E3.3 Excess Air and Combustion

Example 3.3: It is necessary that complete combustion of propane yield a temperature of 1600 °C. The propane supply temperature is 25 °C. Determine the air-to-fuel ratio on a molar basis that must be supplied to the burner to attain the desired temperature. Heat of combustion of propane is 530605.6 cal/mole. Latent heat of vaporization of water at 25 °C is 10500 cal/mole. Mean molal heat capacity of the compounds between 25 °C and 1600 °C is given as below:

C3H8 = 6.601 cal/(mol·°C), O2 = 8.269 cal/(mol·°C), CO2 = 12.75 cal/(mol·°C),

N2 = 7.844 cal/(mol·°C), H2O = 9.95 cal/(mol·°C), air = 7.929 cal/(mol·°C)

Solution: Basis of calculations: 1 mol of gas,

Feed temperature, T_Feed = 298.15 °K,

Product temperature, T_Products = 1873.15 °K.

Enthalpy of feed, SH_Feed = 0

Heat added to the system, q = 0

Standard heat of reaction, ΔH₂₉₈

ΔH₂₉₈ = -530605.6 + 4(10500) = -488605.6 cal

Enthalpy of products, SH_P = SH_F - ΔH₂₉₈ - q = Sn_pC_pm(T_P - T_F) = 488605.6 cal

Excess air, X

(225.67 + 7.929X)(1873.15 - 298.15) = 488605.6

X = 10.664 mol

Stoichiometric air = 5(4.764) = 23.82 mol

Total air requirement = 10.66 + 23.82 = 34.48 mol