Gas Reservoirs: 2021 #20

True statements about drive mechanisms include (select any that apply):
__ From a reservoir viewpoint, dry and wet gas cannot be treated similarly in terms of production characteristics.
__ From a reservoir viewpoint, dry and wet gas cannot be treated similarly in terms of pressure behavior.
__ From a reservoir viewpoint, dry and wet gas cannot be treated similarly in terms of recovery potential.
__ A retrograde-condensate gas reservoir initially contains a single-phase fluid, which changes to two phases (condensate and gas) in the reservoir when the reservoir pressure decreases.
__ The term “condensate” is often applied to light hydrocarbon liquid produced from a gas well.
__ The term “condensate reservoir” should be applied only to situations in which condensate is actually formed in the reservoir because of retrograde behavior.
__ For both wet and dry gasses, reservoir engineering calculations are based on a single-phase reservoir gas.
__ The effect of a weak to moderate waterdrive is often difficult to detect from a simple p/z plot.
__ Depletion behavior of retrograde-condensate reservoirs can be handled through the p/z analyses with the caveat that the z-factor must be the two-phase z factor.

Once again, remember the provided Reference Guide will offer no assistance on these types of word problems. Most of these word-style questions will come from the SPE Handbook Series (as well as the SPE Textbook Series (TS). So a good study plan is to read the parts of the HS you lack experience with.

Click the button for the answer, along with commentary and SPE references sourced. Feel free to ask questions in the comment box below.

__ From a reservoir viewpoint, dry and wet gas can be treated similarly in terms of production characteristics (HS5 P981).
__ From a reservoir viewpoint, dry and wet gas can be treated similarly in terms of pressure behavior (HS5 P981).
__ From a reservoir viewpoint, dry and wet gas can be treated similarly in terms of recovery potential (HS5 P981).
 X A retrograde-condensate gas reservoir initially contains a single-phase fluid, which changes to two phases (condensate and gas) in the reservoir when the reservoir pressure decreases (HS5 P981).
 X The term “condensate” is often applied to light hydrocarbon liquid produced from a gas well (HS5 P982).
 X The term “condensate reservoir” should be applied only to situations in which condensate is actually formed in the reservoir because of retrograde behavior (HS5 P982).
 X For both wet and dry gasses, reservoir engineering calculations are based on a single-phase reservoir gas (HS5 P983).
 X The effect of a weak to moderate waterdrive is often difficult to detect from a simple p/z plot (HS5 P1022).
 X Depletion behavior of retrograde-condensate reservoirs can be handled through the p/z analyses with the caveat that the z-factor must be the two-phase z factor (HS5 P1024).

Gas Reservoirs: 2021 #19

A gas reservoir produced 1 MMscf gas and 13 MSTB water. The current and initial gas formation value factors...reservoir modeling predicts two equally possible scenarios for water influx...The initial gas in place (MMscf) is most likely closest to: A) 27.1 B) 30.1 C) 33.1 D) There is likely not any water influx. 

This problem is fairly simple; just watch the units. It try to crank these out quickly and let the chips fall where they may, so it wouldn't surprise me if I had an error floating around on this one. Just remember on gas reservoir problems, 90% of the errors are units, and the last 10% are just misreading the problem.

Anyway, I'm posting #19 to address any questions, or even to take suggestions on how it could be modified to better prepare one for gas reservoir problems.

Dipping Sand: 2017 #75

Problem 75. A company in the Gulf drills into the top of a trapped sand at 5,162 ft using 9.2 lb/gal mud. This sand is known to dip from 5,162 ft to 9,036 ft, and to be filled with a 0.81 lb/gal gas down to the GWC at about 6,160 ft. The mud weight situation upon entering this sand is closest to:

(A) 350 psi overbalanced
(B) 350 psi underbalanced
(C) 150 psi underbalanced
(D) 150 psi overbalanced


Note: early editions of this problem's answer options were wrong. If your exam shows something other than above for A-D, use this. I'll have an update available to download soon. Solution: see 2 DRL 1.

GWC psi = 0.465(6160) = 2864 psi (pressure 98' below bit using GOM gradient 2 DRL 3).
Subtract gas hydrostatic: 0.052(0.81)998 = 42 psi from GWC psi: 2864 – 42 = 2822 psi (this is psi the bit sees when drilled into top of gas).
Finally, mud weight: 0.052(5162)9.2=2470 psi (MW hydrostatic at depth bit is at).
Balance: 2470-2822= 352 underbalanced (B).

Gas Reservoir MBE: 2016 #17

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).

Gas Recovery: 2005 #47 (similar)

Say one has a volumetric dry gas reservoir with initial/final pressures & z-factors of 4,800/1,500 psia & 0.98/0.90. Pretty standard numbers. What's the recovery factor?

The Guidebook 13 RES 1 & 2 covers dry gas. Since recovery is just Gp/G, it's easy to move the numbers around from the given equations. One way to do this is by finding the difference between the P/z for initial and final conditions and dividing by P/z initial. In this case, it's just (4,900 - 1,700)/4,900 or 65%.

Be careful not to confuse the z-factor with FVF. If given initial and final FVF the problem is much easier as the Guidebook shows: 1-(Bgi/Bgf). It's common to confuse the two numbers because they often look alike.

Dry Gas Reservoir: 2005 #49 (similar)

The gas MBE is very simple; see 13 RES 2:
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You are usually asked to solve for water influx, since this is the only term you don't know from production volumes. For example, given:

G = 10,000 MMCF, Bgi 1 M RB/MMCF
Gp 1,000 MMCF, Bg 1.3 M RB/MMCF
2,200 MSTB water, Bw = 1 RB/STB

G(Bg-Bgi) = GpBg + WpBw - We

10,000 MMCF(1.3 - 1 M RB/MMCF) =
1,000 MMCF(1.3 M RB/MMCF) + 2,200 M STB(1 M RB/M STB) - We

3,000 M RB = 1,300 M RB + 2,200 M RB - We
3,000 M RB = 3,500 M RB - We

We = 500 M RB

With gas it's all about the units.