A gas reservoir produced 1 MMscf gas, 10 MSTB water, and had 1,100 bbl water influx.
Current & initial gas FVFs are...water FVF... IGIP:
These problems are easy minus unit confusion issues. You can't be too careful with units here:
= (1,000 Mscf)(1.1 RB/Mscf ) + 10,000 RB – 1,100 RB
= 1,100 RB + 10,000 RB -1,100 RB = 10,000 RB
= 10,000 RB/(1.1 – 1 RB/Mscf)
= 10,000 RB/(0.1 RB/Mscf)
= 100,000 Mscf, or (B).
Thursday, March 28, 2019
Saturday, March 9, 2019
Drilling Horsepower: 2016 #63
A rig is expected to run at 100 RPM at 14,000 ft depth. What's a back-of-the-envelope estimate of the drilling horsepower required?
For this problem, use a Torque Factor Estimate and the equation F(RPM).
F=1.5 10,000 ft, light), F=1.75 (10,000 ft-15,000 ft, average), and F=2.0 (>15,000 ft, heavy)
Hence, 1.75(100) = 175 hp (A).
For this problem, use a Torque Factor Estimate and the equation F(RPM).
F=1.5 10,000 ft, light), F=1.75 (10,000 ft-15,000 ft, average), and F=2.0 (>15,000 ft, heavy)
Hence, 1.75(100) = 175 hp (A).
I have never seen this pre-drilling estimate in an SPE book (I first saw it in Mian, P288) but I like to use it for a quick check on any horsepower problem, so it's handy to know in real life or even an exam.
Monday, March 4, 2019
Discounted Cash Flow: 2016 #72
Investment of $100M
returns $40M/yr for 3 years when it is sold for $50M. Discount rate 5%. The net discounted cash flow ($1,000/yr)
is closest to?
This is a painfully easy problem. IF you know is that NCF leaves off the initial cost of the item AND it still includes the discounted salvage value. That's the whole ball game.
40(1.05)^-1 + 40(1.05)^-2 + 40(1.05)^-3 +50(1.05)^-3 =
$38.1 + $36.3 + $34.6 + 43.2 = $152 (all values M) or (A).
This is a painfully easy problem. IF you know is that NCF leaves off the initial cost of the item AND it still includes the discounted salvage value. That's the whole ball game.
40(1.05)^-1 + 40(1.05)^-2 + 40(1.05)^-3 +50(1.05)^-3 =
$38.1 + $36.3 + $34.6 + 43.2 = $152 (all values M) or (A).
Saturday, March 2, 2019
Fracture: 2017 #39
As I've said before fracture treatments are simple hydraulics. But I hate them; I get all worked just thinking about the myriad of ways I tend to get them wrong. The Guidebook has a single frac page, 7 PRD 2 in a desperate attempt to keep it simple. But not too simple!
Getting an initial shut-in pressure (ISIP, or just the wellhead pressure) is a common calculation. In this problem ISIP calculates to 1,600 psi (depth 10,000 ft TVD, MW = 8.46 ppg, reservoir pressure 2,800). But ISIP already given, anyway.
All the problem wants? Required (or theoretical) hydraulic pump power (in HP) to frac the reservoir. This is just [ISIP*Qbpm]/40.8 (equation conveniently shown the bottom of the GB frac page). Don't use the starting wellhead pressure. Since the frac is pumped at 50 bbl/min: (1,600)50(1/40.8) = 1,961 hp or (C).
Getting an initial shut-in pressure (ISIP, or just the wellhead pressure) is a common calculation. In this problem ISIP calculates to 1,600 psi (depth 10,000 ft TVD, MW = 8.46 ppg, reservoir pressure 2,800). But ISIP already given, anyway.
All the problem wants? Required (or theoretical) hydraulic pump power (in HP) to frac the reservoir. This is just [ISIP*Qbpm]/40.8 (equation conveniently shown the bottom of the GB frac page). Don't use the starting wellhead pressure. Since the frac is pumped at 50 bbl/min: (1,600)50(1/40.8) = 1,961 hp or (C).
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