141.2 derivation

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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