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.

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.