Combustion Chemistry (Propulsion Bridge)

Q1 (10)

(a) Stoichiometric with air (3): $\mathrm{CH_4 + 2O_2 \to CO_2 + 2H_2O}$ With air, each mole $\mathrm{O_2}$ carries $79/21 = 3.76$ mol $\mathrm{N_2}$: $\mathrm{CH_4 + 2(O_2 + 3.76\,N_2) \to CO_2 + 2H_2O + 7.52\,N_2}$ Balanced eqn (1), air O₂ = 2 (1), N₂ = 7.52 (1).

(b) (3): $\phi = \dfrac{(F/O)_{actual}}{(F/O)_{stoich}} = \dfrac{(O/F)_{stoich}}{(O/F)_{actual}}$. $\phi>1$ fuel-rich (excess fuel, $\phi-1>0$); $\phi<1$ fuel-lean (excess oxidiser). Definition (2), signs (1).

(c) (4): $\phi=0.80$ means less fuel than stoichiometric ⇒ excess air. Actual $\mathrm{O_2} = 2/\phi = 2.5$ mol; air $= 2.5\times4.76 = 11.9$ mol; A/F molar = 11.9 (per mol CH₄). Lean ⇒ excess $\mathrm{O_2}$ appears in exhaust (there is leftover oxidiser). O₂ scaling (1), air ratio (2), excess O₂ identified (1).

Q2 (14)

(a) (3): Adiabatic, constant-P: $H_{reactants}(298) = H_{products}(T_{ad})$. No heat lost. Enthalpy of reaction at 298 goes entirely into sensible heating of products from 298 to $T_{ad}$: $-\Delta H_{rxn}^\circ = \sum n_i c_{p,i}(T_{ad}-298)$. Principle (2), equation form (1).

(b) (4): $\Delta H_{rxn}^\circ = [(-393.5) + 2(-241.8)] - [(-74.9) + 0]$ $= -877.1 + 74.9 = -802.2\ \mathrm{kJ/mol}$ So heat released $= 802.2\ \mathrm{kJ/mol}$. Products sum (2), subtract reactant (1), value (1).

(c) (7): Products per mol CH₄: 1 CO₂, 2 H₂O, 7.52 N₂. $\sum n_i c_{p,i} = 1(54.3)+2(41.2)+7.52(33.6) = 54.3+82.4+252.7 = 389.4\ \mathrm{J/K}$ $T_{ad}-298 = \frac{802200}{389.4} = 2060\ \mathrm{K}$ $T_{ad} \approx 298 + 2060 = 2358\ \mathrm{K} \approx 2.36\times10^3\ \mathrm{K}$ Heat capacity sum (3), rearrange (2), ΔT (1), T_ad (1). (Real value ~2230 K lower due to dissociation — see Q3.)

Q3 (10)

(a) (3): Dissociation reactions (CO₂→CO+½O₂, H₂O→OH+H) are endothermic; they absorb energy that would otherwise raise sensible temperature, so real $T$ < ideal no-dissociation value. Also they increase mole count/species. Endothermic (2), lowers T (1).

(b) (4): For $\mathrm{CO_2 \rightleftharpoons CO + \tfrac12 O_2}$, $K_p = \frac{p_{CO}\,p_{O_2}^{1/2}}{p_{CO_2}} = \frac{x_{CO}\,x_{O_2}^{1/2}}{x_{CO_2}}\left(\frac{P}{P^\circ}\right)^{1/2}$ Net $\Delta n_{gas} = +\tfrac12$, hence the $P^{1/2}$ dependence. Correct ratio (2), pressure exponent ½ (2).

(c) (3): Increasing $P$ shifts equilibrium toward fewer gas moles (toward $\mathrm{CO_2}$), suppressing dissociation. Less dissociation ⇒ less endothermic loss ⇒ hotter chamber. Le Chatelier direction (2), hotter conclusion (1).

Q4 (10)

(a) (4): Deflagration: subsonic (relative to unburned gas), slight pressure drop/near-constant P, driven by thermal conduction + species diffusion. Detonation: supersonic (Mach 4–8), sharp pressure rise (5–30×), driven by a shock wave that compresses/ignites the mixture. 2 marks each mode covering speed+P+mechanism.

(b) (3): C–J condition: the detonation propagates at the minimum wave speed for which the Rayleigh line is tangent to the Hugoniot; at the C–J point the burned-gas velocity relative to the wave equals the local sound speed ($M=1$ downstream). It uniquely selects the steady, self-sustaining detonation velocity. Tangency/sonic condition (2), self-sustaining (1).

(c) (3): In a steady liquid engine, detonation causes destructive pressure spikes/combustion instability that can rupture the chamber. In PDE/RDE the detonation is intentional and confined, giving near-constant-volume (pressure-gain) combustion with higher thermodynamic efficiency than deflagration. Undesirable reason (1.5), exploited efficiency (1.5).

Q5 (10)

(a) (4): $r$ = linear burn rate (mm/s); $a$ = burn-rate coefficient (depends on propellant/temperature); $n$ = pressure exponent (combustion stability requires $n<1$). AP = oxidiser (supplies O), HTPB = polymeric fuel/binder holding grain together, Al = metallic fuel raising flame temperature/$I_{sp}$ and damping instability. Symbols (2), ingredients (2).

(b) (4): Take ratio: $\frac{r_2}{r_1} = \left(\frac{P_2}{P_1}\right)^n \Rightarrow n = \frac{\ln(r_2/r_1)}{\ln(P_2/P_1)} = \frac{\ln(9/6)}{\ln(10/5)} = \frac{\ln1.5}{\ln2} = 0.585$ $a = \frac{r_1}{P_1^n} = \frac{6.0}{5^{0.585}} = \frac{6.0}{2.599} = 2.31\ \mathrm{mm/s/MPa^{n}}$ n derivation (2), a (2).

(c) (2):

import math
def vieille(r1,P1,r2,P2):
    n = math.log(r2/r1)/math.log(P2/P1)
    a = r1/P1**n
    return n, a
print(vieille(6.0,5.0,9.0,10.0))  # (0.585, 2.31)

Correct n & a formulas (2).

Q6 (6)

(a) (3): CEA solves chemical equilibrium (minimising Gibbs free energy) for a given mixture to output equilibrium composition, $T_c$, and rocket performance ($I_{sp}$, $c^*$). Required inputs: (1) propellant identities and O/F ratio (or $\phi$), (2) chamber pressure and problem type (e.g. rocket with an area/pressure ratio for nozzle expansion). What it does (1.5), two inputs (1.5).

(b) (3): NOₓ — increased by high flame temperature (thermal/Zeldovich mechanism, lean-hot zones); Soot — increased by fuel-rich, poorly-mixed (diffusion) combustion; Unburned hydrocarbons (UHC)/CO — increased by low temperature / incomplete combustion (over-lean or quenching). 1 mark each with correct condition.

[
 {"claim":"Q2b heat of reaction magnitude 802.2 kJ/mol","code":"dHr=(-393.5+2*(-241.8))-(-74.9); result=abs(abs(dHr)-802.2)<0.1"},
 {"claim":"Q2c cp sum 389.4 J/K","code":"s=1*54.3+2*41.2+7.52*33.6; result=abs(s-389.36)<0.1"},
 {"claim":"Q2c T_ad approx 2358 K","code":"s=1*54.3+2*41.2+7.52*33.6; T=298+802200/s; result=abs(T-2358)<5"},
 {"claim":"Q5 n=0.585 and a=2.31","code":"import sympy as sp; n=sp.log(sp.Rational(9,6))/sp.log(2); a=6/5**n; result=abs(float(n)-0.585)<0.005 and abs(float(a)-2.31)<0.02"},
 {"claim":"Q1c air-fuel molar ratio 11.9","code":"AF=(2/0.8)*4.76; result=abs(AF-11.9)<0.05"}
]