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21-Mat-A1 Thermodynamics · December 2015

Question 4 of 7: Adiabatic Flame Temperature of Acetylene — Pure Oxygen vs. Air

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2015 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).

Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state 298 K enthalpy/entropy data used in Question 7; supporting 298 K entropies for CO(g), CO₂(g), H₂(g) and H₂O(g) from the NIST–JANAF Thermochemical Tables.

Question 4: Adiabatic Flame Temperature of Acetylene — Pure Oxygen vs. Air (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Species$\Delta H^\circ_f$ (kJ/mol)$C_p(T)$ (J K−1 mol−1)
C₂H₂(g)+226.7—
CO₂(g)−393.5$18.9+7.9\times10^{-2}T$
H₂O(g)−241.8$31.4+0.4\times10^{-2}T$
N₂(g)0$27.9+0.4\times10^{-2}T$

Find. Adiabatic flame temperature $T_{ad}$ for (a) combustion with pure stoichiometric O₂; (b) combustion with the same stoichiometric O₂ supplied as air (21% O₂, 79% N₂).

Approach. The reaction $\text{C}_2\text{H}_2+\tfrac52\text{O}_2=2\text{CO}_2+\text{H}_2\text{O}$ releases $-\Delta H^\circ_{rxn}(298\,\text{K})$ of heat; with no heat loss (adiabatic), that heat raises the product gases from 298 K to $T_{ad}$: $-\Delta H_{rxn}=\int_{298}^{T_{ad}}\sum n_iC_{p,i}(T)\,dT$. Part (b) adds the inert N₂ carried in with the air to the product heat capacity sum, which dilutes and lowers $T_{ad}$.

  1. Heat of reaction at 298 K. $$\Delta H^\circ_{rxn}=\left[2(-393.5)+(-241.8)\right]-\left[226.7\right]=-1028.8-226.7=\boxed{-1255.5\ \text{kJ/mol}}.$$ This is the heat released per mole of C₂H₂ burned, available to heat the products.
  2. (a) Pure O₂: products are 2 mol CO₂ + 1 mol H₂O(g), no excess O₂ or N₂. Summing heat capacities: $$\sum C_p(T)=2(18.9+0.079T)+(31.4+0.004T)=69.2+0.162T\ \text{J/K}.$$ Setting the released heat equal to the sensible-heat integral: $$1{,}255{,}500=\int_{298.15}^{T_{ad}}(69.2+0.162T)\,dT=69.2(T_{ad}-298.15)+0.081\left(T_{ad}^2-298.15^2\right).$$ Solving the resulting quadratic $0.081T_{ad}^2+69.2T_{ad}-1{,}283{,}305=0$: $$\boxed{T_{ad}=3576\ \text{K}\ (3303\,{}^{\circ}\text{C})}.$$
  3. (b) Stoichiometric O₂ supplied as air. $n_{O_2}=2.5$ mol needs $n_{air}=2.5/0.21=11.90$ mol air, carrying $$n_{N_2}=n_{air}-n_{O_2}=11.90-2.5=9.405\ \text{mol N}_2$$ through to the products (no excess O₂, since exactly stoichiometric). Adding N₂'s heat capacity to the sum: $$\sum C_p(T)=69.2+0.162T+9.405(27.9+0.004T)=331.6+0.1996T\ \text{J/K}.$$
  4. Solve the same energy balance with the diluted sum. $$1{,}255{,}500=331.6(T_{ad}-298.15)+0.0998\left(T_{ad}^2-298.15^2\right)\ \Longrightarrow\ 0.0998T_{ad}^2+331.6T_{ad}-1{,}363{,}186=0,$$ $$\boxed{T_{ad}=2391\ \text{K}\ (2118\,{}^{\circ}\text{C})}.$$ The nitrogen — nearly 4 mol of inert gas for every mole of O₂ — absorbs a large share of the released heat without contributing to it, so the air flame runs roughly 1200 K cooler than the pure-oxygen flame; this is exactly why oxy-acetylene (not air-acetylene) torches are used for cutting and welding.
Part$T_{ad}$ (K)$T_{ad}$ (°C)
(a) stoichiometric O₂35763303
(b) stoichiometric O₂ via air23912118