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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?
Show all the steps in the mechanism for the following reaction, When benzene is mixed with deuterated sulfuric acid, deuterium is slowly incorporated onto the ring. Show the mechanism for this reaction and explain how this relates the sulfonation of ..
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