11.07.2015 Views

ISSN: 2250-3005 - ijcer

ISSN: 2250-3005 - ijcer

ISSN: 2250-3005 - ijcer

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

International Journal Of Computational Engineering Research (<strong>ijcer</strong>online.com) Vol. 2 Issue. 8m air = (m fluegas + m ash + m mst ) – m fuel (10)m fuelm flue gasm air Combustion Chamber m mstm ashFig.1 Mass balance in the FurnaceStiochiometric air amount (n=1) can be calculated as follows;m air steo = O 2 required per kilogram of the fuel/23.3% of O 2 in air= m O,H K H – m O,O K O + m O,S K S + m O,C K C /0.233 (11)Where m O,H , m O,O , m O,S , m O,C , are the masses of oxygen in hydrogen,oxygen,sulphur and carbon respectively.328KH KO KS KC120.233m 34.3348K 4.2918K 4.2918K11. 4449Km air,steo =mair. Steo.HOSIssn <strong>2250</strong>-<strong>3005</strong>(online) November| 2012 Page 10C(12) (3K 0.3750K 0.3750K K )11.4449(13)air. steo.HOS CWith excess air ratio,m ( 3K0.3750K 0.3750K K )(11.4449)(1 )(14)airHOSCWhere K denotes the percentage ratio of the element in chemical composition (in %) and m air is the air requirement per kgfuel (kg air/kg fuel). Flue gas amount can be found by Eq. 9Substituting Eq.13 in Eq. 9, knowing that calculations are done for 1 kg fuel, so the equation can be expressed as follows:m 3K 0.3750K 0.3750K K )(11.4449) + (1-K ash -K mst ) (15)fluegas(HOS CEmploying the excess air ratio,mfluegas 3K0.3750K 0.3750K K )(11.4449)(1 ) (1 K K )(16)(HOS Cash mstUsing the elemental composition of waste as shown in figure 1, the calculation of amount of air required and the flue gasproduced can be done considering the above equations.Table 2 Percentage by mass of MSWElement C H O S N Moisture Ashpercentage 35.5 5.1 23.9 0.5 2.4 25 7.6(Source : P.Chattopadhyay, 2006)2.1 Calculation of Combustion air supplyConsidering theoretical combustion reaction for the elemental analysis of MSW showed in table 2.Carbon (C):C+O 2 →CO 212KgC+32KgO 2 →44KgCO 2Oxygen required = 0.355 * (32/12) = 0.947/Kg MSWCarbon dioxide produced = 0.355 * (44/12) = 1.302/Kg MSWHydrogen (H):H 2 + 1 2 O 2 → H 2 O2Kg H 2 + 16Kg O 2 → 18Kg H 2 O1Kg H 2 + 8Kg O 2 → 9Kg H 2 OOxygen required = 0.051 8 = 0.408 Kg/Kg MSWSteam produced = 0.051 9 = 0.459 Kg/Kg MSWSulphur (S):S + O 2 → SO 2

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!