The debate surrounding alternative energy sources has not died down but is becoming more burning with every passing day. This article partly (and maybe directly) discusses the material published in issue No 3 of the “Alternativnaya Energetika i Ekologiya” (“Alternative Energy and Ecology”) magazine in 2005, an article entitled “A new generation of damless hydroelectric stations based on hydro-energy units”.

A group of engineers has constructed a hydraulic turbine to receive energy from a free flow of water (a free flow hydraulic unit). However, when its capacity was measured it was established that it generated more energy than it was designed for. it is well-known that a flow of water has kinetic energy that can be extracted (which is what free-flow turbines do). However, it is impossible to extract all of its kinetic energy. in order to do this, the flow should be stopped completely and then it would cease to be a flow. That is why the velocity of water flow at the exit from a working unit of turbine is slower than its flow at the entrance – it is precisely this difference that defines the efficiency of any facility. Considering the fact that the kinetic energy is known as being proportional to the square of the speed, and that the energy decreases by four times when the speed decreases, it will be easy to calculate that, let’s say, when the water flow speed at the turbine input and output is equal to 1m/sec and 0.5 m/sec, respectively, we will be able to extract 75% of the kinetic energy from the flow.

 Strictly speaking, the power of the free-flow turbine is calculated by a semiempirical formula (1) (this formula can also be applied to calculate the power of wind turbines)

P = K * V ³ * S * p (1)

where

V - incoming flow speed

S - the square of the turbine’s effective cross section across the flow
p - moving medium density

K - constant coefficient that depends on a turbine type and is usually equal to 0.1 - 0.35

 This formula represents the very kinetic energy of the flow per a time unit, because V * S * p is right the water mass that goes through the turbine at one second and the formula (1) takes on the following form, which is familiar to us:

E = m * V ² / 2 = ( V * S * p ) * V ² / 2

However, it should be considered that, according to the flow continuity condition, the flow’s square must increase when the outward flow’s speed drops. This leads to degradation in the flow evenness at the turbine’s outlet and an increase in turbulence, which negatively effects the unit’s efficiency. in order to decrease these factors’ adverse effect in traditional turbines, expanding cones are installed at their outlets, which partly increases the efficiency.

 Because empirical coefficient K of the formula (1) includes the twain from the kinetic energy formula denominator, the hydraulic and mechanical efficiency coefficient of the turbine, losses per irregularity and turbulence in the incoming flow and so on, it accepts values of less than 0.3. This coefficient is measured through an empirical way by means of natural tests of a specific turbine.
This coefficient is often called the WEUC, the watercourse energy utilization coefficient, by an analogy with the wind turbine WEUC – the wind energy utilization coefficient.

 But let’s get back to our machine. As we have already mentioned, this facility produced even a greater amount of energy than the total kinetic energy of the flow.

 Where does this additional energy received from the facility come?

Does the flow of water have kinetic energy only?

(Here we do not consider the internal (thermal) energy of water or the energy of the intermolecular and interatomic bonds of water as a substance.)

Let us try to answer these questions.

Let us take one cubic metre of water (with dimensions of 1m * 1m * 1m) flowing with a velocity of 1 m/s.

There is no doubt about its kinetic energy, which is:

 Ek = m * V ² / 2 = 1000(kg) * 1(m/s) ² / 2 = 500 (Joule ) (2)

However, there is also pressure by the top layers of water on the bottom ones (potential energy). if we let this cube of water spread, then we can extract it. Considering that the gravity centre of the cube is at the middle of its height, that is h = 0.5 m, it is equal to:

Ep = m* g * h = 1000(kg) * 9.8 (m/s²) * 0.5(m) = 4900 (Joule ) (3)

This means that the potential energy of this cubic metre of water is up by almost 10 times on its kinetic energy. it is easy to calculate that, at a speed of 0.5m/sec, this difference will amount to almost 40 times!

it should be noted that in the formula (3) a half of the water column height is taken as h because a separate water volume’s height will decrease from the overall to zero as it will flow. For an infinite water flow with constant depth, which will be reviewed later, the incoming flow’s full depth is taken as water column height.

in other words, we can see that – in addition to the kinetic energy – the flow also has potential energy whose magnitude depends on the flow’s depth. But its exergy (that is the recoverable energy which is able to actually work) is equal to zero at regular conditions. After all, any volume of water is surrounded by water with the same characteristics (depth, speed, temperature). This can also be related to the air. We know that the air surrounding us has a significant amount of energy because the air has non-zero pressure and temperature. But for the same reason mentioned previously, its exergy is equal to zero and it is, therefore, useless from the energy viewpoint (later we will see that it is not useless all the time). (Brodyanskiy V.M “Exergic analysis. Energy: the problem of quality” “Nauka i Zhizn” (“Science and Life”) #3, 1982)

Now let us imagine that we are extracting part of kinetic energy from a cubic metre of water, which is flowing within a current, and use it to “move aside” the cubic metre of water that follows it (downstream). That is we will speed up the downstream cubic metre of water by slowing down the upstream volume of water. As a result, a level difference arises between them and potential energy emerges in the difference between these levels, which can be extracted from the current. The following question arises: will the amount of the extracted potential energy be more, less or equal to the energy used to speed up the second cubic metre of water – or, in other words, the energy expended to increase its kinetic energy?

Let us resort to mathematics.

As an example, we will consider a machine that is shown as a diagram on Picture 1, which makes it possible to speed up the outflowing stream of water by extracting part of the inflowing stream’s energy - that is, a machine with positive feedback between the energies of the inflowing and outflowing streams. By the way, a machine that works on this very principle has been invented. it is this machine that our story started with.

Explanations for Fig. 1:

1 - Working parts of the inflowing stream of water;

2 - Working parts of the outflowing stream of water;

3 - Working parts ensuring positive feedback between the inflowing and outflowing streams of water;

4 - Mark showing the level of the inflowing stream of water;

5 - Mark showing the level of the outflowing stream of water;

6 - Channel bed

H1 – Actual depth of the inflowing stream of water

H2 – Depth of the outflowing stream of water

V1 – Velocity of the inflowing stream of water

V2 – Velocity of the outflowing stream of water

h – Drop between the levels of the inflowing and outflowing streams of water

 

The device works based on the following principle:

The working parts of the inflowing stream 1 extract part of the kinetic energy from the stream and transmit it - with the help of the positive feedback 3 - to the working parts of the outflowing stream 2, which give the outflowing stream additional acceleration.

Because the amount of water entering the device is equal to the amount of outflowing water, and the speed of the outflowing stream is higher than that of the inflowing stream, then the sectional area of the outflowing stream will be less than that of the inflowing stream.

 Therefore, its depth H2 will be less than the depth of the inflowing stream H1 by the value h. As a result of this, potential energy appears between the different levels of the inflowing and outflowing streams.

The device’s energy balance is as follows:

E = K1 + Ph – K2 (4)

The total output of energy from the device is equal to the potential energy of the difference between the marks plus the kinetic energy of the inflowing stream and minus the kinetic energy of the outflowing stream. After omitting all the computations, we have:

E = M * ( g * h + (V1² * (1 - (H1 / (H1 - h)) ² ) / 2 ) (5)

or

E = M * ( g * H1 * (1 - V1 / V2) + (V1 ² - V2 ² ) / 2 ) (6)

 where M is the weight of the water entering the device in a unit of time, which is equal to the density of water multiplied by the active area of the inflowing stream and multiplied by its velocity.

Then the most interesting aspect occurs. it can be seen that the left side of the equation, which is in brackets, will increase in a linear fashion when it depends on h or in a hyperbola when it depends on V2, whereas the right part will decrease, and in a parabola at that. Which side will gain the upper hand?

Let us plot a graph showing energy’s dependence on the drop between the levels h. The graph will be plotted to show the various levels of the inflowing stream’s velocity V1 after designating it as a constant.

 it is a paradox! The graph showing the energy’s dependence on the drop between the levels h has an extremum. On the rising branch of the graph, the energy balance will be positive (the power factor > 1), i.e. the extracted potential energy will be mostly expended as kinetic energy on speeding up the outflowing stream, and the device will self-accelerate until it reaches the maximum.

The energy produced by the device at this point will be several times the kinetic energy of the inflowing stream - and under certain conditions, tens and even hundreds of times!

The speed of the outflowing stream will be significantly higher (2 to 3 times as higher at times) than the speed of the inflowing stream. Therefore, the kinetic energy of the outflowing stream is 4 to 9 times the kinetic energy of the inflowing stream.

Furthermore, the graphs show that that not everything appears to be quite right with the inflowing speed. it also has an extremum. To see this better, let us plot a 3D diagram.

Fig. 3. Energy's dependence on the difference in the levels (left) and exit speed (right).

However paradoxical this may seem at first glance, but the diagrams show there is an optimal speed for the inflowing stream. When it is exceeded, the device’s power capacity will sharply fall. This is due to the fact that a significant amount of energy needs to be spent on speeding up a stream that is flowing fast already.

One parameter has been left unaccounted for in these diagrams, the entry depth H1 to be precise.
But because the diagrams are three-dimensional now, to plot the output energy's dependence on this parameter, too, we will show the sequence of 3-D graphs for various values of the flow's entry depth.

Fig. 4. Energy depends on three parameters: the difference in levels h, the entry speed V1 and the effective entry depth H1 (0.9,  1.2,  1.5  and 1.8 ì)

The diagrams show that depending on the entry depth, the machine's energy output grows in a non-linear fashion, almost in quadratic dependence. Below we will examine what exactly this dependence looks like.

 The question may arise: “How does the outflowing stream, which has a shallower depth, interact with the water flow around it, which has a normal constant depth?” Here we have to recall that the velocity of the outflowing stream is higher than that of the surrounding medium and this creates what is called in hydraulics “hydraulic jump” as a result of ejection effect, which equalises the discrepancy between the kinetic and potential energies of the two flows. This “jump” is in essence surf, a vortex in the flow.

Let us examine in detail what happens with the flow, what the depth and velocity of the outflowing stream depend on, how long the hydraulic jump is and what conditions need to be met to produce such an effect.

In all of the aforementioned computations, only the Bernoulli equation (energy conservation law) and the flow continuity equation (mass conservation law) are used.

Considering the fact that the turbine, which is located on the water flow, extracts some energy from this flow, the Bernoulli generalized equation for two cross-sections of the free voluntary flow – the first one (before incoming the unit) and the second one (at the unit’s outlet), without taking into consideration losses, will take on the following form:

where E - is energy that the turbine takes from the flow.

 Therefore, the energy released at the turbine is equal to

E = MgH1 – MgH2 + MV1²/2 - MV2²/2 (7)

 Let us define H2 = k * H1, where k is a dimensionless coefficient

 Then

E = MgH1 – MgkH1 + MV1²/2 - MV2²/2 (8)

 Let us express V2 in terms of V1 taking into account the flow continuity equation, namely H * V = const  (when the flow’s width is constant in two clear sections). We will get the following:

H1* V1 = H2 * V2 (9)

or V2 = V1 * (H1 / H2 ) and V2 = V1 / k (10)

So the ratio of the velocities of the inflowing and outflowing streams depends only on the ratio of the height (depth) of the streams (with the width being the same).

 Accordingly, the formula (8) assumes the form:

E = MgH1 – MgkH1 + MV1²/2 - MV1²/ 2k² (11)

or

E = MgH1 * (1 - k) + MV1²/2 * (1 - 1 / k² ) (12)

 Let us find the extreme of the energy in relation to k.

 To do this, let us differentiate the formula (11) on k

E’ = 0 - MgH1 + 2MV1²/2k³ = - MgH1 + MV1²/k³ (13)

.By equating (13) to zero, we get

 Hence,

k = ( V1² / gH1 ) ^1/3 (14)

Conclusion: The turbine generates the maximum energy when the ratio of the levels of inflowing and outflowing streams is

therefore H2 = ( H1² V1²/ g ) ^1/3 or     (15)

 If we look up any textbook on hydraulics, for example Gidravlika [Hydraulics] by R. R Chugayev [6] or Gidravlika [Hydraulics] by I. I. Agroskin [7], we will see that formula (15) above agrees with formulas (7-49) [6] or (15-13) [7], which correspond to the so-called “critical depth” of the flow, the depth at which a flow is in the border state between being calm and turbulent.

Hydraulics by R. R Chugayev(page 280)

But why will the depth of the outflowing stream be equal to the critical depth? The thing is that the stream’s energy density is minimal at the critical depth (this is exactly why it is called critical), and, as one can observe, an increase in the velocity of the outflowing stream - with the unit rate of flow being constant and, therefore, its depth decreasing – yields a positive power factor (above 1).

Hydraulics by R. R Chugayev (page 278)

That is to say, the stream releases energy, which is partially spent on the additional acceleration of the outflowing stream through the [positive] feedback ensured by the machine. This process will continue until the power factor becomes equal to 1, that is to say until the stream enters the critical state.

 It is thus possible to conclude that the device described above extracts all the additional energy from a flow by bringing the outflowing stream to the critical state, that is to say to the border state where the flow turns from being calm to being turbulent.

 In accordance with R.R. Chugayev’s Gidravlika (p. 280), the specific velocity head of the flow in the critical state is equal to half of its depth and the energy density of the flow is equivalent to its gross head (the sum of its potential and velocity heads).

 (V k)² /2g = Hk/2 or (V k)² /g = Hk (16)

Ek’ = (V k)² /2g + Hk = 3/2 * Hk (17)

 where Ek’ is the flow’s energy density at the critical depth and velocity.

The flow strength at the effective cross-section is equal to the flow’s energy density multiplied by the weight of the water passing through the effective cross section per unit of time, or unit rate of flow, i.e. V1*H1.

 Taking into account (15) and (17), formula (7) can be re-written as follows:

E’ = E1’ – Ek’ = H1 + V1²/2g – 3/2 * H2 = H1 + V1²/2g – 3/2 * ( H1² V1²/ g ) ^1/3

 And the flow strength at the effective cross section per unit of time is

E = V1*H1*g *(H1 + V1²/2g – 3/2 * ( H1² V1²/ g ) ^1/3) (18)

 Or, ultimately, 

E = V1*H1²*g + H1*V1³/2 – 3/2 * ( H1*V1) ^5/3* g ^2/3 (19)

 Knowing that the depth of the outflowing stream is equal to the critical depth, it is now possible to plot, as a function, the dependence of the generator’s output power on the depth and velocity of the inflowing stream.

Fig. 5. The energy diagram of the calm and turbulent states in relation to the critical state.

The critical flow will be determined as a ratio of the velocity and depth and is shown as the graph (a parabola) lying at the “gulley” of the 3-D diagram. The flow’s specific energy density on this graph has been accepted as zero. The total energy of the calm and turbulent flows are calculated relative to it. in this diagram, all the calm flows are on the left and the all the turbulent flows are on the right.

 The graph and formula (19) show that the output power's dependence on the flow's entry depth is pretty complex. it grows in a quadratic fashion if it depends on the first term in the polynomial formula and in a linear fashion if it depends on the second term. it decreases by the power 5/3 if it depends on the third term

 Let us find the ratio of the inflowing stream’s depth and velocity, i.e. the unit rate of flow, at which it is possible to extract maximum energy from it. To do this, it is necessary to differentiate the function in formula (18) on V and H and to equate the derivations obtained to zero.

 Let us now try to calculate the length of the resulting hydraulic jump. At the moment there is no precise formula for determining it. All available formulas are empirical and very often there is a considerable variance between the results obtained with their help. We will use energy balance for the calculation. Let us examine Fig. 6

Fig. 6. Hydraulic jump at the exit from the device

Explanations for Fig. 6:

 H1, H2 – the depth (potential head) of the inflowing and outflowing streams respectively;

V 1² /2g, V2² /2g – the velocity head of the inflowing and outflowing streams respectively;

∆E’ – the difference between the energy density of the inflowing and outflowing streams;

Lj – the length of the hydraulic jump;

 Fig. 6 shows that there is a lack of energy density in the hydraulic jump area between the original and the emerging (after the jump) flow regimes.

 Therefore, such a machine appears “to cut out” from a flow “a slice of energy” equal to the area of the rectangle limited to the length of the jump Lj and the difference between the energy density within and before the jump area.

 The total energy taken away from the flow in this manner is equal, in accordance with formula (18), to the difference of energy density multiplied by the unit rate of flow and multiplied by the gravitational acceleration:

E = ∆E’ * V1 * H1 * g.

 Therefore, it is evident that the length, in meters, of the hydraulic jump emerging after the device will be numerically equal to the unit rate of flow multiplied by the gravitational acceleration

 To summarize all of the above, it can be concluded that such a machine creates a head of water for itself and is able to extract potential energy from an evenly-flowing stream of water without expenditure of external energy.

 In addition, analyzing the diagram on pictures 3, 4  and 5, a few important conclusions can be drawn.

First of all, one can see on these diagrams that this effect exists and it is very “capricious” – a number of conditions for the flow’s efflux must be strictly met, namely the proportion between the flow’s incoming speed and its depth. Only with the specific combination of these parameters, we can get to the diagram’s peak and extract the maximum power from the flow. At these parameters’ insignificant deviation from the optimal values, the effect either “becomes blurred”,  or disappears an all and it will be very difficult to find it, and it can well be mistaken for measuring errors.

 Secondly, it is interesting that the effect disappears when the flow’s kinetic energy increases (when its speed increases), or in other words, when it approaches critical or turbulent condition. Judging by the diagrams, the optimal speed is the speed of 1 – 1.5 m/sec. However, because the water flow with this speed is considered to be of low potential and it is not often used for extracting energy by free-flow turbines, there have been carried out just few experiments under such conditions and, therefore, it has not been possible to reveal this kind of effect.

 Thirdly, the flow’s incoming effective depth is very critical to the appearance of this effect. it is obvious (the left diagram on picture 4), that if the depth is less than one meter (the overall dimensions of most of the free-flow turbines are usually less than these values) this effect is barely noticeable, commensurate with measurement errors and "spreads" in turbines’ hydraulic and mechanical efficiency.

 Fourthly, in order to reveal this kind of effect it is necessary to use special machines that have feedback between the incoming and outward flows.

 Also, another interesting aspect should be noted. Unlike traditional free-flow turbines, a machine working on a similar principle does not slow down the outward flow by extracting its kinetic energy, but speeds it up by extracting its potential energy. No matter how ironically it appears at first sight, it is a fact…
This partially reminds of the supposition about the existence in the nature of a certain process which was voiced by the English scientist James Maxwell in 1871. Let us remind that according to Maxwell’s hypothesis, in the nature exists a process that makes it possible to extract external energy from gas without expenditure of external energy. it is called Maxwell’s demon.

 Maxwell predicted this process for thermodynamics. The possibility of its existence is still disputed in the scientific community. However, given the fact that Maxwell voiced the theory in such a jocular form – in the form of a vague creature, “a demon” – many always perceived this no other than a joke by the great scientist. Or is it reality after all?

 It is possible that the effect described above is the hydrodynamic interpretation of “Maxwell’s demon”? This has yet to be clarified…

 If it is so indeed, then it is impossible to overestimate the conclusion from the above: in the nature, there exists a process that makes it possible to extract the potential energy from any object that has it, which has been impossible before. And this process has been found!

 This is the principle of positive feedback that makes it possible to transfer energy between different flows of an energy carrier.

 There is a possibility of extracting free and environmentally friendly energy from the environment, which was predicted by the great English scientist James Maxwell as far back as 1871.

 Although not everything is clear with regard to its application in thermodynamics and aerodynamics, but since this process exists in hydrodynamics then it also should exist in any other sphere of physics. There are already some developments in thermo- and aerodynamics. But even if this process is not found for them soon and the search for it drags out for another ten years, then, at least, the use of its hydrodynamic interpretation alone is already promising mankind huge dividends in the form of free energy and clean air.

 In the next article, we will discuss what seems as the utopian idea (possibility) of using this principle of extracting energy on cars, and a hypothetical engine for them.

 Translated by Dilmurad Gulamov, Askar Jumanov, Joanna Lillis

 .08.07 , 15.02.08

1. V. Brodyanskiy “Exergetic analysis. Energy: the Problem of Quality”, Nauka i Zhizn, No 3, 1982
2. 2. N. Shchapov “Turbine Equipment for Hydropower Stations”, Gosenergoizdat, 1961
3. N. Gulia “in Search of an Energy Capsule”, a web publication
4. E. Oparin “Physical Foundations of Fuelless Power-Engineering. The Limitation of the Principle of Entropy increase”, Moscow, URSS, 2004
5. L. Landau, A. Kitaygorodskiy “Physics for Everyone”, Nauka, 1974
6. Chugayev R. R., “Hydraulics” Energoizdat,” 1982
7. Agroskin i.i., Dmitriyev G.T. and Pikalov F.i., “Hydraulics”
8. Gosudarstvennoye energeticheskoye izdatel’stvo, 1954

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