HBX2 Appraisal Well Test
The subsurface team is preparing a well test (DST) for an appraisal well HBX2 in an offshore oil reservoir. The well is slightly deviated with a drilled net reservoir section length L= 187 feet, as shown in the crosssection below.
Table 1 shows the PVT and static data for this well:
————————————–
Fluid VolumeFactor :1.2000 vol/vol
Fluid Viscosity :0.500E+00 CP
Fluid Compressibility :0.170E04 1/PSI
water compressibility :0.300E05 1/PSI
porosity :22.1000 %
water saturation :25.7000 %
rock compressibility :0.400E05 1/PSI
wellbore radius :0.354E+00 FEET
—————————————
The pressure profiles from MDT is presented below:
(click on the plot to enlarge it)
The objectives of the test are to confirm productivity (KH, skin), flow the well at a sustained oil rate of 5,000 bbl/d and prove a connected volume of 10 million barrels of oil at surface.
The completion engineer suggests the following DST string along with a first draft of the test sequence. (click on the plot to enlarge it)
Well Test Design
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Question 1 of 6
1. Question
What is the estimated maximum pressure that could be encountered at surface?




Correct
For the worst case scenario, we assume a gas column with a density of 0.1 psi/ft.
At 3,048mTVDss, we have a pressure P= 6253 psi. So roughtly at surface, P=62533048/0.3048 x 0.1= 5,253 psi.
Pressure may be a bit lower than that due to the use of TVDss.
Incorrect
For the worst case scenario, we assume a gas column with a density of 0.1 psi/ft.
At 3,048mTVDss, we have a pressure P= 6253 psi. So roughtly at surface, P=62533048/0.3048 x 0.1= 5,253 psi.
Pressure may be a bit lower than that due to the use of TVDss.

Question 2 of 6
2. Question
What comment do you have about the above DST string?





Correct
The downhole shutin tool is located at about 150m from the top perforations. At first sight, this seems a bit shallow but we need to check if this distance is in measured depth or true vertical depth.
We should add more pressure gauges below the downhole shutin tool, including some memory gauges as backup.
We should add a gauge carrier mid tubing and two seabed gauges for offshore appraisal well test.
Most importantly, we should try to bring the deepest gauges closer to the perforations.
Incorrect
The downhole shutin tool is located at about 150m from the top perforations. At first sight, this seems a bit shallow but we need to check if this distance is in measured depth or true vertical depth.
We should add more pressure gauges below the downhole shutin tool, including some memory gauges as backup.
We should add a gauge carrier mid tubing and two seabed gauges for offshore appraisal well test.
Most importantly, we should try to bring the deepest gauges closer to the perforations.

Question 3 of 6
3. Question
What is the value of total compressibility?




Correct
Ct= Cr + SoCo + SwCw= 0.400E05 + 0.743x 0.170E04 + 0.257x 0.300E05
Ct= 0.174E04 1/psi (do not forget the unit)
Incorrect
Ct= Cr + SoCo + SwCw= 0.400E05 + 0.743x 0.170E04 + 0.257x 0.300E05
Ct= 0.174E04 1/psi (do not forget the unit)

Question 4 of 6
4. Question
What volume ∆V should we flare from this well to create a level of depletion ∆P= 10psi? How long should the main flow period last for? What methods could we use to challenge and reduce the flaring volume?




Correct
∆V= ∆P xCt xV = 10 x 0.174E04 x 10,000,000 = 1740 bbls of oil
Flowing at 5,000 bbl/d, the main flow period should be 1740/5000= 0.348 days or 8.4 hours long. This is not a very long time to be called “sustained flow”, although some companies may beg to differ when publishing press releases.
We could challenge further this flaring volume by reducing the level of depletion using better quality pressure gauges and by using Deconvolution. If there is, we could connect the well to an existing infrastructure to minimize flaring.
Incorrect
∆V= ∆P xCt xV = 10 x 0.174E04 x 10,000,000 = 1740 bbls of oil
Flowing at 5,000 bbl/d, the main flow period should be 1740/5000= 0.348 days or 8.4 hours long. This is not a very long time to be called “sustained flow”, although some companies may beg to differ when publishing press releases.
We could challenge further this flaring volume by reducing the level of depletion using better quality pressure gauges and by using Deconvolution. If there is, we could connect the well to an existing infrastructure to minimize flaring.

Question 5 of 6
5. Question
Is there any particular feature that we could remove from the test sequence?




Correct
The very first initial flow and shutin before the 2 PBU tests are no longer needed to obtain the initial pressure. This technique will actually bring some uncertainty due to the presence of nonhydrocarbon fluids in the wellbore below the gauge.
Incorrect
The very first initial flow and shutin before the 2 PBU tests are no longer needed to obtain the initial pressure. This technique will actually bring some uncertainty due to the presence of nonhydrocarbon fluids in the wellbore below the gauge.

Question 6 of 6
6. Question
Why two PBU tests?




Correct
For redundancy and for Deconvolution. With 2 PBU tests, we can make sure we recover a reliable deconvolution response and refine initial pressure.
This could also help with the connected volume, if Deconvolution doesn’t work.
Incorrect
For redundancy and for Deconvolution. With 2 PBU tests, we can make sure we recover a reliable deconvolution response and refine initial pressure.
This could also help with the connected volume, if Deconvolution doesn’t work.
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Question 1 of 6
1. Question
What is the estimated maximum pressure that could be encountered at surface?
Correct
For the worst case scenario, we assume a gas column with a density of 0.1 psi/ft.
At 3,048mTVDss, we have a pressure P= 6253 psi. So roughtly at surface, P=62533048/0.3048 x 0.1= 5,253 psi.
Pressure may be a bit lower than that due to the use of TVDss.Incorrect
For the worst case scenario, we assume a gas column with a density of 0.1 psi/ft.
At 3,048mTVDss, we have a pressure P= 6253 psi. So roughtly at surface, P=62533048/0.3048 x 0.1= 5,253 psi.
Pressure may be a bit lower than that due to the use of TVDss. 
Question 2 of 6
2. Question
What comment do you have about the above DST string?
Correct
The downhole shutin tool is located at about 150m from the top perforations. At first sight, this seems a bit shallow but we need to check if this distance is in measured depth or true vertical depth.
We should add more pressure gauges below the downhole shutin tool, including some memory gauges as backup.
We should add a gauge carrier mid tubing and two seabed gauges for offshore appraisal well test.
Most importantly, we should try to bring the deepest gauges closer to the perforations.Incorrect
The downhole shutin tool is located at about 150m from the top perforations. At first sight, this seems a bit shallow but we need to check if this distance is in measured depth or true vertical depth.
We should add more pressure gauges below the downhole shutin tool, including some memory gauges as backup.
We should add a gauge carrier mid tubing and two seabed gauges for offshore appraisal well test.
Most importantly, we should try to bring the deepest gauges closer to the perforations. 
Question 3 of 6
3. Question
What is the value of total compressibility?
Correct
Ct= Cr + SoCo + SwCw= 0.400E05 + 0.743x 0.170E04 + 0.257x 0.300E05
Ct= 0.174E04 1/psi (do not forget the unit)Incorrect
Ct= Cr + SoCo + SwCw= 0.400E05 + 0.743x 0.170E04 + 0.257x 0.300E05
Ct= 0.174E04 1/psi (do not forget the unit) 
Question 4 of 6
4. Question
What volume ∆V should we flare from this well to create a level of depletion ∆P= 10psi? How long should the main flow period last for? What methods could we use to challenge and reduce the flaring volume?
Correct
∆V= ∆P xCt xV = 10 x 0.174E04 x 10,000,000 = 1740 bbls of oil
Flowing at 5,000 bbl/d, the main flow period should be 1740/5000= 0.348 days or 8.4 hours long. This is not a very long time to be called “sustained flow”, although some companies may beg to differ when publishing press releases.
We could challenge further this flaring volume by reducing the level of depletion using better quality pressure gauges and by using Deconvolution. If there is, we could connect the well to an existing infrastructure to minimize flaring.Incorrect
∆V= ∆P xCt xV = 10 x 0.174E04 x 10,000,000 = 1740 bbls of oil
Flowing at 5,000 bbl/d, the main flow period should be 1740/5000= 0.348 days or 8.4 hours long. This is not a very long time to be called “sustained flow”, although some companies may beg to differ when publishing press releases.
We could challenge further this flaring volume by reducing the level of depletion using better quality pressure gauges and by using Deconvolution. If there is, we could connect the well to an existing infrastructure to minimize flaring. 
Question 5 of 6
5. Question
Is there any particular feature that we could remove from the test sequence?
Correct
The very first initial flow and shutin before the 2 PBU tests are no longer needed to obtain the initial pressure. This technique will actually bring some uncertainty due to the presence of nonhydrocarbon fluids in the wellbore below the gauge.
Incorrect
The very first initial flow and shutin before the 2 PBU tests are no longer needed to obtain the initial pressure. This technique will actually bring some uncertainty due to the presence of nonhydrocarbon fluids in the wellbore below the gauge.

Question 6 of 6
6. Question
Why two PBU tests?
Correct
For redundancy and for Deconvolution. With 2 PBU tests, we can make sure we recover a reliable deconvolution response and refine initial pressure.
This could also help with the connected volume, if Deconvolution doesn’t work.Incorrect
For redundancy and for Deconvolution. With 2 PBU tests, we can make sure we recover a reliable deconvolution response and refine initial pressure.
This could also help with the connected volume, if Deconvolution doesn’t work.
RealTime Well Testing Operations
The well HBX2 has now been perforated and we start the cleanup period by gradually increasing the choke size and taking some routine fluid samples at surface.
At the end of the cleanup, the well is closed at the downhole tester valve. As you have access to the realtime data during the operations, the figure below shows the derivative of the first PBU.
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Question 1 of 17
1. Question
What sort of well and reservoir could we see from this loglog plot? (at this stage, we haven’t acquired the 2nd PBU yet)





Correct
1. Slightly deviated or vertical well in a heterogeneous reservoir with some improvement in mobility and/or storativity further away from the well,
2. Slightly deviated or vertical well with limited perforations/partial penetration in a homogeneous reservoir with probably a low vertical permeability,
3. Slightly deviated or vertical well in a homogeneous reservoir with the presence of boundaries (conventional derivative)
4. Slightly deviated or vertical well in a homogeneous multilayer reservoir
Incorrect
1. Slightly deviated or vertical well in a heterogeneous reservoir with some improvement in mobility and/or storativity further away from the well,
2. Slightly deviated or vertical well with limited perforations/partial penetration in a homogeneous reservoir with probably a low vertical permeability,
3. Slightly deviated or vertical well in a homogeneous reservoir with the presence of boundaries (conventional derivative)
4. Slightly deviated or vertical well in a homogeneous multilayer reservoir

Question 2 of 17
2. Question
During the main flow period, the pressure upstream of the choke manifold starts to increase while the pressure downstream of the choke starts to decrease, as shown below.
What could explain these changes in upstream and downstream choke pressure values?




Correct
There seems to be some blockage in the choke (debris, hydrate, etc).
(It can sometimes be an adjustable vibrating shut if it is not locked in place).
Incorrect
There seems to be some blockage in the choke (debris, hydrate, etc).
(It can sometimes be an adjustable vibrating shut if it is not locked in place).

Question 3 of 17
3. Question
At the end of the main flow period, the well is closed at the downhole valve.
The plot below shows a comparison study between gauge 20754 (carrier at 3,020.7mTVDss) and gauge 22818 (carrier at 3,035.7mTVDss).
Below is a comparison study between gauge 20754 (carrier at 3,020.7mTVDss) and gauge 22931 (carrier at 3,035.7 mTVDss).
Could we extract some information about the fluid density between the gauges?




Correct
Incorrect

Question 4 of 17
4. Question
Which gauge would you select for well test analysis?




Correct
Gauge 22818 shows some drift (how convenient that we have two gauges at that depth). It is then between 20754 and 22931 pressure gauges. It would be better to select the deepest reliable one, 22931 as this will result in less uncertainty on pressure and skin.
Incorrect
Gauge 22818 shows some drift (how convenient that we have two gauges at that depth). It is then between 20754 and 22931 pressure gauges. It would be better to select the deepest reliable one, 22931 as this will result in less uncertainty on pressure and skin.

Question 5 of 17
5. Question
After a 12hour shutin for PBU 2, the test sequence is presented below:
What do the different drawdown data sets suggest about well productivity?




Correct
We are producing roughly at the same rate at the end of the cleanup and during main flow period. However, we can see that the drawdown has been reduced, suggesting a decrease in skin and an improvement in productivity.
Incorrect
We are producing roughly at the same rate at the end of the cleanup and during main flow period. However, we can see that the drawdown has been reduced, suggesting a decrease in skin and an improvement in productivity.

Question 6 of 17
6. Question
The PBU data are shown in the loglog plot below:
What observations could we make from this derivative overlay?




Correct
Consistent derivatives.
The shift downwards in DP plots suggests a decrease in skin between PBUs 1 and 2, also confirmed with the different drawdown data sets in the above question.
Incorrect
Consistent derivatives.
The shift downwards in DP plots suggests a decrease in skin between PBUs 1 and 2, also confirmed with the different drawdown data sets in the above question.

Question 7 of 17
7. Question
Both PBU data are presented in the superposition plot below. The reservoir engineer on the rig drew a straight line in the Superposition plot.
An attempt was made to recover a reliable deconvolution response, with two possible options below.
a) Initial pressure Pi= 6,241 psi at deepest gauge depth
b) Initial pressure Pi= 6,236 psi at deepest gauge depth
Which deconvolution and initial pressure would you select, and why?




Correct
We should select the deconvolution with Pi= 6,241psi.
There was a trap here with the superposition plot, which doesn’t show an immediate shift downwards (also P*=6,236 psi is meaningless here as the reservoir is not of infinite acting, as shown in the derivative plot at late times). Also, a pressure of 6,236 psi (which is P* in the superposition plot) is too low when looking at deconvolution, as suggested by the green deconvolved derivative being lower than the blue one.
Incorrect
We should select the deconvolution with Pi= 6,241psi.
There was a trap here with the superposition plot, which doesn’t show an immediate shift downwards (also P*=6,236 psi is meaningless here as the reservoir is not of infinite acting, as shown in the derivative plot at late times). Also, a pressure of 6,236 psi (which is P* in the superposition plot) is too low when looking at deconvolution, as suggested by the green deconvolved derivative being lower than the blue one.

Question 8 of 17
8. Question
As the conditioning coefficient increases, the deconvolved derivative first increases at late times then remains unchanged, as shown below. What would this mean?




Correct
Unlike PBU 1, PBU 2 is bit too short to derive a reliable deconvolution on its own. So we need to use more data to recover a reliable deconvolution.
Incorrect
Unlike PBU 1, PBU 2 is bit too short to derive a reliable deconvolution on its own. So we need to use more data to recover a reliable deconvolution.

Question 9 of 17
9. Question
Do we see a closed reservoir from these dataset?




Correct
Deconvolution doesn’t follow a unitslope straight line at late times, so the reservoir is not closed.
Incorrect
Deconvolution doesn’t follow a unitslope straight line at late times, so the reservoir is not closed.

Question 10 of 17
10. Question
Is the initial pressure from Deconvolution consistent with the MDT data?




Correct
At 3,048mTVDss, we have a pressure P= 6253 psi with the MDT data.
At the gauge 22931 depth (3,035.7 mTVDss), we should have:
P= 6253 – (30483035.7)*1.07 ~ 6,240 psi
Or P=6253 (30483035.7)/0.3048 x 0.325 ~ 6,240 psi.
The initial pressure from Deconvolution is consistent with the MDT data.
Incorrect
At 3,048mTVDss, we have a pressure P= 6253 psi with the MDT data.
At the gauge 22931 depth (3,035.7 mTVDss), we should have:
P= 6253 – (30483035.7)*1.07 ~ 6,240 psi
Or P=6253 (30483035.7)/0.3048 x 0.325 ~ 6,240 psi.
The initial pressure from Deconvolution is consistent with the MDT data.

Question 11 of 17
11. Question
The figure below compares the conventional and deconvolved derivatives:
Why is the deconvolved derivative different from the conventional derivatives at late times?




Correct
While the deconvolved derivative is derived with respect to the log of time, the conventional derivative is derived with respect to the superposition time. This may introduce some error or “distortion” in the conventional derivative, for example when the flow period before the PBU doesn’t end in radial flow regime. (log approx.)
So we see a distortion/error in the conventional derivative due to the way it is calculated (the conventional derivative will be distorted in time, or delayed in time compared to deconvolution).
Deconvolution is a sort of improved derivative that is developed over the entire test sequence, so over a longer time interval. As a result, we see further away in the reservoir.
Incorrect
While the deconvolved derivative is derived with respect to the log of time, the conventional derivative is derived with respect to the superposition time. This may introduce some error or “distortion” in the conventional derivative, for example when the flow period before the PBU doesn’t end in radial flow regime. (log approx.)
So we see a distortion/error in the conventional derivative due to the way it is calculated (the conventional derivative will be distorted in time, or delayed in time compared to deconvolution).
Deconvolution is a sort of improved derivative that is developed over the entire test sequence, so over a longer time interval. As a result, we see further away in the reservoir.

Question 12 of 17
12. Question
What are the possible interpretation models?




Correct
Looking at the deconvolution, it can only be:
1. A well with wellbore storage and skin in a homogeneous reservoir with the presence of boundaries (in this case, at least a sealing fault), or
2. A well with wellbore storage and skin in a heterogeneous reservoir with a decrease in mobility/storativity further away from the well.
Incorrect
Looking at the deconvolution, it can only be:
1. A well with wellbore storage and skin in a homogeneous reservoir with the presence of boundaries (in this case, at least a sealing fault), or
2. A well with wellbore storage and skin in a heterogeneous reservoir with a decrease in mobility/storativity further away from the well.

Question 13 of 17
13. Question
Based on the static data in table 1, what is the permeability value?




Correct
The derivative stabilization from 0.3 to 8 hours could be indicative of radial flow regime. The stabilization level is ∆P’/q= 0.01 psi/stb/d.
We know from the definition of the dimensionless term that PD’= KH ∆P’/(141.2 Bμq). During radial flow regime, PD’= 0.5.
Hence KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ft
H= cos 30 x L= cos 30 x 187 = 162 ft.
The permeability is then K= 4236/ 162= 26.1 mD
Incorrect
The derivative stabilization from 0.3 to 8 hours could be indicative of radial flow regime. The stabilization level is ∆P’/q= 0.01 psi/stb/d.
We know from the definition of the dimensionless term that PD’= KH ∆P’/(141.2 Bμq). During radial flow regime, PD’= 0.5.
Hence KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ft
H= cos 30 x L= cos 30 x 187 = 162 ft.
The permeability is then K= 4236/ 162= 26.1 mD

Question 14 of 17
14. Question
The reservoir engineer on the rig would like to wait longer and spot a clear boundary feature on the conventional derivative.
Do you think it is worth extending PBU 2 test? Or based on the subsurface objectives, could we terminate this well test?




Correct
At first sight, it may not be worth extending PBU 2, looking at the conventional and deconvolved derivatives. However, the subsurface objective is to prove a minimum connected volume of 10 million barrels of oil.
So far with this test sequence, we have seen a minimum porevolume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
64212000 x 0.178 / 1.2 x (10.257) = 7,076,912 bbls.
We should therefore extend PBU 2.
Incorrect
At first sight, it may not be worth extending PBU 2, looking at the conventional and deconvolved derivatives. However, the subsurface objective is to prove a minimum connected volume of 10 million barrels of oil.
So far with this test sequence, we have seen a minimum porevolume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
64212000 x 0.178 / 1.2 x (10.257) = 7,076,912 bbls.
We should therefore extend PBU 2.

Question 15 of 17
15. Question
Unfortunately there was an operation problem at surface and the well had to be left shutin downhole up to 48 hours. The additional data are available below.
Could the shape of the conventional derivative at late times be indicative of a crossflow (communication) from different reservoir layers?




Correct
The Ushape feature that is observed at late times on the conventional derivative is not indicative of a crossflow in the reservoir. This is the distortion due to the way the conventional derivative is calculated. We can see that the conventional derivative is delayed in time, compared to deconvolution.
Incorrect
The Ushape feature that is observed at late times on the conventional derivative is not indicative of a crossflow in the reservoir. This is the distortion due to the way the conventional derivative is calculated. We can see that the conventional derivative is delayed in time, compared to deconvolution.

Question 16 of 17
16. Question
A unitslope straight line was placed on the final point of the deconvolved derivative.
What is the minimum connected volume at surface, after a 48hr PBU2 test?
(hint: bbl =ft³ x 0.178)




Correct
PV x 0.178 x (1Swi) / Bo = 91213000 x 0.178 x (10.257)/1.2 = 10,052,737 bbls
Incorrect
PV x 0.178 x (1Swi) / Bo = 91213000 x 0.178 x (10.257)/1.2 = 10,052,737 bbls

Question 17 of 17
17. Question
Should we leave the well shutin for longer?




Correct
No, we shouldn’t leave the well shutin for longer. We can terminate the test as we reached the subsurface objectives.
Incorrect
No, we shouldn’t leave the well shutin for longer. We can terminate the test as we reached the subsurface objectives.
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Question 1 of 17
1. Question
What sort of well and reservoir could we see from this loglog plot? (at this stage, we haven’t acquired the 2nd PBU yet)
Correct
1. Slightly deviated or vertical well in a heterogeneous reservoir with some improvement in mobility and/or storativity further away from the well,
2. Slightly deviated or vertical well with limited perforations/partial penetration in a homogeneous reservoir with probably a low vertical permeability,
3. Slightly deviated or vertical well in a homogeneous reservoir with the presence of boundaries (conventional derivative)
4. Slightly deviated or vertical well in a homogeneous multilayer reservoirIncorrect
1. Slightly deviated or vertical well in a heterogeneous reservoir with some improvement in mobility and/or storativity further away from the well,
2. Slightly deviated or vertical well with limited perforations/partial penetration in a homogeneous reservoir with probably a low vertical permeability,
3. Slightly deviated or vertical well in a homogeneous reservoir with the presence of boundaries (conventional derivative)
4. Slightly deviated or vertical well in a homogeneous multilayer reservoir 
Question 2 of 17
2. Question
During the main flow period, the pressure upstream of the choke manifold starts to increase while the pressure downstream of the choke starts to decrease, as shown below.
What could explain these changes in upstream and downstream choke pressure values?
Correct
There seems to be some blockage in the choke (debris, hydrate, etc).
(It can sometimes be an adjustable vibrating shut if it is not locked in place).Incorrect
There seems to be some blockage in the choke (debris, hydrate, etc).
(It can sometimes be an adjustable vibrating shut if it is not locked in place). 
Question 3 of 17
3. Question
At the end of the main flow period, the well is closed at the downhole valve.
The plot below shows a comparison study between gauge 20754 (carrier at 3,020.7mTVDss) and gauge 22818 (carrier at 3,035.7mTVDss).
Below is a comparison study between gauge 20754 (carrier at 3,020.7mTVDss) and gauge 22931 (carrier at 3,035.7 mTVDss).
Could we extract some information about the fluid density between the gauges?
Correct
Incorrect

Question 4 of 17
4. Question
Which gauge would you select for well test analysis?
Correct
Gauge 22818 shows some drift (how convenient that we have two gauges at that depth). It is then between 20754 and 22931 pressure gauges. It would be better to select the deepest reliable one, 22931 as this will result in less uncertainty on pressure and skin.
Incorrect
Gauge 22818 shows some drift (how convenient that we have two gauges at that depth). It is then between 20754 and 22931 pressure gauges. It would be better to select the deepest reliable one, 22931 as this will result in less uncertainty on pressure and skin.

Question 5 of 17
5. Question
After a 12hour shutin for PBU 2, the test sequence is presented below:
What do the different drawdown data sets suggest about well productivity?
Correct
We are producing roughly at the same rate at the end of the cleanup and during main flow period. However, we can see that the drawdown has been reduced, suggesting a decrease in skin and an improvement in productivity.
Incorrect
We are producing roughly at the same rate at the end of the cleanup and during main flow period. However, we can see that the drawdown has been reduced, suggesting a decrease in skin and an improvement in productivity.

Question 6 of 17
6. Question
The PBU data are shown in the loglog plot below:
What observations could we make from this derivative overlay?
Correct
Consistent derivatives.
The shift downwards in DP plots suggests a decrease in skin between PBUs 1 and 2, also confirmed with the different drawdown data sets in the above question.Incorrect
Consistent derivatives.
The shift downwards in DP plots suggests a decrease in skin between PBUs 1 and 2, also confirmed with the different drawdown data sets in the above question. 
Question 7 of 17
7. Question
Both PBU data are presented in the superposition plot below. The reservoir engineer on the rig drew a straight line in the Superposition plot.
An attempt was made to recover a reliable deconvolution response, with two possible options below.
a) Initial pressure Pi= 6,241 psi at deepest gauge depth
b) Initial pressure Pi= 6,236 psi at deepest gauge depth
Which deconvolution and initial pressure would you select, and why?
Correct
We should select the deconvolution with Pi= 6,241psi.
There was a trap here with the superposition plot, which doesn’t show an immediate shift downwards (also P*=6,236 psi is meaningless here as the reservoir is not of infinite acting, as shown in the derivative plot at late times). Also, a pressure of 6,236 psi (which is P* in the superposition plot) is too low when looking at deconvolution, as suggested by the green deconvolved derivative being lower than the blue one.Incorrect
We should select the deconvolution with Pi= 6,241psi.
There was a trap here with the superposition plot, which doesn’t show an immediate shift downwards (also P*=6,236 psi is meaningless here as the reservoir is not of infinite acting, as shown in the derivative plot at late times). Also, a pressure of 6,236 psi (which is P* in the superposition plot) is too low when looking at deconvolution, as suggested by the green deconvolved derivative being lower than the blue one. 
Question 8 of 17
8. Question
As the conditioning coefficient increases, the deconvolved derivative first increases at late times then remains unchanged, as shown below. What would this mean?
Correct
Unlike PBU 1, PBU 2 is bit too short to derive a reliable deconvolution on its own. So we need to use more data to recover a reliable deconvolution.
Incorrect
Unlike PBU 1, PBU 2 is bit too short to derive a reliable deconvolution on its own. So we need to use more data to recover a reliable deconvolution.

Question 9 of 17
9. Question
Do we see a closed reservoir from these dataset?
Correct
Deconvolution doesn’t follow a unitslope straight line at late times, so the reservoir is not closed.
Incorrect
Deconvolution doesn’t follow a unitslope straight line at late times, so the reservoir is not closed.

Question 10 of 17
10. Question
Is the initial pressure from Deconvolution consistent with the MDT data?
Correct
At 3,048mTVDss, we have a pressure P= 6253 psi with the MDT data.
At the gauge 22931 depth (3,035.7 mTVDss), we should have:
P= 6253 – (30483035.7)*1.07 ~ 6,240 psi
Or P=6253 (30483035.7)/0.3048 x 0.325 ~ 6,240 psi.
The initial pressure from Deconvolution is consistent with the MDT data.Incorrect
At 3,048mTVDss, we have a pressure P= 6253 psi with the MDT data.
At the gauge 22931 depth (3,035.7 mTVDss), we should have:
P= 6253 – (30483035.7)*1.07 ~ 6,240 psi
Or P=6253 (30483035.7)/0.3048 x 0.325 ~ 6,240 psi.
The initial pressure from Deconvolution is consistent with the MDT data. 
Question 11 of 17
11. Question
The figure below compares the conventional and deconvolved derivatives:
Why is the deconvolved derivative different from the conventional derivatives at late times?
Correct
While the deconvolved derivative is derived with respect to the log of time, the conventional derivative is derived with respect to the superposition time. This may introduce some error or “distortion” in the conventional derivative, for example when the flow period before the PBU doesn’t end in radial flow regime. (log approx.)
So we see a distortion/error in the conventional derivative due to the way it is calculated (the conventional derivative will be distorted in time, or delayed in time compared to deconvolution).
Deconvolution is a sort of improved derivative that is developed over the entire test sequence, so over a longer time interval. As a result, we see further away in the reservoir.Incorrect
While the deconvolved derivative is derived with respect to the log of time, the conventional derivative is derived with respect to the superposition time. This may introduce some error or “distortion” in the conventional derivative, for example when the flow period before the PBU doesn’t end in radial flow regime. (log approx.)
So we see a distortion/error in the conventional derivative due to the way it is calculated (the conventional derivative will be distorted in time, or delayed in time compared to deconvolution).
Deconvolution is a sort of improved derivative that is developed over the entire test sequence, so over a longer time interval. As a result, we see further away in the reservoir. 
Question 12 of 17
12. Question
What are the possible interpretation models?
Correct
Looking at the deconvolution, it can only be:
1. A well with wellbore storage and skin in a homogeneous reservoir with the presence of boundaries (in this case, at least a sealing fault), or
2. A well with wellbore storage and skin in a heterogeneous reservoir with a decrease in mobility/storativity further away from the well.Incorrect
Looking at the deconvolution, it can only be:
1. A well with wellbore storage and skin in a homogeneous reservoir with the presence of boundaries (in this case, at least a sealing fault), or
2. A well with wellbore storage and skin in a heterogeneous reservoir with a decrease in mobility/storativity further away from the well. 
Question 13 of 17
13. Question
Based on the static data in table 1, what is the permeability value?
Correct
The derivative stabilization from 0.3 to 8 hours could be indicative of radial flow regime. The stabilization level is ∆P’/q= 0.01 psi/stb/d.
We know from the definition of the dimensionless term that PD’= KH ∆P’/(141.2 Bμq). During radial flow regime, PD’= 0.5.
Hence KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ftH= cos 30 x L= cos 30 x 187 = 162 ft.
The permeability is then K= 4236/ 162= 26.1 mDIncorrect
The derivative stabilization from 0.3 to 8 hours could be indicative of radial flow regime. The stabilization level is ∆P’/q= 0.01 psi/stb/d.
We know from the definition of the dimensionless term that PD’= KH ∆P’/(141.2 Bμq). During radial flow regime, PD’= 0.5.
Hence KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ftH= cos 30 x L= cos 30 x 187 = 162 ft.
The permeability is then K= 4236/ 162= 26.1 mD 
Question 14 of 17
14. Question
The reservoir engineer on the rig would like to wait longer and spot a clear boundary feature on the conventional derivative.
Do you think it is worth extending PBU 2 test? Or based on the subsurface objectives, could we terminate this well test?
Correct
At first sight, it may not be worth extending PBU 2, looking at the conventional and deconvolved derivatives. However, the subsurface objective is to prove a minimum connected volume of 10 million barrels of oil.
So far with this test sequence, we have seen a minimum porevolume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
64212000 x 0.178 / 1.2 x (10.257) = 7,076,912 bbls.
We should therefore extend PBU 2.Incorrect
At first sight, it may not be worth extending PBU 2, looking at the conventional and deconvolved derivatives. However, the subsurface objective is to prove a minimum connected volume of 10 million barrels of oil.
So far with this test sequence, we have seen a minimum porevolume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
64212000 x 0.178 / 1.2 x (10.257) = 7,076,912 bbls.
We should therefore extend PBU 2. 
Question 15 of 17
15. Question
Unfortunately there was an operation problem at surface and the well had to be left shutin downhole up to 48 hours. The additional data are available below.
Could the shape of the conventional derivative at late times be indicative of a crossflow (communication) from different reservoir layers?
Correct
The Ushape feature that is observed at late times on the conventional derivative is not indicative of a crossflow in the reservoir. This is the distortion due to the way the conventional derivative is calculated. We can see that the conventional derivative is delayed in time, compared to deconvolution.
Incorrect
The Ushape feature that is observed at late times on the conventional derivative is not indicative of a crossflow in the reservoir. This is the distortion due to the way the conventional derivative is calculated. We can see that the conventional derivative is delayed in time, compared to deconvolution.

Question 16 of 17
16. Question
A unitslope straight line was placed on the final point of the deconvolved derivative.
What is the minimum connected volume at surface, after a 48hr PBU2 test?
(hint: bbl =ft³ x 0.178)Correct
PV x 0.178 x (1Swi) / Bo = 91213000 x 0.178 x (10.257)/1.2 = 10,052,737 bbls
Incorrect
PV x 0.178 x (1Swi) / Bo = 91213000 x 0.178 x (10.257)/1.2 = 10,052,737 bbls

Question 17 of 17
17. Question
Should we leave the well shutin for longer?
Correct
No, we shouldn’t leave the well shutin for longer. We can terminate the test as we reached the subsurface objectives.
Incorrect
No, we shouldn’t leave the well shutin for longer. We can terminate the test as we reached the subsurface objectives.
PostOperations Well Test Analysis
The reservoir engineer on the rig sent you the following interpretation.
(click on the plot to enlarge it)
[Mob. Ratio (1) = Mobility of the outer zone / Mobility of the inner zone]
Time limit: 0
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Question 1 of 2
1. Question
What is she suggesting? Is this correct?




Correct
She is using a linear composite model assuming 2 different regions in the reservoir, with an increase in mobility/storativity from 240 ft away from the well in one direction. There is also a sealing fault at about 420 ft. The KH in this model is 4,430mD.ft, resulting in a permeability K= 23.7mD if a net thickness of 187ft is used. The total skin is equal to +3.3.
She didn’t use Deconvolution, and as a result didn’t spot the distortion on the conventional derivative. She assumed a reservoir region at an improved mobility and storativity further away from the well (likely an increase in reservoir thickness) to try to model the Ushape derivative at late times. This is not correct, and this is why the initial pressure is a bit too low too.
Incorrect
She is using a linear composite model assuming 2 different regions in the reservoir, with an increase in mobility/storativity from 240 ft away from the well in one direction. There is also a sealing fault at about 420 ft. The KH in this model is 4,430mD.ft, resulting in a permeability K= 23.7mD if a net thickness of 187ft is used. The total skin is equal to +3.3.
She didn’t use Deconvolution, and as a result didn’t spot the distortion on the conventional derivative. She assumed a reservoir region at an improved mobility and storativity further away from the well (likely an increase in reservoir thickness) to try to model the Ushape derivative at late times. This is not correct, and this is why the initial pressure is a bit too low too.

Question 2 of 2
2. Question
Do you agree with her KH and permeability values?




Correct
We found a KH= 4,236 mD.ft and a permeability of 26.1 mD.
Her KH= 4,434 mD.ft is a bit high compared to our estimate, but still within 10%.
However, her permeability is a bit too low, as she used the measured length L as the reservoir thickness. In her interpretation, H= 187ft. It should be 162 ft.
Incorrect
We found a KH= 4,236 mD.ft and a permeability of 26.1 mD.
Her KH= 4,434 mD.ft is a bit high compared to our estimate, but still within 10%.
However, her permeability is a bit too low, as she used the measured length L as the reservoir thickness. In her interpretation, H= 187ft. It should be 162 ft.
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Question 1 of 2
1. Question
What is she suggesting? Is this correct?
Correct
She is using a linear composite model assuming 2 different regions in the reservoir, with an increase in mobility/storativity from 240 ft away from the well in one direction. There is also a sealing fault at about 420 ft. The KH in this model is 4,430mD.ft, resulting in a permeability K= 23.7mD if a net thickness of 187ft is used. The total skin is equal to +3.3.
She didn’t use Deconvolution, and as a result didn’t spot the distortion on the conventional derivative. She assumed a reservoir region at an improved mobility and storativity further away from the well (likely an increase in reservoir thickness) to try to model the Ushape derivative at late times. This is not correct, and this is why the initial pressure is a bit too low too.Incorrect
She is using a linear composite model assuming 2 different regions in the reservoir, with an increase in mobility/storativity from 240 ft away from the well in one direction. There is also a sealing fault at about 420 ft. The KH in this model is 4,430mD.ft, resulting in a permeability K= 23.7mD if a net thickness of 187ft is used. The total skin is equal to +3.3.
She didn’t use Deconvolution, and as a result didn’t spot the distortion on the conventional derivative. She assumed a reservoir region at an improved mobility and storativity further away from the well (likely an increase in reservoir thickness) to try to model the Ushape derivative at late times. This is not correct, and this is why the initial pressure is a bit too low too. 
Question 2 of 2
2. Question
Do you agree with her KH and permeability values?
Correct
We found a KH= 4,236 mD.ft and a permeability of 26.1 mD.
Her KH= 4,434 mD.ft is a bit high compared to our estimate, but still within 10%.
However, her permeability is a bit too low, as she used the measured length L as the reservoir thickness. In her interpretation, H= 187ft. It should be 162 ft.Incorrect
We found a KH= 4,236 mD.ft and a permeability of 26.1 mD.
Her KH= 4,434 mD.ft is a bit high compared to our estimate, but still within 10%.
However, her permeability is a bit too low, as she used the measured length L as the reservoir thickness. In her interpretation, H= 187ft. It should be 162 ft.
Perforating and testing a new sand
At the end of the 2nd PBU, the team decides to run in hole some perforating guns and perforate a sand layer of net stratigraphic thickness H2=42 feet, above the pay zone.
Then they flow all the zones and perform a shutin (PBU 3). The result is shown below on the loglog plot.
(Click on the plot to enlarge it).
Time limit: 0
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Question 1 of 2
1. Question
Assuming a uniform skin distribution across the layers, what is the permeability of that newly perforated layer?




Correct
The derivative stabilization has shifted downwards as the total mobility has increased.
Now the stabilization level is about ∆P’/q= 0.0065 psi/stb/d.
We know that PD’= 0.5 = KH ∆P’/(141.2 Bμq)
Hence total KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.0065 = 6,516.9 mD.ft
Total KH= (KH)1 + (KH)2 = 4236.0 + (KH)2 = 6,516.9
As a result, (KH)2 = 6,516.9 – 4236.0 = 2,280.9 mD.ft
Using H2=42 feet, then the permeability in the new layer is then K= 2280.9/ 42= 54.3 mD
The assumption of a uniform skin distribution is important here. Otherwise, we could have had a “false” derivative stabilization.
Incorrect
The derivative stabilization has shifted downwards as the total mobility has increased.
Now the stabilization level is about ∆P’/q= 0.0065 psi/stb/d.
We know that PD’= 0.5 = KH ∆P’/(141.2 Bμq)
Hence total KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.0065 = 6,516.9 mD.ft
Total KH= (KH)1 + (KH)2 = 4236.0 + (KH)2 = 6,516.9
As a result, (KH)2 = 6,516.9 – 4236.0 = 2,280.9 mD.ft
Using H2=42 feet, then the permeability in the new layer is then K= 2280.9/ 42= 54.3 mD
The assumption of a uniform skin distribution is important here. Otherwise, we could have had a “false” derivative stabilization.

Question 2 of 2
2. Question
Do we have enough evidence to show that both layers are in communication?




Correct
It could be tempting to say that the decrease in the derivative at late times could be due to a crossflow in the reservoir. However, we know that we have some boundaries or heterogeneities in layer 1, which could also create the same derivative feature.
As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.
Incorrect
It could be tempting to say that the decrease in the derivative at late times could be due to a crossflow in the reservoir. However, we know that we have some boundaries or heterogeneities in layer 1, which could also create the same derivative feature.
As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.
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Question 1 of 2
1. Question
Assuming a uniform skin distribution across the layers, what is the permeability of that newly perforated layer?
Correct
The derivative stabilization has shifted downwards as the total mobility has increased.
Now the stabilization level is about ∆P’/q= 0.0065 psi/stb/d.
We know that PD’= 0.5 = KH ∆P’/(141.2 Bμq)
Hence total KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.0065 = 6,516.9 mD.ft
Total KH= (KH)1 + (KH)2 = 4236.0 + (KH)2 = 6,516.9
As a result, (KH)2 = 6,516.9 – 4236.0 = 2,280.9 mD.ft
Using H2=42 feet, then the permeability in the new layer is then K= 2280.9/ 42= 54.3 mD
The assumption of a uniform skin distribution is important here. Otherwise, we could have had a “false” derivative stabilization.Incorrect
The derivative stabilization has shifted downwards as the total mobility has increased.
Now the stabilization level is about ∆P’/q= 0.0065 psi/stb/d.
We know that PD’= 0.5 = KH ∆P’/(141.2 Bμq)
Hence total KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.0065 = 6,516.9 mD.ft
Total KH= (KH)1 + (KH)2 = 4236.0 + (KH)2 = 6,516.9
As a result, (KH)2 = 6,516.9 – 4236.0 = 2,280.9 mD.ft
Using H2=42 feet, then the permeability in the new layer is then K= 2280.9/ 42= 54.3 mD
The assumption of a uniform skin distribution is important here. Otherwise, we could have had a “false” derivative stabilization. 
Question 2 of 2
2. Question
Do we have enough evidence to show that both layers are in communication?
Correct
It could be tempting to say that the decrease in the derivative at late times could be due to a crossflow in the reservoir. However, we know that we have some boundaries or heterogeneities in layer 1, which could also create the same derivative feature.
As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.Incorrect
It could be tempting to say that the decrease in the derivative at late times could be due to a crossflow in the reservoir. However, we know that we have some boundaries or heterogeneities in layer 1, which could also create the same derivative feature.
As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.