An electrically heated, stirred-tank system is shown in Fig. E11.8. Using the given information, do the following: (a) Draw a block diagram for the case where T3 is the controlled variable and voltage signal V2 is the manipulated variable. Derive an expression for each transfer function. (b) Repeat

Question

An electrically heated, stirred-tank system is shown in Fig. E11.8. Using the given information, do the following:

(a) Draw a block diagram for the case where [latex]T_{3}[/latex] is the controlled variable and voltage signal [latex]V_{2}[/latex] is the manipulated variable. Derive an expression for each transfer function.

(b) Repeat part (a) using [latex]V_{1}[/latex] as the manipulated variable.

(c) Which of these two control configurations would provide better control? Justify your answer.

Available Information

The volume of liquid in each tank is kept constant using an overflow line.
Temperature [latex]T_{0}[/latex] is constant.
A [latex]0.75- gal / min[/latex] decrease in [latex]q_{0}[/latex] ultimately makes [latex]T_{1}[/latex] increase by [latex]3^{\circ} F[/latex]. Two-thirds of this total temperature change occurs in [latex]12 min .[/latex] This change in [latex]q_{0}[/latex] ultimately results in a [latex]5^{\circ} F[/latex] increase in [latex]T_{3}[/latex].
A change in [latex]V_{1}[/latex] from 10 to 12 volts ultimately causes [latex]T_{1}[/latex] to change from 70 to [latex]78^{\circ} F[/latex]. A similar test for [latex]V_{2}[/latex] causes [latex]T_{3}[/latex] to change from 85 to [latex]90^{\circ} F[/latex]. The apparent time constant for these tests is [latex]10 min[/latex].
A step change in [latex]T_{2}[/latex] produces a transient response in [latex]T_{3}[/latex] that is essentially complete in [latex]50(=5 au) min[/latex].
The thermocouple output is amplified to give [latex]V_{3}=0.15 T_{3}+[/latex] 5 , where [latex]V_{3}[=][/latex] volts and [latex]T_{3}[=]^{\circ} F[/latex].
The pipe connecting the two tanks has a mean residence time of [latex]30 s[/latex].

Accepted Answer

The available information can be translated as follows

The outlets of both the tanks have flow rate [latex]q_{0}[/latex] at all times.
[latex]T_{o}(s)=0[/latex]
Since an energy balance would indicate a first-order transfer function between [latex]T_{1}[/latex] and [latex]Q_{0}[/latex],

[latex]\frac{T^{\prime}(t)}{T^{\prime}(\infty)}=1-e^{-t / au_{1}} \quad[/latex] or [latex]\quad \frac{2}{3}=1-e^{-12 / au_{1}}, au_{l}=10.9 \min[/latex]

Therefore

[latex]\begin{aligned}&\frac{T_{1}(s)}{Q_{0}(s)}=\frac{3^{\circ} F /(-0.75 g p m)}{10.9 s+1}=-\frac{4}{10.9 s+1} \&\frac{T_{3}(s)}{Q_{0}(s)}=\frac{(5-3)^{\circ} F /(-0.75 g p m)}{ au_{2} s+1}=-\frac{2.67}{ au_{2} s+1} \quad \quad ext { for } \quad T_{2}(s)=0 \end{aligned}[/latex]

[latex]\frac{T_{1}(s)}{V_{1}(s)}=\frac{(78-70)^{\circ} F /(12-10) V}{10 s+1}=\frac{4}{10 s+1}[/latex]

[latex]\frac{T_{3}(s)}{V_{2}(s)}=\frac{(90-85)^{\circ} F /(12-10) V}{10 s+1}=\frac{2.5}{10 s+1}[/latex]

[latex]5 au_{2}=50 min[/latex] or [latex] au_{2}=10 min[/latex]

Since inlet and outlet flow rates for tank 2 are [latex]q_{0}[/latex] and the volumes of the tanks are equal,

[latex]\frac{T_{3}(s)}{T_{2}(s)}=\frac{\bar{q}_{0} / \bar{q}_{0}}{ au_{2} s+1}=\frac{1}{10.0 s+1}[/latex]

[latex]\frac{V_{3}(s)}{T_{3}(s)}=0.15[/latex]
[latex]T_{2}(t)=T_{1}\left(t-\frac{30}{60} ight)=T_{1}(t-0.5)[/latex]

[latex]\frac{T_{2}(s)}{T_{1}(s)}=e^{-0.5 s}[/latex]

Using these transfer functions, the block diagrams are as follows.

a)

b)

c)

The control configuration in part a) will provide the better control. As is evident from the block diagrams above, the feedback loop contains, in addition to [latex]G_{c}[/latex], only a first-order process in part a), but a second-orderplus-time-delay process in part b). Hence the controlled variable responds faster to changes in the manipulated variable for part a).

Question 11.8: An electrically heated, stirred-tank system is shown in Fig....

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