Difference between revisions of "Lee correlation"

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(Math & Physics)
(Math & Physics)
Line 11: Line 11:
 
:<math>  \rho_g =  \frac{1}{62.428} \times \frac{28.967\ SG_g\ p}{z\ 10.732\ T}</math>
 
:<math>  \rho_g =  \frac{1}{62.428} \times \frac{28.967\ SG_g\ p}{z\ 10.732\ T}</math>
  
:<math>  K = \frac{(0.00094+2\times10^-6\ M_g)\ T^{1.5}}{(209+19M_g+T)}</math>
+
:<math>  K = \frac{(0.00094+2\times10^{-6}\ M_g)\ T^{1.5}}{(209+19M_g+T)}</math>
  
 
:<math> X = 3.5+\frac{986}{T}+0.001M_g </math>
 
:<math> X = 3.5+\frac{986}{T}+0.001M_g </math>

Revision as of 14:22, 2 May 2017

Brief

Lee correlation for viscosity of natural gases.

Math & Physics

 \mu_g = K\ e^{X\ \rho_g^Y} [1]

where

  \rho_g =  \frac{1}{62.428} \times \frac{28.967\ SG_g\ p}{z\ 10.732\ T}
  K = \frac{(0.00094+2\times10^{-6}\ M_g)\ T^{1.5}}{(209+19M_g+T)}
 X = 3.5+\frac{986}{T}+0.001M_g
 Y = 2.4-0.203\ X
 M_g = 28.967\ SG_g

Discussion

Why the Lee correlation?

Application range

  560 \le T < 800R\   or\  100 \le T < 340F
 100 < P \le 8000 psia

Nomenclature

 \rho_g = gas density, g/cm3
 \mu_g = gas viscosity, cp
 M_g = gas molecular weight
 p = pressure, psia
 SG_g = gas specific gravity, dimensionless
 T = temperature, °R
 z = gas compressibility factor, dimensionless

References

  1. Lee, A. B.; Gonzalez, M. H.; Eakin, B. E. (1966). "The Viscosity of Natural Gases". J Pet Technol (SPE-1340-PA).