Pumping: 2021 #29

A casing string is to be cemented at 10,000 ft in a well containing 10-lbm/gal mud, which will be displaced from the annulus by 500 ft of 8 lbm/gal mud flush, 1,500 ft of 12.5 lbm/gal filler cement, and 1,000 ft of 16.5 lbm/gal cement. The pump pressure (Mpsig) required to completely displace the cement from the casing with 9-lbm/gal brine is closest to: A) 1; B) 1.1; C) 1.2; D) 1.3.

Solution:
10000-500-1500-1000 = 7000 ft 9-lbm/gal.
0.052*((500*8)+(1500*12.5)+(1000*16.5)+(10*7000))=5681psig.
10,000*9=4680 psig.
5946 – 4680=1001 psig, “A”.
(TS12 P186; GB 3 HYD 2; RG P41, P67).

Mud Calculations: 2021 #5

A rig has 950 bbl of 9.5 lbm/gal, 5% solids content mud. To deal with a high-pressure zone over the next hole section 800 bbl of 14 lbm/gal, 3% solids content mud is needed. Assuming the rig dumps and dilutes before deciding to use 14 lbm/gal mud, the quantity of barite required (sacks) is closest to: A) 1,800; B) 2,000; C) 2,200; D) 2,400.

Click the button for the answer, commentary, and references sourced. Feel free to leave comments or questions.

This is a real-world problem type similar to one in TS12, P106-107, so know how to do it. There are four basic steps:
1. Dump 150 bbl of the 9.5 lbm/gal mud for the required 800 bbl (note this is still 5% solids content mud).
2. Dilute the mud's solid content from 5% to 3%. Since solids are a mixture the dilution calculation is simple: 800 bbl(0.03/0.05) = 480 bbl (20,160 gal) of 5% and 320 bbls of "something else" to make 800 bbl (33,600 gal) of mud with of 3% solids.
3. Weight up to 14 ppg, so our "something else" is barite and water (that is, 4 additives; i=initial mud, w=water, b=barite, f=final mud).
Equations required: mass and volume balance mi+mw+mb=mf and Vi+Vw+Vb=Vf (units lbm and gal). 2 equations, 2 unknowns, so can be solved.
4. Solving the equations:
20160*9.5+mw+mb=33600*14 and 20160+Vw+Vb=33600.
191520+mw+mb=470400 and Vw+Vb=13440.
mw=278880-mb and mw/8.33ppg+mb/35ppg=13440.
mw=278880-mb and 0.12mw+0.02857mb=13440.
0.12(278880-mb)+0.02857mb=13440.
33466-0.12mb+0.02857mb=13440.
-0.09143mb=-20026.
mb=219,000 lbm = 2190 sks barite or C). [Source: TS 12 P106-107]

Remember to keep three significant digits on mud problems. I made this problem similar to one in TS12 as they solve it using equations that you won't have on this exam, but it's possible a few of you may be steely-eyed missile men and have them memorized. If so, this way you can check your method's results there.

Note this would be a very hard problem for the exam; I would say only 1/10 could get this correct in a reasonable time, so don't despair, keep at it until you understand all the different ways the problem could go. Once you've practiced these types of problems you can crank them out with ease.

Mud: 2021 #4

True statements about drilling fluids include (select any that apply):
__ The majority of wells today are drilled with water-based drilling fluids.
__ The use of pneumatic drilling fluids is limited to depleted zones or areas where the formations are low pressured.
__ Polymers are used to prevent clay dispersion in freshwater or seawater-based drilling fluids.
__ Oil-based fluids typically include 10-40% water.
__ One function of a drilling fluid can be categorized as: ensure maximum logging information.
__ One function of a drilling fluid can be categorized as: support part of the drillstring and casing weight.
__ Poor fluid loss control can cause surge, swab, and circulation-pressure problems.

Click the button for the answer, along with commentary and SPE references sourced. Note that the provided SPE Reference Guide will not help at all on these types of problems. Feel free to ask questions in the comment box below.

True statements about drilling fluids include (select any that apply):
 X  The majority of wells today are drilled with water-based drilling fluids.
 X  The use of pneumatic drilling fluids is limited to depleted zones or areas where the formations are low pressured.
 X  Polymers are used to prevent clay dispersion in freshwater or seawater-based drilling fluids.
 X  Oil-based fluids typically include 10-40% water.
 X  One function of a drilling fluid can be categorized as: ensure maximum logging information.
 X  One function of a drilling fluid can be categorized as: support part of the drillstring and casing weight.
 X  Poor fluid loss control can cause surge, swab and circulation-pressure problems.

Note that this "multi-choice" format in the new EBT exam makes it extremely difficult to know if you have it fully correct. The 2021 practice problem sets have many in this format (likely more than the real exam) because it's a good way to test knowledge.

Source HS2 (P89-93).
The majority of wells are drilled with water-based drilling fluids (P91).
The use of pneumatic drilling fluids (i.e. air, gas, and foam)is limited to depleted zones or areas where the formations are low pressured (P90).
Polymers are used to provide viscosity, fluid-loss control, shale inhibition, and prevention of clay dispersion in freshwater or seawater-based drilling fluids (P91).
OBF: The oil/water ratios typically range from 90:10 to 60:40 (P93).
The function of a drilling fluid can be categorized as: ensure maximum logging information (P89).
The function of a drilling fluid can be categorized as: support part of the drillstring and casing weight (P89). Poor fluid loss can cause surge (an increase in wellbore pressure under a bit from running into the hole), swab (a decrease in wellbore pressure under a bit from running out of the hole), and circulation-pressure problems (known as ECD) (P89).

Well Control: 2021 #3

When drilling an oil well using a rig with a 850 bbl mud system a sand was encountered from 9,000 ft to 10,200 ft. Reservoir pressure was measured at 4,950 psi.

Drilling ahead with a 9.8 lbm/gal mud weight continued until 12,000 ft, at which time the ROP rapidly increased and a kick suspected so the well was shut in. Drillpipe pressure was recorded at 650 psi. Company policy requires a trip margin of 200 psi; the barite (sacks) needed is closest to: A) 700; B) 705; C) 710; D) 715.

Click the button for the answer, commentary, and the SPE Reference Guide pages sourced. Feel free to leave comments or questions.

This problem requires basic drilling knowledge, has three distinct steps with two tricks. But all the needed equations are provided in the SPE Reference. One factor not provided in the Reference is how a "trip margin" fits into the equations; you can explore this subject in TS12 or the Guidebook (3 HYD 5).

FP = (9.8*0.052*12,000)+650 = 6,765 psi. [SPE RG P41, P47]
Note how the first reservoir pressure is a red herring, but many get fooled by this sort of misdirection.

KWM = (6,765 psi + 200 psi)/(0.052 psi-ft/ppg*12,000 ft) = 11.162 ppg. [SPE RG P48]
Trip or kick margins (given in pressure here) can be in ppg as well (the Guidebook has a section on trip and kick margins; these are tricky and you won't get any help from the provided SPE Reference Guide).

Barite = [1470*((11.162-9.8)/(35-11.162))] = 83.989 sks/100 bbl = (84 sks/100 bbl)*(850 bbl)=714 sks. [SPE RG P45]
Never forget this equation uses per 100# sk/100 bbl.
D) 715 sks.

Dipping Sand: 2017 #75

Problem 75. A company in the Gulf drills into the top of a trapped sand at 5,162 ft using 9.2 lb/gal mud. This sand is known to dip from 5,162 ft to 9,036 ft, and to be filled with a 0.81 lb/gal gas down to the GWC at about 6,160 ft. The mud weight situation upon entering this sand is closest to:

(A) 350 psi overbalanced
(B) 350 psi underbalanced
(C) 150 psi underbalanced
(D) 150 psi overbalanced


Note: early editions of this problem's answer options were wrong. If your exam shows something other than above for A-D, use this. I'll have an update available to download soon. Solution: see 2 DRL 1.

GWC psi = 0.465(6160) = 2864 psi (pressure 98' below bit using GOM gradient 2 DRL 3).
Subtract gas hydrostatic: 0.052(0.81)998 = 42 psi from GWC psi: 2864 – 42 = 2822 psi (this is psi the bit sees when drilled into top of gas).
Finally, mud weight: 0.052(5162)9.2=2470 psi (MW hydrostatic at depth bit is at).
Balance: 2470-2822= 352 underbalanced (B).

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

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.