Liquid ethanol (C2H5OH) at 298 K, 1 atm enters a steam generator operating at steady state and burns completely with dry air entering at 400 K, 1 atm. The fuel flow rate is 75 kg/s, and the equivalence ratio is 0.7. If the combustion products exit at 1500 K, 1 atm, what is the rate of heat transfer in W?

Q = 6491.100 kJ/s

Explanation:

Air-Fuel Ratio:

For a combustion reaction the proportion of air that is present in a gaseous substance responsible for the reaction, this proportion is known as air-fuel ratio. The air fuel ratio is calculated using the combustion reaction for the substance.

Considering reaction for the Ethanol as

C₂H₅OH + XO₂ (O₂+3.76N₂) → aCO₂+bH₂O+cN₂

Balancing the equation we get;

a=2,

2b=6

∴ b=3

xO₂=3

The air-fuel ratio

A/F = XO₂+H₂O+xN₂ * mass of N₂/mass (fuel)

3.31*31.9+11.28*28.013/46.069

= 8.943

Equivalent ratio = 0.7,

so, heat transfer

Q = m * Cp*ΔT

= 75*0.7*112.4 (1500-400)

Q = 6491.100 kJ/s

1kJ/s=1000w

∴ Q = 6491100 W