Replies to This Discussion

Permalink Reply by Les Bamburg on April 23, 2011 at 2:42

CU, I would say 50 Bcf per section is still a common estimate of recoverable gas with a range of 32 Bcf to 80 Bcf per section in the primary play area.  This would equate to 50 to 125 MMcf per acre.

 

Of course some areas have both Haynesville Shale and Bossier Shale potential and may approach 100 Bcf per section.

Permalink Reply by Coyote Ugly on April 23, 2011 at 9:05

Thanks Les B.

CU

Permalink Reply by Coyote Ugly on April 23, 2011 at 9:04

Thanks Les B

Permalink Reply by Coyote Ugly on April 23, 2011 at 9:40

Les B

I'm not being picky with their numbers, but there must be something I'm missing, or my math is bad.

If the formula for finding Cft is:  Length X Width X Height, when I calculate 50MMcf per acre this is what I get:

207(L) x 207(W) x 1162 would give about 50,000,000.

Is this HS formation really this thick (1100 ft.)???

And this thickness would be much thicker even, allowing for the shale that is also there right??

 

Please help me understand.

Permalink Reply by Coyote Ugly on April 23, 2011 at 10:07

Les B

I was playing with their numbers and maybe I don't understand something.  I got to wondering how thick the HS would have to be to contain 50MMcf.  According to my calculations (L x W x H), the formation would have to be 1100 ft. thick if there were nothing but gas there.  Is the gas in a compressed state or something?

Please help me to understand.

CU

Permalink Reply by Les Bamburg on April 23, 2011 at 14:42

CU, the natural gas in the formation is at a pressure of ~ 10,000 psia.  The original gas in place is +/- 200 Bcf per section.

 

By the way, the net thickness of the Haynesville Shale is about 150 to 200 ft and the porosity is 8% to 12%.  I do not know an appropriate water saturation level.

Permalink Reply by Coyote Ugly on April 24, 2011 at 1:23

Let's see if I understand.  The space occupied by a container that is 207' x 207' x 200' is 8.6 mil. cu. ft.  The volume of gas that theoretically occupies this same space is 312.5 mil. cu. ft. (200Bcf / 640).

So, is this a true statement?  The reason 312.5 mil. cu. ft. can occupy a space of 8.6 mil. cu. ft. is because of the 10,000 psia.??

Sorry I'm so dense, but I really want to plow through this til I understand it.  So, don't abandon me please.

Thanks again,

CU

 

Permalink Reply by dbob on April 24, 2011 at 3:08

CU, 

 

10,000 psia = about 680 atmospheres of pressure.  every time you double the pressure, you halve the volume a gas occupies.  So, if something is 312.5 mmcf at the surface (1 atm), and you increase the pressure to 2 atm, the gas occupies roughly 156.25 mmcf.  At 4 atm, it occupies 39 mmcf, at 8 atm, 9.7 mmcf, at 16 atm 4.9 mmcf and at 32 atm, 2.4 mmcf of space.  You can keep going down that road, and eventually, you'll  get down to the point that the amount of space the gas occupies is very small.  Purists will note I'm ignoring temperature here..  

 

Anyway, to a large degree, the initial pressure tells a lot about how much gas can expected to be recovered.

Permalink Reply by Lerret on May 4, 2011 at 8:50

I do want to counsel everyone to be a little cautious about calculating reserves.  It has been my experience that production decline curves often do not support the valuations and estimates made from volumetric methods. Even Devon CEO Nichols ruminated about what is the real recovery factor here.

 

But decline curves are very controversial yet. ... re: Arthur Berman's dead on assessment of the methods uses.  Prior to "shale plays" it was considered a breach of good common sense to use a B factor (n in some formulas) in excess of 1.  Today many of the "players" are using a B of 1.5 or even higher. This gives a "fat tail" to the decline meaning the decline is extreme at first but levels out and the EURs go to maybe 100 years before reaching economic limit. Of course on a NPV (net present value basis) anything after the first 20 years is so heavily discounted as to be meaningless almost.

 

With a history of less than 20 years in shales and the most common methods today of doing the long laterals and multistage fracs makes the analysis of the older shorter one and two stage fracs "meaningless"... so we don't have enough history to determine if these plays end up lasting 20 years let alone 60 or more.

Having said that what operators do and what they are saying seems a mite overblown. I don't trust a number that any of them gives me. Clearly after touting it like mad then selling out which is what both Chesapeake and Petrohawk did to the Fayetteville, then I have to question whether they were doing that simply to raise cash for a better deal (Bakken or Eagle Ford?) or where they simply getting out while the getting is good. The first 50% of their projected reserves is going to be much easier and cheaper to develop than the last 50%... in fact, that last 50% may not even be economic to develop... so who gets left with the hot potato?  Exxon?  SEECO?  the Chinese? BP?  Time will tell and old timers like me may have returned to room termperature before we know the real answer.  Beware the B over 1...that's all I can say.

Permalink Reply by Les Bamburg on April 24, 2011 at 4:17

CU, here is an example of calculating original gas in place and estimated recoverable gas.

 

I will start with 80 acres for a single well which is an area of 5280 ft by 660 ft or 3,484,800 sq ft.  Now the net formation thickness is 200 ft with a porosity of 10%.

 

Therefore the open (pore) space in the formation is the area (3,484,800) times the net thickness (200) times the porosity (10%) which gives 69,696,000 cubic feet.

 

Part of the pore space is occupied by water.  Assuming the water saturation is 30% this leaves 48,787,200 cubic feet of the pore space filled with gas (ie 70% times the total pore space).

 

For calculation of gas in the pore space you have adjust for reservoir pressure and temperature versus standard pressure (14.7 psi) and temperature (60F).  For a reservoir at 10,000 psi pressure and 200F temperature the factors are as follows:

 

Pressure: 10000/14.7 = 680.27

Temperature:  (60 + 459.67) / (200 + 459.67) = 0.722

 

So the original gas in place for 80 acres is the gas pore space (48,787,200) times 680.27 times 0.722 or a result of 23,470.9 million standard cubic feet (MMscf) for the 80 acres.

 

A well only recovers a portion of the original gas in place so an estimated well recovery of 6,500 MMscf (6.5 Bscf) would equate to a recovery factor of 27.7%.

 

For my example the original gas in place for the entire section would be 8 times the 80 acres or ~ 187,876 MMscf (187.9 Bscf) per section.    

Permalink Reply by S S on May 30, 2011 at 7:15

Les, I follow the math, but where did you get the figure 459.67 from when figuring the temperature? Why do you add something to the standard temperature and the lower zone temperature? Look forward to your answer.

Permalink Reply by Les Bamburg on May 30, 2011 at 16:05

SS, you have to convert the temperatures from degrees fahrenheit to degrees Rankin in the calculation.  The formula is:

 

Degrees R = Degrees F + 459.67 

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