Feed gas is at (1.0 mathrm{~atm}) and (30.0^{circ} mathrm{C}) and is (90.0 mathrm{~mol} %) air and (10.0
Question:
Feed gas is at \(1.0 \mathrm{~atm}\) and \(30.0^{\circ} \mathrm{C}\) and is \(90.0 \mathrm{~mol} \%\) air and \(10.0 \mathrm{~mol} \%\) ammonia. Flow rate is \(200.0 \mathrm{kmol} / \mathrm{h}\). Ammonia is absorbed at \(1.0 \mathrm{~atm}\) using water at \(25.0^{\circ} \mathrm{C}\) as solvent. We desire outlet ammonia in exiting air that is 0.32 \(\mathrm{mol} \%\) or less. Column is adiabatic.
a. If \(\mathrm{N}=4\), what \(\mathrm{L}\) is required \(( \pm 10.0 \mathrm{kmol} / \mathrm{h})\) ?
b. If \(\mathrm{N}=8\), what \(\mathrm{L}\) is required \(( \pm 10.0 \mathrm{kmol} / \mathrm{h})\) ?
c. If \(\mathrm{N}=16\), what \(\mathrm{L}\) is required \(( \pm 10.0 \mathrm{kmol} / \mathrm{h})\) ?
d. Examine your answers. Why does \(\mathrm{N}=16\) not decrease L more?
Step by Step Answer:
Separation Process Engineering Includes Mass Transfer Analysis
ISBN: 9780137468041
5th Edition
Authors: Phillip Wankat