The dynamic behavior of the heat exchanger shown in Fig. E14.27 can be described by the following transfer functions (H. S. Wilson and L. M. Zoss, ISA J., 9, 59 (1962)): Process: T′/W′s = 2 °F/lb/min /(0.432s + 1)(0.017s + 1) Control valve: X′/P′ = 0.047in/psi /0.083s + 1 W′s/X′ = 112 lb /min in

Question

The dynamic behavior of the heat exchanger shown in Fig. E14.27 can be described by the following transfer functions (H. S. Wilson and L. M. Zoss, ISA J., 9, 59 (1962)):

Process:

[latex]\frac{T^{\prime}}{W_{s}^{\prime}}=\frac{2^{\circ} F / lb / min }{(0.432 s+1)(0.017 s+1)}[/latex]

Control valve:

[latex]\frac{X^{\prime}}{P^{\prime}}=\frac{0.047 in / psi }{0.083 s+1} \quad \frac{W_{s}^{\prime}}{X^{\prime}}=112 \frac{ lb }{ min in }[/latex]

Temperature sensor-transmitter:

[latex]\frac{P^{\prime}}{T^{\prime}}=\frac{0.12 psi /{ }^{\circ} F }{0.024 s+1}[/latex]

The valve lift [latex]x[/latex] is measured in inches. Other symbols are defined in Fig. E14.27.

(a) Find the Ziegler-Nichols settings for a PI controller.

(b) Calculate the corresponding gain and phase margins.

Accepted Answer

a) [latex]G_{v}(s)=\frac{0.047}{0.083 s+1} imes 112=\frac{5.264}{0.083 s+1}[/latex]

[latex]\begin{aligned}G_{p}(s) &=\frac{2}{(0.432 s+1)(0.017 s+1)} \G_{m}(s) &=\frac{0.12}{(0.024 s+1)}\end{aligned}[/latex]

Using Eq. 14-61

[latex]P M riangleq 180+\phi_{g}[/latex] (14-61)

[latex]\begin{aligned}-180^{\circ}=& 0- an ^{-1}\left(0.083 \omega_{c} ight)- an ^{-1}\left(0.432 \omega_{c} ight)- an ^{-1}\left(0.017 \omega_{c} ight) \&- an ^{-1}\left(0.024 \omega_{c} ight)\end{aligned}[/latex]

Solving, [latex]\omega_{c}=18.19 rad / min[/latex]

Using Eq. 14-60

[latex]G M riangleq \frac{1}{A R_{c}}[/latex] (14-60)

[latex]\begin{aligned}1=\left(K_{c u} ight) &\left(\frac{5.624}{\sqrt{\left(0.083 \omega_{c} ight)^{2}+1}} ight) \cdot\left(\frac{2}{\sqrt{\left(0.432 \omega_{c} ight)^{2}+1} \sqrt{\left(0.017 \omega_{c} ight)^{2}+1}} ight) \& \cdot\left(\frac{0.12}{\sqrt{\left(0.024 \omega_{c} ight)^{2}+1}} ight)\end{aligned}[/latex]

Substituting [latex]\omega_{c}=18.19, \quad K_{c u}=12.97[/latex]

[latex]P_{u}=2 \pi / \omega_{c}=0.345 min[/latex]

Using Table 12.4, the Ziegler-Nichols PI settings are

[latex]K_{c}=0.45 K_{c u}=5.84 \quad, \quad au_{I}=P_{u} / 1.2=0.288 min[/latex]

b) Using Eqs. [latex]14-39[/latex] and 14-40

[latex]A R=\left|G_{c}(j \omega) ight|=K_{c} \sqrt{1+\frac{1}{\left(\omega au_{I} ight)^{2}}}=K_{c} \frac{\sqrt{\left(\omega au_{I} ight)^{2}+1}}{\omega au_{I}}[/latex] (14-39)

[latex]\phi=\angle G_{c}(j \omega)= an ^{-1}\left(-1 / \omega au_{I} ight)= an ^{-1}\left(\omega au_{I} ight)-90^{\circ}[/latex] (14-40)

[latex]\begin{aligned}&\varphi_{c}=\angle G_{c}= an ^{-1}(-1 / 0.288 \omega)=-(\pi / 2)+ an ^{-1}(0.288 \omega) \&\left|G_{c} ight|=5.84 \sqrt{\left(\frac{1}{0.288 \omega} ight)^{2}+1}\end{aligned}[/latex]

Then, from Eq. 14-57,

[latex]\phi=\angle G= an ^{-1}(I / R)[/latex] (14-57)

[latex]\begin{aligned}-\pi=&-(\pi / 2)+ an ^{-1}\left(0.288 \omega_{c} ight)- an ^{-1}\left(0.083 \omega_{c} ight)- an ^{-1}\left(0.432 \omega_{c} ight) \&- an ^{-1}\left(0.017 \omega_{c} ight)- an ^{-1}\left(0.024 \omega_{c} ight)\end{aligned}[/latex]

Solving, [latex]\omega_{c}=15.11 rad / min[/latex]

Using 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]\begin{aligned}&A_{c}=\left.A R_{OL} ight|_{\omega=\omega c}=\left[5.84 \sqrt{\left(\frac{1}{0.288 \omega_{c}} ight)^{2}+1} ight] \cdot\left[\frac{5.264}{\sqrt{\left(0.083 \omega_{c} ight)^{2}+1}} ight] \&\cdot\left[\frac{2}{\sqrt{\left(0.432 \omega_{c} ight)^{2}+1} \sqrt{\left(0.017 \omega_{c} ight)^{2}+1}} ight] \cdot\left[\frac{0.12}{\sqrt{\left(0.024 \omega_{c} ight)^{2}+1}} ight] \&=0.651\end{aligned}[/latex]

Using Eq. 14-60, [latex]G M=1 / A_{c}=1.54[/latex]

Solving Eq. 14-56 for [latex]\omega_{g}[/latex] gives

[latex]\left. AR _{O L} ight|_{\omega=\omega_{g}}=1 \quad ext { at } \quad \omega_{g}=11.78 rad / min[/latex]

Substituting into Eq. 14-57 gives

[latex]\begin{aligned}\varphi_{g}=\left.\varphi ight|_{\omega=\omega_{g}}=&-(\pi / 2)+ an ^{-1}\left(0.288 \omega_{g} ight)- an ^{-1}\left(0.083 \omega_{g} ight)- an ^{-1}\left(0.432 \omega_{g} ight) \&- an ^{-1}\left(0.017 \omega_{g} ight)- an ^{-1}\left(0.024 \omega_{g} ight)=-166.8^{\circ}\end{aligned}[/latex]

Using Eq. 14-61

[latex]P M=180^{\circ}+\varphi_{g}=13.2^{\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 14.27: The dynamic behavior of the heat exchanger shown in Fig. E14...

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