Pipeline: 2017 #79

Problem 79. A 26 mile, 18 inches internal diameter pipeline in good condition is flowing methane...50 dF & 500 psig and falls to 100 psig at the sales point; water content is 6 lbm water per MMSCF. Sales (MBOE/D) is best estimated to be: (A) 29. (B) 34. (C) 39. (D) 44.
Note: Older versions erroneously had "millions" in the answer wording (corrected here). 

This problem is a Panhandle problem as per 8 FAC 1.

Methane has a SG of 0.56 and a z of ~ 0.9 for these conditions. The water vapor of 6 lbm/MMSCF is small (sales quality for many places) so should not overly effect the Panhandle equation result of ~240 MMSCF of sales quality gas. Next, we divide by 6 for ~40 MBOE. Again, the water content is sales quality so the BOE calculation should hold within the answer the answer key margins.

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.

Mechanical Energy Balance: 2018 #31

Problem 31. 10 ppg, 20 cp fluid pumped through 697 ft, 3.5-inch OD tubing rising 12 ft...100 ft...down 1.5 ft...203 ft down 10.5 ft into a 10x10x10 ft open tank filling 1 ft/min. Pump pressure (psig)? (A) 800; (B) 810; (C) 820; (D) 830.

This is a basic Mechanical Energy Balance (8 FAC 6) problem. It's simple if you don't forget the kinetic energy; of course, the wrong answer is conveniently waiting for you as a choice if you do forget it (in this case, "B"):

1. HS vertical: 12 – 1.5 – 10.5 = 0 vertical ft
2. Friction: 697 + 100 + 203 = 1,000 ft
3. 34.14 ft/s (2x GB 3 HYD 1) = 0.809 psi/ft(1000 ft) = - 809 psi
4. KE: 8.074E-4(10)(34.14^2-0^2) = - 9.4 psi
5. Total: 818 psi.

NG Processing: 2018 #30

Problem 30. The statement most FALSE about natural gas treating and processing is:

(A) In long distance transmission of sales gas by pipeline, pressure is usually less than 1,000 psig.
(B) A potential cause for treated natural gas “going sour” is too low of an inlet gas rate.
(C) All raw natural gas is fully saturated with water vapor when produced from an...
(D) There are four glycols that are used in removing water vapor from natural gas or in...

This question can be answered quickly knowing the basics of natural gas processing. Or you can quickly glean the answer from HS III, pages 186, 192, and 198. Overall, a pretty easy question, but it can take time if you don't have much experience, or lack the right resources.

Watch the wording like a hawk. It's always best to find a direct quote from an SPE source for these types of problems if the wording seems vague.