Liquid octane (C 8 H 18 Cp= 2.28 kJ/kg-K) is burned in a combustor with a given

Question

Liquid octane (C8H18Cp= 2.28 kJ/kg-K) is burned in a combustor with a given percentage of excess air. The pressure of both the air and fuel is given as 101 kPa while the temperature of each is given. Assume that the mole fractions are 79% nitrogen and 21 % oxygen for air (use M=28.97 kg/kmol and R=0.287 kJ/kg-K) and that water is a vapor in the exhaust. Given the values below determine the following:

--Given Values--
m_fuel (kg/min) = 1.1
T_fuel (C) = 84
T_air (C) = 70
Excess Air (%) = 25%
T_exhaust (C) = 1400

a) Determine the air fuel ratio (kmol_air/kmol_fuel).
b) Determine the rate (kmol/s) that fuel is burned.
c) Determine the rate (kmol/s) of CO2 produced.
d) Determine the rate (kmol/s) of H2O produced
e) Determine the rate (kmol/s) of N2 produced.
f) Determine the rate (kmol/s) of O2 produced.
g) Determine the flow of Enthalpy (kW) of CO2 in the products.
h) Determine the flow of Enthalpy (kW) of H2O in the products.
i) Determine the flow of Enthalpy (kW) of N2 in the products
j) Determine the flow of Enthalpy (kW) of O2 in the products.
k) Determine the Molecular Weight (kg/kmol) of the exhaust gas
l) Determine the specific volume (m^3/kg) of the exhaust gas.

m) Determine the flow of Enthalpy (kW) of the fuel.

n) Determine the flow of Enthalpy (kW) of the inlet air.

O) Determine the total heat transfer (kJ) into the combustion chamber (a negative value indicates heat transfer out of the chamber).

Explanation / Answer

The combustion of liquid octane is given by C8H18+12.5 O2----à8CO2+ 9H2O
1 mole of octane requires 12.5 mole of oxygen and produces 8 moles of CO2 and 9 moles of water.

Moles of fuel burned = mass/molar mass= 1.1/114= 0.00965 kg moles/min= 0.00965/60 kg moles/sec =0.000161 kgmoles/sec

Moles of oxygen = 0.00965*12.5= 0.1206 kg moles/min

Air contains 21% O2 and 79% N2, moles of air required= 0.1206/0.21= 0.5743 kg moles/min. Air is supplied 25% excess, Moles of air supplied= 1.25*0.5743= 0.72 kg moles/min

Air to fuel ratio = 0.72/0.00965= 74.6

Kg Moles of CO2 produced/min= 8*0.00965= 0.0772 and kg moles of water/min= 9*0.00965= 0.087

Kg moles of CO2 produced per sec= 0.0772/60 =0.001287 kg moles/sec and kg moles of water produced per sec = 0.087/60 =0.00145

N2= 0.72*0.79= 0.57 kg moles /min = 0.57/60 kg moles/sec= 0.0095 kgmoles/sec

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