Problem 27. 4,200 psig reservoir; 9,000 ft TVD. 100 bbl of mud is in 5.5 inch 26 lb/ft casing; 0.1 psi/ft gas cushion.
The required height of the gas (ft), mud weight (lb/gal), and surface pressure (psig) for 200 psi underbalance are nearest to, respectively:
(A) 4,100; 10.1; 970
(B) 3,900; 10.3; 950
(C) 4,000; 10.2; 1,100
(D) 4,000; 10.0; 1,010
This one is pretty simple if you don't get cross-threaded. Steps:
1. 100 bbl/0.02 bbl/ft = 5,000 ft of mud.
2. 9,000-5,000 = 4,000 height of gas (this eliminates A & B).
3. 4,000(.1) = 400 psi (gas)
4. Pr = 4,200 psi (reservoir) – 200 psi (underbalance) = 4,000 psi
Using the given mud weights of 10.2 & 10.0, the hydrostatic calcs sum to a surface pressures of 4,152 psi for (C), and 4,010 psi for (D). Since D is only +10, it's the answer.
Could you please explain how do you get 4152 psi/4010 psi. If the mud column is only 5000 ft shouldn't it be 0.052*10ppg*5000ft = 2600 psi. where am I getting it wrong? thanks!
ReplyDeleteYou are right for the mud column. But that's just one piece of the 5-part pressure calc (Pres, Punderbal, Pmud, Pgas, Psurface). 5000 ft mud, 4000 ft of gas (400 psi) then the surface pressure, which all must sum to 200 psi underbalanced of 4200 psi. Does this check out?
DeleteBut if you use A) 4100, 10.1 and 970 .... the BHP is 4006 psi, which is closer to 4000.
DeleteDon't understand your Q what specifically do you not like on the solution?
DeleteDave, i'm still lost on how you are calculating 4152 psi and 4010 psi from the assumed MW's. Could you please write out the complete equation or location in the guidebook to an example of this?
DeleteThanks! makes sense now
ReplyDeleteUnrelated to this but could you make a post covering the API standards and how to use them during the exam? I see them listed in GB but I am at a loss as to how to solve problems involving API standards. thanks.
ReplyDeleteDo you have a solution to #28?
ReplyDeleteReally my confusion is just coming from the Volume calculation in the Cs = Cwb*Vwb step before plugging that into the equation at the bottom of 12 WLT 13.
DeleteOkay if my thinking is correct, you use inside diameter of 4.5 12.6 lb/ft tubing which I got to be 3.958 inches. Divide that by 2 to get radius so r = 1.979 inches. Convert to feet and use V = pir^2h with h being 2630 feet. Convert to barrels. Plug and Chug and I got 11.25hrs. Is this thinking correct?
DeleteI'll put up the solution by tomorrow
Deletethank you!
DeleteCody, does this check out? Let me know if you see something amiss...I went through this without checking...
DeleteYes that is the solution I came up with. Thanks!
Deleteplease who knows the answer ?
ReplyDeleteThe following data are pertinent to current reservoir condition
Buble point pressuer = 4600 psig
Average reservoir pressuer = 4500 psig
Wellbore radius = 0.5 ft
Drainage radius = 3000 ft
Water cut (fw) = 0
Stabilized test data
Pwf (psig) Q (STB/Day)
4000 500
A- Determine the oil production rate under current reservoir condition when following bottom hole pressure is 2000 psig (10 Marks)
B- After performing a hydraulic fracturing job, well started producing 25% water (fw = 0.25). a production test was conducted after the frac job and the following data were obtained (15 Marks)
Pwf (psig) Q (STB/Day)
3662 2000
Determine the anticipated oil production rate (STB/Day) after the frac job at the flowing bottom hole pressure of 2000 psig. Is this a successful frac job? Explain!
Hello David! Quick question, where does the 0.02 bbl/ft come from in line item #1 of the solution? Thanks.
ReplyDelete0.02bbls/ft is the capacity of 5.5" 26# casing which the red book shows has an ID of 4.548". 4.548^2/1029.4 gives the capacity.
Delete