3. Basics of Solar Radiaiton

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Energy Scenario

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Energy demand

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Current energy production status

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Solar energy potential

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Career opportunities

Solar Radiation availability

Optimized system design

Photovoltaic Effect

Selection of Battery, Charge controller, Inverter

Working of Solar Cell

What could be the amount of solar energy 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

~ 0.5% 7.6% 48.4 % 43% ~ 0.5%

E  hv 1.24 E eV  ( m)

Ref: http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/sw_atm.html

Measurements indicate that the energy flow received from sun outside the earth’s atmosphere is essentially constant at particular distance

What is Intensity ? The rate at which energy is received from sun on a unit area perpendicular to the rays of sun is called as Intensity

How does it vary with the distance ?

Radiation is inversely proportional to square of the distance

What will be the average intensity falling on earth ? 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

Assumed to be Only for calculation of average radiation

What will be the actual solar radiation intensity at specific day ?

360n   I  I sc 1  0.033 cos( ) 365   ' sc

Where n is the day of the year

Beam radiations (Direct ) Diffused radiations (Diffuse from sky + Reflected from ground) Global (Beam+Diffused)

PYRANOMETER Measures global or diffuse radiation Principle of ‘heating proportional to radiation’ For measuring Global radiations How does it works !!!

1. 2. 3.

4. 5.

The pyronometer is consist of ‘black surface’ which heats up when exposed to solar radiation It’s temperature increases until the rate of heat gain by solar radiation equals the rate of heat loss. 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. EMF is generated (in range of 0 to 10mV) Integrated over a period of time and is a measure of the global radiation

For measuring diffused radiation PYRANOMETER with shading ring

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

2. Ring is fixed in such a way that it’s plane is parallel to plane of the path of the sun’s daily movement. 3. Hence, the pyranometer measures only the diffused radiation using same principal of thermopile

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

Amount of solar radiation on an object will depend on 

Location

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Day of year

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Time of day

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Inclination of the object

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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!)

Latitude (Φ)

Longitude

Day of the year is characterized by an angle Called as Declination angle (δ)

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

Declination Angle δ This is to take care of daily variation of solar radiations

 360    23.45 sin  (284  n)   365 

n – day of year (=1 for Jan-1)

Graphically

Declination (degree)

30

June 21

20 10 0 -10

0

Mar50 21 100

150

200

250

Sep 21300

350

-20 -30

Dec21

Days of year

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

Dec21

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 ) Hour 15 angle degree per hour

With reference to solar noon

Local apparent time In hour

Tilt of solar collector 

Normal to collector

90O  S Horizontal plane

N O

Solar collector

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

Orientation of object (w.r.t. North-south direction) Normal to the plane

γ

For inclined object

Surface azimuth angle (γ) It can vary from -180O to +180O

South direction (horizontal plane)

Positive if the normal is east of south And Negative if the normal is west of south

For object on the horizontal plane

Normal to the plane

South direction (horizontal plane)

γ=0O

Surface azimuth angle (γ)

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 the direction of the sun to an equivalent value corresponding to the normal direction to the surface.

Equivalent flux falling normal to surface

Ib beam flux

Ibn

θ

I b  I b n cos 

Vertical

Normal to the plane

(Θz = Zenith angle)

Solid lines are reference lines θz

β

θ

γ

South direction (horizontal plane)

θ is affected by five parameters - Latitude of location (φ) - Day of year (δ) - Time of the day (w) - Inclination of surface (β) - Orientation in horizontal plane (γ)

Latitude (φ) – angle of a location on earth w.r.t. to equatorial plane Surface azimuth angle (+90o to -90o, +ve in the north) Declination angle (δ) – 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 (β) – Angle of the surface w.r.t horizontal plane (0 to 180o)

Surface azimuth angle (γ) – angle between surface normal and south direction in horizontal plane, (+180o to -180o, +ve in the east of south)

Angle of Sun rays on collector Incidence angle of rays on collector () (w.r.t. to collector normal)

cos   sin  (sin  cos   cos  cos  cos  sin  )  cos  (cos  cos  cos   sin  cos  sin  )  cos  sin  sin  sin  Latitude (φ) Surface azimuth angle (γ) Hour angle (w) Surface slope (β) Declination angle (δ)

Case-1: i.e. β = 0o. Thus, for the horizontal surface, then : (Slope is zero)

cos  sin  sin   cos  cos  cos   cos z Case-2: =0o, collector facing due south

cos  sin  sin(    )  cos  cos  cos(   ) Will see the significance of special cases in later part

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

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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.

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When sun is coming at us from north (anyways the days are going to be cloudy) so this orientation is not preferred

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Also, the Radiation is symmetric about the true south

Calculate the angle made by beam radiation with the normal to a PV panel on May 1 at 0900 h (Local Apparent time) the panel is located in NEW DELHI (28O35’n,77O21’E). It is tilted at angle of 36O with the horizontal and pointing due south.

Solution Let’s find out all the parameters From given Tilt angle =

ᵦ =36o

ᵠ =28o35’=28.55 Orientation = ᵧ =0 (due facing south) Longitude=

o

Calculation We need to characterize time and day parameter

Hour angle =W = 15 (12-LAT) = 45O

…..For LAT = 9h

Declination angle For May 1 , n=121

 360    23.45 sin  (284  n)   365 

ᵟ

= 14.90o

Result Use all parameters to find cos Θ

cos   sin  (sin  cos   cos  cos  cos  sin  )  cos  (cos  cos  cos   sin  cos  sin  )  cos  sin  sin  sin  Answer

Cos Θ = 0.65

Θ = 48.90O

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 with Total cell area = 15 m2 Efficiency = 12.5%

From last solution – cos Θ = 0.65

We will learn this in later section of course

Power output from array = (Normal incident flux) X Cell area X Efficiency = (1000Xcos Θ) X 15 X 0.125 = (1000X0.65)X 15X0.125 = 1218.75 W

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

Define variables

Call up different parameters

Give input values

Display output

Set formulae

While developing the code  Include declarations of the basic standard library  Use the angle values in radiations

How will the code look like ! #include #include #include using namespace std;

//algorithm in C++ // Output

Now develop the code…

Optimum inclination for fixed collector For the power output to be maximum, the incident radiation must be perpendicular to the panel.

But this will require continuous tracking of position of sun A solar tracker is used to orient the panel such that the incident radiation is perpendicular to the panel.

 Our aim to find out the optimum tilt angle of the panel (β) so that cos ϴ should be maximum

2 tracking modes are usually employed for this.  

Single Axis Double Axis Tracking

I b  I b n cos  Recall

• Continuous tracking of sun will ensure that the sunrays are always perpendicular to the solar panel (tilt angle=β is changed to ensure that incident angle=ϴ = 0)

So let’s find out optimal angle for fixed collectors

Optimum angle for fixed panel

What should be the optimum tilt angle () for south facing fixed collector located in Mumbai?

Optimum angle ᵦ for fixed panel  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

The inclination of the fixed collector (facing South) w.r.t. horizontal at noon time should be Using case 2 - : =0o, collector facing due south

cos   sin  sin(    )  cos  cos  cos(   ) At noon,

 0

For optimal radiations

 0

cos   cos(     )       0     

    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

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)

Average  is Zero over the year Hence  = 

Declination (degree)

30 20 10 0 -10

0

50

100

150

200

-20 -30 Days of year

250

300

350

Optimum Inclination over a Month

Declination (degree)

30 20 10 0 -10

0

50

100

150

200

250

300

350

-20 -30 Days of year

Average inclination over a month (a= is monthly average)

What should be fixed collector inclination in summer?

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.

This can be termed as number of sun shine hours

This will be dependant on Sunrise and Sunset at particular location

How to find Sunshine hours (number of hours for which sun is available) For horizontal collector

cos   sin  sin   cos  cos  cos  For sunrise as well as for sunset the =90o

cos    tan  tan 

From special case 1 β = 0o. Thus, for the horizontal surface

cos    tan  tan  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

S max

2  cos 1 ( tan  tan  ) 15

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)

Calculate the hour angle at sunrise and sunset sun shine hours on January 10 for a horizontal panel and facing due south (γ=0o). The panel is located in Mumbai (19o07’ N,72o51’E) Solution

On January 1, n=10

Declination angle = Latitude =

ᵟ o

= -22.03o

Φ = 19.12

s  cos 1{ tan( ) tan  } = 81.93 O

Maximum number of sunshine hours

S max

2  cos 1 ( tan  tan  ) 15 S max

2  (81.93) 15

Smax = 10.92 h Maximum sunshine hour = 10 h 55min

Local apparent time (LAT)  As sun can’t 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.

LAT  Tst  4( Long st  Longlocal )  Eq. of timecorrection Difference in longitude of location

Correction factors Difference in longitude of location Indian Standard Time (IST) is calculated on the basis of 82.5° E longitude, from a clock tower in Mirzapur (25.15°N, 82.58°E) (near Allahabad in the state of Uttar Pradesh)

Equation of time correction Due to the fact that earth’s orbit and rate of rotation are subject to small variation

Determine the local apparent time (LAT) corresponding to 1430h (Indian Standard Time) at Mumbai (19o07’N , 72o 51’E) on May 1. In India, standard time is based on 82.50oE.

LAT = = = =

1430h = 870min 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

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Measuring instruments

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Parameters that define energy received by a particular object

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Basic codes in simulation software

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Local apparent time

ANY QUESTIONS