Difference between revisions of "141.2 derivation"

Revision as of 05:58, 21 April 2018

141.2 is the well know constant with is used for converting from the Darcy's law units to the field units

Darcy's law:

q = -\frac{kA}{\mu} \frac{dP}{dL}

In Darcy's units:

\frac{[cm^3]}{[sec]} = - \frac{[D][cm^2]}{[cP]} \frac{[atm]}{[cm]}

Converting to the filed units:

\frac{[cm^3] \frac{[ft^3]}{[30.48^3 cm^3]} \frac{[bbl]}{[5.61458333 ft^3]} }{[sec] \frac{[day]}{[86400 sec]}} = - \frac{[D] \frac{[1000 mD]}{[D]}[cm^2] \frac{[ft^2]}{[30.48^2 cm]}}{[cP]} \frac{[atm] \frac{[14.695950253959 psia]}{[atm]}}{[cm] \frac{[ft]}{[30.48 cm]}}

So:

\frac{[bbl]}{[day]} \frac{86400}{30.48^3\ 5.61458333} = - \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]} \frac{1000\ 14.695950253959}{30.48}

@@ Line 12: / Line 12: @@
-:<math> \frac{[cm^3]}{[sec]} = \frac{[D][cm^2]}{[cP]} \frac{[atm]}{[cm]}</math>
+:<math> \frac{[cm^3]}{[sec]} = - \frac{[D][cm^2]}{[cP]} \frac{[atm]}{[cm]}</math>
 Converting to the filed units:
-:<math> \frac{[cm^3] \frac{[ft^3]}{[30.48^3 cm^3]} \frac{[bbl]}{[5.61458333 ft^3]} }{[sec] \frac{[day]}{[86400 sec]}} = \frac{[D] \frac{[1000 mD]}{[D]}[cm^2] \frac{[ft^2]}{[30.48^2 cm]}}{[cP]} \frac{[atm] \frac{[14.695950253959 psia]}{[atm]}}{[cm] \frac{[ft]}{[30.48 cm]}}</math>
+:<math> \frac{[cm^3] \frac{[ft^3]}{[30.48^3 cm^3]} \frac{[bbl]}{[5.61458333 ft^3]} }{[sec] \frac{[day]}{[86400 sec]}} = - \frac{[D] \frac{[1000 mD]}{[D]}[cm^2] \frac{[ft^2]}{[30.48^2 cm]}}{[cP]} \frac{[atm] \frac{[14.695950253959 psia]}{[atm]}}{[cm] \frac{[ft]}{[30.48 cm]}}</math>
 So:
-:<math> \frac{[bbl]}{[day]} \frac{86400}{30.48^3\ 5.61458333}</math>
+:<math> \frac{[bbl]}{[day]} \frac{86400}{30.48^3\ 5.61458333} = - \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]} \frac{1000\ 14.695950253959}{30.48}</math>
 ==See Also==