Heat Transfer: 2018 #69

Problem 69: A 5.8-mile flowline moves 10 MMscf/D of natural gas of SG 0.7. Indirect fired heaters along the flowline to prevent hydrates. Using flowline temperature & pressure sensors and charts, the NG enthalpy should drop from 234 to 200 Btu/lbm between matching flowline heaters. The flowline heat transfer coefficient is 1.2 Btu/hr-ft^2-deg F. The true rating of the line heaters (MBtu/hr) is closest to: (A) 747; (B) 757; (C) 767; (D) 777.

This can be solved using conversions found on 8 FAC 9 & 9 PVT 1. A similar, more difficult problem can be found in Guide to Professional Registration for Petroleum Engineers, 1991.

This problem is merely a heat-transfer calculation, and so many don't find this problem type applicable to the modern petroleum PE exam. I generally agree. However, I include it because a) it's fair game, and b) it's good practice for heaters & enthalpy, which can find their way into facility-type problems. In the solution, you need to calculate q in lbm/hr to use with the given enthapy:

1. m=0.7(28.97 lbm/mol)(10MM scf/D)/(379.49 mol/scf * 24 hr/D) = 22,267 lbm/hr.
2. q = m(h2-h1) = 22,267 lbm/hr (234-200 Btu/lbm) = 757,031 Btu/hr (B).

It's all dimensional analysis; fast work if you know the needed constants on 8 FAC9 & 9 PVT 1.

Underbalanced Drilling: 2018 #68

Problem 68: The following statement about underbalanced drilling most FALSE is: 

(A) During the workover...a suspension plug is installed...after the well is lubricated to kill fluid. 
(B) Advantages include eliminating differential sticking & faster ROP. 
(C) Disadvantages include possible increased torque & drag & poor MWD compatibility. 
(D) Flow control, not pressure control, is the main well control issue.

This problem is referenced on GB 2 DRL 11 or on HS2 P556, P522, P522, and P520. Note the Guidebook should have the suspension plug step listed before the kill fluid; please pencil this in (fixed in V3).

Note these type of questions could come from any direction; for more detailed information on the read HS P556 (it's quite interesting). Note there are a million ways this problem can be worded; the Guidebook can only introduce and guide you to the subject. You may need to dig deeper using the listed references; this mostly depends on your experience.

Casing Design: 2018 #66

Problem 66: The following 5.5-inch casing options are available for a 11,000 ft hole (the casing will be run empty into 11.5 lb/gal mud):

lb/ft Type $/ft.
14.0 J-55 6.50.
15.5 K-55 7.70.
17.0 K-55 8.80.
17.0 C-75 9.70.
20.0 N-80 10.30.
23.0 N-80 11.60.

Considering collapse, tension, and burst (use max. pressure of 5,000 psi) the most economical casing design option using safety factors of 1.25 (and ignoring buoyancy) will use mostly: (A) J-55 (B) K-75 (C) C-75 (D) N-80.

Use Guidebook 3 HYD 2.  Also A Guide to Professional Registration for Petroleum Engineers, 6th Edition 1991 has a similar example. Steps:

1. Pb = 5,000(1.25) = 6,250 psi. Thus J-55, K-55 fail due to burst; others good.
2. pc at bottom of hole = 0.052(11.5 ppg)11,000 ft(1.25) = 8,223 psi.
3. The next least expensive csg is 17 lb/ft C-75 pc = 6,040 psi.
4. Max. set depth of next least expensive csg: 6040/[0.052(11.5)1.25 = 8,080 ft.

So C-75 used from surface to 8,080 ft; more than half of the 11,000 ft thus (C).

This is an ugly problem for two reasons: 1) 6 csg options to work through, and 2) multiple casings could be together (which means a lot of calculations for tension & burst). So you have to be smart about how you work it plus not fall into the trap that only one casing type must be used (which gives the wrong answer of (D).

Weight Up: 2018 #64

Problem 64: Drill 10.1M to 10.2M ft; pressure 5M psig. At 12M ft ROP increases. Surface casing 5M ft; LOT calculated fracture MW 14.5 ppg. Standpipe pressure is 650 psi. Using a 200-psi trip margin the barite (sacks) needed to weight up the 800 bbl mud system from 9.8 ppg to the KMW? 

Res pressure: Pr = 9.8(0.052)12,000 ft + 650 psi = 6,765 psi.
KWM: = (6765+200)/(0.052*12,000) = 11.16 lbm/gal.
(35 – W1)/(35 – W2) = (11.16 – 9.8)/(35 – 11.16) = 1.057 bbl bar/bbl mud.
Volume = 1.057(800) = 845.6 bbl
Sacks: 845.6 bbl (1/sk/100 lbm)(35 lbm/gal)(42 gal/bbl) = 671 sacks (C).

Note this problem has the wrong answer in the answer key, as does the original SPE source for the problem; I should not have been so trusting.... 

UPDATE: The 2019 Update SPE Petroleum Engineering Reference Guide (provided during the exam) shows the following:

 

I've shown traditional way of solving above, but using the provided equation gets the same thing: 

1470*((11.16-9.8)/(35-11.16)) = 83.9*8 = 671 sacks.

 Note I would read TS12 on P105 to get familiar with this sort of problem. One of the issues with the reference guide is it gives only a few specific equation options for calculating and doesn't show how the equation is a function of Wf/Wi...and the exam most certainly could expect you to understand how this all works for any situation: (1. Weight up add V. 2. Weight up same V. 3. Dilute & dump). I'll include a few types of these problems using the Reference Guide on the 2021 problem set. 

Source: GB 1 DRL 1, 4 MUD 1; A Guide to Prof. Reg. for PEs, SPE, 1991.