10.35 Coal having the following analysis on a dry basis and containing 3.9% mois
ID: 702154 • Letter: 1
Question
10.35 Coal having the following analysis on a dry basis and containing 3.9% moisture on a dry basis is burned withe C = 83.05%, H = 4.45%, O = 3.3696, N = 1.0896, S = 0.70% and ash = 7.36%. Air used contains 0.0048 kg of water vapour per kg of dry air. The Orsat analysis of the combustion gases gave the following result: CO2 + SO2-15.4%, CO-O·0,02-4.0% andN2-80.6%. The refuse analyzed 14.096 unburned ash-free coal and the rest ash. Check the consistency of the given data and calculate the percent excess air used. ANSWER: 19.9%Explanation / Answer
Let there be 103.9 atoms initial of all Coal elements
=> Total Molecules of water in coal = 3.9% = 3.9
=>Total atoms in dry basis = 100
=>Total atoms of C = 83.05
=>Total atoms of H=4.45
=>Total atoms of O = 3.36
=>Total atoms of N=1.08
=>Total atoms of S=0.7
=>Total atoms of ash = 7.36
Ash in coal = Ash in refuse
=>Number of atoms in refuse =7.36*100/(100-14)=8.55
=> Atoms of C in refuse = 14/100*8.55=1.197
=>Total C Combusted = 83.05-1.197=81.85
Since no S in refuse, Total S combusted = 0.7
=> Total number of CO2 +SO2 form = 81.85+0.7=82.55
% CO2+SO2 in combustion gas = 15.4%
=> Total number of molecules in combustion gas = 82.55*100/15.4=536.03
% O2 in combustion gas = 4
=>Total Molecules of O2 in combustion gas = 4/100*536.03=21.44
% N2 in combustion gas = 80.6
=>Total Molecules of N2 in combustion gas = 80.6/100*536.03=432.04
Total Moles of N2 in combustion gas from air = Total Molecules of N2 in combustion gas -0.5*Total atoms of N2 in Coal
=>Total Moles of N2 in combustion gas from air=432.04-0.5*1.08=431.5
Air is 21% O2 and 79% N2
=>Total O2 in feed air = 431.5/79*21=114.7
Total required O2 in feed air = Total number of CO2 +SO2 form-0.5* oxygen atom in Coal+0.25* Hydrgen atom in Coal=82.55-0.5*3.36+.25*4.45=81.98
=> %excess air=% excess O2 = (114.7-81.98)/114.7
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