Question 31.6: Ammonia is to be absorbed from air at 293 K and 1.013 × 10^5...
Ammonia is to be absorbed from air at 293 K and 1.013 \times 10^5 \mathrm{~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 m³/s and the inlet water rate will be 203 kg/s. Under these conditions, the overall capacity coefficient, K_Y, may be assumed to be 80 \frac{\text { mole }}{\mathrm{m}^3 \cdot \mathrm{s} \cdot \Delta Y_A}. 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.
The concentrations of three of the streams were given; these may be expressed on a \mathrm{NH}_3 \text {-free } basis by
Y_{\mathrm{NH}_3, 1}=\frac{y_{\mathrm{NH}_3, 1}}{1-y_{\mathrm{NH}_3, 1}}=\frac{0.0825}{0.9175}=0.09Y_{\mathrm{NH}_3, 2}=\frac{y_{\mathrm{NH}_3, 2}}{1-y_{\mathrm{NH}_3, 2}}=\frac{0.003}{0.997}=0.003
and
X_{\mathrm{NH}_3, 2}=0.0The cross-sectional area of the tower is equal to \frac{\pi D^2}{4}=\frac{\pi(0.5 \mathrm{~m})^2}{4}=0.196 \mathrm{~m}^2. The gas enters the exchanger at plane 1, and its flow
G=\frac{V P}{R T} \cdot \frac{1}{\mathrm{~A}}=\frac{\left(0.20 \mathrm{~m}^3 / \mathrm{s}\right)\left(1.013 \times 10^5 \mathrm{~Pa}\right)}{\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^3}{\mathrm{~mol} \cdot \mathrm{K}}\right)(293 \mathrm{~K})} \cdot \frac{1}{\left(0.196 \mathrm{~m}^2\right)}=42.4 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}The \mathrm{NH}_3 \text {-free } gas flow rate is
G_s=G_1\left(1-y_{\mathrm{NH}_3, 1}\right)=42.4 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}(0.9175)=38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}The \mathrm{NH}_3 \text {-free } water flow rate is evaluated using the flow into the exchanger at plane 2.
L_s=(203 \mathrm{~g} / \mathrm{s})\left(\frac{\mathrm{mol}}{18 \mathrm{~g}}\right)\left(\frac{1}{0.196 \mathrm{~m}^2}\right)=57.5 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}The liquid leaves the exchanger at plane 1; its concentration is evaluated, using the countercurrent material balance equation
G_s\left(Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2}\right)=L_s\left(X_{\mathrm{NH}_3, 1}-X_{\mathrm{NH}_3, 2}\right)38.9(0.09-0.003)=57.5\left(X_{\mathrm{NH}_3, 1}-0\right)
X_{\mathrm{NH}_3, 1}=0.059
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
z=\frac{G_S}{K_Y a} \int_{Y_{A_2}}^{Y_{A_1}} \frac{d Y_A}{Y_A-Y_A^*}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}^*}
In Table 31.1, Y_{\mathrm{NH}_3} is the composition at a point on the operating line, and Y_{\mathrm{NH}_3}^* is the composition on the equilibrium line directly below the Y_{\mathrm{NH}_3} value. An example of these compositions is illustrated in Figure 31.21. The integral
\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}^*}is graphically evaluated in Figure 31.23.
| Table 31.1 Example 3 gas compositions | |||
| Y_A | Y_A^* | Y_A-Y_A^* | 1 /\left(Y_A-Y_A^*\right) |
| 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
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}The integral \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}^*} can also be evaluated numerically using either a spreadsheet or a mathematical software program.With the values for G_s, L_s, y_{\mathrm{NH}_3 1} \text { and } X_{\mathrm{NH}_3 1} already determined, the operating line equation becomes
G_S\left(Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, z}\right)=L_S\left(X_{\mathrm{NH}_3, 1}-X_{\mathrm{NH}_3, z}\right)38.9 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}\left(0.09-Y_{\mathrm{NH}_3, z}\right)=57.7 \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}\left(0.059-X_{\mathrm{NH}_3, z}\right)
which upon rearrangement yields
X_{\mathrm{NH}_3, z}=0.6765 \quad Y_{\mathrm{NH}_3, z}-0.0020The equilibrium data for example 3, when fitted to a polynomial of the second order, can be represented by
Y_{\mathrm{NH}_3}^*=-2.6903 X_{\mathrm{NH}_3}^2+1.2953 X_{\mathrm{NH}_3}+0.0003These two equations are used to develop the following table:
| Y_{\mathrm{NH}_3} | X_{\mathrm{NH}_3} | Y_{\mathrm{NH}_3}^* | Y-Y^* | \frac{1}{Y-Y^*} | \left(\frac{1}{Y-Y^*}\right)_{\mathrm{avg}} | \Delta Y_{\mathrm{NH}_3} | \left(\frac{1}{Y-Y^*}\right)_{\mathrm{avg}} \Delta Y_{\mathrm{NH}_3} |
| 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 \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}^*} equals 11.02 and the height of the mass exchanger is
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}^*}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}
