Csg MW Max/Min: 2017 #36

For casing set depth:
MW max is limited by prior casing shoe fracture pressure (minus kick margin).
MW min is limited by BHP (plus trip margin).

In this problem:

3,000 psi = 0.052(MWmax)4,000 ft
...so MW max = 14.4 ppg

 6,000 psi = 0.052(MWmax)9,000 ft
 ...so MW max = 12.8 + 0.2 = 13.0 ppg

Well Communication: 2017 #35

Problem 35. Your company investigates a lease sale containing two wells, A and B. Kelly bushing elevations are 1,000 ft and 0 ft above MSL, respectively, and bottom hole pressures of 3,170 and 3,500 psi, respectively. Both wells are 11,000 ft vertical depth to the perforations with pressure gradients of 0.3 psi per foot. The seller claims the two wells are in communication. Which of the following is most likely TRUE?
 
(A) The seller is incorrect; the wells are about 6 percent different in reservoir pressure and unlikely to be in communication.
(B) The seller is correct; the two wells are about 1 percent different in reservoir pressure and thus likely to be in communication.
(C) The seller is incorrect; the wells are over 10 percent different in reservoir pressure and unlikely to be in communication.
(D) The seller is correct; the two wells are about 4 percent different in reservoir pressure and thus likely to be in communication.

This problem is pretty straightforward and calculates to "B". But it requires practical but diverse petroleum engineering knowledge that trips many up.

Specifically: how much pressure difference is there within a reservoir from well to well? How to calculate pressures in wells that have different MSL depths? Nothing complex, just meat-and-potatoes for the average experienced engineer, but it can take time to think it through and look for possible tricks.

If you have more questions (or notice an error) note it in the comments below. For more help, this type of problem is in 13 RES 8 and SPE TS8 (where I got the idea for it). Myself, I'm used to drilling in the flats...for Rocky Mountain types this sort of problem is probably a yawn...

Reciprocating Compressor VE: 2017 #32

This problem uses the standard reciprocating compressor volumetric efficiency equation (found in the Guidebook or the SPE Handbook Series). It can be difficult to recognize this, though, when you are merely asked for a gas rate.

It's fairly plug-n-chug once you have the necessary equation. The only trick? "L" (gas slippage) is not given so you must draw on experience assuming it's between 0-5%. When I did this problem I used 0%, then 5%, and found both gave the same answer "B": (A) 7.4 SCF/min (B) 8.4 SCF/min (C) 9.4 SCF/min (D) 6.4 SCF/min.

Displacement: 2017 #20

5,000' of 14.00 lb/ft drillpipe (adjusted weight is 1.1 times nominal) and 100 ft of 120 lb/ft drill collars are pulled dry. Fluid volume change?

14 ppf(1.1)5,000' =  77M lb.
120 ppf(100') = 12M lb.
89 M lb/(2751 lb/bbl) = 32 bbl (A).

The trick on this one is to use adjusted weight whenever given; never use OD & ID unless you don't have adjusted weight. Then use the adjusted weight / density equation (see 1 RIG 1). Know how to do this sort of problem fast.

Note some prefer to use 490 lbm/cf x 5.615 cf/bbl rather than 2,751 lb/bbl that I use here since they have 490 &amp & 5.615 memorized.

ESP: 2017 #19

ESPs have many testing options. In this problem, we are given:

1) SG of the working fluid (10% water SG 1.01, 90% oil SPI 40)
...so SG = 0.1(1.01) + 0.9(0.8244) = 0.84

2) Net Lift 3,500 ft, Friction 300 ft, Surface pressure 200 ft
...so TDH = 3,500 + 300 + 200 = 4,000 ft

3) ESP with head capacity 4,000 ft/100 stages and 1 BHP/stage:
...4,000 ft / (4000 ft/100 stages) = 100 stages
...so BHP = (1 BHP/stage)100 stages(0.84) = 84 BHP

Fairly simple. If you aren't familiar with all three steps before seeing the problem it's hard to do it quickly. So be familiar with all three steps.

Displacement: 2017 #18

This is a fairly easy problem. The annulus is 10,000 ft of 10 ppg mud which weighs in at 5,200 psi.

The drillpipe has three layers: 4,000 ft of 8 ppg (1,664 psi), 4,000 ft 9 ppg (1,872 psi), and 2,000 ft 10 ppg mud for a total 4,576 psi. Thus we need another 625 psi for pressure balance.

This problem is easy, but hey not every problem can be hard.