According to equation: \begin{equation} q^F= \dfrac{h^{F,V} - h^F}{h^{F,V}-h^{F,L}} \end{equation}, the molar feed enthalpy has a not insignificant influence on the rectification process.
Where \begin{equation} h^{F,V}-h^{F} \end{equation} is the molar enthalpy difference from feed as saturated vapor and feed in added state,
and \begin{equation} h^{F,V}-h^{F,L} \end{equation} is the molar enthalpy difference from feed as saturated vapor and feed as boiling liquid (equals molar evaporation heat).
Assign the corresponding feed property to the thermal states!
| \begin{equation} h^F=h^{F,L} \end{equation} | \begin{equation} q^F=1 \end{equation}: | Feed as
|
| \begin{equation} h^F=h^{F,V} \end{equation} | \begin{equation} q^F=0 \end{equation}: | Feed as
|
| \begin{equation} h^F>h^{F,V} \end{equation} | \begin{equation} q^F<0 \end{equation}: | Feed as
|
| \begin{equation} h^F<h^{F,L} \end{equation} | \begin{equation} q^F>1 \end{equation}: | Feed as
|
| \begin{equation} h^{F,L}<h^F<h^{F,V} \end{equation} | \begin{equation} 0<q^F<1 \end{equation}: | Feed as
|