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To wet basis...
- Thread starter AndreChE
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Of course humidity can be measured by a variety of instruments.
Otherwise, if you wish to estimate it, assuming perfect and complete combustion with excess (xs) air, knowing the % CO2 on a dry basis you obtain y/x (i.e., the H/C atomic ratio) of your fuel in CxHy.
Since the volumes of CO2 and H2O aren't affected by xs air, and the volume of water from combustion is equal to (y/2x) times the vol of CO2, one can estimate the volume of combustion water from that ratio.
For example, assume one is burning a hydrocarbon fuel with 20% xs air (about 3.5% oxygen in the flue gases) and measures 11% CO2 on a dry basis, it would result in y/x=2.9 (a typical refinery gas). And y/2x=1.45, meaning water volume would be 1.45*11=15.95. Added to 100% dry gas results in 115.95 volumes of wet gas, and any % expressed on a dry basis should be divided by 1.1595 to get the % on a wet basis.
BTW, the xs air on a perfect and complete combustion of a hydrocarbon can be determined just by knowing the % O2 with little appreciable error for any hydrocarbon fuel, as follows:
1% oxygen ~ 5% xs air; 2%, ~10% xs air; 3.5%, ~20% xs air; 5%, ~30% xs air, etc.
Don't forget that any fuel oil atomizing steam would have to be accounted for. I sincerely hope to be understood.![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
Otherwise, if you wish to estimate it, assuming perfect and complete combustion with excess (xs) air, knowing the % CO2 on a dry basis you obtain y/x (i.e., the H/C atomic ratio) of your fuel in CxHy.
Since the volumes of CO2 and H2O aren't affected by xs air, and the volume of water from combustion is equal to (y/2x) times the vol of CO2, one can estimate the volume of combustion water from that ratio.
For example, assume one is burning a hydrocarbon fuel with 20% xs air (about 3.5% oxygen in the flue gases) and measures 11% CO2 on a dry basis, it would result in y/x=2.9 (a typical refinery gas). And y/2x=1.45, meaning water volume would be 1.45*11=15.95. Added to 100% dry gas results in 115.95 volumes of wet gas, and any % expressed on a dry basis should be divided by 1.1595 to get the % on a wet basis.
BTW, the xs air on a perfect and complete combustion of a hydrocarbon can be determined just by knowing the % O2 with little appreciable error for any hydrocarbon fuel, as follows:
1% oxygen ~ 5% xs air; 2%, ~10% xs air; 3.5%, ~20% xs air; 5%, ~30% xs air, etc.
Don't forget that any fuel oil atomizing steam would have to be accounted for. I sincerely hope to be understood.
To convert the xs air -where air components are given in the ratio N2/O2=3.76- and %CO2 content (dry basis), into y/x of your hydrocarbon fuel the equation is:
where
n, is the xs air coefficient, for 10% xs air, 1.1; for 20% xs air, 1.2, and so forth.
p = %CO2, dry basis.
The above formula is derived from the stoichiometric one giving all (wet) products of combustion of CxHy with xs air:
p = 100[÷][4.76n+(1.19n-0.25)(y/x)]
where
n, is the xs air coefficient, for 10% xs air, 1.1; for 20% xs air, 1.2, and so forth.
p = %CO2, dry basis.
The above formula is derived from the stoichiometric one giving all (wet) products of combustion of CxHy with xs air:
xCO2+(y/2)H2O+3.76n(x+y/4)N2+(n-1)(x+y/4)O2
AndreChE,
Summarizing, when the composition of the hydrocarbon fuel CxHy is not known, the (approximate) procedure is as follows:
1. From %O2 (dry basis), A, find the excess
air coefficient n.
2. Measure %CO2 (dry basis) = p.
3. From n and p determine y/x.
4. Find %H2O, H = p(y/2x).
5. Find %O2 (wet basis), B=A/(100+H).
All the % are volume or mol %.
I invite comments from experts in combustion engineering.
Summarizing, when the composition of the hydrocarbon fuel CxHy is not known, the (approximate) procedure is as follows:
1. From %O2 (dry basis), A, find the excess
air coefficient n.
2. Measure %CO2 (dry basis) = p.
3. From n and p determine y/x.
4. Find %H2O, H = p(y/2x).
5. Find %O2 (wet basis), B=A/(100+H).
All the % are volume or mol %.
I invite comments from experts in combustion engineering.
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