Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850...

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Hydraulic fracturing to determine Hydraulic fracturing to determine

Transcript of Hydraulic fracturing to determine 4 - John T. Foster · Second cycle 2050 2000 1950 1900 8 1850...

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

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

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

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

from instantaneous shut-in pressure from instantaneous shut-in pressure

© Cambridge University Press (Fig. 7.3, pp. 213)

S3

Zoback, Reservoir Geomechanics

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Step-rate testStep-rate test

© Cambridge University Press (Fig. 7.5, pp. 216)Zoback, Reservoir Geomechanics

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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

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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

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.