Difference between revisions of "Lee correlation"

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where
 
where
  
−
:<math>  K = \frac{(0.00094+2\times10^{-6}\ M_g)\ T^{1.5}}{(209+19M_g+T)}</math>
+
:<math>  K = \frac{(9.4+0.02\ 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 09:44, 3 May 2017

Brief

Lee correlation for viscosity of natural gases.

Math & Physics

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

where

K = \frac{(9.4+0.02\ 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
\rho_g = \frac{1}{62.428} \times \frac{28.967\ SG_g\ p}{z\ 10.732\ T}

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, dimensionless
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". Journal of Petroleum Technology (SPE-1340-PA). 
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