MMSCFD to CO2 equivalent to MW

Does anyone know how to convert MMSCFD to CO2 to MW?

[Firia]

How to convert MMSCFD to CO2equivalent to MW?

for example, if there is 0,6 MMSCFD gas, how to convert it to CO2equivalent? and how much MW can we get from the CO2equivalent that we have

[JohnS]

Originally Posted by Firia

How to convert MMSCFD to CO2equivalent to MW?

for example, if there is 0,6 MMSCFD gas, how to convert it to CO2equivalent? and how much MW can we get from the CO2equivalent that we have

depends on the composition of the gas.

Thx .

Ok, let's say it is a flare gas, and the composition of the gas is like this,

component mol percent

CO2 8.22
N2 0.02
METHANE 81.31
ETHANE 5.36
PROPANE 3.04
ISO-BUTANE 0.61
BUTANE 0.73
ISO-PENTANE 0.20
n-PENTANE 0.13
HEXANE 0.13
HEPTANE PLUS 0.25
H2S 0.00

and the amount of the gas is 0,6 MMSCFD.
can you help me?

[JohnS]

Originally Posted by Unregistered

Thx .

Ok, let's say it is a flare gas, and the composition of the gas is like this,

component mol percent

CO2 8.22
N2 0.02
METHANE 81.31
ETHANE 5.36
PROPANE 3.04
ISO-BUTANE 0.61
BUTANE 0.73
ISO-PENTANE 0.20
n-PENTANE 0.13
HEXANE 0.13
HEPTANE PLUS 0.25
H2S 0.00

and the amount of the gas is 0,6 MMSCFD.
can you help me?

It is possible to look up heats of combustion for each gas and do weighted average however, the mixture is not terribly atypical for natural gas. Just using a DoE avergae figure of 1089 BTU/ft³
600000 ft³/d x 1 d/24 h x 1089 BTU/ft³ x 1 kWh/3412 BTU = 7979 kW or about 7.98 MW. This is combustion heat; if you are trying to generate electricity, it will depend on the efficiency of your generator system, but maybe 1/3 that.




Assuming "standard" in MMSCFD is measured at 15 °C
you have
600000 ft³ x (0.3048 m/ft)³ x (273.15 +15)/273.15 x 1 kmol/22.414 m³ = 718.554 kmol of gas.

Based on composition analysis, in 100 moles of gas we have 8.22 moles of C from CO2, 81.31 mol from methane, but ethane has 2 mol C/mol of ethane, propane 3, etc. Adding up, 100 mol of gas has 118.96 mol of C which will burn to 118.96 mol CO2 (assuming sufficient combustion air).

718.554 kmol of gas x 118.96 mol CO2/100 mol gas = 854.8 kmol CO2
1 kmol CO2 = 44 kg, so 37611 kg, or 37.6 t CO2 (per day).

The CO2 analysis is as exact as your data. The thermal output is based on "average" gas, not your specific analysis. There is software that will calculate heat of combustion from composition, but I don't have a copy.

thank you so much for your help .
(this site is really great!! tks)

Hello,
I tried the calculation myself for this equation
600000 ft³ x (0.3048 m/ft)³ x (273.15 +15)/273.15 x 1 kmol/22.414 m³ = 718.554 kmol of gas.

but the result i got is about 799 kmol.
did I miscalculcated it?

[JohnS]

Originally Posted by Unregistered

Hello,
I tried the calculation myself for this equation
600000 ft³ x (0.3048 m/ft)³ x (273.15 +15)/273.15 x 1 kmol/22.414 m³ = 718.554 kmol of gas.

but the result i got is about 799 kmol.
did I miscalculcated it?

No, I wrote it out wrong (but the math was correct on my calculator)
The temperature term is written inverted and should be
273.15/(273.15 +15). The "standard" gas volume (at 15 °C) is being converted to "normal" (0 °C), so it is smaller. Sorry.