Mass balance on the rectifying section
Or better said, on the plates of the top section. This balance applies to a binary mixture only. In Fig. 01 a general representation of this is presented.
- All plates are numbered from the top
- $n$ is the last plate on the rectifying section. Of course, $n$ is also a counter going from 1 to the last plate.
- The referred component on the mol fractions $(x,y,z)$ is the most volatile
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| Fig. 01 Mass balance on the rectifying section. |
In this way, a global mass balance can be easily written as,
$G_{n+1}=L_n+D$ Eq. (01)
Notice that, for Eq. (01), it was assumed that a vapor stream $G_{n+1}$ enters the top section from the bottom, while two liquid streams, $L_n$ and $D$, exit the section at the bottom and the top, respectively. On the other hand, a mass balance for the component of interest (say component $A$) on each plate in this section can be easily written as,
$G_{n+1}\, y_{n+1}\, -\, L_n\, x_n \,=\,D\, z_D$ Eq. (02)
Equation (02) was derived based on the sketch in Fig. 02.
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| Fig. 02 Mass balance on a plate |
From Eq. (02), the mole fraction in the vapor entering the nth plate can be isolated to be,
$y_{n+1}=\dfrac{L_n}{G_{n+1}}x_n+\dfrac{D}{G_{n+1}}z_D$ Eq. (03)
One important simplification on Eq. (03) can be made on the nature of the flow rates entering and leaving the plates: the flow rates of the liquid and vapor are constant always. This is a very strong assumption and is at the heart of the method of calculation called after Sorel and Lewis.
In this way, Eq. (03) becomes,
$y_{n+1}=\dfrac{L}{G}x_n+\dfrac{D}{G}z_D$ Eq. (04)
Mass balance on the stripping section
In this case, we just repeat the same process as before. Fig. 02 shows a schematic of this section.
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| Fig. 03 Mass balance on the stripping section. |
In. Fig. 03 variables for vapor and liquid have been barred $(\bar{V},\bar{L})$ to avoid confusion with those of the rectifying section. Also, the counter $m$ has been introduced to number the plates of the stripping section. The global mass balance is,
$\bar{L}_m=\bar{G}_{m+1}+W$ Eq. (05)
In the same way as before, from the mass balance for the component $A$, the following equations arise.
$\bar{L}_m x_m=\bar{G}_{m+1}y_{m+1}+Wx_W$ Eq. (06)
and
$y_{m+1}=\dfrac{\bar{L}_m}{\bar{G}_{m+1}} x_m-\dfrac{W}{\bar{G}_{m+1}}x_W$ Eq. (07)
Again, using the idea of constant liquid and vapor flow rate at each plate, Eq. (07) reduces to,
$y_{m+1}=\dfrac{\bar{L}}{\bar{G}} x_m-\dfrac{W}{\bar{G}}x_W$ Eq. (08)
Recall that Eq. (08) comes from a rough approximation.



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