Basics of Solar Radiation
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Transcript of Basics of Solar Radiation
Energy Scenario
Energy demand
Current energy production status
Solar energy potential
Career opportunities
Solar Radiation availability
Photo-voltaic
Effect
Working of Solar
Cell
Selection of Battery,
Charge controller, Inverter
Optimized system design
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
hvE
eVm
E)(
24.1
~ 0.5% ~ 0.5% 7.6% 48.4 % 43%
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
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 ?
Radiation is inversely proportional to square of the distance
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 ?
Assumed
to be
Only for
calculation of
average radiation
What will be the actual solar radiation intensity
at specific day ?
)
365
360cos(033.01' n
II scsc
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’
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 !!!
For measuring Global radiations -
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
PYRANOMETER with shading ring
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
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!)
Latitude Longitude
(Φ)
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
-30
-20
-10
0
10
20
30
0 50 100 150 200 250 300 350
Days of year
De
clin
ati
on
(d
eg
ree
)
Dec-
21
Sep
21 Mar-
21
Dec-
21
June
21
)284(
365
360sin45.23 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
Graphically
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
angle 15
degree
per hour
With
reference
to solar
noon
Tilt of solar collector
O Horizontal plane S N
Solar collector
90O
Normal to collector
is tilt of collector w.r.t. to horizontal plane
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
For object on the horizontal plane
γ=0O
Normal to
the plane
Surface azimuth angle (γ)
South direction (horizontal plane)
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
cosnbb II
Ib
Ibn
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 (Θz = Zenith angle)
β
γ
South direction (horizontal plane)
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
sinsinsincos
)sincossincoscos(coscos
)sincoscoscoscos(sinsincos
Incidence angle of rays on collector ()
(w.r.t. to collector normal)
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)
z coscoscoscossinsincos
Case-2: =0o, collector facing due south
)cos(coscos)sin(sincos
Will see the significance of special cases in later part
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
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
Tilt angle = ᵦ =36o
Longitude= ᵠ =28o35’=28.55
Orientation = ᵧ =0o (due facing south)
From given -
Let’s find out all the parameters
ᵟ
Hour angle =W = 15 (12-LAT)
= 45O
Calculation -
= 14.90o
Declination angle -
We need to characterize time and day parameter
…..For LAT = 9h
)284(
365
360sin45.23 n
For May 1 , n=121
Θ = 48.90O
Result -
Use all parameters to find cos Θ
sinsinsincos
)sincossincoscos(coscos
)sincoscoscoscos(sinsincos
Answer Cos Θ = 0.65
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
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
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
Set formulae Display output
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 !
Now develop the code…
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.
cosnbb II
Recall
Optimum inclination for fixed collector
But this will require continuous tracking of position of sun
• 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?
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 ᵦ for fixed panel
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
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
De
clin
ati
on
(d
eg
ree
)
Average is Zero over the year
Hence =
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
De
clin
ati
on
(d
eg
ree
)
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
From special case 1
β = 0o. Thus, for the
horizontal surface coscoscossinsincos
tantancos
For sunrise as well as for sunset the =90o
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 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)
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)
On January 1, n=10
ᵟ
}tan)tan({cos 1
s
Latitude = Φ = 19.12o
= -22.03o Declination angle =
Solution
= 81.93 O
)tantan(cos15
2 1
max S
Smax = 10.92 h
Maximum number of sunshine hours
)81.93(15
2max S
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.
tiontimecorrecofEqLongLongTLAT localstst .)(4
Difference in
longitude of location
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)
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.
1430h = 870min
LAT = 870min– 4(82.50 – 72.85)min + (3.5 min)
= 870min – 38.6 min + 3.5 min
= 834.9min
= 1355h
Solar radiations
Measuring instruments
Parameters that define energy received by a particular object
Basic codes in simulation software
Local apparent time
ANY QUESTIONS