Ethane gas is combusted with 130% theoretical air. Assume that the reactants ent

Question

Ethane gas is combusted with 130% theoretical air. Assume that the reactants enter thecombustion chamber at standard conditions and the products exit to the atmosphere at 400 K. Assume the boundary temperature is 350 K.

a. Write and balance the theoretical combustion equation.

b. Write and balance the actual combustion equation.

c. Determine the air-fuel ratio on a mass basis. Comment whether the mixture is rich or lean.

d. Determine the heat transfer for the combustion process [kJ/kmol-K].

e. Determine the entropy generation [kJ/kmol fuel-K] associated with the process if the ethane and air are added to the combustion chamber as separate streams.

f. Determine the entropy generation [kJ/kmol fuel-K] associated with the process if the ethane and air are mixed before addition to the combustion chamber.

g. Explain why results from parts (e) and (f) differ.

h. If the combustion reaction occurred at theoretical conditions, would you expect the Tpto be <, >, or == to 400 K? Explain your reasoning.

Select answers:

c) AF = 20.87

d)-1355440

e) 4196.5

f) 4158.3

Explanation / Answer

a. Theoretical Combustion equation

C2H6 + 3.5O2 -----> 2CO2 + 3H2O

b. Actual Combustion equation

C2H6 + 4.55O2 -----> 2CO2 + 3H2O + 1.05O2

Since 30 % excess air is supplied which means 30 % excess oxygen . 30% of theoretical requirement 3.5 mole is 1.05 mole which comes along with product without contribute in reaction.

c. Air-fuel ratio on a mass basis

Basis 1 mole of fuel (Ethane)

Air contains 79% N2 and 21% O2

Molecular weight of Ethane = 30

Molecular weight of N2 = 28

Molecular weight of O2 = 32

Average Molecular weight of air = (0.79 * 28) + (0.21 * 32) = 28.84

Mass of ethane = Moles * Molecular wt. = 1 * 30 = 30 g

For 4.55 moles of O2, Moles of air supplied = 4.55/0.21 = 21.667 moles

Mass of air = Moles * Molecular wt. = 21.667 * 28.84 = 624.88 g

Air fuel mass ratio = 624.88/30 = 20.839

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