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Capillary pressure (Pc): buoyancy, height, and pore throat radius
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Revision as of 20:42, 31 January 2014
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20:42, 31 January 2014
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math
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'''Answers''' (refer to the table below for values)
'''Answers''' (refer to the table below for values)
−
* <math>\mbox{
A.\quad
P}_{\rm c\ Hg} = \mbox{P}_{\rm c\ brine} \left(\frac{\gamma \cos \Theta \mbox{ for Hg}}{\gamma \cos \Theta \mbox{ for brine}}\right) = \mbox{P}_{\rm c\ brine} \left(\frac{367}{72}\right) = 40 (5.1) = 204 \mbox{ psi}</math>
+
* <math>\mbox{P}_{\rm c\ Hg} = \mbox{P}_{\rm c\ brine} \left(\frac{\gamma \cos \Theta \mbox{ for Hg}}{\gamma \cos \Theta \mbox{ for brine}}\right) = \mbox{P}_{\rm c\ brine} \left(\frac{367}{72}\right) = 40 (5.1) = 204 \mbox{ psi}</math>
−
* <math>\mbox{
B.\quad
Equivalent oil P}_{\rm c} = \mbox{P}_{\rm c\ lab} \left(\frac{\gamma \cos \Theta_{\rm res}}{\gamma \cos \Theta_{\rm lab}}\right) = \mbox{P}_{\rm c\ lab} \left(\frac{26}{367}\right) = \mbox{P}_{\rm c\ lab} (0.071) = 204 (0.071) = 14.5 \mbox{ psi}</math>
+
* <math>\mbox{Equivalent oil P}_{\rm c} = \mbox{P}_{\rm c\ lab} \left(\frac{\gamma \cos \Theta_{\rm res}}{\gamma \cos \Theta_{\rm lab}}\right) = \mbox{P}_{\rm c\ lab} \left(\frac{26}{367}\right) = \mbox{P}_{\rm c\ lab} (0.071) = 204 (0.071) = 14.5 \mbox{ psi}</math>
−
* <math>\mbox{
C.\quad
h}' = \frac{\mbox{P}_{\rm c\ res}}{(\mbox{water gradient} - \mbox{hydrocarbon gradient})} = \frac{14.5}{(0.45 - 0.33)} = 120 \mbox{ ft}</math>
+
* <math>\mbox{h}' = \frac{\mbox{P}_{\rm c\ res}}{(\mbox{water gradient} - \mbox{hydrocarbon gradient})} = \frac{14.5}{(0.45 - 0.33)} = 120 \mbox{ ft}</math>
−
* <math>
\begin{array}{c@{\quad}l}\mbox{D.} amp;
\mbox{r at P}_{\rm c\ Hg} = \displaystyle\frac{-2\gamma \cos \Theta}{\mbox{P}_{\rm c\ lab}} \mbox{ or C} \left(\frac{\gamma \cos \Theta_{\rm lab}}{\mbox{P}_{\rm c\ lab}}\right) = 0.29 \times \frac{367}{204} = 0.52 \mu</math>
+
* <math>\mbox{r at P}_{\rm c\ Hg} = \displaystyle\frac{-2\gamma \cos \Theta}{\mbox{P}_{\rm c\ lab}} \mbox{ or C} \left(\frac{\gamma \cos \Theta_{\rm lab}}{\mbox{P}_{\rm c\ lab}}\right) = 0.29 \times \frac{367}{204} = 0.52 \mu</math>
−
:
<math>
[12pt]&
\mbox{P}_{\rm c\ Hg} \mbox{ is equivalent to P}_{\rm c\ lab
}\end{array
}</math>
+
*
<math>\mbox{P}_{\rm c\ Hg} \mbox{ is equivalent to P}_{\rm c\ lab}</math>
==Conversion variables==
==Conversion variables==
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