ABSTRACTA solar lighting system which can make a 18x3 w lamp glow continuously for

about (9) hours if the battery is fully charged has been constructed. Here, solar energy is

collected with the aid of a solar panel and thus, a battery is charged during day time with

the help of a simple charging circuit. During night time, this stored energy is used

to l ight . The device can be used for small -scale lighting applications in remote areas

that are far away from the power grid. The system has a panel to collect the sun’s

energy, a battery to store that energy and a light source to use the energy. The system

operates like a bank account. Withdrawals from the battery to power the light source

must be compensated for by commensurate deposits of energy form the solar panels.

CHAPTER - 1

INTRODUCTION

Fig.1.1 Solar Panels

These early solar panels were first used in space in 1958.

Solar lighting system is the use of natural light to provide illumination.

Solar lighting system is the technology of obtaining usable energy from the

light ofthe sun using semi conductor materials and this is energy

efficient lighting technology. Solar panels are devices that generate power from

the sun by converting sunlight into electricity with no moving parts, zero emissions

and no maintenance. They are used in residential, commercial, institutional and light

industrial applications.

The construction of a solar lighting system serves as a means of reducing

energy imports and dependence upon oil and gas, which mitigate the risk of fuel-

price volatility and supplies energy for small-scale lighting applications when and

where electricity is mostlimited and most expensive.

CHAPTER 2

BLOCK DIAGRAM OF SOLAR

LIGHTING SYSTEM

Block Diagram Of Solar Lighting System

A Basic Photovoltaic System For Power Generation:

A basic photovoltaic system integrated with the utility grid is shown in fig.It

permits solely generated electrical power to be delivered to a local load. It consists of:

Solar array, large or small, which converts the insulation to useful DC

electrical power.

A blocking diode which lets the array-generated power flow only toward the

battery or grid. Without a blocking diode the battery would discharge back through

the solar array during times of no insulation.

Battery storage, in which the solarly generated electric energy may be stored.

Inverter, usually solid state which converts the battery bus voltage to AC of

frequency and phase to match that needed to integrate with the utility grid. Thus

it is typically a DC,AC inverter. It may also contain a suitable output step up

transformer, perhaps some filtering and power factor correction circuits and perhaps

some power cpnditioning, circuitry to initiate battery charging and to prevent over

charging. Power conditioning may be shown as a separate system functional block. This

block may also be used in figure shown to function as a rectifier to charge the battery

from the utility feeder when needed and when no insulation was present. Appropriate

switches and circuit breakers, to permit isolating parts of the system, as the battery.

One would also want to include breakers and fusing protection between the inverter

output and the utility grid to protect both the photovoltaic system and the grid.

CHAPTER - 3

WORKING PRINCIPLES

AND OPERATION

The solar energy can be directly converted into electrical energy by means of

photovoltaic effect, i.e. conversion of light into electricity. Generation of an

electromotive force due to absorption of ionizing radiation is known as photovoltaic

effect. The energy conversion devices which are used to convert sunlight to electricity

by use of the photovoltaic effect are called solar cells. Photo voltaic energy

conversion is one of the most popular nonconventional energy source. The

photovoltaic cell offers an existing potential for capturing solar energy in a way

that will provide clean, versatile, renewable energy. This simple device has no moving

parts, negligible maintenance costs, produces no pollution and has a lifetime equal to

that of a conventional fossil fuel.

Photovoltaic cells capture solar energy and convert it directly to electrical current by

separating electrons from their parent atoms and accelerating them across a one way

electrostatic barrier formed by the function between two different types of semiconductor

material.

Battery storage: the simplest means of storage on a smaller moderate scale is in

electric storage batteries, especially as solar cells produce the direct electric current

required for battery charging. The stored energy can then be delivered as electricity upon

discharge. The common iead acid storage batteries, such as are used in automobiles, are

not ideal for this purpose, but they are probably the best presently available. Extensive

research in progress should leadto the development of more suitable batteries. A possible

alternative is to use the direct current from solar cells to decompose water into

hydrogen and oxygen gases. These gases would be stored in a suitable form and

utilized as needed to generate electricity in a fuel cell.

Inverters: these are the device susually solid state, which change the array DC

output to AC of suitable voltage, frequency, and phase to feed photo voltaically generated

power into the power grid or local load, as shown in figure. These functional blocksare

sometimes referred to as power conditioning.

A general type of inverter circuit which is found best suitable for the utility application

is shown in fig. the current can be used in two modes: (1) as an inverter changing DC

to AC or (2) as a rectifier changing AC to DC, thus charging the battery. It is clear that

the system photovoltaic offers the option of DC power, AC power, hydrogen and

oxygen fuels in either gas or liquid forms from which electricity can be generated. The

system has many advantages and disadvantages.

CHAPTER 4

SOLAR PHOTOVOLTAIC ARRAY:

Fig 3.1 Solar Photovoltaic Array

The solar photovoltaic array consists of an appropriate number of solar cells

connected in series and or parallel to provide the required current and voltage. The array

is so oriented as to collect the maximum solar radiation throughout the year. There may be

tracking arrays or modules or fixed arrays. A tracking array is defined as one which is

always kept mechanically perpendicular to the sun array line so that all times it intercepts

the maximum isolation. Such arrays must be physically movable by a suitable prime

mover and are generally considerably more complex than fixed arrays. A fixed array is

usually oriented east west and tilted up at an angle approximately equal to the latitude of

the site. Thus the array design falls into two broad classes.

Application of solar photovoltaic system:

Various solar photovoltaic systems have been developed and installed at

different sites for demonstration and field trial purposes. The terrestrial applications of

these include provision of power supply to:

1. Water pumping sets for micro irrigation and drinking water supply,

2. Radio beacons for ship navigation at ports,

3. Community radio and television sets,

4. Cathodic protection of oil pipe lines,

5. Weather monitoring,

6. Railway signaling equipment,

7. Battery charging,

8. Street lighting.

The major application of photovoltaic systems lies in water pumping for

drinking water supply and irrigation in rural areas. The photovoltaic water pumping

system essentially consists of:

(a) A photovoltaic (PV) array, (b)

Storage battery,

(c) Power control equipment, (d)

Motor pump sets, and

(e) Water storage tank.

Solar electric power generation: solar photo-voltaic:

1. The direct conversion of solar energy into electrical energy by means of the

photovoltaic effect that is the conversion of light into electricity. The photovoltaic effect is

defined as the generation of an electromotive force as the result of the absorption on of

ionizing rad ia t ion . Energy convers ion dev ices which a re used to convert sunlight

to electricity by the use of the photovoltaic effect are called solar cells. A single

converter cell is called a solar or, more generally, a photovoltaic cell, and combination

of such cells; designed to increase the electric power output is called a solar

module or solar array.

2. Photovoltaic cells are made of semiconductors that generate

electricity when they absorb light. As photons are received, free electrical charges are

generated that can be collected on contacts applied to the surfaces of the

semiconductors. Because solar cells are not heat engines, and therefore do not need

to operate at high temperatures, they are adopted to the weak energy flux of

solar radiation, operating at room temperature. These devices have theoretical

efficiencies are less than half this value, and decrease fairly rapidly with increasing

temperature.

(I) Flat Plate Arrays

3. Where i n so l a r ce l l s a r e a t t a ched wi th a su i t ab l e adhesive to some kind

of substrate structure usually semi rigid to prevent cells being cracked.

4. This technology springs from the space related photovoltaic technology and

many such arrays have been built in various power sizes.

(II) Concentrating Arrays:

5. Where in suitable optics, e.g. Fresnel lenses, parabolic mirrors are

combined with photovoltaic cells in an array fashion. This technology is relatively new

to photovoltaic in terms of hardware development and comparatively fewer such

arrays have actually been built.

Crystalline silicon modules:

6. Most solar modules are currently produced from silicon photovoltaic cells .

These are typical ly ca t ego r i ze d as mono crystalline or polycrystalline modules.

Solar cell modules (solar photovoltaic arrays):

7. There may be tracking arrays or modules or fixed arrays. A tracking array

is defined as one which is always kept mechanically perpendicular to the sun-array line

so that all times it intercepts the maximum isolation. Such arrays must be physically

movable by a suitable prime mover and are generally considerably more complex then fixed

arrays. A fixed array is usually oriented east west and tilted up at an angle

approximately equal to the latitude of the site. Fixed arrays are mechanically simpler

then tracking arrays. Thus the array designs fall into two broad classes:

8. Flat-plate arrays: Wherein solar cells are attached with a suitable adhesive

to some kind of substrates structure usually semi - rigid to prevent cells being cracked.

This t echno logy spr ings f rom the space -related pho tovo l t a ic technology,

and many such arrays have been built in various power sizes.

9. Concentrating arrays: Wherein suitable option, e.g. Fresnel lenses,

parabolic mirrors, compound parabolic concentrators (CPC), and others, are

combined with photovoltaic cells in an array fashion. This technology is relatively new

to photovoltaic in terms of hardware development, and comparatively fewer such arrays

have actually been built.

The photo-voltaic effect can be observed in nature in a variety of materials, but

the materials that have shown the best performance in sunlight are the semi-conductors

as stated above. When photons from the sun are absorbed in a semiconductor, they

create free electrons with higher energies than the electrons which provide the bonding

in the base crystal. Once these electrons are created, there must be an electric field to

induce these higher energy electrons to flow out of the semi-conductor to do useful

work. The electric field in most solar cells is provided by a junction of materials which

have different electrical properties.

To obtain a useful power output from photon interaction in a semi-conductor

three processes are required.

The photons have to be absorbed in the active part of the material and

result in electrons being excited to a higher energy potential. The electron-hole

charge carrier created by the absorption must be physically separated and moved to the

edge of the cell. The charge carriers must be removed from the cell and delivered to a

useful load before they lose their extra potential.

Advantages and disadvantages of photovoltaic solar energy conversion

Advantages:

Direct room temperature conversion of light to electricity through a simple solid state

device.

Absence of moving parts.

Ability to function unattended for long periods as evidence in space program me.

Modular nature in which desired currents, voltages and power levels can be achieved by mere integration.

Maintenance cost is low as they are easy to operate.They do not create pollution.They have a long effective life.They are highly reliable.

They consume no fuel to operate as the sun’s energy is freeResponse i n ou t pu t t o i np u t r a d i a t i o n changes; no long-time constant is involved, as on thermal systems, before steady state is reached. They are easy to fabricate, being one of the simplest of semiconductor devices. They can be used with or without sun tracking, making possible a wide range of application possibilities. Their principal disadvantages are their high cost, and the fact that, in many applications, energy storage is required because of no isolation at night. Efforts are being made world-wide to reduce costs through various technological innovations.

For completing the above processes, a solar cell consists of:

Semi-conductor in which electron hole pairs are created by absorption of incident solar radiation. Region containing a drift field for charge separation, and Charge collecting front and back electrodes.

CHAPTER - 5

CHARGE CONTROLLER:

Fig:4.1 Charge Controller

Overcharging of some batteries results in lossof electrolytic, corrosion, plate growth and loss of active material from

the plates, causing reduction in battery life. Also, the repeated failure to reach full charge also leads to stratification of electrolyte.

Thus, there is a need of charge regulators to optimize the battery life. Most charge regulators start the charging process with a high current and reduce it to a very low level when a certain battery voltage is reached. A digital based charge regulator monitors the batterycurrent, and voltage computes the levelof charge and regulates the input and output currents so asto avoid both overcharging and excessive discharge.

CHAPTER 6

INVERTER

Inverters

Sometimes some petty things makes big confusion in our mind, One of those petty thing is

calculating our inverter battery Back-up time. 99% of people who own Home inverter would ask

this question to their battery dealers at least once in their life time. But most battery shop owners

don’t let their customers know the simple formula to calculate the Back-up time of inverter

battery. Don’t worry readers, now I will let you know that simple formula!

Types of Power Inverters

Based on the output waveforms, there are three types of Inverters. These are Sine wave,

Modified Sine wave and Square wave inverters.

Sine wave power inverter

Alternating current has continuously varying voltage, which swings from positive to negative.

This has an advantage in power transmission over long distance. Power from the Grid is

carefully regulated to get a pure sine wave and also the sine wave radiate the least amount of

radio power during long distance transmission. But it is expensive to generate sine wave in an

inverter. Its quality is excellent and almost all electrical and electronic appliances work well in

sine wave inverter.

The sine wave is the AC waveform we get from the domestic lines and from the generator. The

major advantage of sine wave inverter is that all of the house hold appliances are designed to

operate in sine wave AC. Another advantage is that the sine wave is a form of soft temporal rise

voltage and it lacks harmonic oscillations which can cause unwanted counter forces on engines,

interference on radio equipments and surge current on condensers.

Modified Sine wave or Quasi Sine wave

Modified sine wave is designed to simulate a sine wave since the generation of sine wave is

expensive. This waveform consists of a Flat Plateau of positive voltage, dropping abruptly to

zero for a short period, then dropping again to a flat plateau of negative voltage. It then go back

to zero again and returning to positive. This short pause at zero volts gives more power to 50 Hz

fundamental frequency of AC than the simple square wave.

Inverters providing modified sine wave can adequately power most house hold appliances. It is

more economical but may present certain problems with appliances like microwave ovens, laser

printers, digital clocks and some music systems. 99% of appliances run happily in modified sine

wave. Instruments using SCR (Silicon Controlled Rectifier) in the power supply section behave

badly with modified sine wave. The SCR will consider the sharp corners of the sine wave as

trashes and shut off the instrument. Many of the Laser printers behave like this and fail to work

in inverters and UPS providing modified sine wave power. Most variable speed fans buzz when

used in modified sine wave inverters.

Square Wave Power Inverter

This is the simplest form of output wave available in the cheapest form of inverters. They can

run simple appliances without problems but not much else. Square wave voltage can be easily

generated using a simple oscillator. With the help of a transformer, the generated square wave

voltage can be transformed into a value of 230 volt AC or higher.

Terms related to Inverter

Watt (W)

Watt is the measure of how much power a device uses when turned on or can supply. If a device

uses 100 watts, it is simply the voltage times the ampere (rate of current).If the device takes 10

Amps at 12 Volt DC, it uses 120 watts power. That is 10A x 12 V = 120 W.

Watt Hour (WH)

A watt hour (or Kilo Watt hour – kWh) is simply how many watt times, how many hours the

device is used. If the device uses 100 watts for 10 hours, it is 1000 watt hour or 1 kWh. The

electricity tariff is based on kWh.

Ampere (A)

It is the measure of electrical current at the moment. Amps are important to determine the wire

size for connecting the inverter to the battery. Low gauge wire will heat up and burn if heavy

current flows through it from the battery.

Ampere Hour (Ah)

Amp-Hour usually abbreviated as Ah is the Amps x Time. Ah is the measure of battery capacity

which determines the backup time of the inverter

Volt Ampere (VA)

It represents the maximum load capacity of the inverter. Commonly available inverters are 500

A, 800 VA, 1000 VA, 1500 VA etc.

Power in Inverter Peak power and Typical or Average power

An inverter needs Peak or Surge power and Typical or Average (Usual) power. Peak power is

the maximum power that an inverter can supply usually for short time.Some heavy current

appliances like motor and refrigerator requires a startup peak power than they require when

running. Typical power is the power that the inverter gives on a steady basis. This is usually

much lower than the peak power. Typical power is useful in estimating the battery capacity.

Therefore inverters must be’ sized’ for the maximum peak load and typical continuous power.

Power Rating of Inverter

Inverters are available in different ‘Size ratings’ from 50 VA up to 50000 VA. Inverters larger

than 11000 VA are seldom used in household applications. The first thing you have to consider

about the inverter system is its maximum peak power or surge power and steady current supply.

The surge rating is usually specified at so many watts for so many seconds. This means that the

inverter will handle an over load of that many watts for a short time. This ‘surge capacity’ will

vary considerably between inverters and even with in the same brand. Generally 3-15 seconds

surge rating is enough to cover 99% of appliances. Inverters with lowest surge rating are the high

speed electronic switching types.

Current consumption

Depends on the wattage of the appliances used, current consumption of the inverter can be

calculated using the formula

I = W / V Where I is the current in amps, W is the wattage of the appliance and V the 12volt

(battery voltage).

Watt

Wattage of an instrument is calculated using the formula

W = V X I Where V is the 230 volt AC and I is the current consumption.

Watt rating is usually printed on the back side of the appliance near the power cord.

VA

It indicates the ‘Size’ (capacity) of the inverter. To select the inverter size, the given formula is

useful

VA = W x Inverter loss. Inverter loss is typical around 1.15. If the total load connected to the

inverter is 400 watts then the minimum inverter size should be 400 x 1.15. That is 460 VA. A

500 VA is suitable for the load.

Ampere Hour (Ah)

The capacity of the battery is represented in Ah. It is the amount of current a battery can give

during one hour of charge / discharge cycle. High capacity batteries (100 Ah, 150 Ah) are used

to power inverters to get sufficient backup time. The formula to select the battery power (Ah) is

Load in watts / Voltage of battery x Backup hour.

For example if you wants to run 400 watts load on 12 volt battery for 3 hours, then the capacity

of the battery should be minimum 100 Ah. Ah = 400 / 12 x 3 = 100 Ah. If the load increases

(within the capacity of the inverter), backup time reduces.

CHAPTER 6

POWER CALCULATION

Power Calculation

Before purchasing an Inverter, it is necessary to calculate the power requirements of the loads

that are going to be connected. Mid size inverter (500VA to 800 VA) will handle most of the low

power household appliances.

If you wish to use 2 Fans, 2 Tube lights and 1 TV, the total power consumption will be around

380 Watts. Power loss in the inverter is estimated as 1.15. So the Inverter capacity should be at

least 380×1.15. That is 437 VA. The suitable available size is 500 VA. Inverter rating is usually

given in VA (Volt Ampere). The backup time of the inverter depends on the type and capacity of

the battery used. Battery capacity is represented in Ah (Ampere hour). To power the whole house

(say 3000 watts), more planning is necessary. A 5 KVA inverter is necessary in this case. But

obviously not every appliance will be ‘ON’ at the same time. So a high capacity inverter is a

good choice because, if the load increases above the rated capacity of the inverter, it can lead to

hazardous results. If a 24 Volt inverter with 24 volt battery is used, backup time can be doubled.

But the cost of the inverter and battery will be high compared to a 12 volt inverter and 12 volt

battery.

Power loss in Inverter and Current consumption during charging.

No power inverter can function efficiently. The working of the inverter depends on many factors

like conducted load, battery efficiency and maintenance. During operation, inverter will heat up

and the transformer will dissipate heat. So some energy will be lost which reduces the efficiency.

Proper charging of the battery and its efficiency to hold charge are two very important aspects.

Input voltage from the AC lines should be close to 230 volts for proper charging of the battery.

Fully charged battery will show 13.8 volts. Inverter should switch on charging immediately

when the battery voltage reduces to 12 volts. Charging current depends on the time taken to

complete the charging process and also the ‘charge condition’ of the battery. If the battery is

discharged to 80% of its efficiency, it will take 5 to 7 Ampere current for charging during the

first few hours. Then the current reduces to 500 milli amperes or less. A fully charged battery

will not take any current. Most inverters have two mode charging-Boosts charging and Trickle

charging. During boost charging, around 5 to 7 ampere current will utilized and during trickle

charging only 25 to 50 milli ampere current will be utilized.

The back bone of the Inverter system is the battery. We use a high current tubular battery to give

a maintenance free performance. The charge in the battery is used to convert DC to AC by the

inverter. Measure of charge is Coulomb and each electron carries 1.602 e- 19 coulomb charge.

As a rule, when 1 amps current passes through a conductor in 1 second, it uses 1 coulomb charge

So charge Q = It

If 1 amps current flows through the conductor in 1 hour, 3600 coulomb charge will be utilized.

The amount of charge in the battery is represented in Amps hour (Ah). That is Amps times Hour.

It is the amount of charge present in the battery. But Amps hours cannot be used to measure the

charge level, since the voltage changes during discharge. So the measure of charge is Watts

Hours. Watt hour can be calculated by multiplying the nominal voltage with the battery capacity

in Amps hours.

E = C x Vg. where C is the capacity of the battery in Ah and Vg is the discharge rate.

Charge requirement of the load

The following equation tells you, how much battery charge is required for your load.

First find out the capacity of the battery. If ‘X amps’ current is drawn by the load in ‘t hours’,

then the capacity of the battery.

C = Xt

Suppose the load is drawing 120 mA current in 24 hours, then the capacity of the battery should

be

C = 0.12A x 24 = 2.88 Ah

It is not a good practice to discharge the battery completely till the load shut off. Stop running

the load, if the battery charge reduces to 20%.

Battery capacity

If the load takes 120 mA current in 1 hour, maximum capacity of the battery should be 0.12A x

24 = 2.88 Ah. Best method to keep the charge/discharge cycles perfect is to stop discharging the

battery till it maintains 20% charge. Hence to retain 20% charge, the capacity of the battery

should be

C / 0.8

If the battery requires 2.88 Ah in one hour, then to keep 20% charge in it, the capacity of the

battery should be

2.88Ah / 0.8 = 3.6Ah.

Discharge Rate

Lead acid batteries have few Amps hour if the discharge rate is fast. Generally, the lead acid

battery is rated for 20 hours discharge rate provided the discharge rate is slow. At high discharge

rate, the capacity of the battery drops steeply. Suppose the battery is 10 Ah and its discharge rate

is 1C. One hour discharge of the battery at the rate of 1C (10 Amps in 1 hour), the capacity

reduces to 5 Ah in one hour. So the following tips will guide you to keep a steady discharge rate

of the battery. Suppose you want to run a load at 20 Amps for 1hour. Then the capacity of the

battery should be

C = It = 20A x 1 Hour = 20Ah.

Keep discharge rate to maximum 80%. Then the battery capacity should be

20Ah / 0.8 = 25 Ah

So a 25 Ah battery can give 20 Amps current for 1 hour to run the load.

But it is better to drain the battery to 50% only then,

25 Ah / 0.5 = 50 Ah.

Hence as a rule of thumb, it is better to use a 50Ah battery to run the load at 20 Amps per Hour

to keep 50% charge in the battery. Suppose the load is not drawing 20 Amps continuously in 1

hour. During the first second, it draws 20 Amps and then 100 mA during the remaining period.

So the average current drawn by the load is

20A x 1/ 3600 + 0.1 / 3600 = 0.1044 A

3600 is the total seconds in 1 hour.

In short, if the load is not drawing the charge in a steady manner, the capacity of the battery will

be increased.

How to calculate the charge?

It is difficult to measure the current drawing by the load at different times. The easy way is to

consider the power rating (Watts) of the load. Suppose the power rating of the load is 250 watts

and is drawing current from the inverter system for 5 hours.

Then, its Watts hour is Watts x Hour = 250 x 5 = 1250 Watts hour.

Consider the efficiency of the inverter as 85 %( no inverter is 100% efficient).

So 1250 / 0.85 = 1470 watts hour

So the load has 1470 watts hour instead of 1250 watts hour.

Then calculate the capacity of the battery

As you know, Watt is Amps x Volt

So if the Watt hour is divided by Voltage of the battery, you will get the Amps hour

Watt hour / Volt = Amps hour

1470 / 12 = 122.5 Ah

Thus to run 1470 watts load, minimum capacity of the battery should be 125Ah to run the load

for 5 hours. As already stated, it is better to use a 150 Ah battery to keep discharge cycle 50%.

How to select an Inverter

Total load to be connected = 500 watts

Power factor = 0.8 (all inverters have a power factor between 0.6 to 0.8)

Inverter VA = 500 / 0.8 = 625 VA

So select 800 VA inverter to run 500 Watts load

How to select the battery

Backup time = Watt / Battery voltage x Hours

500 / 12 x 3Hr = 125 Ah

12 volt tubular battery has a terminal voltage of 14.8 volts in fully charged condition. Usually the

inverter has cutoff facility to protect battery from deep discharge. Most of the inverters are set

for 80 % (Retaining charge) cutoff voltage. That is after 20% discharge, inverter will shut off.

Inverter cutoff voltage = 14.8 x 0.8 = 11.84 volts

Watt hour = watt x hour = 500 watts x 3 hours = 1500 watt hours

Ah of battery = watt hour / volt = 1500 / 11 = 126 Ah.

In short, an 800 VA inverter with 126 Ah battery can power 500 watts load for 3 hours.

Calculating UPS/INVERTER Battery Backup:

Before calculating the Battery Backup, let us know a few factors that vary battery backup.

For and UPS with 875Va we can us a maximum load of 640 watts.  677Va we can us a

maximum load of 540 watts The main thing we have understand is that what ever may be the

UPS wattage the battery backup will not vary. The battery backup will only vary depending on

the battery Ah and the Usage Load.   To calculate UPS backup, We have a simple formula.

 UPS Backup = Battery Ah * (Volts/Load) * (1/Powerfactor)

Example:      

 Let us calculate the backup for a system UPS Load is the usage power, suppose we are running a

PC then the load is around 300 watts Power Factor varies for device to device, the average power

factor is 1.4 Voltage is the voltage of the battery.  For a single battery the voltate is 12v.

 If the batteries are connected in a series   Voltage = 12 * no. of batteries.  Battery Ah is the

battery ampere used for the ups, System UPS battery AH is 7

 Then   Battery Backup =  7 * (12/300) * (1/1.4)

                                                        =  7 * (0.04) * (0.7)

                                                        =   0.19 hours

                                                        =    19 Minutes

Back up Time of Inverter Battery = Battery Volt x Battery AH rating / Total watts on Load

So,

If a Person use 1 ceiling Fan + 1 Tube Light + 2 (15watts) CFL simultaneously with 150 AH

battery, then the backup time will be calculated as

1 ceiling Fan = 75 watts

1 Tube light = 40 watts

2x 15watts CFL= 30 watts

Total = 145 watts

Back-up time of 150 AH Battery = 12 * 150 / 145

= 12.41 hours (approximately)

This is Not Accurate Why?

This calculation shows only approximate value because there will be some loss of energy when

converting 12 v battery power to 220 volts through inverter.

Moreover, we cannot figure out the exact power consumption of ceiling fan, as it has speed

adjustment dimmer switch. The power consumption will be low when the fan runs at slow speed.

Note :( Power consumption of old fans with big sized Regulator remains same at all speed

points)

The alternation of the direction of current in the primary winding of the transformer produces

alternating current (AC) in the secondary circuit. This will be 230 /110 Volt AC and can operate

all the electric devices.

Efficiency of the Power Inverter for Home

The quality of the output wave form (230 volt AC) from the inverter determines its efficiency.

The quality of the inverter output wave form is expressed using Fourier analysis data to calculate

the ‘Total Harmonic Distortion’ (THD). THD is the square root of the sum of the squares of the

harmonic voltage divided by the fundamental voltage.

THD = √ V2 2 + V3 2 + V4 2…………. Vn 2 / V1

Selecting the Best Home Power Inverter

Before selecting the inverter, it important to calculate the total power consumption of the

appliances that are going to be connected to the inverter.

It is important to note that the power consumption (Current drain) increases when the driving

voltage decreases. When the input voltage drops in the domestic supply lines from 230 V AC to

200 VAC, electric devices will consume more current. For example a load rated 300 Watts (like

Fridge) operating in 230 volts uses 1.4 Amps in one hour. The same load takes 1.5 Amps if the

line voltage drops to 200 volt during peak hours (6pm- 9 pm). So it is advisable to switch off

high current devices during peak hours. This conserves energy as well as reduces the electricity

bill. If heavy loads are running in inverter, battery voltage diminishes from 13.5 volts to 12 volts

with in a short time. This reduces the backup time. So it is better to connect only light loads such

as fan, lights and TV to inverter lines.

CHPTER 7

CONCLUSION

ADVANTAGES OF SOLAR LIGHTING SYSTEM:

Solar cells directly convert the solar radiationinto electricity using photovoltaic effect without going through a thermal

process. Solar cells are reliable, modular, durable and generally maintenance

free and therefore, suitable even in isolated and remote areas. Solar cells are quiet, benign, and compatible with almost all environments,

respond instantaneously with solar radiation and have an expected life time of 20 or more years.

Solar cells can be located at the place of use and hence no distribution network is required.

DISADVANTAGES OF SOLAR LIGHTING SYSTEM:

The conversion efficiency of solar cells is limited to 10 percent. Large areas of solar cell modular are required to generate sufficient useful power.

The present costs of solar cells are comparatively high, making them economically uncompetitive with other conventional power generation methods for terrestrial applications, particularly

where the demand of power is very large Solar energy is intermittentand solar cells produce electricity when sun shines and in

proportion to solar intensity. Hence, some kind of electric storage is required making the whole system more costly. However, in large installations, the electricity generated by solar cells can be fed directly into the electric grid system.

Battery charge level maintenance and discharge limit and

life shortened

REFERENCE

1. non-conventional energy sources By - GD RAI

2. non-conventional energy sources By - Atul Prakashan

3. h tt p : / / e n . w i k i pe d i a. o r g / w i k i/ S o l a r _S y s t em

4. h tt p : / / www .s a n j a y m a r k e ti n g .c o m / so l a r l i g h ti n g s y s t e m .h t m

5. h tt p : / / www .dc m s m e . g o v . i n /r e p o r t s / e l ec t r o n i c / S O L A R L I GH T S Y S T EMS . pdf

6. htt p: // mnr e.gov.i n/f il e -manager/ User Fil es/ pdf /Trai ner s% 20T ext book% 20 -% 20Sol ar% 20Li ghti ng% 20Syst ems.pdf