics in most of your posting. But then you go and completely contradict everything you said by proposing exactly what you said is not allowed. <

I think he was just joking.
 
Q

quantumtangles

As a matter of fact it is not a perpetual motion machine. Perpetual motion machines are not possible. But neither was I joking.

Here is a system summary followed by calculations:

System summary:

Two Cylinders A and B, each 25m high and 1m in diameter, stand side by side.

Cylinder A is 90% full of seawater of density 1020kg/m3. It also contains approximately 10% by volume of castor oil (density 961kg/m3) which floats on top of the seawater. A small air gap is left at the top of cylinder A and a pressure relief valve sits at the top of cylinder A.

A siphon of diameter 0.12m leads from the top of cylinder A down into cylinder B. An electric pump primes the siphon and begins fluid flow into cylinder B.

Cylinder B contains only air to begin with, but there is an impulse turbine connected to an alternator motor at the base of cylinder B.

A 3m space at the bottom of tank B (underneath the turbine) is needed to allow tailgate oil to accumulate without interfering with the movement of the turbine.

The siphon (the short end of which ascends from the oil on the surface of tank A) has its longer end in Cylinder B, so that the longer end of the siphon allows oil to flow into tank B and strike the turbine.

The flow rate of the oil is 1 cubic metre per second and it falls 20m (after exiting the nozzle at the long end of the siphon) before striking the turbine.

An electric pump is used to prime the siphon flowing at a flow rate of 1 cubic metre per second. The pump consumes 30kW but requires only pulsed power because once the siphon has started working, it will continue working without help from the electric pump until the level of working fluid in tank A drops below the input nozzle of the siphon.

For reasons that will become clear, the level of the working fluid in tank A cannot fall below the level of the input nozzle of the siphon.

Now we come to the critical factor. An air compressor at the top of tank B (as well as being directed to increase the velocity of the working fluid as it travels down with the help of gravity to strike the turbine) is also used to pressurise the oil that accumulates at the bottom of tank B (in the tailgate area after having struck the turbine).

This higher pressure in tank B is needed to prevent the 350kPA (absolute) pressure at the base of tank A flooding tank B through the lower connecting pipe (also of diameter 0.12m) and it is also needed to force tailgate oil back into tank A (which is full of seawater and oil).

The pressure at the base of tank A is approximately 350,000 Pascals absolute, whereas the pressure at the base of tank B (which only contains a small height of oil) would only be 130,000 Pascals absolute before the air compressor operates.

The air compressor consumes 11kW and can pressurise the volume of air inside tank B (which is hermetically sealed) to 800,000 pascals within 7 minutes.

A constant pressure of at least 350Kpa must be maintained in tank B to prevent water from tank A forcing its way into the turbine tank (into tank B) and flooding the turbine housing.

This pressure is also needed to evacuate tailgate oil from tank B. So we are using cheaply generated pressure to control the flow of fluid, and we know the fluid must flow towards the lower pressure area (in other words it must flow where we want it to flow).

There is a pressure release valve at the top of tank A to prevent P1V1 = P2V2 equilibrium in the air gap at the top of tank A.

Preventing pressure equilibrium is critical as the system will want to equalise fluid and pressure levels immediately (and would most certainly do so were it not for the pulsed power air compressor, the air gap in tank A, and the pressure relief valve at the top of tank A).

The air compressor is float activated, so that when the level of oil at the base of tank B gets too close to the rotating turbine, the air compressor is then triggered, pressurising tank B from 350Kpa to in excess of 350Kpa and forcing tailgate oil through the lower connecting pipe back into tank A where it floats to the surface of the tank.

The flow of oil onto the turbine generates 160kW.

Power (watts) = 20(m) x 961(kg/m3) x 9.81 m/s/s x 0.85 (efficiency fraction)

However the pumps used to recirculate the working fluid (the siphon pump and air compressor) together consume only 41kW if operated continuously.

Note that neither the pump nor the air compressor must work continuously. Both use pulsed power when required. A computer controlled pressure/valve regulator can prevent pressure equalisation and maximise efficiency. I would be grateful if someone would be kind enough to send me a schematic for a board to regulate pressure automatically (can we build it...yes we can).

The siphon at the top of the cylinders minimises (to zero) the work that has to be done to move oil from tank A to tank B.

The principles underlying siphons are well established and do not need expansion here.

However the positive buoyancy of the oil leading it to float back to the top of tank A does not in fact provide energy benefit.

I used the example of a less dense working fluid (castor oil) simply to illustrate that work does not have to be done to 'lift' working fluid to the top of tank A (in the sense of work performed during the lifting process up through the height of tank A).

Of course work has to be carried out to force tailgate fluid back into tank A, but that is actually the only work the system must perform in order to operate. So although gravity is a conservative force and is not path dependent, the recirculation of fluid in this system is path dependent in terms of efficiency,and pressure regulation is an efficient way of carrying it out.

Strong ionic bonds between water molecules, if expressed (allbeit inelegantly) in terms of pressure, attract one another with relative pressure of 3000kpa. In other words, the water molecules are virtually chained together. This enables processes in nature to work as well, including the phenomenon of Giant Redwood trees being able to lift water 115m into the air at a rate of 160 gallons per day.

In these trees, water evaporating from the leaves creates a partial vacuum pulling water up from the roots.

In this system, water leaving tank A in the siphon pulls more water towards the input nozzle of the siphon.

Seawater throughout tank A (and also used as a working fluid) actually performs more efficiently because it has a higher density. We know from Newton's formula (F=m.a) that a fluid of higher density will cause higher energy output from the turbine because the mass in kg will be higher and we will not have to worry about the higher viscosity of oil at low temperatures decreasing the velocity of the working fluid.

My point is that the energy ostensibly gained by having a less dense working fluid (that uses buoyancy to float up a more dense substrate) is matched and neutralised by the lower energy generated by less dense working fluid striking the turbine.

These large cylinders can form the pillars of energy pyramids. Platforms at the top of the cylinders would make excellent locations for wind turbines. Solar panels can cover the pyramids enclosing the cylinders. Geothermal energy may be generated from beneath the cylinders.

Calculations:


Cylinder A = height 25m and diameter 1m
Cylinder B = identical to cylinder A
Working fluid = seawater of density 1020kg/m3
Flow rate of working fluid: 1 cubic meter per second
Height water falls before striking turbine = 20m
Diameter of turbine = 0.9m
Pitch Circle Diameter of turbine = 0.87m

First of all, we must calculate the maximum total electrical power in watts the system is capable of generating.

To do this we multiply the density of the working fluid (seawater = 1020kg/m3) by the height the fluid falls before hitting the turbine (20m), then by acceleration due to gravity (9.81 m/s/s) by the flow rate in cubic metres per second of the seawater (here 1 m3/s) and finally by a unit-less fraction representing turbine efficiency (85% = 0.85).

Pw ([power in watts) = 1020kg/m3 x 20m x 9.81 m/s/s x 1m/3/s (flow rate) x 0.85 (efficiency of turbine)

Pw = 170,105.4 watts = 170.1054 kW

So this is the maximum output we can get out of the turbine if 'the turbine' is 85% efficient. Other inefficiencies in the system must later be taken into account, but this is a good starting point based on head and flow rate.

We can calculate the force applied to the turbine by using Newton's equation F = m.a

If we assume for the moment that the acceleration of the working fluid will be the same as acceleration due to gravity (9.81 m/s/s), then we can calculate the force in Newtons that will be applied to the turbine by the flow rate.

F (Newtons) = Mass (kg/s) x Acceleration (m/s/s)
F = 1020kg/s x 9.81 m/s/s
F = 10006 Newtons

This is a useful piece of information.

We now know that over 10,000 Newtons of force will be applied to the buckets of the turbine by the flow of fluid striking it from above. This is an enormous amount of force.

Once we have decided the diameter of the impulse turbine (which must be less than 1m in diameter to fit inside the cylinder and must be greater than 0.2m in diameter if we are to avoid serious inefficiencies), we can also use this force figure (Fjet) in Newtons to calculate the angular velocity of the turbine in radians per second (which I converted to RPM below).

So we should be able to calculate how fast the turbine will rotate just from knowing the value of Fjet (which we calculated to be 10,006 Newtons) as well as other variables we established earlier.

I have chosen to use a Pelton impulse turbine of diameter 0.9 meters. I could have chosen a smaller diameter. I decided not to. However I could not have chosen a much larger diameter because it would not have fitted inside the system cylinder.

In fact a 0.9m diameter turbine is a giant by Pelton turbine standards. Lets look at the maths.

The equation for determining the mechanical power output in watts of a turbine is as follows:

Pmech (watts) = Fjet x Njet x pi x h x w x d / 60

Explanation:

Pmech = 170,000 watts (from the first calculation of maximum power output)

Fjet = Force in Newtons of the water striking the turbine = 10,006 Newtons (calculated above)

Njet = number of water jets = 1 jet nozzle

pi = 3.141592654

h = efficiency coefficient (unit-less fraction between 0 and 1). This is always going to be a figure between 0.69 and 0.94 for Pelton turbines. The larger they are, the more efficient they become, so 0.85 efficiency for a large turbine is an acceptable estimate = 0.85)

d = pitch circle diameter of the turbine in meters (this will be a slightly smaller diameter than the outer diameter of the turbine = 0.87m)

w = rpm (here not rad/s) which is the mystery value

The mechanical power output in watts is going to be approximately the same as the electrical power output in watts (if we make allowance for heat dissipation and the inefficiency of components other than the turbine itself which we can do at any later point).

Accordingly, applying the Pmech equation:

170,000 (watts) = Fjet (10,006 Newtons) x Njet (1) x pi (3.141592654) x h (0.85) x w (mystery value in rpm) x d (0.87m) / 60

= 10,006 X 1 X Pi x 0.85 x RPM x 0.87 / 60
= 23246 x w / 60
170,000 = 23246w / 60
170,000 = 387.4333w
w = 438.78 RPM

So the RPM figure look reasonable. It we had obtained a very high RPM figure this would been worrying because high RPM values would cause a turbine under this sort of force to fail after a few months of use.

Pressure calculations:

When considering the pressure in the two cylinders, the first thing we need to calculate is the pressure at the base of cylinder A (which is 25m high and full of seawater).

We can ignore the small air gap at the top of cylinder A for the moment because the air gap would reduce base pressure rather than increase it.

The formula for calculating the pressure at the bottom of a cylinder is as follows:

P = height(m) x density(kg/m3) x gravity (9.81m/s/s)
P = 25m x 1020kg/m3 x 9.81m/s/s
P = 250155 Pascals

However this is gauge pressure. We need to add atmospheric pressure to obtain the absolute pressure value of the fluid in the base of cylinder A.

Adding 101,325 Pascals of atmospheric pressure gives us an absolute pressure value at the base of Cylinder A of 351,480 Pascals.

This means that the tailgate water in Cylinder B must somehow force its way back into Cylinder A despite there being a pressure of 351.5 Kpa in cylinder A.

The pressure in Cylinder B, which is hermetically sealed and is not subject to atmospheric pressure, is due only to the height of the tailgate water contained in it. We have not switched on the air compressor yet.

The tailgate water is only 2.5m high. So the pressure at the base of cylinder B is only 25,000 Pascals. Even if it were also subject to atmospheric pressure (which it is not) it would only have a pressure of 125,325 Pascals.

However, the air compressor at the top of cylinder B comes to the rescue. It can pressurize the volume of air in tank B to 800,000 Pascals in 10.58 minutes.

The compressor in question is the Abac Genesis 1108 air compressor which can provide a maximum pressure of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute.

The volume of tank B (h=25m d=1m)
I calculated using the formula V= pi.r2.h
V= pi x (r x r) x h (where r = radius in metres and h= height in metres)
V = 3.141592654 x (0.5 x 0.5) x 20m = 15.7m3.

The volume of air in tank B (after deducting the tailgate water taking up 10% by volume of the cylinder) is 14.13m3.

The air compressor takes just over 8 minutes to pressurize the 14.13m3 of air inside tank B to 800,000 Pa.

In fact, the air compressor does not need to create 800,000 Pascals of pressure inside tank B.

It only needs to exceed the pressure at the base of Cylinder A (351,480 Pascals).

Once 351Kpa pressure has been exceeded, the system will try to equalise pressure in both of the connected cylinders as per the formula:

P1V1 = P2V2

This formula means that the pressure multiplied by the volume in one cylinder (P1 x V1) will always equal P2 x V2 in a connected vessel (unless some force prevents equalisation).

Here the force preventing equalisation is provided by the air compressor. The pressure relief valve in Tank A breaks the equalising pressure circuit from continuing its journey into tank B.

Note that the output of the pressure relief could be used to perform work if I need to make amendments to the schematic.

In any event, once the air compressor kicks in, tailgate water must move from the area of higher pressure (at the base of tank B) into the base of tank A (which has now become the lower pressure area).

So the tailgate water is forced through the lower connecting pipe back into tank A, whereupon the siphon recirculates it back into tank B.

The pressure relief valve at the top of tank A prevents the pressure in the air gap exceeding 350Kpa. So any compressed air or excess fluid forced into tank A (which will try and cause the pressure in tank A to become the same as in tank B) will be released by the pressure relief valve, thus ensuring no equalisation of pressure in the two tanks (or dangerous pressure build up in tank A).

This air compressor consumes 11kW of electricity when operating at maximum capacity.

Maximum capacity involves generating 800,000 Pascals of pressure.

The air compressor should be able to expel tailgate water from tank B at less than 50% of its operating capacity. In other words, it will consume approximately 5.5kW of power to pressurize tank B to just over 351,480 Pascals.

Only when the pressure in tank B falls below 351,480 Pascals will the float trigger the air compressor (only when the water level in tank B nears the turbine will the air compressor be activated).

However, even if the air compressor continuously consumed 11 kW, it would still consume only a small fraction of the 170kW output of the turbine.
 
G

Get A Grip People

> Complicated Stuff ....

>However, even if the air compressor
>continuously consumed 11 kW, it would
>still consume only a small fraction of
>the 170kW output of the turbine.

So what you're proposing is a completely closed system which you put 11Kw into, but get 170kW out of? That's what engineers call a violation of Thermodynamics.

Or have you got a clearer explanation for an idiot like me of where the 159kW of energy actually comes from.

Note to moderators: Peg, maybe its time this thread just got locked.
 
If I understand your proposal correctly, you intend to use your siphon to fill up the bottom three feet of tank B with seawater, but then you will have to pressurize tank B to force out the tailgate seawater, meaning that water will not be flowing from tank A during the pressurization cycle.

So, you'll be moving a volume of seawater 3 meters tall by 1 meter in diameter at a rate of 1 cubic meter per second over your turbine. That means you're going to be moving about 2.35 cubic meters of water in about 2.35 seconds. Your theoretical output of 170 kw from the generator will only take place over those 2.35 seconds, but then your air compressor will have to draw 5.5 kw for at least 8 minutes.

It's been a long time since my high school physics class, but if I'm doing the conversions correctly, and if I'm understanding your plan correctly, your air compressor is going to draw about 7 times the kWh in 8 minutes as your turbine produces in 2.35 seconds. Even that is assuming that your turbine operates at peak efficiency for the entire 2.35 seconds, when I suspect that a good portion of the energy from the water flowing over it will be expended just to get the turbine spinning from a state of rest.

If you think about it, that only makes sense. You are converting the kinetic energy of water dropping 20 meters into electricity, but then you use electricity to raise a 22 meter water column an additional 3 meters (more or less). I understand about the cohesive properties of water that you described (the redwoods and all that), but that doesn't help the fact that you still have to inject all of your tank B water into the bottom of a water column in tank A. Injecting the water against the pressure of that water column is pretty much the same as lifting the water column.

You could probably make your idea closer to efficient by making tank B only 4 meters tall so that you don't have to pressurize as much volume. After all, you only need the pipe to drop 20 meters, not the whole tank. You would still need a valve to close your pipe before you begin pressurization of course. You could also save a bit of energy using a cascading system of air tanks to recapture some of the pressure from Tank B at the end of the pressurization cycle, thus reducing the workload of your air compressor a bit.

Still, you will not get over, or even near, 100% efficiency. Physics just can't get around the fact that you're trying to raise a 22 - 25 meter water column with the energy from a 20 meter waterfall.
 
Q

quantumtangles

It is not a closed system. It is an open system. Both mass and energy may enter and leave the system boundary.

What interests me is why the system gives the appearance of being able to work. Just as perpetual motion wheels give this appearance but cannot work as all forces and all torques in the system are not taken into account by inventors.

Any fool can quote the laws of thermodynamics. But only people who actually understand them they have learnt parrot fashion. But only someone who actually understands them (as opposed to parroting what they think will be well received) will be able to pinpoint the precise reasons why this system does not work.

Doing so would be educational and very welcome to the inventor.

Consider it an IQ test.
 
Q

quantumtangles

Quote
"If I understand your proposal
>correctly, you intend to use your siphon
>to fill up the bottom three feet of tank
>B with seawater, but then you will have
>to pressurize tank B to force out the
>tailgate seawater, meaning that water
>will not be flowing from tank A during
>the pressurization cycle."

That is very interesting. In fact it would be filling the bottom 2.5m of tank B with seawater. But yes you are correct. The water has to be forced out of tank B by pressure generated in tank B by the air compressor.

However the pressure in tank B can maintain over 350Kpa throughout the cycle (spending energy only as is required to maintain higher pressure than tank A).

Flow would not have to stop at any point in the cycle because the pressure relief value at the top of tank A prevents pressure equalisation between the two tanks.

This allows tank B to remain pressurised to over 350Kpa at the same time as tank A always has a pressure of less than 350Kpa.

You make an interesting further point about the energy needed to move the water.

But if the fluid must flow from high pressure to low pressure areas (eg from tank B to tank A) at the bottom of the system, and if the pump assisted siphon keeps water flowing at the top of the system, this is the point that needs clarification.

Many thanks for your objections.
 
S
I suspect that at least one of two things is the case.

Either

1) You will not be able to sustain a siphon from a low-pressure vessel into a high pressure one, even with a starter pump, or

2) You've under-estimated the hp requirement to lift the flow rate at the effective lift to get it back to the full tank.
 
Q

quantumtangles

You have thought the system through carefully and your objections are logically sound.

First objection: (whether the siphon can work if it flows from a low pressure to a high pressure area). The fact it is a siphon (with a longer column of fluid hanging over the high pressure area) may help, but you may right in spite of the powerful pump that assists the siphon.

Your second objection (about needing power to lift water) is also interesting. As devil's advocate, I argue that a castor oil system (where one could use less dense working fluid which would float up through the seawater in tank A because of positive buoyancy) illustrates that work does not have to be done to lift working fluid (though I accept there would be no energy gain by using less fluid). So it is not about lifting working fluid a certain distance. Rather it is all about how much work must be done to pressurise the working fluid so as to force it back into tank A.

However you may have identified the dilemma. One is forcing fluid to enter the base of tank A under high pressure, the same high pressure that may prevent the upper siphon from flowing at the top of the system. You are saying in effect that you cannot have it both ways (from high to low pressure and from low to high pressure) in terms of fluid flow and this is an interesting objection.

Thank you for thinking it through.
 
S
"So it is not about lifting working fluid a certain distance. Rather it is all about how much work must be done to pressurise the working fluid so as to force it back into tank A."

Well, when I said "lifting" I meant in the same more general sense in which you speak. If the liquid ends up at a higher elevation by the time force is no longer being applied to it, it has been "lifted" in that sense. The energy for the lift must be added to the fluid.
 
>But if the fluid must flow from high
>pressure to low pressure areas (eg from
>tank B to tank A) at the bottom of the
>system, and if the pump assisted siphon
>keeps water flowing at the top of the
>system, this is the point that needs
>clarification.

Ok. I think this is your problem then. If you're going to maintain enough pressure in Tank B to force fluid into the bottom of the water column in Tank A, then that same pressure will also force the water in your siphon back into Tank A. The pump on your siphon will have to do just as much work to put fluids into tank B as your air compressor is doing to put fluids into the bottom of Tank A.
 
Q

quantumtangles

Yes I understand. I think your earlier post pinpointed why the system cannot work.

Fluid will indeed flow from the high pressure tank B to the lower pressure tank A, but the problem is in getting this fluid from the low pressure tank it is now in, back into the high pressure of tank B.

Also my calculation of the energy required to generate enough pressure to expel tailgate water is suspect. My figures for the air compressor are in cubic metres per minute, whereas the water flow rate is 1 cubic metre per second.

Accordingly, I have to ask myself whether constant high pressure in tank B would do all the required work (in terms of shifting water in the lower part of the system) or whether a cubic metre of air per second would be needed to expel a cubic metre of water per second (though I don't think the latter would be right as the fluid must flow from the high pressure to the low pressure area regardless of the work that was needed to create the pressure differential).

It is the fact of there being a pressure differential that is critical. How that pressure differential was generated is irrelevant. How much energy was needed to create the pressure difference does not matter (in the sense the fluid must move from the high to the low pressure area).

So the system fails not because the fluid wont move back into tank A via the lower pipe, but because the siphon wont work when tank B is at higher pressure than tank A.

And this is a reason for failure quite distinct from the more pressing issue of energy expenditure.

Thanks again.
 
>Are you familiar with the term 'comic relief'?

Your friendly local moderator here. I've gotten serveral off line comments and complaints about the direction this thread has taken. Comic relief or not, I'm going not going to be posting any more messages to this thread. It think it's time has come and gone judging by the mail in my inbox.

Happy Wednesday everyone!
Peg Ferraro
 
Top