The jacketed vessel in Fig. E6.21 is used to heat a liquid by means of condensing steam. The following information is available: (i) The volume of liquid within the tank may vary, thus changing the area available for heat transfer. (ii) Heat losses are negligible. (iii) The tank contents are well

Question

The jacketed vessel in Fig. E6.21 is used to heat a liquid by means of condensing steam. The following information is available:

(i) The volume of liquid within the tank may vary, thus changing the area available for heat transfer.

(ii) Heat losses are negligible.

(iii) The tank contents are well mixed. Steam condensate is removed from the jacket by a steam trap as soon as it has formed.

(iv) The thermal capacitances of the tank and jacket walls are negligible.

(v) The steam condensation pressure [latex]P_{s}[/latex] is set by a control valve and is not necessarily constant.

(vi) The overall heat transfer coefficient [latex]U[/latex] for this system is constant.

(vii) Flow rates [latex]q_{F}[/latex] and [latex]q[/latex] are independently set and may vary. Derive a dynamic model for this process. The model should be simplified as much as possible. State any additional assumptions that you make.

(a) Find transfer functions relating the two primary output variables [latex]h[/latex] (level) and [latex]T[/latex] (liquid temperature) to inputs [latex]q_{F}, q[/latex], and [latex]T_{S}[/latex]. You should obtain six separate transfer functions.

(b) Briefly interpret the form of each transfer function using physical arguments as much as possible.

(c) Discuss qualitatively the form of the response of each output to a step change in each input.

Accepted Answer

Additional assumptions:

(i) The density [latex] ho[/latex] and specific heat [latex]C[/latex] of the liquid are constant.

(ii) The temperature of steam, [latex]T_{ s }[/latex], is uniform over the entire heat transfer area.

(iii) The feed temperature [latex]T_{F}[/latex] is constant (not needed in the solution).

Mass balance for the tank is

[latex]\frac{d V}{d t}=q_{F}-q[/latex] (1)

Energy balance for the tank is

[latex] ho C \frac{d\left[V\left(T-T_{r e f} ight) ight]}{d t}=q_{F} ho C\left(T_{F}-T_{r e f} ight)-q ho C\left(T-T_{r e f} ight)+U A\left(T_{s}-T ight)[/latex] (2)

where [latex]T_{ ext {ref }}[/latex] is a constant reference temperature and [latex]A[/latex] is the heat transfer area

Eq. 2 is simplified by substituting for [latex]\frac{d V}{d t}[/latex] from Eq. 1. Also, replace [latex]V[/latex] by [latex]A_{T} h[/latex] (where [latex]A_{T}[/latex] is the tank area) and replace [latex]A[/latex] by [latex]p_{T} h[/latex] (where [latex]p_{T}[/latex] is the perimeter of the tank). Then,

[latex]A_{T} \frac{d h}{d t}=q_{F}-q[/latex] (3)

[latex] ho C A_{T} h \frac{d T}{d t}=q_{F} ho C\left(T_{F}-T ight)+U p_{T} h\left(T_{s}-T ight)[/latex] (4)

Then, Eqs. 3 and 4 are the dynamic model for the system.

a) Making a Taylor series expansion of nonlinear terms in (4) and introducing deviation variables, Eqs. 3 and 4 become:

[latex]A_{T} \frac{d h^{\prime}}{d t}=q_{F}^{\prime}-q^{\prime}[/latex] (5)

[latex] ho C A_{T} \bar{h} \frac{d T^{\prime}}{d t}= ho C\left(T_{F}-\bar{T} ight) q_{F}^{\prime}-\left( ho C \bar{q}_{F}+U p_{T} \bar{h} ight) T^{\prime}+U p_{T} \bar{h} T_{s}^{\prime}+U p_{T}\left(\bar{T}_{s}-\bar{T} ight) h^{\prime}[/latex] (6)

Taking Laplace transforms,

[latex]H^{\prime}(s)=\frac{1}{A_{T} s} Q_{F}^{\prime}(s)-\frac{1}{A_{T} s} Q^{\prime}(s)[/latex] (7)

[latex]\left[\left(\frac{ ho C A_{T} \bar{h}}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight) s+1 ight] T^{\prime}(s)=\left[\frac{ ho C\left(T_{F}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight] Q_{F}^{\prime}(s)[/latex]

[latex]+\left[\frac{U p_{T} \bar{h}}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight] T_{s}^{\prime}(s)+\left[\frac{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight] H^{\prime}(s)[/latex] (8)

Substituting for [latex]H^{\prime}(s)[/latex] from (7) into (8) and rearranging gives

[latex]\begin{gathered}{\left[A_{T} s ight]\left[\left(\frac{ ho C A_{T} \bar{h}}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight) s+1 ight] T^{\prime}(s)=\left[\frac{ ho C\left(T_{F}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} A_{T} s ight] Q_{F}^{\prime}(s)} \+\left[\frac{U p_{T} \bar{h} A_{T} s}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight] T_{s}^{\prime}(s)+\left[\frac{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight]\left[Q_{F}^{\prime}(s)-Q^{\prime}(s) ight]\end{gathered}[/latex] (9)

Let [latex] au=\frac{ ho C A_{T} \bar{h}}{ ho C \bar{q}_{F}+U p_{T} \bar{h}}[/latex]

Then from Eq. 7

[latex]\frac{H^{\prime}(s)}{Q_{F}^{\prime}(s)}=\frac{1}{A_{T} s} \quad, \quad \frac{H^{\prime}(s)}{Q^{\prime}(s)}=-\frac{1}{A_{T} s} \quad, \quad \frac{H^{\prime}(s)}{T_{s}^{\prime}(s)}=0[/latex]

And from Eq. 9

[latex]\frac{T^{\prime}(s)}{Q_{F}^{\prime}(s)}=\frac{\left[\frac{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight]\left[\left(\frac{ ho C\left(T_{F}-\bar{T} ight) A_{T}}{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)} ight) s+1 ight]}{\left(A_{T} s ight)( au s+1)}[/latex]

[latex]\frac{T^{\prime}(s)}{Q^{\prime}(s)}=\frac{-\left[\frac{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight]}{\left(A_{T} s ight)( au s+1)}[/latex]

[latex]\frac{T^{\prime}(s)}{T_{s}^{\prime}(s)}=\frac{\left[\frac{U p_{T} \bar{h}}{ ho C \bar{q}_{F}+U p_{T} \bar{h}} ight]}{ au s+1}[/latex]

Note:

[latex] au_{2}=\frac{ ho C\left(T_{F}-\bar{T} ight) A_{T}}{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)} \quad ext { is the time constant in the numerator. }[/latex]

Because [latex]T_{F}-\bar{T}<0[/latex] (heating) and [latex]\bar{T}_{s}-T>0, au_{2}[/latex] is negative, we can show this property by using Eq. 2 at steady state:

[latex] ho C \bar{q}_{F}\left(T_{F}-\bar{T} ight)=-U p_{T} \bar{h}\left(\bar{T}_{s}-\bar{T} ight)[/latex]

or [latex] ho C\left(T_{F}-\bar{T} ight)=\frac{-U p_{T} \bar{h}\left(\bar{T}_{s}-\bar{T} ight)}{\bar{q}_{F }}[/latex]

Substituting

[latex] au_{2}=-\frac{\bar{h} A_{T}}{\bar{q}_{F}}[/latex]

Let [latex]\bar{V}=\bar{h} A_{T}[/latex] so that [latex] au_{2}=-\frac{\bar{V}}{\bar{q}_{F}}=-[/latex] (initial residence time of tank)

For [latex]\frac{T^{\prime}(s)}{Q_{F}^{\prime}(s)}[/latex] and [latex]\frac{T^{\prime}(s)}{Q^{\prime}(s)}[/latex] the "gain" in each transfer function is

[latex]K=\left[\frac{U p_{T}\left(\bar{T}_{s}-\bar{T} ight)}{A_{T}\left( ho C \bar{q}_{F}+U p_{T} \bar{h} ight)} ight][/latex]

and must have the units of temperature/volume. (The integrator [latex]s[/latex] has units of [latex]t^{-1}[/latex] ).

To simplify the transfer function gain, we can substitute

[latex]U p_{T}\left(\bar{T}_{s}-\bar{T} ight)=-\frac{ ho C \bar{q}_{F}\left(\bar{T}_{F}-\bar{T} ight)}{\bar{h}}[/latex]

from the steady-state relation. Then

[latex]\begin{aligned}&K=\frac{- ho C \bar{q}_{F}\left(\bar{T}_{F}-\bar{T} ight)}{\bar{h} A_{T}\left( ho C \bar{q}_{F}+U p_{T} \bar{h} ight)} \& ext { or } \quad K=\frac{\bar{T}-\bar{T}_{F}}{\bar{V}\left(1+\frac{U p_{T} \bar{h}}{ ho C \bar{q}_{F}} ight)}\end{aligned}[/latex]

and the gain is positive since [latex]\bar{T}-\bar{T}_{F}>0[/latex]. Furthermore, it has dimensions of temperature/volume.

(The ratio [latex]\frac{U p_{T} \bar{h}}{ ho C \bar{q}_{F}}[/latex] is dimensionless).

(b) The [latex]h-q_{F}[/latex] transfer function is an integrator with a positive gain. Liquid level accumulates any changes in [latex]q_{F}[/latex], increasing for positive changes and vice-versa.

[latex]h-q[/latex] transfer function is an integrator with a negative gain. [latex]h[/latex] accumulates changes in [latex]q[/latex], in the opposite direction, decreasing as [latex]q[/latex] increases and vice versa.

[latex]h-T_{s}[/latex] transfer function is zero. Liquid level is independent of [latex]T_{s}[/latex] and steam pressure [latex]P_{s}[/latex].

[latex]T-q[/latex] transfer function is second-order due to the interaction with liquid level; it is the product of an integrator and a first-order process.

[latex]T-q_{F}[/latex] transfer function is second-order due to the interaction with liquid level; it has numerator dynamics since [latex]q_{F}[/latex] affects [latex]T[/latex] directly as well if [latex]T_{F} eq \bar{T}[/latex]

[latex]T-T_{s}[/latex] transfer function is first-order because there is no interaction with liquid level.

c) [latex]h-q_{F}: h[/latex] increases continuously at a constant rate.

[latex]h-q: h[/latex] decreases continuously at a constant rate.

[latex]h-T_{s}: h[/latex] stays constant.

[latex]T-q_{F}[/latex] : for [latex]T_{F}<\bar{T}, T[/latex] decreases initially (inverse response) and then increases. After long times, [latex]T[/latex] increases like a ramp function.

[latex]T-q: T[/latex] decreases, eventually at a constant rate.

[latex]T-T_{s}: T[/latex] increases with a first-order response and attains a new steady state.

Question 6.21: The jacketed vessel in Fig. E6.21 is used to heat a liquid b...

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