Enhanced Oil Recovery is a great of interest by researchers which divided to thermal and non-thermal methods. Thermal methods indicate the heat transfer to reservoir in order to decrease the viscosity of fluid reservoir and make it mobile.
4. EOR Production by ProcessTypes
4
Nowadays around 40% of all EOR projects are produced by TEOR (Thermal Enhanced Oil Recovery)
Thermal EOR projects have been concentrated mostly in Canada, U.S, Venezuela, Brazil and China
5. 5
Type of crude oil Conventional oil Heavy oil Extra Heavy Oil
Gravity (ºAPI)
<22 22-10 >10
7. Selecting EOR
7
Do you focus on
offshore or
onshore
reservoirs?
Are the reservoir
heavy oil ?
( more than 200
cp viscosity?
Are the reservoir
deeper than 2700
m / 8860 ft ?
Are the reservoir
temperature above
90 C / 194 F ?
Onshore
Offshore
Yes
No
Gas
Injection
EOR
Chemical
Injection
EOR
Thermal
Injection
EOR
No
Yes
Chemical
Injection
EOR And
8. STEAM STIMULATION
Known as Steam Huff and Puff, Steam Soak, Cyclic Steam Injection (CCS)
• Mechanism:
Stage 1:
High pressure steam injected during several weeks ,heating of the oil, reduction of flow resistance by reducing crude oil
viscosity
Stage 2:
Shutting-in the well to allow the steam to soak the area around the injection well
Soak period during several weeks
Stage 3:
Placing the injection well into production.
Pumping of the oil up to the surface
When production declines: switch back to injection 8
10. STEAM FLOODING
known as steam derive (SD), Continuous steam injection
• Mechanism:
Steam generated at surface and is injected to the reservoir
Steam is injected into several injection wells while the oil is produced from other wells. (diff. from steam stimulation)
The steam moves through the reservoir and comes in contact with cold oil, rock, and water. Then it Heats up the Crude
oil and causes to lower the Viscosity.
Stage 1:
As the steam comes in contact with the cold environment condense is formed.
Stage 2:
Steam condenses into the water (during the steam loses heat to the formation)
condenses and a hot water bank is formed. This hot water bank acts as a water flood and pushes additional oil to the
producing wells.
Stage 3:
Then steam and hot water itself generate behind of the heated oil zone.
Stage 4:
an artificial drive that sweeps oil and condenses toward producing wells.
10
12. STEAM ASSISTED GRAVITY DRAINAGE (SAGD)
• Mechanism:
Heat conduction and gravity forces
One injector (upper well) and one producer (lower well)
steam injection through tubing that has been run to the horizontal well and fluids are returned from the tubing/casing
annulus
It requires comparatively thick reservoirs
12
14. VAPEX
Similar in concept to SAGD, in vapor extraction a solvent vapor is used to reduce viscosity of the heavy oil.
The injected solvent vapor expands and dilutes the heavy oil by contact. The diluted heavy oil will drain by
gravity to the lower horizontal well, to be produced.
Approximately 93% less natural gas is required in VAPEX compared to other steam processes.
Also, almost 93% less fresh water is used in VAPEX compared to other steam processes.
• less greenhouse gas emissions into the atmosphere
• The numbers shown are on a basis of 4,000 cubic meters oil production per day.
14
Process Natural Gas (m^3/day) Fresh Water (m^3/day)
SAGD 1,150,000 85,000
VAPEX 1,700 110
16. HORIZONTAL ALTERNATING STEAM DRIVE (HASD)
• Mechanism:
• The process is like cyclic steam soak and continuous steam drive
• Injector and producer wells are located at the bottom of the reservoir, parallels to each other
• Switching the producers and injectors well after some months
In contrast to SAGD, HASD can be applied to thin reservoirs (6-15 m)
Needs to be drilled less number of wells
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• Hot waterflooding, also known as hot water injection, hot water is injected to the reservoir.
• Typically less effective than a steam-injection process because water has lower heat content than steam.
HotWater Flood
17. Heat orThermal Capacity
• Specific Heat Capacity : it is the amount of energy that
must be added, in the form of heat, to one unit of mass
of the substance in order to cause an increase of one
unit in its temperature. in units of Btu/(lbm-°F).
• Notice that petroleum has a specific heat half of water
17
18. Thermal Conductivity
• Thermal conductivity can be defined in terms of the heat flow across a temperature difference.
• Heat will flow from the hot environment to the cold one in an attempt to equalize the temperature difference
18
k=Q∗L/A(T2−T1)
Q = heat flow (This is quantified in terms of a heat flux)
L = length or thickness
A = surface area
T2−T1 = temperature difference
K = Thermal Conductivity BTU/(h⋅ft⋅°F)
Steam Quality
mv=mass of vapor, lbm
ml=mass of liquid, lbm
fs=steam quality
• The condensation of steam that transfers energy to the oil heats it and reduces its viscosity. If the steam
quality is below 50 percent, steam injection becomes much less effective.
• The major problem with low steam quality is reduction heat transfer efficiency
19. 19
Nps: volume of oil displaced by steam
hn: net reservoir thickness
ht: reservoir thickness
Sors: Residual Oil Saturation by SD
Bos: Formation Volume Factor in zone swept by steam
Bol: FVF before start SD
Vs: Steam Swept Volume
Ws,eq: is the vol. of steam injection as equivalent barrel of water
at temperature of boiler feed water
ms:mass of water injected as steam
HsbA: enthalpy of steam at boiler outlet relative to feed water
temperature
Fsor: Steam Oil Ratio SOR
Myhill and Stegemier (1978):
• One economic parameter in thermal recovery is
steam-oil ratio SOR : which is defined as ratio of
the volume of steam required to produce one unit
volume of oil
20. Problem 1
• Use the following steam-drive data that were reported by Myhill and Stegemeier. The injection rate
and pressure are 850 B/D and of 200 psig, respectively. The temperatures of the reservoir and input
water are 110 and 70F, respectively. The quality of the output steam is 0.8.
Determine the value of SOR using the following information
20
Φ=0.30
ΔS=0.31 : oil saturation at the start of steaming minus oil saturation during steam drive
ms=Mass of water injected after 4.5 years is equal 489.17 * 106 lbm
HsbA= 993.83 Btu/lbm
Assuming that the volume of oil displaced from the steam zone is 331.5 acre-ft
21. Solution
• First step: Calculate the volume of oil produced
• Second step: Calculate the volume of steam injected
• Third step: Calculate SOR
21
22. Reservoir Heating
22
tD=dimensionless time
Ms=volumetric heat capacity of steam zone, Btu/(ft3-°F) [kJ/m3K]
MR=volumetric heat capacity of the reservoir, Btu/(ft3-°F) [kJ/m3K]
α=thermal diffusivity of reservoir, ft2/D [m2/d]
h=enthalpy per unit mass, Btu/lbm [kJ/kg]
t =time, D [d]
ht=gross reservoir thickness, ft [m]
Hs=energy content of steam injection
Hw= enthalpy of produced water and water vapor
Ts=temperature of steam
TR= reservoir temperature
fs=steam quality
λS= thermal conductivity of formation, cp [Pa.s]
ΔT =steam temperature/reservoir temperature, Ts/TR , °F
At=time-dependent heated area, sq ft [m2
Qi= heat injection rate, Btu/D [kJ/d]
ms:mass of water injected as steam
G(tD )=is a function of dimensionless time, tD.
The heated area over time
• Radius of steam zone for constant injection rate
Volumetric Heat Capacity : to estimate how much heat is needed to heat the reservoir
23. Problem 2
• The steam with the quality of 80% at pressure of 500 psig is injected to the reservoir at rate of 500 BWPD
CWE. The reservoir has 25% porosity. The initial saturations of oil and water is Soi =0.2 and Swi=0.8. The
volume occupied by the steam is 40% of the PV.
• Determine the heated area while 14 days is passed from injection.
23
At 500 psig, Ts=470.9 F
HwTR= 77 Btu/lbm(at 80F)
HwTs= 452.9 Btu/lbm(at 470.9F)
λs= 751.4 Btu/lbm
ms= (500 bbl/d) (350 lbm/bbl) (24 h/d) 57292 lbm/h
kh=1.5 Btu/h ft F,
α=0.0482 ft2/h,
MR=32.74 Btu/ft3F,
M(=kh/α)=31.12 Btu/ft3F,
TR= 80F
Thermal diffusivity, α, is the ratio of the thermal conductivity to the volumetric heat capacity
24. Solution
• First step: Calculate tD
• Second step: Calculate energy content of steam
• Third step: Calculate heated area at steam temperature
24
25. Heat Loss
• Heat loss to the formation occurs by a series of heat transfer mechanisms that included through tubing,
annuals, casing steel and cement into formation
• Slow injection rate will result in higher heat loss, therefore steam injection should be as high as possible
and in limited by fracture pressure
Qi= Heat Loss Rate
Qlr is the heat loss through radiation
Qlc is the heat loss through convection
25
26. 26
Natural Convection Heat-Transfer Rate
hnc=heat transfer coefficient for natural convection in the
tubing/casing annuals
knc=equivalent thermal conductivity of annuals fluid
rto=outside radius of tubing
rci=inside radius of casing
Radiation Heat-Transfer
hr=heat transfer coefficient for radiation in the tubing/casing annuals
σ is the Stefan-Boltzmann constant (1.713 *10-9 Btu/h ft2R4)
Ftci= View factor based on outside tubing and inside casing, dimensionless
Tto=outside temperature of tubing
Tci=inside temperature of casing
εto is the emissivity of external tubing surface,
εci is the emissivity of internal casing surface,
T is the absolute T(R),
27. HeatTransfer at Cement-Formation Interface
27
Khf=Thermal conductivity of formation
Th=Temperature at cement/formation interface
Ts=temperature of steam
Te= temperature of earth
f(t)=Time function dimensionless
rci=inside radius of casing
rto=outside radius of tubing
If injection time is long enough, the transient function would be applied
αf is the thermal diffusivity (k/ρc) of formation (ft2/h)
t is time (hour),
rhd is the radius of hole drill (ft)
HeatTransfer-RateThroughWellbore
rhd=radius of drill hole
rco=outside radius of casing
rto=outside radius of tubing
khcem is the thermal conductivity of cement
Tci= inside temperature of casing
Th=Temperature at cement/formation
interface
28. Overall HeatTransfer Coefficient
• To calculate the overall
heat transfer coefficient
use the following Willhite
• In most cases, hf, khtub, and
khcas are so large that
previous equation can be
approximated by
28
Uto= Overal heat transfer coefficient
hf =heat transfer coefficient between fluid and inside of tubing
khtub is the thermal conductivity of tubing
hnc is the natural convection htc in annulus
hr is the radiation htc
khcas is the thermal conductivity of casing
khcem is the thermal conductivity of cement
h is calculated in Btu/h ft2 F, and k is calculated in Btu/h ft F
29. Problem 3
• Consider tubing with 3.5” in which gas in injected at temperature of 600F. The casing is N-80. The air
pressure is 14.7 psia. The casing is cemented in a hole of 12” diameter. The depth of the reservoir is
3000ft.The packer size is 9.625”. The average temperature of subsurface is 100F.
• Determine the overall heat transfer coefficient, the mean temperature of the casing, the value of heat loss
from the wellbore after 21 days of injection.
29
Injecting steam :Ts=600 (F)
Well completion: rto=0.146 (ft), rco=0.4 (ft), rci=0.355 (ft), and rh=0.5 (ft)
Stagnant air at 14.7 psia in annulus
Formation characteristics: Te=100 (F), L=3000 (ft),
Heat transfer parameters: khf=1 (Btu/h ft F), khcem=0.2 (Btu/h ft F),
εto=0.9, αf=0.0286 (ft2/h)
30. Solution
• Calculate f(t)
• CalculateTemperature at cement and
formation interfaceTh
• CalculateTemperature through the
wellboreTci
30
31. • Calculate heat transfer for radiation hr
• Calculate heat transfer for natural
convection hnc
• Calculate overall heat transfer coefficient
Uto
31
32. IN SITU COMBUSTION (ISC)
• Mechanism:
• ISC is a TEOR method in which the air is injected into the reservoir burning the heaviest crude oil
components generating heat and combustion gases that enhance recovery by reducing oil viscosity
• After ignition the generated heat by combustion keeps the combustion front moving toward the producer
well.
• Combustion front burns all the fuel in its way. Approximately 30% of initial in place is used as a fuel and
the rest is going to be produced in the production well( It could be various according to saturation and rock
properties).
• The heat of reaction vaporizes initial water and also the light components of the oil in front of the
combustion front.
• The steam is condensed while distancing from the hot region.
32
33. There are 3 methods of combustion:
Dry forward:
• uses an igniter (air) to set fire to the oil
• Ignition of crude oil downhole.
• Propagation of flame front through the reservoir heating oil
• As the fire progresses the oil is pushed away from the fire toward the producing well (T>600 ºC)
Wet forward :
• Water is injected just behind the front and turned into steam by the hot rock
• Beginning as a dry process
• Once flame front is established, the oxygen stream is replaced by water
• Water meets hot zone left by combustion front
• Turns into steam, and aids the displacement of oil
Reverse:
• Air is injected until it reaches the production well., then the fire is ignited by an electrical system downhole
33
35. Thermal Project:
Duri Steam Flood (DSF) Indonesia
• The largest thermal enhanced oil
recovery (EOR) project in the world
• Second largest field in this country
• Following the decline of it production
yield due to the drop in natural reservoir
pressures, the oil field started using
steam-flood technology
• The DSF Project will develop over 15,000
acres of reservoir. Only about one third
of the field is under active steam flood.
35
1941 : Duri Oil Field discovered
1954 : Production started
1975 : A pilot steam flood project was launched
1985 :The field started using steam flooding technology
36. 36
Geology of Field:
• The field is located in Riau Province, on the Sumatra, in Indonesia.
• Duri oil field is approximately 18km long and 18km wide
• Structurally, the Duri field is a faulted, asymmetric anticline with a north-south trend.
• About 25% of the field has been developed using 5 and 7 spot pattern configurations
• Sandstone Formation :The sands consist mainly of quartz (52-75% by weight), orthoclase feldspar (7 - 12
% by weight), and lesser amounts of other clays, calcite, dolomite, siderite, and other minerals
• Reservoir depth 140 ft to 240 ft
37. • The primary recovery mechanisms were
mostly solution gas drive and compaction drive
with minor support from water influx and
gravity drainage.
• Field performance showed that ultimate
recovery from primary would be only about 8%
of OOIP.
• High oil viscosities and low solution GORs and
were key causes for this low recovery factor
• Reservoir and Fluid Properties:
37
Permeability 1550 mD
Porosity % 36%
Oil API degree 20
Average Initial Oil saturation (%) 53
Temperature (F) 100
Oil viscosity at 100°F 330 cp
Oil viscosity at 300°F 8.2 cp
Solution gas-oil ratio 15 SCF/STB
Rock compressibility 57X1 o-6 psi-1
Average net pay thickness 120ft
Oil formation volume factor 1.02 RB/STB
Reservoir thermal conductivity 27.4 BTU/ft-day-°F
Rock heat capacity 33.2 BTU/ft3-°F
38. • Production to injection ratio is 1.2 bbl production per bcwepd injected.
• Producer completions: 7" casing to top of pay and OHGP across the entire flood zone.
• Injector completions are now usually 7" casings with a single string of 3.12" tubing, packered above the
top perforations.
• Wellhead injection pressure at design rate is 450 psia (limited by fracture pressure).
• Average reservoir pressures increase by 150- 200 psi during the startup of an Area.
38
The steam-flooding technology, by 2008, enhanced the Duri oil production by more than three times
Producing nearly 200,000 BOPD of steam flood production 2008
Recovery Factor 60%
39. References
• Enhanced Oil Recovery, ByWillhite
• Enhanced Oil Recovery Fundamental
• Reservoir Management in the Duri Steam flood
• Petrowiki, Oil and Gas Journals, onepetro and etcetera
• EnhancedOil Recovery andAnalyse Sensitive Parameters, By Arzhang Nabilou
39