The National Academies | 500 Fifth St. N.W. | Washington, D.C. 20001
Copyright © National Academy of Sciences. All rights reserved.
Terms of Use and Privacy Statement



Below are the first 10 and last 10 pages of uncorrected machine-read text (when available) of this chapter, followed by the top 30 algorithmically extracted key phrases from the chapter as a whole.
Intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text on the opening pages of each chapter. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.

Do not use for reproduction, copying, pasting, or reading; exclusively for search engines.

OCR for page 166
Appendix D Calculating the Differential Pressure at the Start of the Negative Test and the Quality of Foam Cement See the well diagram in Figure D-1. 8367' 17483' top of spacer 18037' TOC inside casing 18304' end of casing FIGURE D-1 Well diagram. 1. Pressure differential at the start of the negative test: 166

OCR for page 166
167 Appendix D p = po – pi where p = pressure differential [pounds per square inch (psi)]; po = pressure outside the casing at the bottom (psi), assumed equal to reservoir pressure of 11,892 psi, which is a pore pressure of 12.57 pounds per gallon (ppg) at the bottom of the reservoir at 18,212 feet (true vertical depth); and pi = pressure on the inside above the cement (psi). 0.433 8,367(8.66)  9,116(14.17)  554(14.3)  999 psi p  11,892  8.33 Here the differential is into the casing. The cement is treated as a solid that does not transmit hydrostatic pressure but that must be strong enough to withstand the pressure differential across it. The top of the cement inside the casing is based on the assumption that 2.8 barrels of foam cement flowed back into the casing when the pressure was bled off at the end of the cement job. 2. Foam quality calculations: Foam cement: The purpose in this case is to reduce the bottom hole (in situ) density of the slurry from 16.74 ppg to 14.5 ppg. The bottom hole pressure is the hydrostatic pressure of 14 ppg mud or 13,321 pounds per square inch gauge (psig) at 18,304 feet. The static bottom hole temperature is 245F. s = 16.74fc + NfN where 1 = fc + fN, s = slurry density (lbm/gal), N = nitrogen density (lbm/gal), fc = weight fraction of cement base slurry, and fN = weight fraction of nitrogen. N p ft 3 13,335.7 lb lb N  2.7  2.7(0.9672)  28.9 m  3.86 m 3 zT 1.71(705) ft 7.48 gal gal fN = (14.5 – 16.74)/(–16.74 + 3.86) = 0.174 fc = 0.826

OCR for page 166
168 Macondo Well Deepwater Horizon Blowout where N = specific gravity of nitrogen (compared with air), p = pressure (pounds per square inch absolute), z = gas deviation factor (dimensionless), and T = temperature (degrees Rankine = 460 + degrees Fahrenheit). So, for every in situ gallon of slurry there will be 0.174 gallon of nitrogen mixed with 0.826 gallon of base 16.74-ppg cement slurry. Thus, the in situ foam quality is 17.4 percent. Note that the Chevron tests used a 13 percent quality foam, which corresponds to the weight fraction of nitrogen necessary to create a 14.5 ppg density foam at atmospheric conditions. Therefore, more nitrogen is required to create the same density foam at the much higher pressure and temperature of the bottom of the Macondo well. At the mixer at the surface, the slurry is blended and pumped at about 600 psig. The volume of nitrogen introduced to 0.826 gallons of base cement is the in situ volume increased through the real gas law. 0.979 520 13,335.7 V600  0.174  1.6 gallons 1.71 614.7 705 This is added to 0.826 gallon of base cement. Thus, for every 1 gallon of base cement, 1.94 gallons of N2 at 600 psig is required. This is a 66 percent quality foam. The density of the foam slurry at the mixer will be as follows: 614.7 lb lb N  2.7(0.9672)  3.15 m  0.42 m ft 3 0.979(520) gal lb m s  16.74(0.34)  0.42(0.66)  5.97 gal The previous equations and results can be combined to obtain an equation for the density of the slurry at any depth with a corresponding pressure, temperature, and gas deviation factor. s  16.74(1  f N )  N f N 2.7(0.9672) p p N   0.349 7.48 Tz Tz VN fN  VN  0.826

OCR for page 166
169 Appendix D Tz 13,335.7 Tz VN  0.174  1.925 p 1.71(705) p   Tz Tz 1.925 1.925   p p p s  16.74   0.349 1  1.925 Tz  0.826  Tz 1.925 Tz  0.826     p p    Tz  Tz Tz s 1.925  0.826 16.74 1.925  0.826  1.925  0.349(1.925)    p  p p 14.5 s    Tz 1.925  0.826    p where s, T, p, and z are as previously defined.