Difference between revisions of "Relative Permeability"

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where
 
where
  
:<math> k_{ro}</math> Oil relative permeability, fraction
+
:<math> k_{ro} =</math> Oil relative permeability, fraction
:<math> k_{rw}</math> Water relative permeability, fraction  
+
:<math> k_{rw} =</math> Water relative permeability, fraction  
:<math> k_o</math> Effective water permeability, mD  
+
:<math> k_o =</math> Effective water permeability, mD  
:<math> k_w</math> Effective water permeability, mD
+
:<math> k_w =</math> Effective water permeability, mD
:<math> k_{oS_{wc}} </math> Effective oil permeability at irreducible oil saturation, mD
+
:<math> k_{oS_{wc}} =</math> Effective oil permeability at irreducible oil saturation, mD
:<math> S_{wc} </math> Connate water saturation, fraction
+
:<math> S_{wc} =</math> Connate water saturation, fraction
  
 
==Example==
 
==Example==

Revision as of 16:26, 30 March 2022

Brief

Relative Permeability is the ratio of the effective permeability to base oil permeability measured at connate water saturation.

 k_{ro} = k_o/k_{oS_{wir}}
 k_{rw} = k_w/k_{oS_{wc}}

where

 k_{ro} = Oil relative permeability, fraction
 k_{rw} = Water relative permeability, fraction
 k_o = Effective water permeability, mD
 k_w = Effective water permeability, mD
 k_{oS_{wc}} = Effective oil permeability at irreducible oil saturation, mD
 S_{wc} = Connate water saturation, fraction

Example

Determine the Mobility Ratio using the following data[1]:
Core is at 70% water and 30% oil saturation. Water phase permeability is 248 mD, oil phase permeability is 50 mD. Water viscosity is 1 cP, oil viscosity is 3 cP.

 M = \frac{248/1}{50/3} =15

In this case the mobility of water is 15 times higher than the mobility of water.

See Also

References

  1. Wolcott, Don (2009). Applied Waterflood Field DevelopmentPaid subscription required. Houston: Energy Tribune Publishing Inc.