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Stoich combustion equation
2

Stoich combustion equation

Stoich combustion equation

(OP)
I am looking at a stoich heat equation and can't figure out an example.  It's basic chem/thermo so I was wondering if somebody could help.  I attached the equation and surrounding text to this post as a pdf.  Basically I can't figure out why they are multiplying "v" by 4.762 in the denominator of equation (1.7)  3.762 is the mole ratio of N2 to O2, but what is 4.762? and why woudl you put it in the denominator like that.
Thanks,
Adam

RE: Stoich combustion equation


fraction fuel in feed = (fuel in feed)/(total moles in feed)

    = (fuel)/ [(fuel) + (O2) + (N2)]

    = (1)/ [ (1) + (1*O2) + (3.762*O2)]

    = (1)/ [ (1) + O2[1+3.762]

    = (1)/ [ (1) + O2[4.762]     ------> equation (1.7)

Equation (1.7) is mirrored by the left hand side of equations in Table 1.3.
 

RE: Stoich combustion equation


Simply the molar (ie, volume) ratio of air-to-oxygen is 4.762 (=1/0.21) representing 21% vol of oxygen in air.

RE: Stoich combustion equation

(OP)
Thanks guys.  Very helpful
Adam

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