Recirculating fluid turbine invention

[eisnad karm]

Hi
many thanks for the posts
there are osmotic membranes that now have a high flow rate. I will send some links when I find them.
Will send a longer post latter.
mark

[quantumtangles]

Mathematics

The specifications of the final system are as follows:

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 metres. 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.

Please stick around for part II of the maths relating to the system which I will post shortly.

[quantumtangles]

In part I of the mathematics posting relating to this system, we established electrical output of the turbine would be 170kW using a 0.9 metre diameter turbine rotating at 439 RPM.

This is all very well provided the system can recirculate the working fluid.

If it cannot do so, the turbine will only work for a few seconds before getting flooded by tailgate water that builds up in cylinder B.

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 pressurise 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 pressurise 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 pressurise 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 11kW, it would still consume only a small fraction of the 170kW output of the turbine.

In part III of the system mathematics I want to look at the pressure calculations in more detail.

[quantumtangles]

Here are images of the schematic for the recirculating fluid turbine. The first image shows the entire system. Clearer pictures of part of the system follow as the labels on the main image are hard to read.

[quantumtangles]

We know from sections I and II of the earlier calculations that:
Tank A has base pressure of 351480 Pascals
Tank A water volume = 15.7m3
Tank B has base pressure of 25,000 Pascals.
Tank B water volume = 1.57m3

The system will try to equalise pressure and volume. Ceteris paribus:
P1V1 = P2V2

If P2 is the mystery value
351480 x 15.7 = P2 x 1.57m3
5518236 = P2 X 1.57m3
P2 = 3,514,800 Pascals
This is an enormous figure (3514 Kpa)

Who can explain why tank B does not need to be pressurised to 3514kPa?