For any component, Air flowrate = K*(PPM,in-PPM,out)*10^-6*flowrate(gpm)*.13 SCFM
=>For 1,2 Dichloroethane, Airflowrate = 60*(85-.005)*10^-6*1500*.13 =0.99 SCFM
=>For Trichloroethylene, Air flowrate = 650*(120-.005)*10^-6*1500*.13 =15.2 SCFM
=> For 1,1,1 Trichloroethane, Air flowrate = 275*(145-.2)*10^-6*1500*.13 =7.76 SCFM
Therefore, Air flowrate required is maximum of all three = 15.2 SCFM
=> Concentrations of 1,2 Dichloroethane in Cleaned water = 0 ppm
=>Concentrations of Trichloroethylene in Cleaned water = .005 ppm
=>Concentrations of 1,1,1 Trichloroethane in Cleaned water = 0 ppm
#3 Local groundwater has been contaminated with three industrial solvents. We wish to treat 1500 gpm...
6.11. Stripping of VOCs from groundwater with air. Groundwater at a rate of 1,500 gpm, containing three volatile organic compounds (VOCs), is to be stripped in a trayed tower witlh air to produce drinking water that will meet EPA standards. Rele- vant data are given below. Determine the minimum air flow rate in scfm (60°F, 1 atm) and the number of equilibrium stages required if an air flow rate of twice the minimum is used and the tower operates at...