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cmlpetro (Chemical)
2 Aug 05 17:05
A well im looking at produced from 1980 - 1986, producing 10.3 E6m3. The well has been shut in since 1986. There were two deliverability tests done
1980, P=5380kPa, Z=0.897, CumGas=0.00
2001, P=4490kPa, Z=0.911, CumGas=10.3E6m3, CumWater=40m3
Plotting P/Z vs CumGas gives recoverable gas to be 50 E6m3 when using abandonment pressure of 700 kPa.
Using gas volumetrics I calculate OGIP = 20 E6m3
Why do I get such different estimations? Maybe I just did something wrong, or is there a reason for this?
Thanks.
DrillerNic (Petroleum)
3 Aug 05 8:24
I saw a very similar thing a couple of years ago. We had regular pressure tests and a good P/Z plot on a fault block with a single well, which showed a nice straight line. Then, after the reservoir had been quite delepeted (I don't remember the exact figure) the P/Z plot moved onto a shallower gradient, suggsting an increase in reserves. We looked at the volumetrics, and there was another fault block next to the drained fault block, separated by a sealing fault, that sort of made up the additional gas reserves. The suggestion was that the sealing fault was only partially sealing, and that once the resevoir had been blown down to a particular pressure, gas could move across the fault from the undrained fault block. 字串2
The undrained fault block was then drilled, and was found to be pressure delepted- so the reservoir engineers and production engineers were looking very smug!
So have a look at the difference between the two figures, and see what it could be telling you. The P/Z plot figure is higher than the volumentrics figure; what could be causing this?
1. The volumetrics doesn't include all the gas...undrained faults blocks in the example above, reserves in low net: gross sections that are ignored in the volumetrics, that kind of thing..
2. The P/Z plot is wrong...how good is the Z factor? How good are the pressure measurements? How many points are there on the P/Z plot?
3. (linked to point 2) The P/Z plot isn't applicable.
P/Z plots don't really work if you're getting aquifer support. Where is the water in the 2001 test coming from? If it's aquifer support, then you're blowing down the reservoir, but getting additional pressure support, and the P/Z plot will be a curve, and will overestimate the gas reserves. You have to use the Havellena- Odeh (spelling?) method, where you guess at the aquifer support & try to get a horizontal line, and then, having guessed a value that gives you a horizontal line, do your analysis from that plot.
字串5
cmlpetro (Chemical)
4 Aug 05 10:59
Sorry, just to clarify.
The 10.3E6m3 gas, and 40m3 are the cum production to date at the time of that test. This was not produced during the test.
ppata (Petroleum)
5 Aug 05 8:37
I think you can be in one of those two situation:
1. Your reservoir was initialy overpressurised and, during the explotation, in the moment when the pressure become equal with the normal gradient presure, the aquifer start its activity. If you do not have a constant tracing of the pressure in the begining (30% of reserves), you can loose that efect from p/z and that can offer you wrong results.
2. The second situation was described excellent by DrillerNic with my observation that it is possible to have initial two separate hidrodinamic units and during the exploatation the diferent pressure to "open" the fault (depends on lithology - see EkoFisk, Valhala,...). 字串3
A.
ppata (Petroleum)
6 Aug 05 6:35
What is the depth of the producing reservoir, temperature and gas density. Can you give other p/z=f(Gp) points?
cmlpetro (Chemical)
8 Aug 05 12:25
Depth 635 mKB
Res Temp 299.5 K
SG 0.592
I only have two p/z points.
ppata (Petroleum)
9 Aug 05 4:33
From my point of view:
The material balance is probably overestimated as a result of an undetected active aquifer. Try to find some gas analisys, use a pvt software and recalculate Z and Bg for some pressure steps, assume that at perforated level you have constant flow rate ( Q*Bg=const.), and you can estimate a theoretical decline for your reservoir.
Compare the calculated results with you production data, put them on excel on the same scatter chart, and you will be able to see where the diferences are starting. That is the point when your aquifer begun its activity. Take the pressure, Z factor and cumulative production of that point, and probably you will have a better image on your reservoir. 字串5
After you analise the problem please send me your result and we can discuss on them.
Best regards,
A.
cmlpetro (Chemical)
9 Aug 05 15:38
I dont have time to work on this today but I will try what you have suggested tomorrow.
The two pressure tests I listed in my original post were the only two pressure tests ever performed on the well.
I calculated both of my z factors in my original post from gas analysis that was performed when the well tests were done. I will also do this for formation volume factor.
Sorry, I dont understand what you mean by "assume that at perforated level you have constant flow rate ( Q*Bg=const.), and you can estimate a theoretical decline for your reservoir." How do I estimate a theoretical decline?
I also have some questions concerning water production. At what kind of water production and gas production rates will a well have trouble lifting water up the tubing/annulus. Is there an easy way to tell is a well is loading? 字串8
Thanks for all you help!
ppata (Petroleum)
10 Aug 05 2:19
"assume that at perforated level you have constant flow rate ( Q*Bg=const.), and you can estimate a theoretical decline for your reservoir."
The ideea is to take the initial gas rate, multipy it with Bg and consider that this result is constant for every pressure step(this supposition can be accepted for dry gas). After that, you take for the next step of pressure the calculated Bg from pvt software, and divide the Q*Bg calculated to this Bg. The result will be the gas rate for that step. Do the same for every pressure step. After that you can calculate the p/z for every pressure step.
Here is the problem. The ideea of this calculation is to verify your volumetric with the material balance. Two points are not enough to define the material balance, and for that reason you will take the volumetric value as Gp for pressure=0. Calculate a trendline between your initial point(P=5380kPa, Z=0.897) and that one (p=0), use that equation and calculate the cum production for every pressure step. Now you have a rate(m3/day) and a cumulative production(m3) for every pressure step, and with a simple division you can calculate the number of days neded to fiinish every step. 字串7
Put all the results (calculated and from production data)on an excel chart (Q=f(time)), and analise the diferences. If your calculated curve is under the real one take a bigger resource(for the point p=0 on p/z diagram) and recalculate all (probably that is the best ideea for just 40 m3 water).
The answer for the other problem (water), you can find in technical literature if you search for Begs&Brill and Turner equations.
I hope that will help you,
Best regards,
A.
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