Reservoir Gradient: 2021 #13

Well 34X was drilled and completed, but there is debate about the quality of the wireline logs. During well testing, 34X produced 10,200 SCF of 0.8 specific gravity gas with 20 STB of 30 API oil. If the reservoir has a FVF = 1.4 bbl/STB, the fluid gradient is closest to:

A) 0.28 psi/ft
B) 0.29 psi/ft
C) 0.30 psi/ft
D) 0.31 psi/ft.

This problem can be solved using the new Reference exclusively. A similar problem can be found in the Guidebook 13 RES 9. TS8 P33-35 has some good explanations as well.

Fluid Gradient: 2016 #35

Problem 35. A well produces 100 MSCF of gas with a SG of 0.8 along with 200 STB of 30 API oil. The oil FVF is 1.4 RB/STB. The 200 dF reservoir has a fluid gradient closest to: (A) 0.28; (B) 0.30; (C) 0.32; (D) There is not enough information given to solve the problem.

This is the example problem in the Guidebook on 13 RES 9 in the latest edition, and solves to (B). It's good to think about all the different ways this problem can be written so you won't be blindsided.

In this problem, solve for unitless Rs (use 5.615 SCF/STB constant) to find the density of both the oil & gas (from the given API & SGg) using lbm/SCF. Then find Bo from Standing's equation. I don't think the provided Reference includes Standing's equation or the API/SG conversion, but don't let that lull you into complacency; the exam could just provide the equation with a bunch of other stuff to confuse you. Note that SPE TS8 by Towler has this sort of problem, so it's absolutely fair game.

The constant needed for this problem, however, is in the new Reference Guide (provided on the exam) on page 190 of chapter 7 under volume, but you probably have this conversion memorized.

IMO this multi-step problem is kinda amazing inasmuch as some guy using oil/gas from the separator can calculate his reservoir gradient 10,000 ft below his feet with just the FVF.

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).