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Design a feedforward-feedback control system for the
Chapter 15, Problem 15.9(choose chapter or problem)
Design a feedforward-feedback control system for the blending system in Example 15.5, for a situation in which an improved sensor is available that has a smaller time delay of 0.1 min. Repeat parts (b), (c), and (d) of Example 15.5. For part (c), approximate \(G_{v} G_{p} G_{m}\) with a first-order plus time-delay transfer function, and then use a PI controller with ITAE controller tuning for disturbances (see Table 12.3). For the feedforward controller in (15-34), use \(\alpha=0.1\). Develop a Simulink diagram for feedforward-feedback control and generate two graphs similar to those in Fig. 15.13.
Questions & Answers
QUESTION:
Design a feedforward-feedback control system for the blending system in Example 15.5, for a situation in which an improved sensor is available that has a smaller time delay of 0.1 min. Repeat parts (b), (c), and (d) of Example 15.5. For part (c), approximate \(G_{v} G_{p} G_{m}\) with a first-order plus time-delay transfer function, and then use a PI controller with ITAE controller tuning for disturbances (see Table 12.3). For the feedforward controller in (15-34), use \(\alpha=0.1\). Develop a Simulink diagram for feedforward-feedback control and generate two graphs similar to those in Fig. 15.13.
ANSWER:Step 1 of 10
We are given the following information:
- A blending system with a pneumatic valve and \(I / P\) transducer are used
- To reduce the effect of the disturbances in the feed composition \(x_{1}\) on the product composition x which is the cont lled variable, a feedforward controller is used.
- The manipulated variable is the inlet flow rate \(w_{2}\).
- The internal diameter of the pilot-scale blending tank is 2 m and its height is 3 m.
- The flow rate of the inlet \(w_{1}\) and its composition \(x_{2}\) are constant.
- The following are the conditions of the nominal steady state operation:
\(\begin{array}{l}
\bar{w}_{1}=650\left(\frac{\mathrm{kg}}{\mathrm{min}}\right) \\
\bar{w}_{2}=350\left(\frac{\mathrm{kg}}{\mathrm{min}}\right) \\
\bar{x}=0.34 \\
\bar{x}_{1}=0.2 \\
\bar{x}_{2}=0.6 \\
\bar{h}=1.5(\mathrm{~m}) \\
\rho=1\left(\frac{\mathrm{g}}{\mathrm{cm}^{3}}\right)
\end{array}\)
- The relation between the flow and head of the valve at the exit line is:
\(w=C_{w} \sqrt{h}\)
- The steady state gains are:
\(\begin{array}{l}
K_{I P}=0.75\left(\frac{\mathrm{psi}}{\mathrm{mA}}\right) \\
K_{v}=25\left(\frac{\mathrm{kg}}{\mathrm{min} \cdot \mathrm{Psi}}\right) \\
K_{t}=32(\mathrm{~mA}) \\
K_{p}=2.64 \cdot 10^{-4}\left(\frac{\mathrm{min}}{\mathrm{kg}}\right) \\
K_{d}=0.65
\end{array}\)
- The time constants are:
\(\begin{aligned}
\tau & =4.71(\mathrm{~min}) \\
\tau_{v} & =0.0833(\mathrm{~min})
\end{aligned}\)
- The time delay of the sensor \(\theta\) is 0.1 minutes.
- For the feedforward controller, \(\alpha=\mathbf{0 . 1}\)