Separators: 2021 #1

A separator may contain which of the following to aid in the separating of the gas, oil, and water streams (select any that apply):
__ Inlet diverters
__ Outlet vortex breakers
__ MAWP limit switch
__ Vapor pressure overflow drum
__ Turbo-expander system
__ Sweetening tube (low-psi)
__ Buckets
__ Weirs
__ Mist extractors

Click the button for the answer, with commentary and SPE Reference Guide pages sourced. Feel free to ask questions.

A separator may contain which of the following to aid in the separating of the gas, oil, and water streams (select any that apply):
 X  Inlet diverters
 X  Outlet vortex breakers
__ MAWP limit switch
__ Vapor pressure overflow drum
__ Turbo-expander system
__ Sweetening tube (low-psi)
 X  Buckets
 X  Weirs
 X  Mist extractors

Note that this "multi-choice" format (appearing on the new EBT exam last year) can make a problem extremely difficult, as this one is. The new practice problem sets will have quite a few of these, more than the real exam for sure, merely because it's a good way to test knowledge.

Source HS3 P3: The first step in the process is separating the gas from the liquid and the water from the oil. This is usually done in a separator - a pressure vessel into which the well-stream flows to allow the gas, oil, and water to separate because of gravity. Separators may contain inlet diverters, outlet vortex breakers, buckets, weirs, and mist extractors to aid separating the streams.

2020 PE Exam Comments

 2020 is history! Comments - especially suggestions for blog/Guidebook/Companion improvements - are welcome and appreciated. I enjoy hearing from all.

Please remember the blog rule: prior PE Exam questions, in whole or in part, will NOT be discussed. General topics, resource suggestions, and testing techniques only. Please don't "cross the line" by discussing specific problems from prior exams. Comments like: "...several of the drilling questions with probability...” is crossing the line. Thanks, folk!

There will be a delay between comment submission and when it appears; please be patient.

UPDATE 1: It looks like most people have their results back. Congrats to everyone who passed; you've earned it! There seems general agreement 2020 was more difficult than the 2019 CBT format exam...but it also seems (at first glance) that repeat-testers passed in larger numbers than is usual. So for all of you who did not pass, remember that repeat test takers do have a huge advantage and I encourage you to give it another try.

UPDATE 2: If anyone here did not pass, I'm going to re-vamp the Guidebook and problem sets for the 2021 CBT and I'm looking for a few guys to run problems by. This would probably benefit you for practice and help me in editing. If interested, reach me at mdavidgo on gmail.

 UPDATE 3: For anyone who got bumped to the backup test in January (due to the testing facilities needing more space due to Covid) please comment here when you get your results and mention you took the backup test. Thanks!

Pipe Stretch: 2016 #61

Problem 61: A 1.05 inch tubing string parted at 10,000 ft. When fished with 2-3/8", 4.85 ppf drillpipe, 50" of stretch is measured when 10,000 pounds of tension is applied. Fish length to stuck point?

This problem is solved in the Guidebook 6 DTC 10. The needed equation in the CBT SPE Reference Guide is on page 116.

Steps:

1) Solve for the stretch of the known length of drillpipe.

2) Subtract calculated DP stretch from the total stretch to find the stretch attributed the fish. 

3) Calculate fish length.

Keep in mind this sort of problem could ask you to calculate any step when given the required inputs.

Don't worry about using the Redbook or other similar references when doing these practice problems. On the CBT they could just provide what you need in the problem, most likely in similar format to the SPE material.

Reservoir Volume: 2016 #54

Problem 54. Isolated volumetric reservoir with single new well flows at 100 BOPD until pseudosteady-state reached/maintained & BHP falls 130 psi over 100 hours. Reservoir 0.00001 1/psi & oil FVF 1.3 bbl/STB. Reservoir volume estimate (MMRB): (A) 4.2 (B) Not enough info. (C) 0.42 (D) 2.1.

Same problem shown on 12 WLT 3.  Just watch units:

[0.234(100)1.3]/[1E-5(130/100)] = 2,340 Mcf/5.615 cf/bbl = 416,741 RB (C).

Temperature Force: 2016 #52

3.5 inch, 9.2 lb/ft tubing: 

X-section area = 2.59 sq in (Baker Packer Handbook, or Redbook w/ OD & ID calc.). On the CBT exam, these dimensions would be given in a text box, or even provided in a tricky way as part of a table where you have to be familiar with tubing.

1) ST = 50 dF; BHT = 150 dF: Average = 100 dF.

2) ST = 75 dF; BHT = 75 dF: Average = 75 dF. 

dT = 75 - 100 = - 25 dF.

Force applied by dT: 207(Ats)dT = 207(2.59)-25 = - 13,403 lb tension up. (A)

Hydrates: 2016 #49

A 0.7 SG NG mix at 40 dF can raise to what pressure without anticipating hydrates?

165 dF; see table on 8 FAC 1hydrate tables can be found anywhere (Katz, McCain, etc.) and the Guidebook updates have an entire page on hydrates word problem discussion.

Note that on the CBT you would likely be provided with charts or tables for this sort of problem. The Guidebook uses tables, and I think this would be the most likely approach (other than a word problem, in which the Guidebook summarizes the many approaches).

But I would definitely do practice problems because it's confusing when hydrate problems are worded to make them difficult to understand. No matter what, I would  at least look over the Handbook Series on hydrates using the Guidebook summary as the most relevant time-saving overview I know of. Hydrates are a very big and complicated subject but it is highly unlikely you would see something esoteric on the PE exam and thus I would stick with the summary and the practice problems.

Gas Choke: 2016 #48

A 0.70 SG natural gas mix expands through a 1/2 inch choke. Upstream pressure & temperature are 500 psig & 140 degF. Flow rate is closest to (Mscf/day): (A) 2,500; (B) 2,700; (C) 2,900; (D) 3,100.

8 FAC 2 has the required formula, including the Rawlins constant options where I need z for this gas mix. This takes time and has lots of room for error. So I just solve the problem quickly with z = 1, giving an answer 0 2,830, or (C).

A note in the Guidebook says this sort of estimate might be 10% low (i.e., using z = 1 if z is actually 0.8). So I quickly add 300 for 3,100. Looking over the answer key, that's moving into (D) territory and thus too close to "estimate". So I run to 9 PVT 2 and lock down my true z using the chart & SG 0.7, Ppr=515/663= 0.78,  & Ppr = 600/378 = 1.59. This charts out a  z = 0.94, very close to my original guess. When I use this z I calculate q = 2,922, or (C), like my original calculation.

Were I in a bind for time on this problem I would have just ran with my original calculation with z = 1 and got the right answer. But this is risky and I would do this only if under time constraints, or if the answer key was so broad-spaced I knew could get away with.

Look, knowing what you can get away with, and taking calculated (heh) risks is a big deal on any exam.Get used to life not being perfect and learn to role with what you are given.

Also: on the CBT questions regarding equations in the SPE books not be in the provided reference may focus more "understanding" rather than plug-and-chug. In this example knowing about how much z accuracy matters is just the kind of question I might ask were I writing an CBT question to get everyone flipping through their resources in vain to look for an equation that is in the SPE text but not in the approved reference.. Look, if it's in an SPE book, the concept and understanding are fair game, even if you are not expected to memorize the equation. Personally I see no other way to study for this situation except to continue to do  problems that use resources that won't be provided on the EBT. Focus on learning and understanding, not memorizing things, and never assume just because it's not in the provided resource the concept is off limits.

Rod Pump: 2016 #46

A 160-173-54 PU. 16 SPM. 200 BFPD. Rods have no significant stretch. 1.5 inch diameter pump. Anchored. Fluid SG 0.9 at 4,500 ft. Lift system volumetric efficiency?

The PD equation is on 7 PRD 9 (note we can use S for Sp since there is assumed no rod stretch):
Pump Displacement (ideal) = 0.1166(54 in)(16 spm)(1.5^2 in^2) = 227 BFPD
Pump Displacement (actual) = 200 BFPD 200/227 = 88% (B) 

The provided Reference Guide can only hurt you on this problem; any time spent looking in chapter 4 (rod pumps, pg 124-131) is only wasted time. IMO, this is a big weakness for this reference; what good is a PE reference without the rod pump displacement equation? UPDATE: the SPE Reference Guide does have 9702.03 ci/bbl under Constants. and a % Clearance Volume for a Single-acting Cylinder equation (page 151 under Facilities) that can be manipulated so as to be close to the PD equation. Note 9702.03/127 is 0.1169 not 0.1166, but 1% error is not going to matter.

Regardless: know all terms and how the pump designation works (HS V4 pg 489) plus the 0.1166 volumetric conversion constant (HS V4 pg 471). Note you cannot use a direct volume conversion (cubic inches/bbl & min/day) as the number will be too large. 0.1166 is an easy number to memorize though, and the rest is just pump volume.

Fluid Gradient: 2016 #35

Problem 35. A well produces 100 MSCF of gas with a SG of 0.8 along with 200 STB of 30 API oil. The oil FVF is 1.4 RB/STB. The 200 dF reservoir has a fluid gradient closest to: (A) 0.28; (B) 0.30; (C) 0.32; (D) There is not enough information given to solve the problem.

This is the example problem in the Guidebook on 13 RES 9 in the latest edition, and solves to (B). It's good to think about all the different ways this problem can be written so you won't be blindsided.

In this problem, solve for unitless Rs (use 5.615 SCF/STB constant) to find the density of both the oil & gas (from the given API & SGg) using lbm/SCF. Then find Bo from Standing's equation. I don't think the provided Reference includes Standing's equation or the API/SG conversion, but don't let that lull you into complacency; the exam could just provide the equation with a bunch of other stuff to confuse you. Note that SPE TS8 by Towler has this sort of problem, so it's absolutely fair game.

The constant needed for this problem, however, is in the new Reference Guide (provided on the exam) on page 190 of chapter 7 under volume, but you probably have this conversion memorized.

IMO this multi-step problem is kinda amazing inasmuch as some guy using oil/gas from the separator can calculate his reservoir gradient 10,000 ft below his feet with just the FVF.

DCA: 2016 #28

Problem 28: An oil well was drilled but facility limited to 1,000 STB/D. After 2.5 years it started to decline at 7% per month. Production during the third year (MSTB)? (A) 310; (B) 245; (C) 65; (D) 975.

DCA Decline Curves: 1 Exponential; 2 Hyperbolic; 3 Harmonic. In the Guidebook use 11 DCA 1. In the new Reference Guide (provided on the exam) use pages 29-31 of chapter 1. Practice finding these equations quickly. Using the new reference:

a) Convert to Nominal decline: Ln(1-0.07)=-.07257/mo. 
b) q = qi/exp(d*t) so 1000*(EXP(-0.07257*6)) = 647 BPD (rate after 6 months decline).
c) Np=[(qi-q)/D]*c = [(1000-647.0)/0.07257]*30.4 = 147,874 bbl production second half year 3.
d) 1000*182.5 = 182,500 BO (production first half year 3).
e) 147,874+182,500 = 330,374 BO total production year 3 or (A).

Practice DCA problems on the SPE 2005 Exam, this blog, and the 2016, 2017, and 2018 problem sets until you can do them fast and without error. You really can't practice these enough.

Flow Test: 2016 #22

A flow test...cylindrical reservoir...estimated porosity 0.2 – 0.23, viscosity 0.5 – 0.55 cp, permeability 100 – 110 md. Total compressibility 0.00002 1/psi. How long...to ensure...test reaches 2,000 ft?

See 12 WLT 6*. To "ensure" 2,000 ft, choose the most conservative estimates for the longest time.  Viscosity & porosity directly correlate, so higher numbers mean longer time for them. But permeability is inversely correlated, so use the smaller number. Using the radius of investigation equation (12 WLT 6) we calculate:

t = 948(0.23)0.55(0.00002)(2000^2)/100 = 95.9 (D).

The info on 12 WLT 6 needed for this problem is in the new Reference Guide (provided with the exam) on page 7 of chapter 1. Practice finding these equations quickly.

Displacement: 2016 #20

5,000' of 14.00 lb/ft drillpipe (adjusted weight is 1.1 times nominal) and 100 ft of 120 lb/ft drill collars are pulled dry. Fluid volume change?

14 ppf(1.1)5,000' =  77M lb.
120 ppf(100') = 12M lb.
77M+12M lb = 89 M lb/(490 lb/cf*5.615 cf/bbl) = 32 bbl (A).

The trick on this one is to use adjusted weight if given; never use OD & ID unless you don't have adjusted weight. In that case use the adjusted weight / density equation (see 1 RIG 1). Regardless, know how to do this sort of problem fast.

During the exam, locate the needed constants of 490 lbm/cf & 5.615 cf/bbl are in the new Reference Guide (provided with the exam) on page 190 of chapter 7. Practice finding these numbers quickly.

API Equation: 2016 #14

 Problem 14. An empty N180 pipe has 40,000 psi tension applied to it. The reduced collapse rating is closest to what percent of pre-tension rating? A) 0%, B) 45%, C) 55%, D) 65%.

This is straight out of the Guidebook on 6 DTC 3; the collapse pressure rating is reduced to 65.1%. 

Note the equation in the new  Reference Guide (provided with the exam) is on page 55 of Chapter 2. One really has to get used to this reference, since it's not very intuitive where one would expect to find it. You can waste a lot of time looking for it if you don't have it very well mapped out before the exam.

Note: it was pointed out to me today by a steely-eyed engineer that the Kindle problem asks for "percent reduction" making the answer 1-0.65 = 35% or (B), not (D). I actually like this mistake, since it gets one thinking about how clever and tricky the wording on these problems may be...one can't be too careful here. Read, read again, and then read once more...

Economics: 2016 #10

Problem 10. An investor buys a lease for a 12.5% landowner royalty. An engineer is retained to work the lease for a 5.5% overriding royalty interest. The first investor sells...but keeps a 12% overriding royalty interest with a 25% back-in working interest...net revenue ownership interest for the engineer after payout is closest to: (A) 7.5%; (B) 3.75%; (C) 4.75%; (D) 5.5%

This one is easy; the engineer’s share is an "overriding royalty interest" and thus not effected by anything after his initial 5.5%, hence, 5.5%. See 10 ECN 2 for the definition of ORRI and a more complex example that includes this part.

Directional Drilling: 2016 #4

What is the MD of the Problem #3 well if at a TVD of 6,000 ft drops at 2 deg/100 ft until vertical?

This is easy since in #3 build was 895 ft (MD) at 3 deg/100 ft. If the well then drops at the same rate, we merely add this MD to the sequence: 6,198 + 895 = 7,093 ft (C).

Directional Drilling: 2016 #3

What is the MD at 6,000 ft TVD for a build, hold, and drop well (KOP, TVD & HD at end of hold, and build/drop rates will all be given).

This is a standard directional drilling problem; see Guidebook on 2 DRL 9 or TS2 (note TS12 doesn't have a good section here). First calculate r1 & r2 (2,865 ft); next calculate Theta1 & Theta2 (17.9 deg). Finally, use the MD equation and find the MD of each component of the vertical, build, hold sections:

MD = 1,500 + 895 + 3,803 = 6,198 ft or (C).

Note these equations are all in the GB or TS2; just plug-n-chug through them.

Cement Displacment: 2016 #2

Given: 10,000' casing in 10.7 ppg mud.
Pump into casing, up annulus:
100' 8.5 ppg mud = 44 psi hydrostatic.
1,550' 12.9 ppg cement = 1,040 psi hydrostatic.
1,500' 15.4 ppg cement = 1,201 psi hydrostatic.
9.1 ppg brine until cement is displaced.
Pump pressure required to displace?

Annulus (remaining):
6,850' 10.7 ppg mud = 3,811 psi hydrostatic.
Total hydrostatic pressure in annulus: 6,096 psi.

Hydrostatic pressure in casing: 9.1 ppg brine = 4,732 psi

Pump pressure required to displace?
6,096 psi - 4,732 psi = 1,364 psi (A).

This is a simple displacement problem. It's common in industry and textbooks both (good examples are in TS2, TS12, Mian, etc...and if you have a chance to take Wines' class, Bing has the best examples of this problem type I've seen). I referenced this problem using TS12 but as mentioned there are a dozen just like it. All you need is the basic hydrostatic equation (0.052(D)MW) and to understand how cement is generally pumped:

Pipeline: 2017 #79

Problem 79. A 26 mile, 18 inches internal diameter pipeline in good condition is flowing methane...50 dF & 500 psig and falls to 100 psig at the sales point; water content is 6 lbm water per MMSCF. Sales (MBOE/D) is best estimated to be: (A) 29. (B) 34. (C) 39. (D) 44.
Note: Older versions erroneously had "millions" in the answer wording (corrected here). 

This problem is a Panhandle problem as per 8 FAC 1.

Methane has a SG of 0.56 and a z of ~ 0.9 for these conditions. The water vapor of 6 lbm/MMSCF is small (sales quality for many places) so should not overly effect the Panhandle equation result of ~240 MMSCF of sales quality gas. Next, we divide by 6 for ~40 MBOE. Again, the water content is sales quality so the BOE calculation should hold within the answer the answer key margins.

NPV, dNCF, PVR: 2017 #76

A tool was purchased at the start of 2011 for $240M. It returned $65M each year...sold for half price end of 2016. 6% discount rate. NPV, dNCF, & PVR are closest to ($M)?

Solution:
NPV: -240+61+58+55+51+49+46+85 = 165 (A)
dNCF: 153.8+240 = 404
PVR: 164/240 = 0.68.

Since 2/3 of the choices are closest to A, we select (A). Source: 10 ECN 1.

Dipping Sand: 2017 #75

Problem 75. A company in the Gulf drills into the top of a trapped sand at 5,162 ft using 9.2 lb/gal mud. This sand is known to dip from 5,162 ft to 9,036 ft, and to be filled with a 0.81 lb/gal gas down to the GWC at about 6,160 ft. The mud weight situation upon entering this sand is closest to:

(A) 350 psi overbalanced
(B) 350 psi underbalanced
(C) 150 psi underbalanced
(D) 150 psi overbalanced


Note: early editions of this problem's answer options were wrong. If your exam shows something other than above for A-D, use this. I'll have an update available to download soon. Solution: see 2 DRL 1.

GWC psi = 0.465(6160) = 2864 psi (pressure 98' below bit using GOM gradient 2 DRL 3).
Subtract gas hydrostatic: 0.052(0.81)998 = 42 psi from GWC psi: 2864 – 42 = 2822 psi (this is psi the bit sees when drilled into top of gas).
Finally, mud weight: 0.052(5162)9.2=2470 psi (MW hydrostatic at depth bit is at).
Balance: 2470-2822= 352 underbalanced (B).