Difference between revisions of "141.2 derivation"

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(Math and Physics)
(Math and Physics)
Line 24: Line 24:
 
And:
 
And:
  
:<math> \frac{[bbl]}{[day]} = - const \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]}</math>
+
:<math> \frac{[bbl]}{[day]} = - C_{LF} \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]}</math>
  
 
where  
 
where  
  
:<math> 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</math>
+
:<math> C_{LF} = \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</math>
 +
 
 +
For the radial flow:
 +
 
 +
:<math> C_{RF} =  \frac{C_{LF}}{2\pi} = \frac{887.2201322}{2\pi} = 141.20546964</math>
  
 
==See Also==
 
==See Also==

Revision as of 06:07, 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] \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]} = - C_{LF} \frac{[mD][ft^2]}{[cP]} \frac{[psia]}{[ft]}

where

 C_{LF} = \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

For the radial flow:

 C_{RF} =  \frac{C_{LF}}{2\pi} = \frac{887.2201322}{2\pi} = 141.20546964

See Also

Darcy's law

18.41 derivation