Pipe Collapse: 2017 #14

The problem reads: Given 5-1/2” 26 lb/ft N80 casing with 40,000 psi tension: find the reduced collapse rating (if your older copy is missing pipe weight, please assume it):

The whole problem can be solved never leaving 6 DTC 4.
The 80 clues us to an max axial stress of 80,000 psi. Add the internal pressure to compute: (40,000 + 20,000)/80,000 = 0.75.
Enter this into the GB table (ellipse of plasticity) for -0.385.
Solve using GB equation (with collapse rating from HES Redbook of 12,650 psi) for: 12,650(0.385)+20,000 = 24,875 psi.

Diagenetic Porosity: 2017 #12

Porosity calculations are shown on a single GB page: 15 LOG 4. It starts:

Total Porosity: (measured with nuclear tools).
…equals primary porosity + secondary porosity.

Primary Porosity (apparent, intergranular).
Found from Wylie’s (acoustic) porosity equation below.

Secondary Porosity (isolated pores, vugs, and fractures).
also called diagenetic porosity.
…may be overlooked by acoustic-logs.
…equals total porosity – primary porosity.

At a glance, it's easy to see diagenetic porosity is found by nuclear tool porosity minus sonic porosity.

In this problem, nuclear tool porosity is given: 20 pu, and a sonic tool slowness of 79 microseconds/ft over the zone. The known slowness in sandstone & oil (shown in the GB variable box) are 55.5 & 232 microseconds/ft.

Wylie's Equation is next in the GB:
Por sonic = (dt - dtma)/(dtf - dtma) = (79 - 55.5)/(232 - 55.5) = 13 pu.

Since diagenetic = secondary = total - primary: 20 - 13 = 7 pu.

Economics: 2017 #11

Problem 11. Your company invests $100,000 in operating equipment that returns $40,000 annually for 3 years. The interest rate is 5 percent. The equipment has a salvage value of $50,000. The investment’s discounted net cash flow ($ thousand) is closest to:  (A) $142  (B) $152 (C) $42 (D) $52.

Solved in the Guidebook for Net Present Value (NPV).  This problem, however, asks for discounted net cash flow (NCF). So just ignore the initial investment of $100M: answer (B).

This problem is too easy but I include it to help practice reading questions carefully. Note that the NPV answer is included as an option, and since that's a more common calculation it's the kind of mistake can happen to anyone. So always check work.

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".

Metering: 2017 #5

Gas metering problem; four scenarios are given: 1) orifice plate installed backwards, 2) plate leading edge worn, or 3 & 4) flow computer orifice entry too large/small.

The first situation is described on 8 FAC 3 & 4 (negative bias). This infers the answer to the second situation (worn leading edge is then a negative bias). So the only remaining question: does a larger or smaller computer entry for the orifice provide a positive bias on flow?

The answer: a larger value entered for the orifice ID gives a positive bias to flow. You can see this from the Guidebook's Fb table; a larger orifice ID has a larger Fb constant thus giving a larger q.

Yet be ready for pipe diameter computer entry errors as well. Unlike orifice diameter this has an inverse effect on Fb and thus q (again, this can be seen from the Guidebook table). So be careful when reading the wording of the question.

MD from TVD: 2017 #3 & #4

Guidebook 2 DRL 9 has a quick and simple method of calculating directional drilling problems.

This problem asks for MD given TVD, as well as MD if the well turns vertical. The GB shows both the radius of curvature calculation (2,865 ft) as well as the max inc (17.9 degrees).

The MD calculation is as follows:
1,500 + 17.9/(2/100) + [8,200 - 1,500 -2,865sin(17.9)]/[cos(17.9)]
1,500 - 895 + 6,119 = 8,514 ft (C).

In the second part of the problem, if the well turns vertical from 8,200 ft TVD using the same DLS, the 895 ft is merely inverted. So just take the build calculation (inc/build) already done and use it on the drop section. If we start at 8,514 ft, that's 8,514 + 895 = 9,408 ft (C).

Casing: 2018 #80

Problem 80: The most accurate statement about oilfield casing is: 

(A) API classifies casing by five properties, namely: manner of manufacture, steel grade... 
(B) There are only 12 steel API standard casing grades, which includes L-80 and Q-125. 
(C) If oilfield tubulars experience collapse, most do so in “elastic” & “transition” regimes. 
(D) A tieback provides pressure integrity from liner top to wellhead & can be partially cemented.

This solution is found on Guidebook 6 DTC 11. Further information is in HS II.

(A) Length range, not new or used. So false.
(B) 10 steel API standard casing grades, not 12. Also false.
(C) Most collapse is in “plastic” & “transition” regimes, sans "elastic". False.
(D) True; this is what a "tieback" is, and it can indeed be partially cemented. 

Produced Water: 2018 #74

Problem 74: The following statement is most FALSE: 

(A) Chlorine is the most common chemical biocide used in oilfield surface water-injection systems.
(B) Moderate pressure seawater service most commonly uses FRP though sunlight can degrade...
(C) A solid/liquid hydrocyclone cannot remove both primary & secondary solids from surface water.
(D) Surface-water injection is an attractive option as...very little environmental impact or concern. 

Produced water has many potential questions; the Guidebook has a good summary. My recommendation: take an hour and read the referenced HS section with an eye to areas that seem non-intuitive or are unfamiliar to you.

For more information on the above selection options such as biocide, FRP, and water injection, see HS3 P168, P179, P163, and P159 and help clarify this fairly confusing subject.

Compressors: 2018 #70

Problem 70: The following statement about reciprocating compressors is most FALSE: 

(A) Crankshaft counterweights cannot eliminate imbalance forces that cause mechanical vibration.
(B) ...concrete foundations...should increase depth not length or width to meet weight requirements...
(C) A suction scrubber handles liquids injected into the compressor through the inlet gas stream...
(D) Volumetric efficiency is a function of the properties of the gas being compressed.

This answer is found on 8 FAC 8. For further reading see the Guidebook referenced pages from HS2. But it's really common sense; concrete foundations should increase length or width sans depth to meet weight requirements (for soil stability).