Q3. () Briefly discuss the species mole fraction, temperature profile and specie
ID: 702206 • Letter: Q
Question
Q3. () Briefly discuss the species mole fraction, temperature profile and species molar production rate profiles of a stoichiometric CH4-air premixed flame as shown in Fig. Q3(a) 16 12 10 CHx 10 x (mm) 20 25 3.0 aco 1500 1000 500 0.5 1.0 15 2.0 25 3.0 Axial distance (mm) Fig. Q3(a) (6 marks) (b) Determine the size distribution, the cumulative number and volume fraction distributions, the most probable diameter, the mean diameter, the volume mean diameter and the Sauter mean diameter. The drop size was obtained experimentally as shown in Table Q3(a) Size bin (microns 0-10 11-30 31-50 51-80 81-100 Count 450 320 220 150 100 Table Q3(a) (10 marks)Explanation / Answer
a) Initially CH4 will be incompletely combusted into CO + H2O
Reaction is CH4+ 1.5O2=> CO+2H2O
After that, further CO getting reacted with Oxygen to completely combusted into CO2 by reaction reversibily
CO + 0.5 O2<=> CO2
Therefore, CH4 concentration will diminish as it gets completely reacted instantaneously. H2O mole fraction will increase.
At a down distance, CO will be converted into CO2, Therefore CO concentration will be reduced and CO2 concentration increase. After reaching equilibrium, Their mole fraction started reaching stability.
For reaction rate, At a point, CH4 reacts completely and there it has maximum reaction rate. At that point, CO and H2O production rate will be highest. After that, CH4 is not available to react and hence CH4 consumption rate is zero.H2O is not taking part in any reaction neither is forming in secondary reaction, Therefore, its reaction rate is zero further. For, CO it will continuously react to form CO2 and reach equilibrium. Hence, CO consumption rate will increase and similarly CO2 production rate will increase. After reaching equilibrium, Both CO stops converting and hence reaction rate decays to zero.
b) Cumulative number = summation of counts = 450+320+220+150+100=1240
Most Probable diameter = ( (Average size bin)*(Count))/Cumulative number
e.g. Average size bin of 1st row = (0+10)/2
=> Most probable diameter = (5*450+20*320+40*220+65*150+90*100)/1240=29.19 microns
Mean diameter=(( (Average size bin)^2*(Count))/Cumulative number )^.5 = ((5^2*450+20^2*320+40^2*220+65^2*150+90^2*100)/1240)^.5=39.5 microns
Volume mean diameter=(( (Average size bin)^3*(Count))/Cumulative number )^.33 = ((5^3*450+20^3*320+40^3*220+65^3*150+90^3*100)/1240)^.33=45.46 microns
Sauter mean diameter=(( (Average size bin)^3*(Count))/Cumulative number )/(( (Average size bin)^2*(Count))/Cumulative number ) = (5^2*450+20^2*320+40^2*220+65^2*150+90^2*100)/(5^3*450+20^3*320+40^3*220+65^3*150+90^3*100)=.014
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