ALL CHAPTERS COVERED
,SOLUTIONS TO END-
OF-CHAPTER
PROBLEMS
for
Well Logging for Earth Scientists
by
Darwin V. Ellis and Julian M. Singer
Published by Springer, P.O. Box 17,
3300 AA Dordrecht, The Netherlands
2007
, Solution to Problems
Chapter 2: Introduction to Well Log Interpretation
2.1 47.6%
2.1.1 85.8%
2.1.2 grain size variations, overburden compaction, cementation, clay-plugging
2.2 80.3%
2.3 39%
2.4 –
2.5 2.8 ft or 1.15 ft into formation.
2.6 114.6 ft
Chapter 3: Basic Resistivity and Spontaneous Potential
3.1 –
3.1.1 less saline
3.1.2 greater
3.1.3 Both GR and SP indicate shale, so ϕn > ϕd
3.1.4 1) 9320 – 9362 ft where Rxo ≈ Rt , and 2) 9363 – 9394ft where Rxo > Rt .
Both probably have hydrocarbons because Rt has increased above that of
lower zone. Possible reasons include, decrease in porosity, presence of hy-
drocarbons or a change in water resistivity.
3.1.5 Lower, since Rxo / Rt suggests invasion.
3.2 An exercise in using Chart SP-4. Interpolate between results obtained for
charts with Rxo = Rt and Rxo = 5 Rt in row for Rt /Rm = 5 to obtain
E SP /E SPcorr = 0.675. So E SPcorr = −33.33 mV.
3.3 Rw = 0.172 ohm-m. Using the uncorrected value, Rw = 0.245 ohm-m.
3.4.1 Using relation between resistance and resistivity; 46.5 k-ohm.
3.4.2 24.2 k-ohm. See chart Gen-6 for handy approximation.
3.4.3 1.26 k-ohm
3.5 –
3.5.3 The resistivity changes by nearly a factor of two but the temperature only by
10◦ F, so the salinity must change.
3.6 Deviation below 200◦ F negligible, but ∼50% at 350◦ F.
Chapter 4: Empiricism: The Cornerstone of Interpretation
4.1 –
4.2 –
4.3 1) For any value of Sw the core with the greater porosity should be less
resistive
,SOLUTIONS TO END-
OF-CHAPTER
PROBLEMS
for
Well Logging for Earth Scientists
by
Darwin V. Ellis and Julian M. Singer
Published by Springer, P.O. Box 17,
3300 AA Dordrecht, The Netherlands
2007
, Solution to Problems
Chapter 2: Introduction to Well Log Interpretation
2.1 47.6%
2.1.1 85.8%
2.1.2 grain size variations, overburden compaction, cementation, clay-plugging
2.2 80.3%
2.3 39%
2.4 –
2.5 2.8 ft or 1.15 ft into formation.
2.6 114.6 ft
Chapter 3: Basic Resistivity and Spontaneous Potential
3.1 –
3.1.1 less saline
3.1.2 greater
3.1.3 Both GR and SP indicate shale, so ϕn > ϕd
3.1.4 1) 9320 – 9362 ft where Rxo ≈ Rt , and 2) 9363 – 9394ft where Rxo > Rt .
Both probably have hydrocarbons because Rt has increased above that of
lower zone. Possible reasons include, decrease in porosity, presence of hy-
drocarbons or a change in water resistivity.
3.1.5 Lower, since Rxo / Rt suggests invasion.
3.2 An exercise in using Chart SP-4. Interpolate between results obtained for
charts with Rxo = Rt and Rxo = 5 Rt in row for Rt /Rm = 5 to obtain
E SP /E SPcorr = 0.675. So E SPcorr = −33.33 mV.
3.3 Rw = 0.172 ohm-m. Using the uncorrected value, Rw = 0.245 ohm-m.
3.4.1 Using relation between resistance and resistivity; 46.5 k-ohm.
3.4.2 24.2 k-ohm. See chart Gen-6 for handy approximation.
3.4.3 1.26 k-ohm
3.5 –
3.5.3 The resistivity changes by nearly a factor of two but the temperature only by
10◦ F, so the salinity must change.
3.6 Deviation below 200◦ F negligible, but ∼50% at 350◦ F.
Chapter 4: Empiricism: The Cornerstone of Interpretation
4.1 –
4.2 –
4.3 1) For any value of Sw the core with the greater porosity should be less
resistive
