hydraulic fracturing to determine 4 - john t. foster · second cycle 2050 2000 1950 1900 8 1850...
TRANSCRIPT
![Page 1: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/1.jpg)
Hydraulic fracturing to determine Hydraulic fracturing to determine S3
![Page 2: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/2.jpg)
Hydraulic fracture initiation in vertical wellHydraulic fracture initiation in vertical well
= 3 − − 2 − ΔP − = −σminθθ
Shmin SHmax Pp σΔTT0
![Page 3: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/3.jpg)
Leako� test (mini-frac, FIT, XLOT)Leako� test (mini-frac, FIT, XLOT)
© Cambridge University Press (Fig. 7.2, pp. 211)Zoback, Reservoir Geomechanics
![Page 4: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/4.jpg)
from instantaneous shut-in pressure from instantaneous shut-in pressure
© Cambridge University Press (Fig. 7.3, pp. 213)
S3
Zoback, Reservoir Geomechanics
![Page 5: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/5.jpg)
© Cambridge University Press (Fig. 7.3, pp. 213)Zoback, Reservoir Geomechanics
![Page 6: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/6.jpg)
Step-rate testStep-rate test
© Cambridge University Press (Fig. 7.5, pp. 216)Zoback, Reservoir Geomechanics
![Page 7: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/7.jpg)
Be careful!Be careful!
When When integrate density logs integrate density logs
© Cambridge University Press (Fig. 7.4, pp. 214)
∼S3 Sv
Zoback, Reservoir Geomechanics
![Page 8: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/8.jpg)
What about What about ??
so
or
SHmax
ΔP = −Pb Pp
= 3 − − +SHmax Shmin Pb Pp T0
= 3 − (T = 0) −SHmax Shmin Pb Pp
![Page 9: Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850 1800 1750 1700 Time shut Second cycle Closure pressure (min) 2050 2000 1950 1900 1850](https://reader033.vdocuments.mx/reader033/viewer/2022052613/5f90e2c9bad813567560abe4/html5/thumbnails/9.jpg)
Does it work?Does it work?Consider a system with compressibility βs
=βsΔVs
Vs
1
ΔP
ΔP = Δ1
βsVs
Vs
=ΔP
Δt
1
βsVs
ΔVs
Δt
Answer: Not very well.Answer: Not very well.