(a) Derive the equation for the RTD function E(θ) for constant-density flow of a fluid through two stirred tanks in series; that is, the flow in each tank is BMF. The volume of each tank is V/2, and the steady-state flow rate is q.

(b) On the same plot,graphically compare the result from (a) (for N= 2), with that for
a single stirred tank (N = 1).

(c) What is F(θ) for two stirred tanks in series?

Question

(a) Derive the equation for the RTD function E(θ) for constant-density flow of a fluid through two stirred tanks in series; that is, the flow in each tank is BMF. The volume of each tank is V/2, and the steady-state flow rate is q.

(b) On the same plot,graphically compare the result from (a) (for N= 2), with that for
a single stirred tank (N = 1).

(c) What is F(θ) for two stirred tanks in series?

Accepted Answer

(a) The flow diagram and symbols are shown in Figure 17.3.
To develop E(B) for two CSTRs in series, we use a slightly different, but equivalent, method from that used for a single CSTR in Section 13.4.1.1. Thus, consider a small amount (moles) of tracer [latex] M , n_{Mo} \equiv F_tdt [/latex] , where [latex]F_t[/latex] is the total steady-state molar flow rate, added to the first vessel at time 0. The initial concentration of M is [latex] c_{Mo} \equiv n_{M0} /(V/2) [/latex] . We develop a material balance for M around each tank to determine the time-dependent outlet concentration of M from the second vessel, [latex] c_{M2}(t) [/latex] ). For vessel 1, there is no subsequent input of tracer, the outlet flow rate is [latex] qc_{M1}[/latex], and the accumulation rate is [latex] (V/2) dc_{M1}/dt [/latex] . Thus, at any subsequent time t,

[latex] 0 - qc_{M1} = (V/2) dc_{M1} /dt [/latex] (A)

or

[latex] \int _{c_{Mo}}^{c_{M1}}{ \frac {dc_{M1}}{c_{M1}}} = \overset{\underset{}{ t} }{\underset{I_{0}}{}}{\frac{2q}{V}dt} [/latex]

which integrates to

[latex]c_{M1} = c_{Mo}e^{-2qt/V} [/latex] (B)

For vessel 2, the input rate is [latex]qc_{M1}[/latex], the output rate is [latex]qc_{M2}[/latex], and the accumulation rate is [latex](V/2)dc_{M2}/dt[/latex] . Therefore ,

[latex]qc_{M1} - qc_{M2} = (V/2)dc_{M2} /dt[/latex]

Substituting for [latex]c_{M1}[/latex] from (B), and rearranging in the form of a first-order differential equation, we obtain

[latex]\frac{dc_{M2}}{dt} + \frac{2q}{V}c_{M2} = \frac{2q}{V}c_{Mo}e^{-2qt/V}[/latex] (C)

Integration of (C) from [latex]c_{M2} = 0 at t = 0 to c_{M2} [/latex] at t results in the tracer concentration in vessel 2 as a function of time :

[latex] c_{M2} = (2qt/V)c_{Mo} e^{-2qt/V} [/latex] (D)

To obtain E(t) for the two-vessel configuration, consider fluid leaving vessel 2 between t and t + dt. The total amount is [latex]F_tdt \equiv n_{Mo} [/latex], and the amount of tracer is [latex]c_{M}qdt [/latex]. Thus,

[latex] fraction tracer = \frac{amount of tracer }{ amount of flow } = \frac{c_Mqdt}{n_{Mo}} = \frac{(2qt/V)c_{Mo}e^{-2qt/V}qdt }{n_{Mo}} \ \frac{(2qt/V)[n_{Mo}/(V/2)]e^{-2qt/V} qdt}{n_{Mo}} = 4(q/V)^2te^{-2qt/V}dt[/latex] (E)

This is also the fraction of the stream leaving vessel 2 that is of age between t and t + dt, since only the amount [latex]n_{Mo}[/latex] entered at t = 0. But this fraction is also E(t)dt, from the definition of E(t). Thus, with [latex]V/q = \bar{t} [/latex] the mean residence time for fluid in the two tanks (of total volume V), from (E),

[latex]E(t) = (4t/\bar{t}^2) \exp(-2t/\bar{t}) (BMF , N = 2) [/latex] (17.2-4)

In terms of dimensionless time [latex] heta = t/\bar{t} [/latex] , since [latex] E( heta) = \bar{t} E(t) [/latex] , from equation 13.3-7, equation 17.2-4 becomes

[latex]E( heta) = 4 heta \exp(-2 heta) (BMF ,N = 2) [/latex] (17.2-5)

(b) Figure 17.4 compares the RTD for N = 2, given by equation 17.2-5, with that for
N = 1, E(θ) = exp(-θ), given by equation 13.4-3. The behavior differs in two main
respects:

(1) At 13 = 0, E(θ) = 1 for N = 1, but E(θ) = 0 for N = 2.
(2) For N = 2, E(θ) goes through a maximum value of 0.736 at θ = 0.5 (obtained by setting dE(θ)/dθ = 0, from equation 17.2-5).

(c) From equations 13.3-7, -9, and -10, and 17.2-5,

[latex] F(t) = F(B) [/latex] (13.3-9)

[latex] F(t) = \int_{0}{E(t)dt} [/latex] (13.3-10)

[latex] F( heta) = \overset{\underset{}{ e} }{\underset{I_{0}}{}}{E( heta )d heta } = \overset{\underset{}{ e} }{\underset{I_{0}}{}}{48 \exp(-2 heta )d heta } [/latex]

Integration by parts yields the result

[latex] F( heta) = 1 - e^{-2 heta}( 1 + 2 heta) (BMF , N= 2) [/latex] (17.2-6)

Comparison may be made of the reactor performance of tanks in series and tanks in parallel, in light of the different RTDs shown in Figure 17.4. If the two tanks shown in Figure 17.3 are arranged in parallel, and the flow, q, is split evenly between them, they act together as a single tank of volume [latex]V = V_1 + V_2[/latex] (Section 17.2.1), with the same performance, and the same RTD shown in Figure 17.4 for N = 1. For the same two tanks arranged in series (Figure 17.3), the RTD is different (Figure 17.4 for N = 2), and we expect a different performance. Thus, as discussed in Section 14-4, for positive, power-law (normal) kinetics,the series arrangement gives a better performance (e.g., [latex] f_A [/latex](series) > [latex] f_A [/latex] (parallel), since reaction takes place with a higher rate at the higher reactant concentration in the first tank (the opposite conclusion applies for the case of abnormal kinetics, which is much less common).

Question 17.4: (a) Derive the equation for the RTD function E(θ) for consta...

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