Recirculating fluid turbine invention

[quantumtangles]

Two Cylinders A and B, 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 does not fall below the level of the input nozzle of the siphon.

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).

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.

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.

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.

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


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

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

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 the turbine diameter and other variables we established earlier.

I have chosen to use a Pelton impulse turbine of diameter 0.9 meters.

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 looks reasonable and the turbine should last more than a few months.

Pressure calculations:

For Cylinder A

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.

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.

[rfesalbon]

Congratulations!!! for very innovative invention, you also explained it in great detail. You mentioned about the compressor company can you help us on the company that manufacture the pelton turbine based on your specs.

Thanks..

[Vincevl]

Very interesting!

A very interesting idea! You've given this some serious thought.

As with most turbine generators you need to do this on a large scale, such as your 20m fall. This makes testing theories very difficult.

Do you think the oil will become foamy after a few passes through the turbine due the churning effect and from air introduced via the compressor? This would affect your free fall calcs.

[jibbguy]

A very interesting and novel approach. However, a question/observation about the generation turbine (output). The operation would appear to me to not be "continuous", and that the "horsepower" and kW output would vary depending on the feeding tank level/pressure. Therefor the actual output rating would need to be an average over at least one complete cycle... and that would be difficult to calculate without empirical data.

[quantumtangles]

Turbine manufacture

Congratulations!!! for very innovative invention, you also explained it in great detail. You mentioned about the compressor company can you help us on the company that manufacture the pelton turbine based on your specs.

Thanks..

Apologies for the delayed response.

I had my first Pelton turbine manufactured by a company in the United States. If I can remember their name I will post it.

My second turbine was bought on ebay (there is only one vendor I am aware of).

Both these turbines were for lab use and were much smaller than the turbine referred to in the invention.

A web search will highlight the few companies capable of making impulse turbines. Alternatively you could buy only the turbine cups and attach them to a hub of the required diameter.

The Micro-hydro Pelton Turbine Manual by Jeremy Thake is an excellent source of information about turbines including how to make cast Pelton turbines (a fantastic book but beware errors in equation derivations on pages 58 and 136 (Practical Action Publishing (2000)).

[quantumtangles]

A very interesting idea! You've given this some serious thought.

As with most turbine generators you need to do this on a large scale, such as your 20m fall. This makes testing theories very difficult.

Do you think the oil will become foamy after a few passes through the turbine due the churning effect and from air introduced via the compressor? This would affect your free fall calcs.

Initially I thought building a model of the system could not work because of friction. But I changed my mind. I think a properly constructed model would be capable of demonstrating viability just as well as a full size machine.

Note that the cost of building a high quality model might actually exceed the cost of building a full size machine.

As for using oil, I concluded seawater would work best on its own (not least for the reason you highlight).

[quantumtangles]

A very interesting and novel approach. However, a question/observation about the generation turbine (output). The operation would appear to me to not be "continuous", and that the "horsepower" and kW output would vary depending on the feeding tank level/pressure. Therefor the actual output rating would need to be an average over at least one complete cycle... and that would be difficult to calculate without empirical data.

Quite right. Initially I thought operation could be pulsed. I was wrong. There must be continuous operation of both siphon pump and air compressor.

The weaknesses of the system (leaving aside an unfortunate 'impression' of greater output than input) are that high air pressure in the turbine housing will to some extent impede rotation of the turbine (the turbine must fight through a soup of high pressure air, lowering RPM and therefore lowering power output in Watts).

Secondly, power output calculations in Watts are themselves dependant on on velocity in m/s of the water flow in the siphon.

Velocity in turn depends (per Bernoulli) on pressure in Pascals. So the key question is whether 400kPA or more is possible in the siphon.

If the pressure really can be that high for the energy expenditure calculated, then high water velocity must follow and accordingly there must be high power output in Watts.

These are the best theoretical objections I can think of.

I do not consider 'greater output than input' a valid objection because output does not exceed input. Much more energy is consumed by the machine (when energy provided by the force of gravity is taken into account) than is generated.

[quantumtangles]

My most recent calculations (and revised system specifications) are as follows:

We have two 30m high steel cylinders 1m in diameter.

Tank A is full of seawater. Tank B contains only 10% by volume of seawater. An impulse turbine is placed at the base of tank B above the water line. Two pipes connect the cylinders. An upper siphon leading into tank B from the open surface of tank A, and a lower connecting pipe leading from the base of tank B to the base of tank A. Water flows from tank A into the siphon by means of a water pump. The water enters the siphon and is accelerated by compressed air from the air compressor before it leaves the siphon nozzle. This water strikes the impulse turbine 25m beneath with a flow rate of 1m3/s.

The tailgate water is ejected from tank B into tank A by means of higher pressure in tank B. A pressure relief valve in tank B prevents pressure exceeding 501kPa. Tank B pressure always exceeds tank A base pressure (401kPa). Tank A surface pressure is always Patmos (101325Pa). The fluid recirculates and the turbine produces electricity.

Summary:

Steel Cylinder A: Height 30m Diameter 1m (full of seawater)
Steel Cylinder B: Height 30m Diameter 1m (3m depth of seawater)

Pelton Impulse Turbine: Diameter 0.9m (PCD 0.87m) (Efficiency 0.9)
Abac Genesis 1108 air compressor (11kW 800kPa @ 1.67 m3/m = 0.027833 m3/s)
ESP-10 150 siphon water pump (37.77kW 1m3/s flow rate)
Drawn Copper Siphon Pipe: Diameter 0.38m
Drawn Copper Lower Connecting Pipe: Diameter 0.38m
Relative pipe roughness: 0.0000125

Back flow prevention valve
Supplying the widest range of rainwater harvesting filters and components

Pressure relief valve
Chemical & Process Technology: Useful Documents Related to Pressure Relief Valve (PRV) - Part 1


Working fluid: Seawater of density 1020kg/m3
Fluid viscosity 0.00108 Pa-s

Pressure in Tank A
Surface Pressure = Patmos = 101350 Pa
Base Pressure
= 30m x 1020kg/m3 x 9.81 m/s/s
= 300,186 Pa Gauge Pressure + Patmos (101350 Pa)
= 401,536 Pa Absolute

Operating (Absolute) Pressure of tank B 500kPa

Water Jet Velocity Vjet

Bernoulli's equation gives us the velocity of the water jet applied to the turbine.

P = ½ r . V2

P = Pressure (401,536 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value

401536 = ½ 1020 . V2
401536 = 510 . V2
V2 = 401,536 / 510
V = 28 m/s

Force of water Jet (Fjet)

1 m3/s of seawater (1020kg/m3) represents a flow rate of 1020kg/s.

F = 1020kg/s x 28 m/s/s
= 28560 Newtons

Fjet Momentum Change Calculation

Turbine speed may not exceed 50% of water jet speed

Vjet = 28 m/s
Vrunner = 14 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet – Vrunner)
Delta Mom = 1020kg/s x (28 m/s – 14 m/s)
Delta Mom = 14280 N

Fjet = 14280N.

Turbine RPM

We can now calculate the RPM figure for the turbine based on runner velocity
of 14 m/s.

First we need to know the circumference of the turbine.

Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference

Vrunner = 14 m/s
RPS = 4.951487 revolutions per second x 60
= 297 RPM

Power Output

Applying 297 RPM and Fjet = 14280 Newtons to the Pmech equation:

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 14280N x 1(jet) x pi x 1m3/s x 297RPM x 0.9(eff) x 0.87m / 60
= 173Kw

Cross referencing this output figure with the conventional equation for electrical power output in watts:

Pwatts = h(25m) x g(9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

It would be highly conservative and I think reasonable to take the lower of these two figures to represent maximum electrical output of the turbine (173kW).

Calculations of electrical output from hydroelectric facilities always rely on the second (higher output) equation based on head and flow rate (because it always correctly predicts output).

But being conservative to help you debunk the system, I am happy to give you a 52kW head start.

Maximum total system power output = 173,000 watts

Tank Pressure Dynamics

When the system is operating, tank A base pressure is always 401,350 Pa. Tank A surface pressure is always Patmos (101,350 Pa).

But tank B is the place of interest in terms of pressure dynamics.

The air compressor (Abac Genesis 1108) has output of 800,000 Pa at a rate of 59 cubic feet per minute = 1.67 m3 per minute = 0.027833 m3/s.

The volume of tank B (h=30m d=1m)
V= pi.r2.h
V= pi x (r x r) x h
V = 3.141592654 x (0.5 x 0.5) x 30m
= 23.56m3.

The volume of air in tank B (after deducting the 3m tailgate water taking up 10% by volume of the cylinder = 2.356m3)
= 21.2m3.

The air compressor takes 12.697 minutes to pressurise the 21.2m3 of air inside tank B to 800,000 Pa. We can reduce the volume of tank B by 86% by using just a pipe with a shorter cylinder beneath it, but I have stayed with a full size tank B with 3m of water in the tailgate.

The air compressor has to do the work of exceeding the pressure at the base of tank A (401,536 Pascals) to force fluid out of tank B.

There is a back flow prevention valve in the lower connecting pipe. The working fluid must move water from a high pressure to the lower pressure area.

Once air pressure has built up in tank B, it can only leave tank B through the upper siphon (which would stop the system working) or the lower connecting pipe (which would keep the system working).

The problem seems to be that the siphon may not work because of the high pressure in tank B. Surprisingly this is not the case.

First, the tailgate water (3m in depth) supplies pressure of 30,018.6 Pa.

P = 3m x 1020kg/m3 x 9.81 m/s/s
= 30,018.6 Pa

Secondly and much more importantly the output pipe of the air-compressor is connected to the siphon exit nozzle (this is the only place the air compressor output exits into tank B).

The force applied by the compressed air is not going to increase just because we apply it inside a thin siphon pipe.

Nonetheless pressure increases dramatically. Not because the force has increased, but because the area over which the force is applied in the nozzle has decreased.

The siphon pipe is 0.38m in diameter. Delta pipe (a cross sectional area of the siphon pipe) = pi*r2
Delta pipe = 0.1134 m2

This is the surface area to which pressure from the air compressor is applied to the water exiting the siphon (whilst it is still inside the siphon).

The 800,000 Pascal output of the air compressor (at a air flow rate 0.027833 m3/s) is delivered to the siphon nozzle over an area of 0.1134m2.

We know the area of delta pipe is 0.1134 m2 (the siphon nozzle).

In comparison the surface area of the inside of tank B (height 30m) is:

A = 2*pi*r2 + 2*pi*r*h
A = 1.57 + 94.24778
A = 95.8177 m2

So if the maximum pressure applied by the air compressor to the walls of tank B is 800,000 Pa, the pressure applied over the much smaller area in the siphon nozzle will be 845 times greater.

P = 422500000 Pa = 422500000 N/m2 = 42250 N/cm2.

Not because the force has increased. But because the area has decreased.

In practical terms, the pressure applied to the walls of tank B and the fluid column at the bottom of it will be no more than 500kPa (because of the pressure relief valve built into tank B).

But the pressure relief valve will not prevent very high pressure being generated inside the siphon nozzle because this pressure of 4222500 kPa will decrease as soon as it enters the wider chamber of tank B.

Not because the force has decreased, but because the area has increased.

In summary, the pressure applied in the siphon nozzle will significantly exceed 600Kpa.

So pressure in tank B is 500kPa. Pressure in the siphon nozzle is over 600KPa (actually it would be 4222500kPa) and pressure at the base of tank A is 401350 Pa.

The water must recirculate from tank B into tank A, and the siphon nozzle must continue to operate despite 500kPa pressure in tank B.

Conclusion

Power output is 173kW. The air compressor consumes 11kW. The water pump consumes 37.66kW. Net power output is 124kW.

You may be thinking the maths don't make sense because right at the beginning I assumed water pressure of 401350 Pa for the purpose of calculating fluid velocity.

But if you consider that the the water pump and in particular the air compressor increase siphon nozzle pressure to over 600kPa, velocity per Bernoulli and power output would be even higher:

P = ½ r . V2

P = Pressure (600,000 Pa)
rho = density (1020kg/m3)
V = velocity (m/s) mystery value

600,000 = ½ 1020 . V2
600,000 = 510 . V2
V2 = 600,000 / 510
V = 34.3 m/s

F = m*a
= 1020kg/s x 34.3m/s/s
= 34986 N

Vjet = 34.3 m/s
Vrunner = 17.15 m/s

Delta Mom = mass flow rate x Delta V
Delta Mom = mass flow rate x (Vjet – Vrunner)
Delta Mom = 1020kg/s x (34.3 m/s – 17.15 m/s)
Delta Mom = 17493 N

Fjet = 17493N

Vrunner = 17.15 m/s
Diameter = 0.9m
radius = 0.45m
2.pi.r = 2.827433388m circumference
RPS = 6 revolutions per second x 60
= 360 RPM

Pmech = Fjet x Njet x pi x flowrate x RPM x 0.9 x 0.87m / 60
= 17493N x 1(jet) x pi x 1m3/s x 360RPM x 0.9(eff) x 0.87m / 60
= 258Kw

Cross referencing this output figure with the conventional equation for electrical power output in watts:

Pwatts = h(25m) x g(9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

Whichever way you look at it, output would be over 200kW. Power consumption is less than 25% of output.

It does not breach the laws of thermodynamics because it is an open system in which both mass and energy may pass through the system boundaries. Note also that the process is irreversible, and that entrophy increases due to frictional heat losses.

It is tantamount to a giant electric pressure jet but has a high mass flow rate.

The siphon nozzle has become the nozzle of the water jet. 500kPA pressure in tank B cannot stop the 1020kg/s juggernaut entering tank B because the siphon water is pressurised to over 600kPa by the air compressor (in the narrow confines of the siphon exit nozzle).

Equally, the tailgate fluid must leave tank B through the one way flow valve in the lower connecting pipe because tank A base pressure (401.35kPa) is lower than tank B overall pressure (500kPa).

Interestingly, tank B pressure, once raised by the air compressor, will remain at that pressure despite fluid being evacuated back into tank A. This is because every cubic metre of water that leaves tank B is replaced by another cubic metre of water entering tank B through the siphon nozzle.

Tank B will stay at 500kPa once pressurised (because it has a pressure relief valve triggered at 501kPa).

The pressure will only fall if there is a reason for it to fall. Evacuation of one cubic metre of water per second through the lower connecting pipe is balanced by one cubic metre of water per second flowing from the upper siphon. So the mass flow in itself does not cause air pressure to fall in tank B (in the sense lower water volume reduces pressure).

The gas fluid pressure in tank B will not find its way into tank A because only aqueous fluid is forced into tank A. Not air.

A pressure relief valve on tank A releases any compressed air that manages to get into tank A and also releases any gases released by the agitated water.

Higher fluid pressure in tank A would only help the siphon, not hinder it. A pressure relief valve prevents excessively high water pressure in tank A. But the turbine also removes kinetic energy from the water before it re-enters tank A.

The one way flow valve in the lower connecting pipe prevents the water at the base of tank A flowing into tank B.