Lee correlation

Brief

Lee correlation for viscosity of natural gases.

Math & Physics

$\mu_g = K\ e^{X\ \rho_g^Y}$

where

$\rho = 0.00149406\ P\ \frac{\ M_g}{z\ T}$

$K = \frac{(0.00094+2\times10^-6\ M_g)\ T^{1.5}}{(209+19M_g+T)}$

$X = 3.5+\frac{986}{T}+0.001M_g$

$Y = 2.4-0.203\ X$

$M_g = 28.967\ SG_g$

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