A two-input/two-output process involving simultaneous heating and liquid-level changes is illustrated in Fig. E6.20. Find the transfer function models and expressions for the gains and the time constant τ for this process. What is the output response for a unit step change in Q for a unit step

Question

A two-input/two-output process involving simultaneous heating and liquid-level changes is illustrated in Fig. E6.20. Find the transfer function models and expressions for the gains and the time constant [latex] au[/latex] for this process. What is the output response for a unit step change in [latex]Q[/latex] for a unit step change in [latex]w[/latex] ? Note: Transfer function models for a somewhat similar process depicted in Fig. [latex]6.13[/latex] are given in Eqs. 6-96 through 6-103. They can be compared with your results. For this exercise, [latex]T[/latex] and [latex]h[/latex] are the outputs and [latex]Q[/latex] and [latex]w[/latex] are the inputs.

[latex]\frac{T^{\prime}(s)}{W_{h}^{\prime}(s)}=\frac{\left(\bar{T}_{h}-\bar{T} ight) / \bar{w}}{ au s+1}[/latex] (6-96)

[latex]\frac{T^{\prime}(s)}{W_{c}^{\prime}(s)}=\frac{\left(\bar{T}_{c}-\bar{T} ight) / \bar{w}}{ au s+1}[/latex] (6-97)

[latex]\frac{T^{\prime}(s)}{T_{h}^{\prime}(s)}=\frac{\bar{w}_{h} / \bar{w}}{ au s+1}[/latex] (6-98)

[latex]\frac{T^{\prime}(s)}{T_{c}^{\prime}(s)}=\frac{\bar{w}_{C} / \bar{w}}{ au s+1}[/latex] (6-99)

[latex]\frac{H^{\prime}(s)}{W_{h}^{\prime}(s)}=\frac{1 / A ho}{s}[/latex] (6-100)

[latex]\frac{H^{\prime}(s)}{W_{c}^{\prime}(s)}=\frac{1 / A ho}{s}[/latex] (6-101)

[latex]\frac{H^{\prime}(s)}{T_{h}^{\prime}(s)}=0[/latex] (6-102)

[latex]\frac{H^{\prime}(s)}{T_{c}^{\prime}(s)}=0[/latex] (6-103)

Accepted Answer

The dynamic model for the process is given by Eqs. 2-45 and 2-46, which can be written as

[latex]\frac{d V}{d t}=\frac{1}{ ho}\left(w_{i}-w ight)[/latex] (2-45)

[latex]\frac{d T}{d t}=\frac{w_{i}}{V_{ ho}}\left(T_{i}-T ight)+\frac{Q}{ ho C V}[/latex] (2-46)

[latex]\frac{d h}{d t}=\frac{1}{ ho A}\left(w_{i}-w ight)[/latex] (1)

[latex]\frac{d T}{d t}=\frac{w_{i}}{ ho A h}\left(T_{i}-T ight)+\frac{Q}{ ho A h C}[/latex] (2)

where

h is the liquid-level
A is the constant cross-sectional area

System outputs: h , T
System inputs : w, Q

Assume that [latex]w_{i}[/latex] and [latex]T_{i}[/latex] are constant. In Eq. 2 , note that the nonlinear term [latex]\left(h \frac{d T}{d t} ight)[/latex] can be linearized as

[latex]\bar{h} \frac{d T^{\prime}}{d t}+\frac{d \bar{T}}{d t} h^{\prime}[/latex]

or [latex]\bar{h} \frac{d T^{\prime}}{d t}[/latex] since [latex]\frac{d \bar{T}}{d t}=0[/latex]

Then the linearized deviation variable form of [latex](1)[/latex] and (2)is

[latex]\frac{d h^{\prime}}{d t}=-\frac{1}{ ho A} w^{\prime}[/latex]

[latex]\frac{d T^{\prime}}{d t}=\frac{-w_{i}}{ ho A \bar{h}} T^{\prime}+\frac{1}{ ho A \bar{h} C} Q^{\prime}[/latex]

Taking Laplace transforms and rearranging,

[latex]\frac{H^{\prime}(s)}{W^{\prime}(s)}=\frac{K_{1}}{s}, \frac{H^{\prime}(s)}{Q^{\prime}(s)}=0 \quad, \quad \frac{T^{\prime}(s)}{W^{\prime}(s)}=0 \quad, \quad \frac{T^{\prime}(s)}{Q^{\prime}(s)}=\frac{K_{2}}{ au_{2} s+1}[/latex]

where [latex]K_{1}=-\frac{1}{ ho A} ;[/latex] and [latex]K_{2}=\frac{1}{w_{i} C} \quad, au_{2}=\frac{ ho A \bar{h}}{w_{i}}[/latex]

For an unit-step change in [latex]Q: h(t)=\bar{h} \quad, \quad T(t)=\bar{T}+K_{2}\left(1-e^{-t / au_{2}} ight)[/latex]

For an unit step change in [latex]w: h(t)=\bar{h}+K_{1} t, T(t)=\bar{T}[/latex]

Question 6.20: A two-input/two-output process involving simultaneous heatin...

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