Ammonia is to be absorbed from air at 293 K and 1.013 × 10^5 Pa pressure in a countercurrent, packed tower, 0.5 m in diameter, using ammonia-free water as the absorbent. The inlet gas rate will be 0.2 m^3/s and the inlet water rate will be 203 kg/s. Under these conditions, the overall capacity

Question

Ammonia is to be absorbed from air at 293 K and [latex]1.013 imes 10^5 \mathrm{~Pa}[/latex] pressure in a countercurrent, packed tower, 0.5 m in diameter, using ammonia-free water as the absorbent. The inlet gas rate will be 0.2 m³/s and the inlet water rate will be 203 kg/s. Under these conditions, the overall capacity coefficient, [latex]K_Y[/latex], may be assumed to be [latex]80 \frac{ ext { mole }}{\mathrm{m}^3 \cdot \mathrm{s} \cdot \Delta Y_A}[/latex]. The ammonia concentration will be reduced from 0.0825 to 0.003 mol fraction. The tower will be cooled, the operation thus taking place essentially at 293 K; the equilibrium data of example 3 may be used. Determine the length of the mass exchanger.

Accepted Answer

The concentrations of three of the streams were given; these may be expressed on a [latex]\mathrm{NH}_3 ext {-free }[/latex] basis by

[latex]Y_{\mathrm{NH}_3, 1}=\frac{y_{\mathrm{NH}_3, 1}}{1-y_{\mathrm{NH}_3, 1}}=\frac{0.0825}{0.9175}=0.09[/latex]

[latex]Y_{\mathrm{NH}_3, 2}=\frac{y_{\mathrm{NH}_3, 2}}{1-y_{\mathrm{NH}_3, 2}}=\frac{0.003}{0.997}=0.003[/latex]

and

[latex]X_{\mathrm{NH}_3, 2}=0.0[/latex]

The cross-sectional area of the tower is equal to [latex]\frac{\pi D^2}{4}=\frac{\pi(0.5 \mathrm{~m})^2}{4}=0.196 \mathrm{~m}^2[/latex]. The gas enters the exchanger at plane 1, and its flow

[latex]G=\frac{V P}{R T} \cdot \frac{1}{\mathrm{~A}}=\frac{\left(0.20 \mathrm{~m}^3 / \mathrm{s} ight)\left(1.013 imes 10^5 \mathrm{~Pa} ight)}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^3}{\mathrm{~mol} \cdot \mathrm{K}} ight)(293 \mathrm{~K})} \cdot \frac{1}{\left(0.196 \mathrm{~m}^2 ight)}=42.4 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}[/latex]

The [latex]\mathrm{NH}_3 ext {-free }[/latex] gas flow rate is

[latex]G_s=G_1\left(1-y_{\mathrm{NH}_3, 1} ight)=42.4 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}(0.9175)=38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}[/latex]

The [latex]\mathrm{NH}_3 ext {-free }[/latex] water flow rate is evaluated using the flow into the exchanger at plane 2.

[latex]L_s=(203 \mathrm{~g} / \mathrm{s})\left(\frac{\mathrm{mol}}{18 \mathrm{~g}} ight)\left(\frac{1}{0.196 \mathrm{~m}^2} ight)=57.5 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}[/latex]

The liquid leaves the exchanger at plane 1; its concentration is evaluated, using the countercurrent material balance equation

[latex]G_s\left(Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2} ight)=L_s\left(X_{\mathrm{NH}_3, 1}-X_{\mathrm{NH}_3, 2} ight)[/latex]

[latex]38.9(0.09-0.003)=57.5\left(X_{\mathrm{NH}_3, 1}-0 ight)[/latex]

[latex]X_{\mathrm{NH}_3, 1}=0.059[/latex]

The operating and equilibrium lines are shown in Figure 31.22.

As this is not a case in which the gas and liquid concentrations are dilute enough for us to assume straight equilibrium and operating lines on the plot of y vs. x, the height of the tower must be evaluated by the graphical integration procedure, using

[latex]z=\frac{G_S}{K_Y a} \int_{Y_{A_2}}^{Y_{A_1}} \frac{d Y_A}{Y_A-Y_A^*}[/latex]

[latex]z=\frac{38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}}{80 \frac{\mathrm{mol}}{\mathrm{m}^3 \cdot \mathrm{s} \cdot \Delta Y_A}} \int_{Y_{\mathrm{NH}_3, 2}}^{Y_{\mathrm{NH}_3, 1}} \frac{d Y_{\mathrm{NH}_3}}{Y_{\mathrm{NH}_3}-Y_{\mathrm{NH}_3}^*}[/latex]

In Table 31.1, [latex]Y_{\mathrm{NH}_3}[/latex] is the composition at a point on the operating line, and [latex]Y_{\mathrm{NH}_3}^*[/latex] is the composition on the equilibrium line directly below the [latex]Y_{\mathrm{NH}_3}[/latex] value. An example of these compositions is illustrated in Figure 31.21. The integral

[latex]\int_{Y_{\mathrm{NH}_3, 2}}^{Y_{\mathrm{NH}_3, 1}} \frac{d Y_{\mathrm{NH}_3}}{Y_{\mathrm{NH}_3}-Y_{\mathrm{NH}_3}^*}[/latex]

is graphically evaluated in Figure 31.23.

Table 31.1 Example 3 gas compositions
[latex]Y_A[/latex]	[latex]Y_A^*[/latex]	[latex]Y_A-Y_A^*[/latex]	[latex]1 /\left(Y_A-Y_A^* ight)[/latex]
0.003	0	0.003	333.3
0.01	0.0056	0.0035	296
0.02	0.0153	0.0047	212.5
0.035	0.0275	0.0075	133.3
0.055	0.0425	0.0125	80.0
0.065	0.0503	0.0147	68.0
0.075	0.058	0.017	58.9
0.09	0.0683	0.0217	47.6

The area under the curve is the numerical value of the integral. The resulting length of the mass exchanger is

[latex]Z=\frac{38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}}{80 \frac{\mathrm{mol}}{\mathrm{m}^3 \cdot \mathrm{s} \cdot \Delta Y_A}}(10.95)=5.32 \mathrm{~m}[/latex]

The integral [latex]\int_{Y_{\mathrm{NH}_3, 2}}^{Y_{\mathrm{NH}_3, 1}} \frac{d Y_{\mathrm{NH}_3}}{Y_{\mathrm{NH}_3}-Y_{\mathrm{NH}_3}^*}[/latex] can also be evaluated numerically using either a spreadsheet or a mathematical software program.With the values for [latex]G_s, L_s, y_{\mathrm{NH}_3 1} ext { and } X_{\mathrm{NH}_3 1}[/latex] already determined, the operating line equation becomes

[latex]G_S\left(Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, z} ight)=L_S\left(X_{\mathrm{NH}_3, 1}-X_{\mathrm{NH}_3, z} ight)[/latex]

[latex]38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}\left(0.09-Y_{\mathrm{NH}_3, z} ight)=57.7 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}\left(0.059-X_{\mathrm{NH}_3, z} ight)[/latex]

which upon rearrangement yields

[latex]X_{\mathrm{NH}_3, z}=0.6765 \quad Y_{\mathrm{NH}_3, z}-0.0020[/latex]

The equilibrium data for example 3, when fitted to a polynomial of the second order, can be represented by

[latex]Y_{\mathrm{NH}_3}^*=-2.6903 X_{\mathrm{NH}_3}^2+1.2953 X_{\mathrm{NH}_3}+0.0003[/latex]

These two equations are used to develop the following table:

[latex]Y_{\mathrm{NH}_3}[/latex]	[latex]X_{\mathrm{NH}_3}[/latex]	[latex]Y_{\mathrm{NH}_3}^*[/latex]	[latex]Y-Y^*[/latex]	[latex]\frac{1}{Y-Y^*}[/latex]	[latex]\left(\frac{1}{Y-Y^*} ight)_{\mathrm{avg}}[/latex]	[latex]\Delta Y_{\mathrm{NH}_3}[/latex]	[latex]\left(\frac{1}{Y-Y^*} ight)_{\mathrm{avg}} \Delta Y_{\mathrm{NH}_3}[/latex]
0.003	0	0	0.003	333.3
					305.95	0.007	2.14
0.01	0.00476	0.06411	0.003589	278.6
					236.95	0.01	2.37
0.02	0.01153	0.01488	0.00512	195.3
					170.15	0.01	1.70
0.03	0.01830	0.02310	0.006897	145.0
					128.5	0.01	1.28
0.04	0.02506	0.03107	0.008929	112.0
					100.6	0.01	1.01
0.05	0.03182	0.03879	0.01121	89.2
					81.03	0.01	0.81
0.06	0.03859	0.04628	0.01372	72.9
					66.78	0.01	0.67
0.07	0.04536	0.05351	0.01648	60.66
					55.98	0.01	0.56
0.08	0.05212	0.06050	0.01950	51.29
					47.62	0.01	0.48
0.09	0.05889	0.06725	0.02275	43.96			Total area = 11.02

The integral [latex]\int_{Y_{\mathrm{NH}_3, 2}}^{Y_{\mathrm{NH}_3,}{ }^1} \frac{d Y_{\mathrm{NH}_3}}{Y_{\mathrm{NH}_3}-Y_{\mathrm{NH}_3}^*}[/latex] equals 11.02 and the height of the mass exchanger is

[latex]Z=\frac{G_S}{K_Y a} \int_{Y_{\mathrm{NH}_3, 2}}^{Y_{\mathrm{NH}_3, 1}} \frac{d Y_{\mathrm{NH}_3}}{Y_{\mathrm{NH}_3}-Y_{\mathrm{NH}_3}^*}[/latex]

[latex]Z=\frac{38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}}{80 \frac{\mathrm{mol}}{\mathrm{m}^3 \cdot \mathrm{s}}}(11.02)=5.36 \mathrm{~m}[/latex]

Question 31.6: Ammonia is to be absorbed from air at 293 K and 1.013 × 10^5...

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