This approach will be illustrated using the example of steam gasification of char.

Let A denote the air supply in kg dry air per kg dry fuel, F the amount of dry fuel required to obtain one normal (at 0°C) cubic meter of the gas and Xc the carbon content of the fuel (kg carbon/kg dry fuel). This carbon is split between CO, CO2, and CH4 in the product gas. We know that 1 kmol gas occupies 22.4 Nm3. So, for 1 Nm3 of gas produced, one can write the carbon (moles) balance between inflow and outflow streams corresponding to:

where V represents the volumetric fraction of a constituent of the gas.

Similarly, one can develop three more equations balancing hydrogen moles, oxygen moles, and nitrogen moles.

rx, K^v-4 %,). [169]

2 22.4

cv [2V +V +V +2V )

_ I V W °if, [16.10]

2 22.4

We assume the product gas to be made of CO, CO2, CH4, H2, H2O, O2, and N2. The volume fractions of individual gases make up the total product gas, which is 1 Nm3 for F kg feed. So:

VCO2 + VCO + VH2 + VCH4 + VH2O + VO2 + VN2 = 1 [16.12]

Together one gets five equations. But it is necessary to find eight unknowns: VCO, VCO, VCH, VH, VHiq, VO2, V^, and F, the fuel feed for production of 1 Nm3 of

the product gas. To find the eight unknowns it is necessary to obtain three more equations. These are obtained from reaction kinetics. The following three overall gasification equations are relevant for steam gasification:

Water gas reaction: C + H2O CO + H2 + 131 kJ/mol [16.13]

Methanation reaction: C + 2H2 CH4 — 75 kJ/mol [16.14]

Shift reaction: CO + H2O H2 + CO2 — 41 kJ/mol [16.15]

For oxygen or air-gasification, other relevant sets of equations are to be considered. If allowed, these equations could reach equilibrium when forward and backward reaction rates are equal. Let PCO and PCO be the partial pressure of CO and CO2, respectively, under equilibrium.

The partial pressure of CO, PCO corresponds to the volume fraction of CO. So, PCO = VCO. P when the pressure of the reactor is P.

For the water gas reaction,

[16.16]

’ h, o

For the shift reaction,

P P V V

J, _’h,’co, _ H, CO,

I’pWS — p p ~ у у

COJ HjO CO HjO

For the Methanation reaction the equilibrium equation is:

[16.18]

Values of these equilibrium constants at different temperatures may be taken from Basu (2006, p. 68). Equations [16.8] to [16.12] are solved along with Eq. [16.16] to Eq. [16.18] to get eight unknowns. More details of this method are given in Basu (2006).