Two thousand kilograms per minute of a slurry containing 10 percentage CaCO_3 ar
ID: 1048631 • Letter: T
Question
Two thousand kilograms per minute of a slurry containing 10 percentage CaCO_3 are to be filtered in a rotary vacuum filter. The filter cake from tie filter contains 60 t percentage water. Assume all CaCO_3 from the slurry is recovered in the filter cake. This cake is placed into a drier and dried to a moisture content of 9.09 kg H_2O/100 kg bone dry CaCO_3. The humidity of air entering the drier is 0.005 kg H_2O.1g dry air, and the humidity of the air leaving the drier is 0.015 kg H_2O./kg dry air. What is the quantity of water (kg/minute) removed by the filter? What is the flow rate of air on a water-free basis (kg dry air/minute) through the drier?Explanation / Answer
Basis : 2000 kg/min of slurry
It contains 10 wt% CaCO3. Mass of CaCO3= 2000*0.1= 200 kg/min
Rest is water = 2000*0.9= 1800 kg/min
All the cake is recovered in the filter cake, let y= quantity of filter cake/min
Since filter cake contains 40% CaCO3, y*0.4= 200, y= 200/0.4 =500 kg/min
Water in the filter cake = 60%, which is 500*0.6= 300 kg/min
Water removed = 1800-300 = 1500 kg/min
Water content need to be reduced to 9.09 kg/ 100 kg dry CaCO3
Since there are 300 kg dry CaCO3, moisture content = 9.09*3 = 27.27 kg
Water to be removed in the dryer= 300-27.27= 272.73 kg /min
Humidity of inlet air = 0.005 kg/ dry air
Out let humidity = 0.015 kg/ dry air
Change in humidity = (0.015-0.005) kg H2O/kg dry air =0.01kg water/ kg dry air
Water to be removed = 272.73 kg H2O/min
Amount of dry air/mi 272.73/0.01 kg dry air/min=27273 kg/min
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