In an effort to determine the cause of low yields from a reactor, a tracer study was conducted. An amount mo = 3.80 kg of an inert tracer A was injected into the feed port of the 1 .9-m³ reactor. The volumetric flow rate was constant at q0 = 3.1 L s-¹. The following tracer-response data were

Question

In an effort to determine the cause of low yields from a reactor, a tracer study was conducted. An amount [latex]m_o = 3.80 kg[/latex] of an inert tracer A was injected into the feed port of the 1 .9-m³ reactor. The volumetric flow rate was constant at [latex]q_o = 3.1 L s^{-1}[/latex] . The following tracer-response data were acquired:
tlmin : 0.0 0.2 0.4 2.0 4.0 10.0 20.0 40.0 60.0 80.0
[latex]c_A/Kg m^{-3}[/latex]: 2.00 1.96 1.93 1.642 1.344 0.736 0.268 0.034 0.004 0.000
Calculate values of E(t), [latex]\bar{t}[/latex] ; and [latex]\sigma_t^2[/latex]: for flow through the vessel.

Accepted Answer

Since the response data are discrete, we may use either the histogram method or the trape-zoid rule. We use the trapezoid rule for illustration, but both methods give the same results.

Table 19.1 shows the calculations, in spreadsheet form, for E(t) (from summation form of equation 19.3-4, with C(t) = E(t), and using equation 19.3-15), [latex]\bar{t}[/latex] (equation 19.3-7a), and [latex]\sigma_t^2[/latex] (equation 19.3-8a). (Subscript i has been omitted from the headings for brevity.)

[latex]C(t) = \frac{c(t)}{m_o/q_o} = \frac{c(t)}{\int _{0}{\infty} {c(t)dt} } \simeq \frac {c_i(t) }{\sum_i{c_i(t)\Delta t_i}}[/latex] (19.3-4)

[latex]area = \sum\limits_{i = 1}^{n_c - 1}{c_i(t)_{av} \Delta t_i} = \sum\limits_{ i = 1}^{n_c - 1}{\left(\frac{c_i + c_{i + 1}}{2}\right) }(t_{i + 1} - t_i)[/latex] (19.3-15)

[latex]\bar{t} = \sum\limits_{i = 1}^{n_c - 1}{[t_iE_i{t}]_{av}\Delta t_i} [/latex] (19.3-7a)

[latex]\sigma _t^2 = \sum\limits_{i = 1}^{n_c - 1}{[t_i^2E_i{t}]_{av}\Delta t_i} \underset{I}{} - \bar{t}^2[/latex] (19.3-8a)

The sum from column (6) may be used to check the material balance on the tracer from equation 19.3-3:

[latex]m_o/q_o \simeq \sum\limits_{i}{c_i\Delta t_i} [/latex] (19.3-3)

[latex]m_o/q_o = 3.80 / [3.1(60)/1000] = 20.4 kg min m^{-3}[/latex]

Table 19.1 Spreadsheet calculations for Example 19-1, using the trapezoid rule

(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)	(9)	(10)	(11)
i	t / min	c /	c_av/	Δt/min	c_avΔt/ kg min m^-3	E(t)/ min^-1= (3) /Σ(6)	[latex][tE(t)]_{av}[/latex]	(8) Δt/min	[t²E(t)]_av	Δt/min²
i	t / min	kg m^-3		Δt/min	c_avΔt/ kg min m^-3	E(t)/ min^-1= (3) /Σ(6)	[latex][tE(t)]_{av}[/latex]	(8) Δt/min	[t²E(t)]_av	Δt/min²
1	0.0	2.000				0.0938
			1.980	0.2	0.396		0.00919	0.002	0.002	0.00
2	0.2	1.960				0.0919
			1.945	0.2	0.389		0.02729	0.005	0.009	0.00
3	0.4	1.930				0.0905
			1.786	1.6	2.858		0.0951	0.152	0.161	0.26
4	2.0	1.642				0.0770
			1.493	2.0	2.986		0.2030	0.406	0.658	1.32
5	4.0	1.344				0.0630
			1.040	6.0	6.240		0.2985	1.791	2.229	13.38
6	10.0	0.736				0.0345
			0.502	10.0	5.020		0.2982	2.982	4.245	42.38
7	20.0	0.268				0.0126
			0.151	20.0	3.020		0.1575	3.151	3.792	75.77
8	40.0	0.034				0.0016
			0.019	20.0	0.380		0.0375	0.750	1.614	32.26
9	60.0	0.004				0.0002
			0.002	20.0	0.040		0.0057	0.113	0.342	6.75
10	80.0	0.000				0.0000
∑					21.33		9.35		172.1

from column (6)

[latex]\sum{c_i\Delta t_i} = 21.3 kg min m^{-3} [/latex]

The difference is 4.4%, which may be acceptable.
Column (7) gives the value of E(t) calculated at each of the ten points.
The sum of column (9) gives [latex]\bar{t} = 9.35 min[/latex]. This may be compared with V/q, = 1.9/[3.1(60)/1000] = 10.21 min. The difference of 8.3% may indicate some stagnation within the vessel to account for a lower-than-expected yield. Put another way, the effective volume is less than 1.9 m³. The difference may also indicate that more frequent sampling is needed.
The sum of column (11) gives the first term in equation 19.3-8. Therefore,

[latex]\sigma _t^2 = \left[\int_{0}^{\infty }{t^2E(t)dt} \right] -\bar{t}^2 \simeq \left[\sum\limits_{i}{t_i^2E_i(t)\Delta t_i} \right] -\bar{t}^2[/latex] (19.3-8)

[latex]\sigma_t^2 = 172.1 - 9.35^2 = 84.7 min^2[/latex]

By comparison, estimation of [latex]\bar{t}[/latex] and [latex]\sigma_t^2[/latex] : by nonlinear regression (file ex19- 1 .msp) leads to the following values: [latex]\bar{t} = 9.9 mm[/latex] , and [latex]\sigma_t^2 = 97.7 min^2[/latex] . The only way to determine which parameter set is more “correct” is to predict the experimental concentrations using these parameters in an appropriate mixing model. This procedure is explained in Section 19.4 .

Question 19.1: In an effort to determine the cause of low yields from a rea...

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