Energy balances on reactive processes and adiabatic flame temperature

Why it matters

In a reactor the heat of reaction usually dominates the energy balance. It decides how much cooling an exothermic reactor needs to avoid runaway, how much fuel an endothermic cracker burns, and how hot a flame or an adiabatic bed gets – which limits materials of construction, NOx formation and catalyst life.

Key ideas

Energy balance with reaction. At steady state with negligible kinetic/potential energy and no shaft work: Q = ΔH = Σ n_out·H_out − Σ n_in·H_in. The enthalpies must include the chemical energy change, which is done with one of two equivalent reference choices.

Heat-of-reaction method (best for a single reaction or known extents). References: reactant and product species at 25 °C in the phases used for ΔH_r°. ΔH = ξ·ΔH_r° + Σ n_out·Ĥ_out − Σ n_in·Ĥ_in, where each Ĥ is the sensible (and latent) enthalpy of the species relative to 25 °C. Physically: cool the feed to 25 °C, react at 25 °C, heat the products to the outlet temperature.

Heat-of-formation method (best for multiple or unknown reactions). References: the elements at 25 °C. Each species' enthalpy is Ĥ_i = ΔH_f,i° + (sensible enthalpy from 25 °C to T). Then ΔH = Σ n_out·Ĥ_out − Σ n_in·Ĥ_in, and no extent is needed.

Sign of Q. Q < 0: heat must be removed (exothermic, cooled reactor). Q > 0: heat must be supplied. Q = 0: adiabatic reactor – the outlet temperature is the unknown.

Adiabatic reaction temperature. For an adiabatic reactor, set Q = 0 and solve for the outlet temperature: the heat released by reaction equals the sensible heat absorbed by all outlet species (products, unreacted reactants and inerts). If C_p depends on T, iterate or solve the polynomial; mean C_p values over the range give a quick answer.

Adiabatic flame temperature. The adiabatic reaction temperature for complete combustion. It is the highest temperature a fuel–oxidant mixture can reach, and it is:

• highest with stoichiometric (theoretical) oxidant; excess air adds N₂ and O₂ that absorb heat;
• much higher with pure O₂ than with air (no N₂ diluent);
• higher if reactants are preheated;
• lower if LHV rather than HHV applies (product water stays vapour, which is the normal case). Real flames are cooler still because of heat losses and, above about 2000 K, dissociation of CO₂ and H₂O; simple balances ignore dissociation.

Formulas

• Q = ΔH = ξ·ΔH_r° + Σ n_out·Ĥ_out − Σ n_in·Ĥ_in – heat-of-reaction method; Ĥ relative to 25 °C (kJ/mol), ξ (mol/s).
• Ĥ_i = ΔH_f,i° + ∫ C_p,i dT from 298.15 K to T – heat-of-formation method.
• Q = Σ n_out·Ĥ_out − Σ n_in·Ĥ_in
• Ĥ_i ≈ C_p,mean,i·(T − 298.15) – sensible enthalpy with a mean heat capacity, no phase change.
• Adiabatic, feed at 25 °C: −ξ·ΔH_r° = (Σ n_out·C_p,mean)·(T_ad − 298.15)
• T_ad = 298.15 + (−ξ·ΔH_r°) / Σ(n_out·C_p,mean) – K; for combustion use the LHV (water vapour) when water leaves as vapour.

Worked examples

Use ΔH_c°(CH₄, water vapour) = −802.3 kJ/mol. With 20 % excess air, 1 mol CH₄ gives 1 CO₂, 2 H₂O, 0.4 O₂ and 9.03 N₂ (see the combustion topic).

Example 1 (standard, cooled furnace). Methane and 20 % excess air enter a furnace at 25 °C; flue gas leaves at 300 °C. Mean C_p (25–300 °C, J/(mol·K)): CO₂ 41.0, H₂O(g) 34.5, O₂ 30.3, N₂ 29.4. Find the heat transferred to the process per mol CH₄.

1. Feed is at 25 °C, so Σ n_in·Ĥ_in = 0. ξ = 1 mol.
2. Σ n_out·C_p = 1 × 41.0 + 2 × 34.5 + 0.4 × 30.3 + 9.03 × 29.4 = 41.0 + 69.0 + 12.1 + 265.4 = 387.6 J/K.
3. Sensible enthalpy of flue gas = 387.6 × (300 − 25) = 106.6 kJ.
4. Q = ξ·ΔH_r° + Σ n_out·Ĥ_out = −802.3 + 106.6 = −695.7 kJ.

Answer: 695.7 kJ per mol CH₄ is transferred to the process (about 86.7 % of the LHV)

Example 2 (GATE level, adiabatic flame temperature). Same feed, but the combustor is adiabatic. Mean C_p over 25 °C to the flame temperature (J/(mol·K), data-book values): CO₂ 54.0, H₂O(g) 43.0, O₂ 34.5, N₂ 33.0. Find T_ad.

1. Q = 0: heat released = sensible heat of products.
2. Σ n_out·C_p = 54.0 + 2 × 43.0 + 0.4 × 34.5 + 9.03 × 33.0 = 54.0 + 86.0 + 13.8 + 297.9 = 451.7 J/K.
3. ΔT = 802 300 / 451.7 = 1776 K.
4. T_ad = 298.15 + 1776 = 2074 K (≈ 1801 °C).
5. If C_p were given as polynomials, one would check that the mean values correspond to this temperature range and iterate once.

Answer: T_ad ≈ 2074 K

Common mistakes

• Adding ΔH_r° with the wrong sign, or forgetting to multiply it by the extent.
• Leaving out inerts (N₂) and excess reactant from Σ n·C_p – this overestimates T_ad grossly.
• Using HHV when product water leaves as vapour.
• Using C_p of the reactants or of only one product.
• Mixing the two reference methods within one balance.
• Forgetting that the sensible enthalpy of a feed above 25 °C adds to the heat available.

For GATE CH

Expect NAT questions on heat removed from a reactor for a given conversion, adiabatic outlet temperature with mean C_p, adiabatic flame temperature with excess air, and the effect of preheating or dilution. Practise laying out a table of n_in, n_out, C_p and Ĥ, and stating which reference method you use.

Quick check

1. In an adiabatic reactor, what is Q?
2. Does excess air raise or lower the adiabatic flame temperature?
3. 8 mol/s reacts with ΔH_r° = −60 kJ/mol; outlet ΣnC_p = 2000 W/K; feed at 25 °C. What is the adiabatic outlet temperature?
4. Which reference states does the heat-of-formation method use?
5. Why are real flame temperatures lower than calculated T_ad?

Answers: 1. zero; 2. lowers it; 3. 25 + 240 = 265 °C; 4. elements at 25 °C; 5. heat losses and dissociation of products.