Question 31.5: The ammonia–air feed stream described in example 3 is fed co...
The ammonia–air feed stream described in example 3 is fed cocurrently with an ammonia–free water stream. The ammonia concentration is to be reduced from 3.52 to 1.29 % by volume, using a water stream 1.37 times the minimum. Determine (a) the minimum L_s / G_s ratio, (b) the actual water rate, and (c) the concentration in the exiting aqueous stream.
In example 3, the following compositions were evaluated:
\text { entering } \quad Y_{\mathrm{NH}_3, 1}=0.0365\text { exiting } \quad Y_{\mathrm{NH}_3, 2}=0.0131
\text { entering } \quad X_{\mathrm{NH}_3, 1}=0.0
The moles of G on a solute-free basis were evaluated to be \frac{0.483}{A} \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}. In Figure 31.17, the minimum and actual operating lines are shown. For these operating lines
\left(\frac{L_s}{G_s}\right)_{\min }=\frac{Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2}}{X_{\mathrm{NH}_3, 2}-X_{\mathrm{NH}_3, 2}}=\frac{0.0365-0.0131}{0.01-0}=2.34 \frac{\text { moles } \mathrm{NH}_3 \text {-free } L \text { phase }}{\text { moles } \mathrm{NH}_3 \text {-free } G \text { phase }}
and
\left(\frac{L_S}{G_S}\right)_{\text {actual }}=1.37\left(\frac{L_S}{G_S}\right)_{\min }=1.37(2.34)=3.21 \frac{\text { moles } \mathrm{NH}_3 \text {-free } L \text { phase }}{\text { moles } \mathrm{NH}_3 \text {-free } G \text { phase }}The composition of the exiting stream can be evaluated with the slope of the actual operating line by
\left(\frac{L_S}{G_S}\right)_{\text {actual }}=3.21=\frac{Y_{\mathrm{NH}_3, 1}-Y_{\mathrm{NH}_3, 2}}{X_{\mathrm{NH}_3, 2}-X_{\mathrm{NH}_3, 1}}=\frac{0.0365-0.0131}{X_{\mathrm{NH}_3, 2}-0}or
X_{\mathrm{NH}_3, 2}=\frac{0.0234}{3.21}=0.0073=\frac{\mathrm{mol} \mathrm{NH}_3}{\mathrm{~mol} \mathrm{NH}_3 \text {-free water }}The moles of \mathrm{NH}_3-free water fed to the tower, L_s, is also evaluated using the value of
\left(\frac{L_S}{G_S}\right)_{\text {actual }}=3.21 \frac{\mathrm{mol} \mathrm{NH}_3 \text { free } L \text { phase }}{\mathrm{mol} \mathrm{NH}_3 \text { free } G \text { phase }}Then
L_S=3.21 G_S=3.21\left(\frac{0.481}{A}\right)=\frac{1.55}{\mathrm{~A}} \frac{\mathrm{mol}}{\mathrm{m}^2 \cdot \mathrm{s}}