Solar and Hydroelectric Power
Abbie Thill
Becca Mattson
Grace Nordquist
Keira Jacobs
Miyabi Goedert
Photovoltaic Cell vs Solar Heating Panel
Photovoltaic cells power things such as calculators and
satellites. Photo means light and voltaic means
electricity. Photovoltaic cells convert sunlight into electricity,
hence their name. Many times they are called PV cells. PV cells
are made of semiconductors, the most common being silicon.
When light hits the semiconductor, some of the energy is
absorbed by the semiconductor. Then the energy knocks loose
the electrons so they can flow freely. One or more electric fields
inside the PV cell force the electrons to flow in one direction.
The flow creates a current that can be extracted for external
use. The current and the voltage crated by the electric field
make the photovoltaic cell's power.
Photovoltaic Cell vs Solar Heating Panel
Solar Heating panels work by heating water. The heat that is
gathered by the solar collector is then transferred to the water
supply. The water is then stored until it is used. A conventional
water heater would take hot water from the solar storage tank
instead of using household energy to heat water in the boiler.
This technology is a good source of renewable energy. In the
US, the first commercial solar water heaters were sold in
California in the 1890’s, and today this method is most
commonly used to heat pools.
Solar Heating Panel Photovoltaic Cells
Solar Heating Panel vs. photovoltaic
cells
Solar Thermal
Efficiency: 70% Energy
Production
Cost: ~$6,000 per system
Daily Output: 22.3 kWh requires
2 panels/80 sq ft.
Size: King-sized mattress
Solar Photovoltaic
Efficiency: 12-15% Energy
Production
Cost: ~$48,000 per system
Daily Output: 22.3 kWh
requires 32 panels/456 sq ft.
Size: covers an entire roof
Photovoltaic Cell vs Solar Heating Panel
Even though solar heating panels are much more efficient than
photovoltaic panels, solar heating panels are used for heating
water, mostly in pools and household water heaters. To use the
sun's energy to power other devices photovoltaic cells are used.
Sankey Diagram For Photovoltaic Cell
The photovoltaic cell only uses the light from the sun and
not the heat so it isn't 100% efficient. The Semi-conductor
transforms only about 20% of the sun's energy into current
for outside use.
Sankey Diagram For Solar Heating Panel
The Solar Panel collects the sun's heat and turns it into
energy to heat the water. It uses the sun's heat but not the
light, and the panel cannot absorb all of the heat either. The
panel is about 70% efficient.
Applications of Photovoltaic cells
You want to install photovoltaic cells on the roof of your
cottage. The cottage has roof area of 139 m2. What is the
available electrical power considering the total cell efficiency is
14% and the constant light intensity is 980Wm-2?
Solution:
Power input = (light intensity)*(roof area)
= (980 Wm-2)*(139m2)
= 136 x 103 W
= 136 kW
Since Efficiency = (Power output)/(Power input),
Power output = Efficiency*(Power input)
= 0.14*(136 kW)
= 19 kW of electricity is available
Back to the Future Application
In Back to the Future, the DeLorean needs 1.21 GW in order to operate.
If Doc decides to go green and use photovoltaic cells to power his trip,
(a) what is the area of panels needed? Assume that the efficiency is 16%
and the constant light intensity of 965 Wm-2. (b) How many 1.30 m2 tiles
are needed? (c) Is this realistic?
Solution:
a) Efficiency = (output power)/(input power)
Input power = 1.21*109W/0.16 = 7.56 GW
Area = Power Input/light intensity
Area = 7.56*109/965 Wm-2 = 7.84*106 m2
b) Number of tiles = 7.84*106 m2/1.30 m2 = 6.03 *106 tiles
c) There are 6050 m2 in a football field
7.84*106 m2/6050 m2 = over 1,295 football fields
This is definitely not realistic!
Application of Solar Heating Panels
A solar-powered water heater absorbs solar energy at the rate of 10.2
MJm-2 per day. (a) Calculate the total water energy gain per day if the
collector area is 6.1 m2, the collector efficiency is 65% and the water
volume is 430 dm3. (b) Calculate the daily change in the tank's water
temperature.
Solution:
a) Water energy gain per day = (rate of solar energy received)*
(collector area)
Water energy gain per day = (10.2*106 MJm-2)*(6.1m2) = 6.22*107 J
b) Q = mc∆T, where the density of water is 1.0 kg dm-3,
m = (density of water)*(volume) = (1.0 kg dm-3)*(430 dm3) = 430 kg
∆T = Q/(mc) = 6.22*107 J/(430 kg*4186 J oC-1 kg-1) =35 oC.
The water temperature increases 35 0C daily.
Seasonal variations
in Earth's solar
power incident
-Solar radiation differs
with the seasons because
of a change in the way the
sun's radiation is spread.
-In the summer, the
earth's north pole is tilted
towards the sun, so the
sun's radiation is more
concentrated in the North
and more spread out in
the South.
Regional variations in Earth's solar
power incident.
- The sun's radiation is
most intense when the rays
strike perpendicular to the
surface of the earth.
Therefore, the suns
radiation is strongest at the
equator, and much less at
higher and lower latitudes.
- The reason for this is at
latitudes farther away from
the equator, the same
amount of radiation is
spread over a larger area.
Main Components of Hydroelectric Power
Plants
Reservoir: A place where a natural body of water is stored, at a level higher than
the turbine.
Dam: Obstructs the water flow from the reservoir. Stopping the water helps
harness the energy. The bottom of the dam has gates that can be lifted to let the
water through.
Penstock: Connects the reservoir to the turbine propeller. When the gates are
lifted the force of gravity makes the water flow through the penstock, the potential
energy stored in the dam is converted into kinetic energy.
Turbine: The kinetic energy created by the penstock turns the blades in the
turbine. The turbine has a shaft connected to the generator
Generator: When the blades in the turbine rotate, the shaft turns a motor which
produces an electric current in the generator.
Power Lines: Transports the power produced by the generator to various
stations through the power lines.
Main Components of Hydroelectric
Power Plants
Hydroelectric Schemes
1. Impoundment hydropower projects
2. Run-of-river projects
3. Microhydropower projects
4. Diversion hydropower projects
5. Pumped storage projects
Impoundment Hydropower
The water is
dammed and may
be released to
meet either
changing
electricity
demands or to
maintain a
constant water
level.
Run-of-River
Utilizes the natural
flow of the river and
can be used with either
large flow rates and
low heads or small
flow rates and high
heads.
The head of a hydropower project is the
height of the diversion
structure.
Microhydropower
A hydroelectric plant
which produces 100
kilowatts or less of
power. Used by
isolated home
owners and third
world countries.
Microhydropower
plants can use either
low or high heads.
Diversion Hydropower
Channels a portion
of the river
through a canal or
penstock and may
or may not require
a dam.
A penstock is the
channel that
carries the water to
the turbines.
Pumped Storage
Pumps water from low
levels to high levels
during times when
demand for electricity
is low, and releases it
to fall back down
when demand is high.
Energy transformations in
hydroelectric schemes
The water stored in the reservoir behind the dam has
gravitational potential energy because it is stored at a higher
elevation. When the water falls through the penstock toward
the turbines it gains translational kinetic energy. This kinetic
energy turns the turbines and gives rotational kinetic energy to
the generator. Electromagnets in the generator create electrical
current from the rotational energy gained from the turbine, and
then a transformer turns this AC current into a higher voltage
current which is then sent through power lines to homes and
businesses.
Sankey Diagram For Hydroelectric
Not all of the potential energy from the falling water is turned
into kinetic energy. Not all of the kinetic energy in the turbine
gets turned into current in the generator. Some of it is turned
into heat in the process that's why it is not a 100% efficient
process.
Calculating Power and Flow Rate in
Hydroelectric Schemes
P=(mgh)/t
Q= 0.83((b*d)/144)(100/t)
P= power in kW
m= mass flow rate in
tons/second
g= gravity
h= head height
t= time
Q= flow rate in cubic feet per second
b= average stream width
d= average stream depth
t= time it takes for stream to drift
100 feet
Solving Problems with Hydroelectric
Schemes
How much electrical power per second can a hydroelectric scheme
produce when the net head of the water is 120m high and the mass
flow rate of the water is 5.5 tons/second?
P=(mgh)/t
P=(5.5*9.8*120)/1
P=6500 kW
Calculate the flow rate of a hydroelectric scheme if the river is 240
inches wide with an average depth of 72 inches and it takes 23 seconds
for the river to drift 100 feet.
Q=0.83(bd/144)(100/t)
Q=0.83((240*72)/144)(100/23)
Q=430 cubic feet per second