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.