Difference between revisions of "141.2 derivation"

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(Math and Physics)
(Math and Physics)
Line 14: Line 14:
 
:<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 filed units:
+
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][cm^2]}{[cP]} \frac{[atm]}{[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][cm^2]}{[cP]} \frac{[atm]}{[cm]}</math>

Revision as of 05:51, 21 April 2018

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][cm^2]}{[cP]} \frac{[atm]}{[cm]}

See Also

Darcy's law

18.41 derivation