2018/19 Question Number (2) Workthrough\ a)\ b) Using

C_(3)H_(8)+SO_(2)->3CO_(2)+4H_(2)O

(Assume 100 mol of flue gas)

20kg

?\ Thus, % excess of

O_(2)

air is:

((300)/(103)-(2270)/(1133))/((2270)/(1133))\\\\times 100%~~45.37%

(too big!) Alternatively:

((25)/(11)-(2270)/(1133))/((220)/(1133))\\\\times 100%~~( Actual ~~28.1%)/(13.44% (too small!) )

2018/19 Question Number (2) Workthrough Using C3H8+5SO23CO2+4H2O (Assume 100 mol of flue gas) Theoretical H2moles=0.642 =1.28kmal Use 1115 kmol of C plus 11331450km of H2 2 Requined O2 for complete combution =1115+1133725=11332270kmol of O2 2.00353 T excess of O2 air is: 1133227010330011332270100%45.37% (toobig!) 11332270Actual28.1% Alsernatively: 11332270112511332270100%13.44% (too small!)