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.