**5. The special pistons and their high pressure**

**Figure 8** shows the shape of the special piston that works inside the HPC. The HPC may work either with a massive special piston (piston B) or with one (piston A) where the increase of pressure is achieved mainly with the weight of water obtained from R1.

The pressure PA exerted by the special piston A on its circular bottom of diameter *de* is a function of the geometric parameters illustrated in **Figure 8**. According to results already obtained [2], it is possible to achieve pressures high enough (over

**71**

**Figure 9.** *Underground HPC.*

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational…*

55 bar) to desalinate seawater with a special piston of moderate dimensions (fitting

The expression for pressure PA exerted by the special piston A on its bottom of diameter *de* is given by the total weight of the piston *Pt*, divided by the circular area **Ae**:

Starting with Eq. (1) and knowing that *Pt* is the sum of the weight of steel, *Psteel* plus that of the water *Pwater* in R2 and the filling up material *Pfill*, and that **Ae** = π/4

A similar approach with special piston B leads to the following expression for PB:

A very interesting aspect of using special piston A is that the filling up material may be anything available where the system will operate such as stones, sand, or even recycled metals. This low-cost characteristic is useful for remote areas because

**6. The system's specific energy consumption regarding the RO module**

The analysis of a configuration for an underground HPC (**Figure 9**) led us to some conclusions about the specific consumption of energy [2, 3]. Notice that there is no need to pump water to R1, and the energy consumption is used only to lift the piston.

, after some algebraic manipulation, one gets the following relation [2, 3]:

PA = *Pt*/ **Ae** (1)

]; **ρstl** is the density of steel [kg/m3

]; and **Di = De -2 t** [m], being **t** the thickness of

]; **ρfill** is the density of

(2)

(3)

];

*DOI: http://dx.doi.org/10.5772/intechopen.91135*

where **g** is the acceleration of gravity [m/s2

the of the filling up material [kg/m3

the piston's wall.

**ρwat** is the density of seawater (or brackish water) [kg/m3

it would not be necessary to transport a very heavy structure.

inside a 5-m-height HPC).

*de*<sup>2</sup>

**Figure 8.** *Scheme of the special piston.*

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational… DOI: http://dx.doi.org/10.5772/intechopen.91135*

55 bar) to desalinate seawater with a special piston of moderate dimensions (fitting inside a 5-m-height HPC).

The expression for pressure PA exerted by the special piston A on its bottom of diameter *de* is given by the total weight of the piston *Pt*, divided by the circular area **Ae**:

$$\mathbf{P\_A = P} \mathbf{t} / \mathbf{A\_e} \tag{1}$$

Starting with Eq. (1) and knowing that *Pt* is the sum of the weight of steel, *Psteel* plus that of the water *Pwater* in R2 and the filling up material *Pfill*, and that **Ae** = π/4 *de*<sup>2</sup> , after some algebraic manipulation, one gets the following relation [2, 3]:

$$\mathbf{P\_A} = \frac{\mathbf{g}}{\mathbf{d}\boldsymbol{\theta}^2} \mathbf{l} \begin{bmatrix} \mathbf{\hat{p\_{sat}}} \ \mathbf{\hat{n}} \ \mathbf{\hat{s}} \ \mathbf{\hat{n}} \ \mathbf{\hat{s}} \ \end{bmatrix} \cdot \mathbf{D} \mathbf{\hat{i}}^2 \begin{bmatrix} \mathbf{\hat{p}\_{\parallel}} \ \mathbf{\hat{n}} \ \mathbf{\hat{i}} \ \mathbf{\hat{n}} \ \mathbf{\hat{i}} \end{bmatrix} \cdot \begin{bmatrix} \mathbf{\hat{p\_{sat}}} \ \mathbf{\hat{n}} \end{bmatrix} \cdot \begin{bmatrix} \mathbf{\hat{p\_{sat}}} \ \mathbf{\hat{n}} \end{bmatrix}^2 \mathbf{a} \tag{2}$$

where **g** is the acceleration of gravity [m/s2 ]; **ρstl** is the density of steel [kg/m3 ]; **ρwat** is the density of seawater (or brackish water) [kg/m3 ]; **ρfill** is the density of the of the filling up material [kg/m3 ]; and **Di = De -2 t** [m], being **t** the thickness of the piston's wall.

A similar approach with special piston B leads to the following expression for PB:

$$\mathbf{P\_{B}} = \frac{\mathbf{g}}{\mathfrak{sl}\mathfrak{e}^{2}} \,\mathrm{\mathfrak{f}\_{\mathrm{stl}}} \left( \mathfrak{Do}^{2} \,\mathrm{\mathfrak{b}} \star \mathfrak{sl}\mathfrak{e}^{2} \mathbf{a} \right) \tag{3}$$

A very interesting aspect of using special piston A is that the filling up material may be anything available where the system will operate such as stones, sand, or even recycled metals. This low-cost characteristic is useful for remote areas because it would not be necessary to transport a very heavy structure.

#### **6. The system's specific energy consumption regarding the RO module**

The analysis of a configuration for an underground HPC (**Figure 9**) led us to some conclusions about the specific consumption of energy [2, 3]. Notice that there is no need to pump water to R1, and the energy consumption is used only to lift the piston.

**Figure 9.** *Underground HPC.*

*Electrodialysis*

course.

from R1.

connects the two columns.

course, two other events occur:

• Its CW empties because valve V6 opens.

able to reach 55 bar because of the size of the special piston.

**5. The special pistons and their high pressure**

2.When the piston of column B, raised by the counterweight, reaches the upper

• It is braked by the actuators and will remain in that position until, after filled up with water, it receives the signal sent by the actuators of column A.

An identical process takes place when the piston of column A comes to its upper

The hydraulic automatism guarantees the continuous function of the system as long as there is enough water in the reservoirs R1. The continuous flux of salted water to the modules of membrane is guaranteed by the two-way valve V7 that

So far, a small prototype has been constructed at the polytechnic school of the *Universidade de São Paulo-USP in Brazil*. The prototype has proven that the mechanism of the HPC works. Since osmosis reverse is already a solid technique, the focus of the prototype was not to obtain drinking water because it would not be reason-

Therefore the objective of the prototype was to prove that the HPC pumps water using the gravitational potential energy obtained from the water in R1 and that the necessary increase of pressure is achieved due to the shape of the special piston. The flux under the pressure exerted by the piston was used to drive a small Pelton wheel supported with magnetic levitation. This was done so, also to prove that the system can be used for electricity generation which is to be used to power the ED module.

**Figure 8** shows the shape of the special piston that works inside the HPC. The HPC may work either with a massive special piston (piston B) or with one (piston A) where the increase of pressure is achieved mainly with the weight of water obtained

The pressure PA exerted by the special piston A on its circular bottom of diameter *de* is a function of the geometric parameters illustrated in **Figure 8**. According to results already obtained [2], it is possible to achieve pressures high enough (over

**70**

**Figure 8.**

*Scheme of the special piston.*


### **7. Calculation of the systems' pumping power**

**Figure 10** shows the chain of energy transformations that occur in in the proposed system. Starting from wind energy until the storage of gravitational potential energy (GPE) in the reservoir R1 on top of the hydraulic power column.

In the block diagram of energy transformation, the conversion of wind power Pw into pumping power Pp that will be used to feed R1 with water to be pumped from a well is illustrated. Along those conversions, there are power losses due to the inefficiencies of the devices and mechanical connections (shafts, gears, etc.). Further in this chapter, it will be shown how the pumping power Pp is converted into a volume of water Vw, stored in R1.

The power of wind flow Pw is given by [5]:

$$\mathbf{P\_w = 1/2 \text{ p A V}^3} \tag{4}$$

where **ρ** is the air density; **A** is the frontal area through which the air flows; and **V** is the wind speed.

A wind machine (windmill or wind turbine) can be used to harvest this wind power and convert it into its rotor power Pr. Betz has proven that 59% of Pw is the maximum Pr that an ideal rotor would be able to harvest from the wind [5]. Observing **Figures 10** and **11**, one concludes that for system 1 (windmill), Pr is used to drive a piston pump which will feed R1, while in system 2 (wind turbine), it will be converted into an electric power Pe that will power an electric pump which feeds R1.

Back to the rotor power, its magnitude is:

$$\mathbf{P\_r = C\_p \mathbf{1}/2 \text{ p A V}^3} \tag{5}$$

**73**

rotor.

**Figure 11.**

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational…*

where Cp is the power coefficient of the machine. Cp can generally be expressed

λ = blade tip speed/wind speed = ΩR/V (6)

where Ω is the angular velocity of the wind rotor and R is the radius of the wind

**Figure 11** illustrates the aerodynamic efficiency of a windmill which is characterized by its power coefficient Cp. Resortint to **Figure 11**, a windmill's Cp can be

According to **Figure 10**, taking into account the efficiencies of the gear ηm and

Eq. (8) represents the theoretical pumping power Pp at the end of the chain of conversions shown in **Figure 10** for the system. However, in practice one can resort to other equations that are based on the characteristics of each wind machine used by the system. Since windmills are generally used to pump water, it is sufficient to determine its pumping power Pp. For S2, first it is necessary to determine the electric power Pe delivered by the wind turbine and then estimate the efficiency of

Pp = Cp 1/2 ρ A V<sup>3</sup>

its connection with the electric pump in order to find the pumping power.

**8. From pumping power to potential gravitational energy stored in R1**

After finding the pumping power Pp, one has to find the flux of pumped water to R1 as a function of the manometric height Hm (head) as well as the volume of

Pp = Pr. ηm. η<sup>p</sup> (7)

ηm. η<sup>p</sup> (8)

as a function of the tip speed ratio, λ, defined by [5]:

*Aerodynamic efficiency of wind machines: Source [5] with our drawings.*

compared with Betz's maximum theoretical efficiency.

that of the piston pump ηp, the pumping power Pp is given by [6]:

So, the pumping power of the system with windmill is:

*DOI: http://dx.doi.org/10.5772/intechopen.91135*

**Figure 10.** *Energy conversions of the system.*

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational… DOI: http://dx.doi.org/10.5772/intechopen.91135*

**Figure 11.**

*Electrodialysis*

tion about 3 kWh/ m3

of water Vw, stored in R1.

**V** is the wind speed.

• Considering a frictionless piston with an operating pressure of 20.3 bar, the system would consume around 0.563 kWh for each cubic meter of drinking

• Considering a frictionless piston with an operating pressure of 55 bar, the HPC would consume around 1528 kWh for each cubic meter of drinking water.

• For an operating pressure of 55 bar and taking into account acceptable efficiencies ("actual" values) for the components of the HPC, it has been shown that the predicted "actual" specific energy consumption is about 2811 kWh for each cubic meter of drinking water. This is a very good result compared to the conventional RO systems for which this parameter is generally above 4 kWh/m3

Taking into account the ED module, we can predict an overall energy consump-

**Figure 10** shows the chain of energy transformations that occur in in the proposed system. Starting from wind energy until the storage of gravitational potential

In the block diagram of energy transformation, the conversion of wind power Pw into pumping power Pp that will be used to feed R1 with water to be pumped from a well is illustrated. Along those conversions, there are power losses due to the inefficiencies of the devices and mechanical connections (shafts, gears, etc.). Further in this chapter, it will be shown how the pumping power Pp is converted into a volume

where **ρ** is the air density; **A** is the frontal area through which the air flows; and

A wind machine (windmill or wind turbine) can be used to harvest this wind power and convert it into its rotor power Pr. Betz has proven that 59% of Pw is the maximum Pr that an ideal rotor would be able to harvest from the wind [5]. Observing **Figures 10** and **11**, one concludes that for system 1 (windmill), Pr is used to drive a piston pump which will feed R1, while in system 2 (wind turbine), it will be converted into an electric power Pe that will power an electric pump which feeds R1.

Pw = 1/2 ρ A V<sup>3</sup> (4)

Pr = Cp 1/2 ρ A V<sup>3</sup> (5)

of drinking water.

energy (GPE) in the reservoir R1 on top of the hydraulic power column.

**7. Calculation of the systems' pumping power**

The power of wind flow Pw is given by [5]:

Back to the rotor power, its magnitude is:

.

water which is the minimum allowed by nature [4].

**72**

**Figure 10.**

*Energy conversions of the system.*

*Aerodynamic efficiency of wind machines: Source [5] with our drawings.*

where Cp is the power coefficient of the machine. Cp can generally be expressed as a function of the tip speed ratio, λ, defined by [5]:

$$
\lambda = \text{blade tip speed} / \text{wind speed} = \Omega R / V \tag{6}
$$

where Ω is the angular velocity of the wind rotor and R is the radius of the wind rotor.

**Figure 11** illustrates the aerodynamic efficiency of a windmill which is characterized by its power coefficient Cp. Resortint to **Figure 11**, a windmill's Cp can be compared with Betz's maximum theoretical efficiency.

According to **Figure 10**, taking into account the efficiencies of the gear ηm and that of the piston pump ηp, the pumping power Pp is given by [6]:

$$\mathbf{P\_p} = \mathbf{P\_r} \cdot \eta\_{\mathbf{m}} \cdot \eta\_{\mathbf{p}} \tag{7}$$

So, the pumping power of the system with windmill is:

$$\mathbf{P\_p = C\_p \mathbf{1}/2} \text{ } \mathbf{\color{red}{O}P \text{ } A \text{ V}^3 \text{ } \eta\_m. \text{ } \eta\_p\tag{8}$$

Eq. (8) represents the theoretical pumping power Pp at the end of the chain of conversions shown in **Figure 10** for the system. However, in practice one can resort to other equations that are based on the characteristics of each wind machine used by the system. Since windmills are generally used to pump water, it is sufficient to determine its pumping power Pp. For S2, first it is necessary to determine the electric power Pe delivered by the wind turbine and then estimate the efficiency of its connection with the electric pump in order to find the pumping power.

#### **8. From pumping power to potential gravitational energy stored in R1**

After finding the pumping power Pp, one has to find the flux of pumped water to R1 as a function of the manometric height Hm (head) as well as the volume of

**Figure 12.**

*Storage of gravitational potential energy in reservoir R1.*

water Vw stored in R1 (**Figure 12**). During the pumping process, the work done by the pump is equivalent to energy transferred (Etra) to the water which will be stored in R1 as gravitational potential energy (GPE).

The energy transferred Etra to the water during the pumping process is:

$$\mathbf{E}\_{\rm tra} = \mathbf{P}\_{\rm P} \,\Delta \mathbf{t}. \tag{9}$$

where Δt is time (in seconds) corresponding to the period of operation of the pump [hour(s), day(s), month(s), year(s), etc.].

As mentioned, the transferred energy is stored as potential gravitational energy (GPE) in R1. Therefore one has:

$$\mathbf{E}\_{\rm tra} = \mathbf{GPE},\tag{10}$$

or

$$\mathbf{P}\_{\mathbf{P}} . \,\Delta \mathbf{t} = \mathbf{m} . \,\mathbf{g} . \,\mathbf{H}\_{\mathbf{m}}. \tag{11}$$

where Hm (the head) is the height that the pump must overcome in order to force water from the well to the top reservoir R1 (**Figure 12**).

Since mass of pumped water is the product of the specific mass for its volume (m = ρw V), Eq. (11) becomes:

$$\mathbf{P\_{p}}.\,\Delta\mathbf{t} = \rho\_{\mathbf{w}}\,\mathbf{V\_{w}}\,\mathbf{g}\,\mathbf{H\_{m}}.\tag{12}$$

Solving Eq. (12) to obtain the volume of a pumped water, one has:

$$\mathbf{V}\_{\mathbf{w}} = \mathbf{P}\_{\mathbf{p}} \cdot \Delta \mathbf{t} / (\rho\_{\mathbf{w}} \cdot \mathbf{g} \, \mathbf{H}\_{\mathbf{m}}) \tag{13}$$

**75**

**Figure 13.**

*Feeding of reservoir R1 with windmill and Searaser.*

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational…*

If one considers the time of operation of 1 hour (Δt = 3600 s), Eq. (13) becomes:

Eq. (15) yields the volume of pumped water per hour. Therefore the result of this equation should be multiplied by 24 to obtain the volume of water stored in R1 during a day. Once there is enough water in R1, to make the hydraulic power column work, as already proven with the prototype, the reverse osmosis process will occur as long as the special piston exerts the necessary pressure. For that, the pressure is previously calculated according to the design of the piston and using

**9. Alternatives to get the potential gravitational energy stored in R1**

The HPC is designed so that it works as long as there is enough water stored in the reservoir R1 regardless of how water was put in it. This characteristic yields a certain versatility to the whole system. This means that one could feed R1 with other sources of energy such as wind, sea waves, sun, etc. In this regard, it is interesting to consider the possibility of using a device similar to the HPC's plunger pump to harness sea wave power to pump water to reservoir R1. Such device is currently a reality due to the invention of Alvin Smith who developed the Searaser (**Figure 13**). This feature is especially important if the system is to be designed for seawater desalination with the RO module and brackish water for

Vw = 3600 Pp ./ (ρ<sup>a</sup> .g. Hm) (14)

corresponding to the density of seawater to be pumped to

Vw = Pp/ (2790 Hm) (15)

*DOI: http://dx.doi.org/10.5772/intechopen.91135*

With *ρw = 1025* kg/m3

*g = 9.81*, Eq. (14) reduces to:

R1, and.

Eqs. (2) and (3).

the ED module.

where **Vw** is the volume of pumped water (stored in R1); **Pp** is the pumping power; **ρw** is the water density; **g** is the acceleration of gravity; and **Hm** is the manometric height.

*A Reverse Osmosis and Electrodialysis System Simultaneously Powered by Gravitational… DOI: http://dx.doi.org/10.5772/intechopen.91135*

If one considers the time of operation of 1 hour (Δt = 3600 s), Eq. (13) becomes:

$$\mathbf{V\_w = \mathcal{G}\mathcal{G}\mathcal{O}\mathcal{O} \ P\_\mathbf{p} ./ (\rho\_\mathbf{a} . \mathbf{g} . \mathcal{H}\_\mathbf{m})} \tag{14}$$

With *ρw = 1025* kg/m3 corresponding to the density of seawater to be pumped to R1, and.

*g = 9.81*, Eq. (14) reduces to:

*Electrodialysis*

water Vw stored in R1 (**Figure 12**). During the pumping process, the work done by the pump is equivalent to energy transferred (Etra) to the water which will be stored

where Δt is time (in seconds) corresponding to the period of operation of the

As mentioned, the transferred energy is stored as potential gravitational energy

where Hm (the head) is the height that the pump must overcome in order to force

Since mass of pumped water is the product of the specific mass for its volume

where **Vw** is the volume of pumped water (stored in R1); **Pp** is the pumping power; **ρw** is the water density; **g** is the acceleration of gravity; and **Hm** is the

Solving Eq. (12) to obtain the volume of a pumped water, one has:

Vw = Pp .Δt/

Etra = Pp. Δt. (9)

Etra = GPE, (10)

Pp .Δt = m.g. Hm. (11)

Pp. Δt = ρw Vw g Hm. (12)

(ρ<sup>w</sup> .g. Hm) (13)

The energy transferred Etra to the water during the pumping process is:

in R1 as gravitational potential energy (GPE).

*Storage of gravitational potential energy in reservoir R1.*

pump [hour(s), day(s), month(s), year(s), etc.].

water from the well to the top reservoir R1 (**Figure 12**).

(GPE) in R1. Therefore one has:

(m = ρw V), Eq. (11) becomes:

manometric height.

or

**Figure 12.**

**74**

$$\mathbf{V\_w = P\_p/(2790\ H\_m)}\tag{15}$$

Eq. (15) yields the volume of pumped water per hour. Therefore the result of this equation should be multiplied by 24 to obtain the volume of water stored in R1 during a day. Once there is enough water in R1, to make the hydraulic power column work, as already proven with the prototype, the reverse osmosis process will occur as long as the special piston exerts the necessary pressure. For that, the pressure is previously calculated according to the design of the piston and using Eqs. (2) and (3).

## **9. Alternatives to get the potential gravitational energy stored in R1**

The HPC is designed so that it works as long as there is enough water stored in the reservoir R1 regardless of how water was put in it. This characteristic yields a certain versatility to the whole system. This means that one could feed R1 with other sources of energy such as wind, sea waves, sun, etc. In this regard, it is interesting to consider the possibility of using a device similar to the HPC's plunger pump to harness sea wave power to pump water to reservoir R1. Such device is currently a reality due to the invention of Alvin Smith who developed the Searaser (**Figure 13**). This feature is especially important if the system is to be designed for seawater desalination with the RO module and brackish water for the ED module.

**Figure 13.** *Feeding of reservoir R1 with windmill and Searaser.*
