Seeking help with a drying problem
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I’m very grateful for this section; I plan to take the CPA exam this year, and I’ve benefited a lot from it. There is a problem as follows: A steam mixture of CO2 and H2O flows at a rate of 2500 kg/h through a vessel, with a water vapor content of 60% by mass. At an operating pressure of p = 3 kgf/cm2 (absolute pressure), the temperature of this mixture is reduced to 80°C. What is the amount of water that condenses, in kg/h? (The vapor pressure of water at 80°C is 0.483 kgf/cm2.) It seems not too difficult; the answer is 1421.5 kg/h, but I’m not quite sure how to arrive at that result Can’t do it either
Under the operating conditions, the ratio of the partial pressure of water vapor to the total pressure is equal to the ratio of the moles of water vapor in the gas phase to the total number of moles in the gas phase. Therefore, initially: n_water = 2500*0.6/18 = 83.3, n_CO2 = 2500*0.4/44 = 22.7. After condensation, 0.483/3 = n_water_in_gas_phase / total_moles_in_gas_phase = (83.3 – n_condensed) / (83.3 + 22.7 – n_condensed), which gives n_condensed = 78.94 kmol/h = 1421 kg/h
Reply to 2# kaminocmyhc: This question was nicely posed; I was only thinking about the saturated vapor pressure of water at 3 kgf/cm2 and forgot about the lever principle