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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