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Power Electronics for

RES

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Basic steps to be followed when installing a PV system:

1. Ensure the roof area or other installation site is capable of handling the desiredsystem size.

2. If roof mounted, verify that the roof is capable of handling additional weight of PV

system. Expanded roof structure as necessary.

3. Properly seal any roof penetrations with roofing industry approved sealingmethods.

4. Install equipment according to manufacturer's specifications, using installation

requirements and procedures from the manufacturers' specifications.

5. Properly ground the system parts to reduce the threat of shock hazards andinduced surges.

6. Check for proper PV system operation by following the checkout procedures onthe PV System Installation Checklist.

7. Ensure the design meets local utility interconnection requirements

8. Have final inspections completed by the Authority Having Jurisdiction (AHJ) and

the utility (if required).

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Sizing of Solar Electric System

When we consider using solar electricity, we have to

know exactly how many appliances have to power.

Sizing is about calculating the number of solar modulesand batteries that are needed to run the required number

of appliances.

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Units of Consumption of Electricity

Calculation of Daily requirements of Appliances

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Mr. X decides to install a stand-alone PV system in his house.

The first step that he has to do is to determine the total daily

requirement of appliances of his house.

The appliances in his house are as follow:

- 8 fluorescent tubes, 20 W each (4 hours per day)

- 2 filament bulbs, 50 W each (2 Hours per day)

- One 10 W-DVD player (2 hours per day)

- One 80 W-color television (4 hours per day) - 2 cooling fans,40 W each (6 hours per day)

- Refrigerator, 100 W (24 hours per day)

- Clothes iron, 1KW (30 minutes per day)

- Electric cooker, 3KW (1 hour per day) - Air-conditioner 1.5hp

(8 hours per day)

Calculate the total daily electricity requirement of all the

appliances in Mr. X's house.

(1 hp = 0.7457 KW)

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1.5 hp = 1.5 X 0.7457 K = 1.12KW

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Estimating the consumption for a PV system

The output from a solar cell module of 40 W peak outputcan reach about 150 W h per day.

Recall that the total daily requirement of electricity for Mr. X'shouse is 16.52 KW h.

Therefore at least 110 modules (16.52K / 150 = 110.1

modules) are required to meet a daily requirement of16.52 KW h per day.

This is a very large number of solar modules for just one

home.

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For Mr. X's case, it is very obvious that air-

conditioner is not suitable to be run using solar

electricity.

Therefore, we have to replace the air-conditioner

by adding more low- power cooling fans.

Electric cooker has to be eliminated from this

system as the power consumed is too high.

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Therefore, the total daily requirement of electricity for Mr.

X's house is about 4 KW h.

The requirement has reduced about 75% from 16.52 KW h

to 4 KW h.

If using the solar module that can produce 150 W h perday, the minimum number of solar modules needed is

about 27 (4 K / 150 = 27) for stand-alone photovoltaic

system

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Thank You

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Energy Efficiency& Energy Costs

Light EmittingDiodes (LEDs)

IncandescentLight Bulbs

CompactFluorescents

(CFLs)

Life Span (average) 50,000 hours 1,200 hours 8,000 hoursWatts of electricity used(equivalent to 60 watt bulb).

LEDs use less power (watts) per

unit of light generated

(lumens). LEDs help reduce

greenhouse gas emissions from

power plants and lower electricbills

6 - 8 watts 60 watts 13-15 watts

Kilo-watts of Electricity used(30 Incandescent Bulbs per year

equivalent)

329 KWh/yr. 3285 KWh/yr. 767 KWh/yr.

Annual Operating Cost(30 Incandescent Bulbs per year

equivalent)

$32.85/year $328.59/year $76.65/year

Comparison ChartLED Lightsvs.Incandescent Light Bulbs vs. CFLs

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EnvironmentalImpact

Light Emitting Diodes(LEDs)

IncandescentLight Bulbs

Compact Fluorescents(CFLs)

Contains the TOXIC Mercury No NoYes - Mercury is very toxic

to your health and the

environment

RoHS Compliant

(Restriction of Hazardous SubstancesDirective ) Yes Yes

No - contains 1mg-5mg of

Mercury and is a major riskto the environment

Carbon Dioxide Emissions(30 bulbs per year)

Lower energy consumption decreases: CO2emissions, sulfur oxide, and high-level nuclear

waste.

451 pounds/year 4500 pounds/year 1051 pounds/year

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Important FactsLight Emitting Diodes

(LEDs)IncandescentLight Bulbs

Compact Fluorescents(CFLs)

Sensitive to humidity No Some Yes

On/off CyclingSwitching a CFL on/off quickly, in a closetfor instance, may decrease the lifespan of

the bulb.

No Effect SomeYes - can reduce lifespan

drastically

Turns on instantly Yes Yes No - takes time to warm up

DurabilityVery Durable - LEDs can

handle jarring andbumping

Not Very Durable -glass or filament can

break easily

Not Very Durable - glass canbreak easily

Heat Emitted 3.4 btu's/hour 85 btu's/hour 30 btu's/hour

Failure Modes Not typical SomeYes - may catch on fire,smoke, or omit an odor

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Light OutputLight Emitting Diodes

(LEDs)IncandescentLight Bulbs

Compact Fluorescents(CFLs)

Lumens Watts Watts Watts

450 4-5 40 9-13

800 6-8 60 13-15

1,100 9-13 75 18-25

1,600 16-20 100 23-30

2,600 25-28 150 30-55

S l ti S it bl M d l f th A li ti

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Selecting Suitable Modules for the Application

The number of cells needed in a module depends on the type

of charge regulation to be used and the local temperature.

Self-regulating modules with thirty to thirty-two cells are good

for small solar systems. A separate charge-regulating unit is not

needed, which keeps the system simple and low cost.

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additional of 2 cells is needed for self-regulating crystalline-

type with diode because it has to compensate voltage drop

in the diode. Diode is used to avoid current from flowing

back to the cells when the batteries are fully charged.

Determining the Daily Output from One Module

The daily electrical output from one module in units ofW h per day at 12 Vis

calculated using the following formula

F i l M X h d id d t i t ll

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From previous example, Mr. X has decided to installmodules with a current at load of 4.5 A under STC in hishouse. His house is located in Kuala Lumpur. What willbe the lowest daily electrical output of one module

averaged over a three-month period?

Kuala Lumpur is at latitude of 3 8 ' N. Based on the map, we

can calculate the average

output of a module over a three-month period using above

Equation. The calculations are

as follow:-

March - May

Average output = 4.5 X 5.5 X 12 = 297 W h per day at 12 V.

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June - August

Average output = 4.5 X 4.6 X 12 = 248.4 W h per day at 12

V.

September - November

Average output = 4.5 X 5.0 X 12 = 270 W h per day at 12 V.

December - February

Average output = 4.5 X 5.4 X 12 = 294.6 W h per day at 12

V.

Therefore, we have determined that the lowest average

daily output for Mr. X's module is 248.4 W h per day at 12 V

for June to August. As Kuala Lumpur is located at latitude of

3 8 ' N, the modules should be tilted at 15 from horizontaland facing South to generate electrical output.

Th f l f i i th i i b f d l

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The formula for sizing the minimum number of modules

Calculate the minimum number of modules needed by Mr. X from thevalues that we have determined from previous examples if he uses lead-

acid batteries.

S ol ut i on

Daily requirement of appliances = 3.966 K W h

Daily output of one module = 248.4 W h

Therefore, minimum number of solar modules needed by Mr. X is

= (3.966 K X 100) / (248.4 X 80)

= 19.96

That is 20 modules

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Sizing the Number of Batteries Needed

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Sizing the Number of Batteries Needed

We know that the daily requirement of appliances forMr. X house is 4 KW h per day. Mr. X has decided tochoose batteries which have a capacity of 500 A h.They are lead-acid batteries intended for deep-cycleoperation and can be discharged to adepth of 65 %. What is the smallest number of batteriesthat can be used?

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Sizing the Number of Batteries Needed

Since the estimate of period for storage is not given, we assume that 4

days of storage is adequate in Kuala Lumpur.

Using Equation 7.7, the total usable battery capacity needed i s:

(4 K X 4) / 12 = 1333.33A h at 12 V

Thus using Equation 7.8, the minimum number of batteries needed is:-

(1333.33 X 100) / (500 X 65) = 4.103

Therefore, the minimum number of batteries that are needed is 5.

Si i th N b f B tt i N d d

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Sizing the Number of Batteries Needed

Batteries have energy storage ratings mentioned in Amp-hour (Ah) or

milli-Amp-hour (mAh).

They also have a nominal voltage that they generate (typically deep

discharge batteries are 12 V batteries, cell phone batteries are 5 V

batteries, etc).

To calculate the total energy a battery can store you can use following

formula:

Units = (Volt x Ah) 1000 or (Volt x mAh) 1000000

So assuming we have a 1 kWp system and we assume that on an average

it generates 6 units a day and if we have to buy 12 V battery for it, the Ah

(or storage) of battery required would be:

(6 x1000) 12 = 500 Ah

Si i I t f S l PV t

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Sizing Inverterfor a Solar PV system

The input rating of inverter should never be lower than the total

wattages of the appliances

It is always better to have inverter wattage about 20-25% more than

that of the appliances connected

Most inverters available in market are rated on Kilo Volt Ampere/VoltAmpere.

In ideal situations (power factorof 1) 1 VA = 1 Watt. But in realpower

factorvaries from 0.85 to 0.99

So one can assume 1.18 VA = 1 Watt. So if you have a setup where the

total wattage of the system is 1000 Watts, it means your inverter size

required is more than 1180 VA or 1.18 kVA

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