Economics: 2016 #10

Problem 10. An investor buys a lease for a 12.5% landowner royalty. An engineer is retained to work the lease for a 5.5% overriding royalty interest. The first investor sells...but keeps a 12% overriding royalty interest with a 25% back-in working interest...net revenue ownership interest for the engineer after payout is closest to: (A) 7.5%; (B) 3.75%; (C) 4.75%; (D) 5.5%

This one is easy; the engineer’s share is an "overriding royalty interest" and thus not effected by anything after his initial 5.5%, hence, 5.5%. See 10 ECN 2 for the definition of ORRI and a more complex example that includes this part.

Directional Drilling: 2016 #4

What is the MD of the Problem #3 well if at a TVD of 6,000 ft drops at 2 deg/100 ft until vertical?

This is easy since in #3 build was 895 ft (MD) at 3 deg/100 ft. If the well then drops at the same rate, we merely add this MD to the sequence: 6,198 + 895 = 7,093 ft (C).

Directional Drilling: 2016 #3

What is the MD at 6,000 ft TVD for a build, hold, and drop well (KOP, TVD & HD at end of hold, and build/drop rates will all be given).

This is a standard directional drilling problem; see Guidebook on 2 DRL 9 or TS2 (note TS12 doesn't have a good section here). First calculate r1 & r2 (2,865 ft); next calculate Theta1 & Theta2 (17.9 deg). Finally, use the MD equation and find the MD of each component of the vertical, build, hold sections:

MD = 1,500 + 895 + 3,803 = 6,198 ft or (C).

Note these equations are all in the GB or TS2; just plug-n-chug through them.

Cement Displacment: 2016 #2

Given: 10,000' casing in 10.7 ppg mud.
Pump into casing, up annulus:
100' 8.5 ppg mud = 44 psi hydrostatic.
1,550' 12.9 ppg cement = 1,040 psi hydrostatic.
1,500' 15.4 ppg cement = 1,201 psi hydrostatic.
9.1 ppg brine until cement is displaced.
Pump pressure required to displace?

Annulus (remaining):
6,850' 10.7 ppg mud = 3,811 psi hydrostatic.
Total hydrostatic pressure in annulus: 6,096 psi.

Hydrostatic pressure in casing: 9.1 ppg brine = 4,732 psi

Pump pressure required to displace?
6,096 psi - 4,732 psi = 1,364 psi (A).

This is a simple displacement problem. It's common in industry and textbooks both (good examples are in TS2, TS12, Mian, etc...and if you have a chance to take Wines' class, Bing has the best examples of this problem type I've seen). I referenced this problem using TS12 but as mentioned there are a dozen just like it. All you need is the basic hydrostatic equation (0.052(D)MW) and to understand how cement is generally pumped:

Pipeline: 2017 #79

Problem 79. A 26 mile, 18 inches internal diameter pipeline in good condition is flowing methane...50 dF & 500 psig and falls to 100 psig at the sales point; water content is 6 lbm water per MMSCF. Sales (MBOE/D) is best estimated to be: (A) 29. (B) 34. (C) 39. (D) 44.
Note: Older versions erroneously had "millions" in the answer wording (corrected here). 

This problem is a Panhandle problem as per 8 FAC 1.

Methane has a SG of 0.56 and a z of ~ 0.9 for these conditions. The water vapor of 6 lbm/MMSCF is small (sales quality for many places) so should not overly effect the Panhandle equation result of ~240 MMSCF of sales quality gas. Next, we divide by 6 for ~40 MBOE. Again, the water content is sales quality so the BOE calculation should hold within the answer the answer key margins.

NPV, dNCF, PVR: 2017 #76

A tool was purchased at the start of 2011 for $240M. It returned $65M each year...sold for half price end of 2016. 6% discount rate. NPV, dNCF, & PVR are closest to ($M)?

Solution:
NPV: -240+61+58+55+51+49+46+85 = 165 (A)
dNCF: 153.8+240 = 404
PVR: 164/240 = 0.68.

Since 2/3 of the choices are closest to A, we select (A). Source: 10 ECN 1.

Dipping Sand: 2017 #75

Problem 75. A company in the Gulf drills into the top of a trapped sand at 5,162 ft using 9.2 lb/gal mud. This sand is known to dip from 5,162 ft to 9,036 ft, and to be filled with a 0.81 lb/gal gas down to the GWC at about 6,160 ft. The mud weight situation upon entering this sand is closest to:

(A) 350 psi overbalanced
(B) 350 psi underbalanced
(C) 150 psi underbalanced
(D) 150 psi overbalanced


Note: early editions of this problem's answer options were wrong. If your exam shows something other than above for A-D, use this. I'll have an update available to download soon. Solution: see 2 DRL 1.

GWC psi = 0.465(6160) = 2864 psi (pressure 98' below bit using GOM gradient 2 DRL 3).
Subtract gas hydrostatic: 0.052(0.81)998 = 42 psi from GWC psi: 2864 – 42 = 2822 psi (this is psi the bit sees when drilled into top of gas).
Finally, mud weight: 0.052(5162)9.2=2470 psi (MW hydrostatic at depth bit is at).
Balance: 2470-2822= 352 underbalanced (B).