HBX-2 Appraisal Well Test

The subsurface team is preparing a well test (DST) for an appraisal well HBX-2 in an offshore oil reservoir. The well is slightly deviated with a drilled net reservoir section length L= 187 feet, as shown in the cross-section below.

Table 1 shows the PVT and static data for this well:

————————————–
Fluid Volume-Factor :1.2000 vol/vol
Fluid Viscosity :0.500E+00 CP
Fluid Compressibility :0.170E-04 1/PSI
water compressibility :0.300E-05 1/PSI
porosity :22.1000 %
water saturation :25.7000 %
rock compressibility :0.400E-05 1/PSI
well-bore 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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1. Question 1 of 6

   1. Question

    

   What is the estimated maximum pressure that could be encountered at surface?

   • perhaps 6,450 psi
   • about 6,250 psi
   • without a doubt: 3,001.8 psi
   • roughly 5,250 psi
   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=6253-3048/0.3048 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=6253-3048/0.3048 x 0.1= 5,253 psi.
   Pressure may be a bit lower than that due to the use of TVDss.
2. Question 2 of 6

   2. Question

   What comment do you have about the above DST string?

   • We should add one more pressure gauge in the middle gauge carrier.
   • We should add another downhole sampler for redundancy.
   • The downhole shut-in tool is too shallow.
   • We should add more pressure gauges along the tubing.
   • We should bring the deepest gauges closer to the perforations.
   Correct
   

   The downhole shut-in 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 shut-in tool, including some memory gauges as back-up.
   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 shut-in 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 shut-in tool, including some memory gauges as back-up.
   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.
3. Question 3 of 6

   3. Question

   What is the value of total compressibility?

   • Ct= Cr + SoCo + SwCw + SgCg = 0.400E-05 + 0.743x 0.170E-04 + 0.257x 0.300E-05 + 0 = 0.174E-04 1/psi
   • Ct= Cr + Co + Cw + Cg= 0.400E-05 + 0.170E-04 + 0.300E-05 = 0.240E-04 1/psi
   • Ct= Cr + SoCo + SwCw + SgCg= 0.400E-05 + 0.643x 0.170E-04 + 0.257x 0.300E-05 + 0.1x 1/6300 = 0.316E-04 1/psi
   • Ct= Cr + Co + Cw + Cg= 0.400E-05 + 0.170E-04 + 0.300E-05 = 0.240E-04 1/bar
   Correct
   

   Ct= Cr + SoCo + SwCw= 0.400E-05 + 0.743x 0.170E-04 + 0.257x 0.300E-05
   Ct= 0.174E-04 1/psi (do not forget the unit)

   Incorrect
   

   Ct= Cr + SoCo + SwCw= 0.400E-05 + 0.743x 0.170E-04 + 0.257x 0.300E-05
   Ct= 0.174E-04 1/psi (do not forget the unit)
4. 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?

   • ∆V= ∆P x Ct = 10 x 0.174E-04 = 1.740E-4 million barrels = 174 bbls of oil to be flared for 174/ 24 = 7.25 hours We could open the well as quick as possible and accelerate the clean-up and main flow periods to reduce the flaring volume.
   • ∆V= ∆P xCt xV = 10 x 0.174E-04 x 10,000,000 = 1740 bbls of oil to be flared for 24 hours. We could challenge further this flaring volume by using Deconvolution.
   • Ct=(1/V)x(∆P/∆V) so ∆V=∆P / (V Ct) hence ∆V= 10 / (10 x 0.174E-04) = 57471.3 bbls of oil to be flared for 57471.3 / 5000 = 11.5 days We could connect the well to an existing infrastructure to challenge and reduce the flaring volume.
   • ∆V= ∆P x Ct x V = 10 x 0.174E-04 x 10,000,000 = 1740 bbls of oil to be flared for 1740 / 5000 = 0.4 days We could reduce the level of depletion or use some other techniques.
   Correct
   

   ∆V= ∆P xCt xV = 10 x 0.174E-04 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 0.174E-04 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.
5. Question 5 of 6

   5. Question

   Is there any particular feature that we could remove from the test sequence?

   • We could remove the clean-up period and first PBU test.
   • We could remove the step rate test as this is an oil well and we do not need to calculate the turbulence factor.
   • We could remove the very first initial flow and shut-in.
   • No, this looks good.
   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 non-hydrocarbon 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 non-hydrocarbon fluids in the wellbore below the gauge.
6. Question 6 of 6

   6. Question

   Why two PBU tests?

   • We should have only one PBU test to save rig time and cost.
   • for redundancy and for Deconvolution.
   • to track the dynamic changes.
   • to refine initial pressure and obtain the connected volume.
   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.

Real-Time Well Testing Operations

The well HBX-2 has now been perforated and we start the clean-up period by gradually increasing the choke size and taking some routine fluid samples at surface.

At the end of the clean-up, the well is closed at the downhole tester valve. As you have access to the real-time data during the operations, the figure below shows the derivative of the first PBU.

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2. Review

1. Question 1 of 17

   1. Question

   
    

   What sort of well and reservoir could we see from this log-log plot? (at this stage, we haven’t acquired the 2nd PBU yet)

   • Well with an increase in mobility and storativity further away in the reservoir
   • Well in a homogeneous multi-layer reservoir
   • Well with limited perforations
   • Well with some improvement in mobility and/or storativity further away in the reservoir
   • Well in a homogeneous reservoir with the presence of boundaries
   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 multi-layer 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 multi-layer reservoir
2. 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?

   • The choke is getting plugged.
   • The choke is getting eroded.
   • This is normal behaviour as we start the well, pressure upstream increases while pressure downstream decreases with the test separator.
   • This is due to some changes in fluid density (gas or multiphase)
   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).
3. 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?

   • ∆P ~ 16psi for a vertical distance of 15m. Hence 16/(15/0.3048)= 0.33 psi/ft
   • ∆P ~ 16psi for a vertical distance of 15m. Hence 16/15= 1.07 psi/ft
   • ∆P ~ 11psi for a vertical distance of 15m. Hence 11/15= 0.73 psi/ft
   • ∆P ~ 11psi for a vertical distance of 15m. Hence 11/(15/0.3048)= 0.22 psi/ft
   Correct
   

   Incorrect
   
4. Question 4 of 17

   4. Question

   Which gauge would you select for well test analysis?

   • We should select the deepest one, 22818 or 22931.
   • We should select 20754 as the pressure difference is less.
   • We should select 22931. We could also have selected 20754 as second choice.
   • We should select 22818. as it shows better quality data.
   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.
5. Question 5 of 17

   5. Question

   After a 12-hour shut-in for PBU 2, the test sequence is presented below:
   

   

   What do the different drawdown data sets suggest about well productivity?

   • There doesn't seem to be any depletion trend between both PBUs.
   • Productivity seems low, with PI ~ 5000/1000 = 5 stb/d/psi
   • There is a clear error in rate.
   • There seems to be an increase in productivity, suggesting a decrease in skin.
   Correct
   

   We are producing roughly at the same rate at the end of the clean-up 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 clean-up 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.
6. Question 6 of 17

   6. Question

    

   The PBU data are shown in the log-log plot below:
   

   

   What observations could we make from this derivative overlay?

   • The DP seems to shift down. The skin is decreasing between PBUs 1 and 2.
   • The derivatives are consistent. PBU 2 confirms the derivative stabilization level and shape of the derivative.
   • The DP seems to shift down. The skin is increasing between PBUs 1 and 2.
   • The changes in derivatives explain the change in productivity.
   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.
7. 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?

   • We should select the deconvolution a) with Pi= 6,241psi as both deconvolved derivatives converge at late times due to the boundaries.
   • We should select the deconvolution b) with Pi= 6,236psi because it doesn't show boundary and the initial pressure is consistent with that in the superposition plot.
   • We should select the deconvolution b) with Pi= 6,236psi. Even if both deconvolution responses are not consistent, it could be due to the short PBU duration.
   • We should calculate the initial pressure from our match with the conventional derivative and then calculate Deconvolution with that pressure value.
   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.
8. 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?

   

   • This means that we have some boundaries at late times.
   • This means that we have enough data from PBU 1 and 2 to derive a reliable Deconvolution response.
   • This means that Deconvolution is not reliable.
   • This is a distortion due to the way the derivative is calculated.
   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.
9. Question 9 of 17

   9. Question

   Do we see a closed reservoir from these dataset?

   • Deconvolution doesn’t follow a unit-slope straight line at late times, so the reservoir is not closed.
   • Deconvolution seems to follow a unit-slope straight line at late times, so the reservoir is closed.
   • The derivative and superposition plots do not show the presence of boundaries so the reservoir is not closed.
   • We cannot tell at this moment, we do not have enough data.
   Correct
   

   Deconvolution doesn’t follow a unit-slope straight line at late times, so the reservoir is not closed.

   Incorrect
   

   Deconvolution doesn’t follow a unit-slope straight line at late times, so the reservoir is not closed.
10. Question 10 of 17

    10. Question

    Is the initial pressure from Deconvolution consistent with the MDT data?

    • Yes.
    • No.
    • Maybe.
    • I don't know.
    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 – (3048-3035.7)*1.07 ~ 6,240 psi
    Or P=6253- (3048-3035.7)/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 – (3048-3035.7)*1.07 ~ 6,240 psi
    Or P=6253- (3048-3035.7)/0.3048 x 0.325 ~ 6,240 psi.
    The initial pressure from Deconvolution is consistent with the MDT data.
11. 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?

    • The initial pressure may not be correct.
    • Deconvolution cannot be trusted here. Deconvolution is not "real", it is calculated and as a result it can be wrong.
    • This is an error in the conventional derivative due to the way it is calculated.
    • The two conventional derivatives from PBUs 1 and 2 are consistent as we saw earlier. The fact that Deconvolution is not consistent with the other derivatives means that Deconvolution is not correct.
    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.
12. Question 12 of 17

    12. Question

    What are the possible interpretation models?

    • A well with wellbore storage and skin in a homogeneous reservoir with the presence of boundaries.
    • A well with wellbore storage and skin in a heterogeneous reservoir with a decrease in mobility/storativity further away from the well.
    • A well with wellbore storage and skin in a heterogeneous reservoir with an increase in mobility/storativity further away from the well.
    • A well with limited perforations/partial penetration in a homogeneous reservoir of infinite extent.
    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.
13. Question 13 of 17

    13. Question

    Based on the static data in table 1, what is the permeability value?

    • KH= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ft The permeability is then K= 4236/ 162= 26.1 mD
    • K= 0.5 x 141.2 x 1.2 x 0.5 / 0.01 = 4,236.0 mD.ft The permeability is then K= 4236/ 187= 22.7 mD
    • K= 0.5 x 141.2 x 1.2 / 0.01 = 8,472.0 mD.ft The permeability is then K= 4236/ 187= 45.3 mD
    • From the reservoir engineer on the rig, the radius of investigation Ri is about 400ft. We know that Ri=0.029.[K.Dt/(Phi u Ct)]^0.5 K=3306 x 0.22 x 0.5 /10 = 36.3 mD
    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
14. 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?

    • Deconvolution seems to work and is developed over a longer time interval. We can terminate the test and move on.
    • We need to extend the PBU test to see if we have a closed reservoir.
    • We need to extend the PBU test to prove more volume.
    • We need to terminate the test as we have reached the test objectives and there is no more value here.
    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 pore-volume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
    64212000 x 0.178 / 1.2 x (1-0.257) = 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 pore-volume of 64,212,000 ft3. This converts to a minimum connected oil volume at surface of
    64212000 x 0.178 / 1.2 x (1-0.257) = 7,076,912 bbls.
    We should therefore extend PBU 2.
15. Question 15 of 17

    15. Question

     

    Unfortunately there was an operation problem at surface and the well had to be left shut-in 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?
    

    • No, this is a distortion due to the way the conventional derivative is calculated.
    • No, a U-shape feature in the derivative is a sign of heterogeneities.
    • Yes, a U-shape derivative could a sign of crossflow from a low permeability layer.
    • Yes, a U-shape derivative could a sign of crossflow from a high permeability layer away from the perforations.
    Correct
    

    The U-shape 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 U-shape 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.
16. Question 16 of 17

    16. Question

     

    A unit-slope straight line was placed on the final point of the deconvolved derivative.

    

    What is the minimum connected volume at surface, after a 48-hr PBU2 test?
    (hint: bbl =ft³ x 0.178)

    • PV x 0.178 x (1-Swi) / Bo = 91213000 x 0.178 x (1-0.257)/1.2 = 10,052,737 bbls
    • PV / 0.178 x (1-Swi) / Bo = 91213000 / 0.178 x (1-0.257)/1.2 = 317,281,175 bbls
    • PV x 0.178 x (1-Swi) / Bo = 91213000 x 0.178 x (1-0.257)/(1/1.2) = 14,475,941 bbls
    • PV / 0.178 / Bo = 91213000 / 0.178 /1.2 = 427,027,154 bbls
    Correct
    

    PV x 0.178 x (1-Swi) / Bo = 91213000 x 0.178 x (1-0.257)/1.2 = 10,052,737 bbls

    Incorrect
    

    PV x 0.178 x (1-Swi) / Bo = 91213000 x 0.178 x (1-0.257)/1.2 = 10,052,737 bbls
17. Question 17 of 17

    17. Question

    Should we leave the well shut-in for longer?

    • No, we can terminate the test as we reached the subsurface objectives.
    • No, we can terminate the test as we obtained permeability and skin.
    • We should leave the well shut-in for longer to obtain more value.
    • We should continue the test to learn more about the boundaries.
    Correct
    

    No, we shouldn’t leave the well shut-in for longer. We can terminate the test as we reached the subsurface objectives.

    Incorrect
    

    No, we shouldn’t leave the well shut-in for longer. We can terminate the test as we reached the subsurface objectives.

Post-Operations 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]

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1. Question 1 of 2

   1. Question

   What is she suggesting? Is this correct?

   • Linear composite reservoir with 2 regions in the reservoir, with an increase in mobility/storativity further away from the well and a boundary at about 420ft. All the plots are matched but this doesn't seem to be a correct interpretation model.
   • Linear composite reservoir with 2 regions in the reservoir, with an increase in mobility/storativity further away from the well and a boundary at about 420ft. All the plots are matched and this seems to be a correct interpretation model.
   • Linear composite reservoir with 2 regions in the reservoir, with a decrease in mobility/storativity further away from the well and a boundary at about 420ft. All the plots are matched but this doesn't seem to be a correct interpretation model.
   • Radial composite reservoir with 2 regions in the reservoir, with an increase in mobility/storativity further away from the well and a boundary at about 420ft. All the plots are matched and this seems to be a correct interpretation model.
   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 U-shape 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 U-shape derivative at late times. This is not correct, and this is why the initial pressure is a bit too low too.
2. Question 2 of 2

   2. Question

   Do you agree with her KH and permeability values?

   • The permeability-thickness value KH is correct but not permeability.
   • The permeability K value is correct but not KH.
   • The permeability-thickness KH and permeability values are correct.
   • Her KH and permeability are not correct.
   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 shut-in (PBU 3). The result is shown below on the log-log plot.

(Click on the plot to enlarge it).

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   1. Question

   Assuming a uniform skin distribution across the layers, what is the permeability of that newly perforated layer?

   • KH= 2,280.9 mD.ft Using 42 feet, then the permeability in the new layer is then K= 2280.9/ 42= 54.3 mD
   • KH= 4,434.4 mD.ft Using 42 feet, then the permeability in the new layer is then K= 4434.4/ 42= 105.6 mD
   • KH= 5429.8 mD.ft Using 42 feet, then the permeability in the new layer is then K= 5429.8/ 42= 129.3 mD
   • From a derivative stabilization level at about 6.5E-03 psi/stb, we obtain KH = 6,516.9 mD.ft Using 42 feet, then the permeability in the new layer is then K= 5429.8/ 42= 155.2 mD
   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.
2. Question 2 of 2

   2. Question

   Do we have enough evidence to show that both layers are in communication?

   • We noticed before that the declining trend in the derivative at late times could be due to some boundaries/heterogeneities. As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.
   • We do not see a U-shaped feature in the derivative. As a result, from using only these data, we do not have enough evidence to show that both layers are in communication.
   • The declining trend in the derivative at late times could be a crossflow from the top layer, whose permeability is not negligible. We shouldn't expect to see a U-shaped derivative feature unless the contrast in KH is significant (kappa tending towards 1). As a result, we have some evidence to show that both layers are in communication.
   • The derivative stabilization has shifted downwards, increasing the KH that the well connects to. As a result, we have some 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.

well testing in an appraisal well is divided into 3 sections: well test design, real-time well testing during operations and post-ops well test interpretation

Fracture behavior