A pulse study was performed by adding 16.0 mg of a reactive tracer, A, which has a half-life of 25 s, to an inlet stream flowing at 0.625 L s^-1. Determine E(t), t; and σ²t from the following response data t/s: 0 1 5 10 1 5 20 25 30 35 40 50 60 80 cA/mg L^-1: 0 0.25 0.38 0.55 0.58 0.51 0.39 0.22

Question

A pulse study was performed by adding 16.0 mg of a reactive tracer, A, which has a half-life of 25 s, to an inlet stream flowing at 0.625 L s[latex]^{-1}[/latex]. Determine E(t), [latex]\bar{t} [/latex] ; and [latex]\sigma^2_t [/latex] from the following response data :

t/s: 0 1 5 10 1 5 20 25 30 35 40 50 60 80

[latex]c_A/ mg L^{-1} [/latex] : 0 0.25 0.38 0.55 0.58 0.51 0.39 0.22 0.11 0.05 0.02 0.01 0.00

Table 19.4 Spreadsheet calculations for Example 19-3, using the histogram method

(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)	(9)
i	t/s	[latex] c_A /mg L^{-1} [/latex]	[latex] c_{adj} /mg L^{-1} eq [/latex] (A)	[latex] \Delta t/s [/latex]	[latex] c_{adj} \Delta t /mg L^{-1} s [/latex]	[latex] E(t) /s^{-1} = (4) / \sum(6) [/latex]	[latex] tE(t)\Delta t /s [/latex]	[latex] t^2E(t)\Delta t /s^2 [/latex]
1	0	0.00	0.000	0.5	0.00	0.0000	0.00	0.0
2	1	0.25	0.257	2.5	0.64	0.0100	0.03	0.0
3	5	0.38	0.436	4.5	1.96	0.0170	0.38	1.9
4	10	0.55	0.726	5.0	3.63	0.0283	1.42	14.2
5	15	0.58	0.879	5.0	4.40	0.0342	2.57	54.7
6	20	0.51	0.888	5.0	4.44	0.0346	3.46	69.2
7	25	0.39	0.780	5.0	3.90	0.0304	3.80	95.0
8	30	0.22	0.505	5.0	2.53	0.0197	2.96	88.8
9	35	0.11	0.290	5.0	1.45	0.0113	1.98	69.3
10	40	0.05	0.152	7.5	1.14	0.0059	1.77	70.8
11	50	0.02	0.080	10.0	0.80	0.0031	1.55	77.5
12	60	0.01	0.053	15.0	0.80	0.0021	1.89	113.4
13	80	0.00	0.000	20.0	0.00	0.0000	0.00	0.0
∑					25.69		21.81	654.8

Accepted Answer

Since the tracer reacts as it moves through the vessel, the measured concentration of tracer at the outlet is less than it otherwise would be. The adjusted concentration, [latex]c_{adj}[/latex], is obtained by taking the first-order-decay into account at each time: [latex](-r_A) = k_Ac_A[/latex]. From equation 3.4-14,

[latex]t_{1/2} = (\ln2)/k_A (n = 1)[/latex] (3.4-14)

[latex]k_A = \ln 2/t_{1/2} = \ln 2/25 = 0.0277 s^{-1}[/latex]

Then, from equation 3.4- 10,

[latex]c_A = c_{Ao} \exp(-k_At) (n = 1) [/latex] (3.4-10)

[latex]c_A=c_{adj}e^{-k_At}[/latex]

or

[latex]c_{adj} = c_Ae^{k_At} = c_Ae^{0.0277t}[/latex] (A)

Table 19.4 gives the calculations of E(t), [latex]\bar{t} [/latex] ; and [latex]\sigma ^2_{t} [/latex] based on the histogram method. Column (4) lists the values of [latex]c_{adj}[/latex] calculated from (A). Column (5) gives At, required for the calculations in subsequent columns, from equations 19.3-12 to -14.

[latex]\Delta t _1 \frac{t_1 + t_2}{2} [/latex] (19.3-12)

[latex]\Delta t_i = \frac{(t_{i + 1} - t_i ) + (t_i - t_{i - 1})}{2} = \frac{t_{i + 1} - t_{i - 1}}{2} [/latex] (19.3-13)

[latex]\Delta t_{n_c} = t_{n_c} - t_{n_c-1} (i = n_c)[/latex] (19.3-14)

Column (6) gives values for the tracer material-balance (see below). Column (7) gives values of E(t) from equation 19.3-4 with C(t) = E(t).

[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)

Columns (8) and (9) give values required for the calculation of Tin equation 19.3-7, and of [latex]\sigma^2_t[/latex] in 19.3-8, respectively.

[latex]\bar{t} = \int_{0}^{\infty }{tE(t)dt} \simeq \sum\limits_{i}{t_iE_i(t)\Delta t_i} [/latex] (19.3-7)

[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)

For the tracer material balance 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 = 16.0 /0.625 = 25.6 mg s L^{-1} [/latex]

From the sum of column (6):

[latex]\sum{c_{adj} \Delta t } = 25.7 mg s L^{-1} [/latex]

From the sum of column (8):

[latex]\bar{t} = 21.8 s[/latex]

From this and the sum of column (9):

[latex]\sigma ^2_t = 654.8 - (21.8)^2 = 180 s^2[/latex]

Note that, if the data were not adjusted to account for decay of the tracer, the material balance criterion would not be met, and E(t), [latex]\bar{t}[/latex] and [latex]\sigma^2_t[/latex] would be estimated incorrectly. Values of [latex]\bar{t}[/latex] and [latex]\sigma^2_t[/latex] obtained using the trapezoidal rule are 21.9 s and 161.4 s², respectively. These values are used in Examples 19-8 and -10 below.
Estimation of [latex]\bar{t}[/latex] and [latex]\sigma^2_t[/latex] by nonlinear regression (file exl9-3.msp) leads to the following values: [latex]\bar{t} = 22.2 s [/latex] and [latex]\sigma^2_t = 165 s^2 [/latex]. These results are quite close to the values predicted by the trapezoidal rule approximation; the histogram technique leads to a slightly larger estimate of [latex]\sigma^2_t[/latex] .

Question 19.3: A pulse study was performed by adding 16.0 mg of a reactive ...

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