Your company is having problems with the feed stream to a reactor. The feed must be kept at a constant mass flow rate (w) even though the supply from the upstream process unit varies with time, wi(t). Your boss feels that an available tank can be modified to serve as a surge unit, with the tank

Question

Your company is having problems with the feed stream to a reactor. The feed must be kept at a constant mass flow rate [latex](\bar{w})[/latex] even though the supply from the upstream process unit varies with time, [latex]w_{i}(t)[/latex]. Your boss feels that an available tank can be modified to serve as a surge unit, with the tank level expected to vary up and down somewhat as the supply fluctuates around the desired feed rate. She wants you to consider whether (1) the available tank should be used, or (2) the tank should be modified by inserting an interior wall, thus effectively providing two tanks in series to smooth the flow fluctuations.

The available tank would be piped as shown in Fig. E6.22a:

In the second proposed scheme, the process would be modified as shown in Fig. E6.22b:

In this case, an opening placed at the bottom of the interior wall permits flow between the two partitions. You can assume that the resistance to flow [latex]w_{1}(t)[/latex] is linear and constant [latex](R)[/latex].

(a) Derive a transfer function model for the two-tank process [latex]\left[H_{2}^{\prime}(s) / W_{i}^{\prime}(s)\right][/latex] and compare it to the one-tank process [latex]\left[H^{\prime}(s) / W_{i}^{\prime}(s)\right][/latex]. In particular, for each transfer function indicate its order, presence of any zeros, gain, time constants, presence or absence of an integrating element, and whether it is interacting or noninteracting.

(b) Evaluate how well the two-tank surge process would work relative to the one-tank process for the case [latex]A_{1}=A_{2}=A / 2[/latex] where [latex]A[/latex] is the cross-sectional area of the single tank. Your analysis should consider whether [latex]h_{2}[/latex] will vary more or less rapidly than [latex]h[/latex] for the same input flow rate change, for example, a step input change.

(c) Determine the best way to partition the volume in the two-tank system to smooth inlet flow changes. In other words, should the first tank contain a larger portion of the volume than the second?

(d) Plot typical responses to corroborate your analysis. For this purpose, you should size the opening in the two-tank interior wall (i.e., choose [latex]R[/latex] ) such that the tank levels will be the same at whatever nominal flow rate you choose.

Accepted Answer

a) The two-tank process is described by the following equations in deviation variables:

[latex]\frac{d h_{1}^{\prime}}{d t}=\frac{1}{ ho A_{1}}\left[w_{1}^{\prime}-\frac{1}{R}\left(h_{1}^{\prime}-h_{2}^{\prime} ight] ight.[/latex] (1)

[latex]\frac{d h_{2}^{\prime}}{d t}=\frac{1}{ ho A_{2}}\left[\frac{1}{R}\left(h_{1}^{\prime}-h_{2}^{\prime} ight] ight.[/latex] (2)

Laplace transforming

[latex] ho A_{1} R s H_{1}^{\prime}(s)=R W_{i}^{\prime}(s)-H_{1}^{\prime}(s)+H_{2}^{\prime}(s)[/latex] (3)

[latex] ho A_{2} R s H_{2}^{\prime}(s)=H_{1}^{\prime}(s)-H_{2}^{\prime}(s)[/latex] (4)

From (4)

[latex]\left( ho A_{2} R s+1 ight) H_{2}^{\prime}(s)=H_{1}^{\prime}(s)[/latex] (5)

or

[latex]\frac{H_{2}^{\prime}(s)}{H_{1}^{\prime}(s)}=\frac{1}{ ho A_{2} R s+1}=\frac{1}{ au_{2} s+1}[/latex] (6)

where [latex] au_{2}= ho A_{2} R[/latex]

Returning to (3)

[latex]\left( ho A_{1} R s+1 ight) H_{1}^{\prime}(s)-H_{2}^{\prime}(s)=R W_{i}^{\prime}(s)[/latex] (7)

Substituting (6) with [latex] au_{1}= ho A_{1} R[/latex]

[latex]\left[\left( au_{1} s+1 ight)-\frac{1}{ au_{2} s+1} ight] H_{1}^{\prime}(s)=R W_{i}^{\prime}(s)[/latex] (8)

or

[latex]\left[\left( au_{1} au_{2} ight) s^{2}+\left( au_{1} au_{2} ight) s ight] H_{1}^{\prime}(s)=R\left( au_{2} s+1 ight) W_{i}^{\prime}(s)[/latex] (9)

[latex]\frac{H_{1}^{\prime}(s)}{W_{1}^{\prime}(s)}=\frac{R\left( au_{2} s+1 ight)}{s\left[ au_{1} au_{2} s+\left( au_{1}+ au_{2} ight) ight]}[/latex] (10)

Dividing numerator and denominator by [latex]\left( au_{1}+ au_{2} ight)[/latex] to put into standard form

[latex]\frac{H_{1}^{\prime}(s)}{W_{1}^{\prime}(s)}=\frac{\left[R /\left( au_{1}+ au_{2} ight) ight]\left( au_{2} s+1 ight)}{s\left[\frac{ au_{1} au_{2}}{ au_{1}+ au_{2}} s+1 ight]}[/latex] (11)

Note that

[latex]K=\frac{R}{ au_{1}+ au_{2}}=\frac{R}{ ho A_{1} R+ ho A_{2} R}=\frac{1}{ ho\left(A_{1}+A_{2} ight)}=\frac{1}{ ho A}[/latex] (12)

since [latex]A=A_{1}+A_{2}[/latex]

Also, let

[latex] au_{s}=\frac{ au_{1} au_{2}}{ au_{1}+ au_{2}}=\frac{ ho^{2} R^{2} A_{1} A_{2}}{ ho R\left(A_{1}+A_{2} ight)}=\frac{ ho R A_{1} A_{2}}{A}[/latex] (13)

so that

[latex]\frac{H_{1}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K\left( au_{2} s+1 ight)}{s\left( au_{3} s+1 ight)}[/latex] (14)

and

[latex]\begin{aligned}\frac{H_{2}^{\prime}(s)}{W_{i}^{\prime}(s)}&=\frac{H_{2}^{\prime}(s)}{H_{1}^{\prime}(s)} \frac{H_{1}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{1}{\left( au_{2} s+1 ight)} \frac{K\left( au_{2} s+1 ight)}{s\left( au_{3} s+1 ight)} \&=\frac{K}{s\left( au_{3} s+1 ight)}\end{aligned}[/latex] (15)

Transfer functions (6), (14) and (15) define the operation of the two tank process.

The single-tank process is described by the following equation in deviation variables

[latex]\frac{d h^{\prime}}{d t}=\frac{1}{ ho A} w_{i}^{\prime}[/latex] (16)

Note that [latex]\bar{\omega}[/latex], which is constant, subtracts out.

Laplace transforming and rearranging:

[latex]\frac{H^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{1 / ho A}{s}[/latex] (17)

Again

[latex]K=\frac{1}{ ho A}[/latex]

[latex]\frac{H^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K}{s}[/latex] (18)

which is the expected integral relationship with no zero.

b) For [latex]A_{1}=A_{2}=A / 2[/latex]

[latex]\left.\begin{array}{l} au_{2}= ho A R / 2 \ au_{3}= ho A R / 4\end{array} ight\}[/latex] (19)

Thus [latex] au_{2}=2 au_{3}[/latex]

We have two sets of transfer functions:

[latex]\begin{array}{ll}\underline{ ext{One-Tank Process}} &\underline{ ext{Two-Tank Process}} \\frac{H^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K}{s} & \frac{H_{i}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K\left(2 au_{3} s+1 ight)}{s\left( au_{3} s+1 ight)} \& \frac{H_{2}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K}{s\left( au_{3} s+1 ight)}\end{array}[/latex]

Remarks:

- The gain (K =1/ρA) is the same for all TFs.

- Each TF contains an integrating element.

- However, the two-tank TF's contain a pole [latex]\left( au_{3} s+1 ight)[/latex] that will "filter out" changes in level caused by changing [latex]w_{i}(t)[/latex].

- On the other hand, for this special case, we see that the zero in the first tank transfer function [latex]\left(H_{i}^{\prime}(s) / W_{i}^{\prime}(s) ight)[/latex] is larger than the pole:

[latex]2 au_{3}> au_{3}[/latex]

Thus we should make sure that amplification of changes in [latex]h_{1}(t)[/latex] caused by the zero do not more than cancel the beneficial filtering of the pole so as to cause the first compartment to overflow easily.

Now look at more general situations of the two-tank case:

[latex]\frac{H_{1}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K\left( ho A_{2} R s+1 ight)}{s\left(\frac{ ho R A_{1} A_{2}}{A} s+1 ight)}=\frac{K\left( au_{2} s+1 ight)}{s\left( au_{3} s+1 ight)}[/latex] (20)

[latex]\frac{H_{2}^{\prime}(s)}{W_{i}^{\prime}(s)}=\frac{K}{s\left( au_{3} s+1 ight)}[/latex] (21)

For either [latex]A_{1} ightarrow 0 ext { or } A_{2} ightarrow 0[/latex],

[latex] au_{3}=\frac{ ho R A_{1} A_{2}}{A} ightarrow 0[/latex]

Thus the beneficial effect of the pole is lost as the process tends to look more like the first-order process.

c) The optimum filtering can be found by maximizing [latex] au_{3}[/latex] with respect to [latex]A_{1}[/latex] (or [latex]A_{2}[/latex])

[latex] au_{3}=\frac{ ho R A_{1} A_{2}}{A}=\frac{ ho R A_{1}\left(A-A_{1} ight)}{A}[/latex]

Find max [latex] au_{3}: \frac{\partial au_{3}}{\partial A_{1}}=\frac{ ho R}{A}\left[\left(A-A_{1} ight)+A_{1}(-1) ight][/latex]

[latex]\begin{aligned} ext{Set to 0: }\quad\quad&A-A_{1}-A_{1}=0 \&2 A_{1}=A \&A_{1}=A / 2\end{aligned}[/latex]

Thus the maximum filtering action is obtained when [latex]A_{1}=A_{2}=A / 2[/latex].

The ratio of [latex] au_{2} / au_{3}[/latex] determines the "amplification effect" of the zero on [latex]h_{1}(t)[/latex]

[latex]\frac{ au_{2}}{ au_{3}}=\frac{ ho A_{2} R}{\frac{ ho R A_{1} A_{2}}{A}}=\frac{A}{A_{1}}[/latex]

As [latex]A_{1}[/latex] goes to [latex]0, \frac{ au_{2}}{ au_{3}} ightarrow \infty[/latex]

Therefore, the influence of changes in [latex]w_{i}(t)[/latex] on [latex]h_{1}(t)[/latex] will be very large, leading to the possibility of overflow in the first tank.

Summing up:

The process designer would like to have [latex]A_{1}=A_{2}=A / 2[/latex] in order to obtain the maximum filtering of [latex]h_{1}(t)[/latex] and [latex]h_{2}(t)[/latex]. However, the process response should be checked for typical changes in [latex]w_{i}(t)[/latex] to make sure that [latex]h_{1}[/latex] does not overflow. If it does, area [latex]A_{1}[/latex] needs to be increased until it is not a problem.

Note that [latex] au_{2}= au_{3}[/latex] when [latex]A_{1}=A[/latex], thus a careful study (simulations) should be made before designing the partitioned tank. Otherwise, leave wellenough alone and use the non-partitioned tank.

Question 6.22: Your company is having problems with the feed stream to a re...

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