Question

Determine the enthalpy of foA novel gaseous hydrocarbon fuel CxHy is proposed for use in spark-ignition engines. An analysis of a sample of this fuel revealed that its molecular weight is 140 and its molar H/C ratio is 2.0. In order to evaluate some of its properties, the fuel was burned with stoichiometric standard air in a constant pressure, steady-flow reactor. The fuel and air entered the reactor at 25°C, and the products of complete combustion were cooled to 25°C. At the exit condition, water in the products was a liquid. It was measuredrmation of the fuel at 25°C

Answer 1

a. The fuel formula is C10H20 (dec-1-ene)

b. The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel

c. LHV = 44.36 KJ/kg

d. ΔH (formation) = -6669.6 KJ/mol

Step-by-step explanation:

Molecular weight = 140, Molar H/C ratio = 2.0

a) To determine the fuel formula, we represent the information we have as equations

Molecular weight of C = 12, Molecular weight of H = 1

CxHy = 140g

12 · x + 1 · y = 140 ·········Eqn 1

y = 2x ·········Eqn 2

Substitute y = 2x into Eqn 1, we have:

12 (x) + 1 (2x) = 140

12x + 2x = 140

14x = 140

x = 10

Substitute x into Eqns 2, we have:

y = 2 x 10 = 20

∴ the fuel formula is C10H20 (dec-1-ene)

b) The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel . HHV refers to the amount of heat given-off in the combustion of a specified amount of fuel which was initially at 25°C and cooled back to a temperature of 25°C.

c) We will use LHV formula given as:

LHV = HHV – [2.441 (0.09H + M) ÷ 1000]

where:

LHV = lower heating value of fuel in MJ/kg

HHV = higher heating value of fuel in MJ/kg

M = Weight % of moisture in fuel

H = Weight % of hydrogen in fuel

HHV = 47.5 MJ/kg, M = 0, H = mass of hydrogen ÷ total mass of fuel

⇒ H = 20 ÷ 140 = 0.143 * 100 = 14.3 kg per 100 kg, Latent heat of steam (at 25°C) = 2441 KJ/kg

LHV = 47.5 - [2441 * (0.09 * 14.3 + 0) ÷ 1000]

LHV = 47.5 - 3.14 = 44.36

LHV = 44.36 KJ/kg

d) To get the enthalpy of the combustion of the fuel, we have:

fuel + air - - - - > carbon dioxide + water

C10H20 + 15 O2 - - - - > 10 CO2 + 10 H20

Enthalpy (formation) = Enthalpy (product) - Enthalpy (reactant)

ΔH (CO2) = -393.5 kJ/mol, ΔH (H2O) = -285.8 kJ/mol, ΔH (C10h20) = - 123.4KJ/mol

ΔH (formation) = 10 * (-393.5) + 10 * (-285.8) - (-123.4)

ΔH (formation) = -3935 - 2858 + 123.4

ΔH (formation) = -6669.6 KJ/mol