Weight Up: 2018 #64

Problem 64: Drill 10.1M to 10.2M ft; pressure 5M psig. At 12M ft ROP increases. Surface casing 5M ft; LOT calculated fracture MW 14.5 ppg. Standpipe pressure is 650 psi. Using a 200-psi trip margin the barite (sacks) needed to weight up the 800 bbl mud system from 9.8 ppg to the KMW? 

Res pressure: Pr = 9.8(0.052)12,000 ft + 650 psi = 6,765 psi.
KWM: = (6765+200)/(0.052*12,000) = 11.16 lbm/gal.
(35 – W1)/(35 – W2) = (11.16 – 9.8)/(35 – 11.16) = 1.057 bbl bar/bbl mud.
Volume = 1.057(800) = 845.6 bbl
Sacks: 845.6 bbl (1/sk/100 lbm)(35 lbm/gal)(42 gal/bbl) = 671 sacks (C).

Note this problem has the wrong answer in the answer key, as does the original SPE source for the problem; I should not have been so trusting.... 

UPDATE: The 2019 Update SPE Petroleum Engineering Reference Guide (provided during the exam) shows the following:

 

I've shown traditional way of solving above, but using the provided equation gets the same thing: 

1470*((11.16-9.8)/(35-11.16)) = 83.9*8 = 671 sacks.

 Note I would read TS12 on P105 to get familiar with this sort of problem. One of the issues with the reference guide is it gives only a few specific equation options for calculating and doesn't show how the equation is a function of Wf/Wi...and the exam most certainly could expect you to understand how this all works for any situation: (1. Weight up add V. 2. Weight up same V. 3. Dilute & dump). I'll include a few types of these problems using the Reference Guide on the 2021 problem set. 

Source: GB 1 DRL 1, 4 MUD 1; A Guide to Prof. Reg. for PEs, SPE, 1991.

Permeablity & Compression: 2018 #61

Problem 61. A new 4.892-inch diameter well flows at 439 BOPD for 72 hours, after which a PBU is performed. Oil viscosity is 0.32 cp. The formation is 68 ft thick, with 28% porosity and 27% water saturation. Compressibility (water and formation) are 2.57E-6 and 3.05E-6 1/ psi, respectively. 

The FVF is 1.369, 1.376, 1.362, and 1.360 at pressures of 2230, 2572, 2910, and 3000 psia, respectively. PBU pressures are measured at 0, 0.25, 0.5, 1, 2, 4, 8, 24, and 48 hours showing 2234, 2342, 2456, 2582, 2748, 2825, 2840, 2871, and 2882 psia, respectively. 

The reservoir permeability (md) and total compression (1/psi) are most nearly: (A) 5.7 & 16E-6; (B) 6.7 & 15E-6; (C) 7.7 & 16E-6; (D) 8.7 & 15E-6.

Solution: Guidebook 12 WLT 12, 13 RES 3; A Guide to Professional Registration for PEs, SPE, 1991.
dt = 4, 8, 24, 48 becomes t+dt/dt = 19, 10, 4, 2.5…at….2825, 2840, 2871, 2882 psia.

1. p* = 2910 from graph
2. pave=0.5(p*+pwf)=0.5(2910+2234)=2572.
3. Bo=B at pave or 2572=1.369 (given).
4. k=(162.6q uo Bo)/(mh) = [162.6(439)0.322(1.369)]/[(69)68] = 6.7 md.
5. co = (Bo-Boi)/(Boi(Pr-pwf)) = (1.376 - 1.362)/[1.362(2910 - 2234)].
    co = 0.014/[(1.362)(676)] = 15.21E-6.
6. ct = coSo+cwSw+cf = (15.21*0.73)+(2.57*0.27)+3.05][10E-6]=14.85E-6 1/psi.

This one is long and ugly; I might skip it on an exam. Making good judgements here is often the difference between P or F.

Nodal Analysis: 2018 #60

Problem 60: The statement about well total systems analysis that is most FALSE is: 

(A) If the inflow and outflow curves do not intersect, the well will not flow. 
(B) If the inflow & outflow curves have two intersection points, the lower rate is not a stable solution and is meaningless (please correct this typo in the Guidebook; it has been corrected in V3).
(C) The first step in applying systems analysis is to select a node to divide the system. 
(D) The node is typically selected to be at the choke to isolate the inflow performance from the flow behavior in the tubing.

Reference Guidebook 7 PRD 1 and HS4 29, 30, 32, 32. The answer is (D) as the node typically selected is the perforations.

IPR: 2018 #58

Problem 58: Well: 2,200 psia reservoir produced through a workstring at 50% drawdown produces 600 BOPD & 192 BWPD.

Analysis of the well’s planned tubing predicts bottomhole pressures (psia) of 1386, 1296, 1236, 1196, 1166, 1146, 1136, 1146 at corresponding oil rates of 100, 200, 300, 400, 500, 600, 700, 800 bbl/day. Water production rates are 32% of oil production.

Applying a well-known, easy-to-use three-phase flow model, the fluid production (BFPD) through the new tubing string will be closest to:

7 PRD 1 or HS4 P19. Wiggins is the only well-known 3-phase model. Steps:

1. Calculate water rates 32% of oil rate.
2. Calculate qomax & qwmax (968 qo & 337 qw) using Wiggins from rates (600 qo & 192 qw) at 50% DD.

...Given pwf: 1546, 1386, 1296, 1236, 1196, 1166, 1146, 1136, 1146.
3. Sum qo+qw at pwf (bfpd): 466, 132, 264, 396, 528, 660, 792, 924, 1056 for TBG curve.
4. Calculate Wiggins IPR (oil & water; sum for bfpd): 613, 672, 709, 734, 753, 765, 771, 765.
Crossover is found at 762 bfpd (note underlines at 753 & 765).
On an exam skip for central points & round your answers. It's easy to make an error if you don't have Excel handy...I sure did trying it fast by hand. But that's how you will want to practice; how you practice is how you will test.

Sorry for the delay on this solution; The answer (A) was not an exact answer yet this problem is tough enough I updated it for an exact solution for future users. Note there should be nothing like this level of difficulty on the CBT, since those equations are not in the approved text and could not easily be given in the problem. But a word question could be asked, so reviewing this problem is a good idea.

Economics: 2018 #57

Problem 57. Two 20-year leases are considered for investment:
Type/Investment: Gas/$450M; Oil/$720M.
Type/Annual Revenues: Gas/$90M; Oil/$135M.
Type/Annual Expenses: Gas/$9M; Oil/$11.7M.
Type/Salvage value: Gas/$45M; Oil/$67M.

If the cost of unlimited capital is 15%, the statement most TRUE regarding NPV analysis is:

(A) The difference between the oil and gas leases NPV is between $0 and $5,000.
(B) ONLY the oil lease investment may be chosen.
(C) ONLY the gas lease investment may be chosen.
(D) NEITHER the oil or gas lease investment may be chosen.

Guidebook 10 ECN 1 and Petroleum Engineering Handbook I, Mian, PennWell (1983) can help on this problem. Once you know about NPV, the difficulty becomes apparent. Which is: 20 years is difficult to calculate by hand for the two scenarios. It can be done, but time is precious. Use PWF (present worth factor) for the salvage & SCAF (series compound amount factor) for the revenue: 
-450M+(90M-9M)(SCAF)+45,000(PWF)
-450M+(90M-9M)(6.2593)+45,000(0.0611) = $59,753
-720M+(135M-11.7M)(6.2593)+67.5M(0.0611) = $55,896

UPDATE: It seems most don't have access to the tables so I'm going to show the calculations for SCAF for a number of years (n) at an interest rate (i) is:  
SCAF = {[1 - (1 + i)^(-n)] / i} = (1 - 1.15^(-20)) = 6.2593.
And of course PWF is in the Guidebook:
PWF = (1 + i)^(-n) = (1.15^(-20) = 0.0611

Anon below points out the formula (arranged differently) is in HS1 P777. It's called the "Series CAF".

Vogel: 2018 #55

Problem 55: A saturated 2,200 psi sandstone reservoir has 13 producing wells yielding 18 MBOPD. One of these wells produces 400 BOPD with a flowing bottom hole pressure of 1800 psig; this well has a SSSV at 1,000 ft TVD with a 3-inch ID.

Oil viscosity, the formation volume factor, oil saturation, and permeability are 3.1 cp, 1.2 bbl/STB, 0.77, and 0.82 millidarcies, respectively.

Simulations predict the reservoir pressure will fall 400 psi in 3 years and remain saturated; oil viscosity, FVF, saturation, and permeability should remain roughly the same. The maximum oil rate (STB/day) for this well after 3 years will be closest to: (A) 1010; (B) 1060; (C) 1090; (D) 1120

Guidebook 7 PRD 1. Saturated? Think Vogel.

qo max = qo current/(1-0.2*(pwf/pr)-0.8*(pwf/pr)^2) = STB/day
qo max = 400/(1-0.2*(1800/2200)-0.8*(1800/2200)^2) = 1330 STB/day
qo max future = qo max(pr f/pr p) = 1330/(1800/2200) = 1088 STB/day

Update: this is a "future" well performance problem. First do the Vogel calculation for "current" qo max (using current qo, pf, & pr). Second find the "future" qo max based on future pr. Note this is a simple, linear relationship only if oil fluid properties & permeabilities "remain roughly the same". For more details, see the Guidebook's references on 7 PRD 1. 

Well on Choke: 2018 #53

Problem 53: You are bringing on a well on choke with a subsea stack. The shut-in casing pressure is 800 psig. The drill pipe is 5.5-inch and held in tension with 500 ft of 8-by-3-inch drill collars.

The 12.25-inch PDC bit has 3 13/32-inch jets and is currently at 5,000 ft.
The pump pressure is 450 psi at 30 strokes per minute (SPM) on the pump.

When closing the BOP and opening the choke line to the poor boy degasser, the pressure is 750 psig when the pump rate is 30 SPM. When bringing the well on choke, the reduced casing pressure (psig) is closest to: (A) 484 (B) 494 (C) 504 (D) 514.

This one is quick if you know what everything is definition-wise; expect simple problems occasionally and to waste time searching for a non-existent trick. Note the answer choices are not exact for this reason, to leave you wondering what you are missing. Solution: Guidebook 3 HYD 2; FC P179-180 SICP – Choke line loss = reduced csg pressure so 800 – (750 – 450) = 500.

MBE: 2018 #51

Problem 51: A 150-degF gas reservoir, originally 4,000 psia...known water drive...10 Mbbl of water and 2 MMMCF of gas has been produced...the percentage recovered to date is closest to: 

This problem is wordy (I've cut most for the blog post) but is just plug -and-chug. Using 13 RES 2:

Bi = 5.04(150+460)0.9/(4000) = 0.692
Bf = 5.04(150+460)0.85/(2500) = 1.045
Delta B = 0.354 bbl/MCF
G = [GpBf – We + BwWp]/(delta B) = ((2000000*1.0453)-1000000+(10000*1))*(1/0.354)
G = [(2 MMMCF(1.0453 bbl/MCF) – 1 MMbbl + (10 Mbbl*1 bbl/STB)]*(1/0.354 bbl/MCF = 3.1 MMMCF
2/3.1 = 64%

Emissions & Flares: 2018 #50

Problem 50: Which of the following statements concerning flaring is most FALSE? 

(A) All hydrocarbons with a C-to-H ratio of greater than 0.33 tend to soot. 
(B) Combustion efficiency of 98% is equivalent to a destruction efficiency of 96.5%. 
(C) Combustion efficiency is the % of HC in flare vent gas completely converted to CO2 & H2O. 
(D) Petro refinery flare: NHV of gas in flare combustion zone is >= 270 Btu/ft3 for 98% destruction.

This answer are easily found on 8 FAC 9. On (B), the 96.5% & 98% are merely reversed.

Emissions: 2018 #49

Problem 49. Flare 4.9 MSCF/D sour gas 4% H2S. SO2 emissions TPY? (A) 5.5; (B) 6;(C) 6.5; (D) 7?

Solved using 8 FAC 9 of the 2018 V2 Guidebook. Note emission calculations are rarely found in PE textbooks (they aren't in any of mine). But they are pretty much just dimensional analysis. However, it isn't the thing one can do without knowing the lingo.

Steps:
1. Convert lbs of NG to SCF w/Ideal Gas conversion: 379.3 SCF/lb-mole SO2 Emissions (lb/hr).
2. Flare gas vol.(scf/hr)*(1/379 scf/lb-mole)(64 lb/lb-mole)(%H2S).
3. (4,900 scf/d)(1 d/24 hr)(1/379/scf/lb-mole)(64 lb/lb-mole)(4/100) = 1.38 lb/hr SO2.
4. Emissions (TPY) = (1.38 lb/hr SO2)(8,760 hr/yr)(1 ton/2,000 lb) = 6.04 TPY SO2.

Note the Guidebook walks you through this type of calculation step-by-step so the process is second nature under time pressure.

OCTG: 2018 #48

Problem 48. The statement regarding OCTG corrosion most TRUE is: 

(A) Higher fluid flow rates lower corrosion rates because the pipe is kept cleaner. 
(B) High temperatures and higher-stress states accelerate hydrogen embrittlement. 
(C) Grades such as C-090 and T-95 are especially susceptible to sulfide stress cracking. 
(D) Carbon dioxide alone is a is noncorrosive gas. 

The first thing to figure out: what is "OCTG"? If you know, great. If not, look it up fast:

1. Guidebook TOC: not there (the GB rarely has definitions).
2. Dictionary under O: not there (this surprises me, actually).
3. HS Index: not there (again this surprises me).
4. TS12 Index: I choose TS12 because the context suggests pipe/casing related: Bingo; P385 & 395. It stands for Oil Country Tubular Goods.

Note that TS2, the other SPE drilling book, does NOT have this in the index. TS2 simply doesn't cut it anymore, you must have TS12 as well (in my humble opinion).

Now that I confirm the subject I use 6 DTC and TS12 to discover:

A: High flow removes protective film. So this is false.
B: Low temp actually accelerates...  So again this is false.
C: These grades were developed to resist SSC. Once again false.
D: CO2 is very corrosive with water but alone is noncorrosive. True, and so the answer.

Get used to this way of testing. Never panic on an exam if you see terms you don't know, just march through your resources like a machine, never hesitating to quit if it's not there. Time is of the essence; don't spend more than 30 minutes on 5 problems is a good rule of thumb.

ESP: 2018 #43

Problem 43. The following statement concerning troubleshooting an ESP most FALSE is:

(A) The major source of info troubleshooting an ESP...
(B) Gas locking is marked by amperage decline...
(C) Solids are spotted by amperage fluctuation...
(D) Fluid pumpoff is detected by slow amperage decline..

This is a fairly tough problem to my mind. Let's count the ways:

1. It claims "the" major source for ESP troubleshooting is the ammeter? Really? Come on, the major source? I can imagine quite a few sources, and "the major source" a definite statement. But this is also a direct quote from an SPE source (Bradley) so consider it gospel. It's also in the Guidebook.

2. The gas locking question is fair, not hard to find. You can get that even if you don't know anything about ESPs.

3. "C" is a little harder, but again with a few minutes of good sources you can find this one.

4. "Fluid pumpoff" is indeed detected by slow amperage decline often due to oversized pump (not an undersized one). Again, this is a direct quote from the Guidebook (and that same pesky Bradley source). But it's a fair question; one should know this if you understand ESPs.

Note it's easy to misread this kind of problem because it gives a correct fact first and only then gives slightly incorrect second part.

ESP: 2018 #42

Problem 42. Designing an ESP installation from given data, the TDH (ft) is closest to: 
Current production: 500 BFPD at 500 psi drawdown.
Desired production: 4X (negligible free gas, annulus friction). 
Pressures (psi) at 2M BFPD: Pump-intake/wellhead = 1M/200. 
Depth (ft) of pump at perfs = 6M; Friction loss = 41.7 ft/1M ft tbg. 
WC: 80%, 1.0625 SG; OC: 20%, 0.91 SG & 24 API. 
(A) 4,460 (B) 4,660 (C) 4,860 (D) 5,060 

See Guidebook 7 PRD 3-4, or the ESP Sizing Guide reference listed there.
Note: this problem was updated on Kindle after 10/12/2018 as the original given well data was nonsensical (even though it led to the same answer selection). Please use the well info shown above:

1. PI = dq/dp  = 500/500 = 1 bfpd/psi.
2. SGw = 0.8(1.0625) = 0.850; SGo = 0.2(.91) = 0.182; so SGf = 0.85 + 0.182 = 1.032.
4. NDL = Pdepth - PIP(2.31)/SGliq = (6,000 – (1,000*2.31)/1.032 = 3,762 psi ft
5. Hf = 6,000 ft and 41.7 ft/1,000 ft = 250 ft.
6. Hwh = 200 * 2.31 / 1.032 = 448 ft.
7. TDH = 3762 + 250 + 448 = 4460 ft (A).

Because the answer is the lowest TDH option given, any lower value one calculates will give the correct answer.

Probability: 2018 #38

Problem 38. When investigating a lease your geologist estimates there is a 40% chance of finding marketable oil, an 80% chance of finding marketable gas, and a 30% chance of finding marketable oil with marketable gas. The estimated probabilities of finding any marketable hydrocarbon, only marketable gas, and no marketable hydrocarbons at all (respectively) are:

(A) 90%, 50%, 10%
(B) 85%, 60%, 15%
(C) 90%, 60%, 10%
(D) 80%, 40%, 20%

For many, even most, this problem is difficult to do in six minutes. Why? It's uncommon, not intuitive, and working out the logic takes too much time to learn on the spot. However, the problem style is in the Guidebook as well as Mian's (I think), and since I also remember this sort of thing in the SPE HS (I think) it's definitely fair game. Note I cranked this one out fast, so please let me know if you find a typo!

Solution: A=oil, B=gas
P(A or B) = P(A) + P(B) – P(A and B) = 0.4 + 0.8 – 0.3 = 0.9
P(B only) = P(B) – P(A and B) = 0.8 – 0.3 = 0.5
P(neither A or B) = 1 – P(A or B) = 1 – 0.9 = 0.1
Key 0.4+0.8-0.3=0.9; 0.8-0.3=0.5; 1-0.9=0.1 (A)

Separator: 2018 #35

Problem 35. A two-phase 24-inch ID horizontal separator needs to handle 1,000 bbl/d with a liquid retention rate of... The difference between the effective, manufactured, and seam-to-seam lengths of the separator in the two GOR scenarios calculates respectively closest to (ft): (A) 0, 5, 1; (B) 0, 2.5, 1; (C) 0, 0, 2; (D) 1, 2.5, 0. 

This is a similar problem as shown in the Guidebook example. It just uses 2X liquid retention and 2X fraction full. I like the problem because it gives a person experience in calculating each type of length. This problem thus calculates to:
1. Effective length: both GOR options give the same Le, so 3.4 – 3.4 = 0 ft.
2. Manufactured length: liquid/gas dominate so 7.5 – 5 = 2.5 ft.
3. Seam-to-seam length: liquid/gas dominate so 5.4/4.55 so 5.41 – 4.55 = 0.86 ft.

The Guidebook presentation of separator problems is a highly-organized single page for rapid use.

Mechanical Energy Balance: 2018 #31

Problem 31. 10 ppg, 20 cp fluid pumped through 697 ft, 3.5-inch OD tubing rising 12 ft...100 ft...down 1.5 ft...203 ft down 10.5 ft into a 10x10x10 ft open tank filling 1 ft/min. Pump pressure (psig)? (A) 800; (B) 810; (C) 820; (D) 830.

This is a basic Mechanical Energy Balance (8 FAC 6) problem. It's simple if you don't forget the kinetic energy; of course, the wrong answer is conveniently waiting for you as a choice if you do forget it (in this case, "B"):

1. HS vertical: 12 – 1.5 – 10.5 = 0 vertical ft
2. Friction: 697 + 100 + 203 = 1,000 ft
3. 34.14 ft/s (2x GB 3 HYD 1) = 0.809 psi/ft(1000 ft) = - 809 psi
4. KE: 8.074E-4(10)(34.14^2-0^2) = - 9.4 psi
5. Total: 818 psi.

NG Processing: 2018 #30

Problem 30. The statement most FALSE about natural gas treating and processing is:

(A) In long distance transmission of sales gas by pipeline, pressure is usually less than 1,000 psig.
(B) A potential cause for treated natural gas “going sour” is too low of an inlet gas rate.
(C) All raw natural gas is fully saturated with water vapor when produced from an...
(D) There are four glycols that are used in removing water vapor from natural gas or in...

This question can be answered quickly knowing the basics of natural gas processing. Or you can quickly glean the answer from HS III, pages 186, 192, and 198. Overall, a pretty easy question, but it can take time if you don't have much experience, or lack the right resources.

Watch the wording like a hawk. It's always best to find a direct quote from an SPE source for these types of problems if the wording seems vague.

Skin From Drawdown: 2018 #29

This is the exact problem found on 12 WLT 4. The skin calculates to 3.0.

It's pretty simple; I include it to show a few testing tricks:
1. FVF is given as "shrinkage factor" (STB/bbl); merely invert for 1.2 bbl/STB.
2. The answer 3.0 splits the difference between options (A) 3.2 & (B) 2.7; the closest answer is "A".

This, as I said, is simple. But these tricks can unsettle anyone under stress, so they are ideal to practice on "plug-and-chug" problems like this.

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.

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).