(9.18) NH?. +1.86302.+ 0.098CO.. 0.0196C:H:N02+0.98No, 3 (9.19) +0.0941H20+1.98H
ID: 1732675 • Letter: #
Question
(9.18) NH?. +1.86302.+ 0.098CO.. 0.0196C:H:N02+0.98No, 3 (9.19) +0.0941H20+1.98H Problem 3. Nitrification: The influent into secondary treatment for a wastewater treatment plant has 40 mg of NH4'/ L. The nitrification process removes the NH4 by using lithotrophic bacteria to convert it to NO3. The flowrate is 100,000 L/day and the design specification is to remove all NH4-. a) Determine the minimum mass flowrate of oxygen (kg/day) and the associated volumetric flowrate of air (m3/day) if all of the NH4 is converted to NO3 via reaction 9.18 in the textbook. Assume that the air has the following properties: pressure of 1 atm, temperature of 20° C, and a molar oxygen content of 21% b) In reality, some of the N is incorporated into biomass, as shown in reaction 9.19 in the textbook. For this reaction, determine the rate of oxygen (kg/day) and inorganic carbon (kg/day) consumption.Explanation / Answer
Influent Ammonium ion concentration = 40 mg/l
Flow rate of influent = 100,000 l/d
Thus flow rate of ammonium ion = 40*100000= 4000000 mg/d
Molecular weight of ammonium ion = 18 g
Thus moles of ammonium ion in equation 9.18 is = 4000000/18000= 222.223 moles
Thus moles of oxygen required = 2*222.223= 444.446
Thus mass flow rate of oxygen = 444.446*32 g/d = 14222.272 g/d = 14.22 kg/d ( roundoff to two decimal)
Molar oxygen content in air = 21%
Thus moles of air = 444.446*100/21 = 2116.41 moles
Applying gas equation PV=nRT
T= 20°= 293.15 K , P= 1atm , R= 0.082057 L-atm/mol-K ,n= 2116.41
Thus V = 2116.41*0.082057*293.15= 50910.2628 L
Thus volume flow rate of air = 50.91 m3/d
Part (b)
Actual oxygen consumption per day = 1.863*222.223moles = 1.863*222.223*32/1000=13.248 kg/d
Amount of inorganic carbon = 0.098*222.223 moles = 0.098*222.223*12/1000= 0.261 kg/d
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