The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that ?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements with atomic numbers less than roughly 24, which means that the element has one neutron for every proton. Cesium, with an atomic number of 55, has considerably more neutrons than protons in its nucleus. Therefore when measuring electron density, which is then used to calculate bulk density, the algorithm should show an under-estimation of density due to this large ‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over- estimated,” he continues. So what was the explanation? “When going through the data I noticed an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher photoelectric effect. This increases the absorption of gamma rays in the formation so that the electron density appears higher than it is and physical density is therefore over estimated. This far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now StatoilHydro understood why the logging results were wrong when interpreted using methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way of interpreting the numbers accurately. The simplicity of the solution was stunning. After close examination of the calibration data, a linear relationship was found between the photoelectric factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is that it relies only on the Pef measurements, which are logged simultaneously with bulk density, and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model The logged photoelectric factor isn’t only useful for correcting the density curve and estimating saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the photoelectric factor and its high vertical resolution with resistivity measurements from both the drill pass and ream pass produces a very reliable and accurate net reservoir de?nition. Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all production wells high-quality resistivity image logs can be run. These logs provide important information on structural dip, depositional environment, sedimentary features, facies and geological correlations, all used by geomodelers to produce better reservoir models. “The results match cor.

1. There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)