stirred-tank heat exchanger with a bypass stream is shown in Fig. E18.17 with the available control valves. The possible manipulated variables are mass flow rate w2, valve stem positions xc and x3, and f, the fraction of mass flow rate w1 that bypasses the tank before being added to the exit stream

Question

A stirred-tank heat exchanger with a bypass stream is shown in Fig. E18.17 with the available control valves. The possible manipulated variables are mass flow rate [latex]w_{2}[/latex], valve stem positions [latex]x_{c}[/latex] and [latex]x_{3}[/latex], and [latex]f[/latex], the fraction of mass flow rate [latex]w_{1}[/latex] that bypasses the tank before being added to the exit stream. Using the information given here, do the following:

(a) Derive a dynamic model of the stirred-tank system. Define any additional symbols that you introduce.

(b) Determine the degrees of freedom for control that are available. Allocate the degrees of freedom by specifying manipulated variables and variables that are determined by the environment.

(c) Select controlled variables and briefly justify your choice.

(d) Suppose that only [latex]T_{4}[/latex] and [latex]h[/latex] are to be controlled by using [latex]x_{2}[/latex] and [latex]f[/latex] as the manipulated variables. (Valve stem positions, [latex]x_{c}[/latex] and [latex]x_{3}[/latex], are held constant.) Derive an expression for the relative gain array for this control configuration.

(e) It has been proposed that [latex]x_{2}[/latex] be replaced by [latex]x_{3}[/latex] in the control problem of (d). Briefly analyze this proposal. (It is not necessary to perform another RGA analysis.)

Available Information

(i) The tank is perfectly mixed, and the temperature changes are relatively small so that constant physical properties can be assumed. Mass flow rates are denoted by [latex]w_{1}[/latex] to [latex]w_{4}[/latex] and temperatures by [latex]T_{1}[/latex] to [latex]T_{4}[/latex].

(ii) The exit flow rate [latex]w_{3}[/latex] depends on the pressures upstream and downstream of the control valve, and the valve stem position [latex]x_{3} .[/latex] The following empirical relation is available where [latex]C_{1}[/latex] and [latex]C_{2}[/latex] are constants:

[latex]w_{3}=x_{3}\left(C_{1} h-C_{2} f w_{1}\right)[/latex]

(iii) The overall heat transfer coefficient for the cooling coil [latex]U[/latex] depends on the velocity of the coolant in the line and hence on the valve stem position [latex]x_{c}[/latex], according to the relation below where [latex]C_{3}[/latex] is a constant:

[latex]U=C_{3} x_{c}[/latex]

(iv) The pump on the bypass line operates in the "flat part" of the pump curve so that the mass flow rate, [latex]f w_{1}[/latex], depends only on the control valve.

(v) The following process variables remain constant: [latex]T_{1}, w_{1}[/latex], [latex]T_{2}[/latex], and [latex]T_{c}[/latex].

Accepted Answer

a) Dynamic Model:

Mass Balance:

[latex] ho A \frac{d h}{d t}=(1-f) w_{1}+w_{2}-w_{3}[/latex] (1)

Energy Balance: [latex]\left(T_{r e f}=0 ight)[/latex]

[latex] ho C_{p} A \frac{d\left(h T_{3} ight)}{d t}=C_{p}(1-f) w_{1} T_{1}+C_{p} w_{2} T_{2}-C_{p} w_{3} T_{3}-U A_{c}\left(T_{3}-T_{c} ight)[/latex] (2)

Mixing Point:

[latex]w_{4}=w_{3}+f w_{1}[/latex] (3)

Energy Balance on Mixing Point:

[latex]C_{p} w_{4} T_{4}=C_{p} w_{3} T_{3}+C_{p} f w_{1} T_{1}[/latex] (4)

Control valves:

[latex]\begin{aligned}&U=C_{3} X_{c} (5) \&w_{3}=x_{3}\left(C_{1} h-C_{2} f w_{1} ight) (6)\end{aligned}[/latex]

b) Degrees of freedom:

Variables: 14

[latex]h, w_{1}, w_{2}, w_{3}, w_{4}, T_{1}, T_{2}, T_{3}, T_{4}, T_{ c }, x_{ c }, x_{3}, f, U[/latex]

Equations: 6

Degrees of freedom [latex]=N_{V}-N_{E}=8[/latex]

Specified by the environment: [latex]4\left(T_{ c }, w_{1}, T_{1}, T_{2} ight)[/latex]

Manipulated variables: [latex]4\left(f, w_{2}, x_{ c }, x_{3} ight)[/latex]

c) Controlled variables:

[latex]h[/latex] Guidelines #2 and 5 (i.e., G2 and G5)

[latex]T_{4}[/latex] G3 and G5

[latex]w_{4}[/latex] G3 and G5

[latex]w_{2}\left( ight.[/latex] or [latex]\left.T_{3} ight)[/latex] G4 and G5 (or G2 and G5)

d) RGA

At steady state, (1) and (2) become:

[latex]0=(1-f) w_{1}+w_{2}-w_{3}[/latex] (7)

[latex]0=C_{p}(1-f) w_{1} T_{1}+C_{p} w_{2} T_{2}-C_{3} w_{3} T_{3}-U A_{c}\left(T_{3}-T_{c} ight)[/latex] (8)

Rearrange (8) and substitute (5),

[latex]T_{3}=\frac{C_{p}(1-f) w_{1}+C_{p} w_{2} T_{2}-C_{3} x_{c} A_{c} T_{c}}{C_{3} w_{3}+C_{3} x_{c} A_{c}}[/latex] (9)

Rearrange (7)

[latex]w_{3}=(1-f) w_{1}+w_{2}[/latex] (10)

Substitute (10) into (9),

[latex]T_{3}=\frac{C_{p}(1-f) w_{1}+C_{p} w_{2} T_{2}+C_{3} x_{c} A_{c} T_{c}}{C_{3}(1-f) w_{1}+C_{3} w_{2}+C_{3} x_{c} A_{c}}[/latex] (11)

Substitute (10), (3) and (11) into (4),

[latex]\left(w_{3}+f w_{1} ight) T_{4}=w_{3} T_{3}+f w_{1} T_{1}[/latex] (12)

or

[latex]\begin{aligned}&{\left[(1-f) w_{1}+w_{2}+f w_{1} ight] T_{4}=f w_{1} T_{1}+} \&+\left[(1-f) w_{1}+w_{2} ight]\left[\frac{C_{p}(1-f) w_{1}+C_{p} w_{2} T_{2}-C_{3} x_{c} A_{c} T_{c}}{C_{3}(1-f) w_{1}+C_{3} w_{2}+C_{3} x_{c} A_{c}} ight]\end{aligned}[/latex] (13)

Rearrange,

[latex]T_{4}=\frac{f w_{1} T_{1}}{w_{1}+w_{2}}+\left[\frac{(1-f) w_{1}+w_{2}}{w_{1}+w_{2}} ight]\left[\frac{C_{p}(1-f) w_{1}+C_{p} w_{2} T_{2}-C_{3} x_{c} A_{c} T_{c}}{C_{3}(1-f) w_{1}+C_{3} w_{2}+C_{3} x_{c} A_{c}} ight][/latex] (14)

Rearrange (6),

[latex]h=\frac{w_{3}+x_{3} C_{2} f w_{1}}{x_{3} C_{1}}[/latex] (15)

Substitute (10) into (15),

[latex]h=\frac{(1-f) w_{1}+w_{2}+x_{3} C_{2} f w_{1}}{x_{3} C_{1}}[/latex] (16)

Rewrite (14) as,

[latex]T_{4}=\frac{f w_{1} T_{1}}{w_{1}+w_{2}}+\left[\frac{E_{1}+E_{8} f+w_{2}}{w_{1}+w_{2}} ight]\left[\frac{E_{2} f+E_{3} w_{2}+E_{4}}{E_{5} f+E_{6} w_{2}+E_{7}} ight][/latex] (17)

where:

[latex]\left.\begin{array}{lcl}E_{1}=w_{1} & E_{2}=-C_{p} w_{1} & E_{3}=C_{p} T_{2} \E_{4}=C_{3} X_{c} A T_{c}+C_{p} w_{1} && E_{5}=-C_{3} w_{1} \E_{6}=C_{3} & E_{7}=C_{3} X_{c} A+C_{3} w_{1} & E_{8}=-w_{1}\end{array} ight\}[/latex] (18)

Can write (17) as

[latex]\begin{aligned}&T_{4}=\frac{f w_{1} T_{1}}{w_{1}+w_{2}}+\&+\frac{\overbrace{E_{8} E_{2} f^{2}+\left(E_{3} E_{8}+E_{2} ight) f w_{2}+\left(E_{1} E_{3}+E_{4} ight) w_{2}+\left(E_{1} E_{2}+E_{8} E_{4} ight) f+E_{1} E_{4}}^{F_{1}}}{\underbrace{E_{6} w_{2}^{2}+\left(w_{1} E_{6}+E_{7} ight) w_{2}+w_{1} E_{5} f+E_{5} w_{2} f+E_{7} w_{1}}_{F_{2}}}\end{aligned}[/latex] (19)

Thus

[latex]\begin{aligned}\frac{\partial T_{4}}{\partial f}&=K_{11}=\frac{w_{1} T_{1}}{w_{1}+w_{2}}+\frac{2 E_{B} E_{2} f+\left(E_{3} E_{8}+E_{2} ight) w_{2}+E_{1} E_{2}+E_{8} E_{4}}{\left[F_{2} ight]} \&-\frac{\left(F_{1} ight)\left[w_{1} E_{5}+E_{5} w_{2} ight]}{F_{2}^{2}}\end{aligned}[/latex] (20)

Similarly

[latex]\frac{\partial T_{4}}{\partial f}=K_{12}[/latex]

From (16)

[latex]\begin{aligned}&\frac{\partial h}{\partial f}=K_{21}=\frac{x_{3} C_{2} w_{1}-w_{1}}{x_{3} C_{1}} \&\frac{\partial h}{\partial w_{2}}=K_{22}=\frac{1}{x_{3} C_{1}}\end{aligned}[/latex]

Then

[latex]\Lambda=\left[\begin{array}{cc}\lambda & 1-\lambda \1-\lambda & \lambda\end{array} ight][/latex]

where

[latex]\lambda=\frac{1}{1-\frac{K_{12} K_{21}}{K_{11} K_{22}}}[/latex]

e) It will be difficult to control [latex]T_{4}[/latex] because neither [latex]x_{3}[/latex] nor [latex]f[/latex] has a large steadystate effect on [latex]T_{4}[/latex]

Question 18.17: stirred-tank heat exchanger with a bypass stream is shown in...

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