3. Basics of Solar Radiaiton

57 

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(1)
(2)

 Energy Scenario  Energy demand

 Current energy production status  Solar energy potential

(3)

Solar Radiation availability Photo-voltaic Effect Working of Solar Cell Selection of Battery, Charge controller, Inverter Optimized system design

(4)

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 = 10

18

J

Energy from sun

on the earth

in 1 hour

Energy used by whole world

in 1 year

(5)

hv

E

eV

m

E

)

(

24

.

1

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

(6)
(7)

Measurements indicate that the energy flow received from sun

outside the earth’s 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 ?

(8)
(9)

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/m

2

= I

sc

What will be the average intensity falling on earth ?

Assumed to be

Only for calculation of average radiation

(10)

What will be the actual solar radiation intensity

at specific day ?

 

)

365

360

cos(

033

.

0

1

'

n

I

I

sc

sc

(11)

Beam radiations (Direct )

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

(12)

PYRANOMETER

Measures global or diffuse radiation

Principle of ‘heating proportional to radiation’

1. The pyronometer is consist of ‘black surface’ which heats up when exposed to solar radiation

2. It’s 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 !!!

(13)

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 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

(14)

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.

(15)
(16)

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)

(17)

Latitude Longitude

(18)

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 (+23.45

o

to -23.45

o

)

(19)

Declination Angle

δ

-30 -20 -10 0 10 20 30 0 50 100 150 200 250 300 350 Days of year D e c li n a ti o n (d e g re e ) Dec-21 Sep 21 Mar-21 Dec-21 June 21

(

284

)

365

360

sin

45

.

23

n

n – day 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

(20)

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 360

o

corresponds to 24 hours

15

o

corresponds to 1 hour

W = 15 (12 - LAT )

Local apparent time In hour Hour angle 15 degree per hour With reference to solar noon

(21)

Tilt of solar collector

O Horizontal plane S N Solar collector 90O Normal to collector 

(22)

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

Surface azimuth angle (γ)

γ

Normal to the 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

(23)

For object on the horizontal plane

γ=0

O

Normal to the plane

Surface azimuth angle (γ)

(24)

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.

θ

beam flux Equivalent flux falling normal to surface

cos

n

b

b

I

I

I

b

I

bn

(25)

Normal to the plane

θ

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)

θz

Solid lines are reference lines

Vertical

z = Zenith angle)

β

γ

(26)

Latitude (φ)

– angle of a location on earth w.r.t. to equatorial plane

Surface azimuth angle (+90

o

to -90

o

, +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 (+23.45

o

to -23.45

o

)

Hour angle (w)

– angular measure of time w.r.t. noon (LAT), 15

o

per hour,

(+180

o

to -180

o

, +ve in the morning)

Surface slope (β

) – Angle of the surface w.r.t horizontal plane (0 to 180

o

)

Surface azimuth angle (γ)

– angle between surface normal and south

direction in horizontal plane, (+180

o

to -180

o

, +ve in the east of south)

(27)

Angle of Sun rays on collector

sin

sin

sin

cos

)

sin

cos

sin

cos

cos

(cos

cos

)

sin

cos

cos

cos

cos

(sin

sin

cos

Incidence angle of rays on collector (

)

(w.r.t. to collector normal)

Latitude (φ)

Surface azimuth angle (γ)

Hour angle (w)

Surface slope (β)

(28)

Case-1: i.e. β = 0

o

. Thus, for the horizontal surface, then :

(Slope is zero)

z

sin

sin

cos

cos

cos

cos

cos

Case-2:

=0

o

, collector facing due south

)

cos(

cos

cos

)

sin(

sin

cos

(29)
(30)

India, being in the Northern Hemisphere, experiences a

sun 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 not

preferred

(31)

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

(28

O

35’n,77

O

21’E). It is tilted at angle of 36

O

with the

horizontal and pointing due south.

Solution

Tilt angle

= ᵦ =

36

o

Longitude

= ᵠ

=28

o

35’=28.55

Orientation =

=0

o

(due facing south)

From given -

(32)

Hour angle =W = 15 (12-LAT)

= 45

O

Calculation -

= 14.90o

Declination angle -

We need to characterize time and day parameter

…..For LAT = 9h

(

284

)

365

360

sin

45

.

23

n

For May 1 , n=121

(33)

Θ = 48.90

O

Result -

Use all parameters to find cos Θ

sin

sin

sin

cos

)

sin

cos

sin

cos

cos

(cos

cos

)

sin

cos

cos

cos

cos

(sin

sin

cos

Answer Cos Θ = 0.65

(34)

Calculate the power output of array at location and conditions given

at last problem.

The beam radiations in direction of the rays (I

bn

) is 1000W/m

2

with

Total cell area = 15 m

2

Efficiency = 12.5%

From last solution –

cos Θ = 0.65

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 will learn this in later section of course

(35)

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 !

(36)

Define variables Call up different parameters

Give input values

Set formulae Display output

(37)

While developing the code

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

#include <iostream> #include <math.h> #include <iomanip> using namespace std; //algorithm in C++ // Output

How will the code look like !

(38)

 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

For the power output to be maximum, the incident

radiation must be perpendicular to the panel.

A solar tracker is used to orient the panel such that the incident radiation is perpendicular to the panel.

cos

n b b

I

I

Recall

Optimum inclination for fixed collector

(39)

• Continuous tracking of sun will ensure that the sunrays are

always perpendicular to the solar panel

(40)

So let’s find out optimal angle for fixed

collectors

(41)

Optimum angle for fixed panel

What should be the optimum tilt angle (

) for south facing fixed

collector located in Mumbai?

(42)

 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

(43)

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 -

:

=0

o

, collector facing due south

)

cos(

cos

cos

)

sin(

sin

cos

)

cos(

cos

At noon,

0

0

For optimal radiations

(44)

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 D e c li n a ti o n (d e g re e )

Average

is Zero over the year

(45)
(46)

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 D e c li n a ti o n (d e g re e )

(47)

Summer and winter orientation for

maximum energy production

best winter performance collector should be mounted at

+15

o

.

(48)

This can be termed as number of sun shine hours

This will be dependant on Sunrise and Sunset at

particular location

(49)

How to find Sunshine hours

(number of hours for which sun is available)

For horizontal collector

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

sin

sin

cos

cos

cos

cos

tan

tan

cos

(50)

tan

tan

cos

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

)

tan

tan

(

cos

15

2

1 max

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)

(51)

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 (19

o

07’ N,72o51’E)

On January 1, n=10

}

tan

)

tan(

{

cos

1

s

Latitude =

Φ

= 19.12o

=

-22.03

o

Declination angle =

Solution

= 81.93

O

(52)

)

tan

tan

(

cos

15

2

1 max

S

S

max

= 10.92 h

Maximum number of sunshine hours

)

81.93

(

15

2

max

S

(53)

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, 1

o

longitude

difference = 4 min.

tion

timecorrec

of

Eq

Long

Long

T

LAT

st

4

(

st

local

)

.

Difference in longitude of location

(54)

Correction factors

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

variation

Equation of time correction 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)

(55)

Determine the local apparent time (LAT) corresponding

to 1430h (Indian Standard Time) at Mumbai (19

o

07’N ,

72

o

51’E) on May 1. In India, standard time is based

on 82.50

o

E

.

1430h = 870min

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

= 870min – 38.6 min + 3.5 min

= 834.9min

= 1355h

(56)

Solar radiations

Measuring instruments

Parameters that define energy received by a particular object

Basic codes in simulation software

(57)

Figure

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