Rod Pump Counterbalance: 2016 #45

A 160-173-54 pumping unit runs at 16 SPM producing 179 BOPD and 21 BWPD out of 2 inch anchored tubing. The 7/6 rods ...counterweight required? 

Counterbalance problems are meat & potatoes on exams. Why? No fancy charts are needed like most rod pump problems. This can be solved using just two pages in the Guidebook and on the EBT any equations/data not found in the provided reference are easy to give for the specific problem. Now that's not saying there isn't plenty of room for error. I make mistakes on these problems all the time. For this one:

1) 7/6 1.5” rods; Wr = 1.833 lb/ft (7 PRD 10) 
2) W = 1.833(5,000) = 9,165 lbs 
3) Wfr = W(1-0.128G) = 9,165(1-(0.128*0.9)) = 8,109 (7 PRD 9) 
4) Fo=0.34*0.9*1.5^2*4500 = 3098 
5) CBE = 1.06(8,109 + (0.5*3,098)) = 10,238 (C).

Flanges: 2016 #41

Question: What flanges classes can be used in a 200 F & 1,450 psig flowstream (SF = 2, or 2,900 psi)?

I just glanced at the cover TOC and scanned to Facilities (FAC). "Flange Ratings" was 8 FAC 7. Less than 60 seconds to size up the problem and find my page with all the needed tables. Note these tables are also in HS3-356 (tab this page).

Answers (C) and (D) gave API Class 2000; 2,000 psi. So they are eliminated.
ANSI Class 900?  2,025 psi. So it can't be (A) either. This has taken me two minutes.
API Class 3000? 3000 psi. ASME 1500? 3,375 psi. So we have a winner, (B).

Less than 3 minutes; I take another minute to review for tricks, finding none, I move on.

Note on the new EBT exam they can merely provide the needed charts, much like the Guidebook does. 

Frac Gradient: 2016 #39

A frac with 8.5 lb/gal fluid, tubing pressure loss of 500 psia, and perforations centered at 5,000 TVD ft had negligible pressure losses in the prefs. ISIP = 1,525 psig.  The frac gradient (psi/ft) was most nearly: (A) 0.65 (B) 0.70 (C) 0.75 (D) 0.80. (Note: I've left out all the unneeded information, which makes the problem much harder).

Note this sort of equation is unlikely to be provided in the reference, but it is simple enough an engineer should know it. 

Solution by using equations on 7 PRD 2:
1) Find hydrostatic pressure at perfs: 0.052(8.5)5000 = 2210 psi.
2) Add the ISIP for total pressure at perfs: 2210 + 1525 = 3735 psi.
3) Calculate the gradient to perfs: 3735/5000 = 0.75 psi/ft or (C).

LOT: 2016 #37

Problem 37. Which statement below is most TRUE? 

(A) When a LOT or FIT fails a cement squeeze is necessary before resuming drilling.
(B) A LOT and a FIT are two different names for the same test.
(C) A LOT tests the rock for failure against a preset pressure; an FIT tests...without a preset pressure.
(D) A FIT measures...when the rock starts to fracture...a LOT measures...when...completely fails.

Choice A; the answer, has been edited due to lack of clarity on the original. It is now nearly a direct quote from the Guidebook/HS2.

Diagenetic Porosity: 2016 #12

An oil reservoir has average diagenetic porosity of 10% and a CNL log measures a porosity of 15%. The oil reservoir is 200 acres, 10 ft thick, with residual oil saturation & initial water saturation of 25% each. The initial oil formation volume factor is 1.2 RB/STB. The maximum oil production from primary porosity would be closest to: (A) 320 MSTB (B) 340 MSTB (C) 620 MSTB (D) 640 MSTB

Definitions from 15 LOG 4:
Total Porosity: (measured with nuclear tools) …equals primary porosity + secondary porosity.
Primary Porosity (apparent, intergranular) used for reserves or maximum producible oil…equals total porosity – secondary porosity.
Secondary Porosity (isolated pores, vugs, and fractures) also called diagenetic porosity …may be overlooked by acoustic-logs …equals total porosity – primary porosity

Therefore, for this problem, primary porosity is: 0.15 – 0.10 = 0.05 pu. And maximum oi production from primary porosity is: 7758(200)10(1-.25-.25)0.05(1/1.2) = 323MSTB or (A).

2019 CBT PE Exam

The 2019 Petroleum PE Exam is now CBT and closed book except for a single reference: 2019 Update - SPE Petroleum Engineering Certification and PE License Exam Reference Guide (Ghalambor).* This has just become available from SPE.

I have read, reviewed, and used the 2014 Reference Guide (see the upper right blog link for my review). I have also purchased the 2019 Update, and I'm impressed with the improvements! It's now a good book in its own right, one worth keeping, and not just for the exam.

Bottom line? Yes, the exam has changed and people can no longer claim it is a "battle of references". But honestly I believe it has been this way at least since 2015, with pass rates being inversely proportional to the number of pages turned during the exam. This is not a bug, it's a feature. The CBT merely formalizes this reality.

So how to study for the CBT? I don't see much change except to know the 2019 Update well. Also know the SPE Petroleum Handbook (for word problems) and the SPE Textbook Series (for word and calculation problems). I think my Guidebook and Guidebook Companion problems still hold up well as a primary framework to study from, since the Guidebook has summaries of both series.

However, we have definitely moved into a digital world, both in real life and on the PE exam. So I think it's time for me to update the Guidebook into a digital format, especially for practical day-to-day use in the field. This will be my next project.

The 2019 Update will be provided in digital format for the exam. The search options will will be important; I'm guessing it will have a "control f" search function. Assuming so, studying from it in digital format will be useful.

Test Taking Strategy - 2019 CBT

Note: I will not talk about specific problems from any prior exam.

First: Get your materials; review the 2019 Update - SPE Petroleum Engineering Certification and PE License Exam Reference Guide (Ghalambor) digital version. Next, learn the Petroleum Engineering Guidebook in detail along with SPE Textbook Series #1, #2, #12, and #4 as time allows. Have a good Petroleum Dictionary to use when you study. Have the Halliburton Red Book or equivalent. Invest in the 7 volume Petroleum Engineering Handbook and read it, focusing on weak areas.

Second: Do the Petroleum PE Problems 2021: 1-43 to get used to the exam style, using the Ghalambor reference that you will have on the real exam. Next, do the Petroleum PE Problems 2018: 1-40 or the 2005 SPE Sample Exam 1-40. It's critical to do these practice problems under actual test conditions. Lock yourself in a room for four hours with your calculator, the approved CBT text, and whatever you will eat and drink. You will need to use resources for pipe sizes, etc., but assume those will be given on the exam problem itself. It is important to start off with timed problems to get a feel for the exam before you invest a lot of time studying. If you lack four uninterrupted hours try 10 problems per 1 hour for the same effect.

Third: Take, practice, and completely understand every problem on the 2005 SPE Sample Exam 1-80. This is your calculation problem study time. Do every problem as many times as needed until each one takes less than 6 minutes each (on average).

Fourth: Review whatever reservoir textbook you are most familiar with. Also, if you have the time, do and completely understand every problem on the 2014 SPE Practice Exam (this isn't like the actual exam, it's mostly easy and fast, but worth a look if you have extra time since it has a few challenging reservoir problems). Study any weak areas.

Fifth: Do the 2016 Petroleum PE Problems 2018: 41-80. If you don't have it, use the 2005 Practice Exam 41-80. This is your final check. If you can get 70% under exam conditions, you will probably do all right on the real thing. Use this test to hone your test-taking skills under time constraints.

Sixth: Sometime in the month before the exam, do the Petroleum PE Problems 2017 and 2016: 1-80. Take them as if it's real, including food and restroom breaks, but again use other resources that have information that would be provided on the problem. Regroup and study as needed. If you have the opportunity to take review courses (Bing is highly recommended) do so. The lectures are invaluable, especially for engineers without a petroleum degree.

Regarding studying: I didn't find practicing calculation problems to help much past a certain point. If you can do every problem on the 2005 exam within time constraints (average 6 minutes per problem), you should be good to go on calculations.

Regarding the futility of studying calculation problems to prepare for the PE Exam? Read David Vaucher who after taking the 2014 exam said, "I think I did well. I’m confident because I spent ample time preparing..." and "...I am certain that I got all the calculation questions right...". But to his chagrin he then learned he did not pass. Think about that; he believes he didn't miss a single calculation problem yet did not pass! This demonstrates the danger of relying on your calculation acumen to carry you.

So how should one prepare? Achieve real understanding and hone your test-taking skills. This was how I generated my study notes (now Guidebook) during the years I waited to take the exam. These notes (later "Guidebook") summarized the main points from SPE textbooks. And little else. Know the basics well.

You have probably heard the joke that to avoid being eaten by a charging bear you don't have to outrun the bear, just the other guy? Well you don't need to get every answer right on this exam. You just need more answers correct than about half the other guys. Think like this when you study.

PE Testing

An timed exam has predictable methods for targeting a 40-60% error rate:

1. Potential subjects too broad to fully study.
2. Problem has an uncommon reference.
3. Misleading anchoring.
4. Confusing or complex wording.
5. Irregular problem difficulty (easy...or missing the trick?).
6. Shortcut required or problem too long.
7. False answer match
8. Unexpected numerical precision.
9. Inexact correct answer.

Most of these are self-explanatory, but #3 is a bit cryptic. "Anchoring" is a cognitive bias toward a particular solution path (the incorrect one) resulting in mentally excluding all others (including the right one) and so freezing up.

Regarding #1 and #2 it's very easy to waste time on a problem you don't really understand rather than cutting one's losses and spending the extra time checking problems one does know how to do. Ironically, limiting the open-book option with the single allowed reference on the CBT will help avoid this problem.

Gaseous Equivalent: 2018 #14

Problem 14. A 50 API retrograde gas-condensate reservoir at 2,500 psia & 225 dF has a gaseous equivalent (scf) closest to: (A) 720 (B) 770 (C) 820 (D) 870?

Straightforward 3-step solution; 13 RES 11 has all needed equations on a single page:
1. Mo = 5954/(api-8.8) = 145.
2. SGo=(1.008*145)/(42.43 + 145) = 0.78.
3. 133,300(0.78/145) = 719 or (A).

Note this problem is more wordy and confusing than is shown above; for the answer here I merely remove all the unneeded information for clarity. But I strongly recommend doing these practice problems in test format. Practicing the confusing nature of problem wording, under time constraint, is critical.

Cement Mix: 2018 #13

Problem 13. ...mix...sack of Class A cement slurry...4% bentonite...standard water...weight of the slurry (pounds): (A) 161; (B) 159; (C) 157; (D) 155.

Slurry volume & density calculations for this exact situation are on 5 CMT 2. Note there are 4 components: cement, cement water, additive, and additive water. They combine for 11.34 gal/sk & 14.18 lbm/gal as shown on the page. Multiply them for 161 lbm/sk (A).

Gas Reservoir MBE: 2016 #17

A gas reservoir produced 1 MMscf gas, 10 MSTB water, and had 1,100 bbl water influx. 
Current & initial gas FVFs are...water FVF... IGIP:

These problems are easy minus unit confusion issues. You can't be too careful with units here:

= (1,000 Mscf)(1.1 RB/Mscf ) + 10,000 RB – 1,100 RB
= 1,100 RB + 10,000 RB -1,100 RB = 10,000 RB
= 10,000 RB/(1.1 – 1 RB/Mscf)
= 10,000 RB/(0.1 RB/Mscf)
= 100,000 Mscf, or (B).

Drilling Horsepower: 2016 #63

A rig is expected to run at 100 RPM at 14,000 ft depth. What's a back-of-the-envelope estimate of the drilling horsepower required?

For this problem, use a Torque Factor Estimate and the equation F(RPM).
F=1.5 10,000 ft, light), F=1.75 (10,000 ft-15,000 ft, average), and F=2.0 (>15,000 ft, heavy)
Hence, 1.75(100) = 175 hp (A).

I have never seen this pre-drilling estimate in an SPE book (I first saw it in Mian, P288) but I like to use it for a quick check on any horsepower problem, so it's handy to know in real life or even an exam.

Discounted Cash Flow: 2016 #72

Investment of $100M returns $40M/yr for 3 years when it is sold for $50M. Discount rate 5%.  The net discounted cash flow ($1,000/yr) is closest to?

This is a painfully easy problem. IF you know is that NCF leaves off the initial cost of the item AND it still includes the discounted salvage value. That's the whole ball game.

40(1.05)^-1 + 40(1.05)^-2 + 40(1.05)^-3 +50(1.05)^-3 =
$38.1 + $36.3 + $34.6 + 43.2 = $152 (all values M) or (A).

Fracture: 2017 #39

As I've said before fracture treatments are simple hydraulics. But I hate them; I get all worked just thinking about the myriad of ways I tend to get them wrong. The Guidebook has a single frac page, 7 PRD 2 in a desperate attempt to keep it simple. But not too simple!

Getting an initial shut-in pressure (ISIP, or just the wellhead pressure) is a common calculation. In this problem ISIP calculates to 1,600 psi (depth 10,000 ft TVD, MW = 8.46 ppg, reservoir pressure 2,800). But ISIP already given, anyway.

All the problem wants? Required (or theoretical) hydraulic pump power (in HP) to frac the reservoir. This is just [ISIP*Qbpm]/40.8 (equation conveniently shown the bottom of the GB frac page). Don't use the starting wellhead pressure. Since the frac is pumped at 50 bbl/min: (1,600)50(1/40.8) = 1,961 hp or (C).

Guidebook exchange/update

Congratulations to all who took the Petroleum PE Exam in 2018 this fall. Pass or fail, you are the elite. Most petroleum engineers never take this plunge. And honestly, can you blame them? The petroleum PE exam is a monster, pure and simple. I'd rather take the bar!

The 2018 #3 Guidebook (3rd printing, Oct. 2018) is the latest version (what you get from Amazon). I'm happy to exchange any older (non-plagiarized) Guidebooks. Even if you passed the exam don't hesitate to exchange because many continue to use it in the years to come. See exchange instructions below.

My purpose in this space has always been to help fellow engineers (via blog, Guidebook, and problems) as time allows because there are so few resources out there. I hope to continue this tradition and stay available for questions/comments (use a recent comment section or email to reach me).

I've also now 240 practice problems on Amazon Kindle (written on request from blog readers). They are organized into groups of 40 so you can get only what you need. They are digital to keep them inexpensive plus available for immediate download. And if on a budget most of these problems and solutions can accessed for free on the left side of this blog. On the right side of the blog are even more free problems solved using the Guidebook similar to the 2005 SPE practice exam.


TO EXCHANGE: First contact me (use a recent blog comment or email) so I can have a copy ready & give you an address. Then mail the following:

1. Self-addressed, USPS stamped 9x12 envelope (no UPS or FedX). 
2. Old Guidebook (*or just include the front & back covers, copyright plus few odd pages).
3. Where you got it.
4. Toss in $5 cash to cover printing (if on a budget, don't worry about it).
5. No guarantees. I'll return a copy ASAP (first-come-first-serve; for fast/guaranteed, use Amazon).

2014 Practice Exam - Reader Question

In a prior comment box somebody asked: "What is the solution to SPE 2014 practice test question 43 or 92?"

I haven't looked at this practice exam closely so can't answer him. Does anyone have an answer they can put below in this comment box?

2018 PE Exam Comments

2018 is history. Leave any comments (and suggestions for blog, Guidebook, and Guidebook Companion improvements) in the comments below. 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, such as resources, testing techniques, or general problems only!

UPDATE: PLEASE remember the blog rule!!! I'm having to delete entire comments below (much with good commentary) because they "cross the line" into problem discussion.

For example, comments like: "several of the...questions with probability” is crossing the line (as I've just been informed by a person who decides where that "line" is).

Come on folk; I'm going to start moderating comments before they post to respect the integrity of the test. Or just remove these type of posts (or even the blog) altogether. Let's help everyone out here so people can have a free resource to study from (as well as become better engineers).

One more clarification: if your comment got deleted, please note I'm NOT claiming it was improper; I'm sure I nuked a lot of good, legit comments that were mixed into other comments. I'm just trying to be very conservative and fast at the same time.

Separator: 2016 #59

Problem 59. A 24 inch dia horizontal separator designed for 1M STB OPD...liquid retention rate is 1.1 minutes with a desired operating level of 40%. The manufactured seam-to-seam length: (A) 3 ft; (B) 5 ft; (C) 7 ft; (D) 9 ft.

This is merely the Guidebook example. Le in this case is 3.4 ft. Next one decides if gas or liq dominates and do the second calc for 5.4 ft. Knowing separtors must be 5 ft min and rounding up by 2.5 ft, seam-to-seam length is 7.5. Nearest answer? (C).

Cement: 2005 #73 (similar)

Time for a simple cement calculation problem. Using random numbers:

16" 65 lbm/ft casing set to 2,000 ft in 20" hole (float collar 100 ft above float shoe).
Use cement with a yield of 1.32 cf/sack. Slurry sacks w/ 5% excess?

Use the Redbook for 20/16 hole/casing for 0.7854 cf/ft.
Annulus cf: 0.7854 cf/ft(1.05 excess)2,000 ft = 1,649 cf.
1.2684 cf/ft(100 ft) = 127 cf.
Total = 1,776 cf/1.32 cf/sack = 1,346 sacks (round up).

Note that if using a log to locate surface water for regulatory reasons any water signature on the resistivity curve must be considered "water" (to be conservative). For example, if there was a dip in resistivity in a 1,500 to 1,550 sand, you would have to use 1,550 ft as the lower boundary of the sand.

Underbalanced Drilling: 2018 #68

Problem 68: The following statement about underbalanced drilling most FALSE is: 

(A) During the workover...a suspension plug is installed...after the well is lubricated to kill fluid. 
(B) Advantages include eliminating differential sticking & faster ROP. 
(C) Disadvantages include possible increased torque & drag & poor MWD compatibility. 
(D) Flow control, not pressure control, is the main well control issue.

This problem is referenced on GB 2 DRL 11 or on HS2 P556, P522, P522, and P520. Note the Guidebook should have the suspension plug step listed before the kill fluid; please pencil this in (fixed in V3).

Note these type of questions could come from any direction; for more detailed information on the read HS P556 (it's quite interesting). Note there are a million ways this problem can be worded; the Guidebook can only introduce and guide you to the subject. You may need to dig deeper using the listed references.