Using the data from Example 19-1, recalculate values of E(t), t; and σ²t by assuming that significant “tailing” begins at tT= 10 min.

Question

Using the data from Example 19-1, recalculate values of E(t), [latex]\bar{t} [/latex] ; and [latex]\sigma^2_t[/latex] by assuming that significant “tailing” begins at [latex]t_T = 10 min [/latex].

Accepted Answer

In Figure 19.10, the experimental response data for t ≥ 10 min are plotted as indicated in Figure 19.9(b). The semilogarithmic plot is linear with a slope - a = - 0.104 min -¹ (i.e.; a = 0.104) . Therefore, since [latex] c(t_T) = c(10) = 0.736 kg m^{-3} [/latex] (from data given in Example 19-l), the tail region is represented by [latex] c(t) = 0.736 \exp[-0.104 (t - 10)]kg m^{-3} [/latex] . The revised calculations are given in Table 19.5, which may be compared with Table 19.1. The calculations for points 1 to 6 are the same as before, but note that the normalizing

Table 19.5 Calculations for Example 19-4

(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)	(9)	(10)	(11)
i	t/l	c /	[latex] c_{a u}/[/latex]	Δt/mm	[latex]c_{a u} \Delta t/ kg m^{-3}[/latex]	[latex]E(t) /min^{-1} = (3)/\sum{(6)}[/latex]	[latex][tE(t)]_{a u}[/latex]	(8) Δt/ mm	[latex][t^2E(t)]_{a u} /mm[/latex]	(10) Δt/ min²
i	t/l	[latex]kg m^{-3}[/latex]		Δt/mm	[latex]c_{a u} \Delta t/ kg m^{-3}[/latex]	[latex]E(t) /min^{-1} = (3)/\sum{(6)}[/latex]	[latex][tE(t)]_{a u}[/latex]	(8) Δt/ mm	[latex][t^2E(t)]_{a u} /mm[/latex]	(10) Δt/ min²
1	0.0	2.000				0.1000
			1.980	0.2	0.396		0.0098	0.00	0.0020	0.00
2	0.2	1.960				0.0980
			1.945	0.2	0.389		0.0291	0.01	0.0097	0.00
3	0.4	1.930				0.0965
			1.786	1.6	2.858		0.1014	0.16	0.1719	0.27
4	2.0	1.642				0.0821
			1.493	2.0	2.986		0.2165	0.43	0.7018	1.41
5	4.0	1.344				0.0672
			1.040	6.0	6.240		0.3184	1.91	2.3776	14.30
6	10.0	0.736				0.0368
above totals					12.87			2.51		15.98
7	20.0	0.268			[latex] from^''[/latex] eqn. 19.3-21 area = 7.08	0.0134		[latex] from^b[/latex] eqn. 19.3-24 area = 6.94		[latex] from^'[/latex] eqn. 19.3-25 area = 168.9

8	40.0	0.034				0.0017
9	60.0	0.004				0.0002

10	80.0	0.000				0.0000
∑					20.0			9.45		184.8
[latex]^ac(t_T)/a = 0.736/0.104 = 7.08.[/latex] [latex]^bE(t_T)(at_T + 1)/a^2 = 0.0368[0.104(10) + 1]/0.104^2 = 6.94. [/latex] [latex]^cE(t_T)(a^2t^2_T + 2at_T + 2)/a^3 = 168.9 .[/latex]

[latex]m_o/q_o=\sum\limits_{i=1}^{n_T}c_i(t)\Delta t_i+\frac{c(t_T)}{a}[/latex] (19.3-21)

[latex]\overline{t}=\sum\limits_{i=1}^{n_T}t_iE_i(t)\Delta t_i+\frac{E(t_T)}{a^2}(at_T+1)[/latex] (19.3-24)

factor to convert c(t) to E(t) is 20.0 instead of 21.3. The contributions to the areas required for calculation of E(t), [latex]\bar{t}[/latex], and [latex]\sigma ^2_t[/latex] in columns (6), (9), and (ll), respectively are from the treatment of the “tail” region by means of the equations indicated in the table.
For the tracer material balance corresponding to 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 = 20.4 kg min m^{-3} [/latex] (from Example 19.1)

[latex] \int _{0}^{ \infty} c(t)dt = 20.0[/latex] (from column (6))

The difference is 2.0%, which is less than the value of 4.4% obtained in Example 19-1. From Table 19.5, column (9), [latex]\bar{t} = 9.45 min [/latex], which is slightly greater than the value in Example 19-1 (9.35 min), suggesting slightly less dead space. From column (11) in Table 19.5 and equation 19.3-25

[latex]\sigma ^2_t = \sum\limits_{i = 1}^{n_T}{t^2_iE_i(t)\Delta t_i} + \frac{E(t_T)}{a^3}(a^2t^2_T + 2at_T + 2) - \bar{t}^2 [/latex] (19.3-25)

[latex]\sigma ^2_t = 184.8 - (9.45)^2 = 95.5 min^2 [/latex]

This is somewhat different from the value in Example 19-1 (84.7 min²). However, these results for [latex]\bar{t}[/latex], and [latex]\sigma ^2_t[/latex] are closer to the results obtained by nonlinear regression using E-Z Solve (9.9 min, 97.7 min2; file ex19-1,msp). One advantage of nonlinear regression is that it automatically accounts for the “long tail,” without requiring additional treatment of the data. Apart from the material balance on the tracer, we cannot tell, from the data provided, which set of values for E(t), [latex]\bar{t}[/latex], and [latex]\sigma ^2_t[/latex] is more correct.

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 /	[latex]c_{av}/ [/latex]	[latex]\Delta t / min [/latex]	[latex]c_{av}\Delta t / kg min m^{-3}[/latex]	[latex]E(t)/ min^{-1} = (3) / \sum(6)[/latex]	[latex][tE(t)]_{av}[/latex]	[latex](8) \Delta t/min[/latex]	[latex][t^2E(t)]_{av}[/latex]	[latex]\Delta t/min^2[/latex]
i	t / min	[latex] kg m^{-3}[/latex]		[latex]\Delta t / min [/latex]	[latex]c_{av}\Delta t / kg min m^{-3}[/latex]	[latex]E(t)/ min^{-1} = (3) / \sum(6)[/latex]	[latex][tE(t)]_{av}[/latex]	[latex](8) \Delta t/min[/latex]	[latex][t^2E(t)]_{av}[/latex]	[latex]\Delta t/min^2[/latex]
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

Question 19.4: Using the data from Example 19-1, recalculate values of E(t)...

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