23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2014
Question 4 of 7: Adiabatic Flame Temperature of a Fuel Gas
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A, 15 marks), one of Q3–Q4 (Part B, 25 marks) and two of Q5–Q7 (Part C, 30 marks each); four questions total 100 marks. All seven are solved below for completeness. Property data are stated explicitly in each Given block; units follow the paper (mixed SI and US/older conventions).
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances, single-phase systems, combustion and recycle; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — VLE and excess-property (Margules) models, compressor work, and reaction equilibrium; supporting property data from Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook (Antoine constants, C₀p polynomials, standard enthalpies and Gibbs energies of formation).
Question 4: Adiabatic Flame Temperature of a Fuel Gas (Part B, 25 marks)
Given. Basis 100 mol fuel: 50 CO, 30 CO₂, 20 N₂. Reaction $CO + \tfrac12 O_2 \to CO_2$, $\Delta H^{\circ}_{r,298}=-283.0$ kJ/mol. Fuel enters at 100 °C, air at 200 °C; 205% excess air; only half the CO burns; adiabatic ($Q=0$).
Find. the product-gas temperature $T$.
Figure 3 — Adiabatic chamber: warm fuel gas and hot excess air enter; half the CO oxidises and the released heat raises the product gas to T.
Approach. Size the air from the excess-air definition, fix the product amounts from the 50%-CO reaction, then impose the adiabatic energy balance about a 25 °C reference: reaction heat plus reactant sensible heat equals product sensible heat.
Theoretical and supplied oxygen. Complete combustion of all 50 mol CO needs $O_2^{\text{theo}} = \tfrac12(50)=25$ mol. With 205% excess,
$$O_2^{\text{sup}} = 25(1+2.05) = 76.25\ \text{mol}, \qquad N_2^{\text{air}} = 76.25\cdot\tfrac{79}{21} = 286.9\ \text{mol}.$$
Reaction and product amounts. Only half the CO burns: CO reacted $=25$ mol, so $O_2$ consumed $=12.5$ mol. The product gas is
$$CO:25,\quad CO_2:30{+}25=55,\quad O_2:76.25{-}12.5=63.75,\quad N_2:20{+}286.9=306.9\ \text{mol}\;(\Sigma=450.6).$$
Energy balance (reference 25 °C, $Q=0$). With extent $\xi = 25$ mol,
$$\xi\,\Delta H^{\circ}_{r} + \underbrace{\sum_{\text{prod}} n_i\!\int_{298}^{T}\!C_{p,i}\,dT}_{\text{products, }25^{\circ}\text{C}\to T} - \underbrace{\sum_{\text{in}} n_i\!\int_{298}^{T_{\text{in},i}}\!C_{p,i}\,dT}_{\text{reactants sensible}} = 0.$$
The reaction term is $25(-283.0)=-7075$ kJ; the reactant sensible heat (fuel 25→100 °C, air 25→200 °C) is $\approx +2.9\times10^{3}$ kJ.
Solve for $T$. Using SVA ideal-gas $C_p/R = A + BT + D/T^2$ polynomials and root-finding on the product enthalpy,
$$\boxed{T \approx 922\ \text{K} = 649\ ^{\circ}\text{C}}.$$
The large mass of hot excess air and inert N₂/CO₂ ballast (over 90% of the 450 mol product) is what holds the flame temperature down to ~650 °C despite the exothermic oxidation.
