Antoine constant's conversion

Abstract

Sometimes we need a vapor pressure equation of one specific compound in certain units, and we find it in another. It's actually easy use the equation as it is, converting input variable (generally temperature, but it could be pressure) into default units, and then converting the output variable into the desired units.

But, some times, for programming reasong and even using some specific software, one wants to have all the constants in certain specific units.

Procedure

Lets develope a general conversion of the units of the well-known Anotoine's vapor pressure equation. \[\log_a \left(p^\text{v}_0\right) = A_0 - \frac{B_0}{T_0 + C_0} \tag{1} \]

where \(p^\text{v}_0\) is the vapor pressure in units of PU0, and \(T\) is the temperature in units of TU0. \(A_0\), \(B_0\), and \(C_0\) are the initial values of Antoine's equations, from whom \(B_0\) and \(C_0\) must be in TU0. Moreover, the logarithm of pressure is in a generic a base because it could be decimal or natural. 

The conversion will be done with the following relations:
\[p^\text{v}_0 =\small{ \left[\frac{\text{PU}_0}{\text{PU}_1}\right]} p^\text{v}_1 \tag{2}\]
\[T_0 = \small{\left[\frac{\text{TU}_0}{\text{TU}_1} \right]} T_1 + Y_{01} \tag{3}\]
\[\log_a x = \frac{\log_b x}{\log_b a} \tag{4} \]

From (1) and (4), the conversion from base a to b will be

\[\log_b \left({p^\text{v}_0} \right) = A_0 \log_b a - \frac{B_0 \log_b a }{T_0 + C_0} \tag{5} \]

Adding (2), we obtain 

 \[\log_b \left(p^\text{v}_1 \left[\frac{\text{PU}_0}{\text{PU}_1}\right]\right) = A_0 \log_b a - \frac{B_0 \log_b a }{T_0 + C_0} \]

\[\Rightarrow\log_b \left(p^\text{v}_1\right) = A_0 \log_b a - \log_b{ \left[ \frac{\text{PU}_0}{\text{PU}_1} \right]} - \frac{B_0 \log_b a }{T_0 + C_0} \tag{6} \]

Now, finally, lets add the most complicated (not that much) unit conversion: temperature. Since it could be absolute of relative, it uses to give some headaches. Applying (3) into (6)

\[\Rightarrow\log_b \left(p^\text{v}_1\right) = A_0 \log_b a - \log_b { \left[ \frac{\text{PU}_0}{\text{PU}_1} \right]} - \frac{B_0 \log_b a \small{ \left[\frac{\text{TU}_1} {\text{TU}_0} \right]}}{T_1 +(Y_{01} + C_0) \small{ \left[ \frac{\text{TU}_1}{\text{TU}_0} \right]}} \tag{7} \]

From (7), it is clear that the transformed constants are

\[A_1 = A_0 \log_b a - \log_b \left[ \frac{\text{PU}_0}{\text{PU}_1} \right] \tag{8} \]
\[B_1 = B_0 {\left[\frac{\text{TU}_1}{\text{TU}_0} \right]} \log_b a \tag{9} \]
\[C_1 = (Y_{01} + C_0) {\left[\frac{\text{TU}_1}{\text{TU}_0} \right]} \tag{10} \]

Note that \(\small{\left[\frac{\text{TU}_1}{\text{TU}_0} \right]} = \small{\left[\frac{\text{TU}_0}{\text{TU}_1} \right]}^{-1} \).

Example

According to NIST, water has the following Antoine constants in the range of 379 - 573 K:

\(A_0\) = 3.55959

\(B_0\) = 643.748 K

\(C_0\) = -198.043 K

with \(p\) in [bar] with base 10 logarithm. Now, lets suppose that we need our units in  natural logarithm, \(p\) in [psia] and \(t\) in [°F]. So we have

\[ \left[ \frac{\text{PU}_0}{\text{PU}_1} \right] = 0.0689476 \frac{\text{bar}}{\text{psia}}\]

\[ \left[\frac{\text{TU}_0}{\text{TU}_1} \right] = \frac{5 \text{K}}{9\text{°F}} \]

\[Y_{01} = 32 \text{°F}\times \frac{5 \text{K}}{9 \text{°F}} + 273.15 \text{K} = 255.37 \text{K} \]

Thus, 

\[A_1 = 3.55959 \ln 10- \ln 0.0689476 = 10.87067 \]

\[ B_1 =643.748 \text{K} \times  \frac{9\text{°F}}{5 \text{K}} \ln 10 = 823.491 \text{°F}   \]

and

\[ C_1 = (-198.043 \text{K} +255.37 \text{K}) \times  \frac{9\text{°F}}{5 \text{K}} = 103.19\text{°F}\]

And that's it.

Final comments

The reader should note that we've calculated the log of a number with units which weren't previously canceled. The units of that logarithm cancels out with the units of each \(\ln p^\text{v}\). A more strict derivation could had been done showing this; but, nonetheless, requiring larger expressions while the final result is the same.

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