Disadvantages of Renewable Energy Sources

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.

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.

The Proposed System: Pump Hydro Storage System

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 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.

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.

Comparison with Existing System

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.

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 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.

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. Block diagram of a solar irrigation system

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, 

Where,

P = hydraulic power in (Kilowatt)

?= density   in (Kilogram per cubic meter)

g = acceleration due to gravity   in   (meters per seconds square)

h = height in meters

?= volume flow rate in (meter cube per second) 

Shaft Pump Power 

The shaft pump power required can be calculated using the given formula,

Ps = shaft power in KW

P = pumping power required in KW

 = efficiency 

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.

Design and Evaluation Process

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.

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. 

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 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)

Water Pump Selection

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

G refers to the Generator

The design of the system consists of selection of each and every parts with a particular requirement criteria.

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,

                    in   KW

Where,

P = hydraulic power in (Kilowatt)

?= density   in (Kilogram per cubic meter)

g = acceleration due to gravity   in   (meters per seconds square)

h = height in meters

?= volume flow ratein (meter cube per second)

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,

Ps = shaft power in KW

P = pumping power required in KW

 = 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.

Pump Power Requirements

The motor power is given mathematically as,

Where,

P_m = motor power in Kilowatts

P_s = Shaft power in Kilowatts

 = mechanical efficiency

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,

Where,

P_G= Generator power in Kilowatts

P_s = shaft power  inKilowatts

 = mechanical efficiency 

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,

Ps = shaft power in KW

P = pumping power required in KW

 = efficiency

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.

Shaft Pump Power

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 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,

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 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, 

Where,

P = power in Kilowatt

 in Kilowatts

I_N = I_bm cos θ   

I_{bm =} Power density in Kilowatt per meter cube.

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

Inverter and its Function

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

The pumping power required can be calculated using the formula, 

Where,

P = hydraulic power in (Kilowatt)

?= density   in (Kilogram per cubic meter)

g = acceleration due to gravity   in   (meters per seconds square)

h = height in meters

?= volume flow rate in (meter cube per second) 

Flow rate, Q= 5.5 m³/s

Density ?: 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 = (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

The Area of solar panel required can be calculated with the following formula as,

To calculate annual output of a solar cell: 

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. 

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.

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, 

Now, 

So the speed needed to generate a power of 80KW is 600 Rpm 

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 

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. 

Importance of Pump Hydro Storage System

The water pump selected is 80kW 8 inch double suction split case water pump with motor. Which is commercially available in the stores. 

Water pump: 80kW 8 inch double suction split case.

Quantity: 1

Price: 80,000 rupees

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

Price: 72,000 rupees

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 

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 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)

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 lakhs)	0.2
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

So, the time vs cost for solar system is 

Cost vs year graph for solar system 

Schematic Diagram of Pump Hydro Storage 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, 

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 

Cost vs year graph for solar system 

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. 

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

Graph comparing PHS vs Solar battery systems 

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.

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.

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 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.

References 

Barbir, F., Veziro?lu, T. N., &Plass Jr, H. J. (1990). Environmental damage due to fossil fuels use. International journal of hydrogen energy, 15(10), 739-749. 

Su, W., Wang, J., &Roh, J. (2014). Stochastic energy scheduling in microgrids with intermittent renewable energy resources. IEEE Transactions on Smart Grid, 5(4), 1876-1883. 

Bueno, C., &Carta, J. A. (2006). Wind powered pumped hydro storage systems, a means of increasing the penetration of renewable energy in the Canary Islands. Renewable and Sustainable Energy Reviews, 10(4), 312-340. 

Winston, R. (1974). Principles of solar concentrators of a novel design. Solar Energy, 16(2), 89-95. 

Sah, C. T., Noyce, R. N., & Shockley, W. (1957). Carrier generation and recombination in pn junctions and pn junction characteristics. Proceedings of the IRE, 45(9), 1228-1243. 

Robinson, R. (1993). Cost-benefit analysis. BMJ, 307(6909), 924-926.