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21-Mat-A1 Thermodynamics · December 2017

Question 4 of 7: Adiabatic Flame Temperature of Acetylene Combustion

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

Notes on this paper

National Exams — December 2017 — 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.) — source of the first-law/second-law relations used throughout and of the attached Ellingham diagram (Fig. 9-3) and its underlying 298 K formation data used in Question 7; supporting 298 K entropies for Ca(s), CaO(s), CO(g), CO₂(g), H₂(g) and H₂O(g) from the NIST–JANAF Thermochemical Tables.

Question 4: Adiabatic Flame Temperature of Acetylene Combustion (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.

Given.

QuantityValue
Reaction$C_2H_2+2.5\,O_2=2CO_2+H_2O$, reactants and initial products at 298 K
$\Delta H^\circ_f(C_2H_2)$+226.7 kJ mol−1
$\Delta H^\circ_f(CO_2)$−393.5 kJ mol−1
$\Delta H^\circ_f(H_2O,g)$−241.8 kJ mol−1
$C_p(CO_2),\,C_p(H_2O),\,C_p(N_2)$$18.9+7.9\times10^{-2}T$; $31.4+0.4\times10^{-2}T$; $27.9+0.4\times10^{-2}T$

Find. (a) $T_{flame}$ burning with pure stoichiometric $O_2$. (b) $T_{flame}$ burning with stoichiometric $O_2$ supplied as air (21% $O_2$/79% $N_2$).

Approach. Compute $\Delta H^\circ_{rxn}(298\text{K})$ from formation enthalpies; the heat released is entirely absorbed in heating the product gases (plus any inert $N_2$) from 298 K to the adiabatic flame temperature, so solve $-\Delta H_{rxn}=\int_{298}^{T_f}\sum n_iC_{p,i}\,dT$ for $T_f$.

  1. Heat of reaction at 298 K. $$\Delta H^\circ_{rxn}=\big[2(-393.5)+1(-241.8)\big]-\big[1(226.7)\big]=-1028.8-226.7=\boxed{-1255.5\ \text{kJ}\,\text{mol}^{-1}}\ (=-1{,}255{,}500\ \text{J}).$$
  2. (a) Product heat-capacity sum, stoichiometric $O_2$. Products are 2 mol $CO_2$ + 1 mol $H_2O$ (no leftover $O_2$, no $N_2$): $$\textstyle\sum n_iC_{p,i}=2(18.9+0.079T)+(31.4+0.004T)=69.2+0.162T\ \text{J}\,\text{K}^{-1}.$$
  3. (a) Energy balance. $-\Delta H_{rxn}=\displaystyle\int_{298}^{T_f}(69.2+0.162T)\,dT=69.2(T_f-298)+0.081(T_f^2-298^2)$. Setting this equal to $1{,}255{,}500$ J gives the quadratic $$0.081\,T_f^2+69.2\,T_f-1{,}283{,}314.7=0\ \Rightarrow\ T_f=\dfrac{-69.2+\sqrt{69.2^2+4(0.081)(1{,}283{,}314.7)}}{2(0.081)}=\boxed{3576\ \text{K}}\ (3303\,{}^\circ\text{C}).$$
  4. (b) Nitrogen carried in with the air. For 2.5 mol $O_2$ supplied as 21% of air, the accompanying inert $N_2$ is $$n_{N_2}=2.5\times\dfrac{79}{21}=\boxed{9.405\ \text{mol}}.$$
  5. (b) Product heat-capacity sum, combustion in air. Now 2 mol $CO_2$ + 1 mol $H_2O$ + 9.405 mol $N_2$ (all inert and product gas share the same flame temperature) all get heated from 298 K: $$\textstyle\sum n_iC_{p,i}=69.2+0.162T+9.405(27.9+0.004T)=331.6+0.1998T\ \text{J}\,\text{K}^{-1}.$$
  6. (b) Energy balance. $-\Delta H_{rxn}=331.6(T_f-298)+0.0998(T_f^2-298^2)=1{,}255{,}500$ J gives $$0.0998\,T_f^2+331.6\,T_f-1{,}363{,}178.2=0\ \Rightarrow\ T_f=\dfrac{-331.6+\sqrt{331.6^2+4(0.0998)(1{,}363{,}178.2)}}{2(0.0998)}=\boxed{2391\ \text{K}}\ (2118\,{}^\circ\text{C}).$$ The huge quantity of inert $N_2$ absorbs much of the released heat, so burning in air rather than pure $O_2$ drops the flame temperature by over 1100 K — the physical reason oxy-acetylene torches (pure $O_2$) run far hotter than acetylene burned in open air.
QuantityValue
$\Delta H^\circ_{rxn}(298\text{K})$−1255.5 kJ mol−1
(a) $T_{flame}$, stoichiometric $O_2$3576 K (3303°C)
(b) $n_{N_2}$ carried in with air9.405 mol
(b) $T_{flame}$, combustion in air2391 K (2118°C)