Energy Scenario Energy demand

Current energy production status Solar energy potential

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 = 10

18_{ J}

**Energy from sun **

**on the earth **

** in 1 hour **

**Energy used by whole world **

**in 1 year **

*hv*

*E*

###

*eV*

*m*

*E*

### )

### (

### 24

### .

### 1

###

###

###

~ 0.5% 7.6% 48.4 % 43% ~ 0.5%### 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 ? **

### 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

### What will be the actual solar radiation intensity

### at specific day ?

###

###

###

###

###

###

###

### )

### 365

### 360

### cos(

### 033

### .

### 0

### 1

### '

*n*

*I*

*I*

_{sc}

_{sc}

_{sc}

_{sc}

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 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

**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.

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)

Latitude 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 (+23.45

o_{ to -23.45}

o_{) }

**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

### 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
**Tilt of solar collector **

###

O Horizontal plane S N Solar collector 90O Normal to collector ### 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 +180}O

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

**For object on the horizontal plane **

### γ=0

### O

Normal to the plane

### 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.

### θ

beam flux Equivalent flux falling normal to surface###

### cos

*n*

*b*

*b*

*I*

*I*

###

### I

_{b}

### I

_{bn }

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)

θ_{z }

Solid lines are reference lines

Vertical

(Θ_{z }= Zenith angle)

### β

### γ

### 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) }

**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)

### 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

###

###

###

###

###

###

###

###

###

###

###

###

###

### 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

### 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 =

### ᵧ

### =0

o_{ (due facing south) }

### From given -

### ᵟ

### 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### 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**

### 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

### 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 !

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 !

** 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 **

**• Continuous tracking of sun will ensure that the sunrays are **

**always perpendicular to the solar panel **

### 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

###

###

###

###

###

** 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

###

###

###

###

###

###

**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 **

** 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**
**)**

**Summer and winter orientation for **

**maximum energy production **

**best winter performance collector should be mounted at **

###

**+15**

**o**

_{. }

_{. }

### 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

###

###

###

###

###

###

### sin

### sin

### cos

### cos

### cos

### cos

###

###

###

###

###

### tan

### tan

### cos

###

###

###

###

###

### tan

### tan

### cos

###

###

This equation yields a positive and a negative value for w_{s }
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*

S_{max }(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 (19

### o

### 07’ N,72o51’E)

### On January 1, n=10

### ᵟ

### }

### tan

### )

### tan(

### {

### cos

1###

###

###

###

###

*s*

### Latitude =

### Φ

### = 19.12o

### =

### -22.03

### o

### Declination angle =

**Solution **

### = 81.93

O### )

### tan

### tan

### (

### cos

### 15

### 2

_{1}max

###

###

###

###

*S*

**S**

_{max }** = 10.92 h**

**Maximum number of sunshine hours**

### )

### 81.93

### (

### 15

### 2

max###

*S*

**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**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)