DOFPro Team

The Most Annoying Equation Conversion is an episode of the Degrees of Freedom Project, Just the Facts videos on chemical and thermal principles. If you want The Full Story version with an in-depth discussion, use the link in the description. Definitions and additional details can be found in the web-page link in the description as well.
An empirical equation for the enthalpy of nitrogen
\(\hat{H} = 1.04 T - 26\)
where \(\hat{H}\) is in \(\mathrm{kJ/kg}\) and \(T\) is in \(^\circ \mathrm{C}\).
You’re doing work for a company that always uses gas enthalpies in \(\text{Btu/lb-mol}\), and temperature in \(^\circ \mathrm{F}\). They want an equation that will give
\(\hat{h} = f(t)\),
where \(\hat{h}\) is in \(\text{Btu/lb-mol}\), and \(t\) is in \(^\circ \mathrm{F}\). How do you transform your equation into such an equation?
There are many methods to derive \(f(t)\), but the following is a relatively easy and guaranteed method.
Start with the variable in the target units (in our case \(\text{Btu/lb-mol}\) and \(^\circ \mathrm{F}\)).
Convert them using dimensional equations and/or temperature equations to the original units (in our case \(\mathrm{kJ/kg}\) and \(^\circ \mathrm{C}\)).
Plug the converted expressions into the original equation.
Simplify.
\(\hat{h}\ \frac{\mathrm{Btu}}{\text{lb-mol}} \vert \frac{1.054\ \mathrm{kJ}}{\mathrm{Btu}} \vert \frac{\text{lb-mol}}{28.01\ \mathrm{lb_m}} \vert \frac{2.205\ \mathrm{lb_m}}{\mathrm{kg}} = 0.08297\ \frac{\mathrm{kJ}}{\mathrm{kg}}(\hat{h})\)
\(\frac{5}{9}(t-32)\)
\((0.08297 \hat{h}) = 1.04 \left[\frac{5}{9}(t-32)\right] - 26\)
\(0.08297 \hat{h} = 1.04 \left(\frac{5t}{9}-\frac{160}{9}\right) - 26\)
\(\hat{h} = \frac{1}{0.08297} \left[\left(\frac{5.20t}{9}-\frac{166.4}{9}\right) - 26\right]\)
\(\boxed{\hat{h} = 6.964t-536.2}\) ,
where \(\hat{h}\) in \(\text{Btu/lb-mol}\) and \(t\) is in \(^\circ \mathrm{F}\).
You can verify the procedure by calculating a few values of enthalpy for values of \(T\) with the original equation, convert the values to \(\text{Btu/lb-mol}\) and \(^\circ \mathrm{F}\) and compare with what you get for the converted equation.
If you don’t follow this method, you are almost guaranteed to get things inverted or just plain wrong
The Antoine equation, as discussed in Wikipedia, is used to calculate the vapor pressure of a pure species. The Antoine equation in Wikipedia is:
\(\log_{10}p^* = A - \frac{B}{T + C}\)
where \(p^*\) is in \(\mathrm{mmHg}\) and \(T\) is in \(^\circ \mathrm{C}\). Convert the equation to the form
\(\ln P^{sat} = a - \frac{b}{t + c}\)
where \(P^{sat}\) is in \(\mathrm{kPa}\) and \(t\) is in \(\mathrm{K}\).
\(\ln (P^{sat}\ \mathrm{kPa})\), \(t\ \mathrm{K}\)
\(\log_{10} e \cdot \ln \left(P^{sat} \frac{\mathrm{kPa}}{} \vert \frac{760\ \mathrm{mmHg}}{101.325\ \mathrm{kPa}} \vert \right)\) , \((t\ \mathrm{K}) - 273.15\)
\(\log_{10} e \cdot \ln \left(P^{sat} \frac{\mathrm{kPa}}{} \vert \frac{760\ \mathrm{mmHg}}{101.325\ \mathrm{kPa}} \vert \right) = A - \frac{B}{(t\ \mathrm{K}) - 273.15 + C}\)
\(\ln(7.50062 P^{sat}\ \mathrm{kPa}) = \frac{1}{\log_{10} e}\left[A - \frac{B}{(t\ \mathrm{K}) - 273.15 + C}\right]\)
\(\ln(7.50062)+\ln(P^{sat}\ \mathrm{kPa}) = 2.30259 A - \frac{2.30259 B}{(t\ \mathrm{K}) + C- 273.15}\)
\(\ln(P^{sat}\ \mathrm{kPa}) = 2.30259 A - 2.01499 - \frac{2.30259 B}{(t\ \mathrm{K}) + C- 273.15}\)
\(a = 2.30259 A - 2.01499\) ,
\(b = 2.30259 B\) ,
\(c = C -273.15\)
To convert an equation (as opposed to a value or a variable) from one set of units to another set of units, do the following steps:
Thanks for watching!
The Full Story companion video is in the link in the upper left. The next video in the series is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.

