(Outcome K) Toluene has contaminated a city\'s drinking water supply. A reactor
ID: 806443 • Letter: #
Question
(Outcome K) Toluene has contaminated a city's drinking water supply. A reactor is constructed to remove the toluene from the water using powdered activated carbon (PAC). If the influent toluene concentration is 1000 ppb and the treatment objective is 5 ppb. what rate of addition of PAC is needed (in g/min) to treat a flow of 378 L/min? The Freundlich parameter, Kf for the PAC that you are using is Kt = 26.1 (mg/g)(L/mg)^0.44. Hint: Draw a diagram for this system and look at the loads of contaminant going into and out of the system.Explanation / Answer
x/m = KC1/n
K = 26.1 and 1/n = 0.44 (from the units given for K)
one min amout of contaminated water = 378 L
In - Out gives amout of toluene to be adsorbed. (378*995/109) = 376110 x 10-9 L. Using the density of tolune (0.867 g/cc), this is converted into weight which is 0.326 g or 326 mg.
We can assume the equlibrium concentration toluene in water is zero as they are immisscible.
therefore, x/m = 26.1 and x = 326 mg. Therefore, m, the mass of PAC = 326/26.1 = 12.5 g
PAC addition rate is to be 12.5 g/min.
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