Rocket Propulsion

Q1 (12)

(a) At time $t$: momentum $p_1 = mv$. At $t+dt$: rocket $(m+dm)$ with $dm<0$ moving at $v+dv$; ejected mass $(-dm)$ moves at $(v - v_e)$ in ground frame. $p_2 = (m+dm)(v+dv) + (-dm)(v-v_e)$ No external force ⇒ $dp = p_2 - p_1 = 0$. (4) — (correct sign of ejected mass velocity = key.)

(b) Expand, drop second-order $dm\,dv$: $m\,dv + v_e\,dm = 0 \Rightarrow dv = -v_e\frac{dm}{m}$ Integrate $m_0\to m_f$: $\Delta v = -v_e\int_{m_0}^{m_f}\frac{dm}{m}= v_e\ln\frac{m_0}{m_f}$ (4)

(c) Reasons (any two, 2 marks): $\Delta v$ grows only as $\ln$ of mass ratio, so doubling $\Delta v$ squares the mass ratio → exponentially more propellant; structural mass sets a floor on $m_f$ so achievable $\Delta v$ is capped; high $\Delta v$ demands either huge propellant or high $v_e$. (2) Computation: $m_0/m_f = e^{9.4/3.0} = e^{3.1333} \approx 22.9$. (2)

Q2 (10)

(a) CV momentum: rate of momentum out − in = net force. Steady thrust reaction: $F = \dot m\,v_e + (P_e - P_a)A_e$ First term = momentum thrust (mass flux × exhaust velocity); second = pressure thrust (net pressure force over exit plane, positive when $P_e>P_a$). (5)

(b) Effective exhaust velocity: $c \equiv F/\dot m = v_e + (P_e-P_a)A_e/\dot m$. Specific impulse $I_{sp} \equiv F/(\dot m g_0)=c/g_0 \Rightarrow c = I_{sp}g_0$. (3)

(c) $F = 250\times3200 + (15000-0)\times1.8 = 800000 + 27000 = 827000\ \text{N} = 827\ \text{kN}$. $c = F/\dot m = 827000/250 = 3308\ \text{m/s}$; $I_{sp}=c/9.81 = 337.2\ \text{s}$. (2)

Q3 (12)

(a) Steady-flow energy (adiabatic, no work): $h_c + \tfrac12 v_c^2 = h_e + \tfrac12 v_e^2$; assume $v_c\approx0$ (large chamber), perfect gas $h=c_pT$, isentropic $T_e/T_c=(P_e/P_c)^{(\gamma-1)/\gamma}$, $c_p=\gamma R/(\gamma-1)$: $\tfrac12 v_e^2 = c_p(T_c-T_e)=\frac{\gamma R T_c}{\gamma-1}\left[1-\left(\tfrac{P_e}{P_c}\right)^{\frac{\gamma-1}{\gamma}}\right]$ $\Rightarrow v_e=\sqrt{\frac{2\gamma R T_c}{\gamma-1}\left[1-(P_e/P_c)^{(\gamma-1)/\gamma}\right]}$ Assumptions (each named): perfect gas, isentropic (adiabatic+reversible), negligible chamber velocity, constant $\gamma$. (6)

(b) At $P_e=P_a$ the pressure term vanishes but momentum thrust is maximal for that ambient — any other exit pressure either wastes expansion (too high $P_e$, gas could do more work) or over-expands (recompression losses). Net thrust for the given altitude is maximised at matched exit pressure. (3)

(c) Over-expanded: $P_e<P_a$ → oblique shocks form in the plume (flow separation possible), efficiency loss. Under-expanded: $P_e>P_a$ → Prandtl–Meyer expansion fans outside nozzle, exhaust energy not fully converted axially. (3)

Q4 (10)

(a) At the throat flow is choked (M=1). Mass flow $\dot m = \rho^\ast v^\ast A^\ast$. Using choked-flow relations and $P_c,T_c$: $\dot m = \frac{P_c A^\ast}{\sqrt{T_c}}\sqrt{\frac{\gamma M}{R_u}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}=\frac{P_c A^\ast \Gamma}{\sqrt{R T_c}}$ with $R=R_u/M$. Then $c^\ast \equiv \frac{P_c A^\ast}{\dot m}=\frac{\sqrt{R T_c}}{\Gamma}$ Because $c^\ast$ contains only chamber/throat quantities ($P_c,T_c,\gamma,M$) and no nozzle exit geometry, it isolates combustion/injector quality from nozzle performance. (6)

(b) $c^\ast \propto \sqrt{T_c/M}$ (equivalently $\sqrt{R T_c}$). Higher $T_c$ ↑, higher $M$ ↓. LOX/LH2 has low $M$ (light H2O + excess H2, ~ 10–13 g/mol) which dominates over its lower flame temperature, giving higher $c^\ast$ and $I_{sp}$ than LOX/RP1 (heavier CO2/H2O products). (4)

Q5 (8)

(a) Staged combustion routes all propellant through preburner(s) and injects turbine exhaust into the main chamber → no propellant dumped, high chamber pressure, high $I_{sp}$; but plumbing (esp. oxidizer-rich or full-flow) is complex/hot. Gas generator burns a small propellant fraction to drive the turbine then dumps that turbine exhaust overboard at low $I_{sp}$ → simple but the dumped flow is dead weight → $I_{sp}$ penalty. (4)

(b) Expander cycle uses the heat picked up by regenerative cooling (usually H2) to expand and drive the turbine — no combustion in preburner; but the drivable power is limited by nozzle/chamber surface area for heat transfer, capping thrust/chamber pressure (square-cube limit). Electric pump-fed uses batteries + electric motors to drive pumps — decouples pump power from thermodynamics, simple control, but battery mass is dead weight. (4)

Q6 (8)

(a) With equal $v_e$ and Lagrange optimisation of $\Delta v=\sum v_e\ln(1/(\text{eff mass ratio}))$ under fixed structural coefficients, the symmetric solution gives equal stage mass ratios; hence each stage contributes equally: $\Delta v_1=\Delta v_2=\Delta v/2 = 5\ \text{km/s}$. (4)

(b) Per-stage: $m_0/m_f = e^{5.0/3.4}=e^{1.4706}\approx4.35$. Overall effective ratio $=4.35^2\approx18.9$. A single stage would need $e^{10/3.4}=e^{2.941}\approx18.9$ as one structure — impossible because one stage cannot shed structural mass, so its $m_f$ (with tanks/engines) makes such a ratio unachievable; staging discards spent structure, making the high overall ratio feasible. (4)

[
{"claim":"Q1c mass ratio for dv=9.4, ve=3.0","code":"import sympy as sp\nr=sp.exp(sp.Rational(94,10)/3)\nresult=abs(float(r)-22.95)<0.5"},
{"claim":"Q2c thrust and Isp","code":"F=250*3200+15000*1.8\nIsp=(F/250)/9.81\nresult=(abs(F-827000)<1) and (abs(Isp-337.2)<0.5)"},
{"claim":"Q6b per-stage mass ratio ~4.35 and square ~18.9","code":"import sympy as sp\nr=float(sp.exp(5.0/3.4))\nsingle=float(sp.exp(10.0/3.4))\nresult=(abs(r-4.35)<0.05) and (abs(r*r-single)<0.2)"}
]