Combustion Chemistry (Propulsion Bridge)

Question 1

(a) Stoichiometric combustion (4) $\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}$ With air, 2 mol O₂ carries $2\times\frac{79}{21}=7.524$ mol N₂: $\text{CH}_4 + 2\,\text{O}_2 + 7.524\,\text{N}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O} + 7.524\,\text{N}_2$ (2 marks equation, 2 marks N₂ = 7.52 mol.)

(b) No-dissociation adiabatic flame temperature (10)

Heat of reaction at 298 K (products − reactants; O₂, N₂ elemental = 0): $\Delta_r H^\circ = [(-393.5) + 2(-241.8)] - (-74.9) = -802.2\ \text{kJ}.$ (3 marks)

Adiabatic ⇒ heat released raises product temperature: $-\Delta_r H^\circ = \Big(\sum n_i \bar c_{p,i}\Big)(T_{ad}-298).$ $\sum n\bar c_p = 1(56.0)+2(43.0)+7.524(33.5) = 56.0+86.0+252.1 = 394.1\ \text{J/K}.$ (4 marks) $T_{ad}-298 = \frac{802200}{394.1}=2035\ \text{K}\ \Rightarrow\ T_{ad}\approx 2333\ \text{K}.$ (3 marks — accept 2300–2350 K.)

(c) Dissociation bookkeeping (10)

Start from 1 mol CO₂; fraction $\alpha$ dissociates: $\text{CO}_2\to\text{CO}+\tfrac12\text{O}_2$.

• $n_{\text{CO}_2}=1-\alpha$, $n_{\text{CO}}=\alpha$, $n_{\text{O}_2}=\alpha/2$, $n_{\text{H}_2\text{O}}=2$, $n_{\text{N}_2}=7.524$. (4)
• Total $n_{tot}= (1-\alpha)+\alpha+\alpha/2+2+7.524 = 10.524 + \alpha/2$. (2)
• With $p_i = (n_i/n_{tot})P$: $K_p=\frac{(p_{\text{CO}})(p_{\text{O}_2})^{1/2}}{p_{\text{CO}_2}} = \frac{\alpha\,(\alpha/2)^{1/2}}{(1-\alpha)}\left(\frac{P}{n_{tot}}\right)^{1/2}.$ (4 marks.)

(d) Why dissociation lowers $T_{ad}$ (8)

CO₂ dissociation is endothermic (bond breaking; reverse of formation is exothermic). At high T it proceeds, absorbing part of the released chemical energy, so less energy heats the gas ⇒ lower $T_{ad}$. It also increases mole number (more heat-capacity sink). (3)

Enthalpy of the dissociation step at 298 K: $\Delta H_{diss}= \Delta_fH(\text{CO}) - \Delta_fH(\text{CO}_2)= -110.5-(-393.5)= +283.0\ \text{kJ/mol}.$ For $\alpha=0.05$ mol dissociating, energy diverted $=0.05\times283.0=14.15$ kJ. (3)

First-order temperature drop (using $\sum n\bar c_p\approx394$ J/K): $\Delta T \approx \frac{14150}{394}\approx 36\ \text{K}.$ So $T_{ad}\approx 2333-36 \approx 2297\ \text{K}$. (2 marks — accept 30–40 K.)

(e) Iterative scheme pseudocode (8)

Given: reactants @298K, P
guess T
repeat:
    Kp = Kp_of_T(T)                       # from ΔG°(T) = -RT ln Kp
    solve  f1(α) = α√(α/2)·(P/n_tot(α))^0.5/(1-α) - Kp = 0   (Newton on α)
    compute product enthalpy H_prod(T, α)
    residual  f2(T) = H_prod(T,α) - H_react(298) = 0
    update T via Newton:  T <- T - f2/(dH_prod/dT)
until |ΔT| < tol and |f1| < tol
return T_ad, α

Two coupled residuals: energy balance $f_2(T,\alpha)=0$ and equilibrium $f_1(\alpha,T)=0$. (4 marks pseudocode structure, 2 marks naming residuals, 2 marks noting $K_p$ from $\Delta G^\circ(T)$.)

Question 2

(a) (8) $p$–$v$ diagram: initial point $(v_1,p_1)$. Rayleigh lines = straight chords through the initial state with slope $-\dot m^2$. Hugoniot = curve of allowed burnt states from energy+mass+momentum. Two tangency points: upper-left = C–J detonation, lower-right = C–J deflagration. Region between the two tangents (the "forbidden" arc between the Hugoniot's intercepts) has no real solution. C–J condition: burnt-gas velocity relative to the wave is sonic, $M_2 = u_2/a_2 = 1$. (4 sketch/labels, 2 tangency identification, 2 C–J condition.)

(b) (10) Wave-fixed frame; mass $\rho_1 u_1=\rho_2 u_2=\dot m$; momentum $p_1+\rho_1 u_1^2 = p_2+\rho_2 u_2^2$. With $\rho=1/v$, $\dot m = u_i/v_i$: $p_1+\dot m^2 v_1 = p_2+\dot m^2 v_2 \Rightarrow \dot m^2=\frac{p_2-p_1}{v_1-v_2}.$ (6) Since $\dot m^2>0$: if $p_2>p_1$ then $v_2<v_1$ (compression) → detonation (high-$p$ branch); if $p_2<p_1$ then $v_2>v_1$ (expansion) → deflagration (low-$p$ branch). Slope must be negative on the $p$–$v$ chart. (4)

(c) (8) Sound speed of burnt gas: $a_2=\sqrt{\gamma R T_2/M}=\sqrt{\frac{1.20\times 8.314\times 3500}{0.018}}.$ Numerator $=1.20\times8.314\times3500=34{,}919$; $/0.018 = 1.940\times10^6$; $a_2=1393\ \text{m/s}$. (5) At the C–J point $u_2=a_2=1393$ m/s. The detonation (wave) velocity $D=u_1$ follows from mass continuity $u_1=u_2\,(v_1/v_2)=a_2\,\rho_2/\rho_1$; since $\rho_2>\rho_1$ the wave speed $D>a_2$ (typically $D\sim 2800$ m/s for H₂/O₂). The momentum/Rayleigh relation fixes the density ratio, so $D=a_2(\rho_2/\rho_1)$. (3 — accept $a_2\approx1.39$ km/s and correct statement that $D=u_1=a_2\rho_2/\rho_1>a_2$.)

(d) (4) Premixed: fuel + oxidiser mixed before reaction; flame propagates as a wave at the laminar burning velocity; reaction-zone-controlled. Diffusion: fuel and oxidiser separate, burn where they interdiffuse to stoichiometric; mixing-rate controlled, sooty. A detonation requires a premixed reactant field (uniform reactive mixture). (2+2.)

Question 3

(a) (8) $\ln r = \ln a + n\ln p$. Two points: $n=\frac{\ln(9.60/7.00)}{\ln(10.0/5.0)}=\frac{\ln 1.3714}{\ln 2}=\frac{0.3159}{0.6931}=0.4557\approx 0.456.$ $a=\frac{r}{p^{n}}=\frac{7.00}{5.0^{0.456}}=\frac{7.00}{2.104}=3.33\ \text{(mm/s at 1 MPa)}.$ (6) Since $n<1$ (well below the instability threshold $n=1$), the motor is stable: a pressure perturbation raises burn rate less than proportionally, so chamber pressure self-corrects. (2)

(b) (4) At 7 MPa: $r=3.33\times7^{0.456}=3.33\times2.44=8.13\ \text{mm/s}$. $t=\frac{\text{web}}{r}=\frac{8\ \text{mm}}{8.13\ \text{mm/s}}=0.984\ \text{s}\approx 0.98\ \text{s}.$ (accept 0.95–1.0 s.)

(c) (8)

• AP (ammonium perchlorate): crystalline oxidiser supplying O for fuel-binder oxidation. (2)
• HTPB: hydroxyl-terminated polybutadiene, the fuel/binder giving the grain mechanical integrity and structural rubbery matrix. (2)
• Al (aluminium): metallic fuel additive raising flame temperature and density-specific impulse via highly exothermic oxidation. (2)
• Adding Al raises $T_c$ and $I_{sp}$, but produces condensed Al₂O₃ (smoky exhaust); trade-off: two-phase flow losses (particle lag, slag) and IR/visible signature. (2)

(d) (6)

import numpy as np
 
def burn_rate_fit(p, r):
    p = np.asarray(p, float); r = np.asarray(r, float)
    n, lna = np.polyfit(np.log(p), np.log(r), 1)  # slope, intercept
    return np.exp(lna), n          # (a, n)
 
def burn_time(a, n, p, web):
    r = a * p**n                   # mm/s
    return web / r                 # web in mm -> seconds
 
a, n = burn_rate_