the geothermal energy2
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The Geothermal EnergyFuture: Possibilities and
Issues
STUDENT: IMAN KAHROBAIESUPERVISOR:DR.M KHASHECHI
In the name of god
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Geopressured resourcesMagnitude of The resource
•fluid pressure exceeded that expected simple hydrostatic gradient
•associated with oil and gas fields
•temperature range of 110°C to 150°C
•the recoverable thermal energy in the northern Gulf of Mexico Basin, a region of geopressured resources, is between 270 × and 2800 × J
•The total capacity for electrical power generation is estimated to be greater than 100,000 MW (Green and Nix 2006).
•have high methane concentrations associated with them
• hydrocarbon gas is an additional resource with an estimated recoverable energy content of between 1 ×and 1640 × J (Westhusing 1981; Garg 2007).
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GeopressuredMap (us)
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Why Geopressured reservoirs form•reduced permeability
•Recrystallization and growth of new minerals
•deposition of carbonate minerals, such as calcite and dolomite, and silica minerals
•concentrations of dissolved solids, with salinities occasionally exceeding 200,000 mg/l
•a reservoir can be up to 4 km3
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Dep
th (
m)
2000
2200 Clay m in e ra l au th ig en esis
2400
2600
2800 0
0.689 1.379 2.068
Over pressure M P a
Norwegian continental shelf
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challenges To development
Fluid chemistry
•highly saline, with dissolved loads as high as 200,000 mg/l
•significant concentrations of CO2
Reinjection
• Separate the dissolved solute load from the aqueous phase while minimizing the loss of thermal energy.
• Separate and capture the dissolved methane gas phase from the aqueous phase.
• Efficiently extract the thermal energy and kinetic energy from the fluid while maintaining sufficient pressure and flow rates.
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Hea
t of s
olut
ion
(J/k
g)
35,000
30,000
25,000
20,000
15,000
10,000
5,000
0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7 .0
NaCl (m o les /k g )
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Enhanced Geothermal systems (EGS)magnitude of The resource
•Temperature greater than about 130°C
•can be found at depths between 5 and 10 km under half the area of the United States
Qex = V × ρ × Cp × ∆T. Qex= function of the heat Cp =capacity of the rock (J/m3-K)∆T=the number of degrees by which the temperature is decreased in the power production cycleρ = the density of the rock V= the rock volume
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Joul
es
1018
1017
1016
1015
1014
Total available en e rg y
1 % ex trac tio n 0 50 100 150 200 250
T e m p e ra tu re difference (°C )
The amount of thermal energy that could be extracted from 1 km3 of rock
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Enhanced geothermal system1. Reservoir 2. Pump house 3. Heat exchanger4. Turbine hall5. Production well6. Injection well7. Hot water to district
heating8. Porous sediments 9. Observation well10. Crystalline bedrock
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Exaj
oule
s ( =
1018
Joul
es)
108
Total annual United States en erg y co n su m p tio n is ~ 100 ex a jou les
107
106
105 0 20 40 60 80 100
Percentage of EGS re so u rce
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Technological requirements
• hydro fractured or stimulated
EGS efforts To date
some of the key challenges that these efforts have identified
drilling and downhole equipment
• circulation and integrity of the drilling fluid
drilling Fluids
• high permeability allows drilling fluids to escape to the surrounding rock
high-Temperature downhole equipment
• EGS components need to survive temperatures of 225–250°C
reservoir engineering
• The ability to assess the orientation and properties of fractures
• Measuring the orientation and magnitude of subsurface stresses at high temperatures
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• response of the rock mass to changes in pressure• pumping rate, and fluid properties reservoir management for sustainability
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P ro d u c tio n
well In jec tio n
well P ro d u c tio n
well
10 0 ° C
7 k m 15 0 ° C
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surface area of Fractures (m2) for the Indicated dimensions
length (m) 50 m 100 m 1000 m 5000 m
2 200 400 4000 20000
4 400 800 8000 40,000
6 600 1200 12,000 60,000
8 800 1600 16,000 80,000
10 1000 2000 20,000 100,000
20 2000 4000 40,000 200,000
50 5000 10,000 100,000 500,000
100 10,000 20,000 200,000 1,000,000
distance from Injection well (m)
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Wel
l sep
arat
ion
(m)
Vol
umet
ric f
low
(m3 /h
r)
1 ,2 0 0 500
1 ,0 0 0
800
600
400
200
Well sep a ra tio n
Volumetric flow
400 300 200 100
0 .0 0 .0 0 50 100 150 200
Time (y rs)
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Tb= is the time (hr.)γt= is the heat capacity (J/m3K) of the reservoir γf= is the heat capacity of the fluid (J/m3K) d =is the distance between wells (m) t= is the reservoir thickness (m)v= is the flow rate (m3/hr.)
Tb = (π × γt × d2 × t)/(3 × γf × v),
Qcv/Qcd = (h × A × dT)/[(k × A) × dT/dx].
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The end