Chapter 3: Solar Energy pps

Solar Energy IIn this chapter, we will find out how to get the most solar energy out of the sun by considering the time, the altitude angle, the radiation and the photovoltaic PV materia

Solar Energy I

In this chapter, we will find out how to get the most solar energy out of the sun

by considering the time, the altitude angle, the radiation and the photovoltaic (PV) materials A simple PV model helps to understand the mechanism of energy conversion

- solar to electricity We will further use a simple equivalent circuit to represent a PV cell to investigate its current-voltage characteristics

Insolation - Incident Solar radiation

Ecliptic Plane - The plane swept out by the earth in its orbit

Equinox - One of the two days in a year when we have equal day and night in length Summer Solstice - The day of the year when we have the most hours of daylight Winter Solstice - The day of the year when we have the fewest hours of daylight Extraterrestrial (ET) - Something happening, existing, or coming from somewhere beyond the planet Earth

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Figure 3.1: A fixed earth and a sun that moves up and down Source: Masters [1].

Figure 3.2: The altitude angle of the sun at solar noon Source: Masters [1]

Fig 3.1 shows how the positions between the sun and the earth during a year The earth is taken as the fixed reference point The sun makes different solar declination δ

to the earth, between ±23.45◦ The declination can be expressed in terms of day in a 365-day year

δ = 23.45 · sin[360

365(n − 81)] (3.1) where the sine function is measured in degrees and n = 81 is the approximated spring equinox (Mar 21) With the help of Fig 3.1 and (3.1) we may gain some insight into what might be a good tilt angle for a solar collector In general, to face the collector toward the equator and tile an angle equal to the local altitude is a good average position to receive sun ray

Altitude angle βN, as depicted in Fig 3.2, is solar angle of the sun at solar noon

s

Equator

L

β N

. .

.

Zenith

Figure 3.3: Tilt angle making sun’s rays perpendicular to the PV module The relationship between altitude angle, latitude and declination is given by

βN = 90◦− L + δ (3.2)

Although (3.2) can be applied anywhere on the earth, it is more convenient to have another altitude equation for places below the equator It is given by

βN = 90◦+ L − δ (3.3)

Example: Find the optimum tilt angle for a north-facing photovoltaic module in Camperdown, NSW (latitude -33.54◦ [2]) at solar noon on November 1

Solution: November 1 is the 304th day of the year so the solar declination in (3.1) is

δ = 23.45 · sin[360

365(304 − 81)] = −15.1

From (3.2), the altitude angle of the sun is equal to

βN = 90◦− 33.54◦− (−15.1◦) = 71.56◦

The tilt angle should be facing north and at an angle equal to

Tilt = 90 − βN = 90 − 71.56 = 18.44◦

Figure 3.4: Altitude and its azimuth angle of the sun’s position Source: Masters [1].

The location of the sun can be found by the altitude angle β and its azimuth angle φS, where the subscript of φ, S, indicates the angle of the sun Fig 3.4 shows the sun’s position described by the altitude angle and its azimuch angle They are given by

β = sin−1(cosLcosδcosH + sinLsinδ) (3.5)

φS = sin−1(cosδsinH

Note that reference point of φS is at noon φS is negative when the sun goes towards west and positive when towards east Suppose that the sun rotates a 360◦ in 24 hours

or (360/24=)15◦ per hour, therefore the hour angle of the sun can be expressed as

Hour angle H = ( 15◦

hour) · (hours before solar noon) (3.7) For example, at 9:00A.M solar time H would be +45◦ But during autumn and winter (in Australia) in the early morning and late afternoon, the magnitude of the sun’s azimuth is likely to be more than 90◦ away from the south We may use the following test conditions to find out the azimuth angle

if cos H ≥ tantanδ

L, then |φS| ≤ 90◦ otherwize |φS| ≥ 90◦ (3.8)

Example: Find the altitude angle and azimuth angle for the sun at 2:00P.M solar time in Camperdown, NSW (latitude -33.54◦) on the summer solstice

Solution: The solar declination δ in summer solstice is -23.45◦ From (3.7) we otbain the hour angle at

H = 15◦

h · (−2h) = −30

◦

Using (3.5), the altitude angle is

β = sin−1[cos(−33.54)cos(−23.45)cos(−30) + sin(−33.54)sin(−23.45)]

= sin−1(0.8821)

= 61.90◦

From (3.6), the azimuth angle is

φS= sin−1( cos(−23.45)cos61.90sin(−30))

= sin−1(−0.9739)

= −76.88◦or = 180 − (−76.88) = 256.88◦

Applying (3.8) we have the following

cosH = cos(−30) = 0.866 and tantanβL = tantan(−23.45)(−33.54) = 0.654

Therefore, cosH > tantanLβ

We may conclude that the azimuth angle is

φS= −76.88◦

With the help of altitude and azimutch angles, we may able to draw a sun path diagram for a given latitude and locate the sun for maximum sun’s ray onto the collector

of the PV panels And we may able to estimate any shading at a site at particular time

An extraterrestrial (ET) solar insolation, I0, describes the sun beam passing perpen-dicularly through an imaginary surface just outside of the earth’s atmosphere It is expressed as

I0 = SC · [1 + 0.034cos(360n365 )] (W/m2) (3.9) where SC is the solar constant and n is the day number

The predicted value of a clear sky beam radiation at the earth’s surface can be modeled as

Figure 3.5: Optical depth, apparent extraterrestrial flux and sky diffuse factor for the 21st Day of Each Month Sources: Masters [1] and ASHRAE [3]

where IB is the beam portion of the radiation reaching the earth’s surface (normal to the rays), A is an “apparent” extraterrestrial flux, and k is a dimensionless factor called the optical depth The air mass ratio m is expressed as

m = 1

where β is the altitude angle of the sun

Fig 3.5 shows a table with values of A and k used in the American Society of Heating, Refrigerating, and Air Conditioning Engineers Clear Day Solar Flux Model

We can write equations representing A and k in the table using curve fitting method

A = 1160 + 75sin[360365(n − 275)] W/m2 (3.12)

k = 0.174 + 0.035sin[360

365(n − 100)] (3.13) Example: Find the direct beam solar radiation normal to the sun’s rays at solar noon on a clear day in Atlanta (latitude 33.7◦) on Mar 21

Solution: Mar 21 is day number 81 From (3.12) the apparent extraterrestrial flux

A is

A = 1160 + 75sin[360

365(81 − 275)] = 1175W/m

2

From (3.13), the optical depth is

k = 0.174 + 0.035sin[360

365(81 − 100)] = 0.178

On Mar 21, the solar declination from (3.1) is 0◦, so from (3.2) the altitude angle of the sun at solar noon is

βN = 90◦− 33.7 + 0 = 56.3◦

Figure 3.6: Illustrating the collector azimuth angle φC and tilt angle Σ along with the solar azimuth angle φS and altitude angle β Source: Masters[1]

The air mass ratio (3.11) is

m = 1 sin 56.3◦ = 1.202 Using (3.10), the predicted value is

IB= 1175e−0.178·1.202= 949W/m2

3.5.1 Direct-beam on Collector Surface

Fig 3.6 introduces the angles when a collector is tilted The incidence angle is given by

cosθ= cosβcos(φS− φC)sinΣ + sinβcosΣ (3.14)

Example: At solar noon in Atlanta (latitude 33.7◦) on Mar 21 the altitude angle

of the sun was found to be 56.3◦ and the clear-sky beam insolation was found to be

949 W/m2 Find the beam insolation at that time on a collector that faces 20◦ toward the southeast if it is tipped up at a 52◦ angle

Solution:

cosθ = cos56.3cos(0 − 20)sin52 + sin56.3cos52 = 0.923

IBC = IBcosθ = 949W/m2· 0.923 = 876W/m2

3.5.2 Diffuse Radiation on Collector Surface

Diffuse radiation, which is due to scattered radiation on clouds, moisture, etc., on

a collector is more difficult to calculate We may relate the diffuse radiation on a horizontal surface IDH to the direct-beam radiation by a constant C, such that

where C is a sky diffuse factor and it can be approximated from Fig 3.5 as

C = 0.095 + 0.04sin[360

365(n − 100)] (3.16) The diffuse radiation on the collector is related by

IDC = IDH(1 + cosΣ

2 ) = CIB(

1 + cosΣ

Example: Recall the last example, find the diffuse radiation on the collector Solution: From (3.16), we have

C = 0.095 + 0.04sin[360

365(81 − 100)] = 0.099 And from (3.17)

IDC = CIB(1 + cosΣ

2 ) = 0.099 · 949W/m

2(1 + cos52

2 ) = 75.90W/m

2

So the total beam on the collector including diffuse radiation is (949+75.9) W/m2

= 1024.5 W/m2

3.5.3 Reflected Radiation on Collector Surface

The sun beam can be reflected on a bright surface like snow and strike onto the collector

to boost its performance The reflected radiation can be approximated as

IRC = ρ(IBH+ IDH)(1 − cosΣ

2 ) = ρIB(sinβ + C)(

1 − cosΣ

2 ) (3.18)

Example: Recall the last example, find the reflected radiation on the collector if the reflectance of the surfaces in front of the panel is 0.15

Solution:

IRC = ρIB(sinβ +C)(1 − cosΣ

2 ) = 0.15·949(sin56.3+0.099)(

1 − cos52

2 ) = 25.47W/m

2

Therefore the total insolation on the collector is therefore

IC = IBC + IDC+ IRC = 949 + 75.9 + 25.47 = 1050W/m2

3.6 Photovoltaic Materials and Electrical Characterisitics

A material or device that is capable of absorbing and converting energy contained in the photons of light into electrical current and voltage is called photovoltaic (PV) The

PV cells can be made up of mixture of silicon, boron and phosphorus, gallium and arsenic (GaAs), and cadmium and tellurium (CdTe)

Electric current occurs when the electrons in a material is free to flow Semicon-ductors like silicon is a perfect electrical insulator at 0K and its conductivity increases

a little at room temperature Energy of the electrons can be described using energy-band diagram The top energy-band region is called the conduction energy-band where the electrons contributes to the electric current flow The electrons, say in silicon, need to absorb sufficient energy to jump from lower energy band to higher energy band Just beneath the conduction band is the forbidden band, and below is the filled band In silicon, the band-gap energy, Eg, is 1.12eV (electron-volts) to jump from filled band to conduction band Where does the energy come from? In the PV cell, the energy comes from the photons of the sun When a solar cell absorbs a photon with high band-gap energy such that the electron can jump to the conduction band, the electron leaves a hole in the nucleus of the cell An adjacent electron fills this hole without change of energy level and creates another hole This results in a flow of holes as well

The energy of photons E to create the electron-hole pair in an semiconductor is characterized by the following equations:

E = hv = hc

where c is the speed of light (3 × 108m/s), v is the frequency (hertz), λ is the wavelength (m), E is the energy of a photon (J) and h is the Planck’s constant (6.626 × 10−34J-s) Example: What maximum wavelength can a photon have to create hole-electron pairs in GaAs (Band energy = 1.42eV)? What minimum frequency is that? Given 1

eV = 1.6 × 10−19J

Solution: From (3.20), the wavelength must be smaller than

λ ≤ hc

E =

6.626 × 10−34J· s × 3 × 108m/s 1.42eV × 1.6 × 10−34J/eV = 0.875 × 10

−6m = 0.875µm

and from (3.19) the frequency must be greater than

v ≥ c

λ =

3 × 108m/s 0.875 × 10−6m = 3.43 × 10

14Hz

From this example it is shown that photon energy less than than 1.42eV is unable to excite an electron in GaAs cell And photons with wavelength shorter than 0.875 µm has more than enough sufficient energy to excite an electron In both cases, energy is wasted

Without an electric field drive, the leaving electron can possibly go back to its original nucleus and recombine the hole it left Then the hole-electron pair disappears

To create this electric field, we make make the semiconductor electrically “unstable”

It is achieved by deliberately insert different impurities to the pure silicon One region

is doped with very small amount of element from column V of the periodic table/chart; the other region is doped with atoms from column V of the table Since silicon has

4 electrons at its outer orbit while, say phosphorus, has 5 electrons at its outer orbit, when they combined one electron is loosely bound and free to move around So the column V element is called donor atom and the semiconductor doped with this element

is called n-type material Similarly, if we doped the silicon with column III element, a free hole will be created Since it accepts electrons, it is called acceptor This kind of semiconductor doped with column III element is called p-type material

If we put these two doped materials together, a junction called depletion region

is formed It happens when the electrons in donors diffuse across the junction to recombine the holes in the acceptors, leaving positively-charged, immobile carriers in

6 I

SC

-?

I d

I

+

−

R L V o

Figure 3.7: A simple equivalent circuit for a PV cell

the n-type region The same is for the p-type region The depletion region widens and the process stops until the recombined atoms build up a strong barrier of electric field

to forbid holes and electrons to pass through it

When the PV material is exposed to sunlight, photons are absorbed and hole-electron pairs may be formed The photon energy push hole-electrons to n-side and holes

to p-side If we connect a wire across the semicondutor terminals, electrons from n-side will flow through the wires and recombine the holes in p-n-side A flow of electrons occurs By convention, we define postive current flows in the direction opposite to electron flow

An simple equivalent circuit for a photovolatic cell is shown in Fig 3.7 The PV cell consists of a current source driven by sunlight and a diode in parallel demonstrating the V-I characteristics Two important parameters of a PV cell are short-circuit current and open-circuit voltage, which are defined by

I = ISC|Vo =0 (3.21)

Vo = VOC|I=0 (3.22)

Using the KCL, we may relate the currents in the equivalent circuit

I = ISC− Id

= ISC− I0(eqV /kT − 1)

(3.23)

where I0 is the reverse saturation current of the diode, Id is the diode forward current,

V is diode forward voltage drop, q is the electron charge (1.602 × 10−19C), k is

When the PV cell is open or RL = ∞, I = 0, we may solve (3.23) to get the open-circuit voltage VOC

VOC = kT

q ln(

ISC

Example: Consider a 100cm2 photovoltaic cell with reverse saturation current I0

= 10−12 A/cm2 In full sun, it produces a short-circuit current of 30mA/cm2 at 25◦C Find the open-circuit voltage at full sun and again for 50% sunlight Plot the results Solution: The reverse saturation current I0 is 10−12 A/cm2× 100cm2 = 1 ×

10−10A At full sun ISCis 0.030A/cm2× 100cm2 = 3.0A From (3.24) the open-circuit voltage is

VOC = 0.0257ln( 3.0

10−10 + 1) = 0.62V Since the short-circuit current is proportional to solar intensity, at half sun ISC= 1.5A and the open voltage is

VOC = 0.0257ln( 1.5

10−10 + 1) = 0.602V Plot of (3.23) we obtain the following V-I curves of the PV cell

Figure 3.8:

[1] G M Masters, Renewable and Efficient Electric Power Systems, John Wiley & Sons, 2004

[2] Infoplease, Latitude and Longitude [Online] Available: http://www infoplease.com/atlas/latitude-longitude.html

[3] ASHRAE, Handbook of Fundamentals, American Society of Heating, Refrigera-tion and Air CondiRefrigera-tioning Engineers, Atlanta, 1993

[4] P Kruger, Alternative Energy Resources, John Wiley & Sons, 2004

[5] G Boyle, Renewable Energy: Power for a Sustainable Future, Oxford University Press, 2004

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