Air stripping is a process used to remove volatile compounds from the liquid pha

Question

Air stripping is a process used to remove volatile compounds from the liquid phase. A wastewater is contaminated with toluene at 50 ppm (in liquid, remember ppm units are a function of phase). If air stripping is used to remove the toluene, and we design our system to reduce toluene to 1 ppm (liquid), what is its concentration in the air that leaves the system? The toluene in the air at the exit will be in equilibrium with the toluene concentration in the liquid entering the system. What is the concentration of toluene in the effluent gas in mg/L (assume effluent gas at 1 atm and 25 degree C)? What volume of air/d (in m^3/d) is required? Use the USEPA website to determine the toluene Henry constant at 25 degree C with results from the OSWER method.

Explanation / Answer

Air Stripping - Process that helps to eliminate the volatile compounds from liquid phase

Total Toluene - 50 ppm

Toluene after air stripping reduced to ---- 1 ppm

Concentration of Toluene in the air that leaves the system???

Toluene Concentration in the air at exit is in equilibrium to the liquid entering the system

Concentration of Toluene in effluent gas in mg/L???

Effluent Gas ---> pressure, P = 1 atm and T = 25oC

Solution:

The total concentration of toluene in the air that leaves the system = 50 ppm or 50 mg/L (because 1 ppm = 1 mg/L)

The reason is since the concentration of toluene in the air at exit remains the same to that when it enters the system.

Now, the concentration in the effluent gas = Mass of solute / volume of solution

According to Ideal Gas Law, pV = nRT

Value of P = 1 atm, V = ?, n = ?, R = 0.0821 atm L / mol K, T = 25oC (or 298.15 K)

Molar Mass of toluene = 92.14 g/mol

Therefore,

1 atm x V = 1 mol x 0.0821 atm L/mol K x 298.15 K

V = 0.0821 L x 298.15 = 24.478 L

Therefore, concentration of toluene, C in effluent gas = n/V

= 1000/ 24.478 =400.8 mg/L

Water Flow Rate = 0.5 MGD or 2273.045 m3/d

Thus, Volume of air/d = ?

Flow Rate, Q = Volume, V/ Time, t

2273.045 = V/ t

Thus, V = 2273.045 m3/d

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