A control scheme is to be developed for the evaporator shown in Fig. E18.20. The feed and product streams are mixtures of a solute and a solvent, while the vapor stream is pure solvent. The liquid level is tightly controlled by manipulating the feed flow rate, wF. The product composition, xp, and

Question

A control scheme is to be developed for the evaporator shown in Fig. E18.20. The feed and product streams are mixtures of a solute and a solvent, while the vapor stream is pure solvent. The liquid level is tightly controlled by manipulating the feed flow rate, $w_{F} .$ The product composition, $x_{p}$ , and the feed flow rate, $w_{F}$ , are to be controlled by manipulating the product flow, $w_{p}$ , and the steam flow rate, $w_{s} .$ The evaporator “economy” is approximately constant, because $E kg$ of solvent are evaporated for each $kg$ of steam. The flow rates have units of $kg / min$ , while the compositions are expressed in weight fraction of solute.

Derive an expression for the relative gain array for this system.

Accepted Answer

Since level is tightly controlled, there is a no accumulation, and a material balance yields:

Overall: [latex]w_{F}-E w_{S}-w_{P} \approx 0[/latex] (1)

Solute: [latex]\quad w_{F} x_{F}-w_{P} x_{P} \approx 0[/latex] (2)

Controlled variable: [latex]x_{P}^{\prime}, w_{F}^{\prime}[/latex]

Manipulated variables: [latex]w_{P}^{\prime}, w_{s}^{\prime}[/latex]

From (1):

[latex]w_{F}=w_{S} E+w_{P}[/latex]

From (2):

[latex]x_{P}=\frac{x_{F}}{w_{P}} w_{F}=\frac{x_{F}}{w_{P}}\left(w_{s} E+w_{P} ight)[/latex] (3)

Using deviation variables:

[latex]w_{F}^{\prime}=w_{s}^{\prime} E+w_{P}^{\prime}[/latex]

Linearizing (3):

[latex]\begin{aligned}&x_{P}=\bar{x}_{P}+\left.\frac{\partial x_{P}}{\partial w_{P}} ight|_{\bar{w}_{P}, \bar{w}_{s}}\left(w_{P}^{\prime} ight)+\left.\frac{\partial x_{P}}{\partial w_{s}} ight|_{\bar{w}_{P}, \bar{w}_{s}} \&x_{P}^{\prime}=\left\lgroup\frac{-x_{F} E \bar{w} _{s}}{\bar{w}_{P}^{2}} ight group w_{P}^{\prime}+\left\lgroup\frac{x_{F} E}{\bar{w}_{P}} ight group w_{s}^{\prime} (5)\end{aligned}[/latex]

Then the steady-state gain matrix is

[latex]\Lambda = \begin{matrix} x_{P}^{\prime} \ w_{F}^{\prime} \end{matrix} \overset{\begin{matrix} w_{P}^{\prime} &&&&&& w_{s}^{\prime} \end{matrix} }{ \left\lgroup \begin{matrix} \left\lgroup\frac{-x_{F} E \bar{w_{s}}}{\bar{w}_{P}^{2}} ight group &\left\lgroup\frac{x_{F} E}{\bar{w}_{P}} ight group \ 1 & E \end{matrix} ight group }[/latex]

By using the formula in Eq.18-34, we obtain

[latex]\lambda_{11}=\frac{1}{1-\frac{K_{12} K_{21}}{K_{11} K_{22}}}[/latex] (18-34)

[latex]\begin{gathered}\lambda_{11}=\frac{1}{1+\frac{\bar{w}_{P}}{E w_{s}}}=\frac{E \bar{w}_{s}}{E \bar{w}_{s}+\bar{w}_{p}}=\lambda_{22} \\lambda_{12}=\lambda_{21}=1-\lambda_{11}=\frac{\bar{w}_{p}}{E \bar{w}_{s}+\bar{w}_{p}}\end{gathered}[/latex]

So the RGA is

[latex]\Lambda =\left[\begin{array}{cc}\frac{E \bar{w}_{s}}{E \bar{w}_{s}+\bar{w}_{p}} & \frac{\bar{w}_{p}}{E \bar{w}_{s}+\bar{w}_{p}} \\frac{\bar{w}_{p}}{E \bar{w}_{s}+\bar{w}_{p}} & \frac{E \bar{w}_{s}}{E \bar{w}_{s}+\bar{w}_{p}}\end{array} ight][/latex]

So, if [latex]E \bar{w}_{s}>\bar{w}_{P}[/latex], the pairing should be [latex]x_{P}^{\prime}-w_{P}^{\prime} / w_{F}^{\prime}-w_{s}^{\prime}[/latex] So, if [latex]E \bar{w}_{s}<\bar{w}_{P}[/latex], the pairing should be [latex]x_{P}^{\prime}-w_{s}^{\prime} / w_{F}^{\prime}-w_{P}^{\prime}[/latex]

Question 18.20: A control scheme is to be developed for the evaporator shown...

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