The reason for replying to you is that I do not believe your example is a suitable analogue which represents well M-2. For example, there is no mention of the length of open hole interval in your 'comparable case'. There is no mention of fluid loss approximating 30 bbl/hr, neither is there mention of a total mud loss approximating 17,000 bbls over 24 days. If you stop posting misleading examples on HotCopper then I shall surely agree to stop sending out posts regarding what I believe to be more realistic estimates.
As discussed below, there should be much less mud loss volume in the sands that have poor/mediocre quality compared to trying to sample in the high perm (high fluid loss) zones, therefore I would expect the sampling to take place in zones that may have less than 5 md/cp mobility. Remember 10-12% porosity is the average for M-2, so the low end might be 8-9% porosity? In addition, of course we don't know if we are comparing 'applies with apples' regarding the total amount of mud loss in your example case compared to M-2? Well M-2 has a very long hole section (approx 2000m), which is estimated to have lost about 17,000 bbls of mud fitrate (by the end of this week). How much mud loss did your example well experience?
An alternative way of looking at the issue is to consider three possible scenarios:a) Assume that 100 bbl/metre is distributed evenly in view of the fact that the Pebbly Arkose has already received fluid loss pills (LCM), or b) perhaps assume that the high-perm zones have taken 10-times more mud filtrate than the low perm zones, or c) at least 3-times more mud filtrate than the low perm zones.Potentially, there might be a) 100 bbl/metre fluid loss, 25 bbl/m fluid loss or 9 bbl/m fluid loss per pump-out operation.Would the probe sample from a) a full one metre of mud filtrate, b) half or c) a third of that height as estimated height of fluid being drawn into a probe? Potentially, the highest volume of mud filtrate for a given pump-out operation (avoiding high perm zones, i.e., high volumes of filtrate) might be 100 bbl, and the minimum might be 3 bbls [9*(1/3 m) = 3 bbls].If the maximum pump-out rate is only 0.017 bbl/min, the time required to recover 100 bbls or 3 bbls would be about 4 days maximum and 3 hours minimum, respectively.Given that they would be using a probe in 10%-12% porosity rock, you have said it can take up to 5-6 hours pump-out duration in tight/poor reservoir rock. So, let’s say the minimum duration is about 5 hours per station.They may try to fill 12 bottles instead of 18, so if all of them are priority samples, the 6 pump-out operations might take at least 5*6 = 30 hours in total.If half of the 12 bottles are considered priority samples, there would be 9 pump-out operations, which could take 5*9 = 45 hours (nearly 2 days just to pump out in total). Additional time will be required for taking several pre-test pressures as well as time to RIH/POOH, therefore, rather than have 6 or 9 pump-out operations, they might decide to restrict to 4 or 5 pump-out operations?These estimates are all based on the assumption there is low risk of the tool getting differentially stuck whilst spending at least 3 hours, maybe 5 hours on each pump-out operation. Let's hope for the best!
