Wellbore Storage: 2018 #28

Problem 28. Well: Perfs: at 2,630 ft. Reservoir:10 ft thick; k = 40 md; skin = 2. Tbg: 4.5 inch, 12.6 lb/ft. Fluid vis & compressibility 1 cp & 0.0005/psi. PBU wellbore storage time (hr): (A) 5.5; (B) 18.5; (C) 21.5; (D) 11.5?

Solve using Guidebook 12 WLT 13.
Wellbore V = 2,630 ft (0.0152 bbl/ft) = 40 bbl.
Cs CwbVwb = 0.0005 1/psi(40 bbl) = 0.02 bbl/psi
t (wbs) = [170M(Cs)(e^(0.14*s))]/[kh/u]
t (wbs) = [170M(0.02)(e^(0.14*2))]/[(40*10)/1] = ~11.5 hr.

Hydrostatic Pressure: 2018 #27

Problem 27. 4,200 psig reservoir; 9,000 ft TVD. 100 bbl of mud is in 5.5 inch 26 lb/ft casing; 0.1 psi/ft gas cushion.

The required height of the gas (ft), mud weight (lb/gal), and surface pressure (psig) for 200 psi underbalance are nearest to, respectively:

(A) 4,100; 10.1; 970
(B) 3,900; 10.3; 950
(C) 4,000; 10.2; 1,100
(D) 4,000; 10.0; 1,010

This one is pretty simple if you don't get cross-threaded. Steps:
1. 100 bbl/0.02 bbl/ft = 5,000 ft of mud.
2. 9,000-5,000 = 4,000 height of gas (this eliminates A & B).
3. 4,000(.1) = 400 psi (gas)
4. Pr = 4,200 psi (reservoir) – 200 psi (underbalance) = 4,000 psi

Using the given mud weights of 10.2 & 10.0, the hydrostatic calcs sum to a surface pressures of 4,152 psi for (C), and 4,010 psi for (D). Since D is only +10, it's the answer.

Cementation Factor: 2018 #26

A 20 ft...homogeneous reservoir has a porosity of 10%...core is taken...(see problem)...cementation factor?  (A) 1.95; (B) 2.00; (C) 2.05 ; (D) 2.10

Solution: 15 LOG 1
1. F = Ro/Rw = 42/.42 = 100
2. F = a/por^m = 0.9/0.10^m
3. (m)log(por) = log(a/F)
4. m = log(a/F)/log(por)
5. m = log(0.9/100)/log(.1)
6. m = -2.05/-1 = 2.05 (C).

Combo Stress: 2018 #25

Problem 25. A 6 in., N-80 23 lb/ft casing has 20M psi of internal pressure. If 20M psi of total axial stress is applied to the casing, the annulus pressure (Mpsi) closest to the pipe's collapse rating is: (A) 22.8; (B) 28.8; (C) 25.8; (D) 31.8.

This is a standard combo stress problem; 6 DTC 4 has a cheat sheet that is faster to use than the ellipse of plasticity, and this page walks you through the calculation as well:

1) (σz + pi)/σyield = (40M + 20M)/80M = 0.75
2) Chart: --> 0.75 ---> -0.385 = (pi - pcrr)/pcr (note negative sign for collapse)
3) pcrr = pi - (-0.385)pcr) = pi + 0.385(pcr) = 20M + 0.385(pcr)
4) Redbook: 6 in. N-80 collapse rating of 7.180M psi.
5) pcrr = pi - (-0.385)pcr) = pi + 0.385(7,180) = 20M + 2.764M = 22.8 M (C).

Reservoir Simulation: 2018 #20

Problem 20. The top 10 golden rules for reservoir simulation studies does NOT include (per Aziz):

(A) Question Data Adjustments for History Matching.
(B) Smooth Extremes.
(C) Keep It Simple.
(D) Do Trust Your Judgement.

This solution if found in the Guidebook on 13 RES 11 (as well as TS7 P360). The answer is of course (B); "don't" smooth extremes regarding reservoir simulation. Note the "golden rules" have  a predominate section in TS7.

Drill Collar Compression: 2018 #18

Problem 18. Drilling...760 ft 2-1/4 x 7-3/4 DC...12.3 ppg mud...50M lb WOB. 50 deg hole...friction factor negligible...DC compression (%)?

Solution:
1. 12.3 ppg is 0.812 BF (Redbook S240 P13).
2. 2.25”x7.75” DC is 147 ppf (Guidebook 1 RIG 8, 2 DRL 6).
3. 50M/[760 ft(147 lb/ft)0.812(cos50 deg)] = 0.86 or (A).

Note extra information was removed for clarity; it's not as easy at it looks in this blog post; extra information is a key confusion-creator on a timed exam, and this problem is no exception.

Cement Plug: 2018 #12

Problem 12. ...balanced cement plug 50 ft above and below perf...10% excess...volume of displacement...: (A) 30, (B) 31, (C) 32, (D) 33 bbl.

First, note what is asked: displacement volume (that is, mud volume behind plug). Nothing more.
Next, organize data & your mind: 7" CSG, plug 3,500 - 4,000 ft. Use 3.5" TBG to pump.
Run your eye down the Guidebook cover to "Cement". "Plug" is on pg 5 CMT 6. It has 5 steps:

1) Redbook for CSG, ANN, TBG cf/ft (0.22, 0.154 & 0.0488). Note on the CBT the Redbook data would merely be given on the problem. Don't think just because these resources are not available to bring to the exam they won't need to be "looked up" on tables. In fact, I could imagine the tables being all grainy and hard to read.
2) Vplug slurry calc: 500 ft plug(0.22 cf/ft) = 110 cf.
3) Plug height w/ TBG in hole: Vplug/(Cann + Ctbg) = 110 cf/(0.154 + 0.0488 cf/ft) = 540 ft.
4) Displacement height: 4,000 - 540 = 3,460 ft (total - plug height w/ TBG in hole = mud height).
5) Displacement volume: 3,460(0.0488 cf/ft) = 169 cf = 30 bbl. Add 10% excess: 3 bbl for 33 bbl.

This is quick if you follow the 5 steps. But beware: there are a million ways to go wrong, both in the question wording and calculations! I just whipped this one out randomly in under 6 minutes like everyone else must on an exam so it wouldn't surprise me if I had an error lurking. Get used to doing these problems fast: the pace you practice is how you will test, and you won't get every one correct.

CBT

Below is a summary of an excellent blog comment discussing the CBT.

1. The only thing you can bring into the room is an approved calculator (no cover/case) and the clothes you wear. No pens/paper/food/drinks/keys. They pat you down, scan your glasses, and empty pockets (they may have spare approved calculator for loan).

2. The reference book is on the screen beside the exam. It has search capability. The errors were still in it but they have no impact as everything needed (and more) was usually given in the problem making the reference rarely needed.

3. A 7 page dry erase tablet with gridlines and a marker for plotting is provided. No eraser. You may ask for another tablet if you fill it up. 

Basically, it's as predicted: everything needed is usually in the problem itself making the provided reference fairly irrelevant (and even a distraction). A good study plan should therefore focus on 1) learning the material in the SPE Handbook, and 2) practicing problems for speed. Avoid getting bogged down with a study plan focusing on looking things up. You will either know it or not.

2019 PE Exam Comments

2019 is history! Comments - especially suggestions for blog/Guidebook/Companion improvements - are most welcome and appreciated. I would enjoy hearing from anyone and everyone.

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

UPDATE 1: There is a delay between comment submission and when it appears; please be patient. Don't hesitate to gmail me at mdavidgo with any issues.

UPDATE 2: I am being extremely conservative in rejecting comments. So if your comment doesn't appear after 24 hours, it's likely this is why. Please understand there is nothing technically "wrong" with these comments, I just don't want there to be any possible question. So if this happens to your comment please re-work it and resubmit. Your suggestions are valuable, so I hope you re-post!

UPDATE 3: Since this is the first year of the CBT, it would be nice if anyone who has taken the previous style of exam and the CBT both to weigh in, especially suggested changes to the Guidebook and companion problems.

MBE Drive Indices: 2018 #10

Problem 10. Material balance equations are used to derive indices to compare the relative strengths of drive mechanisms. The statement regarding these indices most FALSE is: 

(A) Drive indices definitions are subjective and arbitrary. 
(B) Water drive indices must use...
(C) The Sills drive indices attempted to define the source...
(D) When comparing Pirson and Sills drive indices...

The  solution to this problem (or any indices problem you will likely see) are found on 13 RES 5 of the Guidebook, where I source both Slider and Towler. The key point? If  (A) is true, this makes (B) obviously false; how can one thing be "subjective and arbitrary" yet require something else?

Towler explains this well in SPE TS8 P48. Having read quite a few reservoir texts over the years, be warned that while SPE TS8 is not popular it is the "official" SPE reservoir reference, so it's good to know for these kinds of tricky word problems. However, now extra resources are not allowed it's important to learn the material rather than just locate it for future use.
 

Welge's Graphical Method: 2018 #9

Problem 9. The following statement most TRUE concerning modifying the saturation distribution with Welge’s graphical method is:

(A) For gas displacement, the tangent to the fractional flow curve is drawn from the irreducible water saturation.
(B) In actual fact, the saturation front tends to be smeared by gravity at the trailing edge and smeared by capillary action at the leading edge.
(C) For waterdrive reservoirs, a line from the irreducible-water saturation tangent to the fractional-flow curve is drawn; the saturation at the tangent is the flood trailing edge.
(D) For waterdrive reservoirs, a line drawn from the irreducible-water saturation tangent to the fractional-flow curve may be extended to an oil saturation of 1 to find the average water saturation at breakthrough.

This problem is a confusing word garbage dump; it looks like it's written by a lawyer. Be ready for this sort of thing.

So I don't read the details yet. Once I know the theme (Welge) I just run to the Guidebook TOC and scan for Welge. It's on 14 WFL 4. At 10 seconds, I've eliminated (A), since it says that for gas reservoirs you start the line at fw = 0. In the next 30 seconds, (C) and (D) are as easy to eliminate just by looking at the graph and sorting things out.

So by process of elimination I'm selecting (B). And if you understand this whole capillary action, leading edge, and saturation front business, you can see how it makes sense. Need more confirmation? I prefer it. There must be an SPE direct quote here somewhere...

The Handbook Series? I check the index, and I get 1106-1108. Sorting through this mess I find nothing helpful, but do finally find Welge mentioned on page 1108 without any clue for this problem. So I cut my losses and move on.

The Textbook Series? It's a waterflood question so I check TS3 index and find Welge on page 122. But again, this page merely refers me to another section, and neither section gets answers. Ouch.

But it's also a reservoir question. So I check TS8, which has an index entry for Welge with two reference pages (161-162). After several minutes of scanning, I find (B) is a direct quote from page 161! Bingo. As I mentioned last post: be warned that while SPE TS8 is not popular, it is the "official" SPE reservoir reference, and it's got some gems that few other texts have, like hydrodynamic traps and Welge. Of course, by now I'm 5+ minutes down (at least).

The Guidebook would have got you through this one in a minute or less. My general approach? Scan the Guidebook TOC (the cover) and review the relevant page. If I strike out here and have time later on I move to the HS index. Finally, I'm in the TS series, checking books I think relevant.

Honestly, you wouldn't even need to be a petroleum engineer, or even really understand much, to solve this one with the Guidebook. But many, many engineers get flustered by this sort of problem. Just reading the dang thing is a frustrating experience. 

Note: now we have the CBT and no extra resources allowed, one would have to just work through the solution as they best knew how. However, I doubt the CBT will have this level of difficulty for this reason. Only time will tell, but it doesn't hurt to study and understand difficult problems!

Fractional Flow Water: 2018 #8

An oil reservoir dipping at 20 degrees has a flow area of 1 MM square ft and a total fluid flow of 2,830 barrels per day. The oil has a permeability, relative permeability, viscosity, and specific gravity of 40 md, 0.4 md, 1.5 cp, 0.8, respectively. The water has a relative permeability of 0.02 md, with a water/oil viscosity ratio is 0.1. Assuming standard water density and specific gravity, and that the formation permeability averages 100 md, the fractional flow of water is closest to: (A) 0.2 (B) 0.3 (C) 0.4 (D) 0.5.

This problem is nearly the same as the Guidebook example on 14 WLF 2; only the flow area has doubled. The computation then changes to (1-0.4)/(1+2) = 0.6/3 = 0.2, or (A).

To better understand fractional flow in a dipping reservoir, see page 158 of TS8 (Towler). Slider has some good practice problems as well. These problems are not common, but being directly from SPE Textbook Series reservoir text, it's fair game.

Casing/Hole Sizes: 2018 #7

This is a standard problem, with the answer found on 6 DTC 7. The chart is found in many other places but the Guidebook version uses square, evenly spaced formatting for speed (other versions I've seen get fancy with the pipe more "real-looking" and slower to use, at least for me).

Note the answer (B) had incorrect wording on the original publication (it was missing the "do not require" part) but should now be corrected. Just re-download the new version to correct this as desired.

Core Permeability: 2018 #5

Problem 5. The statement most FALSE regarding permeability determination (especially core permeability determination) is:

(A) Klinkenburg used gas in place of nonreacting fluids, but he corrected for slippage.
(B) Water is the most frequently occurring reactive liquid relating to permeability determination.
(C) Slippage occurs when the diameter of the capillary openings approaches the mean free path of the gas.
(D) Carbon dioxide should not be used to find the equivalent liquid permeability of a core.

The TOC shows "Core" on 12 WLT 1. This page clearly states "any gas" can be used. Since CO2 is indeed a gas "D" is plainly false. Done.

Confirmation is nice, so I check out the listed reference: Petroleum Engineering Handbook, P26-18, Bradley. Sure enough, CO2 is shown as a test gas. Being an SPE reference, this locks it.

Reservoir Radius: 2018 #2

Problem 2. A well is centered in an isolated, homogeneous, cylindrical 9 ft thick reservoir of unknown radius. Formation* Reservoir compressibility is expected to be 1E-5/psi, average porosity is 20%, and the formation shrinkage factor (calculated from test well production) is roughly 0.83 STB/bbl. The well flows pseudosteady-state at 10 STB/day for two whole days over which time the reservoir pressure falls a surprising 3 psi. The best estimate of reservoir radius (ft) to report to your company is: (A) 300 (B) 600 (C) 900 (D) 1,200.

This type of problem is in the Guidebook on 12 WLT 3. For this particular example the numbers are adjusted in such a way the answer remains 900 ft. But it's a tough calculation to to in a hurry, so be careful. Although this example is fairly "plug-and-chug" the challenge is to recognize the problem type and solution method fast enough to finish out the calculations quickly. The key word is pseudosteady-state, and a quick overview of the Guidebook cover brings you to the right equation quickly. 

*typo.

Choke: 2018 #1

Problem 1. NG, SG 0.70 through ½" choke at 500 psig, 140 dF. 2,900 MCF/day is closest to: (A) 20% high (B) 10% high (C) Accurate (D) 10% low.

This solution is shown in the Guidebook 8 FAC 1-2. At said conditions, z = ~0.95 (find using 9 PVT 2 with preferred z chart; the Guidebook walks one through the psuedocritcal calculations).

Once you have z, flow computes to 2,900 MCF/day or (C). Note the Guidebook comment how estimating z = 1 gets you close, but this problem demands an exact answer, so you can't risk estimating.

Cash Break Even (CBE): 2016 #80

An investment had net cash losses of $5 million and $12 million in the first and second years, respectively, but gained $4 million, $8 million, and $15 million in years three, four, and five, again respectively. The number of years required for cash flow breakeven was closest to:  (A) 4.6 (B) 4.5 (C) 4.4 (D) 4.3.

On this problem, NCF is calculated for each year until the accumulated amount is passes from negative to positive. One then iterates to find the exact time. To do this quickly, make a table (below):

Year 1 2 3 4 4.33 5
NCF (5) (12) 4 8 15
Cum (5) (17) (13) (5) 0 10
CBE yr 1 2 3 4 4.33 5

This problem is pretty easy since NCF is given directly and need not be calculated. Cumulative cash flow goes from (5) to 10 in years 4 and 5, respectively. So going 1/3 of the way between 4 and 5 gives about 4.33 This is closest to (D).

Waterflood: 2016 #70

A 2000 acre, 15 ft thick, 15% porosity waterflood reservoir with a Sgi of 20% and a producer/injector distance of 330 ft is started on injection.

How much water volume (barrels) must be injected until interference within a single 5-spot pattern?
Wii = [3.14159/(2*5.615)](330^2)15(0.15)(.2) = 13,710 bbl (B)

How much total volume to pattern's fill-up?
Wif = (2/5.615)](330^2)15(0.15)(.2) = 17,455 bbl (B)

Pipe Stress via Injection: 2016 #65

8.681 ID casing string rated for 437,000 lbf tension. 46% of the casing’s tension rating is applied. What is the maximum pumping pressure before the casing exceeds the tension rating?

Turn to 6 DTC 1 (casing, stress are the key words).
Find remaining tension allowed: (437,000)0.54 =  236,000 lbf
Find allowed pressure using 236,000 lbf = P(0.7854)D^2 lbf, or:
P = 3987 psi. (A)

This is a 3 minute problem if you know what you are doing. And it's realistic. And it's in TS 2. Know how to do problems like this quickly.

Reserves: 2018 #65

Problem 65: When estimating primary oil and gas reserves, the LEAST accurate statement is: 

(A) Drive mechanism is more important than well spacing for recovery efficiency. 
(B) A gas deviation factor estimated using empirical correlations...is acceptably accurate...
(C) The differential rather than flash formation volume factor should be used.
(D) ...capillary-pressure method...OWC...must be a robust correlation between porosity & perm. 

This solution (C) is found in the Guidebook 13 RES 13. Keep in mind, this sort of question you just have to understand each part, and the CBT forces you to know it. So read the Guidebook notes and learn them for the exam.

Also, read about this subject in detail in the HS5 P1509, P1509, P1512, and P1606. P1512 has a long discussion of the need for flash FVF use, well worth your time.