Klinkenburg: 2021 #18

This is a fairly tough word problem, and it shows how difficult reasonable questions can be when mixed in with so many different subjects on an 8 hour exam. To properly prepare one would have to read hundreds of pages out of the Handbook Series, have a very good memory, and then hope for the best. 

This problem, in contrast, culls any direct SPE quotes that could "reasonably" be asked on the subject. So: simply read this problem set and thus get a basic understanding and be fully armed with little effort. 

Make a point to avoid getting bogged down on the details, details that simply cannot be reasonably asked on this type of exam.

Core Testing: 2021 #17

Effective liquid permeability is found in the lab by graphing gas permeability versus the reciprocal mean flowing pressure and extrapolating the reciprocal mean pressure to zero. In this problem, a rock core filled with gas “A” has a permeability of 40 md with an average flowing pressure of 1.25 atm and has a permeability of 30 md when said mean flowing pressure is doubled. If this same rock core is filled with gas “B” and then has a permeability of 40 md with a flowing pressure of 2.5 atm, the permeability (md) for gas “B” at a flowing pressure of 5 atm is closest to: A) 30 B) 25 C) 20 D) 15.

kg=30 md, “A”.
(GB 12 WLT 1; HS5 P994-995).

Equivalent Liquid Perm: 2017 #7

A gas filled rock core has an average flowing pressure of 1.67 atm with a corresponding gas permeability of 35 md. When the mean flowing pressure is raised to 2.5 atm the gas permeability is 40 30 md. This core’s equivalent liquid permeability is nearest to:

This problem is just like the GB example. We find mx + b then project to the x axis:
1) m = (30 - 35)/(1/2.5 - 1/1.67)
    m = -5/(0.4 - 0.6)
    -m = -5/0.2 = -25 (this is the slope).
2) -25(0.6) = -15.
3) -15 + 35 =  20 or (D).

The inverse slope makes the math awkward, so I recommend doing a quick graph sketch (not on graph paper, you won't have any). On this problem the graph is identical to the Guidebook's so it's easy. If you had a hard time with this one, redo it over and over until it makes intuitive sense. For more examples check out Bradley, problems online, or the 2016 Practice Exam. Fair warning: many smart guys unexpectedly miss this problem type under time pressure. Don't trust your math; graph it out and ensure the answer "makes sense".

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).

Core Permeability: 2018 #5

Problem 5. The statement most FALSE regarding permeability determination (especially core permeability determination) is:

(A) Klinkenburg used gas in place of nonreacting fluids, but he corrected for slippage.
(B) Water is the most frequently occurring reactive liquid relating to permeability determination.
(C) Slippage occurs when the diameter of the capillary openings approaches the mean free path of the gas.
(D) Carbon dioxide should not be used to find the equivalent liquid permeability of a core.

The TOC shows "Core" on 12 WLT 1. This page clearly states "any gas" can be used. Since CO2 is indeed a gas "D" is plainly false. Done.

Confirmation is nice, so I check out the listed reference: Petroleum Engineering Handbook, P26-18, Bradley. Sure enough, CO2 is shown as a test gas. Being an SPE reference, this locks it.