Question 5 in the afternoon session of the 2016 Chemistry and Engineering case study
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Water is used to absorb ammonia from an air-ammonia mixture; when the total pressure of the mixture is 101 kPa, the partial pressure of ammonia is 1300 Pa. It is known that this system obeys Henry’s law. And there is the following equilibrium relationship: when 2 g of ammonia is present in 100 g of water, the partial pressure of ammonia is 1600 Pa. What is the minimum water consumption (kg/h) required to handle 1000 kg of mixed gas per hour, with an ammonia recovery rate of 95%? (A) 236 (B) 262 (C) 445 (D) 488. Find the process.
The partial pressure of ammonia is proportional to the molar fraction of ammonia in the solution. By determining the molar fraction of ammonia at 1300 Pa and calculating the amount of ammonia recovered (converted to kmol), the total volume of the solution can also be determined. For many questions, I know how to approach them, but there are always problems with the calculation process; I have to redo the calculations twice or three times. I hope to be in better form during the exam and not make so many calculation mistakes.
Lmin=G*m*η. m = E/P = 1.6 / 101 = 0.7639; M_avg = 17 * 1.3 / 101 + 29 * (1 – 1.3 / 101) = 28.85 kg/kmol; G = 1000 / 28.85 * (1 – 1.3 / 101) = 34.22 kmol/h; L_min = 34.22 * 0.7639 * 0.95 = 24.83 kmol/h = 446.95 kg/h