Question

A fuel gas containing 75.0 mole% methane and the balance ethane is burned completely with 35.0% excess air.

The stack gas leaves the furnace at 800.0°C and is cooled to 350.0°C in a waste-heat boiler, a heat exchanger in which heat lost by cooling gases is used to produce steam from liquid water for heating, power generation, or process applications.

Take as a basis 100.0 mol of the fuel gas fed to the furnace.
The amounts of each of the following gases leaving the furnace at 800.0°C:

O2	83.1 mol
N2	1205 mol
CO2	125 mol
H2O	225 mol

Taking as a basis of calculation 100.0 mol of the fuel gas fed to the furnace, calculate the amount (magnitude) of heat (MJ) that must be transferred from the gas in the waste heat boiler to accomplish the indicated cooling. ? MJ

Answer 1

We calculate the mole fractions of each component:

Total moles = 83.1 + 1205 + 125 + 225 = 1638.1 mol

x_O2 = 83.1/1638.1 = 0.050703

Similarly,

x_N2 = 0.7356

x_CO2 = 0.076308

x_H2O = 0.1374

From internet, we find specific heat values:

C_p,O2 = 34.08 J/mol-K

C_p,N2 = 37.49 J/mol-K

C_p,CO2 = 52.228 J/mol-K

C_p,H2O = 39.276 J/mol-K

Taking 1 mole of gas as basis, since mole fractions are calculated by this, Heat transfer

Q = (x_O2*C_p,O2 + x_N2 *C_p,N2 + x_CO2 *C_p,CO2 + x_H2O *C_p,H2O)*(800 - 350)

    = 38.6875 * (800-350)

    = 17409.39 J   --- for 1 mole

For 100 mol,

Q = 17409 * 100 = 1740900 J

= 1.7409 MJ --- Ans