[thrice_a]

Hi all. For a fuel, such as methanol or methane, how would one calculate the energy output and carbon intensity with varying excess air and combustion efficiency? for example, a combustion efficiency of 80% and excess air of 5% for methanol (CH3OH) will result in what energy output and what carbon intensity? Thanks in advance, andy.

[Gerard]

Hi all. For a fuel, such as methanol or methane, how would one calculate the energy output and carbon intensity with varying excess air and combustion efficiency? for example, a combustion efficiency of 80% and excess air of 5% for methanol (CH3OH) will result in what energy output and what carbon intensity? Thanks in advance, andy.

it really kinda depends if the combustion is complete or not
if complete combustion then it will result in carbion dioxide and water if the combustion is incomplete, the result is tha same but with the addition of carbon monoxide...
i usually use stoichiometry with the help of dulong's and calder-wood equation...
ill try to review my combustion chemistry....

[Gerard]

Here is the notes i promised you:
THEORITICAL WEIGHT OF AIR:
if the Ultimate analysis is available:
Wta=11.5C+64.5(H-(O/8))+4.3 S in kg air per kg fuel
if ultimate analysis is not available:
Wta=Qh(higher heating value of the fuel)/3117 in kg air per kilogram fuel

ACTUAL WEIGHT OF AIR

[Gerard]

Here is the notes i promised you:
THEORITICAL WEIGHT OF AIR:
if the Ultimate analysis is available:
Wta=11.5C+64.5(H-(O/8))+4.3 S in kg air per kg fuel
if ultimate analysis is not available:
Wta=Qh(higher heating value of the fuel)/3117 in kg air per kilogram fuel

ACTUAL WEIGHT OF AIR:
Waa=(1/32.72)(CxN2/(CO2+CO))kg air per kg fuel
Waa=Wta(1+%excess air) kg air per kg fuel

Analysis of Flue Gas
CO2+CO+02+N2=100%

WEIGHT OF FLUE GAS
wFG=Wof air+W of fuel
Wieght of dry flue gas:

Wdg=Waa=1-(so2+H2O+ashloss)
where so2=(2*%S)/100
h20=(9*H)/100
ashloss=%ash/100

Combustion of hydrocarbon
fuel+air=products of combustion

CnHm+O2+3.76aN2=bCO2+cH2O+3.76aN2
where a,b and c is the number of moles in the balanced equation
EXAMPLE

COMPOSITION OF AIR:
BY WEIGHT
02(23%)+N2(77%)=100%

BY VOLUME
O2(21%)+N2(79%)=100%

MOLAL RATIO
MOLES N2/MOLES O2=79/21=3.76

[Gerard]

THEORITICAL WEIGHT OF AIR:
if the Ultimate analysis is available:
Wta=11.5C+64.5(H-(O/8))+4.3 S in kg air per kg fuel
if ultimate analysis is not available:
Wta=Qh(higher heating value of the fuel)/3117 in kg air per kilogram fuel

COMBUSTION OF HYDROCARBON
Cn+Hm+O2+3.76aN2=bCO2+cH2O+3.76aN2
example:
octane is being used as fuel in ideal proportion

[technologist]

I can remember that Van Ness Smith - Thermodynamics book is having one similar example for you. You need to put in only combustion efficiency factor into that.

Rest can be easily calculated based on material balance.

[Gerard]

i have that book even staneky sandler's book