Uploaded by Igi Perez

Geothermal Power Plant

advertisement
Geothermal
Power Plant
Geothermal
Energy
Resource in the
Philippines
https://www.geoenergymarketing.com/ener
gy-blog/geothermal-country-overviewphilippines/
Geothermal Energy
A form of energy conversion in which heat energy from within Earth is
captured and harnessed for cooking, bathing, space heating, electrical
power generation, and other uses.
Three Types of Geothermal Energy:
• Liquid-Dominated and Vapor-Dominated Reservoir
• Thermal Energy
• Enhanced Geothermal
Liquid-Dominated and Vapor-Dominated Reservoir
• Liquid-Dominated (LDRs) and
Vapor-Dominated Reservoirs
(VDRS) were common with
temperatures greater than
200 °C (392 °F) and are found
near
young
volcanoes
surrounding the Pacific Ocean
• Flash Plants are commonly
used to generate electricity
from LDR and VDR. Most wells
generate 2-10 MWe. Steam is
separated from liquid via
cyclone separators, while the
liquid is returned to the
reservoir for reheating.
Pacific Ring of Fire
https://www.mirror.co.uk/science/what-pacific-ring-fire-facts-12342864
Thermal Energy
• Heat for these purposes can
be extracted from cogeneration at a geothermal
electrical plant.
• Heating is cost-effective at
many more sites than
electricity generation. At
natural hot springs or
geysers, water can be piped
directly into radiators. In hot,
dry ground, earth tubes or
downhole heat exchangers
can collect the heat.
https://www.slideshare.net/shahbazshaikh21/geothermalenergy-74763210
Enhanced Geothermal
• Enhanced Geothermal Systems (EGS)
actively inject water into wells to be
heated and pumped back out. The
water is injected under high pressure
to expand existing rock fissures to
enable the water to freely flow in and
out.
• The Technique was adapted from oil
and gas extraction, but the geologic
formations are deeper, and no toxic
chemicals are used, reducing the
possibility of environmental damage.
https://doi.org/10.1016/j.rser.2017.09.065
Geothermal Power Plant
Geothermal Power Plant uses Hydrothermal Resources that have both
water (hydro) and heat (thermal).
Three types of Geothermal Power Plant
1. Dry steam plants
2. Binary cycle power plants
3. Flash steam plants
Efficiency of Geothermal Plant compared to
other Power Plants
Moon H., Gawell K., Zarrouk S. J., "EFFICIENCY OF GEOTHERMAL POWER PLANTS: A WORLDWIDE REVIEW," in
New Zealand Geothermal Workshop Proceedings, Auckland, New Zealand, 2012.
Dry Steam Plants
• A Vapor - Dominated
geothermal
resource
produce dry steam at
Temperature above 200 0C.
• The dry steam from the
geothermal reservoir is
extracted to turn the
generator turbines.
• These plants emit only
excess steam and very
minor amounts of gases
https://openei.org/wiki/Geothermal_Steam_Power_Plant
Binary Cycle Power Plants
• This plant transfers the
heat from the low to
moderately heated (below
200°C) geothermal fluid to
a secondary fluid with a
much lower boiling point
that water pass through a
heat exchanger. The heat
causes the second liquid to
turn to steam, which is
used to drive a generator
turbine.
https://geothermalenergyosu.weebly.com/different-types-ofgeothermal-energy.html
Flash Steam Plants
This plant extracts the high-pressure
hot water from deep inside the
earth that is converted to steam by
decreasing the pressure. This is
done by spraying fluid into a tank
held at a much lower pressure than
the hot water, causing some of the
hot water to rapidly vaporize, or
"flash." The produced steam then
drives a turbine, which drives a
generator. When the steam cools, it
condenses to water and is injected
back into the ground to be used
again.
https://ww2.energy.ca.gov/almanac/renewables_data/geother
mal/types.html
Schematic Diagram of Geothermal Power Plant
T-S Diagram of a Geothermal Power Plant
Performance of Geothermal Power Plant
1. Mass Flow Rate of Steam, ms :
π‘šπ‘  = π‘₯ ∗ π‘šπ‘”π‘€
Where: π‘šπ‘”π‘€ = π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ π‘€π‘Žπ‘‘π‘’π‘Ÿ
π‘₯ = π‘žπ‘’π‘Žπ‘™π‘–π‘‘π‘¦ π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘Ÿπ‘œπ‘‘π‘™π‘–π‘›π‘” π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘ 
Solving for steam quality (π‘₯) and considering a throttling like
process during the extraction of ground water:
β„Ž1 = β„Ž2 = β„Žπ‘“2 + π‘₯(β„Žπ‘“π‘”2 )
2. Turbine Output
π‘Šπ‘‘ = π‘šπ‘  (β„Ž3 − β„Ž4 ) ∗ πœ‚π‘‘
Where: πœ‚π‘‘ = π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
3. Generator Output
π‘Šπ‘” = π‘‡π‘’π‘Ÿπ‘π‘–π‘›π‘’ 𝑂𝑒𝑑𝑝𝑒𝑑 ∗ πΊπ‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
π‘Šπ‘” = π‘šπ‘  β„Ž3 − β„Ž4 (πœ‚π‘‘ )(πœ‚π‘” )
4. Heat Rejected in the Condenser
𝑄𝑅 = π‘šπ‘  β„Ž4 − β„Ž5
5. Over-all Plant Efficiency
πœ‚π‘œπ‘£π‘’π‘Ÿ−π‘Žπ‘™π‘™
π‘‡π‘’π‘Ÿπ‘π‘–π‘›π‘’ 𝑂𝑒𝑑𝑝𝑒𝑑 × πœ‚π‘”
=
(π‘šπ‘”π‘€ )(β„Ž1 )
πœ‚π‘œπ‘£π‘’π‘Ÿ−π‘Žπ‘™π‘™
π‘šπ‘  (β„Ž3 − β„Ž4 )(πœ‚π‘‘ ) πœ‚π‘”
=
(π‘šπ‘”π‘€ )(β„Ž1 )
Sample Problems
A flash steam geothermal power plant is located where underground
hot water is available as saturated liquid at 700 kPa. The ground hot
water flow rate is maintained at 29.6 kg/s at the well head with a
pressure of 600 kPa. The flashed steam enters a turbine at 500 kPa and
expands to 15 kPa upon entry to the condensed. Considering an 88%
turbine efficiency and 95% generator efficiency, determine:
a) The power produced by the turbine, kW
b) The power output of the Power Plant, kW
c) The Over-all Plant Efficiency, %
Given:
π‘˜π‘”
π‘šπ‘”π‘€ = 29.6
𝑠
𝑃1 = 700 π‘˜π‘ƒπ‘Ž
𝑃3 = 500 π‘˜π‘ƒπ‘Ž
𝑃4 = 15 π‘˜π‘ƒπ‘Ž
Required:
a) π‘Šπ‘‘
b) π‘Šπ‘”π‘’π‘›
c) πœ‚π‘œπ‘£π‘’π‘Ÿ−π‘Žπ‘™π‘™
Solution:
from steam table:
β„Ž1 = β„Žπ‘“@ 0.70 π‘€π‘ƒπ‘Ž
=
π‘˜π½
697.22
π‘˜π‘”
at 500 kPa;
β„Žπ‘“ =
β„Žπ‘“π‘” =
π‘˜π½
640.23
π‘˜π‘”
π‘˜π½
2108.5
π‘˜π‘”
β„Ž3 = β„Žπ‘”@ 0.50 π‘€π‘ƒπ‘Ž
β„Ž3 =
π‘˜π½
2748.7
π‘˜π‘”
Enthalpy of steam upon entry
to the condenser is taken from
Mollier Chart:
Solving for π‘₯2 :
β„Ž1 = β„Ž2 = β„Žπ‘“ + π‘₯2 β„Žπ‘“π‘”
π‘˜π½
697.22
π‘˜π‘”
=
π‘˜π½
640.23
π‘˜π‘”
+
π‘˜π½
π‘₯2 2108.5
π‘˜π‘”
π‘₯2 = 0.027
Solving for π‘šπ‘  :
π‘šπ‘  = 0.027 × π‘šπ‘”π‘€
π‘˜π‘”
29.6
𝑠
= 0.027
π‘šπ‘  = 0.80 π‘˜π‘”/𝑠
π‘˜π½
β„Ž4 = 2211
π‘˜π‘”
a) Turbine Output
π‘Šπ‘‘ = π‘šπ‘  (β„Ž3 − β„Ž4 ) ∗ πœ‚π‘‘
=
π‘˜π‘”
0.80
𝑠
π‘˜π½
2748.7
π‘˜π‘”
−
π‘˜π½
2211
π‘˜π‘”
= 378.54 π‘˜π‘Š
b) Power Plant Output
π‘ƒπ‘™π‘Žπ‘›π‘‘π‘œπ‘’π‘π‘’π‘‘ = π‘šπ‘  β„Ž3 − β„Ž4 ∗ πœ‚π‘‘ ∗ πœ‚π‘”
= 378.54 π‘˜π‘Š ∗ 0.95
= 359.61 π‘˜π‘Š
∗ 0.88
c) Over-all Plant Efficiency
πœ‚π‘œπ‘£π‘’π‘Ÿ−π‘Žπ‘™π‘™
π‘ƒπ‘™π‘Žπ‘›π‘‘ 𝑂𝑒𝑑𝑝𝑒𝑑
=
× 100%
(π‘šπ‘”π‘€ )(β„Ž1 )
359.61 π‘˜π‘Š
=
29.6 π‘˜π‘”/𝑠
= 1.74%
π‘˜π½
697.22
π‘˜π‘”
× 100%
Download