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.