Rocket Propulsion

Q1. (4 marks) $\Delta v = v_e \ln\!\left(\frac{m_0}{m_f}\right)$

• $\Delta v$ = velocity change achievable (1)
• $v_e$ = effective exhaust velocity; $m_0$ = initial mass, $m_f$ = final/burnout mass (2)
• Assumption: no external forces (gravity, drag) — force-free space (1) Why: the log form arises from integrating $m\,dv = -v_e\,dm$ with only the thrust term.

Q2. (6 marks) (a) Mass ratio $= m_0/m_f = 12000/3000 = 4$ (1) (b) $\Delta v = 2800\ \ln 4 = 2800 \times 1.3863 = 3881.6\ \text{m/s}$ ≈ 3882 m/s (3: formula 1, substitution 1, answer 1) (c) Because $\Delta v$ scales with the logarithm of the mass ratio; doubling the ratio adds only $v_e\ln 2$, so ever-larger propellant fractions yield progressively smaller gains. (2)

Q3. (5 marks) (a) $I_{sp} = v_e/g_0$; unit quoted in seconds (s) (2) (b) $v_e = I_{sp}\,g_0 = 311 \times 9.81 = 3050.9\ \text{m/s}$ ≈ 3051 m/s (2) (c) LOX/RP-1 (kerosene) (1)

Q4. (6 marks) $F = \dot m\, v_e + (P_e - P_a)A_e$

• $\dot m v_e$ = momentum thrust (2)
• $(P_e - P_a)A_e$ = pressure thrust (2)
• Thrust is maximised when $P_e = P_a$ (pressure thrust term optimally balanced with expansion), called optimum / perfect expansion (2) Why: differentiating thrust w.r.t. expansion shows the maximum occurs at matched exit/ambient pressure.

Q5. (6 marks) (a) $F = \dot m v_e + (P_e - P_a)A_e$ $= 250(2900) + (70000 - 101000)(0.9)$ $= 725000 + (-31000)(0.9) = 725000 - 27900 = 697100\ \text{N}$ ≈ 697.1 kN (3) (b) $c = F/\dot m = 697100/250 = 2788.4\ \text{m/s}$ ≈ 2788 m/s (2) (c) $P_e < P_a$ → over-expanded (1)

Q6. (5 marks) (a) $\varepsilon = A_e/A^*$ = ratio of nozzle exit area to throat area (2) (b) In vacuum $P_a \to 0$, so the gas can expand to a much lower $P_e$ before matching ambient; a larger $\varepsilon$ extracts more momentum thrust and gives higher $I_{sp}$. A sea-level engine with too large $\varepsilon$ would be badly over-expanded (flow separation/shocks). (3)

Q7. (5 marks) $r = a\,P^n$

• $r$ = linear burn rate, $P$ = chamber pressure, $a$ = temperature-dependent coefficient, $n$ = pressure exponent (2)
• $n$ measures the sensitivity of burn rate to chamber pressure (1)
• If $n \geq 1$: a pressure rise increases mass generation faster than the nozzle can vent it → runaway (unstable). $n<1$ gives a self-stabilising equilibrium chamber pressure. (2)

Q8. (6 marks) — 1.5 marks each (a) Gas generator (open) cycle: small preburner drives turbine, exhaust dumped overboard; simple but small $I_{sp}$ penalty. (b) Expander cycle: cryogenic fuel (H₂) heated in the regenerative jacket drives the turbine; very clean/reliable, no preburner. (c) Pressure-fed cycle: tank pressure alone forces propellant into chamber, no turbopumps; simplest, high reliability (upper stages). (d) Electric pump-fed cycle: battery-driven electric motors run the pumps; simple, throttleable, decoupled pump from combustion (e.g. Rutherford).

Q9. (4 marks) Electric propulsion achieves very high $I_{sp}$ (~thousands of s) by accelerating ions to high $v_e$, but produces tiny thrust (mN–N) because mass flow is minute and thrust is power-limited: $F \propto P/v_e$, so high $I_{sp}$ means low thrust for fixed power. (3) Thrust-to-weight is far below 1, so it cannot overcome gravity to lift off — only usable in space/vacuum. (1)

Q10. (3 marks) $c^* = \frac{P_c A^*}{\dot m}$ (1) Depends primarily on chamber (flame) temperature $T_c$ (higher → higher $c^*$) and molecular weight of combustion products $M$ (lower → higher $c^*$), i.e. $c^* \propto \sqrt{T_c/M}$. (2)

[
  {"claim":"Q2b: dv = 2800 ln(4) ≈ 3881.6","code":"import sympy as sp; dv=2800*sp.log(4); result = abs(float(dv)-3881.6) < 1.0"},
  {"claim":"Q3b: ve = 311*9.81 = 3050.91","code":"result = abs(311*9.81 - 3050.91) < 0.1"},
  {"claim":"Q5a: thrust = 697100 N","code":"F=250*2900+(70000-101000)*0.9; result = abs(F-697100) < 1"},
  {"claim":"Q5b: c = F/mdot = 2788.4","code":"F=250*2900+(70000-101000)*0.9; c=F/250; result = abs(c-2788.4) < 0.1"}
]