On Passing the Petroleum Professional Engineering Exam
Tuesday, December 22, 2020
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.
If efficiency is intended, don't you retain as much 9.5 ppg mud as possible? That is to say retain (.03/.05)*950=570 bbls, dumping 380 bbl. I believe this requires only 230 bbl for the "something else" dilution and only 2080 sks of barite, leading to answer B).
Thanks for the comment; efficient isn't a good word because we don't know the rig situation nor what is to be efficient - time, materials, process. I've fixed it to make it more clear, let me know if that check out for you (the problem is hard enough already without making it clear how to approach it, that's for sure!).
If efficiency is intended, don't you retain as much 9.5 ppg mud as possible? That is to say retain (.03/.05)*950=570 bbls, dumping 380 bbl. I believe this requires only 230 bbl for the "something else" dilution and only 2080 sks of barite, leading to answer B).
ReplyDeleteThanks for the comment; efficient isn't a good word because we don't know the rig situation nor what is to be efficient - time, materials, process. I've fixed it to make it more clear, let me know if that check out for you (the problem is hard enough already without making it clear how to approach it, that's for sure!).
DeleteHow do you know 219,000 lbm converts to 2190 sks of barite? Where do we know that 100 lbm = 1 sack of barite? Thanks.
ReplyDelete