Download Renewable Energy, Part 1

Renewable Energy
Part 1
Professor Mohamed A. El-Sharkawi
Renewable Energy
• Solar
• Wind
• Fuel Cell
• Small Hydro
• Geothermal
• Tidal
• Biomass
Solar Energy
Solar Power Density
  o cos   dt   wa  p
• : solar power density on earth in kW/m2
• o: extraterrestrial power density (1.353 kW/m2)
• : zenith angle (angle from the outward normal on the earth
surface to the center of the sun)
• dt: direct transmittance of gases except for water (the
fraction of radiant energy that is not absorbed by gases)
• p: is the transmittance of aerosol
• wa: water vapor absorptions of radiation.
Zenith Angle
Center
of Sun

Center of Earth
Solar Energy (Whr/m2/day)
Daily Solar Power Density
Density ratio
100%
80%
2
60%
(t t o ) 2
  max e
2 2
40%
20%
0%
0
2
4
6
8
t:hour of the day using the 24 hr clock
max: the maximum solar power density
to: time at max (noontime in the equator)
 : standard deviation
10
12
14
16
18
20
22
Time
24
Solar Efficiency ()
  o cos   dt   wa  p


0
  cos   dt   wa   p  5  70%
Example
An area located near the equator has the following
parameters:
 dt  80%,  p  95%,  wa  2%
Assume that the standard deviation of the solar distribution
function is 3.5hr. Compute the solar power density and
solar efficiency at 3:00 PM.
Solution
At noon
max  o cos   dt  wa   p
max  1353* cos( 0 )* 0.8  0.02* 0.95  1.0 kW/m
At 3:00PM
 ( t  t o )2
   max e
2 2
 1.0* e
( 1512 )2
2*( 3.5 )2
 0.693 kw/m
  693 / 1353  51.2%
2
2
Types of Solar Systems
– Passive Solar System
– Active Solar System (Photovoltaic or PV)
Passive Solar System
New supply of cold water
Warm water to the house
Warm
water
Lens
Tank
Sun rays
Collector
Cold water
back
to solar
collector
Passive Solar
Integrated Solar Combined
Cycle System (ISCCS)
Receiver
Collector
mirror
Integrated Solar Combined Cycle System (ISCCS)
Stirling System
Trough System
Solar Chimney
Active Solar Cell
(Photovoltaic PV)
Silicon
Empty
space for
extra
electron
Nucleus
Electrons
Silicon Atom
Silicon Crystal
Silicon
• Silicon is a good insulator
• To make the silicon more conductive
electrically, additives (impurities) are
added (doping)
– Phosphorus (P), which has 5 electrons in
its outer shell
– Boron (B), which has 3 electrons in its
outer shell
P-N Material
Electron without
bonding
SI
SI
SI
SI
P
Extra
space for
electron
SI
SI
n-type
SI
SI
SI
SI
B
SI
SI
SI
SI
SI
p-type
SI
Lens
-
N-Type
I
Load
-
P-Type
Base
Active Solar Cell (PV)
• PV cell is built like a diode out of semiconductor
material.
• Sunlight is composed of photons, or particles of
solar energy. Photons are the energy byproducts of
the nuclear reaction in the sun.
• When photons strike a PV cell, some of the
photons energy is absorbed by the semiconductor
material of the PV cell.
Active Solar Cell (PV)
• With this extra energy, the electrons in the
semiconductor material become excited and
break lose, and eventually begin an electric
current.
• Because PV cells are built like diodes, free
electrons are forced to flow in only one
direction
– the current is DC.
Main Parts of PV
Glass cover or lens
Antireflective coating
Contacts grid
n-type material
p-type material
Base
Structure of Solar System
• PV cell: 4X4 inches. The cell can produce
about 1 watt which is enough to run a
calculator.
Structure of Solar System
• Panel or Module: To increase
its energy rating, the PV cells
are connect together in parallel
and series.
• Parallel cells increase the
current rating
• Series cells increase the
voltage rating.
Structure of Solar System
• Array: PV panels connect together in parallel
and series to form a high power system.
Example
• Estimate the maximum power, current, and voltage ratings
of the panel and array in the figure. Assume that each PV
cell produce a maximum power of 2.5 W at the best solar
conditions
Solution
• The panel has 9 series cells. Assume that
the voltage of each cell is 0.5 V, the total
voltage of the panel is
V panel  0.5 *9  4.5 V
The panel has a total of 36 cells, the power
of the panel is
Ppanel  2.5 * 36  90 W
Total current of panel
I panel 
Ppanel
V panel
90

 20 A
4.5
The array consists of 2 columns of 4 series modules.
The total voltage of the array is
Varray V panel * 4  4.5 * 4  18 V
Total power of the array is
Parray  Ppanel * 8  90 *8  720 W
I array 
Parray
Varray
720

 40 A
18
Computation of PV Energy
Density ratio
100%
80%
2
60%
40%
20%
Solar power density
(t t o )
  max e
2 2
0%
2
0
2
4
6
8
10
12
14
16
18
20
22
Time
24
Computation of PV Energy
Linear relationship
Panel power
Solar power density
( t  t o ) 2
(t t o ) 2
  max e
2
Ppanel  Pmax e
2
24
Panel Energy
E panel   Pmax e
0
2 2
( t  t o ) 2
2 2
dt  Pmax 2  
Example
  max e
 ( t 12) 2
12.5
Pmax  100 W
Compute the daily energy produced by a PV panel.
Solution
2 12.5
2
E panel  Pmax
12.5

 2.5
2
2    100 * 2  * 2.5  627
Wh
Example
A 2 m2 panel of solar cells is installed in the Nevada’s
area. The efficiency of the solar panel is 10%.
 max  1.0 kw/m 2
  3.5 h
1. Compute the electrical power of the panel at
2:00PM when the solar power density is 685.5
W/m2
2. Assume the panel is installed on a
geosynchronous satellite. Compute its electrical
power output.
Solution
1.
Psun   * A  685.5* 2 1.371 kW
Ppanel  Psun  0.1* 1371 137.1 W
2.
Psun  o * A 1353* 2  2.706 kW
Ppanel  Psun  0.1* 2706  270.6 W
Solar array
House
Stand Alone
PV System
Charger
Discharger
Battery
Converter
dc current
Local load
ac current
PV System
without battery
Solar
array
ac current
Converter
dc current
House
Local load
Meter To utility
Solar System With Battery
• Battery: To store the energy when
the PV power is not fully utilized by
the load.
– The battery power is later used when the
PV power is less than the demand.
– These batteries are deep cycle types
• Charger: To charge the battery by the
PV
Solar System With Battery
• Inverter: To invert the dc power of
the battery to the 60Hz power used in
homes.
• Synchronizer: Used to connect the
PV system to the power grid.
– DC/AC converter.
Ideal PV Model: P-N Junction
Cathode (K)
Cathode (K)
n
p
Anode (A)
Vd
I
Anode (A)
Ideal PV Model: P-N Junction
Vd
Vs
+
I
Vl
I
R
Io
Forward biased
region
Reverse saturation current
Vd
Reverse biased
region
Ideal PV Model: P-N Junction

I  Io  e


Vd
VT

 1


kT
VT 
q
I
Forward
biased
region
Io
Reverse saturation
current
Io: reverse saturation current
Reverse biased
region
VT: thermal voltage
q: elementary charge constant, i.e. charge of one electron
(1.602 10-19 Coulomb)
k: Boltzmann’s constant (1.380 x 10-23 J/K)
T: absolute temperature in Kelvin (K).
Vd
PV Model
-
-
N-Type
-
P-Type Id
Load
I
Lens
Id
V=Vd
I
Is
Load
Base
PV Model
Is
Id
V=Vd
Load
Is  I  Id
Solar Cell
I
Is: the solar current (is a nonlinear variable that changes
with light density (irradiance)
Id: the forward current through the diode.
PV Characteristics
Is
Id
Io
Vd
Vd
I  Is  Id
I
QII
QIII
QI
QIV
Vd
PV Power Characteristics
Pout  VI
Solar Cell
I  Is  Id
 VVd

I d  I o  e T  1




Is
Id
V=Vd
Vd

 V

Pout  V I  Vd  I s  I o  e T  1





Load
V  Vd
I
PV Power Characteristics
 VVd

T

I  I s  I o e  1




I

 VVd

T

Pout  Vd  I s  I o e  1





Isc
Imp
Pmax
P
Vmp
Voc
Vd
PV Power Characteristics
• Main variables
– Short Circuit Current (Isc)
– Open Circuit Voltage (Voc)
– Operating Point at Maximum Power (Pmax,
Vmp, Imp)
Short Circuit PV
Is
Id=0
Isc=Is
I sc  I s
Open Circuit PV
Is
Id=Is
Voc


I d  I s  I o  e  1




 Is

Voc  VT * ln   1
 Io

Voc
VT
Example
• An ideal PV cell with reverse saturation
current of 1nA is operating at 30oC. The
solar current at 30oC is 1A. Compute
the output voltage and output power of
the PV cell when the load draws 0.5A.
Solution
k T 1.38*10 23 *(30  273.15)
3
VT 

 26.11*10 V
19
q
1.602 *10
 VV

I  I s  I o  e T  1




V


9
0
.
02611
0.5  1  10 *  e
 1




V  ln 1  0.5*10  1 *VT  0.523 V
9
P  V I  0.523 * 0.5  0.2615 W
Example
• An ideal solar cell with reverse
saturation current of 1nA is operating at
20oC. The solar current at 20oC is 0.8A.
Compute the voltage and current of the
solar cell at the maximum power point.
P  VI
Solution
P
I
 Vmp
I 0
V
V
 VV

I  I s  I o  e T  1




I o V / VT
I

e
V
VT
At maximum Power
 Vmp  Vmp /VT
P
 I o e
 I s  I o   1 
0
V
 VT 
Solution
k T 1.38*10 23 *(20  273.15)
3
VT 


25
.
25
*
10
V
19
q
1.602 *10
 Vmp  Vmp / VT I s  I o
1 
 e

Io
 VT 
Vmp  Vmp / 25.25

1 
 e
 0.8 *109
 25.25 
Vmp  443.8479 mV
Solution


VT

 I s  I o e  1




Vmp
I mp
I mp

 0.8  10  e

9
443.8479
25.25

 1  0.7569 A

Pmax  Vmp * I mp  443.8479 * 0.7569  335.948 mW
Operating Point of PV
• The operating point of the
solar cell depends on the
magnitude of the load
resistance R
• The intersection of the PV
cell characteristic with the
load line is the operating
point of the PV cell.
Is
Id
V=Vd
Load
• The load resistance is the
output voltage V divided by
the load current I.
Solar Cell
I
Operating Point of PV
I
R1
R1<R2<R3
1
Load lines
R2
2
3
R3
Voc V
Example
• For the solar cell in the previous example, compute
the load resistance at the maximum output power.
Solution
From the previous example
Vmp  443.8479 mV
I mp  0.7569 A
Rmp 
Vmp
I mp
443.8479

 0.5864 
756.9
Example
• An ideal PV cell with a reverse saturation current of 1nA is
operating at 30oC. The solar current at 30oC is 1A. The cell
is connected to a 10 resistive load. Compute the output
power of the cell
Solution
23
k T 1.38*10 * (30  273.15)
3
VT 


26
.
1
*
10
19
q
1.602 *10
 V
 V
I  I s  I o  eVT  1 

 R


 VV

V  I s R  I o R  e T  1




V


8
0
.
0261
V  10 10  e
 1


Iteratively, solve for V
V  0.4722 V
2
2
V
0.4722
P

 22.29 mW
R
10
Effect of Irradiance 
1<2<3
3
I
3
2
1
Load line
2
1
Voc
V
Effect of Irradiance 
P
1<2<3
1
2
3
V
Effect of Temperature T
I
T1
T1>T2>T3
Load line
1
2
3
T2
T3
Voc
V
Effect of Temperature T
P
T1>T2>T3
T1
T2
T3
V
Id1
V1
Load
Is1
Is2
Id2
V2
V=Vd1+Vd2
Is=Is1=Is2
Load
PV Module (Series
Connection)
Id1
V1
V=Vd1=Vd2
Load
Is1
Is2
Id2
V2
Is=Is1+Is2
Load
PV Module (Parallel
Connection)
Example
• An ideal PV module is composed of 50
solar cells connected in series. At 20oC,
the solar current of each cell is 1A and
the reverse saturation current is 10nA.
Draw the I-V and I-P characteristics of
the module.
Solution
k T 1.38*10 23 *(20  273.15)
VT 

 25.25 mV
19
q
1.602 *10
V
 VV



8
0
.
02525


I d  I o e  1 10 * e
 1






T
Pcell  Vcell I cell
Vmod ule  n *Vcell
Pmod ule  n * Pcell
Module Current and Power
Vcell


8
0.02525

I cell  I s  I d 1  10 *  e
 1


20
Current
15
Power
10
5
0
0
5
10
15
Module Voltage
20
25
Losses of PV Cell
• Irradiance Losses
• Electrical Losses
Irradiance Losses
1. Due to the reflection of the solar radiation at
the top of the PV cell.
2. The light has photons with wide range of
energy levels
– Some don’t have enough energy to excite the
electrons.
– Others have too much energy that is hard to
capture by the electrons.
• These two scenarios account for the loss of
about 70 percent of the solar energy
Losses of PV Cell
(Electrical Losses)
• The resistances of the collector traces at the
top of the cell.
• The resistance of the wires connecting cell to
load.
• The resistance of the semiconductor crystal
Real PV Model
• To account for the electrical losses only
Rs
Solar
Cell
Is
Id
Vd
Rp
Ip
Rs : Resistance of wires and traces
Rp : internal resistance of the cell
V
Load
I
I
Efficiency of PV Cell
Sun power converted to electricit y Pse Vd * I s
 irradiance 


Sun power
Ps
A
Pout
Output power of the cell
V *I
e 


Sun power converted to electricit y Pse Vd * I s
Pse Pout Pout V * I
   irradiance e 


Ps Pse
Ps
A
Example
• A 100 cm2 solar cell is operating at 30oC
where the output current is 1A, the load
voltage is 0.4V and the saturation current of
the diode is 1nA. The series resistance of the
cell is 10 m and the parallel resistance is
1k. At a given time, the solar power density
is 200W/m2. Compute the irradiance
efficiency and the overall efficiency.
Solution
Rs
Solar
Cell
Is
Id
Vd Rp
Ip
V
Load
I
I
k T 1.38*10 23 * (30  273.15)
3
VT 


26
.
1
*
10
V
19
q
1.602 *10
Vd  V  IRs  0.4  1* 0.01  0.41 V
0.41
 VVd



9
0
.
0261
T


I d  I o e  1 10 *  e
 1  6.64 mA






Solution
Rs
Solar
Cell
Is
Id
Vd Rp
Ip
V
Load
I
I
Vd
0.41
Ip 

 0.41 mA
R p 1000
I s  I  I d  I p 1  0.00664  0.00041  1.00705 A
Vd * I s 0.41*1.00705
 irradiance 

 0.205
A
200 * 0.01
Solution
Rs
Solar
Cell
Is
Id
Vd Rp
V
Ip

Peloss  I Rs  I R p  1.0 * 0.01  0.41*10
2
e 
2
p
2
Load
I
I
 *1000  10.168
3 2
mW
Pout
Pout
VI
0.4 *1.0



 0.975
Pse Pout  Peloss VI  Peloss 0.4 *1.0  0.010168
  irradiancee  0.205 * 0.975  0.20
Conclusion: Most of the losses are irradiance
Assessment of PV Systems
$6.0
$5.0
$5.0
$/kWh
$4.0
$3.0
$2.0
$1.5
$1.0
$0.0
1970
$0.6
1980
1990
Year
$0.4
2000
$0.3
2010
Solar Power and the
Environment
• 6kW from a photovoltaic system
instead of a thermal power plant can
reduce annual pollution by
– 3 lbs. of NOx (Nitrogen Oxides),
– 10 lbs. of SO2 (Sulfur Dioxide), and
– 10530 lbs. of CO2 (Carbon Dioxide).
Assessment of PV Systems
• Consumer products (calculators, watches,
battery chargers, light controls, and flashlights)
are the most common applications
• Larger PV systems are extensively used in
space applications (such as satellites)
• In higher power applications, three factors
determine the applicability of the PV systems
1. the cost and the payback period of the system
2. the accessibility to a power grid
3. the individual inclination to invest in environmentally
friendly technologies.
Assessment of PV Systems
• In remote areas without access to power
grids, the PV system is often the first choice
among the available alternatives.
Assessment of PV Systems
• By the end of the 20th century, the PV
systems worldwide had the capacity of more
than 900 GWh annually
– this PV energy is enough for about 70,000 homes
in the USA, or about 4 million homes in developing
countries.
• The largest PV plant in the world is 60MW in
Spain
• 75 MW PV plant is being built in Cle Elum,
Washington
Assessment of PV Systems
• To manufacture the solar cells, arsenic and
silicon compounds are used
– Arsenic is odorless and flavorless semi-metallic
chemical that is highly toxic
– Silicon, by itself, is not toxic. However, when additives
are added to make the PV semiconductor material, the
compound can be extremely toxic.
– Since water is used in the manufacturing process, the
runoff could cause these material to reach local
streams
– Should a PV array catch fire, these chemicals can be
released into the environment.
Assessment of PV Systems
• Solar power density can be intermittent due to
weather conditions
• PVs are limited exclusively to daytime use
• For high power PV systems, the arrays spread
over a large area.
• The PV systems are considered by some to be
visually intrusive
• The efficiency of the solar panel is still low, making
the system expensive and large
• Solar systems require continuous cleaning of their
surfaces