The geothermal energy2

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Information about The geothermal energy2

Published on September 22, 2016

Author: ImanKahrobaiy

Source: slideshare.net

1. The Geothermal Energy Future: Possibilities and Issues STUDENT: IMAN KAHROBAIE SUPERVISOR:DR.M KHASHECHI In the name of god 1

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). 2

3. Geopressured Map (us) 3

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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 5

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7. Depth(m) 2000 2200 Clay mineral authigenesis 2400 2600 2800 0 0.689 1.379 2.068 Over pressure MPa Norwegian continental shelf 7

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. 8

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) 9

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 10

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 12

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 13

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 14

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 16

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) 17

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

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