Difference between revisions of "141.2 derivation"

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Converting to the field units:
 
Converting to the field 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^2]}}{[cP]} \frac{[atm] \frac{[14.695950253959 psia]}{[atm]}}{[cm] \frac{[ft]}{[30.48 cm]}}</math>
  
 
So:
 
So:
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:<math> \frac{1}{C_{RF}} =  \frac{1}{141.20546964} = 7.08E-03 </math>
 
:<math> \frac{1}{C_{RF}} =  \frac{1}{141.20546964} = 7.08E-03 </math>
 
==See Also==
 
[[Darcy's law]]<BR/>
 
[[18.41 derivation]]
 
  
 
== Nomenclature  ==
 
== Nomenclature  ==
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:<math> \mu </math> = [[Darcy's law]] oil viscosity, cp
 
:<math> \mu </math> = [[Darcy's law]] oil viscosity, cp
 
:<math> \mu_o </math> = oil viscosity, cp
 
:<math> \mu_o </math> = oil viscosity, cp
 +
 +
==See Also==
 +
[[Darcy's law]]<BR/>
 +
[[141.2 derivation]]<BR/>
 +
[[18.41 derivation]]
  
 
==References==
 
==References==
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|description=141.2 derivation is the well know constant which is used for converting from the Darcy's law units to the field units in the well's inflow equations.
 
|description=141.2 derivation is the well know constant which is used for converting from the Darcy's law units to the field units in the well's inflow equations.
 
}}
 
}}
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<div style='text-align: right;'>By Mikhail Tuzovskiy on {{REVISIONTIMESTAMP}}</div>

Latest revision as of 12:33, 12 July 2023

Brief

141.2 is the well know constant which is used for converting from the Darcy's law units to the field units in the well's inflow equations.

For example Darcy's law for the single-phase flow is as follows[1]:

 q_o = \frac{1}{141.2} \times \frac{k_oh}{B_o\mu_o} \times \Delta P \times J_D = 7.08 \times 10^-3 \times \frac{k_oh}{B_o\mu_o} \times \Delta  P \times J_D

The derivation of the 141.2 constant is given below.

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 field 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^2]}}{[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

One can be familiar with the inverse of the 141.2 constant:

 \frac{1}{C_{RF}} =  \frac{1}{141.20546964} = 7.08E-03

Nomenclature

 A = Darcy's law cross-sectional area, cm2
 B_o = oil formation volume factor, bbl/stb
 C_{LF} = linear flow units conversion constant
 C_{RF} = radial flow units conversion constant
 h = effective feet of oil pay, ft
 J_D = dimensionless productivity index, dimensionless
 k = Darcy's law permeability, d
 k_o = effective permeability to oil, md
 L = Darcy's law length, cm
 P = Darcy's law pressure, atm
 \Delta P = drawdown, psia
 q = Darcy's law flow rate, cm3/sec
 q_o = oil flow rate, stb/d

Greek symbols

 \mu = Darcy's law oil viscosity, cp
 \mu_o = oil viscosity, cp

See Also

Darcy's law
141.2 derivation
18.41 derivation

References

  1. Brown, Kermit (1984). The Technology of Artificial Lift Methods. Volume 4. Production Optimization of Oil and Gas Wells by Nodal System Analysis. 4. Tulsa, Oklahoma: PennWellBooks. 
By Mikhail Tuzovskiy on 20230712123301