Question

In: Chemistry

The conversion of solid wastes to innocuous gases can be accomplished in incinerators in an environmentally...

The conversion of solid wastes to innocuous gases can be accomplished in incinerators in an environmentally acceptable fashion. However, the hot exhaust gases often must be cooled or diluted with air. An economic feasibility study indicates that solid municipal waste can be burned to a gas of 9.2% CO2, 1.5% CO, 6.5%O2, 78% N2 and rest water on a molar basis. You can neglect any energy effects resulting from the mixing of gaseous components. What is the enthalpy difference if 300 m^3/h of the stack at 290C is cooled to 30C?

Solutions

Expert Solution

Basis : 1 mole of mixture

Moles : CO2= 0.092, CO= 0.015, O2= 0.065 , N2= 0.78 and H2O= 1-0.952=0.048

Mass = Moles* molecular weights , Molecular weights : CO2= 44, CO= 28, N2= 28 and H2O= 18

Mass( gms) : CO2= 0.092*44=4.048, CO= 0.015*28=0.42, O2= 0.065*32=2.08gm N2= 0.78*28=21.84 and H2O= 0.048*18=0.864

Total mass= 4.048+0.42+2.08+21.84+0.864=29.252 gms

1 Kgmole of any gas at STP occupies 22.4 m3

hence moeles of mixture= 300/22.4 kMol/hr=13.39 kmoles/hr= 13.39*1000 gmoles/hr=13390 gmoles/hr

Enthalpy change ( without phase change)= moles*specific heat* temperature difference

This is valid for CO2, CO, O2 and N2

hence for these gases , total enthalpy change = sum of enthalpy changes of all these gases

Specific heat ( average) CO2= 7.017 Kcal/kgmole.Deg.c, CO= 11.92 Kcal/kg.mole.Deg.C, O2= 7.041 Kcal/kg.mole. deg.c and N2= 7.507 Kcal/kg.mole.deg.c

=(290-25)*(13.39*0.092*7.017+13.39*0.015*11.92+13.39*0.065*7.041+13.39*0.78*7.057)=24081 Kcal/hr

when cooled to 100 deg.c, water starts condensing and hence for water the following enthalpy changes are to be consindered

1. Sensible heat of vapor ( from 290 to 100 deg.c)= 13.39*0.048*18 kgs*0.48Kcal/kg.deg.c *(290-100)=1055 Kcal/hr

2. Latent heat at 100 deg.c = 13.39*0.048*544KCal/Kg.mole=350 Kcal/hr

3. Sensible heat from 100 deg.c to 30 deg.c= 13.39*0.048*18*1*(100-30) Kcal/hr=810 Kcal/hr

Total enthalpy change of water= 1+2+3= 1055+350+810=2215 Kcal/hr

total enthalpy change of exhaust = 24081+2215 "Kcal/hr=26296 Kcal/hr


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