141.2 derivation

Derivation of the 141.2 constant

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

Math and Physics

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:

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

And:

$\frac{[bbl]}{[day]} = - const \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]}$

where

$const = \frac{1000\ 14.695950253959}{30.48} \frac{30.48^3\ 5.61458333}{86400} = \frac{1000\ 14.695950253959\ 30.48^2\ 5.61458333}{86400} = 887.2201322$

See Also

Darcy's law

18.41 derivation