Physics:Lee–Kesler method

Short description: Method to estimate saturated vapor pressure

The Lee–Kesler method allows the estimation of the saturated vapor pressure at a given temperature for all components for which the critical pressure P_c, the critical temperature T_c, and the acentric factor ω are known.

Equations

[math]\displaystyle{ \ln P_{\rm r} = f^{(0)} + \omega \cdot f^{(1)} }[/math]

[math]\displaystyle{ f^{(0)}=5.92714 - \frac{6.09648}{T_{\rm r}} - 1.28862 \cdot \ln T_{\rm r} + 0.169347 \cdot T_{\rm r}^6 }[/math]

[math]\displaystyle{ f^{(1)}=15.2518 - \frac{15.6875}{T_{\rm r}}-13.4721 \cdot \ln T_{\rm r} + 0.43577 \cdot T_{\rm r}^6 }[/math]

with

[math]\displaystyle{ P_{\rm r}=\frac{P}{P_{\rm c}} }[/math] (reduced pressure) and [math]\displaystyle{ T_{\rm r}=\frac{T}{T_{\rm c}} }[/math] (reduced temperature).

Typical errors

The prediction error can be up to 10% for polar components and small pressures and the calculated pressure is typically too low. For pressures above 1 bar, that means, above the normal boiling point, the typical errors are below 2%.

Example calculation

For benzene with

• T_c = 562.12 K
• P_c = 4898 kPa
• T_b = 353.15 K
• ω = 0.2120

the following calculation for T=T_b results:

• T_r = 353.15 / 562.12 = 0.628247
• f⁽⁰⁾ = -3.167428
• f⁽¹⁾ = -3.429560
• P_r = exp( f⁽⁰⁾ + ω f⁽¹⁾ ) = 0.020354
• P = P_r * P_c = 99.69 kPa

The correct result would be P = 101.325 kPa, the normal (atmospheric) pressure. The deviation is -1.63 kPa or -1.61 %.

It is important to use the same absolute units for T and T_c as well as for P and P_c. The unit system used (K or R for T) is irrelevant because of the usage of the reduced values T_r and P_r.

See also

• Vapour pressure of water
• Antoine equation
• Tetens equation
• Arden Buck equation
• Goff–Gratch equation