blending system is shown in Fig. E18.13. Liquid level h and exit composition c3 are to be controlled by adjusting flow rates q1 and q3. Based on the information below, do the following: (a) Derive the process transfer function matrix, Gp(s). (b) If a conventional multiloop control system is used,

Question

A blending system is shown in Fig. E18.13. Liquid level [latex]h[/latex] and exit composition [latex]c_{3}[/latex] are to be controlled by adjusting flow rates [latex]q_{1}[/latex] and [latex]q_{3}[/latex]. Based on the information below, do the following:

(a) Derive the process transfer function matrix, [latex]G _{ p }(s)[/latex].

(b) If a conventional multiloop control system is used, which controller pairing should be used? Justify your answer.

(c) Obtain expressions for the ideal decouplers [latex]T_{21}(s)[/latex] and [latex]T_{12}(s)[/latex] in the configuration of Fig. [latex]18.9[/latex].

Available Information

(i) The tank is [latex]3 ft[/latex] in diameter and is perfectly mixed.

(ii) Nominal steady-state values are

[latex]\begin{aligned}\bar{h} &=3 ft & \bar{q}_{3} &=20 ft ^{3} / min \\bar{c}_{1} &=0.4 mole / ft ^{3} & & \bar{c}_{2} =0.1 mole / ft ^{3} \\bar{q}_{1} &=10 ft ^{3} / min & &\end{aligned}[/latex]

(iii) The density of each process stream remains constant at [latex] ho=60 lb / ft ^{3}[/latex].

(iv) The primary disturbance variable is flow rate [latex]q_{2}[/latex].

(v) Inlet compositions [latex]c_{1}[/latex] and [latex]c_{2}[/latex] are constant.

(vi) The transmitter characteristics are approximated by the following transfer functions with time constants in minutes:

[latex]\begin{aligned}G_{m 11}(s) &=\frac{4}{0.1 s+1}( mA / ft ) \G_{m 22}(s) &=\frac{100}{0.2 s+1}\left( m A ft ^{3} / ext { mole } ight)\end{aligned}[/latex]

(vii) Each control valve has a gain of [latex]0.15 ft ^{3} / min mA[/latex] and a time constant of [latex]10 s[/latex].

Accepted Answer

a) Material balances for the tank,

[latex]\begin{aligned}&A \frac{d h}{d t}=q_{1}+q_{2}-q_{3} (1) \&\frac{d\left(A h c_{3} ight)}{d t}=c_{1} q_{1}+c_{2} q_{2}-c_{3} q_{3} (2)\end{aligned}[/latex]

Substituting for [latex]d h / d t[/latex] from (1) into (2) and simplifying

[latex]A h \frac{d C_{3}}{d t}=\left(c_{1}-c_{3} ight) q_{1}+\left(c_{2}-c_{3} ight) q_{2}[/latex] (3)

Linearizing, using deviation variables, and taking the Laplace transform

[latex]A \bar{h} s C_{3}^{\prime}(s)=\left(\overline{c_{1}}-\overline{c_{3}} ight) Q_{1}^{\prime}(s)-\overline{q_{1}} C_{3}^{\prime}(s)+\left(\overline{c_{2}}-\overline{c_{3}} ight) Q_{2}^{\prime}(s)-\overline{q_{2}} C_{3}^{\prime}(s)[/latex]

Since [latex]\overline{q_{1}}+\overline{q_{2}}=\overline{q_{3}}[/latex], this becomes

[latex]\left[\left(\frac{A \bar{h}}{\overline{q_{3}}} ight) s+1 ight] C_{3}^{\prime}(s)=\left(\frac{\overline{c_{1}}-\overline{c_{3}}}{\overline{q_{3}}} ight) Q_{1}^{\prime}(s)+\left(\frac{\overline{c_{2}}-\overline{c_{3}}}{\overline{q_{3}}} ight) Q_{2}^{\prime}(s)[/latex] (4)

Similarly from (1),

[latex]A s H^{\prime}(s)=Q_{1}^{\prime}(s)+Q_{2}^{\prime}(s)-Q_{3}^{\prime}(s)[/latex] (5)

Therefore,

[latex]\underset{=}{G} (s)=\left[\begin{array}{cc}\frac{H^{\prime}(s)}{Q_{1}^{\prime}(s)} & \frac{H^{\prime}(s)}{Q_{3}^{\prime}(s)} \\frac{C_{3}^{\prime}(s)}{Q_{1}^{\prime}(s)} & \frac{C_{3}^{\prime}(s)}{Q_{3}^{\prime}(s)}\end{array} ight]=\left[\begin{array}{cc}\frac{1}{A s} & -\frac{1}{A s} \\frac{\left(\overline{c_{1}}-\overline{c_{3}} ight) / \overline{q_{3}}}{\left(\frac{A \bar{h}}{\overline{q_{3}}} ight) s+1} & 0\end{array} ight][/latex]

Substituting numerical values

[latex]\underset{=}{G}(s)=\left[\begin{array}{cc}\frac{0.1415}{s} & \frac{-0.1415}{s} \\frac{0.0075}{1.06 s+1} & 0\end{array} ight][/latex]

For the control valves

[latex]G_{v}(s)=\frac{0.15}{\left(\frac{10}{60} ight) s+1}=\frac{0.15}{0.167 s+1}[/latex] (6)

Thus,

[latex]\underset{=}{G_{P}}(s)=G_{v}(s) \underset{=}{G}(s)=\left[\begin{array}{cc}\frac{0.0212}{s(0.167 s+1)} & \frac{-0.0212}{s(0.167 s+1)} \\frac{0.0011}{(1.06 s+1)(0.167 s+1)} & 0\end{array} ight][/latex]

b) Since [latex]C_{3}^{\prime}(s) / Q_{3}^{\prime}(s)=0, c_{3}[/latex] is not affected by [latex]q_{3}[/latex] and must be paired with [latex]q_{1}[/latex]. Thus, the pairing that should be used is [latex]h-q_{3}, c_{3}-q_{1}[/latex].

c) For the pairing determined above, Fig. [latex]18.9[/latex] can be used with [latex]Y_{1} \equiv H^{\prime}, Y_{2} \equiv[/latex] [latex]C_{3}^{\prime}, U _{1} \equiv Q_{3}^{\prime}, U _{2} \equiv Q_{1}^{\prime} .[/latex] Notice that this pairing requires [latex]G_{p}(s)[/latex] above the switch columns. Then using Eqs. [latex]18-78[/latex] and [latex]18-80[/latex],

[latex]T_{21}=-\frac{G_{p 21}}{G_{p 22}}[/latex] (18-78)

[latex]T_{21}=-\frac{K_{p 21}}{K_{p 22}} \quad ext { and } \quad T_{12}=-\frac{K_{p 12}}{K_{p 11}}[/latex] (18-80)

[latex]T_{21}(s)=-\frac{G_{P_{21}}(s)}{G_{P_{22}}(s)}=-\frac{0}{\left[\frac{0.0011}{[1.06 s+1)(0.167 s+1)} ight]}=0[/latex]

[latex]T_{12}(s)=-\frac{G_{P_{12}}(s)}{G_{R_{1}}(s)}=-\frac{0.0212 /[s(0.167 s+1)]}{-0.0212 /[s(0.167 s+1)]}=1[/latex]

Question 18.13: blending system is shown in Fig. E18.13. Liquid level h and ...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions