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.
when it says it is an 800 bbl mud system, is that not a fixed volume? Looking at it that way requires discarding 43.2 bbls and adding 635 sx of barite. Will it specifically state that it is a fixed volume or else assume unlimited with additives?
ReplyDeleteIt's a good question. Based on the problem wording it's 800 bbl fixed weighting up directly and just roll with this. SPE gave this type of problem with similar wording for 1991 practice & they solved it like I do here. Check?
DeleteI carried more significant digits, KMW = 11.16 => volume =43.2 bbls => 634.7 sx. This makes it closer to A than B. What is your recommendation on significant digits? Got impression from Bing that more is better, but he did say triple digit accuracy. So, do mud weights in XX.x accuracy?
DeleteGreat catch Dave I'll fix the solution key.
DeleteMy problem wasn't so much triple digit but I didn't even to double :-).
I must say I am not a driller so excuse my noob question. I would think that the well would fracture right below the shoe of the surface casing.... and I would use the depth of 5000 ft to solve this problem. But of corse I dont know where the LOT was taken at the new reservoir or at the shoe of the casing. My understanding is they take LOTs at the shoe to ensure they dont go over the limit as they go down. Please correct me if I am wrong.
ReplyDeleteThanks
I understand your confusion. I wrote 6 DTC 6 to clarify this exact issue.
Delete6 DTC 6 shows two equations: MW_max & MW_min.
1) max MW before fracking formation. 2) min MW before getting a kick.
On this problem it only asks for kill mud weight, which is #2. It doesn't ask if KMW will exceed your LOT limits. The extra info is just there to confuse.
Make sense? Or do you see something I'm missing? I just whipped this one out...
Warning though; be VERY careful about the trip margin or the kick margin. They do different things and have opposite signs. 6 DTC 6 is very good for quick use here, so you don't have to think much in the heat of an exam...
Thank you for explaining this David, it makes total sense now that you explaind it. I did confuse the 2 (KM/TM) and did not think to double check which one is for what before doing this problem. I thought LOT must be the first scenario. Thanks again
ReplyDeleteIs there a mistake in the exam reference guide where the denominator is 35-Pm2 and looks like it should be 35-Pm1?
ReplyDeleteIsn't one sack= 94 lbm?
ReplyDeleteIt sounds you are right. In Ali's book it is 35-Pm2. so answer is 671.
ReplyDeleteSame issue, so many errors in the reference guide. Do you think they will be corrected once the version we look at on the CBE?
ReplyDeleteI wouldn't sweat it. Trust me, any errors that find their way into solutions will be caught upon grading and just voided out. But what I wouldn't do is spend a lot of time trying to figure something out I thought was and error in the reference. Rather, I would just assume I was missing something, do another check or two, and move on. 99% of the time it will be your mistake due to a nasty trick the problem has. I found mistakes I made after the 3rd pass, when I was so SURE I was right...only to spot my mistake at the last minute. But hey, I stayed up the night before and forgot my NoDoz, so I was pretty punchy and they were probably easy :-).
DeleteI think Ali's formula is right and the formula above is wrong and should be 671 sacks, can you please confirm.
ReplyDeleteThis comment has been removed by the author.
DeleteHugo, I think you are talking about Ali's Drilling Engineering Page 45: "Mud Weight Increase With 100# Sacks of Barite" 1470 [(p2-p1)/(35-p2)] sk/100 bbl?
DeleteHugo, I think you are correct, and the SPE problem solution I used is wrong and Ali's formula is correct for this use. I will be fixing it. Thanks! BTW a good set of examples on these kind of problems is in TS12.
DeleteHello Dave, per Ali's guidebook, the KWM formula is: (SIDPP / (0.052 x TVD)) + OMW. Is there a reason why your equation doesn't incorporate the original mud weight? Thanks
ReplyDelete