3. basics of solar radiaiton

7/28/2019 3. Basics of Solar Radiaiton

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Energy Scenario

Energy demand

Current energy production status

Solar energy potential

Career opportunities


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SolarRadiation

availability

Photo-voltaic

Effect

Workingof Solar

Cell

Selectionof Battery,

Chargecontroller,Inverter

Optimizedsystem

design


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What could be the amount of solarenergy impacting the surface of earth ?

The total solar energy absorbed by Earth's atmosphere,

oceans and land masses is approximately

3,850,000 exajoule (EJ) per year.1 EJ = 1018 J

Energy from sun

on the earth

in 1 hour

Energy used by whole world

in 1 year


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hvE

eVm

E)(

24.1

~ 0.5% ~ 0.5%7.6% 48.4 % 43%
http://upload.wikimedia.org/wikipedia/en/f/f1/EM_spectrum.svg


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Ref: htt ://a ollo.lsc.vsc.edu/classes/met130/notes/cha ter2/sw atm.html


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Measurements indicate that the energy flow received from sun

outside the earths atmosphere is essentially constant at particular

distance

The rate at which energy is received from sun on a unit

area perpendicular to the rays of sun is called as Intensity

What is Intensity ?

How does it vary with the distance ?


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Radiation is inversely proportional to square of the distance


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At the mean distance of sun and earth, rate at which energy is

received from sun on unit area perpendicular to rays of sun is solar

constant

Its value is 1367 W/m2 = Isc

What will be the average intensity falling on earth ?

Assumedto be

Only for

calculation of

average radiation


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What will be the actual solar radiation intensity

at specific day ?

)365

360cos(033.01'

nII scsc

Where n is the day of the year


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Beam radiations (Direct )

Diffused radiations (Diffuse from sky + Reflected from ground)

Global (Beam+Diffused)


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PYRANOMETER

Measures global or diffuse radiation

Principle of heating proportional to radiation

1. The pyronometeris consist of black surface which heats up when

exposed to solar radiation

2. Its temperature increases until the rate of heat gain by solar radiation

equals the rate of heat loss.

3. The hot junction of a thermopile are attached to the black surface,

while cold junctions are located on side plate so they do not receive the

radiation directly.

4. EMF is generated (in range of 0 to 10mV)

5. Integrated over a period of time and is a measure of the global

radiation

How does it works !!!

For measuring Global radiations -
http://images.google.co.in/imgres?imgurl=http://www.allweatherinc.com/images/3022.gif&imgrefurl=http://www.allweatherinc.com/meteorological/3022.html&h=223&w=315&sz=13&hl=en&start=6&tbnid=07_XjT2FNroMDM:&tbnh=83&tbnw=117&prev=/images?q=pyranometer&gbv=2&svnum=10&hl=en


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For measuring diffused radiation

1. This is done by mounting it at the centre of a

semicircular shading ring.

2. Ring is fixed in such a way that its plane is

parallel to plane of the path of the suns daily

movement.

3. Hence, the pyranometer measures only the

diffused radiation using same principal of

thermopile

PYRANOMETER with shading ring


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PYRHELIOMETER

Measures beam (direct) solar radiation,

principle similar to Pyranometer is used,

but only direct radiation falls on the detector

In contrast to a pyrnometer, the black absorber plate

(with the hot junction of thermopile attached to it)

is located at base.

The direct (beam radiation) can be measured


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Amount of solar radiation on an object will depend on

Location

Day of year

Time of day

Inclination of the object

Orientation of object (w.r.t. North-south direction)

Here the Object is solar panel, but it is true of any object (For solar thermal also!)


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Latitude Longitude

()


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Day of the yearis characterized by an angle

Called as Declination angle ()

Angle made by line joining center of the sun and the earth w.r.t to

projection on equatorial plane (+23.45o to -23.45o)

Animated video


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Declination Angle

-30

-20

-10

0

10

20

30

0 50 100 150 200 250 300 350

Days of year

Declination(degree)

Dec-

21

Sep

21Mar-

21

Dec-

21

June

21

)284(

365

360sin45.23 n nday of year (=1 for Jan-1)

n=1 Jan 1, n=335 Dec 1, for June-21, what would be n?

This is to take care of daily variation of solar radiations

Graphically


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Time of the day

Time is based on the rotation of the Earth with respect to the Sun

It is characterized by Hour angle (w)

It is angular measure of time w.r.t. solar noon (LAT),

Since 360o corresponds to 24 hours

15o corresponds to 1 hour

W = 15 (12 - LAT )

Local

apparent

time

In hour

Hour

angle15

degree

per hour

With

reference

to solar

noon


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Tilt of solar collector

O Horizontal planeSN

Solar collector

90O

Normal to collector

is tilt of collector w.r.t. to horizontal plane


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Orientation of object (w.r.t. North-south direction)

Surface azimuth angle ()

Normal tothe plane

South direction (horizontal plane)

For inclined object

It can vary from -180O to +180O

Positive if the normal is east of south

And Negative if the normal is west of south


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For object on the horizontal plane

=0O

Normal to

the plane




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In order to find the beam energy falling on a surface

having any orientation,

it is necessary to convert the value of the beam flux coming from thedirection of the sun to an equivalent value corresponding to the normal

direction to the surface.

beam flux

Equivalent fluxfalling normal

to surfacecosnbb

II

Ib

Ibn


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Normal to

the plane

is affected by five parameters

- Latitude of location ()

- Day of year ()

- Time of the day (w)

- Inclination of surface ()

- Orientation in horizontal plane ()

zSolid lines are reference lines

Vertical(z = Zenith angle)



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Latitude ()angle of a location on earth w.r.t. to equatorial plane

Surface azimuth angle (+90o to -90o, +ve in the north)

Declination angle ()Angle made by line joining center of the sun and the

earth w.r.t to projection on equatorial plane (+23.45o

to -23.45o

)

Hour angle (w)angular measure of time w.r.t. noon (LAT), 15o per hour,

(+180o to -180o, +ve in the morning)

Surface slope ()Angle of the surface w.r.t horizontal plane (0 to 180o)

Surface azimuth angle ()angle between surface normal and south

direction in horizontal plane, (+180o to -180o, +ve in the east of south)


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Angle of Sun rays on collector

sinsinsincos

)sincossincoscos(coscos)sincoscoscoscos(sinsincos

Incidence angle of rays on collector ()

(w.r.t. to collector normal)

Latitude ()


Hour angle (w)

Surface slope ()

Declination angle ()


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Case-1:i.e. = 0o. Thus, for the horizontal surface, then :

(Slope is zero)

z coscoscoscossinsincos

Case-2: =0o, collector facing due south

)cos(coscos)sin(sincos

Will see the significance of special cases in later part


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India, being in the Northern Hemisphere, experiences asun that is predominantly coming at us from the South.

There is of course deviance throughout the seasons,

but ideally solar panels should be facing as close to

true South as possible to reduce the impact that the

Winter seasons have on efficiency.

When sun is coming at us from north (anyways the days

are going to be cloudy) so this orientation is notpreferred

Also, the Radiation is symmetric about the true south


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Calculate the angle made by beam radiation with the

normal to a PV panel on May 1 at 0900 h (LocalApparent time) the panel is located in NEW DELHI

(28O35n,77O21E). It is tilted at angle of 36O with the

horizontal and pointing due south.

Solution

Tilt angle =

=36o

Longitude= =28o35=28.55Orientation = =0o (due facing south)

From given -

Lets find out all the parameters


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Hour angle =W = 15 (12-LAT)

= 45O

Calculation -

= 14.90o

Declination angle -

We need to characterize time and day parameter

..For LAT = 9h

)284(365360sin45.23 n

For May 1 , n=121


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= 48.90O

Result -

Use all parameters to find cos

sinsinsincos)sincossincoscos(coscos

)sincoscoscoscos(sinsincos

Answer Cos = 0.65


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Calculate the power output of array at location and conditions given

at last problem.

The beam radiations in direction of the rays (Ibn ) is 1000W/m2

withTotal cell area = 15 m2

Efficiency = 12.5%

From last solution

cos = 0.65

Power output from array = (Normal incident flux) X Cell area X Efficiency

= (1000Xcos ) X 15 X 0.125= (1000X0.65)X 15X0.125

= 1218.75 W

We will learn this in later section of

course


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We need to calculate this angle each time we

find the energy output at particular time period

We will develop a code for these calculations

Such code is actually used in many simulation software !

Can be used to develop your own software


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Define variablesCall up different

parameters

Give inputvalues

Set formulaeDisplay output


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While developing the code

Include declarations of the basic standard library

Use the angle values in radiations

#include

#include

#include

using namespace std;

//algorithm in C++

// Output

How will the code look like !

Now develop the code


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Our aim to find out the optimum tilt angle of the panel () so

that cos should be maximum

2 tracking modes are usually employed for this.

Single Axis

Double Axis Tracking

For the power output to be maximum, the incident

radiation must be perpendicular to the panel.

A solar trackeris used to orient the panel such that the incident radiation

is perpendicular to the panel.

cosnbb II

Recall

Optimum inclination for fixed collector

But this will require continuous tracking of position of sun


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Continuous tracking of sun will ensure that the sunrays are

always perpendicular to the solar panel

(tilt angle= is changed to ensure that incident angle= = 0)


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So lets find out optimal angle for fixed

collectors


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Optimum angle for fixed panel

What should be the optimum tilt angle () for south facing fixed

collector located in Mumbai?


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Collector should be perpendicular to the sun rays

If collector is not moving, it should be perpendicular to sun

rays at noon time.

is tilt of collector w.r.t. to horizontal plane

Optimum angle for fixed panel


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The inclination of the fixed collector (facing South)w.r.t.

horizontal at noon time should be

Under this condition at noon time Sun rays will be perpendicular to the

collector

One need to estimate declination angle for a given day, when

optimum inclination is to be estimated

Using case 2 - : =0o, collector facing due south

)cos(coscos)sin(sincos

)cos(cos At noon, 0

0 For optimal radiations

0


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Optimum Inclination over a Year

The noon position of the sun is changes throughout the year

What is optimum position of collector for whole year(we need to estimate average value of declination angle over year)

-30

-20

-10

0

10

20

30

0 50 100 150 200 250 300 350

Days of year

Declination(de

gree)

Average is Zero over the yearHence =


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What should be fixed collector inclination in summer?

Average inclination over a month

(a= is monthly average)

Optimum Inclination over a Month

-30

-20

-10

0

10

20

30

0 50 100 150 200 250 300 350

Days of year

Declination(degree)


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Summer and winter orientation for

maximum energy production

best winter performance collector should be mounted at +15o.best summer performance collector should be mounted at -15o.


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This can be termed as number of sun shine hours

This will be dependant on Sunrise and Sunset at

particular location


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How to find Sunshine hours

(number of hours for which sun is available)For horizontal collector

From special case 1

= 0o. Thus, for the

horizontal surface coscoscossinsincos

tantancos

For sunrise as well as for sunset the =90o


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tantancos

This equation yields a positive and a negative value for ws

Positive corresponds to Sunrise

And negative corresponds to sunset

Since 360o corresponds to 24 hours

15o corresponds to 1 hour

Corresponding day length will be

)tantan(cos15

2 1max

S

Smax (day length or maximum number of sunshine hours)

And this will be used in simulation in the form of (Horizon) in later classes

Similarly, it can be found out for inclined surface (Home assignment)


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Calculate the hour angle at sunrise and sunset sun shine hours

on January 10 for a horizontal panel and facing due south

(=0o). The panel is located in Mumbai (19o07 N,72o51E)

On January 1, n=10

}tan)tan({cos 1 s

Latitude = = 19.12o

= -22.03oDeclination angle =

Solution

= 81.93 O


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)tantan(cos15

2 1max

S

Smax = 10.92 h

Maximum number of sunshine hours

)81.93(152max S

Maximum sunshine hour = 10 h 55min


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Local apparent time (LAT)

As sun cant be exactly overhead for all location at same time

Due to difference in location there is difference in actual time

Normally the standard time for a country is based on a noon (overhead

Sun position) at a particular longitude

Correction in the real noon time by considering the difference in the

longitude w.r.t. standard longitude of that country, 1o longitude

difference = 4 min.

tiontimecorrecofEqLongLongTLAT localstst .)(4

Difference in

longitude of location


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Correction factors

Due to the fact that earths orbit and

rate of rotation are subject to smallvariation

Equation of time correctionDifference in longitude of location

Indian Standard Time (IST)

is calculated on the basis of

82.5 E longitude, from a clock

tower in

Mirzapur (25.15N, 82.58E)

(near Allahabad in the stateof Uttar Pradesh)


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Determine the local apparent time (LAT) correspondingto 1430h (Indian Standard Time) at Mumbai (19o07N ,

72o51E) on May 1. In India, standard time is based

on 82.50oE.

1430h = 870min

LAT = 870min 4(82.50 72.85)min + (3.5 min)

= 870min 38.6 min + 3.5 min

= 834.9min

= 1355h


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Solar radiations

Measuring instruments

Parameters that define energy received by a particular object

Basic codes in simulation software

Local apparent time


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ANY QUESTIONS

3. basics of solar radiaiton

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