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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.
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.
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