Question

100 tons of MSW is combusted per day at a W.T.E. facility.

Assume the MSW is 100% dry volatile solids with the following molar ratio: C₉H₂O₄

Calculate the total oxygen demand of this 100 tons of waste in whole standard cubic feet per minute.

Answer 1

Solution :

• 100 tons of MSW is combusted per day
• mass rate of MSW = 100000 kg/day = 100000 kg/(24 * 60 minutes) = 69.444 kg/min
• molecular weight of MSW = 9*12 + 2*1 + 4*16 = 174 gm/mol

mol rate of MSW = (mass rate of MSW)/(molecular weight of MSW)

= 69444 / 174

= 399.1 mol/min   

Combustion reacyion of C9H2O4 is

C9H2O4 + 7.5 O2 = 9 CO2 + H2O

from the reaction

for 1 mol of MSW O2 required =  7.5 mol

for 399.1 mol of MSW O2 required =  7.5 * 399.1 /1 = 2993.25 mol/min

O2 required mole rate = 2993.25 mol/min

air required mass rate = 2993.25/0.21 = 14253.57 mol/min

air required mass rate = 14253.57 * 29 gm/min = 413353.57 gm/min

density of air = 1.225 kg/m^3

air required volumetric rate =  413353.57 /1225 = 337.43 m^3/min

O2 required volumetric rate = 0.21*337.43 = 70.86 m^3/min

O2 required volumetric rate = 70.86 *3.28^3 ft^3/min

O2 required volumetric rate = 2500.5 ft^3/min