A hot-air dryer is used to reduce the moisture content of 1500 kg/min of wet wood pulp from 0.75 kg H20/hg dry pulp to 0.15 wt% H20. Air is drawn from the atmosphere at 28 deg. C, 760 mmHg, and 50% relative humidity, sent through a blower-heater, and then fed to the dryer. The air leaves the dryer at 80 deg. C and 10 mm Hg (gauge). A sample of the exit air is drawn into a chamber containing a mirror and cooled slowly, keeping the guage pressure at 10 mm Hg. A mist is observed to form on the mirror at a temp of 40 deg. C. Calculate the mass of water removed from the pulp (Kg/min) and the volumetric flow rate of air entering the system (m^3/min).

What I've done so far: Drew and labelled a flow chart. I found that with 50% humid. we can find y (fraction) of H20 in to be 0.0187 using tabluated data for p*(28C). I can also find the y2 (fraction) of H20 using raoults law. Where do I go from there?