Consider proportional-only control of the level control system in Fig. 11.16. The level transmitter has a span of 10 ft. and a zero of 15 ft. Recall that the standard instrument ranges are 4 to 20 mA and 3 to 15 psia. The nominal design conditions are h = 20 ft. The controller has a gain of 5 while

Question

Consider proportional-only control of the level control system in Fig. 11.16. The level transmitter has a span of $10 ft$ . and a zero of $15 ft$ . Recall that the standard instrument ranges are 4 to $20 mA$ and 3 to 15 psia. The nominal design conditions are $h=20 ft$ . The controller has a gain of 5 while the gain for the control valve is $K_{v}=0.4 cfm / psi$ , respectively. The time constant for the tank is $\tau=5 min$ . After the set-point is changed from 20 to $22 ft$ , the tank level eventually reaches a new steady-state value of $21.92 ft$ .

(a) What is the offset?

(b) What are the gains $K_{m}$ and $K_{p}$ in Fig. 11.16? (Give their units also)

(c) How could the controller be modified to obtain zero offset?

(a) What is the offset?

(b) What are the gains [latex]K_{m}[/latex] and [latex]K_{p}[/latex] in Fig. 11.16? (Give their units also)

(c) How could the controller be modified to obtain zero offset?

Accepted Answer

(a)

Offset [latex]=h_{s s}-h_{ ext {final }}=22.00-21.92=0.08 ft[/latex]

(b)

[latex]\begin{aligned}&K_{m}=\frac{20-4 mA }{10 ft }=1.6 mA / ft \&K_{I P}=\frac{15-3}{20-4}=0.75 \quad ext { psi/mA } \&K_{V}=0.4 \quad cfm / psi ext{ and } K_{c}=5 \end{aligned}[/latex]

We have:

[latex]K_{O L}=K_{m} K_{c} K_{I P} K_{V} G_{p}=1.6 imes 5 imes 0.75 imes 0.4 K_{p}=2.4 K_{p}[/latex]

Offset equals to:

[latex]o f f s e t=\frac{M}{1+K_{O L}}=\frac{22-20}{1+2.4 K_{p}}=0.08[/latex]

[latex]K_{p}=10 ft / cfm[/latex]

(c)
Add integral action to eliminate offset.

Question 11.23: Consider proportional-only control of the level control syst...

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