Consider the storage tank with sightglass in Fig. E14.22. The parameter values are R1 = 0.5 min/ft^2, R2 = 2 min/ft^2, A1 = 10 ft^2, Kv = 2.5 cfm/mA, A2 = 0.8 ft^2, Km = 1.5 mA/ft, and τm = 0.5 min. (a) Suppose that R2 is decreased to 0.5 min/ft^2. Compare the old and new values of the ultimate

Question

Consider the storage tank with sightglass in Fig. E14.22. The parameter values are [latex]R_{1}=0.5 min / ft ^{2}, R_{2}=2 min / ft ^{2}[/latex], [latex]A_{1}=10 ft ^{2}, \quad K_{v}=2.5 cfm / mA , \quad A_{2}=0.8 ft ^{2}, \quad K_{m}=1.5 mA / ft[/latex], and [latex] au_{m}=0.5 min[/latex].

(a) Suppose that [latex]R_{2}[/latex] is decreased to [latex]0.5 min / ft ^{2}[/latex]. Compare the old and new values of the ultimate gain and the critical frequency. Would you expect the control system performance to become better or worse? Justify your answer.

(b) If PI controller settings are calculated using the ZieglerNichols rules, what are the gain and phase margins? Assume [latex]R_{2}=2 min / ft .[/latex]

Accepted Answer

From modifying the solution to the two tanks in Section 6.4, which have a slightly different configurations,

[latex]G_{p}( s )=\frac{R_{1}}{\left(A_{1} R_{1} A_{2} R_{2} ight) s^{2}+\left(A_{1} R_{1}+A_{2} R_{1}+A_{2} R_{2} ight) s+1}[/latex]

For [latex]R_{1}=0.5, R_{2}=2, A_{1}=10, A_{2}=0.8[/latex]

[latex]G_{p}( s )=\frac{0.5}{8 s^{2}+7 s+1}[/latex] (1)

For [latex]R_{2}=0.5, \quad G_{p}( s )=\frac{0.5}{2 s^{2}+5.8 s+1}[/latex] (2)

a) [latex]\quad[/latex] For [latex]R_{2}=2[/latex]

[latex]\angle G_{p}= an ^{-1}\left[\frac{-7 \omega_{c}}{1-8 \omega_{c}^{2}} ight] \quad, \quad\left|G_{p} ight|=\left(\frac{0.5}{\sqrt{\left(1-8 \omega_{c}^{2} ight)^{2}+\left(7 \omega_{c} ight)^{2}}} ight)[/latex]

[latex]K_{c u}[/latex] and [latex]\omega_{c}[/latex] are obtained using Eqs. 14-7 and 14-8:

[latex]A R=\frac{\hat{A}}{A}=|G|=\sqrt{R^{2}+I^{2}}[/latex] (14-7)

[latex]\phi=\angle G= an ^{-1}(I / R)[/latex] (14-8)

[latex]-180^{\circ}=0+0+ an ^{-1}\left[\frac{-7 \omega_{c}}{1-8 \omega_{c}^{2}} ight]- an ^{-1}\left(0.5 \omega_{c} ight)[/latex]

Solving, [latex]\omega_{c}=1.369 rad / min[/latex]

[latex]1=\left(K_{c u} ight)(2.5)\left(\frac{0.5}{\sqrt{\left(1-8 \omega_{c}^{2} ight)^{2}+\left(7 \omega_{c} ight)^{2}}} ight)\left(\frac{1.5}{\sqrt{\left(0.5 \omega_{c} ight)^{2}+1}} ight)[/latex]

Substituting [latex]\omega_{c}=1.369 rad / min , K_{c u}=10.96, \omega_{c} K_{c u}=15.0[/latex]

For [latex]R_{2}=0.5[/latex]

[latex]\angle G_{p}= an ^{-1}\left[\frac{-5.8 \omega_{c}}{1-2 \omega_{c}^{2}} ight] \quad, \quad\left|G_{p} ight|=\left(\frac{0.5}{\sqrt{\left(1-2 \omega_{c}^{2} ight)^{2}+\left(5.8 \omega_{c} ight)^{2}}} ight)[/latex]

For [latex]G_{v}=K_{v}=2.5, \quad \varphi_{v}=0, \quad\left|G_{v} ight|=2.5[/latex]

For [latex]G_{m}=\frac{1.5}{0.5 s+1}, \quad \varphi_{m}=- an ^{-1}(0.5 \omega), \quad\left|G_{m} ight|=\frac{1.5}{\sqrt{\left(0.5 \omega_{c} ight)^{2}+1}}[/latex]

[latex]-180^{\circ}=0+0+ an ^{-1}\left[\frac{-5.8 \omega_{c}}{1-2 \omega_{c}^{2}} ight]- an ^{-1}\left(0.5 \omega_{c} ight)[/latex]

Solving, [latex]\omega_{c}=2.51 rad / min[/latex]

Substituting [latex]\omega_{c}=2.51 rad / min , K_{c u}=15.93, \omega_{c} K_{c u}=40.0[/latex]

b) From part a), for [latex]R_{2}=2[/latex],

[latex]\begin{aligned}&\omega_{c}=1.369 rad / min , \quad K_{c u}=10.96 \&P_{u}=\frac{2 \pi}{\omega_{c}}=4.59 min\end{aligned}[/latex]

Using Table [latex]12.4[/latex], the Ziegler-Nichols PI settings are

[latex]K_{c}=0.45 K_{c u}=4.932 \quad, \quad au_{I}=P_{u} / 1.2=3.825 min[/latex]

Using Eqs. [latex]13.63[/latex] and [latex]13-62[/latex],

[latex]\begin{aligned}&\varphi_{c}=- an ^{-1}(-1 / 3.825 \omega) \&\left|G_{c} ight|=4.932 \sqrt{\left(\frac{1}{3.825 \omega} ight)^{2}+1}\end{aligned}[/latex]

Then, from Eq. 14-56

[latex]A R_{O L}\left(\omega_{c} ight)=\left|G_{O L}\left(j \omega_{c} ight) ight|=1[/latex] (14-56)

[latex]-180^{\circ}= an ^{-1}\left[\frac{-1}{3.825 \omega_{c}} ight]+0+ an ^{-1}\left[\frac{-7 \omega_{c}}{1-8 \omega_{c}^{2}} ight]- an ^{-1}\left(0.5 \omega_{c} ight)[/latex]

Solving, [latex]\omega_{c}=1.086 rad / min[/latex]

Using Eq. 14-57

[latex]\phi_{O L}\left(\omega_{c} ight)=\angle G_{O L}\left(j \omega_{c} ight)=-180^{\circ}[/latex] (14-57)

[latex]\begin{aligned}&A_{c}=\left.\operatorname{AR}_{O L } ight|_{\omega=\omega c}= \&=\left(4.932 \sqrt{\left(\frac{1}{3.825 \omega_{c}} ight)^{2}+1} ight)(2.5)\left(\frac{0.5}{\sqrt{\left(1-8 \omega_{c}{ }^{2} ight)^{2}+\left(7 \omega_{c} ight)^{2}}} ight)\left(\frac{1.5}{\sqrt{\left(0.5 \omega_{c} ight)^{2}+1}} ight) \&=0.7362\end{aligned}[/latex]

Therefore, gain margin [latex]G M=1 / A_{c}=1.358[/latex]

Solving Eq.(14-16) for [latex]\omega_{g}[/latex]

[latex]G(j \omega)=\frac{G_{a}(j \omega) G_{b}(j \omega) G_{c}(j \omega) \cdots}{G_{1}(j \omega) G_{2}(j \omega) G_{3}(j \omega) \cdots}[/latex] (14-16)

[latex]\left. AR _{O L} ight|_{\omega=\omega c}=1 \quad[/latex] at [latex]\quad \omega_{g}=0.925[/latex]

Substituting into Eq. 14-57 gives

[latex]\varphi_{g}=\left.\varphi ight|_{\omega=\omega g}=-172.7^{\circ}[/latex]

Therefore, phase margin [latex]P M=180+\varphi_{g}=7.3^{\circ}[/latex]

Table 12.4 Controller Design Relations Based on the ITAE Performance Index and a First-Order-plus-Time-Delay Model (Lipták, 2006)[latex]{ }^{*}[/latex] ,[latex]{ }^{†}[/latex]
Type of Input	Type of Controller	Mode	A	B
Disturbance	PI	P	0.859	−0.977
		I	0.674	−0.680
Disturbance	PID	P	1.357	−0.947
		I	0.842	−0.738
		D	0.381	0.995
Set point	PI	P	0.586	−0.916
		I	1.03[latex]{ }^{†}[/latex]	−0.165[latex]{ }^{†}[/latex]
Set point	PID	P	0.965	−0.85
		I	0.796[latex]{ }^{†}[/latex]	−0.1465[latex]{ }^{†}[/latex]
		D	0.308	0.929

[latex]{ }^{*}[/latex] Design relation: [latex]Y=A( heta / au)^{B}[/latex] where [latex]Y=K K_{c}[/latex] for the proportional mode, [latex] au / au_{I}[/latex] for the integral mode, and [latex] au_{D} / au[/latex] for the derivative mode.

[latex]{ }^{†}[/latex]For set-point changes, the design relation for the integral mode is [latex] au / au_{I}=A+B( heta / au)[/latex].

Question 14.22: Consider the storage tank with sightglass in Fig. E14.22. Th...

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