Rod Pump Counterbalance: 2016 #45

A 160-173-54 pumping unit runs at 16 SPM producing 179 BOPD and 21 BWPD out of 2 inch anchored tubing. The 7/6 rods ...counterweight required? 

Counterbalance problems are meat & potatoes on exams. Why? No fancy charts are needed like most rod pump problems. This can be solved using just two pages in the Guidebook and on the EBT any equations/data not found in the provided reference are easy to give for the specific problem. Now that's not saying there isn't plenty of room for error. I make mistakes on these problems all the time. For this one:

1) 7/6 1.5” rods; Wr = 1.833 lb/ft (7 PRD 10) 
2) W = 1.833(5,000) = 9,165 lbs 
3) Wfr = W(1-0.128G) = 9,165(1-(0.128*0.9)) = 8,109 (7 PRD 9) 
4) Fo=0.34*0.9*1.5^2*4500 = 3098 
5) CBE = 1.06(8,109 + (0.5*3,098)) = 10,238 (C).

Flanges: 2016 #41

Question: What flanges classes can be used in a 200 F & 1,450 psig flowstream (SF = 2, or 2,900 psi)?

I just glanced at the cover TOC and scanned to Facilities (FAC). "Flange Ratings" was 8 FAC 7. Less than 60 seconds to size up the problem and find my page with all the needed tables. Note these tables are also in HS3-356 (tab this page).

Answers (C) and (D) gave API Class 2000; 2,000 psi. So they are eliminated.
ANSI Class 900?  2,025 psi. So it can't be (A) either. This has taken me two minutes.
API Class 3000? 3000 psi. ASME 1500? 3,375 psi. So we have a winner, (B).

Less than 3 minutes; I take another minute to review for tricks, finding none, I move on.

Note on the new EBT exam they can merely provide the needed charts, much like the Guidebook does. 

Frac Gradient: 2016 #39

A frac with 8.5 lb/gal fluid, tubing pressure loss of 500 psia, and perforations centered at 5,000 TVD ft had negligible pressure losses in the prefs. ISIP = 1,525 psig.  The frac gradient (psi/ft) was most nearly: (A) 0.65 (B) 0.70 (C) 0.75 (D) 0.80. (Note: I've left out all the unneeded information, which makes the problem much harder).

Note this sort of equation is unlikely to be provided in the reference, but it is simple enough an engineer should know it. 

Solution by using equations on 7 PRD 2:
1) Find hydrostatic pressure at perfs: 0.052(8.5)5000 = 2210 psi.
2) Add the ISIP for total pressure at perfs: 2210 + 1525 = 3735 psi.
3) Calculate the gradient to perfs: 3735/5000 = 0.75 psi/ft or (C).

LOT: 2016 #37

Problem 37. Which statement below is most TRUE? 

(A) When a LOT or FIT fails a cement squeeze is necessary before resuming drilling.
(B) A LOT and a FIT are two different names for the same test.
(C) A LOT tests the rock for failure against a preset pressure; an FIT tests...without a preset pressure.
(D) A FIT measures...when the rock starts to fracture...a LOT measures...when...completely fails.

Choice A; the answer, has been edited due to lack of clarity on the original. It is now nearly a direct quote from the Guidebook/HS2.

Diagenetic Porosity: 2016 #12

An oil reservoir has average diagenetic porosity of 10% and a CNL log measures a porosity of 15%. The oil reservoir is 200 acres, 10 ft thick, with residual oil saturation & initial water saturation of 25% each. The initial oil formation volume factor is 1.2 RB/STB. The maximum oil production from primary porosity would be closest to: (A) 320 MSTB (B) 340 MSTB (C) 620 MSTB (D) 640 MSTB

Definitions from 15 LOG 4:
Total Porosity: (measured with nuclear tools) …equals primary porosity + secondary porosity.
Primary Porosity (apparent, intergranular) used for reserves or maximum producible oil…equals total porosity – secondary porosity.
Secondary Porosity (isolated pores, vugs, and fractures) also called diagenetic porosity …may be overlooked by acoustic-logs …equals total porosity – primary porosity

Therefore, for this problem, primary porosity is: 0.15 – 0.10 = 0.05 pu. And maximum oi production from primary porosity is: 7758(200)10(1-.25-.25)0.05(1/1.2) = 323MSTB or (A).