Problem 18. Drilling...760 ft 2-1/4 x 7-3/4 DC...12.3 ppg mud...50M lb WOB. 50 deg hole...friction factor negligible...DC compression (%)?
Solution:
1. 12.3 ppg is 0.812 BF (Redbook S240 P13).
2. 2.25”x7.75” DC is 147 ppf (Guidebook 1 RIG 8, 2 DRL 6).
3. 50M/[760 ft(147 lb/ft)0.812(cos50 deg)] = 0.86 or (A).
Note extra information was removed for clarity; it's not as easy at it looks in this blog post; extra information is a key confusion-creator on a timed exam, and this problem is no exception.
buoyancy factor is 1 - 12.3 / 490 = 0.975
ReplyDeletecorrect me if I'm wrong.
Good question! I have to plunder my memory here:
Delete1. I think you are confusing ppg (65.5 for steel) with ppcf (490 lb/cf of steel).
2. Guidebook Pg V shows BF = 1-12.3ppg mud/65.5 ppg steel = 0.812
3. Guidebook Pg V shows all these conversions; take a look at them because trust me it gets confusing on the exam.
4. I don't know if the new Reference has buoyancy tables? If not, get used to doing this calc using the correct units.