Step1/4

We have to Calculate the adiabatic flame temperature for acetylene

Basis : 1 mole of Acetylene
Combustion reaction: $C_{2} H_{2} (g)^{+} \frac{5}{2} O_{2} (g) ⟶ 2 CO_{2} (g) + H_{2} O (l)$
Considering the heat of the formations of the components
Heat of formation of C₂H₂= 228.3 kJ/mol
Heat of formation of O₂= 0 KJ/mol
Heat of formation of CO₂= -393.5 KJ/mol
Heat of formation of H₂O= -286 KJ/mol
Heat of reaction at 25^oC for the above reaction is calculated as:
Heat of reaction = heat of formation of products - heat of formation of reactants
>>> $[2 (- 393.5) + 1 (- 286)] - [1 (228.3) + \frac{5}{2} (0)]$
>>> $- 1073 - 228.3 = - 1301.3$ KJ/mol of C₂H₂ converted
Negative symbol represents exothermic heat of reaction.

Total heat liberated = 1301.3 x 1 $= 1301.3$ KJ
Considered that initial temperature is 25^oC, enthalpy of reactants at 25^oC = 0 KJ
Total enthalpy of product stream = enthalpy of reactants + heat added by reaction = 0 + 1301300 = 1301300 J

Explanation:

We know that the adiabatic flame temperature is the maximum temperature of the combustion gas that can be reached by the combustion reaction under the no heat loss to the surroundings in the adiabatic condition.

Step2/4

a. combustion with air: (No excess air)

Constituents of product stream are estimated as follows:

CO₂ in the product stream = 2 mol

H₂O in the product stream = 1mol

O₂ theoretically needed = 1 x $\frac{5}{2}$ = $\frac{5}{2}$ mol.
O₂actually entering (No excess) = $\frac{5}{2}$ = 2.25 mol
N₂ entering along with O₂ in the air = 2.25 x 79/21 =8.46 mol

O₂ in the leaving gases (assuming complete combustion of CO to CO₂) = 2.25 - 2.25 = 0 mol

N₂ in the leaving gases =8.46 mol.

Adiabatic temperature or Maximum temperature (T) is calculated as follows:
Heat capacities in J / mol .K are CO₂ = 37.14, O₂ = 29.39, H₂O = 35.82, C₂H₂= 44.04
Initial temperature (in K)= 25 (⁰ C)+273.15= 298.15 K

>>> 1301300 = 2 x 37.14 x (T - 298.15) + 1 x 35.83 x (T - 298.15) + 0 x 29.39 x (T - 298.15)
>>> 1301300 = [74.28 + 35.83] (T - 298.15)
>>> 1301300 = 110.11 (T - 298.15)
>>> 11818.18 = T - 298.15
>>> T = 11818.18 + 298.15 =12116.33 K
>>> T = 12116.33 - 273.15 = 11,843.18 ^oC
Maximum temperature of the final stage ( after the combustion) =11,843.18 ^oC

Explanation: Please refer to solution in this step. Step3/4

b. combustion with oxygen( No excess Oxygen)

Constituents of product stream are estimated as follows:
CO₂ in the product stream = 2 mol
H₂O in the product stream = 1mol
O₂ theoretically needed = 1 x $\frac{5}{2}$ = $\frac{5}{2}$ mol.
O₂actually entering = $\frac{5}{2}$ mol
O₂ in the leaving gasses (assuming complete combustion of C₂H₂ to CO₂) = $\frac{5}{2} - \frac{5}{2}$ = 0 mol
Adiabatic temperature or Maximum temperature (T) is calculated as follows:
Heat capacities in J / mol .K are CO₂ = 37.14, O₂ = 29.39, H₂O = 35.82, C₂H₂= 44.04
Initial temperature (in K)= 25 (⁰ C)+273.15= 298.15 K

>>> 1301300 = 2 x 37.14 x (T - 298.15) + 1 x 35.83 x (T - 298.15) + 0 x 29.39 x (T - 298.15)
>>> 1301300 = [74.28 + 35.83] (T - 298.15)
>>> 1301300 = 110.11 (T - 298.15)
>>> 11818.18 = T - 298.15
>>> T = 11818.18 + 298.15 =12116.33 K
>>> T = 12116.33 - 273.15 = 11,843.18 ^oC
Maximum temperature of the final stage ( after the combustion) at 1 bar =11,843.18 ^oC

Explanation: Please refer to solution in this step. Step4/4

adiabatic flame temperature of acetylene be like if it were burned with an excess of air (Assuming 150% excess)

Constituents of product stream are estimated as follows:
CO₂ in the product stream = 2 mol
H₂O in the product stream = 1mol

O₂ theoretically needed = 1 x $\frac{5}{2}$ = $\frac{5}{2}$ mol.
O₂actually entering (150% excess) = $\frac{5}{2}$ x $(\frac{150}{100} + 1)$ = 6.25 mol
N₂ entering along with O₂ in the air = 6.25 x 79/21 = 23.51 mol
O₂ in the leaving gases (assuming complete combustion of CO to CO₂) = 6.25 - 2.25 = 4 mol
N₂ in the leaving gases = 23.51 mol.

We know that the adiabatic flame temperature is the maximum temperature of the combustion gas that can be reached by the combustion reaction under the no heat loss to the surroundings in the adiabatic condition.
Adiabatic temperature or Maximum temperature (T) is calculated as follows:
Heat capacities in J / mol .K are CO₂ = 37.14, O₂ = 29.39, H₂O = 35.82, C₂H₂= 44.04
Initial temperature (in K)= 25 (⁰ C)+273.15= 298.15 K

>>> 1301300 = 2 x 37.14 x (T - 298.15) + 1 x 35.83 x (T - 298.15) + 4 x 29.39 x (T - 298.15)
>>> 1301300 = [74.28 + 35.83 + 117.56 ] (T - 298.15)
>>> 1301300 = 227.67 (T - 298.15)
>>> 5715.72= T - 298.15
>>> T = 5715.72 + 298.15 = 6013.87 K
>>> T = 6013.87 - 273.15 = 5,740.72 ^oC
Maximum temperature of the final stage ( after the combustion) at 1 bar =5,740.72 ^oC

Explanation: Please refer to solution in this step.

Calculate the adiabatic flame temperature for acetylene considering the following two cases: a. combustion with air b. combustion with oxygen What would the adiabatic flame temperature of acetylene be like if it were burned with an excess of air? Justify your answer.

Public Answer

Calculate the adiabatic flame temperature for acetylene considering the following two
cases:
a. combustion with air
b. combustion with oxygen

What would the adiabatic flame temperature of acetylene be like if it were burned with an excess of air?
Justify your answer.