A mixing process consists of a single stirred-tank instrumented as shown in Fig. E11.11. The concentration of a single species A in the feed stream varies. The controller attempts to compensate for this by varying the flow rate of pure A through the control valve. The transmitter dynamics are

Question

A mixing process consists of a single stirred-tank instrumented as shown in Fig. E11.11. The concentration of a single species A in the feed stream varies. The controller attempts to compensate for this by varying the flow rate of pure A through the control valve. The transmitter dynamics are negligible.

(a) Draw a block diagram for the controlled process.

(b) Derive a transfer function for each block in the block diagram.

Process

(i) The volume is constant [latex]\left(5 m ^{3} ight)[/latex].

(ii) The feed flow rate is constant [latex]\left(\bar{q}_{F}=7 m ^{3} / min ight)[/latex].

(iii) The flow rate of the A stream varies but is small compared to [latex]\bar{q}_{F}\left(\bar{q}_{A}=0.5 m ^{3} / min ight)[/latex].

(iv) [latex]\bar{c}_{F}=50 kg / m ^{3}[/latex] and [latex]\bar{c}_{A}=800 kg / m ^{3}[/latex].

(v) All densities are constant and equal.

Transfer Line

(i) The transfer line is [latex]20 m[/latex] long and has [latex]0.5 m[/latex] inside diameter.

(ii) Pump volume can be neglected.

Composition Transmitter Data

[latex]c\left( kg / m ^{3} ight)[/latex]	[latex] c_{m}( m A )[/latex]
0	4
200	20

Transmitter dynamics are negligible.

PID Controller

(i) Derivative on measurement only (cf. Eq. 8-17)

[latex]p(t)=\bar{p}+K_{c}\left[e(t)+\frac{1}{ au_{I}} \int_{0}^{t} e\left(t^{*} ight) d t^{*}- au_{D} \frac{d y_{m}(t)}{d t} ight][/latex] (8-17)

(ii) Direct or reverse acting, as required

(iii) Current (mA) input and output signals

I/P Transducer Data

p( m A )	[latex]p_{t}( psig )[/latex]
4	3
20	15

Control Valve

An equal percentage valve is used, which has the following relation:

[latex]q_{A}=0.17+0.03(20)^{\frac{p_{v}-3}{12}}[/latex]

For a step change in input pressure, the valve requires approximately 1 min to move to its new position.

Accepted Answer

a)

b) Transfer Line:

Volume of transfer line [latex]=\pi / 4(0.5 m )^{2}(20 m )=3.93 m ^{3}[/latex]

Nominal flow rate in the line [latex]=\bar{q}_{A}+\bar{q}_{F}=7.5 m ^{3} / min[/latex]

Time delay in the line [latex]=\frac{3.93 m ^{3}}{7.5 m ^{3} / min }=0.52 min[/latex]

[latex]G_{T L}(s)=e^{-0.52 s}[/latex]

Composition Transmitter:

[latex]G_{m}(s)=K_{m}=\frac{(20-4) ma }{(200-0) kg / m ^{3}}=0.08 \frac{ ma }{ kg / m ^{3}}[/latex]

Controller

From the ideal controller in Eq. 8-14

[latex]\frac{P^{\prime}(s)}{E(s)}=K_{c}\left[1+\frac{1}{ au_{I} s}+ au_{D} s ight][/latex] (8-14)

[latex]P^{\prime}(s)=K_{c}\left(1+\frac{1}{ au_{I} s} ight) E(s)+K_{c} au_{D} s\left[\widetilde{C}_{s p}^{\prime}(s)-C_{m}^{\prime}(s) ight][/latex]

In the above equation, set [latex]\widetilde{C}_{s p}^{\prime}(s)=0[/latex] in order to get the derivative on the process output only. Then,

[latex]G_{P I}(s)=K_{c}\left(1+\frac{1}{ au_{I} s} ight)[/latex]

[latex]G_{D}(s)=-K_{c} au_{D} s[/latex]

with [latex]K_{c}>0[/latex] as the controller should be reverse-acting, since [latex]P(t)[/latex] should increase when [latex]C_{m}(t)[/latex] decreases.

I/P transducer

[latex]K_{I P}=\frac{(15-3) psig }{(20-4) ma }=0.75 \frac{ psig }{ ma }[/latex]

Control valve

[latex]G_{v}(s)=\frac{K_{v}}{ au_{v} s+1}[/latex]

[latex]\begin{aligned}&5 au_{v}=1 \quad, \quad au_{v}=0.2 min \&K_{v}=\left.\frac{d q_{A}}{d p_{v}} ight|_{p_{v}=\bar{p}_{v}}=0.03(1 / 12)(\ln 20)(20)^{\frac{\bar{p}_{v}-3}{12}}\end{aligned}[/latex]

[latex]\begin{aligned}&q_{A}=0.5=0.17+0.03(20)^{\frac{\bar{p}_{v}-3}{12}} \&0.03(20)^{\frac{\bar{p}_{v}-3}{12}}=0.5-0.17=0.33 \&K_{v}=(1 / 12)(\ln 20)(0.33)=0.082 \frac{ m ^{3} / min }{ psig } \&G_{v}(s)=\frac{0.082}{0.2 s+1}\end{aligned}[/latex]

Process

Assume [latex]c_{A}[/latex] is constant for pure A. Material balance for A:

[latex]V \frac{d c}{d t}=q_{A} \bar{c}_{A}+\bar{q}_{F} c_{F}-\left(q_{A}+\bar{q}_{F} ight) c[/latex] (1)

Linearizing and writing in deviation variable form

[latex]V \frac{d c^{\prime}}{d t}=\bar{c}_{A} q_{A}^{\prime}+\bar{q}_{F} c_{F}^{\prime}-\left(\bar{q}_{A}+\bar{q}_{F} ight) c^{\prime}-\bar{c} q_{A}^{\prime}[/latex]

Taking Laplace transform

[latex]\left[V s+\left(\bar{q}_{A}+\bar{q}_{F} ight) ight] C^{\prime}(s)=\left(\bar{c}_{A}-\bar{c} ight) Q_{A}^{\prime}(s)+\bar{q}_{F} C_{F}^{\prime}(s)[/latex] (2)

From Eq. 1 at steady state, [latex]d c / d t=0[/latex]

[latex]\bar{c}=\left(\bar{q}_{A} \bar{c}_{A}+\bar{q}_{F} \bar{c}_{F} ight) /\left(\bar{q}_{A}+\bar{q}_{F} ight)=100 kg / m ^{3}[/latex]

Substituting numerical values in Eq. 2 ,

[latex]\begin{aligned}&{[5 s+7.5] C^{\prime}(s)=700 Q_{A}^{\prime}(s)+7 C_{F}^{\prime}(s)} \&{[0.67 s+1] C^{\prime}(s)=93.3 Q_{A}^{\prime}(s)+0.93 C_{F}^{\prime}(s)}\end{aligned}[/latex]

[latex]\begin{aligned}&G_{p}(s)=\frac{93.3}{0.67 s+1} \&G_{d}(s)=\frac{0.93}{0.67 s+1}\end{aligned}[/latex]

Question 11.11: A mixing process consists of a single stirred-tank instrumen...

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