Rocket Propulsion

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Question 1

(a) Derivation (6 marks)

At time $t$: rocket mass $m$, velocity $v$; total momentum $p(t) = mv$. (1)

In interval $dt$ the rocket ejects mass $dm_e = -dm$ ($dm<0$) at exhaust velocity $v_e$ relative to rocket, i.e. absolute velocity $v - v_e$. At $t+dt$: rocket mass $m+dm$ at $v+dv$; ejected element at $v-v_e$. (1)

$p(t+dt) = (m+dm)(v+dv) + (-dm)(v-v_e).$

Field-free ⇒ momentum conserved: $p(t+dt)=p(t)$. (1)

Expand: $mv + m\,dv + v\,dm + dm\,dv - v\,dm + v_e\,dm = mv.$ The second-order term $dm\,dv$ is negligible (drop it — justification: product of two infinitesimals). (1)

$m\,dv + v_e\,dm = 0 \;\Rightarrow\; dv = -v_e\frac{dm}{m}.$ (1)

Integrate $m_0\to m_f$, $0\to\Delta v$, $v_e$ constant: $\Delta v = v_e\ln\frac{m_0}{m_f}. \qquad\blacksquare$ (1)

(b) Optimal staging proof + numbers (10 marks)

$v_e = I_{sp}g_0 = 340\times9.80665 = 3334.3\ \text{m/s}$. (1)

Total $\Delta v = v_e(\ln n_1 + \ln n_2)$ where $n_i = m_{0,i}/m_{f,i}$ is stage mass ratio. (1)

Proof of equal ratios: Minimise total mass subject to fixed $\sum \ln n_i$. With identical $v_e$ and $\varepsilon_s$, the payload-fraction contribution of each stage is a function only of its own $n_i$, and the objective is symmetric/separable. Using a Lagrange multiplier on the constraint $\ln n_1+\ln n_2 = \Delta v/v_e$ with the (symmetric) per-stage mass function, the stationarity conditions for both stages are identical, forcing $n_1=n_2=n$. (2) (Full: $\partial/\partial(\ln n_i)$ of $\ln(\text{mass function}) - \lambda\ln n_i$ gives same transcendental equation ⇒ equal.)

Hence $2\ln n = \frac{\Delta v}{v_e} = \frac{9500}{3334.3} = 2.849 \;\Rightarrow\; \ln n = 1.4245,\; n = e^{1.4245}=4.156.$ (2)

Per-stage payload ratio: a stage of ratio $n$ with structural coefficient $\varepsilon_s$ carries an "upper mass" (payload for that stage) $m_u$ where $n = \frac{m_u + m_p + m_s}{m_u + m_s},\quad m_s = \varepsilon_s(m_p+m_s).$ Solving gives the stage payload fraction $\pi = \frac{m_u}{m_0} = \frac{1-\varepsilon_s n}{n(1-\varepsilon_s)}.$ (1)

With $n=4.156,\ \varepsilon_s=0.08$: $\pi = \frac{1-0.08\times4.156}{4.156(1-0.08)} = \frac{1-0.3325}{3.824}=\frac{0.6675}{3.824}=0.17456.$ (1)

Overall payload fraction $\lambda = \pi^2 = 0.03047$. Thus $m_0 = \frac{m_{pl}}{\lambda} = \frac{1200}{0.03047} \approx 3.94\times10^4\ \text{kg} \approx 39{,}400\ \text{kg}.$ (1)

(c) Payload fraction & single-stage comparison (6 marks)

Two-stage $\lambda = 0.0305$ (3.0%). (1)

Single stage: need full $\ln n_{single} = 2.849 \Rightarrow n=17.28$. (1) Then $\varepsilon_s n = 0.08\times17.28 = 1.382 > 1$, giving $\pi = (1-1.382)/(17.28\times0.92) < 0$. (2)

⇒ Physically impossible: the required propellant mass ratio exceeds the structural limit ($n_{max}=1/\varepsilon_s=12.5$), so no positive payload exists for a single stage. (1)

Staging helps because spent structural (tank/engine) mass is discarded, so the remaining stages do not pay the $\ln$-penalty of accelerating dead mass; each stage sees a smaller $m_f$, raising the achievable $\Delta v$ for given structure. (1)

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Question 2

(a) Thrust equation (6 marks)

Control-volume momentum balance around engine, steady state: net axial force = momentum outflow rate. (1) Momentum flux at exit $=\dot m v_e$. (1) Pressure forces: exit plane $P_e A_e$ pushing forward, ambient $P_a$ acting on projected exit area $A_e$ pushing back (net over closed CV the ambient acts everywhere except exit plane). (1) Sum: $F = \dot m v_e + (P_e-P_a)A_e. \qquad\blacksquare$ (1)

Effective exhaust velocity: $F \equiv \dot m c \Rightarrow c = v_e + \dfrac{(P_e-P_a)A_e}{\dot m}$. (1)

Thrust coefficient: $C_F \equiv \dfrac{F}{P_c A^*}$ — nondimensionalises thrust by chamber pressure × throat area; measures nozzle expansion quality (independent of $c^*$). (1)

(b) Exit Mach & pressure (8 marks)

Solve $\varepsilon = 16$ for supersonic $M_e$ with $\gamma=1.2$: $16 = \frac{1}{M_e}\left[\frac{2}{2.2}\left(1+0.1 M_e^2\right)\right]^{\frac{2.2}{0.4}}, \quad \text{exponent} = 5.5.$ (2)

Numerical root (supersonic): $M_e \approx 4.15$. (2) (Check: RHS at $M_e=4.15$: inner $=0.9091(1+1.722)=2.475$, $2.475^{5.5}\approx 66.4$, $/4.15\approx16.0$ ✓.)

Exit pressure via isentropic relation: $\frac{P_e}{P_c}=\left(1+\frac{\gamma-1}{2}M_e^2\right)^{-\frac{\gamma}{\gamma-1}} = \left(1+0.1\times17.22\right)^{-6} = (2.722)^{-6}.$ (2) $(2.722)^{6}\approx 407$, so $P_e/P_c = 2.46\times10^{-3}$, $P_e = 7.0\times10^6 \times2.46\times10^{-3} \approx 1.72\times10^4\ \text{Pa} = 17.2\ \text{kPa}$. (2)

(c) Altitude matching (4 marks)

Optimum ($P_e=P_a$) occurs at ambient $P_a \approx 17.2\ \text{kPa}$, i.e. roughly 13–14 km altitude. (1)

At sea level $P_a = 101.3\ \text{kPa} > P_e = 17.2\ \text{kPa}$ ⇒ over-expanded. (1) Higher ambient pressure compresses the plume: oblique shocks form at/inside the nozzle exit; if severe, flow separates. (1) The $(P_e-P_a)A_e$ term is negative, reducing thrust; shocks and separation cause total-pressure loss and non-axial momentum, lowering efficiency. (1)

(d) Numerical scheme (4 marks)

# Solve area-Mach for supersonic root
gamma = 1.20; eps = 16.0
def AR(M):
    t = (1 + 0.5*(gamma-1)*M**2)
    return (1/M)*( (2/(gamma+1))*t )**((gamma+1)/(2*(gamma-1)))
# Bracket supersonic root: M in [1.0001, 20], AR-eps changes sign
def f(M): return AR(M) - eps
# Bisection (guaranteed for monotonic supersonic branch)
lo, hi = 1.0001, 20.0
while hi-lo > 1e-8:
    mid = 0.5*(lo+hi)
    if f(lo)*f(mid) <= 0: hi = mid
    else: lo = mid
Me = 0.5*(lo+hi)

Method: bisection on the monotonic supersonic branch (robust, no derivative). (1) Bracket $[1^+, 20]$ because $A/A^*$ increases monotonically with $M>1$. (1) Convergence when $|hi-lo|<10^{-8}$ (or $|f|<$ tol). (1) Newton–Raphson with initial guess from a correlation is a faster alternative. (1)

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Question 3

(a) c* derivation (6 marks)

Choked throat: $\dot m = P_c A^* \sqrt{\dfrac{\gamma}{R T_c}}\left(\dfrac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$ with $R=R_u/\mathcal M$. (2) Then $c^* = \frac{P_c A^*}{\dot m} = \frac{1}{\sqrt{\gamma}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}}\sqrt{\frac{R_u T_c}{\mathcal M}} = \frac{1}{\Gamma}\sqrt{\frac{R_u T_c}{\mathcal M}}.$ (2)

Physical point: $c^* \propto \sqrt{T_c/\mathcal M}$. LOX/LH2 flame temp is lower than RP-1's, but its exhaust mean molecular mass ($\mathcal M \approx 10$–$13$ g/mol, water + excess H₂) is far lower than RP-1's (~22–24 g/mol). The $\sqrt{T_c/\mathcal M}$ gain from small $\mathcal M$ dominates, giving higher $c^*$ and $I_{sp}$. (2)

(b) Cycles (6 marks)

Staged combustion: preburner exhaust (fuel- or ox-rich) is routed into the main chamber after driving the turbine — all propellant is burned at full chamber pressure, so no energy is dumped; highest $I_{sp}$. (2) Gas generator: turbine-drive gas is exhausted overboard (or dumped) at low pressure and low $I_{sp}$; this parasitic ~2–7% flow doesn't contribute full thrust ⇒ few-second $I_{sp}$ penalty, but plumbing is simpler/lower pressure. (2) Expander cycle: turbine is driven only by heat picked up in the regenerative jacket. Heat flux scales with wall area, but flow (and thrust) scales with volume/throughput; as the chamber grows, available heat/power cannot rise fast enough (surface-to-vol