Problem 28: An oil well was drilled but facility
limited to 1,000 STB/D. After 2.5 years it started to decline at 7% per month. Production during the
third year (MSTB)? (A) 310; (B) 245; (C) 65; (D) 975.
DCA Decline Curves: 1 Exponential; 2 Hyperbolic; 3 Harmonic. In the Guidebook use 11 DCA 1. In the new Reference Guide
(provided on the exam) use pages 29-31 of chapter 1. Practice finding these equations quickly. Using the new reference:
a) Convert to Nominal decline: Ln(1-0.07)=-.07257/mo.
b) q = qi/exp(d*t) so 1000*(EXP(-0.07257*6)) = 647 BPD (rate after 6 months decline).
c) Np=[(qi-q)/D]*c = [(1000-647.0)/0.07257]*30.4 = 147,874 bbl production second half year 3.
d) 1000*182.5 = 182,500 BO (production first half year 3).
e) 147,874+182,500 = 330,374 BO total production year 3 or (A).
Practice DCA problems on the SPE 2005 Exam, this blog, and the 2016, 2017, and 2018 problem sets until you can do
them fast and without error. You really can't practice these enough.
David can you expand on part d? I don't understand why your adding the 1000 and 647 rates?
ReplyDeleteFacility limited to 1,000 STBD. Check?
DeleteUnderstood now. I was trying to use the Cumulative prod equation to solve this whereas i can just calculate the area under the curve (integrate).
DeleteI'm convinced DCA problems are some of the hardest problems just because they are seemingly simple yet there are a dozen ways to go wrong. So I never overlook them and always check three times. When I took the exam I remember a DCA problem I solved twice and was absoluetly sure I had cold then only at the last second noticed an error I had by misreading the problem. So be afraid!
DeleteDavid - can you please elaborate on parts d and e? I am struggling with these steps even after reading the previous comments. In part d, why is (1,000 bopd + 647 bopd) multiplied by 0.5? The 6 is for 6 months right? In part e, why is the first half of year 3 at 1,000 bopd if production declined from 1,000 bopd to 647 bopd from year 2.5 to year 3?
DeleteAnon, Do each step looking closely at the units in each variable as you go. Remember, you go fro "month" to "day" moving through the steps, and end with "BO" without any time variable. Check?
DeleteI'm still not getting this one. Why is 1,000 added to 647 in step d? Also, why does the first half of year 3 use a production rate of 1,000 bopd if the well started declining 7% per month after 2.5 years? Sorry if I am missing something simple.
DeleteUsing the SPE reference guide
ReplyDeleteYear 3 = 6 months at 1000 bpd + 6 months at 7% decline.
a) Convert to Nominal decline: Ln(1-0.07)=-.07257/mo
b) q = qi*e^(-d*t) = 1000 * e^(-0.073*6) = 645 bpd
c) Np=[(qi-q)/D]*c = [(1000-645)/0.073]*30.4 = 147,835 bbl
d) First half of year 3: 1000*30.4*6 = 182,400 bbl.
e) Total production for all 12 months: 147,835+182,400 = ~330,200 bbl (A).