Economic Feasibility Study of Using Existing Water Wells for Energy Storage in Agricultural Lands

This project evaluates the economic feasibility of replacing solar battery systems with a Pump Hydro Storage system for energy storage in agricultural lands. The proposed system deals with the intermittent nature of renewable energy sources and calculates costs such as initial, operating, labor, maintenance, and replacement costs. The study compares the proposed system with the existing solar system and discusses the final results.

Economic feasibility study of using
existing water wells in agricultural
lands to store energy

Abstract:
In this project economic feasibility study of replacing the existing solar battery system
with a Pump hydro storage system carried out, which includes detailed design and evaluation of
the proposed system with various aspects and conditions are carried out. The intermittent nature
of the Renewable Energy Source is a challenging factor, the proposed system also deals with this
factor. Various costs such as initial cost, operating cost, labor cost, maintenance cost, and
replacement cost of both the systems are calculated with proper assumption and methodology.
The final results obtained are discussed and the graphs are plotted.
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Abstract
Table of contents
1. Introduction
1.1 Objective
2. Existing system
2.1 Design of the existing system
3. Proposed Approach
3.1 Design
3.2 Calculations
3.3 Hardware Description
4. Cost Benefit Analysis
4.1 Cost
4.2 Benefits
4.3 Comparison with existing solar system
5. Results and Discussions
6. Conclusion
References
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1.0 Introduction
Energy is the major building block of modern society, energy is present in all sectors of
sectors in our society such as engineering, business, production, manufacturing, labor
environment and also in our own personal livings which includes food, transportation, shelter,
comfort, security, and all other things which are essential and improves the quality of our living.
Most of these energy we use are derived from fossil fuels such as coal, oil and natural gas, these
are nonrenewable sources of energy which means that they do not replenish themselves, if we
could use them all it will be exhausted one day. As the fossil fuels emit more harmful gases such
as carbon dioxide, carbon monoxide, nitrogen oxides and sulphuric oxides while burning it leads
tomany environmental issues such as global warming, acid rain, air pollution, etc. Which causes
many harmful effects on both the environment and human health . (Plass Jr, H. J., 1990).As the
fossil fuels tend to exhaust over time and causes harmful effects on the environment and human
health it is important that we start switching to other energy sources like Renewable energy
sources.Renewable energy sources are sources of energy which constantly replenish such as
wind energy, solar energy, water, geothermal and biomass. Renewable energy sources do not
exhaust in time, they are environmental friendly and emits very few or no harmful gases.
Although renewable energy sources have many pros it also has many cons such as initial cost,
maintenance cost, requirement of large area for the layout of power plant, intermittent nature,
etc.
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Global RES usage
The important and the major disadvantage of renewable energy source is that its
intermittent nature which means the renewable energy sources cannot match the energy demand
of day to day life, that is the energy is not available all the time in a consistent manner rather
there may be energy surplus sometimes and energy shortage for some period time. (Roh, J.,
2014)
In case of solar water pumps used for irrigation, which converts the solar energy into
electrical energy and stores it in the battery which will be used to operate a pump to pump the
water from the well to use it for irrigation purposes. The output power of photovoltaic systems
and wind turbines change with the fluctuation of solar radiation and wind speed. Various
methods and various kinds of energy storage systems have been developed and tested to solve
this problem. A system will be developed in this project to overcome the intermittent nature of
the Renewable Energy Sources with the existing infrastructures that are available in rural areas
the main goal of the project is to provide a more reliable electricity for customers.
This can be achieved with a Pump hydro storage system (PHS), in which there will be a
water reservoir, and a tank place at some high head difference, a turbine and a pump. During the
day time the solar cells provides electrical energy to pump the water from the well to the
reservoir with the help of solar PV cells. By this technique the potential energy to generate
electricity at night time is stored in form of water inside the tank. During the night time the water
in the tank will be let to flow from the tank to the reservoir through a water turbine which will
rotate when the water flows from top to bottom and generates electricity with the help of an
electric generator coupled to it. This acts as an energystorage medium which provides reliable
energy, with the help of the renewable sources. This might also overcome the intermittent nature
of the Renewable energy sources and provide reliable energy. (Bueno, C., & Carta, J. A., 2006)
The existing PHS system consists of a solar cell and battery combination, in the proposed
system the batteries are eliminated in order to reduce the cost in various aspects. The proposed
system uses the PHS and the solar cells without the battery and inverter, in this system the solar
cells provides electricity to pump the water at day times and the water which is being pumped at
the day time will be released to flow through the turbine during the night time, in order to
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generate electricity. In this proposed system due to the absence of the battery cost is reduced, as
batteries are not a good source to store the energy they tend to deteriorate over a period of time,
the life cycle of the battery is also limited and replacement of the battery is costly. The proposed
system reduces the major cost of a Pump hydro system which is in the form of battery. The
design, hardware consideration and the energy calculation for the proposed system are done.
Also the cost benefit analysis which compares the proposed system with the existing pump hydro
system is also done.
1.1 Objective
The main and primary objective of this system is to eliminate the batteries to store
the energy and bridge the gap between supply and demand in other words, to overcome the
intermittent nature of the Renewable Energy Sources. This can be achieved by introducing a
system which will store the energy in the form of potential energy due to existence of head
difference from the well to the reservoir which is stored during the day time and this can be
released to produce electrical energy during the night time. This is accomplished by using a
Pump Hydro storage System which is combine with a solar cells.
The secondary objective is to make the system more cost effective than the
existing solar system. This can be done by eliminating the batteries. Also proper and flawless
design and evaluation of the proposed Pump Hydro Systemis required. Selection of proper
turbine and pumping system with low maintenance and initial cost, selection of good storage
system, selection of solar cells and analysis of effective generator, etc. makes up the design
process. Calculation of labor cost, maintenance cost, and working cost is also required to be
calculated properly and clearly and to compare them with the existing system in order to perform
a cost benefit analysis and to implement the system in rural areas.
2.0 Existing system
The existing system consists of solar cells, pump, reservoir, batteries, inverters and some
charging circuits. This system is most widely used in rural areas for the irrigation of water. In
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this system the solar cells combined with batteries provides the electrical energy required for the
pumping of the water from the well to the reservoir, from which the water required for the
irrigation is distributed. Then the surplus energy generated during the day time is stored in a
electric battery which stores the electrical energy for later use.
2.1 Design of the Existing system
Existing system
The existing system is shown in the above figure with various parts that make up the
system such as solar cells, battery, inverter, charging circuits, plumbing includes pipes and
valves, water pump, and the upper basin or reservoir as such more than nine parts makes up the
system.
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Block diagram of a solar irrigation system
2.1.1 water Pump
Centrifugal pumps are generally used to pump low viscosity fluids in large flow
rate, low pressure installations, which marks them perfect for applications that require the
pump to deal with large volumes. Thus centrifugal pumps are chosen for this particular
operation which requires pumping high volume of low viscous liquid in high flow rate.
The pumping power required can be calculated using the formula,
P= ρghQ
Where,
P = hydraulic power in (Kilowatt)
𝜌= density in (Kilogram per cubic meter)
g = acceleration due to gravity in (meters per seconds square)
h = height in meters
𝑄= volume flow rate in (meter cube per second)
Shaft Pump Power
The shaft pump power required can be calculated using the given formula,
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P s (kW )=P (kW )/η
Ps = shaft power in KW
P = pumping power required in KW
η = efficiency
2.1.2 Shaft power selection table
Shaft power required
From the above table the shaft power required for the pump can be calculated which can be used
for the proper selection of batteries and other auxiliary equipment.
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2.1.3 Battery requirement
As per the pumping power requirements we can select the number of
batteries required, if the power requirement is 18 KW we would need a total of three
batteries which can store a capacity of 6 Kwh per battery. More the requirements more
the amount of batteries and other storage system usage.
2.1.4 Inverters
An inverter, is an electronic device that changes the direct current (DC) to
alternating current (AC). It can also influence on the input voltage, output voltage and
frequency, and overall power handling rest on on the design of the exact device or
electrical structure.
Block diagram of an inverter
3.0 Proposed Approach
The Pump hydro system is an irrigation system which is combine with wind, solar,
nuclear or any other renewable power generation sources. It is mainly introduced to overcome
the initial cost, maintenance cost, and the environmental damages that the battery system for
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storing the energy causes and to overcome the intermittent nature of the Renewable Energy
Sources. These hybrid systems are to be implemented in rural areas for irrigation purpose.
This system consists of a Pump, turbine, generator, motor, valves, and solar cells the
main and major difference between the proposed and the existing system is the energy storage
system, in the existing system the energy is stored in the battery and then converted with an
inverter from DC to AC then the energy is supplied for the usage, here in the proposed system
the energy storage is accomplished with water tank only due to the potential energy of the water
when it is stored in a high head, this potential energy can be converted into electrical energy with
the help of a turbine coupled with a electricity generator.
This system makes use of a turbine to generate electricity. The energy is stored as a
potential energy of the water at the reservoir. During the daytime the water is pumped to the
reservoir with a pump, the electrical energy required for pumping the water is provided by solar
cells, which converts the solar energy into electrical energy with the help of Photo Voltaic cells.
During the night time the water stored inside the tank is allowed to flow from the reservoir to the
water well, when the water flows from some height the potential energy of the water is converted
into kinetic energy which in turn helps to rotate the turbine, the turbine is coupled with a
generator which generates electricity.
Solar cells have disadvantages such as the solar cells need to be facing the sun in a proper
manner to extract energy efficiently, they have intermittent nature, they cannot provide supply
during a cloudy day or the energy supply may be intervened during a cloudy day. In order to
overcome these, a reliable system such as pumped hydro storage system is very much important
in order to achieve a stable and reliable flow of energy. (Winston, R., 1974)
3.1 Design of the system
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The proposed Pump Hydro System consists of a Pump, turbine, generator, motor,
valves, and solar cells. A schematic diagram of the existing Pump Hydro System is shown in the
figure.
Schematic of a PHS
The schematic diagram of a pumped hydro storage System which is combined with a
solar cells.
In the schematic diagram,
P refers to the pump
T refers to the Turbine
Mrefers to the Motor
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G refers to the Generator
The design of the system consists of selection of each and every parts with a particular
requirement criteria.
3.1.1 Water Pump
Water pump is the most important part of this irrigation system, a water
pump pumps the water from the well to the water reservoir which is located at some distance
from the well. There are many types of water pumps commercially available in the market such
as Dynamic pumps and positive displacement pumps. Pumps like Centrifugal, vertical
centrifugal, horizontal centrifugal, submersible types comes under the Dynamic type and pumps
like diaphragm, gear pump, lobe and piston pumps comes under the displacement pumps.
For this project we will choose dynamic pumps under that centrifugal pump is chosen
because, centrifugal pumps are generally used to pump low viscosity fluids in large flow rate,
low pressure installations, which marks them perfect for applications that require the pump to
deal with large volumes. Thus centrifugal pumps are chosen for this particular operation which
requires pumping high volume of low viscous liquid in high flow rate.
The pump power requirements can be calculated using,
The pumping power required can be calculated using the formula,
P= ρghQ in KW
Where,
P = hydraulic power in (Kilowatt)
𝜌= density in (Kilogram per cubic meter)
g = acceleration due to gravity in (meters per seconds square)
h = height in meters
𝑄= volume flow ratein (meter cube per second)
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The shaft power plays an important role in the calculation of pump, the shaft power is the power
supplied to the pump by the motor
The shaft power required can be calculated using the given formula,
P s (kW )=P (kW )/η
Ps = shaft power in KW
P = pumping power required in KW
η = efficiency
The motor power:
The motor power is the amount of power that is consumed by the pump shaft from the
pump motor. The motor power is the ratio of shaft power over the mechanical efficiency of the
motor, in other words the motor power is the summation of the shaft power and the power losses
in mechanical work doing capacity of the motor as the conversion of electrical energy into
mechanical energy always have some losses.
The motor power is given mathematically as,
Pm= Ps
ηm
Where,
Pm = motor power in Kilowatts
Ps = Shaft power in Kilowatts
ηm = mechanical efficiency
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3.1.2 Generator
A generator is an electrical device which will convert the mechanical
power into electrical energy. The conversion of electrical energy in this project takes place in
four steps, which starts from the water reservoir. The water inside the reservoir when it is
allowed to flow through the conduit the Potential energy of the water is converted into kinetic
energy which helps to rotate the rotor which in turns converts into Mechanical energy then the
mechanical energy is converted into electrical energy with the help of this generator.
The Generator power is the amount of power that generated by the generator. The
Generator Power is the ratio of shaft power over the mechanical efficiency of the generator, in
other words the Generator power is the summation of the shaft power and the power losses in
mechanical work doing capacity of the generator as the conversion ofmechanical energy into
electrical energy always have some losses.
The generator power is given mathematically as,
PG = Ps
ηm
Where,
PG= Generator power in Kilowatts
Ps = shaft power inKilowatts
ηm = mechanical efficiency
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The shaft power is the power delivered from the turbine to the shaft of the generator. The shaft
power generation is directly proportional to the rotation of the turbine blades as the turbine blade
rotates it rotates the shaft of the generator which when creates the electricity.
The shaft power required can be calculated using the given formula,
P s (kW )= P ( kW )
η
Ps = shaft power in KW
P = pumping power required in KW
η = efficiency
3.1.3 Solar cells
Solar cells are made of semiconducting materials, the materials which
conducts electricity, when an impurity is introduced are known as semiconductors. Silicon and
Germanium are the most widely used semiconductors. One of the properties of semiconductors
that makes them most useful is that their conductivity may easily be modified by adding
impurities into their crystal lattice.
Let us consider a Photo Voltaic cell made up of silicon semiconductor, which has four
valence electrons, to improve its conductivity on one side of the cell, the impurities, which are
phosphorus atoms with five valence electrons (n-donor), are added this creates excess negative
charge carriers. On the other side, Boron atoms with three valence electrons (p-donor) are added
to create a positive junction with lack of electrons. The p-type conductor is allowed to make
contact with the n-type silicon semiconductor, thus a p-n junction is established.
Diffusion of electrons occurs from the region of high electron concentration (the n-type
side) into the region of lower electron concentration (p-type side). When the electrons diffuse
across the p-n junction, they recombine with holes on the p-type side. Though, the diffusion of
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carriers does not occur forever, because the inequity of charge immediately on either sides of the
junction originates an electric field. This electric field forms a diode that promotes current to
flow in only one direction. (Sah, C. T., Noyce, R. N., & Shockley, W., 1957)
When the cells are exposed to the solar energy the photon particles in the light transfer
some amount of energy to the Semiconducting material, the electron starts to jump across the PN
junction, thus a flow of electricity is achieved.
Total power in a solar cell is given by the equation,
P=IV =I so V [exp ( eV
kBT )−1]−I ph V In KW
So now the total solar panel area can be calculated regarding to the power supply
required which is obtained from the pump equation.
To calculate annual output of a solar cell:
E=( A)(r )(H )(PR)
E is the Energy in kWh
A is the Total solar panel Area in square meter
r is the solar panel yield or efficiency in (%)
H is the Annual average solar radiation on panels
PR is the Performance ratio, coefficient for losses
Calculation of amount of power collected on a surface:
The amount of power collected for a given surface area can be calculated using the
formula,
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P=I N ( A)
Where,
P = power in Kilowatt
I N =Incident power in Kilowatts
IN = Ibm cos θ
Ibm = Power density in Kilowatt per meter cube.
3.2calculations
Let us calculate the amount of power required, the amount of water flow
required, the volume of the reservoir, capacity of motor required, requirement of the solar panel
area and the plumbing pipes and valves requirements. The calculation are done using the
formulae that are defined previous sections.
Let us calculate the values needed for the irrigation of an acre of land let say if we can
plant a sum of 800 trees or plants in an acre it may vary according to the size of the plant, row
and column space requirement. For on tree roughly needs 20 liters of water for the amount of
800 trees we need to have at least 16,000 liters of water for the irrigation so now the reservoir
capacity must be at least 20,000 Liters.
Thus,
Area: one acre
Reservoir capacity needed: 20,000 Liters
Now the pump that required for pumping the water from the well need to be specified.
The amount of water required is 20,000 liters that much amount of water needs to be pumped
from the well of let say 30 meter long so the pumping capacity of the water can be calculated as,
So now the capacity of reservoir is 20,000 liters in order to pump 20,000 liters in an hour we
need to have flow rate which is calculated as 5.55 meter cube per seconds
3.2.1 Water pump
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The pumping power required can be calculated using the formula,
P= ρghQ
Where,
P = hydraulic power in (Kilowatt)
𝜌= density in (Kilogram per cubic meter)
g = acceleration due to gravity in (meters per seconds square)
h = height in meters
𝑄= volume flow rate in (meter cube per second)
Flow rate, Q= 5.5 m3/s
Density 𝜌: 1000 Kilogram per meter cube (Density of water)
Acceleration due to gravity: 9.8 m/s
Height, h: 30 m
Now,
Power of pump required is,
P= ρghQ
P = (997) (9.81)(15) (5.5)
P = 80.5 KW
So the pump power requirement is 80.5 KW
3.2.2 Solar panel requirement
We have calculated power needed is 80.5KW for an hour run so we know that energy
requirement is 80.5 KWh
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The Area of solar panel required can be calculated with the following formula as,
To calculate annual output of a solar cell:
E=( A)(r )(H )(PR)
E is the Energy in kWh.
A is the Total solar panel Area in square meter.
r is the solar panel yield or efficiency in (%)
H is the Annual average solar radiation on panels.
PR is the Performance ratio, coefficient for losses.
80.5=A ( 0.22 ) (220)(0.45)
Area A = 3.7 m2
Taking the values of efficiency, annual solar radiation fall and performance ratio from
various sources then the value of area is calculated as 3.7 square meter.
3.2.3 Turbine calculation:
The turbine should produce an amount of 80KW of power which
needed to begenerated via the turbine generator, thus a turbine which can
rotate at rpm can be calculated as,
P= 2 ( Π ) NT
60
Now,
N= 60 P
6.28(1500)
N=600 RPM
So the speed needed to generate a power of 80KW is 600 Rpm
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3.2.4 Generator
The generator need to produce the amount of energy that is
required by the pump to run forat least 5 hours of time which is equivalent
to the energy of 400 KWh, Thus the generator capacity should be
minimum of 80 KWh.
Generator capacity: 400 KWh
3.3 Hardware Description
In this section, the hardware that are selected for this project according to
the design requirements are explained and elaborated with their purpose and usage.
3.3.1 Water pump selection
The water pump selected is 80kW 8 inch double suction split case
water pump with motor. Which is commercially available in the stores.
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Selected water pump
Water pump: 80kW 8 inch double suction split case.
Quantity: 1
Price: 80,000 rupees
3.3.2 Solar cells
The solar cells need to cover an area of 3.7 square meter which is
not available as a whole so we need to buy a separate cells of available size and arrange them
regarding to our needs.
The commercially available size of a solar cell is 1.4 square meter. The
requirement is approximately 4 square meter of solar panel area. Thus 1.4 square meter solar
panels are bought in quantity of three.
Solar panel size: 1.4 m2
Quantity: 3
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Price: 72,000 rupees
3.3.3 Turbine
Choosing a turbine is the trickiest part, we need to have a turbine
which should help to generate at least 400 kWh of energy for this purpose, the hydrocoil 600
turbine is selected which works on low head and high flow and generates the required amount of
energy is selected.
Hydrocoil 600
Turbine: Hydrocoil 600
Configuration: Vertical
Price: 90,000 rupees
4.0 Cost Benefit Analysis
The Cost–benefit analysis (CBA) is a systematic approach toestimate the short and long
term consequences measuring all costs and all possible profits and benefits from an investment
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project proposal. It is also sometimes called benefit cost analysis (BCA).Cost benefit Analysis is
theevaluation of projects for a success level.(Robinson, R., 1993)
4.1 Cost of proposed project
The various cost such as initial cost, operating cost, labor cost, maintenance cost
and replacement parts costs are discussed and shown in this chapter.
Initial cost: It is the cost that is initially required to set up the plant in full working
condition.
Labor cost: It is the cost of labor that are required to operate this system this will be
calculated per year.
Maintenance cost: Maintenance cost includes various maintenance operation costs such
as cost for the lubrication, inspection, yearlypainting, etc. will be calculated per year.
Replacement cost: Replacement cost is the cost of a replacement part when it is worn out
or damaged due to unexpected failure.
Cost for the proposed system
S.no Description Cost (lakhs per year)
1 Initial cost 2.92
2 Labor cost 1
3 Maintenance cost 0.3
4 Operating cost 0.3
4 Total cost 4.2
Cost that is spent on an existing solar battery systemis,
S.no Description Cost (per year) in lakhs
1 Initial cost 3.5
2 Labor cost 1
3 Maintenance cost 1
4 Replacement cost of batteries(for 5 years 1 0.2
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lakhs)
5 Operating cost 0.5
6 Total cost 6.2
Now, cost for operating the solar system for 5, 10, and 15 years
For 5 years,
S.no Years Cost (for 5 years) in lakhs
1 Initial cost 3.5
2 Labor cost 5
3 Maintenance 5
4 Operating cost 2.5
5 Total cost 16
For 10 years, here replacement cost for batteries will also be added.
S.no Description Cost (for 10 years) in lakhs
1 Labor cost 10
2 Maintenance 10
3 Replacement 5
4 Operating cost 5
5 Total cost 30
For 15 Yrs,
S.no Description Cost (for 15 years) in lakhs
1 Labor cost 15
2 Maintenance 15
3 Replacement 10
4 Operating cost 7.5
5 Total cost 47.5
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So, the time vs cost for solar system is
5 years 10 years 15 years
0
5
10
15
20
25
30
35
40
45
50
Chart Title
Time in years
Cost in lakhs
Cost vs year graph for solar system
So, in our proposed system the cost for 5, 10 and 15 years can be calculated as
For 5 years,
S.no Years Cost (for 5 years) in lakhs
1 Initial cost 2.92
2 Labor cost 5
3 Maintenance 5
4 Operating cost 1.5
5 Total cost 14.42
For 10 years,
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S.no Years Cost (for 10 years) in lakhs
1 Labor cost 10
2 Operating cost 3
3 Maintenance 10
4 Total cost 23
For 15 years,
S.no Years Cost (for 15 years) in lakhs
1 Labor cost 15
2 Maintenance 10.5
3 Operating cost 4.5
4 Total cost 30
Graph: Cost Vs Time
5 years 10 years 15 years
0
5
10
15
20
25
30
35
Chart Title
Time in years
Cost in Lakhs
Cost vs year graph for solar system
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We can clearly see that the graph is not a straight line as in the case of solar system, it is
rather going down after 10 years of period.
4.2 Benefits of the proposed project
Although the cost of this proposed project is high the benefits are comparatively higher as
the proposed project does not makes use of a battery which have only a life time of
approximately 5 years. This project makes use of Pump Hydro storage system which can give a
life time of about 30-40 year. This system is also environment friendly as it does not have any
chemicals to work with.
Benefits of system compared with the solar battery system
Let us compare the proposed and the eixisting system cost for 5, 10, and 15 years.
S.no Years Solar battery PHS
1 5 16 14.42
2 10 30 23
3 15 47.5 30
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5 years 10 years 15 years
0
5
10
15
20
25
30
35
40
45
50
Chart Title
PHS Solar battery nil
Time in Years
Cost in Lakhs
Graph comparing PHS vs Solar battery systems
4.3 Comparison with existing system
We can clearly see that the proposed system has the following advantages over
the existing system:
 No battery, so maintenance and replacement cost is drastically reduced
 Less complicated structure
 Longer life span than the battery system as the life span of battery is only 5 years
this system has a life span up to 30 years.
 Cheaper than the battery system.
 Lower space is required as battery and inverter is eliminated
PHS system has many advantages over the existing system.
5.0 Results and Discussions:
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So from the comparison of the PHS system with the existing solar battery system in
various aspects the PHS system is proven to be the best, simpler, cheaper, efficient and the most
economic system.
We can clearly see from the Time VS Cost graph that the PHS system is more cost
efficient, as the cost gradually decreases over time, but in case of solar battery system the cost
increases with time, thus the PHS system is the economical among these two.
5 years 10 years 15 years
0
5
10
15
20
25
30
35
40
45
50
Chart Title
PHS Solar battery nil
Time in Years
Cost in Lakhs
Graph comparing PHS vs Solar battery systems
6.0 Conclusion
Various factors included in the Replacement of the existing solar battery system with the
PHS for irrigation in rural areas are discussed and the design of the PHS system is explained and
the hardware selection is discussed. Various Economic aspects such as initial cost, operating
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cost, labor cost, maintenance cost, and replacement cost of both the systems are calculated with
proper assumption and methodology. The result that the PHS system is more economical is
shown graphically and discussed briefly with appropriate manner.
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Bueno, C., &Carta, J. A. (2006). Wind powered pumped hydro storage systems, a means of
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Winston, R. (1974). Principles of solar concentrators of a novel design. Solar Energy, 16(2), 89-
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