Future: Possibilities and Issues

1. The Geothermal Energy
Future: Possibilities and Issues
STUDENT: IMAN KAHROBAIE
SUPERVISOR:DR.M KHASHECHI
In the name of god
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2. Geopressured resources
Magnitude 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 × 1018 and 2800 × 1018 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 ×1018
and 1640 × 1018
J (Westhusing 1981; Garg 2007).
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3. Geopressured
Map (us)
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4. 4

5. 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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6. 6

7. Depth(m)
2000
2200
Clay mineral
authigenesis
2400
2600
2800
0 0.689 1.379 2.068
Over pressure MPa
Norwegian continental shelf
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8. 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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9. Heatofsolution(J/kg)
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 (moles/kg)
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10. 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 cycle
ρ = the density of the rock V= the rock
volume
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11. Joules
1018
1017
1016
1015
1014
Total available energy
1% extraction
0 50 100 150 200 250
Temperature difference (°C)
The amount of thermal energy that could be extracted from 1 km3 of rock 11

12. Enhanced geothermal
system
1. Reservoir
2. Pump house
3. Heat exchanger
4. Turbine hall
5. Production well
6. Injection well
7. Hot water to district
heating
8. Porous sediments
9. Observation well
10. Crystalline bedrock
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13. Exajoules(=1018Joules)
108
Total annual United States energy
consumption is ~ 100 exajoules
107
106
105
0 20 40 60 80 100
Percentage of EGS resource
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14.  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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15. 15
• response of the rock mass to changes in pressure
• pumping rate, and fluid properties
 reservoir management for sustainability

16. Production
well
Injection
well
Production
well
100° C
7 km 150° C
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17. surface area of Fractures (m2) for the Indicated dimensions
length (m)50 m100 m1000 m5000 m
2200400400020000
4400800800040,000
6600120012,00060,000
8800160016,00080,000
101000200020,000100,000
202000400040,000200,000
50500010,000100,000500,000
10010,00020,000200,0001,000,000
distance from Injection well (m)
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18. 18
Wellseparation(m)
Volumetricflow(m3/hr)
1,200 500
1,000
800
600
400
200
Well separation
Volumetric flow
400
300
200
100
0.0 0.0
0 50 100 150 200
Time (yrs)

19. 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),
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Qcv/Qcd = (h × A × dT)/[(k × A) × dT/dx].

20. 20
The end